Downloaded from @Freebooksforjeeneet
EBD_8344
Downloaded from @Freebooksforjeeneet
Downloaded from @Freebooksforjeeneet
EBD_8344
Downloaded from @Freebooksforjeeneet
Downloaded from @Freebooksforjeeneet
EBD_8344
Downloaded from @Freebooksforjeeneet
Downloaded from @Freebooksforjeeneet
EBD_8344
Downloaded from @Freebooksforjeeneet
1
Sets
TOPIC Ć
1.
2.
Set A has m elements and set B has n elements. If the total
number of subsets of A is 112 more than the total number
of subsets of B, then the value of m×n is ______.
[Sep. 06, 2020 (I)]
Let S = {1, 2, 3, ... , 100}. The number of non-empty subsets
A of S such that the product of elements in A is even is :
[Jan. 12, 2019 (I)]
(
3.
5.
7.
8.
)
50 50
(a) 2100 - 1
(b) 2 2 - 1
(c) 250 – 1
(d) 250 + 1
Let S = {x Î R : x ³ 0 and
2|
(a)
(b)
(c)
(d)
4.
6.
Sets, Types of Sets, Disjoint Sets,
Complementary Sets, Subsets,
Power Set, Cardinal Number of
Sets, Operations on Sets
x - 3 | + x ( x - 6) + 6 = 0 . Then S :
contains exactly one element.
contains exactly two elements.
contains exactly four elements.
is an empty set.
æ1ö
If f(x) + 2f ç ÷ = 3x , x ¹ 0 and
èxø
S = {x Î R : f(x) = f(–x)}; then S:
(a) contains exactly two elements.
(b) contains more than two elements.
(c) is an empty set.
(d) contains exactly one element.
Let P = {q : sinq – cosq =
cosq =
(b) Q Ë P
(c) P = Q
(Y, Z) that can formed such that Y Í X , Z Í X and Y Ç Z
is empty is :
[2012]
(a) 52
(b) 35
(c) 25
(d) 53
If A, B and C are three sets such that A Ç B = A Ç C and
[2009]
A È B = A È C , then
(a) A = C
(b) B = C
(c) A Ç B = f
(d) A = B
Telegram @unacademyplusdiscounts
[2018]
TOPIC n
9.
[2016]
2 cosq} and Q = {q : sinq +
2 sinq} be two sets. Then:
[Online April 10, 2016]
(a) P Ì Q and Q - P ¹ f
A relation on the set A = {x : |x| < 3, x ÎZ },
where Z is the set of integers is defined by
R = {(x, y) : y = |x|, x ¹ – 1}. Then the number of elements in
the power set of R is:
[Online April 12, 2014]
(a) 32
(b) 16
(c) 8
(d) 64
Let X ={1,2,3,4,5}. The number of different ordered pairs
10.
11.
Venn Diagrams, Algebraic
Operations on Sets, De Morgan’s
Law, Number of Elements in
Different Sets
A survey shows that 73% of the persons working in an
office like coffee, whereas 65% like tea. If x denotes the
percentage of them, who like both coffee and tea, then x
cannot be :
[Sep. 05, 2020 (I)]
(a) 63
(b) 36
(c) 54
(d) 38
A survey shows that 63% of the people in a city read
newspaper A whereas 76% read newspaper B. If x% of the
people read both the newspapers, then a possible value of
x can be :
[Sep. 04, 2020 (I)]
(a) 29
(b) 37
(c) 65
(d) 55
Let
50
n
i =1
i =1
U X i = U Yi = T , where each Xi contains 10 elements
and each Yi contains 5 elements. If each element of the set
T is an element of exactly 20 of sets Xi's and exactly 6 of
sets Yi's, then n is equal to
[Sep. 04, 2020 (II)]
(a) 15
(b) 50
(c) 45
(d) 30
(d) P Ë Q
Downloaded from @Freebooksforjeeneet
EBD_8344
M-2
12.
13.
14.
15.
Mathematics
Let X = {n Î N: l £ n £ 50}. If
A = {n Î X: n is a multiple of 2} and
B = {n Î X: n is a multiple of 7}, then the number of
elements in the smallest subset of X containing both A
and B is __________.
[Jan. 7, 2020 (II)]
16.
2
Let Z be the set of integers. If A = {xÎZ : 2(x + 2) ( x – 5x + 6) = 1}
and B = {x Î Z : –3 < 2x – 1< 9}, then the number of subsets
of the set A × B, is :
[Jan. 12, 2019 (II)]
(a) 215
(b) 218
(c) 212
(d) 210
In a class of 140 students numbered 1 to 140, all even
numbered students opted Mathematics course, those
whose number is divisible by 3 opted Physics course and
those whose number is divisible by 5 opted Chemistry
course. Then the number of students who did not opt for
any of the three courses is:
[Jan. 10, 2019 (II)]
(a) 102
(b) 42
(c) 1
(d) 38
f
¹
A
Ç
B Í C . Then
Let A, B and C be sets such that
which of the following statements is not true ?
[April 12, 2019 (II)]
(a) B Ç C ¹ f
(b) If ( A - B) Í C , then A Í C
17.
Two newspapers A and B are published in a city. It is
known that 25% of the city population reads A and 20%
reads B while 8% reads both A and B. Further, 30% of
those who read A but not B look into advertisements and
40% of those who read B but not A also look into
advertisements, while 50% of those who read both A and
B look into advertisements. Then the percentage of the
population who look into advertisements is:
[April. 09, 2019 (II)]
(a) 13.9
(b) 12.8
(c) 13
(d) 13.5
In a certain town, 25% of the families own a phone and
15% own a car; 65% families own neither a phone nor a
car and 2,000 families own both a car and a phone. Consider
the following three statements : [Online April 10, 2015]
(A) 5% families own both a car and a phone
(B) 35% families own either a car or a phone
(C) 40,000 families live in the town
Then,
(a) Only (A) and (C) are correct.
(b) Only (B) and (C) are correct.
(c) All (A), (B) and (C) are correct.
(d) Only (A) and (B) are correct.
(c) (C È A) Ç (C È B) = C
(d) If (A - C) Í B , then A Í B
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M-3
Sets
1.
(28) 2m = 112 + 2n Þ 2m - 2n = 112
m -n
Þ 2 (2
n
5.
- 1) = 2 (2 - 1)
4
3
\ m = 7, n = 4 Þ mn = 28
2.
3.
7.
Case-II: x Î[9, ¥]
2( x - 3) + x - 6 x + 6 = 0
8.
2
-x
x
f (x) = f (- x) Þ
x2 = 2
(b) Q
B = ( B Ç A) È B
= ( A È B) Ç ( C È B)
= ( A È C) Ç ( B È C)
æ 1ö
(a) f (x) + 2f ç ÷ = 3x ....(1)
è xø
3
æ 1ö
f ç ÷ + 2f (x) =
....(2)
è xø
x
Adding (1) and (2)
Þ f (x) =
(b) A = {x : |x | < 3, x Î Z }
A = {–2, – 1, 0, 1, 2}
R = {(x, y) : y = |x |, x ¹ -1}
R = {(–2, 2), (0, 0), (1, 1), (2, 2)}
R has four elements
Number of elements in the power set of R
= 24 = 16
(b) Let X = {1, 2, 3, 4, 5}
n(x) = 5
Each element of x has 3 options. Either in set Y or set Z or
none. (Q Y Ç Z = f)
So, number of ordered pairs = 35
= ( A Ç C) È B
Since x Î[9, ¥]
\ x = 16
Hence, x = 4 & 16
1
æ1ö
Þ f (x) + f ç ÷ = x +
...(3)
x
x
è ø
Substracting (1) from (2)
æ1ö 3
Þ f (x) - f ç ÷ = - 3x ...(4)
èxø x
On adding (3) and (4)
2 sinq
\ P= Q
6.
Þ x - 8 x + 12 = 0 Þ x = 4, 2
Þ x = 16, 4
4.
2 cosq
Þ sinq + cosq =
2(3 - x ) + x - 6 x + 6 = 0
Þ x - 4 x = 0 Þ x = 16,0
Þ sinq = cosq +
Þ ( 2 - 1) sinq = cosq
(b) Case-I: x Î[0,9]
Since x Î[0,9]
\ x=4
2 cosq
æ 2 -1 ö
= ( 2 + 1) cosq = ç
÷ cosq
è 2 -1 ø
(b) Q Product of two even number is always even and
product of two odd numbers is always odd.
\ Number of required subsets
= Total number of subsets – Total number of subsets
having only odd numbers
= 2100 – 250 = 250(250 – 1)
(c) sinq – cosq =
= ( A Ç B) È C
= ( A Ç C) È C
=C
9.
(b) Given, n(C ) = 73, n(T ) = 65, n(C Ç T ) = x
\ 65 ³ n(C Ç T ) ³ 65 + 73 - 100
Þ 65 ³ x ³ 38 Þ x ¹ 36.
10.
(d) Let n(U ) = 100, then n(A) = 63, n(B) = 76
n( A Ç B ) = x
Now, n( A È B ) = n( A) + n( B) - n( A Ç B) £ 100
2
-2
2
-x =
+xÞx=
x
x
x
= 63 + 76 - x £ 100
Þ x ³ 139 - 100 Þ x ³ 39
or x = 2, - 2
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EBD_8344
M-4
Mathematics
Q = {6n: n Î N, 1 £ n £ 23} – P
Q n( A Ç B) £ n( A)
Þ n(Q) = 19
Þ x £ 63
R = {15n: n Î N, 1 £ n £ 9} – P
\ 39 £ x £ 63
11.
(d)
50
n
i =1
i =1
Þ n(R) = 5
U X i = U Yi = T
S = {10n: n Î N, 1 £ n £ 14} – P
Þ n(S) = 10
Q
n( X i ) = 10, n (Yi ) = 5
So,
U X i = 500, U Yi = 5n
50
n(T) = 70 – n(P) – n(Q) – n(S) = 70 – 33 = 37
n(V) = 46 – n(P) – n(Q) – n(R) = 46 – 28 = 18
n
i =1
n(W) = 28 – n(P) – n(R) – n(S) = 28 – 19 = 9
i =1
Þ Number of required students
500 5n
Þ
=
Þ n = 30
20
6
12.
= 140 – (4 + 19 + 5 + 10 + 37 + 18 + 9)
= 140 – 102 = 38
(29) From the given conditions,
15.
n(A) = 25, n(B) = 7 and n(A Ç B) = 3
In (3) option,
n(A È B) = n(A) + n(B) – n(A Ç B)
If A = C then A – C = f
= 25 + 7 – 3 = 29
13.
(d) (1), (2) and (4) are always correct.
Clearly, f Í B but A Í B is not always true.
(a) Let x Î A, then
Q 2( x+2)( x2 -5 x+6) = 1 Þ (x + 2)(x – 2)(x – 3) = 0
x = –2, 2, 3
16.
(a)
A
B
25
20
8
A = {–2, 2, 3}
Then, n(A) = 3
Let x Î B, then
–3 < 2x – 1 < 9
% of people who reads A only = 25 – 8 = 17%
–1 < x < 5 and x Î Z
% of people who read B only = 20 – 8 = 12%
\ B = {0, 1, 2, 3, 4}
% of people from A only who read advertisement
n(B) = 5
= 17 × 0.3 = 5.1%
n(A ´ B) = 3 ´ 5 = 15
% of people from B only who read advertisement
Hence, Number of subsets of A ´ B = 2
= 12 × 0.4 = 4.8%
15
% of people from A & B both who read advertisement
14.
(d)
= 8 × 0.5 = 4%
Maths
T
Q
Physics
W
S
\ total % of people who read advertisement
Chemistry
P
R
V
17.
= 5.1 + 4.8 + 4 = 13.9%
(c) n(P) = 25%
n(C) = 15%
n ( P ¢ È C ¢ ) =65%
Þ n(P È C)¢ = 65%
n ( P È C ) = 35%
P = {30, 60, 90, 120}
Þ n(P) = 4
n ( P Ç C) = n ( P ) + n (C) - n ( P È C)
25 + 15 – 35 = 5%
x × 5% = 2000
x = 40,000
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2
Relations and
Functions
TOPIC Ć
1.
The domain of the definition of the function
6.
1
+ log10 ( x3 - x ) is:
4 - x2
(a) (–1, 0) È (1, 2) È (3, ¥)
(b) (–2, –1) È (–1, 0) È (2, ¥)
(c) (–1, 0) È (1, 2) È (2, ¥)
(d) (1, 2) È (2, ¥)
The range of the function
f ( x) =
Let R1 and R2 be two relations defined as follows :
R1 = {(a, b) Î R 2 : a 2 + b 2 Î Q} and
R2 = {(a, b) Î R 2 : a 2 + b2 Ï Q} , where Q is the set of all
rational numbers. Then :
[Sep. 03, 2020 (II)]
(a) Neither R1 nor R2 is transitive.
(b) R2 is transitive but R1 is not transitive.
(c) R1 is transitive but R2 is not transitive.
(d) R1 and R2 are both transitive.
2.
5.
Relations, Domain, Codomain and
Range of a Relation, Functions,
Domain, Codomain and Range of a
Function
| x | +5 ö
The domain of the function f ( x) = sin -1 æç
÷ is
è x2 +1 ø
(-¥, - a] È [a, ¥]. Then a is equal to :
f ( x) =
(a) R
3.
4.
17
2
(b)
17 - 1
1 + 17
(d)
(c)
2
2
Let f : R ® R be defined by f ( x ) =
the range of f is :
x
, x Î R. Then
1 + x2
[Jan. 11, 2019 (I)]
é 1 1ù
(a) ê - , ú
ë 2 2û
(b) R – [–1, 1]
é 1 1ù
(c) R - ê - , ú
ë 2 2û
(d) (–1, 1) – {0}
(c) R – {0}
(d) [– 1, 1]
1
The domain of the function f ( x) =
8.
(a) (0, ¥ )
(b) (– ¥ , 0)
(c) (– ¥ , ¥ ) – {0}
(d) (– ¥ , ¥ )
Domain of definition of the function
f ( x) =
17
+1
2
If R = {(x, y) : x, y Î Z, x2 + 3 y 2 £ 8} is a relation on the
set of integers Z, then the domain of R–1 is :
[Sep. 02, 2020 (I)]
(a) {–2, –1, 1, 2}
(b) {0, 1}
(c) {–2, –1, 0, 1, 2}
(d) {–1, 0, 1}
(b) (– 1, 1)
[Online May 7, 2012]
7.
[Sep. 02, 2020 (I)]
(a)
x
, x Î R , is
1+ x
[April. 09, 2019 (II)]
3
4 - x2
is
+ log10 ( x 3 - x) , is
(a) ( -1,0) È (1,2) È ( 2, ¥)
(b) (a, 2)
(c) ( -1,0) È ( a,2)
(d) (1,2) È (2, ¥)
TOPIC n
9.
x -x
[2011]
[2003]
Even and Odd Functions, Explicit
and Implicit Functions, Greatest
Integer Function, Periodic
Functions, Value of a Function,
Equal Functions, Algebraic
Operations on Functions
Let [t] denote the greatest integer £ t. Then the equation
in x, [x]2 + 2[x + 2] – 7 = 0 has :
[Sep. 04, 2020 (I)]
(a) exactly two solutions
(b) exactly four integral solutions
(c) no integral solution
(d) infinitely many solutions
Downloaded from @Freebooksforjeeneet
EBD_8344
M-6
10.
11.
12.
13.
14.
Mathematics
Let f (x) be a quadratic polynomial such that f (–1) + f (2) =
0. If one of the roots of f (x) = 0 is 3, then its other root lies
in :
[Sep. 02, 2020 (II)]
(a) (–1, 0) (b) (1, 3)
(c) (–3, –1) (d) (0, 1)
15.
Then f(x) at x = -
x[ x]
,
1 + x2
where [x] denotes the greatest integer £ x. Then the range
of f is:
[Jan. 8, 2020 (II)]
Let f (1, 3) ® R be a function defined by f (x) =
æ 2 3ö æ 3 4 ö
(a) çè , ÷ø È çè , ø÷
5 5
4 5
æ 2 1 ö æ 3 4ö
(b) çè , ÷ø È çè , ø÷
5 2
5 5
æ 2 4ö
(c) çè , ÷ø
5 5
æ 3 4ö
(d) çè , ÷ø
5 5
æ 2x ö
1- x ö
If f(x) = loge æç
, |x| < 1, then f ç
÷ is equal to :
÷
è 1+ x2 ø
è 1+ x ø
[April 8, 2019 (I)]
(a) 2f(x)
(b) 2f(x2)
(c) (f(x)) 2
(d) –2f(x)
Let f(x) = ax (a > 0) be written as f(x) = f1(x) + f2(x), where
f1(x) is an even function and f2(x) is an odd function. Then
f1(x + y) + f1(x – y) equals :
[April. 08, 2019 (II)]
(a) 2f1(x) f1(y)
(b) 2f1(x + y) f1(x – y)
(c) 2f1(x)f2(y)
(d) 2f1(x + y) f2(x – y)
é 1 3n ù
n , where [n] denotes the greatest
Let f ( n ) = ê +
ë 3 100 úû
56
integer less than or equal to n. Then
å f ( n) is equal to:
n=1
(a) 56
(b) 689
Let f be an odd function defined on the set of real numbers
such that for x ³ 0, f(x) = 3 sin x + 4 cos x.
(a)
11p
is equal to: [Online April 11, 2014]
6
3
+2 3
2
3
(b) - + 2 3
2
3
3
-2 3
(d) - - 2 3
2
2
A real valued function f (x) satisfies the functional equation
f (x – y) = f (x) f (y) – f (a – x) f (a + y)
where a is a given constant and f (0) = 0, f (2a – x) is equal
to
[2005]
(a) – f (x)
(b) f (x)
(c) f (a) + f (a – x)
(d) f (– x)
The graph of the function y = f(x) is symmetrical about the
line x = 2, then
[2004]
(c)
16.
17.
18.
(a)
f ( x ) = - f (- x)
(b) f (2 + x) = f (2 - x)
(c)
f ( x ) = f ( - x)
(d) f ( x + 2) = f ( x - 2)
If f : R ® R satisfies f ( x + y ) = f ( x) + f ( y ) , for all x,
n
y Î R and f(1) = 7, then S f (r ) is
r=1
(a)
7 n (n + 1)
2
(b)
(c)
7 (n + 1)
2
(d) 7 n + ( n + 1)
7n
2
[Online April 19, 2014]
(c) 1287
(d) 1399
Downloaded from @Freebooksforjeeneet
[2003]
M-7
Relations and Functions
1.
Þ 1 ³ 4y2
1/ 4
(a) For R1 let a = 1 + 2, b = 1 - 2, c = 8
Þ |y| £
bR1c Þ b2 + c 2 = (1 - 2)2 + (81/ 4 )2 = 3 Î Q
1
1
Þ - £ y£
2
2
aR1c Þ a 2 + c2 = (1 + 2)2 + (81/ 4 )2 = 3 + 4 2 Ï Q
é 1 1ù
Þ The range of f is ê - , ú .
ë 2 2û
\ R1 is not transitive.
For R2 let a = 1 + 2, b = 2, c = 1 - 2
5.
aR2b Þ a + b = (1 + 2) + ( 2) = 5 + 2 2 Ï Q
2
2
2
2
bR2 c Þ b 2 + c 2 = ( 2)2 + (1 - 2)2 = 5 - 2 2 Ï Q
aR2c Þ a 2 + c2 = (1 + 2) 2 + (1 - 2) 2 = 6 Î Q
| x | +5 ö
(c) Q f ( x) = sin -1 æç
è x 2 + 1 ÷ø
\ -1 £
| x | +5
x2 + 1
£1
6.
Þ| x | +5 £ x2 + 1
[Q x 2 + 1 ¹ 0]
x
1+ x
which is not defined for x = – 1
æ
1 - 17 ö æ
1 + 17 ö
Þ ç| x | ÷ ç| x | - 2 ÷ ³ 0
2
è
øè
ø
x
1- x
which is not defined for x = 1
Thus f(x) defined for all values of R except 1 and – 1
Hence, range = (– 1, 1).
If x < 0, | x | = – x Þ f ( x ) =
æ
ö
1 + 17 ö é1 + 17
Þ x Î ç -¥, Èê
, ¥÷
÷
2 ø ë 2
è
ø
3.
1 + 17
2
(b)
f ( x) =
8.
(a)
f ( x) =
\ R = {(1, 1), (2, 1), (1, - 1), (0, 1), (1, 0)}
4.
(a) f(x) =
+ log10 ( x 3 - x)
x ¹ ± 4 and - 1 < x < 0 or 1 < x < ¥
x
Let, y =
1 + x2
–
Þ yx2 – x + y = 0 Þ x =
3
4 - x2
4 - x 2 ¹ 0; x 3 - x > 0;
x
, x ÎR
1 + x2
Þ 1 – 4y2 ³ 0
1
, f (x) is define if | x | – x > 0
x -x
Þ | x | > x, Þ x < 0
Hence domain of f (x) is (– ¥, 0)
7.
(d) Since, R = {(x, y) : x, y Î Z, x2 + 3 y 2 £ 8}
Þ DR-1 = {-1, 0, 1}
x
(b) f ( x) = 1 + x , x Î R
If x > 0, | x | = x Þ f ( x ) =
Þ x2 - | x | -4 ³ 0
\a =
(c) To determine domain, denominator ¹ 0 and x3 – x > 0
i.e., 4 – x2 ¹ 0 x ¹ ±2
...(1)
and x (x – 1) (x + 1) > 0
–
+
–
+
x Î (– 1, 0) È (1, ¥)
...(2)
Hence domain is intersection of (1) & (2).
i.e., x Î (–1, 0) È (1, 2) È (2, ¥)
\ R2 is not transitive.
2.
1
2
aR1b Þ a 2 + b2 = (1 + 2)2 + (1 - 2) 2 = 6 Î Q
1 ± 1- 4 y2
2
+
–
+
–1
0
1
\ D = ( -1, 0) È (1, ¥) -
{ 4}
D = ( -1, 0) È (1, 2) È (2, ¥).
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EBD_8344
M-8
9.
Mathematics
(d) The given equation
2
æ 1 + x2 - 2 x ö
æ 1- x ö
÷
= log çç
=
log
2
ç 1+ x ÷
÷
è
ø
è 1+ x + 2x ø
[ x]2 + 2[ x] + 4 - 7 = 0
Þ [ x]2 + 2[ x] - 3 = 0
æ 1- x ö
= 2 log ç
÷ = 2f (x)
è 1+ x ø
Þ [ x]2 + 3[ x] - [ x] - 3 = 0
Þ ([ x] + 3)([ x] - 1) = 0 Þ [ x] = 1 or –3
13.
Þ x Î[ -3, - 2) È [1, 2)
\ The equation has infinitely many solutions.
10.
æ a x + a- x ö æ a x - a-x ö
÷+ç
÷
f (x) = ax = çç
÷ ç
÷
2
2
è
ø è
ø
(a) Let f ( x) = ax 2 + bx + c
Given : f ( -1) + f (2) = 0
where f1(x) =
a - b + c + 4a + 2b + c = 0
Þ 5a + b + 2c = 0
Þb=
f2(x) =
11.
(b)
æ a x+ y + a - x - y ö æ a x - y + a - x + y
÷+ç
= çç
÷ ç
2
2
è
ø è
c -6
and a = 3
=
a
5
-2
Î ( -1, 0)
5
ì x
ïï 2 ; x Î (1, 2)
x +1
f ( x) í
ï 2 x ; x Î[2,3)
îï x 2 + 1
14.
1é x y
a (a + a - y ) + a - x (a y + a - y ) ù
û
2ë
=
(a x + a - x )(a y + a - y )
= 2f1(x) . f1(y)
2
é 1 3n ù
n
(d) Let f (n) = ê +
ë 3 100 úû
where [n] is greatest integer function,
So,
æ 2 1 ö æ 6 4ù
y Îç , ÷ Èç , ú
è 5 2 ø è 10 5 û
å
n =1
f (n) = 1 (23) + 1 (24) + ... + 1 (55) + 2(56)
= (23 + 24 + ... + 55) + 112
=
æ 2 1 ö æ 3 4ù
Þ y Îç , ÷ È ç , ú
è 5 2 ø è 5 5û
33
[46 + 32] + 112
2
33
(78) + 112 = 1399.
2
(c) Given f be an odd function
f (x) = 3 sin x + 4 cos x
Now,
=
æ1- x ö
(a) f (x) = log ç
÷ ,|x|<|
è 1+ x ø
2x
æ
ç 11 + x2
æ 2x ö
ç
fç
=
log
÷
ç 1+ 2x
è 1 + x2 ø
ç
è 1 + 2 x2
=
56
\ f(x) is a decreasing function
12.
ö
÷
÷
ø
3n ù
é
n
= ê0.33 +
100 úû
ë
For n = 1, 2, ..., 22, we get f (n) = 0
and for n = 23, 24, ..., 55, we get f (n) = 1 × n
For n = 56, f (n) = 2 × n
ì 1 - x2
; x Î (1, 2)
ï
ï 1+ x2
f ¢( x ) í
2
ï1 - 2 x
; x Î [2,3)
ï
2
î 1+ x
\
a x + a-x
is even function
2
a x - a-x
is odd function
2
Þ f1(x + y) + f1(x – y)
...(i)
and f (3) = 0 Þ 9a + 3b + c = 0
...(ii)
From equations (i) and (ii),
a
b
c
a
b c
=
=
Þ
= =
1 - 6 18 - 5 15 - 9
-5 13 6
Product of roots, ab =
(a) Given function can be written as
15.
ö
÷
÷
÷
÷
ø
æ -11p ö
æ -11p ö
æ -11p ö
3sin ç
+ 4cos ç
fç
è 6 ÷ø =
è 6 ÷ø
è 6 ÷ø
Downloaded from @Freebooksforjeeneet
M-9
Relations and Functions
æ -11p ö
pö
pö
æ
æ
fç
è 6 ÷ø = 3sin çè -2p + 6 ÷ø + 4cos çè -2p + 6 ÷ø
17.
(b) Given that a graph symmetrical. with respect to line
x = 2 as shown in the figure.
Y
–x
æ -11p ö
ì æ
pö ü
ì æ
pöü
fç
è 6 ÷ø = 3sin íî - çè 2p - 6 ÷ø ýþ + 4cos íî - çè 2p - 6 ÷ø ýþ
x
sin( -q) = - sin q
ì
ü
í For odd functions
ý
and cos( -q) = - cos qþ
î
pö
pö
æ
æ
æ -11p ö
\ fç
= -3sin ç 2p - ÷ - 4cos ç 2p - ÷
è
è
è 6 ÷ø
6ø
6ø
From the figure
p
æ pö
æ -11p ö
Þ fç
= +3sin çè ÷ø - 4cos
è 6 ÷ø
6
6
1
3
æ -11p ö
Þ fç
= 3´ - 4´
è 6 ø÷
2
2
-11p ö 3
or f æç
= -2 3
è 6 ÷ø 2
16.
X
x=2
f ( x1 ) = f ( x2 ), where x1 = 2 - x and x2 = 2 + x
\ f (2 - x) = f (2 + x)
18.
(a) f ( x + y ) = f (x ) + f ( y ) .
Q f (1) = 7
f (2) = f (1 + 1) = f (1) + f (1) = 14
f (3) = f (1 + 2) = f (1) + f (2) = 21
(a) Given that f(0) = 0 and put
x = 0, y = 0,
f (0) = f 2 (0) - f 2 (a)
Þ f 2 (a) = 0 Þ f (a) = 0
f (2a – x) = f (a – (x – a))
= f (a) f (x – a) – f (0) f (x)
= f (a) f (x –a) – f (x) = – f (x)
x2
x1
n
\ å f (r ) = 7 (1 + 2 + 3..... + n )
r =1
=
7 n ( n + 1)
2
Þ f (2a - x) = - f ( x )
Downloaded from @Freebooksforjeeneet
EBD_8344
M-10
Mathematics
3
Trigonometric Functions
TOPIC Ć
1.
Circular System, Trigonometric
Ratios, Domain and Range of
Trigonometric Functions,
Trigonometric Ratios of Allied
Angles
æp pö
For any q Î ç , ÷ the expression
è4 2ø
3(sinq –
cosq)4 +
6(sinq +
cosq)2 +
The expression
5.
can be written as :
[2013]
(a) sinA cosA + 1
(b) secA cosecA + 1
(c) tanA + cotA
(d) secA + cosecA
The value of cos 255 + sin 195 is[Online May 26, 2012]
(a)
4sin6q equals:
2 2
(b)
(d)
3 -1
2
3 +1
2
4cos2q
6sin2qcos2q
(c) 13 –
4cos2q
+
6cos4q
Statement 1: f ( x ) £ g ( x ) for x in ( 0, ¥ )
(d) 13 –
4cos4q
+
2sin2qcos2q
Statement 2: f(x) £ 1 for x in (0, ¥) but g(x) ® ¥ as x ® ¥.
[Online May 7, 2012]
(a) Statement 1 is true, Statement 2 is false.
(a) 13 –
+
(b) 13 – 4cos6q
6.
)
(
1
Let f k ( x ) = sin k x + cos k x where x Î R and k ³ 1.
k
Then f 4 ( x ) - f 6 ( x ) equals
3.
3 -1
æ 3 -1ö
÷÷
(c) - çç
è 2 ø
[Jan. 9, 2019 (I)]
2.
tan A
cot A
+
1 - cot A 1 - tan A
4.
1
4
(b)
1
12
(c)
1
6
(d)
1
3
pö
æ
If 2cos q + sin q = 1 ç q ¹ ÷ ,
2ø
è
then 7 cos q + 6 sin q is equal to:[Online April 11, 2014]
1
(a)
2
(c)
11
2
(b) 2
(d)
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation for Statement 1.
[2014]
(a)
46
5
Let f(x) = sin x, g(x) = x.
(c) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation for Statement 1.
(d) Statement 1 is false, Statement 2 is true.
7.
A triangular park is enclosed on two sides by a fence
and on the third side by a straight river bank. The two
sides having fence are of same length x. The maximum
area enclosed by the park is
[2006]
x3
8
(a) 3 x 2
2
(b)
(c) 1 x 2
2
(d) px 2
Downloaded from @Freebooksforjeeneet
M-11
Trigonometric Functions
13.
TOPIC n
8.
Trigonometric Identities,
Conditional Trigonometric
Identities, Greatest and Least
Value of Trigonometric Expressions
æ 3p ö
æ 3p ö
3æ pö
3æ pö
The value of cos ç ÷ .cos ç ÷ + sin ç ÷ .sin ç ÷ is
è8ø
è 8 ø
è8ø
è 8 ø
[Jan. 9, 2020 (I)]
(a)
(c)
9.
1
2
(b)
1
2
(d)
cos210
(a)
14.
1
2 2
15.
æ pö
æ pö
M = cos ç ÷ - sin 2 ç ÷ , then : [Sep. 05, 2020 (II)]
è 16 ø
è 8ø
11.
12.
(b) 5( 3 + 1)
5
(2 + 3)
2
(d) 10( 3 - 1)
The value of sin 10 sin 30 sin 50 sin 70 is:
1
16
1
(c)
18
1
32
1
(d)
36
(a)
16.
2
p
1
1
p
1
(b) L =
- cos
+ cos
4
8
2
8
4
2
2 2
1
p
1
(d) M =
+ cos
8
2 2 2
(b) 3/4
[April. 09, 2019 (II)]
æ pö
æpö
If L = sin 2 ç ÷ - sin 2 ç ÷ and
16
è ø
è8ø
p
1
+ cos
(c) M =
8
4 2 4
3
+ cos20
4
is :
3
(1 + cos20 )
(d) 3/2
2
Two poles standing on a hori ontal ground are of heights
5m and 10m respectively. The line oining their tops makes
an angle of 15 with the ground. Then the distance (in m)
between the poles, is:
[April. 09, 2019 (II)]
(c)
[Jan. 8, 2020 (II)]
1
– cos10 cos50 +
(a) 5(2 + 3)
1
4
2 sin a
1
1 - cos 2b
1
,
= and
=
If
2
1 + cos 2a 7
10
(a) L = -
[April 9, 2019 (II)]
cos250
(c)
æ pö
a , b Îç 0, ÷ , then tan (a + 2b) is equal to _____.
è 2ø
10.
The value of
1
(b)
3
5
p
, sin(a – b) =
and 0 < a, b < , then
5
13
4
tan(2a) is equal to :
[ April 8, 2019 (I)]
If cos (a + b) =
(a)
63
52
(b)
63
16
(c)
21
16
(d)
33
52
which the points (1, 2) and (sin q, cos q) lie on the same
If sin 4a + 4 cos4b + 2 = 4 2 sin a cos b ; a, bÎ[0, p], then
cos(a + b) – cos(a – b) is equal to : [Jan. 12, 2019 (II)]
(a) 0
(b) – 1
side of the line x + y = 1 is :
(c)
The set of all possible values of q in the interval (0, p) for
17.
[Sep. 02, 2020 (II)]
æ pö
(a) ç 0, ÷
è 2ø
æ p 3p ö
(b) ç ,
÷
è4 4 ø
æ 3p ö
(c) ç 0,
÷
è 4 ø
æ pö
(d) ç 0, ÷
è 4ø
The angle of elevation of the top of a vertical tower standing
on a hori ontal plane is observed to be 45o from a point A
on the plane. Let B be the point 30 m vertically above the
point A. If the angle of elevation of the top of the tower
from B be 30o, then the distance (in m) of the foot of the
tower from the point A is:
[April 12, 2019 (II)]
(a) 15(3 + 3)
(b) 15(5 - 3)
(c) 15(3 - 3)
(d) 15(1 + 3)
18.
2
Let f k ( x ) =
(
(d) – 2
)
1
sin k x + cosk x for k = 1, 2, 3, ... Then for
k
all x Î R, the value of f 4 ( x ) – f 6 ( x ) is equal to :
[Jan. 11, 2019 (I)]
(a)
1
12
(b)
1
4
–1
5
(d)
12
12
The value of
[Jan. 10, 2019 (II)]
p
p
p
p
cos 2 × cos 3 × .... × cos 10 × sin 10
2
2
2
2
1
1
(a)
(b)
512
1024
1
1
(c)
(d)
256
2
(c)
19.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-12
20.
Mathematics
If 5(tan2x – cos2x) = 2cos 2x + 9, then the value of cos 4x is
:
[2017]
(a) -
7
9
26.
3
5
(b) -
If cos (b – g) + cos (g – a) + cos (a – b) = -
1
2
(d)
3
9
If m and M are the minimum and the maximum values of
(c)
21.
1
4 + sin 2 2x - 2 cos 4 x, x Î R, then M – m is equal to :
2
[Online April 9, 2016]
(a)
(c)
22.
23.
9
4
(b)
7
4
(d)
(a)
1
4
1
2
(c)
28.
(b)
If 0 < x < p and cos x + sin x =
(b)
7
5
(a) (1 - 7 )
4
(c)
4
5
(d)
8
5
(c) –
p+q
p-q
(c)
æp qö
then cot ç + ÷ is
è 4 2ø
( p ¹ q ¹ 0) ,
(4 + 7 )
3
(d)
(1 + 7 )
4
If u = a 2 cos 2 q + b 2 sin 2 q + a 2 sin 2 q + b 2 cos 2 q
then the difference between the maximum and minimum
values of u2 is given by
[2004]
q
p
(c) (a + b)2
(d) 2(a 2 + b2 )
pq
(d) pq
13
£ A £1
(a)
16
(b) 1 £ A £ 2
3
13
£ A£
(c)
4
16
3
£ A £1
(d)
4
30.
π
where 0 £ α, β £ . Then tan 2a =
4
[2011]
value of cos
31.
[2010]
19
12
32.
25
(d)
16
Let a, b be such that p < a - b < 3p .
If sin a + sin b = -
5
4
Let cos (α + β) = and sin (a - b) = ,
13
5
20
(c)
7
(b) (4 - 7 )
3
(b) 2 a 2 + b2
(b)
(b)
1
, then tan x is[2006]
2
(a) (a - b)2
p
q
56
33
2
[Online April 9, 2014]
If A = sin2 x + cos4x, then for all real x :
(a)
29.
1
(d) 2.
2
3
5
If cosec q =
3
, then :[2009]
2
(a) A is false and B is true
(b) both A and B are true
(c) both A and B are false
(d) A is true and B is false
If p and q are positive real numbers such that p2 + q2 = 1,
then the maximum value of (p + q) is
[2007]
(a)
(a)
25.
27.
15
4
3
1
If cos a + cos b =
and sin a + sinb
and q is the
2
2
the arithmetic mean of a and b , then sin 2q + cos 2q
is equal to :
[Online April 11, 2015]
equal to:
24.
Let A and B denote the statements
A : cos a + cos b + cos g = 0
B : sin a + sin b + sin g = 0
21
27
and cos a + cos b = - , then the
65
65
a -b
2
[2004]
(a)
-6
65
(b)
(c)
6
65
(d) -
3
130
3
(
130
)
The function f ( x) = log x + x 2 + 1 , is
(a) neither an even nor an odd function
(b) an even function
(c) an odd function
(d) a periodic function.
The period of sin 2 q is
(b) p
(a) p 2
(c) 2 p
(d) p /2
Downloaded from @Freebooksforjeeneet
[2003]
[2002]
M-13
Trigonometric Functions
33.
Which one is not periodic?
| sin3x | +sin 2x
(a)
(c) cos 4x + tan2x
TOPIC Đ
34.
[2002]
(b) cos x + cos2x
(d) cos2x + sinx
æ 5
ö
(a) ç - , - 1÷
è 4
ø
1ù
é
(b) ê -1, - ú
2û
ë
1ù
æ 1
(c) ç - , - ú
2
4û
è
5ù
é 3
(d) ê - , - ú
4û
ë 2
35.
The number of distinct solutions of the equation,
36.
log1/2|sin x| = 2 – log1/2|cos x| in the interval [0, 2p], is _____.
[Jan. 9, 2020 (I)]
The number of solutions of the equation
é 5p 5p ù
1 + sin 4x = cos2 3x, x Î ê - , ú is : [April 12, 2019 (I)]
ë 2 2û
37.
Let S = {qÎ[–2 p, 2p] : 2 cos2q + 3 sinq = 0}.
Then the sum of the elements of S is: [April 9, 2019 (I)]
(a)
Solutions of Trigonometric
Equations
If the equation cos 4 q + sin 4 q + l = 0 has real solutions
for q, then l lies in the interval :
[Sep. 02, 2020 (II)]
(a) 3
(c) 7
39.
p
, then the number of values of x for which
2
40.
If 0 £ x <
41.
sin x – sin 2x + sin 3x = 0, is:
[Jan. 09, 2019 (II)]
(a) 3
(b) 1
(c) 4
(d) 2
The number of solutions of sin 3x = cos 2x, in the interval
42.
æp ö
ç 2 , p ÷ is
è
ø
(a) 3
(b) 4
(c) 2
(d) 1
[Online April 15, 2018]
If sum of all th e solution s of th e equation
æ
æp
ö
æp
ö 1ö
8cos x × ç cos ç + x ÷ × cos ç - x ÷ - ÷ - 1 in [0, p] is kp,
è6
ø
è6
ø 2ø
è
(b) 5
(d) 4
then k is equal to :
Let S be the set of all a Î R such that the equation, cos 2x
+ a sin x = 2a –7 has a solution. Then S is equal to :
(a) R
(b) [1, 4]
(c) [3,7]
(d) [2, 6]
If [x] denotes the greatest integer < x, then the system of
linear equations
[sin q] x + [–cos q] y = 0
[cot q] x + y = 0
[April 12, 2019 (II)]
æ p 2p ö and
(a) have infinitely many solutions if qÎ ç ,
÷
è2 3 ø
43.
44.
p 2p ö æ 7 p ö
(b) has a unique solution if q Î æç ,
÷ È ç p,
÷.
è2 3 ø è 6 ø
45.
æ p 2p ö
(c) has a unique solution if qÎ ç ,
÷ and have
è2 3 ø
æ 7p ö
infinitely many solutions if qÎ ç p,
÷.
è 6 ø
(d) have infinitely many solutions if
[2018]
(a)
13
9
(b)
8
9
(c)
20
9
(d)
2
3
If 0 £ x < 2p , then the number of real values of x, which
satisfy the equation
cos x + cos 2x + cos 3x + cos 4x = 0 is:
[2016]
(a) 7
(b) 9
(c) 3
(d) 5
The number of x Î [0, 2p] for which
2 sin 4 x + 18cos 2 x - 2 cos 4 x + 18sin 2 x = 1 is
æ 7p ö
has a unique solution if qÎ ç p,
÷.
è 6 ø
æ p 2p ö æ 7 p ö
qÎ ç ,
÷ È ç p,
÷
è2 3 ø è 6 ø
(b) 5p
3
(d) p
(c) 2p
[April 12, 2019 (II)]
38.
13p
6
46.
[Online April 9, 2016]
(a) 2
(b) 6
(c) 4
(d) 8
The number of values of a in [0, 2p] for which
2sin3a – 7 sin2 a + 7 sin a = 2, is: [Online April 9, 2014]
(a) 6
(b) 4
(c) 3
(d) 1
Let A = {q : sin(q) = tan(q)} and B = {q : cos(q) = 1} be two
sets. Then :
[Online April 25, 2013]
(a) A = B
(b) A Ë B
(c) B Ë A
(d) A Ì B and B - A ¹ f
Downloaded from @Freebooksforjeeneet
EBD_8344
M-14
47.
48.
49.
Mathematics
The number of solutions of the equation
sin 2x – 2 cos x + 4 sin x = 4 in the interval [0, 5p] is :
[Online April 23, 2013]
(a) 3
(b) 5
(c) 4
(d) 6
Statement-1: The number of common solutions of the
trigonometric equations 2 sin 2 q – cos 2q = 0 and
2 cos2 q – 3 sin q = 0 in the interval [0, 2p] is two.
Statement-2: The number of solutions of the equation, 2
cos2 q – 3 sin q = 0 in the interval [0, p] is two.
[Online April 22, 2013]
(a) Statement-1 is true; Statement-2 is true; Statement-2
is a correct explanation for statement-1.
(b) Statement-1 is true; Statement-2 is true; Statement-2
is not a correct explanation for statement-1.
(c) Statement-1 is false; Statement-2 is true.
(d) Statement-1 is true; Statement-2 is false.
The equation esinx – e–sinx – 4 = 0 has :
[2012]
(a) infinite number of real roots
(b) no real roots
(c) exactly one real root
(d) exactly four real roots
50.
The possible values of q Î( 0, p) such that
sin ( q) + sin ( 4q) + sin ( 7q) = 0 are
(a)
π 5π π 2π 3π 8π
,
, ,
,
,
4 12 2 3 4 9
(b)
2p p p 2p 3p 35p
, , , , ,
9 4 2 3 4 36
(c)
2p p p 2 p 3p 8p
, , , , ,
9 4 2 3 4 9
(d)
2π π 4π π 3π 8π
, ,
, ,
,
9 4 9 2 4 9
[2011RS]
51.
The number of values of x in the interval [0, 3p] satisfying
52.
the equation 2sin 2 x + 5 sin x - 3 = 0 is
[2006]
(a) 4
(b) 6
(c) 1
(d) 2
The number of solution of tan x + sec x = 2cos x in [0, 2 p )
is
[2002]
(a) 2
(b) 3
(c) 0
(d) 1
Downloaded from @Freebooksforjeeneet
M-15
Trigonometric Functions
1.
(b) 3(sin q – cos q)4 + 6(sin q + cos q)2 + 4sin6 q
2
æ 4ö
2
Now, sin2q + cos2q = 1 Þ sin q + ç ÷ = 1
è 5ø
= 3(1 – 2sin q cos q)2 + 6(1 + 2sin q cos q) + 4sin6 q
= 3(1 + 4sin2 q cos2 q – 4sin q cos q) + 6
–12sin q cos q + 4sin6 q
2
2
6
= 9 + 12sin q cos q+ 4 sin q
= 9 + 12cos2 q(1– cos2q) + 4(1 – cos2 q)3
= 9 + 12cos2q – 12cos4 q + 4(1 – cos6 q – 3cos2 q + 3cos4 q)
= 9 + 4 – 4cos6 q
= 13 – 4cos6 q
2.
1
(sin k x + cosk x )
k
1
4
4
Consider f 4 ( x) - f 6 ( x) = (sin x + cos x)
4
1
- (sin 6 x + cos6 x)
6
1
1
= [1 - 2sin 2 x cos 2 x] - [1 - 3sin 2 x cos 2 x]
4
6
4
16
9
2
=
= 1 Þ sin q = 1 5
25 25
3
sin q = ±
...(2)
5
3ö
3ö
æ
æ
Taking çè sin q = + ÷ø because çè sin q = - ÷ø
5
5
satisfy the given equation.
Therefore; 7 cosq + 6 sinq
sin2q +
(b) Let f k ( x) =
4
3 28 18 46
+6´ =
+ =
5
5 5 5
5
(b) Given expression can be written as
= 7´
4.
sin A
sin A
cos A
cos A
´
+
´
cos A sin A - cos A sin A cos A - sin A
æ
Q
ç
ç
ç
è
1 1 1
- =
4 6 12
(d) Given 2 cos q + sin q = 1
Squaring both sides, we get
=
3.
=
(2 cos q + sin q)2 = 12
Þ 4 cos 2 q + sin 2 q + 4sin q cos q = 1
Þ 3cos2 q + (cos 2 q + sin 2 q) + 4sin q cos q = 1
Þ 3cos2 q + 1 + 4sin q cos q = 1
-3
-3
= tan q = sec 2 q - 1 =
4
4
(
5.
Q tan q = sec q - 1
9
æ -3ö
2
Þ sec q - 1 = ç ÷ =
è 4ø
16
or cos q =
9
25
5
+1 =
Þ sec q =
16
4
16
4
5
sin 2 A + sin A cos A + cos 2 A
sin A cos A
= 1 + sec A cosec A
(c) Consider cos 255 + sin 195
= cos (270 – 15 ) + sin (180 + 15 )
= – sin 15 – sin 15
2
)
6.
æ 3 -1 ö
æ 3 -1 ö
= – 2 sin 15 = -2 çç
÷÷ = - çç 2 ÷÷
2
2
è
ø
è
ø
(c) Let f(x) = sin x and g(x) = x
Statement-1: f ( x) £ g ( x) "x Î( 0, ¥)
2
Þ sec2q =
ìï sin 3 A - cos3 A ïü
1
í
ý
sin A - cos A îï cos A sin A þï
=
Þ 3 cos 2 q + 4 sin q cos q = 0
Þ cos q(3cos q + 4sin q) = 0
Þ
sin A
ö
and
÷
cos A
÷
cos A
÷
cot A =
ø
sin A
tan A =
Q a3 - b3 = (a - b)(a 2 + ab + b 2 )
Þ 3 cos q + 4 sin q = 0 Þ 3cos q = -4 sin q
...(1)
cannot
i.e., sin x £ x"x Î ( 0, ¥ )
which is true
Statement-2: f ( x) £ 1 " x Î( 0, ¥)
i.e., sin x £ 1 " x Î( 0, ¥)
It is true and
g(x) = x ® ¥ as x ® ¥ also true.
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EBD_8344
M-16
7.
Mathematics
(c) Area =
1 2
x sin q
2
x
10.
q
Amax =
8.
(b)
p
2
pé
p
pù
cos ê4cos3 - 3cos ú
8ë
8
8û
pö
p
1æ 1
1
1
+ cos ÷ =
+ cos
M = ç
2è 2
8ø 2 2 2
8
11.
\ f (1, 2) × f (sin q, cos q) > 0
Þ 2[sin q + cos q - 1] > 0
éæ
p
p öù
= 4 êç cos 2 - sin 2 ÷ ú
8
8 øû
ëè
1
pö
æ
Þ sin q + cos q > 1 Þ sin ç q + ÷ >
è
4ø
2
éæ 4 p
4p
2p
2 p öù
êç sin 8 + cos 8 + sin 8 cos 8 ÷ ú
è
øû
ë
pé æ
p
pö ù
= cos ê 4 ç1 - sin 2 cos 2 ÷ - 3ú
4ë è
8
8ø û
9.
2 sin a 1
= and
2 cos a 7
Þ
2 sin b
1
=
2
10
12.
p æ p 3p ö
æ pö
Î ç , ÷ Þ q Î ç 0, ÷
è 2ø
4 è4 4ø
(a) Let the height of the tower be h and distance of the
foot of the tower from the point A is d.
By the diagram,
Q
1 - cos 2 b 1
=
2
10
B
30°
h
30 m
45°
1
1
tan a = and sin b =
7
10
tan b =
\
Þ q+
1 é 1ù
1
1- ú =
ê
2
2ë
û 2 2
(1)
\
(a) Let f ( x, y ) = x + y - 1
Given (1, 2) and (sin q, cos q) are lies on same side.
p
p
p
p
- 4sin 6 - 3cos 4 + 3sin 4
8
8
8
8
éæ
p
p öæ
p
p öù
-3 êç cos2 - sin 2 ÷ç cos 2 + sin 2 ÷ ú
8
8 øè
8
8 øû
ëè
...(ii)
pö
p
1æ 1
1
1
- cos ÷ =
- cos and
L= ç
2è 2
8ø 2 2 2
8
1 2
x
2
pé
p
pù
+ sin 3 ê3sin - 4sin 3 ú
8ë
8
8û
=
...(i)
and L – M = - cos
3
= 4cos6
1
p
p
= cos =
8
4
2
p
8
From equations (i) and (ii),
x
Maximum value of sinq is 1 at q =
2
(d) L + M = 1 - 2sin
1
3
A
tan 45 =
1 2
2 tan b
3
tan 2b =
= 3 =3=
2
1
8
4
1 - tan b 1 9 9
P
d
h
=1
d
h=d
2.
tan 30 =
...(i)
h - 30
d
3(h - 30) = d
...(ii)
tan a + tan 2b
tan(a + 2b) =
1 - tan a tan 2b
Put the value of h from (i) to (ii),
1 3
4 + 21
+
7
4
=
= 28 = 1
1 3
25
1- .
7 4
28
d=
3d = d + 30 3
30 3
= 15 3
3 -1
(
)
(
3 + 1 = 15 3 + 3
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)
M-17
Trigonometric Functions
13.
(b) cos2 10 – cos10 cos50 + cos2 50
2
12
æ5ö
cos (a – b) = 1 - ç ÷ =
13
è 13 ø
æ 1 + cos 20° ö æ 1 + cos100° ö 1
=ç
÷+ç
÷ - (2cos10° cos50°)
2
2
è
ø è
ø 2
5
12
Now, tan 2a = tan ((a + b) + (a – b))
Þ tan (a – b) =
1
1
= 1 + (cos 20° + cos100°) - [cos60° + cos 40°]
2
2
æ 1ö 1
= ç1 - ÷ + [cos20° + cos100° - cos 40°]
è 4ø 2
14.
=
3 1
+ [2cos 60° ´ cos 40° - cos 40°]
4 2
=
3
4
4 5
+
3 12 = 63
tan(a + b) + tan(a - b)
=
=
4 5 16
1 - tan(a + b).tan(a - b)
1- .
3 12
17.
sin4 a + 4 cos4 b + 2 = 4 2 sin a × cos b, a, b Î [0, p]
Then, by A.M., G.M. ineqality;
A.M. ³ G.M.
(a)
5
15º
(d) Q The given equation is
sin 4 a + 4cos 4 b + 1 + 1
³ sin 4 a × 4 cos 4 b × 1 ×1
4
(
10
)
1
4
sin4a + 4cos4b + 1 + 1 ³ 4 2 sin a× |cos b |
5
Inequality still holds when cosb < 0 but L.H.S. is positive
than cosb > 0, then
L.H.S. = R.H.S
15º
d
By the diagram,
tan15 =
(
)
5 3 +1
5
5
Þd =
=
d
tan15
3 -1
(
5 4+2 3
=
15.
\
2
Þ a=
) =5 2+ 3
(
)
\
1
sin3A
4
\ sin10 sin50 sin70 = sin10 sin(60 – 10)
1
sin30
4
1
1
sin230 =
4
16
(b) Q a + b and a – b both are acute angles.
16.
2
cos (a + b) =
tan (a + b) =
3
4
æ3ö
, then sin (a + b) = 1 - ç ÷ =
5
5
è5ø
4
3
And sin (a – b) =
5
, then
13
p
p
and b =
2
4
cos (a + b) – cos (a – b)
= –sinb – sinb = -2sin
18.
Þ sin10 sin30 sin50 sin70 =
1
4
æp
ö
æp ö
= cos çè + b÷ø - cos çè - bø÷
2
2
(a) Q sin(60 + A).sin(60 – A) sinA =
sin(60 + 10) =
sin4 a = 1 and cos4 b =
p
=- 2
4
1
k
k
(a) fk(x) = (sin x + cos x )
k
1
4
4
f4(x) = [sin x + cos x]
4
1é
(sin 2 x) 2 ù
2
2
2
ú
= 4 ê(sin x + cos x ) 2
ë
û
1 é (sin 2 x )2 ù
ú
= 4 ê1 2
ë
û
1
6
6
f6(x) = [sin x + cos x ]
6
1é
3
2
2
2
2ù
= ê(sin x + (cos x) - (sin x) ú
6ë
4
û
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EBD_8344
M-18
Mathematics
1é 3
2ù
= ê1 - (sin 2 x) ú
6ë 4
û
Now f4(x) – f(6)(x) =
=
19.
1 1 (sin 2 x )2 1
- + (sin 2 x )2
4 6
8
8
2
ìïæ
17
1ö
17 üï 1
³ -4 íç cos 2 x - ÷ - ý ³
è
4
4ø
16 ïþ 2
îï
1
12
(a) A = cos
p
p
p
p
.cos 3 ... cos 10 .sin 10
22
2
2
2
p
p
p
pö
1æ
= çè cos 2 .cos 3 ... cos 9 sin 9 ÷ø
2
2
2
2
2
=
1æ
p
pö 1
p
ç cos 2 .sin 2 ÷ø = 9 sin
2
28 è
2
2
2
22.
1
512
(a) We have
5 tan2 x – 5 cos2 x = 2 (2 cos2 x –1 ) + 9
Þ 5 tan2 x – 5 cos2 x = 4 cos2 x –2 + 9
Þ 5 tan2 x = 9 cos2 x + 7
Þ 5 (sec2 x – 1) = 9 cos2 x + 7
Let cos2 x = t
5
Þ - 9t - 12 = 0
t
Þ 9t2 + 12t – 5 = 0
Þ 9t2 + 15t – 3t – 5 = 0
Þ (3t – 1) (3t + 5) = 0
2
ìïæ
1ö
17 üï
- 4 íç cos 2 x - ÷ - ý
è
4ø
16 ïþ
îï
2
0 < cos x < 1
-
1
1 3
£ cos2 x - £
4
4 4
2
1ö
9
æ
0 £ ç cos2 x - ÷ £
4ø
16
è
1
2
17 2 15
- =
4 4 4
(b) Let cos a + cos b =
3
2
a+b
a -b 3
cos
=
2
2
2
...(i)
1
2
a +b
a -b 1
cos
=
2
2
2
On dividing (ii) by (i), we get
Þ 2sin
...(ii)
æ a + bö
1
tan ç
è 2 ÷ø = 3
a+b
Þ 2q = a + b
2
Consider sin 2q + cos 2q = sin (a + b) + cos (a + b)
Given : q =
2
1
16 8 7
3 +
9
+ =
=
=
1
1 10 10 5
1+
1+
9
9
2
cos2 x
1 1 ïü
ïì
- 4 ícos 4 x -1 + - ý
2
16
16 þï
îï
m=
and sin a + sin b =
æ1ö
1
cos 2x = 2 cos2 x – 1 = 2 ç ÷ – 1 = –
3
è3ø
21.
17
4
Þ 2 cos
1
5
Þ t = as t ¹ – .
3
3
7
æ 1ö
cos 4x = 2 cos2 2x – 1 = 2 ç - ÷ - 1 = 9
è 3ø
1
(b) 4 + sin2 2x – 2 cos4 x
2
4 + 2 (1 – cos2 x) cos2 x – 2 cos4 x
M=
M–m=
=
20.
2
17 æ 2
1 ö 17 9 17
£ ç cos x - ÷ - £ 16 è
4 ø 16 16 16
23.
(b) cosecq =
p-q
p+q
, sin q =
p+q
p-q
2
2 pq
æ p - qö
=
cos q = ± 1 - sin 2 q = 1 - ç
è p + q ø÷
( p + q)
æ p qö
cot ç + ÷ =
è 4 2ø
p
q
q
cot cot - 1
cot - 1
4
2
2
=
p
q
q
cot + cot
cot + 1
4
2
2
q
q
cos - sin
2
2
=
q
q
cos + sin
2
2
On rationali ing denominator, we get
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M-19
Trigonometric Functions
q
qöæ
q
qö
æ
cos - sin
cos + sin
ç
2
2÷ ç
2
2÷
ç
q
q֍
q
q÷
çè cos + sin ÷ø çè cos + sin ÷ø
2
2
2
2
=
cos q
q
q
q
2q
sin
+ cos + 2sin cos
2
2
2
2
(d)
pq
=
p
- a 2 + b2 £ a cos q + b sin q £ a 2 + b 2
q
p
A = sin 2 x + cos 4 x
\ – 2 £ cos q + sin q £ 2
Hence max. value of p + q is
28.
= sin 2 x + cos 2 x(1 - sin 2 x )
cos(a + b) =
sin(a - b) =
Þ
1
1
Þ 1 + sin 2 x =
2
4
3
,
4
...(i)
3
4
1 + tan x
2
Þ 3 tan x + 8 tan x + 3 = 0
2
\ tan x =
for
(a)
-8 ± 64 - 36
-4 ± 7
=6
3
p
< x < p , tan x < 0
2
\ tan x =
29.
=-
-4 - 7
3
u 2 = a2 + b2 + 2
tan 2a = tan [ (a + b) + (a - b)]
26.
3 5
+
tan(a + b) + tan(a - b)
56
=
= 4 12 =
1 - tan(a + b ) tan(a - b ) 1 - 3 . 5 33
4 12
(b) Given that
3
2
Þ 2 [cos (b – g ) + cos ( g – a) + cos (a – b)] + 3 = 0
Þ 2 [cos (b – g ) + cos ( g – a) + cos (a – b)]
+ sin2 a + cos2 a + sin2 b + cos2 b
+ sin2 g + cos2 a = 0
2
2
2
Þ [sin a + sin b + sin g + 2 sin a sin b
+ 2 sin b sin g + 2 sin g sina ] + [cos2a + cos2 b
+ cos2 g + 2cosa cosb + 2 cos b cos g
+ 2cos g cos a] = 0
cos (b – g ) + cos ( g – a) + cos (a – b) = -
2
p
<x£p
2
2 tan x
4
3
Þ tan(a + b) =
5
4
5
5
Þ tan(a - b) =
13
12
cos x + sin x =
\ p < 2x < 2p
1
1
Þ 0 ³ – sin 2 (2 x ) ³ 4
4
1
Þ 1 ³ 1 - sin 2 (2 x ) ³ 1 - 1
4
4
3
Þ1 ³ A ³
4
(a)
(c)
Þ sin 2 x = -
1
= sin 2 x + cos 2 x - (2 sin x.cos x) 2
4
1 2
= 1 - sin (2 x )
4
Q –1 £ sin 2x £ 1
Þ 0 £ sin 2 (2 x) £ 1
25.
[sina + sin b + sin g ]2 + (cos a + cos b + cos g )2 = 0
sina + sin b + sin g = 0 and cos a + cos b + cos g = 0
A and B both are true.
Given that p2 + q2 = 1
p = cos q and q = sin q satisfy the given equation
Then p + q = cos q + sin q
We know that
2
2 pq / ( p + q )
cos q
=
=
=
( p - q)
1 + sin q
1+
p+q
24.
27.
Þ
Þ
\
(c)
\
( a 4 + b 4 ) cos 2 q sin 2 q
+ a 2b 2 (cos 4 q + sin 4 q)
… (1)
Now, (a 4 + b4 ) cos 2 q sin 2 q + a 2 b2 (cos4 q + sin 4 q )
= (a4 + b4 ) cos 2 q sin 2 q + a 2 b2 (1 - 2cos 2 q sin 2 q )
= (a4 + b4 - 2a 2b 2 ) cos 2 q sin 2 q + a 2 b2
= (a 2 - b 2 )2 .
sin 2 2q
+ a 2b2
4
…(2)
Q 0 £ sin 2 2q £ 1
Þ 0 £ (a 2 - b 2 )2
sin 2 2q (a 2 - b 2 ) 2
£
4
4
Þ a 2b2 £ (a 2 - b 2 )2
sin 2 2q
+ a 2b2
4
1
£ (a 2 - b 2 )2 . + a 2b2
4
[Q cos(A - B) = cosA.cosB + sinA.sinB]
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....(3)
EBD_8344
M-20
Mathematics
From (1)
a 2 + b 2 + 2 a 2b 2 £ u 2 £ a 2 + b 2 +
(
(a + b)2 £ u 2 £ 2 a 2 + b2
30.
)
2
2
( a 2 + b2 )
Þ as sin 2 2q Î[0, 1]
2
\ Max. value – Min. value
(8) log1/2 |sinx| = 2 – log1/2 |cosx|
Þ log1/2 |sinx cosx| = 2
= 2(a 2 + b2 ) - (a + b 2 ) = (a - b)2
Þ |sinx cosx| =
(d)
Þ
35.
p < a - b < 3p
p a - b 3p
a-b
<
<
Þ cos
<0
2
2
2
2
sin a + sin b = -
Þ 2 sin
Þ sin2x = ±
....(1)
21
a+b
a-b
cos
=2
2
65
27
65
36.
27
a+b
a -b
cos
=2
2
65
Squaring and adding (2) and (3), we get
4 cos 2
1170
a - b (21) 2 + (27) 2
=
=
2
2
65 ´ 65
(65)
\ cos 2
9
3
a -b
a -b
=
Þ cos
=2
130
2
130
....(3)
[from (1)]
37.
(c) Given f ( x ) = log( x + x 2 + 1)
{
}
f (- x) = log - x + x2 + 1
ìï x 2 - x 2 + 1 üï
= log í
ý
2
îï x + x + 1 þï
(b) We know that sin 2 q =
Þ sin x =
1 - cos 2q
;
2
2p
=p
2
2
Hence period of sin q is also p.
Since period of cos 2q =
(b) we know that cos x is non periodic
\ cos x + cos 2 x can not be periodic.
34.
(b)
sin 4 q + cos 4 q = -l
Þ (sin 2 q + cos2 q) 2 - 2sin 2 q× cos 2 q = -l
Þ 1 - 2sin 2 q cos 2 q = -l
Þl=
(sin 2q)
-1
2
2
Hence, total number of solutions = 8.
(c) Consider equation, 1 + sin 4x = cos23x
L.H.S. = 1 + sin4 x and R.H.S. = cos23x
QL.H.S. ³ 1 and R.H.S. £ 1 Þ L.H.S. = R.H.S. = 1
sin4x = 0, and cos23x = 1
Þ sin x = 0 and (4cos2x – 3)2 cos2x = 1
Þ sin x = 0 and cos2x = 1 Þ x = 0, ±p, ±2p
Hence, total number of solutions is 5.
(d) Given equation is, cos 2x + a sin x = 2a – 7
1 – 2sin2x + a sin x = 2a – 7
2sin2x – a sinx + (2a – 8) = 0
Þ sin x =
= - log( x + x 2 + 1) = - f ( x )
Þ f(x) is an odd function.
33.
1
2
....(2)
Þ 2 cos
32.
1
4
21
65
cos a + cos b = -
31.
-1ù
é
Þ l Î ê -1, ú
2û
ë
Q
a ± a 2 - 8(2a + 8)
4
a ± (a - 8)
a-4
Þ sin x =
4
4
[sin x = 2 (reected)]
equation has solution, then
a-4
Î [-1,1]
4
Þ a Î [2,6]
38.
(a) According to the question, there are two cases.
æ p 2p ö
Case 1 : qÎ ç , ÷
è2 3 ø
In this interval, [sin q] = 0, [– cos q] = 0 and [cot q] = – 1
Then the system of equations will be ;
0 . x + 0 . y = 0 and – x + y = 0
Which have infinitely many solutions.
æ 7p ö
Case 2 : qÎ ç p, ÷
è 6 ø
In this interval, [sin q] = – 1 and [– cos q] = 0,
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M-21
Trigonometric Functions
Then the system of equations will be ;
– x + 0 . y = 0 and [cot q] x + y = 0
Þ
æ1
ö
8cos x ç - 1 + cos 2 x ÷ = 1
4
è
ø
Þ
3ö
æ
8cos x ç cos 2 x - ÷ = 1
4ø
è
Þ
æ 4cos3 x - 3cos x ö
8ç
÷ =1
ç
÷
4
è
ø
Þ
2(4 cos3 x - 3cos x) = 1
Þ
2cos3x = 1 Þ cos3x =
\
3x = 2np ±
Þ
x=
Clearly, x = 0 and y = 0 which has unique solution.
39.
(c) 2cos2q + 3sinq = 0
(2sinq + 1) (sinq – 2) = 0
Þ sinq = -
1
or sin q = 2 ® Not possibe
2
p
6
p
6
pö æ
pö æ pö æ
pö
æ
= ç p + ÷ + ç 2p - ÷ + ç - ÷ + ç -p + ÷ = 2p
6
6
6
6ø
è
ø è
ø è
ø è
(d) sinx – sin2x + sin3x = 0
Þ sinx – 2 sinx.cosx + 3 sinx – 4 sin3x = 0
Þ 4 sinx – 4
43.
– 2 sinx.cosx = 0
13
æ1 2 1 2 1ö
= ç + + + - ÷p = p
9
3
9
3
9
9
è
ø
(a) cos x + cos 2x + cos 3x + cos 4x = 0
Þ 2 cos 2x cos x + 2 cos 3x cos x = 0
5x
xö
æ
Þ 2cos x ç 2cos cos ÷ = 0
è
2
2ø
Þ 2 sinx.cos2x – sinx.cosx = 0
Þ sinx.cosx(2 cosx – 1) = 0
cos x = 0, cos
sinx = 0, cosx = 0, cosx =
1
2
é pö
p
Q x Î ê0, ÷ø
3
ë 2
(d) sin 3x = cos 2x
Þ 3 sin x – 4 sin3 x = 1 – 2 sin2 x
Þ 4 sin3 x – 2 sin2 x – 3 sin x + 1 = 0
\
x = 0,
-2±2 5
Þ sin x = 1,
8
-2+2 5
æp ö
In the interval ç , p ÷ , sin x =
2
8
è
ø
So, there is only one solution.
42.
2np p
±
3
9
Þ 2 sinx(1– sin2x) – sinx.cosx = 0
\
41.
sin3x
p
, n Î1
3
p 2p p 2p p
In x Î [0, p]: x = , + , - , only
9 3 9 3 9
Sum of all the solutions of the equation
The required sum of all solutions in [–2p , 2p] is
40.
1
2
p
1
(a) Q 8cos x æç cos 2 - sin 2 x - ö÷ = 1
6
2ø
è
Þ
æ3 1
ö
8cos x ç - - sin 2 x ÷ = 1
è4 2
ø
Þ
æ1
ö
8cos x ç - (1 - cos 2 x) ÷ = 1
è4
ø
5x
x
= 0 , cos = 0
2
2
p 3p p 3p 7p 9p
x = p, , , , , ,
2 2 5 5 5 5
44.
2sin 4 x + 18cos 2 x - 2cos 4 x + 18sin 2 x = 1
(d)
2sin 4 x + 18cos2 x - 2 cos 4 x + 18sin 2 x = ± 1
45.
2sin 4 x + 18cos 2 x = ±1 + 2cos4 x + 18sin 2 x
by squaring both the sides we will get 8 solutions
(c) 2 sin3a – 7sin2a + 7 sina – 2 = 0
Þ 2 sin2a (sina – 1) – 5 sina (sina – 1)
+ 2 (sina – 1) = 0
Þ (sin a – 1) (2 sin2 a – 5 sina + 2) = 0
Þ sin a – 1 = 0 or 2 sin2 a – 5 sina + 2 = 0
sin a = 1 or sin a =
a=
p
2
or sin a =
5 ± 25 - 16 5 ± 3
=
4
4
1
,2
2
Now, sin a ¹ 2
Downloaded from @Freebooksforjeeneet
EBD_8344
M-22
Mathematics
49.
1
for, sin a =
2
p 2p
,
3 3
There are three values of a between [0, 2p]
(b) Let A = {q : sin q = tan q}
a=
46.
(b) Given equation is esinx – e–sinx – 4 = 0
Put esin x = t in the given equation, we get
t2 – 4t – 1 = 0
Þ
=
and B = {q : cos q = 1}
47.
48.
sin q ü
ì
Now, A = íq : sin q =
ý
cos
qþ
î
= {q : sin q (cos q – 1) = 0}
= {q = 0, p, 2p, 3p,.....}
For B : cos q = 1 Þ q = p, 2p, 4p,......
This shows that A is not contained in B. i.e. A Ë B. but
B Ì A.
(a) sin 2x – 2 cos x + 4 sin x = 4
Þ 2 sin x . cos x – 2 cos x + 4 sin x – 4 = 0
Þ (sin x – 1) (cos x – 2) = 0
Q cos x – 2 ¹ 0, \ sin x = 1
Þ
Þ4
\ q=
4± 2 5
= 2± 5
2
esin x = 2 ± 5 (Q t = esin x )
Þ
esin x = 2 - 5 < 0
and sin x = ln(2 + 5) > 1
50.
So, reected.
Hence, given equation has no solution.
\ The equation has no real roots.
(d) sin 4q + 2sin 4q cos 3q = 0
sin 4q (1 + 2cos 3q) = 0
sin 4q = 0
or
cos 3q = -
1
2
4q = n p ; n Î I
or 3q = 2np ±
q=
1
= 1 Þ sin q = ±
2
p p 3p
, ,
4 2 4
2p
, n ÎI
3
or
q=
2p 8p 4 p
, ,
9 9 9
p 3p 5p 7 p
, ,
,
, q Î [0, 2 p]
4 4 4 4
y=
51.
But sin q = – 2, is not possible
sin q =
1
, –2
2
Þ q=
p 5p
,
6 6
Hence, there are two common solution, there each of the
statement-1 and 2 are true but statement-2 is not a correct
explanation for statement-1.
x' O
3p
Þ sin x =
1
2
x
1
2
y = sin x
2sin 2 x + 5 sin x - 3 = 0
Þ (sin x + 3)(2sin x - 1) = 0
52.
1
, -2
2
(a)
y'
Þ sin q (2 sin q – 1) + 2 (2 sin q – 1) = 0
Þ sin q =
[Q q, Î (0, p)]
y
p 5p 7 p 11p
,
\ q= , ,
6 6 6 6
Now 2 cos2 q – 3 sin q = 0
Þ 2 (1 – sin2 q) – 3 sin q = 0
Þ – 2 sin2 q – 3 sin q + 2 = 0
Þ – 2 sin2 q – 4 sin q + sin q + 2 = 0
Þ 2 sin2 q – sin q + 4 sin q – 2 = 0
\
4 ± 16 + 4 4 ± 20
=
2
2
sin x
= 2 - 5 and esin x = 2 + 5
Þ e
p 5p 9 p
\ x= , ,
2 2 2
(b) 2 sin2 q – cos 2q = 0
Þ 2 sin2 q – (1 – 2 sin2 q) = 0
Þ 2 sin2 q – 1+ 2 sin2q = 0
sin2q
t=
and sin x ¹ -3
\ In [0, 3p] , x has 4 values.
(b) Q tan x + sec x = 2 cos x;
Þ sin x + 1 = 2cos2 x
Þ sin x + 1 = 2(1 – sin 2 x);
Þ 2sin2x + sin x – 1= 0;
Þ (2sin x – 1)(sin x + 1) = 0
1
, –1.;
2
Þ x = 30 , 150 , 270 .
Number of solution = 3
Þ sin x =
Downloaded from @Freebooksforjeeneet
M-23
Principle of Mathematical Induction
4
Principle of
Mathematical Induction
TOPIC Ć
1.
2.
(a) Principle of mathematical induction can be used to
prove the formula
Problems Based on Sum of Series,
Problems Based on Inequality and
Divisibility
Consider the statement: “P(n) : n2 – n + 41 is prime.” Then
which one of the following is true? [Jan. 10, 2019 (II)]
(a) Both P(3) and P(5) are true.
(b) P(3) is false but P(5) is true.
(c) Both P(3) and P(5) are false.
(d) P(5) is false but P(3) is true.
Let S ( K ) = 1 + 3 + 5... + (2 K - 1) = 3 + K 2 . Then which of
the following is true
[2004]
(b) S ( K ) Þ S ( K + 1)
(c) S ( K ) Þ
/ S ( K + 1)
(d) S (1) is correct
3.
If an = 7 + 7 + 7 + ... ... having n radical signs then by
methods of mathematical induction which is true [2002]
(a) an > 7 " n ³ 1
(b) an < 7 " n ³ 1
(c) an < 4 " n ³ 1
(d) an < 3 " n ³ 1
Downloaded from @Freebooksforjeeneet
EBD_8344
M-24
1.
2.
Mathematics
(a)
Þ
&
\
(b)
P(n) = n2 – n + 41
P(3) = 9 – 3 + 41 = 47 (prime)
P(5) = 25 – 5 + 41 = 61 (prime)
P(3) and P(5) are both prime i.e., true.
S(K) = 1+3+5+...+(2K – 1) = 3 + K2
Adding 2K + 1 on both sides
Þ 1 + 3 + 5.... + (2 K - 1) + 2 K + 1
= 3 + K 2 + 2K + 1 = 3 + (K + 1)2 = S ( K + 1)
\ S ( K ) Þ S ( K + 1)
S (1) :1 = 3 + 1, which is not true
Q S (1) is not true.
\ P.M.I cannot be applied
Let S(K) is true, i.e.
S(K) : 1 + 3 + 5.... + (2 K - 1) = 3 + K 2
3.
(b) For n = 1, a1 =
Then am + 1 =
7 < 7. Let am < 7.
7 + am
Þ a2m + 1 = 7 + am < 7 + 7 < 14.
Þ am + 1 < 14 < 7; So, by the principle of mathematical
induction an < 7, " n.
Downloaded from @Freebooksforjeeneet
M-25
Complex Numbers and Quadratic Equations
5
Complex Numbers and
Quadratic Equations
TOPIC Ć
1.
If
Integral Powers of lota, Algebraic
Operations of Complex Numbers,
Conjugate, Modulus and Argument
or Amplitude of a Complex Number
[Jan. 7, 2020 (II)]
-1 æ 4 ö
(a) p - tan çè ÷ø
3
(a)
4ö
(d) tan çè ÷ø
3
If the four complex numbers z, z , z - 2Re( z ) and
z - 2Re( z ) represent the vertices of a square of side
4 units in the Argand plane, then |z| is equal to :
[Sep. 05, 2020 (I)]
(c) 2 2
(b) 4
8.
(d) 2
4.
5.
æ -1 + i 3 ö
The value of çç
[Sep. 05, 2020 (II)]
÷÷ is :
è 1- i ø
(a) – 215 (b) 215 i
(c) – 215 i (d) 65
æ1+ iö
If ç
è 1 - i ÷ø
m/2
æ1+ iö
=ç
è i - 1÷ø
9.
= 1, (m, n Î N ) , then the
(a)
3
(b) 2 3
(c)
3
2
(d)
1
3
10
(c)
(d)
7
8
2z - n
= 2i - 1 for
2z + n
[April 12, 2019 (II)]
Let z Î C with Im(z) = 10 and it satisfies
The equation z - i = z - 1 , i = -1 , represents:
1
(a) a circle of radius .
2
(b) the line through the origin with slope 1.
(c) a circle of radius 1.
(d) the line through the origin with slope – 1.
greatest common divisor of the least values of m and n is
_________.
[Sep. 03, 2020 (I)]
If z1, z2 are complex numbers such that Re(z1) = |z1 – 1|,
2
(b)
[April 12, 2019 (I)]
n/3
p
Re(z2) = |z2 – 1| and arg( z1 - z2 ) = , then Im( z1 + z2 ) is
6
equal to :
[Sep. 03, 2020 (II)]
17
2
[Jan. 9, 2020 (II)]
some natural number n. Then :
(a) n = 20 and Re(z) = –10
(b) n = 40 and Re(z) = 10
(c) n = 40 and Re(z) = –10
(d) n = 20 and Re(z) = 10
30
3.
If z be a complex number satisfying |Re(z)| + |Im(z)| = 4,
then |z| cannot be:
-1 æ
(c) - tan çè ÷ø
4
(a) 4 2
7.
-1 æ 3 ö
(b) p - tan çè ÷ø
4
-1 æ 3 ö
z -i
=1
z + 2i
Let z be a complex number such that
5
and z = . Then the value of |z + 3i| is :
2
[Jan. 9, 2020 (I)]
7
15
(a) 10
(b)
(c)
(d) 2 3
2
4
3 + i sin q
, q Î [0, 2p], is a real number, then an argument
4 - i cos q
of sinq + icosq is:
2.
6.
10.
If a > 0 and =
equal to :
1 3
(a) - - i
5 5
(c)
1 3
- i
5 5
2
(1 + i )2
, has magnitude
, then is
5
a -i
[April 10, 2019 (I)]
3 1
(b) - - i
5 5
1 3
(d) - + i
5 5
Downloaded from @Freebooksforjeeneet
EBD_8344
M-26
11.
Mathematics
If z and w are two complex numbers such that zw = 1 and
arg(z) – arg(w) =
12.
p
, then:
2
(c) zw = -i
1- i
(d) z w =
2
[April 09, 2019 (II)]
(b) 4 Im (w) > 5
(d) 5 Im (w) < 1
z -a
(a Î R ) is a purely imaginary number and | | = 2,
z +a
then a value of a is :
[Jan. 12, 2019 (I)]
1
(d) 2
2
Let 1 and 2 be two complex numbers satisfying | 1| = 9
and | 2 | – |3|–|4i||=|4. Then the minimum value of
| 1 – 2| is :
[Jan. 12, 2019 (II)]
(b) 1
(b)
(c) 1
2
34
(a)
3
Let 1 and
that 3 |
20.
21.
5
41
(c)
(d)
4
4
2 be any two non- ero complex numbers such
=
3
2
1
2
+
2
3
2
(b) | | =
Then the sum of the elements in A is:
5p
6
(b) p
(c)
3p
4
(d) –5
1
:
2
[2014]
(a) is strictly greater than
5
2
(b) is strictly greater than
3
5
but less than
2
2
22.
5
2
(d) lie in the interval (1, 2)
For all complex numbers of the form 1 + ia, a Î R , if
2 = x + iy, then
[Online April 19, 2014]
(a) y2 – 4x + 2 = 0
(b) y2 + 4x – 4 = 0
(c) y2 – 4x – 4 = 0
(d) y2 + 4x + 2 = 0
23.
Let
5
2
(d) Im( ) = 0
(c) –2
(c) is equal to
then:
1
[Online April 11, 2015]
If z is a complex number such that z ³ 2, then the minimum
[Jan. 11, 2019 (II)]
¹ – i be any complex number such that
-i
is a
+i
purely imaginary number.
ì æ p ö 3 + 2isin q
ü
is purely imaginary ý .
Let A= íqÎ ç - , p ÷ :
2
1
2isin
q
ø
î è
þ
(a)
is :
(b) –4
value of z +
+ =3+i
[Jan. 10 2019 (II)]
17.
(lmz )5
(a) –1
5
(b)
3
1 17
(c) | | =
2 2
If
-1 æ 1 ö
(b) sin ç
÷
è 3ø
lmz 5
)
(a) Re( ) = 0
2 + 3isin q
is purely imaginary, is:
1 - 2isin q
[2016]
p
p
(d)
3
6
is a non-real complex number, then the minimum
value of
-1 .
1 | = 4 | 2 |. If
1 + (1 – 8a) z
is a purely
1– z
(d) equal to R
A value ofqfor which
(c)
(d) 2
Then | | is equal to :
1ü
ý
4þ
-1 æ 3 ö
(a) sin çç 4 ÷÷
è
ø
(c)
Let be a complex number such that
( where i =
16.
19.
If
(a) 0
15.
ì 1
(c) í0, , –
î 4
5 + 3z
Let z Î C be such that |z| < 1. If w = 5(1 - z ) , then :
(a) 2
14.
-1 + i
2
(b) z w =
The set of all a Î R, for which w =
imaginary number, for all z Î C satisfying |z| = 1 and
Re z ¹ 1, is
[Online April 15, 2018]
(a) {0}
(b) an empty set
[April 10, 2019 (II)]
(a) zw = i
(a) 5 Re (w) > 4
(c) 5 Re (w) > 1
13.
18.
[Jan. 9 2019 (I)]
(d)
2p
3
Then
(a)
(b)
(c)
(d)
+
1
is:
[Online April 12, 2014]
ero
any non- ero real number other than 1.
any non- ero real number.
a purely imaginary number.
Downloaded from @Freebooksforjeeneet
M-27
Complex Numbers and Quadratic Equations
24.
If 1,
then
2 and 3, 4 are 2 pairs of complex conugate numbers,
æ ö
æ ö
arg ç 1 ÷ + arg ç 2 ÷ equals:
è 4ø
è 3ø
25.
31.
26.
{
:
) , for some real number k, is
= 1}
(c)
{
: ¹ 1}
(b)
{
: =
(d)
{
:
= 1, ¹ 1}
æ 1+ z ö
argument q, then arg ç
equals:
è 1 + z ÷ø
(a) –q
27.
}
If z is a complex n umber of unit modulus and
p
–q
2
(b)
(c) q
(d) p – q
p
3
[Online April 25, 2013]
Statement 1 is true Statement 2 is true; Statement 2 is
a correct explanation for Statement 1.
Statement 1 is false; Statement 2 is true
Statement 1 is true, Statement 2 is false.
Statement 1 is true; Statement 2 is true; Statement 2 is
not a correct explanation for Statement 1.
(b)
(c)
(d)
æ 1+ z
Let a = Im ç
ç 2iz
è
2
33.
34.
[Online April 23, 2013]
36.
Z2
Z1
2Z1 + 3Z 2
is a purely imaginary number, then 2Z - 3Z is equal to:
1
2
(a) 2
(b) 5
(c) 3
[Online April 9, 2013]
(d) 1
2
(b) 2 z1 + z2
2
z1 z2
2
)
2
Statement 2: Z1 + Z 2 £ Z1 + Z 2 [Online May 7, 2012]
(a) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation of Statement 1.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation of Statement 1.
(c) Statement 1 is true, Statement 2 is false.
(d) Statement 1 is false, Statement 2 is true.
The number of complex numbers such that
|z – 1| = |z + 1| = |z – i| equals
[2010]
(a) 1
(b) 2
(c) ¥
(d) 0
The conugate of a complex number is
–1
i –1
(b)
1
i +1
(c)
–1
i +1
1
ö
÷ , where z is any non- ero complex
÷
ø
If Z1 ¹ 0 and Z2 be two complex numbers such that
(
)
Statement 2: |Z| = |W|, implies arg Z – arg W = p.
(a) Statement 1 is true, Statement 2 is false.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation for Statement 1.
(c) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation for Statement 1.
(d) Statement 1 is false, Statement 2 is true.
Let Z1 and Z2 be any two complex number.
(a)
The set A = {a : | z | = 1 and z ¹ ±1 } is equal to:
(a) (– 1, 1) (b) [– 1, 1] (c) [0, 1)
(d) (– 1, 0]
29.
is equal to [Online May 26, 2012]
complex number is
35.
number.
2
Statement 1: Z1 - Z 2 ³ Z1 - Z 2
Let z satisfy| z | = 1 and = 1– z .
Statement 1 : z is a real number.
(a)
28.
32.
[2013]
Statement 2 : Principal argument of is
+ z1 - z2
Statement 1:If arg Z + arg W = p, then Z = -W .
[Online April 9, 2014]
(a)
2
(d) z1 + z2
Let Z and W be complex numbers such that |Z| = |W|, and
arg Z denotes the principal argument of Z.
[Online May 19, 2012]
(c)
(b)
w - w = k (1 -
z1 + z2
(a) 2 ( z1 + z2
[Online April 11, 2014]
p
3p
(c)
(d) p
2
2
Let w (Im w ¹ 0) be a complex number. Then the set of all
complex number satisfying the equation
(a) 0
30.
1
then that
i –1
[2008]
(d)
æx
1
i –1
yö
If z = x - i y and z 3 = p + iq, then çè p + q ø÷ ( p + q )
is equal to
[2004]
(a) –2
(b) –1
(c) 2
(d) 1
Let and w be complex numbers such that z + i w = 0
and arg zw = p. Then arg equals
[2004]
(a)
5p
4
(b)
p
2
(c)
3p
4
(d)
2
2
p
4
x
37.
æ1+ i ö
÷ = 1 then
è1- i ø
If ç
(a)
(b)
(c)
(d)
x = 2n + 1 , where n is any positive integer
x = 4n , where n is any positive integer
x = 2n , where n is any positive integer
x = 4n + 1 , where n is any positive integer..
Downloaded from @Freebooksforjeeneet
[2003]
EBD_8344
M-28
38.
Mathematics
If z and w are two non- ero complex numbers such that
p
zw = 1 and Arg ( z ) - Arg (w ) = , then zw is equal to
2
[2003]
(a) – 1
(b) 1
(c) – i
(d) i
39. If | – 4 | < | – 2 |, its solution is given by
[2002]
(a) Re(z) > 0
(b) Re(z) < 0
(c) Re(z) > 3
(d) Re(z) > 2
40. z and w are two non ero complex numbers such that
| z | = | w| and Arg + Arg w = p then equals [2002]
(a) w
(b) – w
(c) w
(d) – w
TOPIC n
41.
Rotational Theorem, Square Root
of a Complex Number, Cube Roots
of Unity, Geometry of Complex
Numbers, De-moiver’s Theorem,
Powers of Complex Numbers
(a) line, y = –x
(c) line, y = x
2
(b) straight line whose slope is - .
3
(c) straight line whose slope is
(d) circle whose diameter is
47.
-1 , then z lies on the:
[Sep. 06, 2020 (II)]
(b) imaginary axis
(d) real axis
-1 + i 3
, then a + b is equal to :
2
[Sep. 04, 2020 (II)]
(b) 24
(c) 33
(d) 57
2p
2p ö
æ
1 + sin
+ i cos
ç
9
9 ÷ is :
The value of ç
2p
2p ÷
- i cos ÷
ç 1 + sin
è
9
9 ø
1
(1 - i 3)
2
(b)
1
(c) - ( 3 - i )
2
44.
1
( 3 - i)
2
1
(d) - (1 - i 3)
2
(b) -2 6
(c) 6
100
45.
50.
51.
The imaginary part of (3 + 2 -54)1/ 2 - (3 - 2 -54)1/ 2
can be :
[Sep. 02, 2020 (II)]
(a) - 6
x + iy
1 ö
æ
i = -1 , where x and y are real
Let ç -2 - i ÷ =
3 ø
27
è
numbers then y – x equals :
[Jan. 11, 2019 (I)]
(a) 91
(b) – 85
(c) 85
(d) – 91
(
)
5
æ 3 iö æ 3 iö
Let z = çç
+ ÷÷ + çç
- ÷÷ . If R(z) and I(z)
2ø è 2
2ø
è 2
respectively denote the real and imaginary parts of z,
then:
[Jan. 10, 2019 (II)]
(a) I(z) = 0
(b) R(z) > 0 and I(z) > 0
(c) R(z) < 0 and I(z) > 0 (d) R(z) = – (c)
n
[Sep. 02, 2020 (I)]
(a)
5
.
2
5
49.
3
43.
3
.
2
If z = 3 + i ( i = -1 ) , then (1 + iz + z5 + iz8)9 is equal
2 2
to:
[April 08, 2019 (II)]
(a) 0
(b) 1
(c) (– 1 + 2i)9
(d) – 1
3
48.
If a and b are real numbers such that (2 + a ) = a + ba ,
(a) 9
[Jan. 7, 2020 (I)]
æ 1 3ö
(a) circle whose centre is at èç - , - ÷ø .
2 2
4
where a =
æ z -1 ö
If Re çè
÷ = 1, where z = x + iy, then the point (x, y) lies
2z + i ø
on a:
Let z = x + iy be a non- ero complex number such that
z2 = i |z|2, where i =
42.
46.
(d)
6
100
2k
3k
-1 + i 3
. If a = (1 + a) å a and b = å a ,
2
k =0
k =0
then a and b are the roots of the quadratic equation:
[Jan. 8, 2020 (II)]
(a) x2 + 101x + 100 = 0
(b) x2 – 102 x + 101 = 0
(c) x2 – 101x + 100 = 0
(d) x2 + 102x +101 = 0
52.
æ1+ i 3 ö
The least positive integer n for which çç
÷÷ = 1, is
è1 – i 3 ø
[Online April 16, 2018]
(a) 2
(b) 6
(c) 5
(d) 3
The point represented by 2 + i in the Argand plane moves
1 unit eastwards, then 2 units northwards and finally from
there 2 2 units in the south–westwards direction. Then
its new position in the Argand plane is at the point
represented by :
[Online April 9, 2016]
(a) 1 + i
(b) 2 + 2i
(c) –2 – 2i (d) –1 – i
A complex number is said to be unimodular if | | = 1.
Let a =
Suppose
1 and 2 are complex numbers such that
is unimodular and
lies on a:
1 -2
2- 1
2
2
2 is not unimodular. Then the point 1
Downloaded from @Freebooksforjeeneet
[2015]
M-29
Complex Numbers and Quadratic Equations
(a) circle of radius 2.
(b) circle of radius 2.
(c) straight line parallel to x-axis
(d) straight line parallel to y-axis.
53.
TOPIC Đ
2
If z ¹ 1 and z is real, then the point represented by the
z -1
complex number z lies :
[2012]
(a) either on the real axis or on a circle passing through
the origin.
(b) on a circle with centre at the origin
(c) either on the real axis or on a circle not passing through
the origin.
(d) on the imaginary axis.
54.
If w( ¹ 1) is a cube root of unity, and (1 + w )7 = A + Bw.
55.
Then (A, B) equals
[2011]
(a) (1, 1) (b) (1, 0)
(c) (–1, 1)
(d) (0, 1)
If | z + 4 | £ 3, then th e maximum value of
| z + 1 | is
[2007]
(a) 6
(b) 0
(c) 4
(d) 10
56.
If w =
z
1
z- i
3
and | w | = 1, then lies on
62.
p
2
(b) – p
(c) 0
(d)
63.
(a)
64.
-p
2
65.
66.
(d) – 1, 1 + 2 w , 1 + 2 w
25
81
(c)
5
27
(d)
25
9
a
1- a
2
+
b
1 - b2
is equal to :
(b)
1
24
(c)
3
8
(d)
27
16
2z + i
, z = x + iy and k > 0. If the curve
z - ki
represented by Re(u) + Im(u) = 1 intersects the y-axis at the
points P and Q where PQ = 5, then the value of k is :
[Sep. 04, 2020 (I)]
(a) 3/2
(b) 1/2
(c) 4
(d) 2
Let u =
Let l ¹ 0 be in R. If a and b are roots of the equation,
bg
is equal to :
l
[Sep. 04, 2020 (II)]
(a) 27
(b) 18
(c) 9
(d) 36
If a and b are the roots of the equation x2 + px + 2 = 0 and
3 x 2 - 10 x + 27 l = 0, then
2
If | z 2 - 1|=| z |2 +1, then lies on
[2004]
(a) an ellipse
(b) the imaginary axis
(c) a circle
(d) the real axis
The locus of the centre of a circle which touches the circle
| z – z1 | = a and | – 2 | = b externally (z, z1 & z2 are complex
numbers) will be
[2002]
(a) an ellipse
(b) a hyperbola
(c) a circle
(d) none of these
27
32
x 2 - x + 2l = 0 and a and g are the roots of the equation,
(c) – 1, 1 – 2 w , 1 – 2 w2
60.
(b)
If a and b are the roots of the equation, 7 x 2 - 3 x - 2 = 0,
(a)
[2005]
(a) –1, –1 + 2 w , – 1 – 2 w2
(b) –1, – 1, – 1
59.
5
9
[Sep. 05, 2020 (II)]
If the cube roots of unity are 1, w , w 2 then the roots of
the equation ( x –1)3 + 8 = 0, are
1
(a) 2
(b) 3
(c) 1
(d) 4
If a and b are the roots of the equation 2x(2x + 1) = 1, then
b is equal to:
[Sep. 06, 2020 (II)]
(a) 2a(a + 1)
(b) –2a(a + 1)
(c) 2a(a – 1)
(d) 2a2
The product of the roots of the equation 9x2 – 18| x | + 5 = 0,
is :
[Sep. 05, 2020 (I)]
(b) a circle
(d) a parabola
| z1 + z2 | = | z1 | + | z2 | , then arg z1 – arg z2 is equal to
[2005]
58.
1
æ a 3 ö 8 æ b3 ö 8
Then the value of ç ÷ + ç ÷ is: [Sep. 06, 2020 (I)]
è b5 ø
è a5 ø
If z1 and z2 are two non- ero complex numbers such that
(a)
If a and b be two roots of the equation x2 – 64x + 256 = 0.
the the value of
(a) an ellipse
(c) a straight line
57.
[2005]
61.
Solutions of Quadratic Equations,
Sum and Product of Roots, Nature
of Roots, Relation Between Roots
and Co-efficients, Formation of an
Equation with Given Roots.
67.
1
1
and are the roots of the equation 2x2 + 2qx + 1 = 0,
b
a
1 öæ
1 öæ
1 öæ
1ö
æ
then ç a - ÷ ç b - ÷ ç a + ÷ ç b + ÷ is equal to :
a øè
b øè
b øè
aø
è
Downloaded from @Freebooksforjeeneet
[Sep. 03, 2020 (I)]
EBD_8344
M-30
Mathematics
(a)
9
(9 + q 2 )
4
(b)
9
(9 - q 2 )
4
9
9
(9 + p 2 )
(9 - p 2 )
(d)
4
4
The set of all real values of l for which the quadratic
77.
(c)
68.
78.
equations, (l 2 + 1) x 2 - 4lx + 2 = 0 always have exactly
one root in the interval (0, 1) is :
[Sep. 03, 2020 (II)]
(a) (0, 2) (b) (2, 4]
(c) (1, 3]
(d) (–3, –1)
69.
Let a and b be the roots of the equation, 5 x 2 + 6 x - 2 = 0.
79.
Let p, q Î R. If 2 – 3 is a root of the quadratic equation,
x2 + px + q = 0, then:
[April 9, 2019 (I)]
(a) p2 – 4q + 12 = 0
(b) q2 – 4p – 16 = 0
(c) q2 + 4p + 14 = 0
(d) p2 – 4q – 12 = 0
If m is chosen in the quadratic equation
(m2 + 1) x2 – 3x + (m2 + 1)2 = 0
such that the sum of its roots is greatest, then the absolute
difference of the cubes of its roots is: [April 09, 2019 (II)]
(a) 10 5 (b) 8 3
(c) 8 5
(d) 4 3
The sum of the solutions of the equation
If Sn = an + bn , n = 1, 2,3, ..., then : [Sep. 02, 2020 (I)]
(a) 6S6 + 5S5 = 2S4
70.
71.
x -2 + x
(b) 6S6 + 5S5 + 2S4 = 0
(c) 5S6 + 6S5 = 2S4
(d) 5S6 + 6S5 + 2S4 = 0
The number of real roots of the equation,
e4x + e3x – 4e2x + ex + 1 = 0 is:
[Jan. 9, 2020 (I)]
(a) 1
(b) 3
(c) 2
(d) 4
The least positive value of ‘a’ for which the equation,
33
= 2a has real roots is _______.
2
[Jan. 8, 2020 (I)]
If the equation, x2 + bx + 45 = 0 (b Î R) has conugate
80.
73.
74.
complex roots and they satisfy |z + l| = 2 10 , then:
[Jan. 8, 2020 (I)]
(a) b2 – b = 30
(b) b2 + b = 72
(c) b2–b = 42
(d) b2 + b = 12
Let a and b be the roots of the equation x2 – x – l = 0. If
pk = (a)k + (b)k, k ³ l, then which one of the following
statements is not true ?
[Jan. 7, 2020 (II)]
(a) p3 = p5 – p4
(b) P5 = 11
(c) (p1 + p2 + p3 + p4 + p5) = 26
(d) p5 = p2 × p3
Let a and b be two real roots of the equation (k +1) tan2x
81.
12
(a)
82.
83.
(b)
2
(sin q + 8)12
212
26
(d)
6
(sin q - 8)
(sin q + 8)12
The number of real roots of the equation
(c)
76.
5 + 2 x - 1 = 2 x (2 x - 2) is:
(a) 3
84.
(b) 2
[April 10, 2019 (II)]
(c) 4
(d) 1
1
= 1, is :
l
[Jan. 12, 2019 (I)]
(b) 4 - 3 2
(c) –2 + 2
(d) 4 - 2 3
If one real root of the quadratic equation 81x2 + kx + 256 = 0
is cube of the other root, then a value of k is :
[Jan. 11, 2019 (I)]
(a) –81
(b) 100
(c) 144
(d) – 300
Consider the quadratic equation (c – 5)x2 – 2cx + (c – 4) = 0,
c ¹ 5. Let S be the set of all integral values of c for which
one root of the equation lies in the interval (0, 2) and its
other root lies in the interval (2, 3). Then the number of
elements in S is:
[Jan. 10, 2019 (I)]
(a) 18
(b) 12
(c) 10
(d) 11
The value of l such that sum of the squares of the roots
of the quadratic equation, x2 + (3 – l)x + 2 = l has the
least value is:
[Jan. 10, 2019 (II)]
15
4
(b) 1
(c)
(d) 2
8
9
Let a and b be two roots of the equation x2 + 2x + 2 = 0, then
a15 + b15 is equal to:
[Jan. 9, 2019 (I)]
(a) – 256 (b) 512
(c) – 512
(d) 256
The number of all possible positive integral values of a
for which the roots of the quadratic equation,
6x2 – 11x + a = 0 are rational numbers is:
[Jan. 09, 2019 (II)]
(a) 3
(b) 2
(c) 4
(d) 5
(a)
12
2
(sin q - 4)12
æaö
the least value of n for which ç ÷ = 1 is :
èbø
[April 8, 2019 (I)]
(a) 2
(b) 5
(c) 4
(d) 3
If l be the ratio of the roots of the quadratic equation in x,
3m2x2 + m(m – 4)x + 2 = 0, then the least value of m for
(a) 2 - 3
(a) 10 2 (b) 10
(c) 5
(d) 5 2
If a and b are the roots of the quadratic equation, x2 + x sin q
a12 + b12
æ pö
– 2sinq = 0 0, qÎ ç 0, ÷ , then
is
(a -12 + b-12 )(a - b)24
è 2ø
equal to :
[April 10, 2019 (I)]
[April 8, 2019 (I)]
(a) 9
(b) 12
(c) 4
(d) 10
If a and b be the roots of the equation x2 – 2x + 2 = 0, then
which l +
– 2 . ltan x = (1 – k), where k(¹ –1) and l are real numbers.
If tan2(a + b) = 50, then a value of l is: [Jan. 7, 2020 (I)]
75.
)
x - 4 + 2 = 0, (x > 0) is equal to:
n
2x2 + (a – 10)x +
72.
(
85.
86.
Downloaded from @Freebooksforjeeneet
M-31
Complex Numbers and Quadratic Equations
87.
88.
If both the roots of the quadratic equation x2 – mx + 4 = 0
are real and distinct and they lie in the interval [1, 5],
then m lies in the interval:
[Jan. 09, 2019 (II)]
(a) (– 5, – 4)
(b) (4, 5)
(c) (5, 6)
(d) (3, 4)
Let z0 be a root of the quadratic equation,
x2 + x + 1 = 0. If z = 3 + 6i z081 – 3i z093, then arg z is
equal to:
[Jan. 09, 2019 (II)]
97.
2
(x 2 - 5x + 5) x + 4x -60 = 1 is :
98.
1
1
1
+
=
are equal in
x+ p x+q r
magnitude but opposite in sign, then the sum of squares
of these roots is equal to.
[Online April 16, 2018]
(a) p2 + q2 + r2
(b) p2 + q2
p2 + q2
(c)
(d)
2
If an angle A of a D ABC satisfies 5 cos A + 3 = 0, then the
roots of the quadratic equation, 9x2 + 27x + 20 = 0 are.
[Online April 16, 2018]
(a) sin A, sec A
(b) sec A, tan A
(c) tan A, cos A
(d) sec A, cot A
If tan A and tan B are the roots of the quadratic equation,
3x2 – 10x – 25 = 0 then the value of
3 sin2 (A + B) – 10 sin (A + B). cos (A + B) – 25 cos2 (A + B)
is
[Online April 15, 2018]
(a) 25
(b) – 25
(c) – 10
(d) 10
If f (x) is a quadratic expression such that f (a) + f (b) = 0,
and – 1 is a root of f (x) = 0, then the other root of f (x) = 0
is
[Online April 15, 2018]
2 (p2 + q2)
90.
91.
92.
(a) 93.
5
8
(b) -
8
5
(c)
5
8
(d)
8
5
[2018]
94.
(a) 0
(b) 1
(c) 2
(d) – 1
If, for a positive integer n, the quadratic equation,
95.
x(x + 1) + (x + 1) (x + 2) + ..... + (x + n - 1 ) (x + n) = 10n
has two consecutive integral solutions, then n is equal to:
[2017]
(a) 11
(b) 12
(c) 9
(d) 10
The sum of all the real values of x satisfying the equation
(
96.
)
(a) 6
(b) 5
(c) 3
If x is a solution of the equation,
(d) – 4
3
1
(b)
(c) 2 2
(d) 2
4
2
Let a and b be the roots of equation x2 – 6x – 2 = 0. If an =
(a)
99.
an – bn, for n ³ 1, then the value of
a10 - 2a 8
is equal to:
2a 9
[2015]
(a) 3
(b) – 3
(c) 6
(d) – 6
100. If the two roots of the equation, (a – 1)(x4 + x2 + 1) +
(a + 1)(x2 + x + 1)2 = 0 are real and distinct, then the set
of all values of ‘a’ is :
[Online April 11, 2015]
æ 1ö
(a) ç 0, ÷
è 2ø
æ 1 ö æ 1ö
(b) ç - , 0 ÷ È ç 0, ÷
è 2 ø è 2ø
æ 1 ö
(c) ç - , 0 ÷
(d) (– ¥, –2) È (2, ¥)
è 2 ø
101. If 2 + 3i is one of the roots of the equation 2x3 – 9x2 + kx – 13
= 0, k Î R, then the real root of this equation :
[Online April 10, 2015]
(a) exists and is equal to –
1
.
2
1
.
2
(c) exists and is equal to 1.
(d) does not exist.
102. If a Î R and the equation
(b) exists and is equal to
If a, b Î C are the distinct roots, of the equation
x 2 - x + 1 = 0 , then a101 + b107 is equal to :
[2016]
1ö
æ
2x + 1 - 2x - 1 = 1, ç x ³ ÷ , then
4x 2 - 1 is equal
è
2ø
to :
[Online April 10, 2016]
p
p
p
(a)
(b)
(c)
(d) 0
4
6
3
89. Let p, q and r be real numbers (p ¹ q, r ¹ 0), such that the
roots of the equation
(a) p(b) = 11
(b) p(b) = 19
(c) p(–2) = 19
(d) p(–2) = 11
The sum of all real values of x satisfying the equation
( x -1) x 2 +5x -50
2
= 1 is :
[Online April 9, 2017]
(a) 16
(b) 14
(c) – 4
(d) – 5
Let p(x) be a quadratic polynomial such that p(0)=1. If p(x)
leaves remainder 4 when divided by x–1 and it leaves
remainder 6 when divided by x+1; then :
[Online April 8, 2017]
- 3 ( x - [ x ]) + 2 ( x - [ x ]) + a 2 = 0
2
(where [x] denotes the greatest integer £ x ) has no integral
solution, then all possible values of a lie in the interval:
[2014]
(a)
( -2, -1)
(b)
( -¥, -2 ) È ( 2, ¥ )
(c)
( -1, 0 ) È ( 0,1)
(d)
(1, 2 )
3x 2 + x + 5 = x - 3 , where x is real, has;
[Online April 19, 2014]
(a) no solution
(b) exactly one solution
(c) exactly two solution (d) exactly four solution
103. The equation
Downloaded from @Freebooksforjeeneet
EBD_8344
M-32
Mathematics
104. The sum of the roots of the equation,
x2 + |2x – 3| – 4 = 0, is:
[Online April 12, 2014]
(d) - 2
(a) 2
(b) – 2
(c)
2
105. If a and b are roots of the equation,
x 2 - 4 2 kx + 2e 4 lnk - 1 = 0 for some k, and
a2 + b2 = 66, then a3 + b3 is equal to:
[Online April 11, 2014]
=
+ + ( )=
+
(a) 248 2 (b) 280 2 (c) -32 2 (d) -280 2
1
1
106. If
and
a
b
are the roots of the equation,
ax2 + bx + 1 = 0 (a ¹ 0, a, b, Î R), then the equation,
(
) (
)
x x + b3 + a 3 - 3abx = 0 as roots :
[Online April 9, 2014]
3
(a) a
3
(c)
ab and ab
2
and b
2
(b) ab
(d)
a
-
1
3
2
2
and a
1
2b
3
and b- 2
107. If p an d q are non- ero real numbers and
a3 + b3 = - p, ab = q, then a quadratic equation whose
roots are
a 2 b2
is :
,
b a
(a) px2 – qx + p2 = 0
(c) px2 + qx + p2 = 0
[Online April 25, 2013]
(b) qx2 + px + q2 = 0
(d) qx2 – px + q2 = 0
108. If a and b are roots of the equation x 2 + px +
3p
= 0,
4
such that | a - b |= 10, then p belongs to the set :
[Online April 22, 2013]
(a) {2, – 5} (b) {– 3, 2} (c) {– 2, 5} (d) {3, – 5}
109. If a complex number z statisfies the equation
z + 2 | z + 1| +i = 0 , then | z | is equal to :
[Online April 22, 2013]
(a) 2
(b) 3
(c) 5
(d) 1
110. Let p, q, r Î R and r > p > 0. If the quadratic equation
px2 + qx + r = 0 has two complex roots a and b, then |a| +
|b| is
[Online May 19, 2012]
(a) equal to1
(b) less than 2 but not equal to 1
(c) greater than 2
(d) equal to 2
111. If the sum of the square of the roots of the equation
x2 – (sina – 2) x – (1 + sina) = 0 is least, then a is equal to
[Online May 12, 2012]
(a)
p
6
(b)
p
4
(c)
p
3
(d)
p
2
112. The value of k for which the equation
(k – 2)x2 + 8x + k + 4 = 0 has both roots real, distinct and
negative is
[Online May 7, 2012]
(a) 6
(b) 3
(c) 4
(d) 1
113. Let for a ¹ a1 ¹ 0,
f ( x ) = ax 2 + bx + c, g ( x) = a1 x 2 + b1 x + c1an
1
=
+ c1and p ( x ) = f ( x ) - g ( x ) .
If p ( x ) = 0 only for x = -1 and p (– 2) = 2, then the value
of p (b) is :
[2011 RS]
(a) 3
(b) 9
(c) 6
(d) 18
114. Sachin and Rahul attempted to solve a quadratic equation.
Sachin made a mistake in writing down the constant term
and ended up in roots (4,3). Rahul made a mistake in writing
down coefficient of x to get roots (3,2). The correct roots
of equation are :
[2011 RS]
(a) 6, 1
(b) 4, 3
(c) – 6, – 1 (d) – 4, – 3
115. Let a, b be real and z be a complex number. If
z2 + az + b = 0 has two distinct roots on the line Re z =1,
then it is necessary that :
[2011]
(b) b = 1
(a) b Î (-1, 0)
(c) b Î (1, ¥)
(d) b Î (0,1)
116. If a and b are the roots of the equation
x2 – x + 1 = 0, then a2009 + b2009 =
[2010]
(a) –1
(b) 1
(c) 2
(d) –2
117. If the roots of the equation bx2 + cx + a = 0 be imaginary,
then for all real values of x, the expression 3b2x2 + 6bcx +
2c2 is :
[2009]
(a) less than 4ab
(b) greater than – 4ab
(c) 1ess than – 4ab
(d) greater than 4ab
118. If the difference between the roots of the equation
x2 + ax + 1 = 0 is less than
values of a is
5 , then the set of possible
[2007]
(a) (3, ¥) (b) (- ¥, - 3) (c) (– 3, 3) (d) (-3, ¥)
119. All the values of m for which both roots of the equation
x 2 - 2mx + m 2 - 1 = 0 are greater than – 2 but less than 4,
lie in the interval
[2006]
(a) -2 < m < 0
(b) m > 3
(c) -1 < m < 3
(d) 1 < m < 4
120. If the roots of the quadratic equation
x 2 + px + q = 0 are tan30 and tan15 ,
respectively, then the value of 2 + q – p is
(a) 2
(b) 3
(c) 0
(d) 1
[2006]
121. If z 2 + z + 1 = 0 , where is complex number, then the value
2
2
of
1ö
æ 2 1ö
æ 3 1ö
æ
çè z + ÷ø + çè z + 2 ÷ø + çè z + 3 ÷ø
z
z
z
2
+ ......... + æç z 6 +
è
1ö
÷
z6 ø
2
is
[2006]
(a) 18
(b) 54
(c) 6
Downloaded from @Freebooksforjeeneet
(d) 12
-
M-33
Complex Numbers and Quadratic Equations
p
æ Pö
. If tan ç ÷ and
122. In a triangle PQR, Ð R =
è 2ø
2
æ Qö
– tan ç ÷ are the roots of ax 2 + bx + c = 0, a ¹ 0 then
è 2ø
[2005]
(a) a = b + c
(c) b = c
(b) c = a + b
(d) b = a + c
131. Difference between the corresponding roots of x2+ax+b=0
[2002]
and x2+bx+a=0 is same and a ¹ b, then
(a) a + b + 4 = 0
(b) a + b – 4 = 0
(c) a – b – 4 = 0
(d) a – b + 4 = 0
132. If a ¹ b but a2 = 5a – 3 and b2 = 5b – 3 then the equation
having a/b and b/a as its roots is
[2002]
(a) 3x2 – 19x + 3 = 0
(b) 3x2 + 19x – 3 = 0
(c) 3x2 – 19x – 3 = 0
(d) x2 – 5x + 3 = 0.
123. If the roots of the equation x 2 – bx + c = 0 be two
consecutive integers, then b 2 – 4c equals
(a) – 2
(b) 3
(c) 2
(d) 1
[2005]
124. If one root of the equation x 2 + px + 12 = 0 is 4, while the
equation x 2 + px + q = 0 has equal roots , then the value
of ‘q’ is
[2004]
(a) 4
(b) 12
(1 - p)
125. If
(c) 3
(d)
is a root of quadratic equation
[2004]
(a) –1, 2
(b) –1, 1
(c) 0, –1
(d) 0, 1
126. The number of real solutions of the equation
x 2 - 3 x + 2 = 0 is
[2003]
(a) 3
(b) 2
(c) 4
(d) 1
127. The value of 'a' for which one root of the quadratic equation
2
2
(a - 5a + 3) x + (3a - 1) x + 2 = 0 is twice as large as the
other is
[2003]
(a) -
1
3
(b)
2
3
(c) -
2
3
(d)
1
3
128. Let Z1 and Z 2 be two roots of the equation
Z 2 + aZ + b = 0 , Z being complex. Further , assume that
the origin, Z1 and Z 2 form an equilateral triangle. Then
[2003]
(a) a 2 = 4b
133. If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r
cannot be equal to:
[Jan. 10, 2019 (I)]
3
5
7
3
(b)
(c)
(d)
4
4
4
2
134. Let a, b Î R, a ¹ 0 be such that the equation,
ax2 – 2bx + 5 = 0 has a repeated root a, which is also a root
of the equation, x2 – 2bx – 10 = 0. If b is the other root of
this equation, then a2 + b2 is equal to :[Jan. 9, 2020 (II)]
(a) 25
(b) 26
(c) 28
(d) 24
135. If lÎ R is such that the sum of the cubes of the roots of the
equation, x2 + (2 – l) x + (10 – l) = 0 is minimum, then the
magnitude of the difference of the roots of this equation is
[Online April 15, 2018]
(a)
49
4
x 2 + px + (1 - p) = 0 then its root are
TOPIC Ė
Condition for Common Roots,
Maximum and Minimum value of
Quadratic Equation, Quadratic
Expression in two Variables,
Solution of Quadratic Inequalities.
(a) 20
(b) 2 5
(c) 2 7
(d) 4 2
136. If | z – 3 + 2i | £ 4 then the difference between the greatest
value and the least value of | z | is [Online April 15, 2018]
(a)
(b) 2 13
(d) 4 + 13
(c) 8
(a) 2
(b) 3
(c) 3
(d) 2
138. If non- ero real numbers b and c are such that
min f(x) > max g(x), where f(x) = x2 + 2bx + 2c2 and
g(x) = – x2 – 2cx + b2 (x Î R);
(b) a 2 = b
(c) a 2 = 2b
(d) a 2 = 3b
129. If p and q are the roots of the equation
x2 + px + q = 0, then
[2002]
(a) p = 1, q = –2
(b) p = 0, q = 1
(c) p = –2, q = 0
(d) p = – 2, q = 1
130. Product of real roots of the equation
t2 x2 + | x | + 9 = 0
[2002]
(a) is always positive
(b) is always negative
(c) does not exist
(d) none of these
13
137. If the equations x2 + bx – 1 = 0 and x2 + x + b = 0 have a
common root different from –1, then |b| is equal to :
[Online April 9, 2016]
then
c
lies in the interval:
b
[Online April 19, 2014]
æ 1ö
(a) ç 0, ÷
è 2ø
é1 1 ö
(b) ê ,
÷
ë2 2 ø
é 1
ù
2ú
(c) ê
ë 2, û
(d)
(
Downloaded from @Freebooksforjeeneet
2, ¥
)
EBD_8344
M-34
Mathematics
139. If equations ax2 + bx + c = 0 ( a, b, c Î R, a ¹ 0 ) and
2x2 + 3x + 4 = 0 have a common root, then a : b : c equals:
[Online April 9, 2014]
(a) 1 : 2 : 3 (b) 2 : 3 : 4 (c) 4 : 3 : 2 (d) 3 : 2 : 1
140. If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0,
a,b,c Î R, have a common root, then a : b : c is [2013]
(a) 1 : 2 : 3 (b) 3 : 2 : 1 (c) 1 : 3 : 2 (d) 3 : 1 : 2
141. The least integral value a of x such that
x-5
x 2 + 5x - 14
143. If z -
(a) 1
(d) a £ -2
4
= 2 , then the maximum value of | | is equal to :
z
(b) 4
(c) 3
145. If x is real, the maximum value of
>0,
satisfies :
[Online April 23, 2013]
(a) a2 + 3a – 4 = 0
(b) a2 – 5a + 4 = 0
(c) a2 – 7a + 6 = 0
(d) a2 + 5a – 6 = 0
142. The values of ‘a’ for which one root of the equation
x2 – (a + 1 ) x + a2 + a – 8 = 0 exceeds 2 and the other is
lesser than 2, are given by :
[Online April 9, 2013]
(a) 3 < a < 10
(b) a ³ 10
(c) -2 < a < 3
144. The quadratic equations x 2 – 6x + a = 0 and
x2 – cx + 6 = 0 have one root in common. The other roots
of the first and second equations are integers in the ratio
4 : 3. Then the common root is
[2009]
(a)
1
4
(b) 41
(d) 2
3x 2 + 9 x + 17
3x 2 + 9 x + 7
(c) 1
5 + 1 (b) 2
(c) 2 + 2
(d)
17
7
2
2
146. If both the roots of the quadratic equation x - 2 kx + k
+ k – 5 = 0 are less than 5, then k lies in the interval
[2005]
(a) (5, 6]
(b) (6, ¥ )
(c) (– ¥ , 4) (d) [4, 5]
147. The value of a for which the sum of the squares of the
roots of the equation x2 – (a – 2) x – a – 1 = 0 assume the
least value is
[2005]
(a) 1
(b) 0
(c) 3
(d) 2
[2009]
(a)
(d)
is [2006]
3 +1
Downloaded from @Freebooksforjeeneet
M-35
Complex Numbers and Quadratic Equations
1.
(b) Let z =
z=
3 + i sin q
, after rationalising
4 - i cos q
æ (1 + i )2 ö
Þç
÷
è 2 ø
(3 + i sin q) (4 + i cos q)
´
(4 - i cos q) (4 + i cos q)
3
4
5.
D(z–2Re(z))
(b) Let z1 = x1 + iy1 and z2 = x2 + iy2
Þ ( x1 - 1) 2 + y12 = x12
æ 4ö
æ4ö
= p + tan -1 ç - ÷ = p - tan -1 ç ÷
è 3ø
è3ø
(c)
=1
Q | z1 - 1 | = Re( z1 )
æ cos q ö
arg(sinq + icosq) = p + tan–1 ç sin q ÷
è
ø
2.
n /3
m (least) = 8, n (least) = 12
GCD (8, 12) = 4.
Þ tanq = –
3cosq + 4sinq = 0
æ (1 + i )2 ö
=ç
÷
è -2 ø
Þ i m / 2 = ( -i ) n / 3 = 1
As z is purely real
Þ
m/2
Þ y12 - 2 x1 + 1 = 0
...(i)
| z2 - 1|= Re( z2 ) Þ ( x2 - 1)2 + y22 = x22
C(z –2Re(z ))
Þ y22 - 2 x2 + 1 = 0
...(ii)
From eqn. (i) – (ii),
y12 - y22 - 2( x1 - x2 ) = 0
æ x -x ö
Þ y1 + y2 = 2 ç 1 2 ÷
è y1 - y2 ø
A(z)
Q arg( z1 - z2 ) =
B(z)
Then, | z - z | = 4 Þ | 2iy | = 4 Þ | y | = 2
Also, | z - ( z - 2 Re( z )) | = 4
Þ
y1 - y2
1
=
x1 - x2
3
Þ
2
1
=
y1 + y2
3
Þ | 2 Re( z ) | = 4 Þ | 2 x | = 4 Þ | x | = 2
\| z | = x + y = 4+ 4 = 2 2
2
2p
3.
30
æ -1 + 3i ö
\ çç
÷÷
è 1- i ø
30
æ 2p p ö ö
æ
ç + ÷i
= ç 2eè 3 4 ø ÷
ç
÷
è
ø
=2
15
4.
i
(c) Q -1 + 3i = 2 × e 3 and 1 - i = 2 × e
p
- i
×e 2
m/2
ip
4
é
y1 - y2
2 ù
=
ê From,
ú
x1 - x2 y1 + y2 û
ë
\ y1 + y2 = 2 3 Þ lm( z1 + z2 ) = 2 3
6.
(b) Let z = x + iy
z -i
= 1 Þ x2 + (y – 1)2
z + 2i
= x2 + (y + 2)2 Þ –2y + 1 = 4y + 4
= -2 × i.
æ 1+ i ö
=ç
è i - 1ø÷
-
Then,
15
(4)
æ 1+ i ö
Given that ç
è 1 - i ÷ø
p
6
æ y - y2 ö p
Þ tan -1 ç 1
=
è x1 - x2 ÷ø 6
Let z = x + iy
Q Length of side of square = 4 units
2
Þ
6y = – 3 Þ y = -
Q
|z| =
n /3
=1
...(iii)
1
2
5
25
Þ x2 + y2 =
2
4
Downloaded from @Freebooksforjeeneet
EBD_8344
M-36
Mathematics
x2 =
\
z = x + iy
| + 3i| =
7.
24
=6
4
Þ
Since, it is given that | z | =
Þ z= ±
6-
i
2
Then, from equation (i),
2
2
=
5
1 + a2
25
49
6+
=
4
4
Now, square on both side; we get
7
Þ |z + 3i| =
2
(c) z = x + iy
|x| + |y| = 4
Þ 1 + a2 = 10 Þ a = ± 3
Since, it is given that a > 0 Þ a = 3
Then, z =
z = x2 + y 2
=
Minimum value of
|z| = 2 2
Maximum value of
|z| = 4
11.
So, | | can’t be 7 .
(c) Let Re (z) = x i.e., z = x + 10i
2i (3 + i ) -1 + 3i
=
10
5
(2x – n) + 20i = (2i – 1) ((2x + n) + 20i)
-1 3
- i
5 5
(c) Given | zw | = 1
...(i)
æzö p
and arg ç ÷ =
èwø 2
...(ii)
\
2z – n = (2i – 1) (2z + n)
é
ù
æzö
êQ Re ç w ÷ = 0 ú
è
ø
ë
û
z z
+ =0
w w
Þ zw = - z w
On comparing real and imaginary parts,
– (2x + n) – 40 = 2x – n and 20 = 4x + 2n – 20
from equation (i), zz ww = 1 [using zz =| z |2 ]
Þ 4x = – 40 and 40 = – 40 + 2n
( z w)2 = -1 Þ z w = ± i
Þ x = – 10 and n = 40
from equation (ii), - arg( z ) - arg w =
Hence, Re(z) = – 10
9.
(1 + i )2 1 + i 2 + 2i
2i
=
=
a -i
3-i
3-i
Hence, z =
z Î éë 8, 16 ùû
8.
2
5
(b) Given equation is, | z – 1 | = | z – i |
Þ
(x – 1)2 + y2 = x2 + (y – 1)2
Þ
1 – 2x = 1 – 2y Þ x – y = 0
[Here, z = x + iy]
z=
a2 + 1
=
2
4(1 + a 2 )
2 2
(1 + a )
=
2
1 + a2
5 ( w - 1)
3 + 5w
Þ 25 ( ww - w - w + 1) < 9 + 25ww + 15w + 15w
a2 + 1
æ -2 ö æ 2a ö
ç 2
÷ +ç
÷ =
è a + 1 ø è a2 + 1 ø
Þ|z|=
5 + 3z
Þ 5w - 5wz = 5 + 3z
5 - 5z
Q | z | < 1, \5|w – 1| < |3 + 5w|
2ai - 2
2
|z|=
w=
Þ 5w - 5 = z ( 3 + 5w) Þ z =
(1 + i )2 a + i
´
10. (a) z =
a -i a +i
(1 - 1 + 2i )(a + i )
Hence, zw = -i
12. (c)
Hence, locus is straight line with slope 1.
p
-p
- arg( z w) =
2
2
(Q z
4 + 4a 2
2
(a + 1)
2
...(i)
Þ 16 < 40w + 40w Þ w + w >
Þ Re ( w) >
2
2
2
Þ 2Re(w) >
5
5
1
5
Downloaded from @Freebooksforjeeneet
=zz
)
M-37
Complex Numbers and Quadratic Equations
13. (a) Let t =
z-a
z+a
3z1 2 z2 3r1 i (q -f ) 2 r2 i (f -q )
+
e
z = 2 z + 3 z = 2r e
3 r1
2
1
2
Q
t is purely imaginary number.
\
t + t =0
Þ
z -a z -a
+
=0
z +a z +a
Þ
=
2 3
´ [cos(q - f) - i sin(q - f)]
3 4
(z – a)( z + a) + ( z – a)(z + a) = 0
– a2 + z z
Þ
zz
Þ
z z – a2 = 0
Þ
|z|2 – a2 = 0
Þ
a2 = 4
Þ
a= ±2
14. (a) |
3 4
´ (cos(q – f) + i sin(q – f)) +
2 3
1ö
1ö
æ
æ
z = èç 2 + ø÷ cos(q - f) + i èç 2 - ÷ø sin(q - f)
2
2
– a2 = 0
\
1 | = 9, | 2 – 3 – 4i | = 4
2 lies on a circle with centre C2(3, 4) and radius r 2 = 4
So, minimum value of |
(i.e. A)
1 – 2 | is
ero at point of contact
17.
25
9
cos2 (q - f) + sin 2 (q - f)
4
4
16
9
3
5
cos 2 (q - f) +
Þ £ |z| £
2
2
4
4
=
1 lies on a circle with centre C1(0, 0) and radius r 1 = 9
| z|=
(d) Suppose z =
3 + 2i sin q
1 - 2i sin q
Since, z is purely imaginary, then z + z = 0
Þ
3 + 2i sin q 3 - 2i sin q
+
=0
1 - 2i sin q 1 + 2i sin q
Þ
(3 + 2i sin q)(1 + 2i sin q) + (3 - 2i sin q)(1 - 2i sin q)
1 + 4sin 2 q
=0
sin2 q =
Þ
p p 2p
q= - , ,
3 3 3
15. (b) Since, |z| + z = 3 + i
Let z = a + ib, then
|z| + z = 3 + i Þ
a 2 + b 2 + a + ib = 3 + i
Compare real and imaginary coefficients on both sides
b = 1,
a2 + b2 + a = 3
Now, the sum of elements in A = -
a2 + 1 = a2 + 9 – 6a
4
6a = 8 Þ a =
3
Then,
p p 2p 2p
+ +
=
3 3 3
3
18. (a) Q |z| = 1 & Re z ¹ 1
Suppose z = x + iy Þ x2 + y2 = 1 .....(i)
Now, w =
a 2 +1 = 3 – a
3
3
Þ sin q =
4
2
Þ
1 + (1 – 8a ) z
1– z
Þ w=
1 + (1 – 8a ) ( x + iy)
1 – ( x + iy )
Þ w=
1 + (1 – 8a) ( x + iy ) )((1 – x) + iy)
1 – ( x + iy))((1 – x) + iy)
2
|z| =
16
5
æ 4ö
çè 3 ÷ø +1 = 9 + 1 = 3
16. (none) Let z1 = r1eiq and z2 = r2eif
3|z1| = 4|z2| Þ 3r1 = 4r2
é (1 + x (1 – 8a ) (1 – x) – (1 – 8a) y 2 ù
û
Þ w= ë
(1 – x )2 + y 2
+i
[ (1 + x (1 – 8a)) y – (1 – 8a) y (1 – x)]
(1 – x )2 + y 2
Downloaded from @Freebooksforjeeneet
EBD_8344
M-38
Mathematics
As, w is purely imaginary. So,
é(1 + x (1 – 8a ) )(1 – x) – (1 – 8a) y 2 ù
ë
û
Re w =
=0
(1 – x)2 + y 2
Þ (1 – x) + x (1 – 8a) (1 – x) = (1 – 8a) y2
Þ (1 – x) + x (1 – 8a) – x2 (1 – 8a) = (1 – 8x)y2
Þ (1 – x) + x (1 – 8a) = 1 – 8a [From (i), x2 + y2 = 1]
Þ 1 – 8a = 1
Þa=0
\ a Î {0}
19. (b) Rationali ing the given expression
(2 + 3isin q)(1 + 2isin q)
1 + 4sin 2 q
For the given expression to be purely imaginary, real part
of the above expression should be equal to ero.
1
= 0 Þ sin 2 q =
3
1 + 4 sin 2 q
1
1
æ ö
Þ sin q = ±
Þ q = sin -1 ç ÷
è 3ø
3
20. (b) Let = reiq
Im
Consider
5
(Im )
r (sin q )
5
=
sin 5 q
5
sin q
5
5 cosec4 q
20sin 3 q
+
sin q
5
– 20 cosec2
5sin q
sin 5 q
q + 16
Im 5
(Im )5
is – 4.
21. (d) We know minimum value of |Z1 + Z2| is
| |Z1| – |Z2||. Thus minimum value of Z +
£ Z+
1
1
£| Z | +
2
2
Since, | Z |³ 2 therefore
2-
1
1
1
< Z + < 2+
2
2
2
Þ
3
1
5
< Z+ <
2
2
2
x2 + y2 -1
x + ( y + 1)
2
x 2 + ( y + 1) 2
sin q
minimum value of
x 2 + ( y + 1) 2
x2 + y2 -1
5
–
x 2 - 2ix ( y + 1) + xi ( y - 1) + y 2 - 1
2
-
2 xi
x + ( y + 1) 2
2
for pure imaginary, we have
16sin5 q - 20sin3 q + 5sin q
16sin5 q
=
=
=
z-i
is purely imaginary means its real part is ero.
z+i
=
(Q eiq = cos q + i sinq)
sin 5q
= 4a 2 + 4 - 4a 2 - 4
= 0 = R.H.S.
Hence, y2 + 4x – 4 = 0
23. (c) Let z = x + iy
=
r5 (sin 5q)
=
5
2
2
LHS : y 2 + 4 x - 4 = (2a ) + 4(1 - a ) - 4
x + iy - i
x + i ( y - 1) x - i ( y + 1)
´
=
x + i ( y + 1) x - i ( y + 1)
x + iy + i
2 - 6 sin 2 q
Þ
22. (b) Let z = 1+ ia, a Î R
z2 = (1 + ia) (1 + ia)
x + iy = (1 + 2ia – a2)
On comparing real and imaginary parts, we get
x = 1 – a2, y = 2a
Now, consider option (b), which is
y2 + 4x – 4 = 0
1
1
is | Z | 2
2
=0
Þ
Þ
x2 + y2 = 1
(x + iy) (x – iy) = 1
Þ
x + iy =
and
1
= x – iy
z
z+
1
=z
x - iy
1
= ( x + iy ) + ( x - iy ) = 2 x
z
1ö
æ
çè z + ÷ø is any non- ero real number
z
æz ö
æz ö
24. (a) Consider arg ç 1 ÷ + arg ç 2 ÷
è z4 ø
è z3 ø
= arg( z1 ) - arg( z4 ) + arg( z2 ) - arg( z3 )
= (arg( z1 ) + arg( z2 )) - (arg( z3 ) + arg( z4 ))
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æ z2 = z1 &ö
given ç
è z4 = z3 ÷ø
M-39
Complex Numbers and Quadratic Equations
= (arg( z1 ) + arg( z1 )) - (arg( z3 ) + arg( z3 ))
ìalso (arg( z1 ) = - arg( z1 )ü
í
ý
îarg( z3 ) = - arg( z3 )
þ
28. (a) Let z = x + iy Þ z2 = x2 – y2 + 2ixy
Now,
= (arg( z1 ) - arg( z1 )) - (arg( z3 ) - arg( z3 ))
1 + z 2 1 + x 2 - y 2 + 2ixy ( x 2 - y 2 + 1) + 2ixy
=
=
2iz
2i ( x + iy )
2ix - 2 y
( x2 - y 2 + 1) + 2ixy -2 y - 2ix
=
´
-2 y + 2ix
-2 y - 2ix
=0–0=0
25. (d) Consider the equation
=
w - wz = k (1 - z ), k Î R
a=
w - wz
Clearly z ¹ 1 and
is purely real
1- z
Þ w - wz - wz + wzz = w - wz - wz + wzz
2
2
Þ w+w| z| = w+w| z|
Þ ( w - w)(| z |2 ) = w - w
Þ |z|2 = 1 (QIm w ¹ 0)
Þ |z| = 1 and z ¹ 1
\ The required set is {z : |z| = 1, z ¹ 1 }
26. (c) Given | | = 1, arg z = q
1
z
æ 1+ z ö
æ 1+ z ö
= arg
\ arg ç
= arg (z) = q.
ç 1÷
è 1 + z ÷ø
çè 1 + ÷ø
z
27. (b) Let z = x + iy, z = x – iy
Now, z = 1 – z
Þ x + iy = 1 – (x – iy)
1
Þ 2x = 1 Þ x =
2
Now, | z | = 1 Þ x2 + y2 = 1 Þ y2 = 1 – x2
Þ
3
y =±
2
Now, tan q =
y
(q is the argument)
x
3 1
¸
2 2
= 3
(+ve since only principal argument)
=
p
3
Hence, z is not a real number
So, statement-1 is false and 2 is true.
Þ
q = tan -1 3 =
x ( x 2 + y 2 + 1)
2( x 2 + y 2 )
x2 + y 2 = 1
Þ x2 + y2 = 1
w - wz
w - wz
=
1- z
1- z
Þ z=
2( x 2 + y 2 )
Since, | z | = 1 Þ
w - wz
w - wz
\
=
1- z
1- z
Þ
y ( x 2 + y 2 - 1) + x ( x 2 + y 2 + 1)i
x (1 + 1)
=x
2 ´1
Also z ¹ 1 Þ x + iy ¹ 1
\ A = (– 1, 1)
29. (d) Let z1 = 1 + i and z2 = 1 – i
\ a=
z2 1 - i (1 - i ) (1 - i )
=
=
=-i
z1 1 + i (1 + i ) (1 - i )
æz ö
2 + 3ç 2 ÷
2 z1 + 3z2
è z1 ø = 2 - 3i
=
2 z1 - 3z2
æ z ö 2 + 3i
2-3ç 2 ÷
è z1 ø
2 z1 + 3z2
2 - 3i
2 - 3i
=
=
2 z1 - 3z2
2 + 3i
2 + 3i
4+9
=
=1
4+9
30. (b) z1 + z2 2 + z1 - z2 2
= z1 2 + z2
2
2
2
+ 2 z1 z2 + z1 + z2
= 2 z1 + 2 z2
2
é
êQ
ë
2
z1
| z |ù
= 1 ú
z2
| z2 | û
- 2 z1 z2
2
2
= 2 éê z1 + z2 ùú
ë
û
31. (a) Let |Z|= |W| = r
Þ Z= reiq, W = reif
where q + f = p
\
–if
W = re
Now, Z = rei(p – f) = reip × e–if = – re–if
= –W
Thus, statement-1 is true but statement-2 is false.
32. (b) Statement - 1 and 2 both are true.
It is fundamental property.
But Statement - 2 is not correct explanation for Statement - 1.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-40
Mathematics
33. (a) Let z = x + iy
38. (a)
Arg( z w) = arg( z ) + arg(w)
z - 1 = z + 1 Þ ( x -1)2 + y2 = ( x + 1)2 + y2
Þ x=0
p
2
[Q arg ( z ) = - arg( )]
Þ Re z = 0
= - arg( z ) + arg w = -
z - 1 = z - i Þ ( x - 1)2 + y 2 = x2 + ( y - 1)2
Þ x= y
2
2
2
z + 1 = z - i Þ ( x + 1) + y = x + ( y - 1)
\ z w = -1
39. (c) Given that | z – 4 | < | z – 2 |
Let z = x + iy
Þ | (x – 4) + iy) | < | (x – 2) + iy |
Þ (x – 4)2 + y2 < (x – 2)2 + y2
Þ x2 – 8x + 16 < x2 – 4x + 4 Þ 12 < 4x
Þ x > 3 Þ Re(z) > 3
40. (b) Let | z | = | w | = r
\ z = reiq, w = reif where q + f = p.
2
Þ x = –y
Only (0, 0) will satisfy all conditions.
Þ Number of complex number z = 1
34. (c)
| z w |=| z || w |=| z || w |=| zw |= 1 [Q z = z ]
æ 1 ö
1
1
–1
çè i – 1÷ø = (i - 1) = – i – 1 = i + 1
1
\ z = rei(p–f) = reip . e–if = –re–if = – w .
35. (a) Given that z 3 = p + iq
[Q eip = –1 and w = re–if]
Þ z = p3 + (iq )3 + 3 p (iq )( p + iq )
41. (c) Let z = x + iy
Þ x - iy = p3 - 3 pq 2 + i (3 p 2 q - q 3 )
Q z 2 = i | z |2
Comparing both side, we get
\ x 2 - y 2 + 2ixy = i( x 2 + y 2 )
x
\ x = p3 - 3 pq 2 Þ = p 2 - 3q2
p
Þ x 2 - y 2 = 0 and 2 xy = x2 + y 2
...(i)
Þ ( x - y )( x + y ) = 0 and ( x - y)2 = 0
Þx=y
y
and y = q - 3 p q Þ = q 2 - 3 p 2 ...(ii)
q
3
2
Adding (i) and (ii), we get
\
42. (a) Given that, a =
æ x yö
x y
+ = -2 p 2 - 2q2 \ ç + ÷ ( p 2 + q 2 ) = -2
è p qø
p q
36. (c) Given that arg zw = p
Þ arg z + arg w = p
\ (2 + w)4 = a + bw Þ (4 + w2 + 4w)2 = a + bw
Þ (w2 + 4(1 + w)) 2 = a + bw
...(i)
z + iw = 0 Þ z = -iw
Replace i by –i, we get
p
\ z = iw Þ arg z = + arg w
2
p
Þ arg z = + p - arg z (from (i))
2
3p
\ arg =
4
37. (b) Given that
Þ (w 2 - 4w 2 ) 2 = a + bw
[Q1 + w = -w 2 ]
Þ ( -3w 2 )2 = a + bw Þ 9w 4 = a + bw
x
x
é (1 + i )2 ù
æ 1+ i ö
ú =1
çè
÷ø = 1 Þ ê
2
1- i
êë 1 - i úû
æ 1 + i 2 + 2i ö
x
+
ç
÷ = 1 Þ (i ) = 1; \ x = 4n ; n Î I
1
1
+
è
ø
(Q w3 = 1)
Þ 9w = a + bw
On comparing, a = 0, b = 9
Þ a + b = 0 + 9 = 9.
43. (c)
x
-1 + 3i
=w
2
5p
5p ö
æ
1 + cos + i sin
ç
18
18 ÷
ç
5p
5p ÷
ç 1 + cos - i sin ÷
è
18
18 ø
3
5p
5p
5p ö
æ
2cos 2
+ i 2sin × cos
ç
36
36
36 ÷
=ç
5p
5p ÷
2 5p
- i 2sin × cos ÷
ç 2cos
è
36
36
36 ø
3
Downloaded from @Freebooksforjeeneet
M-41
Complex Numbers and Quadratic Equations
3
5p
5p ö
æ
6
cos + i sin
ç
36
36 ÷ = æ cos 5p + i sin 5p ö
=ç
ç
÷
5p
5p ÷
è
36
36 ø
- i sin ÷
ç cos
è
ø
36
36
5p ö
5p ö
5p
5p
æ
æ
= cos ç 6 ´ ÷ + i sin ç 6 ´ ÷ = cos + i sin
è
è
36 ø
36 ø
6
6
=44. (b)
Let
3 1
1
+ i = - ( 3 - i)
2
2
2
3 + 2 -54 = 3 + 6 6i
3 + 6 6i = a + ib
Þ a 2 - b2 = 3 and ab = 3 6
Þ a 2 + b2 = (a 2 - b 2 ) 2 + 4a 2b 2 = 15
So, a = ±3 and b = ± 6
47. (d)
æ 1
3 i
3 ö
+ = -i ç - +
i ÷ = -iw
ç
÷
2 2
è 2 2 ø
where w is imaginary cube root of unity.
Now, (1 + i + 5 + i 8)9
= (1 + w – iw2 + iw2)9 = (1 + w)9
= (– w2)9 = – w18 = – 1
(Q 1 + w + w2 = 0)
48. (a) –(6 + i)3 = x + iy
Þ –[216 + i3 + 18i(6 + i)] = x + iy
Þ –[216 – i +108i – 18] = x + iy
Þ –216 + i – 108i + 18 = x + iy
Þ –198 – 107i = x + iy
Þ x = – 198, y = –107
Þ y – x = –107 + 198 = 91
5
æ 3 iö æ 3 iö
+ ÷ +ç
- ÷
49. (a) z = ç
è 2 2ø è 2 2ø
5
5
p
pö æ
p
pö
æ
= ç cos + i sin ÷ + ç cos - i sin ÷
è
6
6ø è
6
6ø
3 + 6 6 i = ± (3 + 6 i)
Similarly, 3 - 6 6 i = ± (3 - 6 i)
5
lm ( 3 + 6 6i - 3 - 6 6i ) = ±2 6
45. (b) Let a = w, b = 1 + w3 + w6 + ..... = 101
a = (1 + w) (1 + w2 + w4 + ..... w198 + w200)
= (1 + w)
Þ
a =
(
1 - (w2 )101
)
(w + 1)(w202 - 1)
=
2
1- w
(1 + w)(1 - w)
1 - w2
(w2 - 1)
=1
Required equation = x2 – (101 + 1)x + (101) × 1 = 0
Þ
x2 – 102x + 101 = 0
46. (d) Q
z = x + iy
5
5
æ ip ö
æ -i p ö
p
6
6
= ç e ÷ + ç e ÷ = 2 cos 6 = 3
è ø
è
ø
Þ
I(z) = 0, Re(z) = 3
æ1+ i 3 ö
50. (d) Let l = çç
÷÷ .
è1– i 3 ø
æ1 + i 3 ö æ1 + i 3 ö
\ l = çç
÷÷ ´ çç
÷÷
è1 – i 3 ø è1 + i 3 ø
æ – 2 + i2 3 ö æ 1 – i 3 ö
=ç
÷÷ = çç
÷÷
ç
4
è
ø è –2 ø
æ1 + i 3 ö æ1 – i 3 ö
Also, l = çç
÷÷
÷÷ ´ çç
è1 – i 3 ø è1 – i 3 ø
æ z - 1 ö ( x - 1) + iy
ç 2 z + i ÷ = 2( x + iy ) + i
è
ø
æ
ö æ –2 ö
4
=ç
÷=ç
÷
è – 2 – i2 3 ø è 1 + i 3 ø
( x - 1) + iy
2 x - (2 y + 1)i
=
´
2 x + (2 y + 1)i 2 x - (2 y + 1)i
æ z + 1 ö 2 x( x - 1) + y (2 y + 1)
Re ç
=1
÷=
è 2z + i ø
(2 x ) 2 + (2 y + 1) 2
2
2
2 æ
1ö æ
3ö
5ö
æ
÷÷ .
Þ ç x + ÷ + ç y + ÷ = çç
2
4
4
è
ø è
ø
è
ø
3
æ1+ i 3 ö æ1+ i 3 ö æ1 + i 3 ö æ1+ i 3 ö
Now, ç
÷ =ç
÷´ç
÷´ç
÷
è1 – i 3 ø è1 – i 3 ø è1 – i 3 ø è1 – i 3 ø
æ1+ i 3 ö æ – 2 ö æ1 – i 3 ö
=ç
÷´ç
÷ =1
÷´ç
è1 – i 3 ø è1 + i 3 ø è – 2 ø
\ least positive integer n is 3.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-42
Mathematics
54. (a) (1 + w)7 = A + Bw
(–w2)7 = A + Bw (Q w14 = w12.w2 = w2)
– w2 = A + Bw
1 + w = A + Bw
Þ A = 1, B = 1.
55. (a) |z + 1 | = | z + 4 – 3 | £ | z + 4 | + | –3 | £ | 3 | + | – 3|
(3,3)
51. (a)
2 2
(1,1)
Final
(3,1)
(2,1) 1
position
Þ | z + 1 | £ 6 Þ | z + 1|max = 6
56. (c) Given that w =
So new position is at the point 1 + i
52. (a)
1 - 2 2 =1
2- 1 2
1 -2 2
Þ
2
Þ |w|=
= 2-
2
( 1 - 2 2 )( 1 - 2 2 ) = (2 - 1 2 )(2 - 1 2 )
Þ
( 1 - 2 2 )( 1 - 2 2 ) = (2 - 1 2 )(2 - 1 2 )
Þ
(
1 1) - 2 1 2
-2
-2
+
= 4-2
1 2
2
Þ
1
Þ
1
2
1 2
2
+4
2
+4
2
2
+4
1 2
2 2
2
2
1
1
57. (c)
2
2
2
–4–
(z
1
Q
2
2
1
Þ
1
Þ
Point
53. (a)
2
)(
- 4 1 - z2
¹1
\
| z1 + z2 | = | z1 | + | z2 | Þ z1 and z2 are collinear
and are to the same side of origin; hence arg z1 – arg z2 = 0.
=0
2
é z1
z ù
= 1ú
êQ
z2 û
ë z2
|z|
=1
1
|z- i|
3
æ 1ö
Þ distance of from origin and point ç 0, ÷ is same
è 3ø
hence lies on bisector of the line oining points (0, 0) and
(0, 1/3).
Hence lies on a straight line.
1 1 2 2
= 4+
1
z- i
3
1
Þ z = z- i
3
1 2
Þ
z
2
)=0
58. (c) Q ( x - 1)3 + 8 = 0 Þ ( x - 1) = (-2) (1)1/ 3
Þ x – 1 = – 2 or -2w or - 2w 2
or x = – 1 or 1 – 2 w or 1 – 2 w2 .
=4
59. (b) Given that | z 2 - 1 |=| z |2 +1 Þ| z 2 - 1|2 = ( zz + 1) 2
=2
2
[Q z =
1 lies on circle of radius 2.
z2
z 2 é æ z1 ö z1 ù
=
= ú
êQ
z - 1 z - 1 ë çè z2 ÷ø z2 û
2
2
Þ zzz - z = z. z . z - z
Þ z 2 .z - z 2 = z 2 .z - z 2
Þ z 2 ( z - z ) - ( z - z )( z + z ) = 0
(
)
2
Þ (z - z ) z -(z + z ) = 0
Either z - z = 0 or z 2 - ( z + z ) = 0
Either z = z Þ real axis
or z 2 = z + z Þ zz - z - z = 0
]
Þ(z2 -1)(z 2 -1) = (zz +1)2 (Qz1 - z2 = z1 - 2)
Þ
2 2
-
2
-
2
+1 =
2 2
+2
+1
Þ z 2 + 2 zz + z 2 = 0
Þ ( z + z )2 = 0 Þ z = - z
Þ z is purely imaginary
60. (b) Let the circle be |z – z0| = r. Then according to given
conditions |z0 – z1| = r + a
...(i)
|z0 – z2|= r + b
...(ii)
Subtract (ii) from (i)
we get |z0 – z1| – |z0 – z2| = a – b.
\ Locus of centre z0 is |z – z1| –|z – z2|
= a – b, which represents a hyperbola.
represents a circle passing through origin.
Downloaded from @Freebooksforjeeneet
M-43
Complex Numbers and Quadratic Equations
Imaginary part of u
61. (a) Q a + b = 64, ab = 256
a 3/8
b3/8
a +b
64
=2
5/8
5/8
5/8
8 5/8
32
b
a
(ab)
(2 )
62. (b) Let a and b be the roots of the given quadratic
equation,
+
=
64
=
=
2x2 + 2x - 1 = 0
...(i)
Then, a + b = -
1
Þ -1 = 2a + 2b
2
-2 x( y - K ) + x (2 y + 1)
x 2 + ( y - K )2
Q Re(u ) + Im(u ) = 1
Þ 2 x2 + 2 y 2 - 2 Ky + y - K - 2 xy + 2 Kx + 2 xy + x
= x 2 + y 2 + K 2 - 2 Ky
Since, the curve intersect at y-axis
[Q a is root of
and 4a 2 + 2a - 1 = 0
eq. (i)]
= Im(u ) =
\x = 0
Þ y 2 + y - K ( K + 1) = 0
Þ 4a 2 + 2a + 2a + 2b = 0 Þ b = -2a (a + 1)
Let y1 and y2 are roots of equations if x = 0
Q y1 + y2 = -1
63. (b) Let | x | = y then
9 y 2 - 18 y + 5 = 0
y1 y2 = -( K 2 + K )
Þ 9 y 2 - 15 y - 3 y + 5 = 0
\ ( y1 - y2 )2 = (1 + 4 K 2 + 4 K )
Þ (3 y - 1)(3 y - 5) = 0
Given PQ = 5 Þ| y1 - y2 |= 5
Þ y=
1
5
1
5
or Þ| x |= or
3
3
3
3
Roots are ±
3 x 2 - 10 x + 27l = 0
25
81
64. (d) Let a and b be the roots of the quadratic equation
=
=
a
+
\ 3a 2 - 10a + 27l = 0
3a 2 - 3a + 6l = 0
b
Þ a + b = 1 Þ 3l +
1 - b2
a - ab(a + b) + b
2
3
2
1
= 1 Þ l = and
3
9
ag = 9l Þ 3l × g = 9l Þ g = 3
\
(a + b) - ab(a + b)
1 - (a + b) 2 + 2ab + (ab) 2
bg
= 18
l
67. (d)
3 2 3
+ ´
27
7
7 7
=
=
-2 4 16
9
+
1- + 2 ´
49
7 49
a ×b = 2 and a + b = - p also
1 1
+ = -q
a b
Þ p = 2q
2( x + iy ) + i 2 x + i (2 y + 1)
u=
=
( x + iy ) - ki
x + i( y - k )
Real part of u = Re(u ) =
...(ii)
Now, ab = 2l Þ 3l ×b = 2l Þ b =
1 - (a 2 + b 2 ) + (ab) 2
65. (d)
...(i)
\ On subtract, we get a = 3l
7 x2 - 3x - 2 = 0
3
-2
\ a + b = , ab =
7
7
1 - a2
as K > 0, \ K = 2
66. (b) Since a is common root of x 2 - x + 2l = 0 and
1
5
and ±
3
3
\ Product =
Now,
Þ 4 K 2 + 4 K - 24 = 0 Þ K = 2 or – 3
2 x + ( y - K )(2 y + 1)
2
1ö æ
1ö æ
1ö æ
1ö
æ
Now çè a - a ÷ø èç b - bø÷ èç a + bø÷ çè b + a ÷ø
é
ù
1 a bùé
1
= êab +
- - ú êab +
+ 2ú
ab b a û ë
ab
ë
û
x 2 + ( y - K )2
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EBD_8344
M-44
Mathematics
=
9 é 5 a2 + b2 ù 9
2
ê ú = [5 - ( p - 4)]
2 ë2
2 û 4
9
= (9 - p 2 )
[Q a 2 + b2 = (a + b) 2 - 2ab]
4
68. (c) The given quadratic equation is
(l + 1) x - 4lx + 2 = 0
2
2
Q One root is in the interval (0, 1)
\ f (0) f (1) £ 0
Þ 2(l 2 + 1 - 4l + 2) £ 0
Þ 2(l 2 - 4l + 3) £ 0
71. (8) Since, 2x2 + (a – 10) x +
\
D³0
Þ
æ 33
ö
(a - 10)2 - 4(2) ç - 2a ÷ ³ 0
è 2
ø
Þ (a – 10)2 – 4(33 – 4a) ³ 0
Þ a2 – 4a – 32 ³ 0
Þ (a – 8) (a + 4) ³ 0
Þ a£–4Èa³8
Þ aÎ (– ¥, – 4] È [8, ¥)
72. (a) Let z = a ± ib be the complex roots of the equation
So, sum of roots = 2a = – b and
Product of roots = a2 + b2 = 45
(a + 1)2 + b2 = 40
(l - 1)(l - 3) £ 0 Þ l Î[1, 3]
But at l = 1, both roots are 1 so l ¹ 1
\ l Î (1, 3]
69. (c) Since, a and b are the roots of the equaton
5x2 + 6x – 2 = 0
Then, 5a 2 + 6a - 2 = 0 , 5b 2 + 6b - 2 = 0
5a 2 + 6a = 2
5S6 + 6S5 = 5(a 6 + b6 ) + 6(a 5 + b5 )
Given, | z + 1|= 2 10
Þ (a + 1)2 – a2 = – 5
[Q b2 = 45 – a2]
Þ 2a + 1 = – 5
Þ
2a = – 6
2
Hence, b = 6 and b – b = 30
73. (d) a5 = 5a + 3
b5 = 5b + 3
p5 = 5(a + b) + 6 = 5(1) + 6
[Q from x2 – x – 1 = 0, a + b =
= (5a 4 + 6a 5 ) + (5b6 + 6b5 )
= a 4 (5a 2 + 6a ) + b 4 (5b 2 + 6b)
= 2(a 4 + b4 ) = 2S4
70. (a) Let ex = t Î (0, ¥)
Given equation
t4 + t3 – 4t2 + t + 1 = 0
Þ
Þ
tan a × tan b =
\
Þ
Þ
(k + 1) tan 2 x - 2l tan x + ( k - 1) = 0
tan a + tan b =
æ 2 1 ö æ 1ö
çt + 2 ÷ + çt + t ÷ - 4 = 0
ø
t ø è
è
y2 + y – 6 = 0
y = – 3, 2
1
Þ y=2 Þ t + = 2
t
Þ ex + e–x = 2
x = 0, is the only solution of the equation
Hence, there only one solution of the given equation.
-b
= 1]
a
p5 = 11 and p5 = a2 + b2 = a + 1 + b + 1
p2 = 3 and p3 = a3 + b3 = 2a + 1 + 2b + 1
= 2(1) + 2 = 4
p2 × p3 = 12 and p5 = 11 Þ p5 ¹ p2 × p3
74. (b)
1 1
t2 + t – 4 + + 2 = 0
t t
1
Let t + = y
t
2
(y – 2) + y – 4 = 0
y2 + y – 6 = 0
33
= 2a has real roots,
2
2l
k +1
k -1
k +1
[Sum of roots]
[Product of roots]
2l
k
+ 1 = 2l = l
tan(a + b) =
k -1
2
2
1k +1
tan 2 (a + b) =
l2
= 50
2
l = 10.
75. (b) Given equation is, x2 + x sin q – 2 sin q = 0
a + b = – sin q and ab = – 2 sin q
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M-45
Complex Numbers and Quadratic Equations
(a12 + b12 )a12b12
(a12 + b12 )(a - b) 24
=
80. (c) The given quadratic equation is x2 – 2x + 2 = 0
(ab)12
(a - b) 24
Then, the roots of the this equation are
\ | a - b |= (a + b)2 - 4ab = sin 2 q + 8sin q
12
\
(ab)
( a - b)
24
=
12
(2sin q)
12
12
sin q(sin q + 8)
=
12
2
Now,
a 1 - i (1 - i ) 2
=
=
=i
b 1 + i 1 - i2
or
a
a 1 - i (1 - i ) 2
=
=
= – i So, = ±i
2
b
b 1+ i 1- i
(sin q + 8)12
76. (d) Let 2x – 1 = t
5 + | t | = (t + 1) (t – 1) Þ | t | = t2 – 6
When t > 0, t2 – t – 6 = 0 Þ t = 3 or – 2
t = – 2 (reected)
When t < 0, t2 + t – 6 = 0 Þ t = – 3 or 2 (both reected)
\ 2x – 1 = 3 Þ 2x = 4 Þ x = 2
77. (d) Since 2 - 3 is a root of the quadratic equation
x2 + px + q = 0
Þ x 2 + px + q = [ x - (2 - 3)[ x - (2 + 3)]
= x 2 - (2 + 3)x - (2 - 3)x + (22 - ( 3) 2 )
Þ
81. (b) Let roots of the quadratic equation are a, b.
a
a b
1
and l + = 1 Þ + = 1
b
b a
l
(a + b) 2 - 2ab
= 1...(i)
ab
\
Now, by comparing p = –4, q = 1
Þ p2 – 4q – 12 = 16 – 4 – 12 = 0
78. (c) Sum of roots =
m2 + 1
Now, a + b = 3, ab = 1 Þ | -a - b | = 5
(
a+b=
m(4 - m) 4 - m
2
and ab =
=
2
m
3
3m
3m 2
Put these values in eq (1),
3
Q sum of roots is greatest. \ m = 0
Hence equation becomes x2 – 3x + 1 = 0
)
a3 - b3 = (a - b) a 2 + b2 + ab = 5 ( 9 - 1) = 8 5
x =a
given equation will become:
79. (d) Let
æ 4 - mö
çè
÷
3m ø
2
m
3 2
Þ
| a – 2 | + a2 – 4a + 4 – 2 = 0
| a – 2 | + (a – 2)2 – 2 = 0
2
=3
(m – 4)2 = 18 Þ m = 4 ± 18
Therefore, least value is 4 – 18 = 4 - 3 2
82. (d) Let a and b be the roots of the equation,
81x2 + kx + 256 = 0
| a – 2 | + a (a – 4) + 2 = 0
Þ
Þ
n must be a multiple of 4.
Hence, the required least value of n = 4.
The quadratic equation is, 3m2x2 + m(m – 4) x + 2 = 0
= x 2 - 4x + 1
\
n
æaö
Now, ç ÷ = 1 Þ (±i)n = 1
èbø
Given, l =
\ 2 + 3 is the other root
1
Given (a) 3 = b Þ a = b3
256
81
Let | a – 2| = y (Clearly y ³ 0)
Þ y + y2 – 2 = 0
Q
Product of the roots =
Þ
Þ
\
(a)(b) =
Þ
4
64
æ 4ö
b4 = ç ÷ Þ b = Þ a =
è 3ø
3
27
y = 1 or – 2 (reected)
| a – 2 | = 1 Þ a = 1, 3
When
x = 1Þx=1
x = 3Þx=9
Hence, the required sum of solutions of the equation
= 10
When
2 ± -4
=1± i
2
256
81
4
Q Sum of the roots = -
k
81
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EBD_8344
M-46
Mathematics
a+ b=-
\
k
k
4 64
=Þ +
81
3 27
81
Þ k = –300
83. (d) Consider the quadratic equation
(c – 5) x2 – 2cx + (c – 4) = 0
Now, f(0).f(3) > 0 and f(0).f(2) < 0
Þ (c – 4) (4c – 49) > 0 and (c – 4) (c – 24) < 0
Þ
æ 49 ö
c Î(–¥, 4) È ç , ¥ ÷ and c Î (4, 24)
è 4
ø
Þ
æ 49
ö
c Î ç , 24 ÷
è 4
ø
\ S = {13, 14, ..., 23}
84. (d) The given quadratic equation is
x2 + (3 – l) x + 2 = l
Sum of roots = a + b = l – 3
Product of roots = ab = 2 – l
a2 + b2 = (a + b)2 – 2ab
= (l – 3)2 – 2 (2– l)
= l2 – 4l + 5
= (l – 2)2 + 1
For least (a2 + b2) l = 2.
85.
(a) Consider the equation
x2 + 2x + 2 = 0
-2 ± 4 - 8
= -1 ± i
2
Let a = –1 + i, b = –1 – i
a15 + b15 = (–1 + i)15 + (–1 – i)15
x=
æ
=ç
è
÷
ø
( 2)
15
=
=
=
(
3p 15
-i ö
2e 4
÷
ø
- i 45p ù
é i 45 p
êe 4 + e 4 ú
êë
úû
2 ) .2 cos
15
-2
2
= -2
æ
+ç
è
45p ( )15
3p
= 2 .2cos
4
4
( 2)
15
( 2)
14
\
Discriminant D must be perfect square number.
D = (–11)2 – 4 × 6 × a
= 121 – 24a must be a perfect square
Hence, possible values for a are
a = 3, 4, 5.
\ 3 positive integral values are possible.
87. (b) Given quadratic equation is: x2 – mx + 4 = 0
Both the roots are real and distinct.
So, discriminant B2 – 4AC > 0.
æ 49
ö
Integral values in the interval ç , 24 ÷ are 13, 14, ..., 23.
4
è
ø
3p 15
i ö
2e 4
86. (a) The roots of 6x2 – 11x + a = 0 are rational numbers.
\
m2 – 4 × 1 × 4 > 0
\
(m – 4) (m + 4) > 0
\
m Î (–฀, –4) È (4, ฀)
Since, both roots lies in [1, 5]
-m
Î (1, 5)
2
\
-
Þ
m Î (2, 10)
...(ii)
And 1 × (1 – m + 4) > 0 Þ m < 5
\
m Î (–฀, 5)
...(iii)
And 1 × (25 – 5m + 4) > 0 Þ m <
\
29 ö
æ
m Îç - ฀ ÷
è
5ø
29
5
...(iv)
From (i), (ii), (iii) and (iv), m Î (4, 5)
88. (a) Q z0 is a root of quadratic equation
x2 + x + 1 = 0
\
z0 = w or w2 Þ z03 = 1
\
z = 3 + 6i z081 – 3i z093
= 3 + 6i((z0)3)27 – 3i((z0)3)31
= 3 + 6i – 3i
= 3 + 3i
\
89. (b)
æ 3ö p
=
arg(z) tan–1 çè 3÷ø 4
1
1
1
+
=
x+ p x+q r
x+ p+x+q
( x + p) ( x + q)
= -256
...(i)
=
1
r
(2x + p + q) r = x2 + px + qx + pq
x2 + (p + q – 2r) x + pq – pr – qr = 0
Let a and b be the roots.
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M-47
Complex Numbers and Quadratic Equations
\ a + b = – (p + q – 2r)
.... (i)
& ab = pq – pr – qr
.... (ii)
Q a = – b (given)
\ in eq. (1), we get
Þ – (p + q – 2r) = 0
... (iii)
Now, a2 + b2 = (a + b)2 – 2ab
= (– (p + q –2r))2 – 2 (pq –pr – qr) .... (from (i) and (ii))
= p2 + q2 + 4r2 + 2pq – 4pr – 4qr – 2pq + 2pr + 2qr
= p2 + q2 + 4r2 – 2pr – 2qr
= p2 + q2 + 2r (2r – p – q)
... (from (iii))
= p2 + q2 + 0
= p2 + q2
90. (b) Here, 9x2 + 27x + 20 = 0
– b ± b – 4ac
and f (2) = 6 – 3a
As, f (1) + f (2) = 0 Þ 2 – 2a + 6 – 3a = 0 Þ a =
8
5
93. (b) a, b are roots of x2 – x + 1 = 0
Therefore, the other root is
\
a = -w and b = -w2
where w is cube root of unity
\ a101 + b107 = (–w)101 + (–w)107
= –[w2 + w] = –[–1] = 1
n
94. (a) We have,
2
\ x=
Þ x=
n
å
2a
2
Þ
4
5
,–
3
3
\ sec A =
3
5
1
5
=–
cos A
3
Here, A is an obtuse angle.
\ tan A = – sec 2 A – 1 = –
4
.
3
Hence, roots of the equation are sec A and tan A.
91. (b) As tan A and tan B are the roots of 3x2 – 10x – 25 = 0,
tan A + tan B
So, tan (A + B) =
=
1 – tan A tan B
10
3 = 10 / 3 = 5
25
28 / 3 14
1+
3
Now, cos2 (A + B) = – 1 + 2 cos2 (A + B)
=
1 – tan 2 ( A + B )
1 + tan 2 ( A + B )
Þ cos 2 ( A + B ) =
196
221
\ 3sin2 (A + B) – 10 sin (A + B) cos (A + B) – 25 cos2 (A + B)
= cos2 (A + B) [3 tan2 (A + B) – 10 tan (A + B) – 25]
=
(x + r - 1)(x + r) = 10n
r =1
(x 2 + xr + (r –1)x + r 2 - r) = 10n
n
Þ
2´9
Given, cos A = –
å
r =1
– 27 ± 27 – 4 ´ 9 ´ 20
Þ x=–
8
5
75 – 700 – 4900 196
5525 196
´
=–
´
= – 25
196
221
196 221
92. (d) If a and – 1 are the roots of the polynomial, then we
get
f (x) = x2 + (1 – a) x – a.
\ f (1) = 2 – 2a
å
(x 2 + (2r - 1)x + r(r - 1)) = 10n
r =1
nx2 + {1 + 3 + 5 + .... + (2n – 1) }x
+ {1.2 + 2.3 +.... + (n – 1) n} = 10 n
(n - 1) n(n + 1)
= 10n
Þ nx2 + n2 x +
3
n 2 - 31
=0
Þ x2 + nx +
3
Let a and a + 1 be its two solutions
(Q it has two consequtive integral solutions)
Þ a + (a + 1) = – n
-n - 1
Þ a=
...(i)
2
n 2 - 31
Also a (a+1) =
...(ii)
3
Putting value of (i) in (ii), we get
2
æ n + 1ö æ 1 - n ö n - 31
-ç
=
÷
ç
÷
è 2 øè 2 ø
3
Þ
95. (c)
Þ
Þ
n2 = 121 Þ n = 11
(x – 1) (x2 + 5x – 50) = 0
(x – 1) (x + 10) (x – 5) = 0
x = 1, 5, –10
Sum = – 4
96. (c) Let p (x) = ax2 + bx + c
Q p (0) = 1 Þ c = 1
Also, p (1) = 4 & p ( -1) = 6
Þ
Þ
Þ
a+ b+1= 4&a –b+1=6
a +b=3&a–b=5
a = 4 & b = –1
p (x) = 4x2 – x + 1
p (b) = 16 – 2 + 1 = 15
p (–2) = 16 + 2 + 1 = 19
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EBD_8344
M-48
Mathematics
(x 2 - 5 x + 5) x
97. (c)
2
+ 4x - 60
=1
Case I
x2 – 5x + 5 = 1 and x2 + 4x – 60 can be any real number
Þ x = 1, 4
Case II
x2 – 5x + 5 = –1 and x2 + 4x – 60 has to be an even number
Þ x = 2, 3
where 3 is reected because for x = 3,
x2 + 4x – 60 is odd.
Case III
x2 – 5x + 5 can be any real number and
x2 + 4x – 60 = 0
Þ x = –10, 6
Þ Sum of all values of x
= –10 + 6 + 2 + 1 + 4 = 3
2x + 1 - 2 x -1 = 1
98. (a)
.....(i)
Þ
2x + 1 + 2x – 1 – 2 4 x 2 - 1 = 1
Þ
Þ
Þ
4x – 1 = 2 4 x 2 - 1
16x2 – 8x + 1 = 16x2 – 4
8x = 5
Þ
5
x = which satisfies equation (i)
8
So,
100. (b) (a – 1) (x4 + x2 + 1) + (a + 1) (x2 + x + 1)2 = 0
Þ (a – 1) (x2 + x + 1) (x2 – x + 1) + (a + 1) (x2 + x + 1)2 = 0
Þ (x2 + x + 1) [(a – 1) (x2 – x + 1) + (a + 1) (x2 + x + 1)] = 0
Þ (x2 + x + 1) (ax2 + x + a) = 0
For roots to be distinct and real, a ¹ 0 and 1 – 4a2 > 0
Þ a ¹ 0 and a2 <
æ 1 ö æ 1ö
Þ a Î çè - , 0÷ø È çè 0, ÷ø
2
2
101. (b) a = 2 + 3i; b = 2 – 3i, g = ?
abg =
13
1
Þ g=
2
2
102. (c) Consider –3(x – [x])2 + 2 [x – [x]) + a2 = 0
Þ 3{x}2 – 2{x} –a2 = 0
(Q x – [x] = {x})
Þ
2
æ
ö
3 ç {x}2 - {x} ÷ = a2 , a ¹ 0
3
è
ø
Þ
2ö
æ
a 2 = 3{x} ç {x} - ÷
è
3ø
3
4 x2 - 1 = 4
(
\ an = 3 + 11
1/3
) – (3 – 11)
n
n
–1/3
a10 – 2a 8
2a 9
10
10
8
8
3 + 11 ) – ( 3 – 11 ) – 2 ( 3 + 11 ) + 2 ( 3 - 11 )
(
=
9
9ù
é
2 ê( 3 + 11) - ( 3 - 11 ) ú
ë
û
(
) (
)
(
)
(
)
2
8é
2ù
é
ù
3 + 11 ê 3 + 11 – 2 ú + 3 - 11 ê 2 - 3 - 11 ú
ë
û
ë
û
9
9ù
é
2 ê 3 + 11 - 3 - 11 ú
ë
û
8
(
) (
)
( 3 + 11)8 ( 9 + 11 + 6 11 – 2) + (3 - 11)8 ( 2 - 9 - 11 + 6 11)
=
9
9ù
é
2 ê( 3 + 11 ) – ( 3 - 11 ) ú
ë
û
(
) – 6 (3 - 11) 6
= =3
9
9ù
é
2
2 ê( 3 + 11 ) - ( 3 - 11 ) ú
ë
û
6 3 + 11
=
dù
13 é
êësince product of roots = a úû
2
Þ ( 4 + 9) g =
6 ± 36 + 8
99. (a) a, b =
= 3 ± 11
2
a = 3 + 11 , b = 3 - 11
=
1
4
9
Now, {x} Î (0,1) and
2/3
-2
£ a2 < 1
3
Since , x is not an integer
\
a Î (-1,1) - {0}
Þ a Î (-1, 0) È (0,1)
103. (a) Consider 3 x 2 + x + 5 = x – 3
Squaring both the sides, we get
3x2 + x + 5 = (x – 3)2
Þ 3 x2 + x + 5 = x2 + 9 - 6 x
Þ 2x2 + 7 x - 4 = 0
9
Þ 2 x2 + 8 x - x - 4 = 0
Þ 2 x( x + 4) - 1( x + 4) = 0
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(by graph)
M-49
Complex Numbers and Quadratic Equations
\ a3 + b3 = (4 2k ) [66 – (2k4 – 1)]
1
or x = – 4
2
1
and x = – 4
For x =
2
Þ x=
Putting k = – 2, (k = +2 cannot be taken because it does
not satisfy the above equation)
\ a3 + b3 = (4 2( -2))[66 - 2( -2) 4 - 1]
L.H.S. ¹ R.H.S. of equation, 3 x 2 + x + 5 = x – 3
Also, for every x Î R, LHS ¹ RHS of the given equation.
\ Given equation has no solution.
104. (c)
x2 + 2 x - 3 - 4 = 0
ì
ïï (2 x - 3) if
|2x – 3| = í
ï - (2 x - 3) if
ïî
a3 + b3 = ( -8 2) (66 – 32 + 1) = ( -8 2) (35)
\ a3 + b3 = -280 2
1
106. (a) Let
3
2
3
x<
2
x>
a
1
+ b
a
1
1
a b
3
for x > , x 2 + 2 x - 3 - 4 = 0
2
x2 + 2x – 7 = 0
x=
b= -
-2 ± 4 + 28
-2 ± 4 2
= -1 ± 2 2
=
2
2
Here x = 2 2 - 1
3ü
ì
í2 2 - 1 < ý
2þ
î
1
Þ a = ab
a
a+ b
)
x( x + b3 ) + (a3 - 3abx) = 0
Þ
x2 + (b3 - 3ab) x + a3 = 0
(
)
3
+3
(
ab
)(
)
a + b ù + (ab)3/ 2 = 0
ûú
Þ x 2 - éëa3/2 + b3/2 + 3 ab ( a + b ) - 3 ab ( a + b ) ùû x + (ab)3/2 = 0
2± 4+ 4 2± 2 2
=
= 1± 2
x=
2
2
Þ
x2 - (a 3/ 2 + b3/ 2 ) x + a 3/ 2b3/ 2 = 0
Roots of this equation are a 3/ 2 , b3/ 2
107. (b) Given a3 + b3 = – p and ab = q
3ü
ì
í(1 - 2) < ý
2þ
î
Sum of roots : (2 2 - 1) + (1 - 2) =
105. (d)
æ a + bö
b
÷ = -a
= ç
ab ø
è
x2 + é - a + b
ëê
3
2
x2 – 2x + 3 – 4 = 0
Þ x2 – 2x – 1 = 0
Here x = 1 - 2
be the roots of ax 2 + bx + 1 = 0
b
Putting values of a and b, we get
for x <
Þ
(
=
1
and
Let
a2
b2
and
be the root of required quadratic equation.
b
a
So,
a 2 b2 a3 + b3 - p
+
=
=
b
a
ab
q
2
x 2 - 4 2kx + 2e 4 ln k - 1 = 0
or, x 2 - 4 2kx + 2k 4 - 1 = 0
a + b = 4 2k and a.b = 2k4 – 1
Squaring both sides, we get
(a + b)2 = (4 2k )2 Þ a2 + b2 + 2ab = 32k2
66 + 2ab = 32k2
66 + 2 (2k4 – 1) = 32k2
66 + 4k4 – 2 = 32k2 Þ 4k4 – 32k2 + 64 = 0
or, k4 – 8k2 + 16 = 0 Þ (k2)2 – 8k2 + 16 = 0
Þ (k2 – 4) (k2 – 4) = 0 Þ k2 = 4, k2 = 4
Þk= ± 2
Now, a3 + b3 = (a + b) (a2 + b2– ab)
and
a 2 b2
´
= ab = q
b
a
Hence, required quadratic equation is
æ -p ö
x2 - ç
÷x+q =0
è q ø
Þ x2 +
p
x + q = 0 Þ qx2 + px + q2 = 0
q
108. (c) Given quadratic eqn. is
x 2 + px +
3p
=0
4
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EBD_8344
M-50
Mathematics
So, a + b = – p, ab =
Now, By putting
3p
4
a=
Now, given | a – b | = 10
Þ a – b = ± 10
Þ (a – b)2 = 10 Þ a2 + b2 – 2ab = 10
Þ (a + b)2 – 4ab = 10
We get least value of x12 + x22 at
3p
= 10 Þ p2 – 3p – 10 = 0
4
Þ p = – 2, 5 Þ p Î {– 2, 5}
109. (c) Given equation is
z + 2 | z + 1| + i = 0
put z = x + iy in the given equation.
2 | x + iy + 1 | + i = 0
2
2ù
é
Þ x + iy + 2 ê ( x + 1) + y ú + i = 0
ë
û
Now, equating real and imaginary part, we get
p
2
112. (b) (k – 2) x2 + 8x + k + 4 = 0
If real roots then,
82 – 4(k – 2) (k + 4) > 0
Þ k2 + 2k – 8 < 16
Þ k2 + 6k – 4k – 24 < 0
Þ (k + 6) (k – 4) < 0
Þ –6<k<4
If both roots are negative
then ab is +ve
Þ
x + 2 ( x + 1) 2 + y 2 = 0 and
Also,
y + 1= 0 Þ y = – 1
Þ x + 2 ( x + 1) 2 + (-1)2 = 0
Þ
(Q y = – 1)
2 ( x + 1)2 + 1 = - x
Þ 2[(x + 1)2 + 1] = x2
Þ x2 + 4x + 4 = 0
Þx=–2
Thus, z = – 2 + i(– 1) Þ | z | = 5
110. (c) Given quadratic equation is
px2 + qx + r = 0
...(i)
D = q2 – 4pr
Since a and b are two complex root
\ b = a Þ |b| = | a | Þ |b| = |a|
(Q| a |=| a|)
Consider
|a| + |b| = |a| + |a|
(Q |b| = |a|)
= 2 |a| > 2.1 = 2
(Q |a| > 1 )
Hence, |a| + |b| is greater than 2.
111. (d) Given equation is
x2 – (sina – 2)x – (1 + sina) = 0
Let x1 and x2 be two roots of quadratic equation.
\ x1 + x2 = sina – 2 and x1x2 = – (1 + sina)
(x1 + x2)2 = (sina – 2)2 = sin2a + 4 – 4 sina
Þ
x12 + x22 = sin 2 a + 4 - 4sin a - 2 x1 x2
= sin2a + 4 – 4 sina + 2 (1 + sina)
= sin2a – 2 sina + 6
...(i)
p
2
Hence, a =
Þ p2 – 4 ×
(x + iy) +
p
p
p
p
, a = , a = and a = in (i) one by one
2
6
4
3
k+4
> 0 Þk>–4
k -2
k -2
> 0 Þk>2
k+4
Roots are real so, – 6 < k < 4
So, 6 and 4 are not correct.
Since, k > 2, so 1 is also not correct value of k.
\k=3
113. (d) p (x) = 0
Þ
f ( x ) = g ( x)
Þ
ax2 + bx + c = a1x 2 + b1 x + c1
Þ
(a - a1 ) x2 + (b - b1 ) x + (c - c1 ) = 0.
It has only one solution, x = – 1
Þ
b - b1 = a - a1 + c - c1
Sum of roots
Þ
...(i)
-(b - b1 )
= -1 - 1
(a - a1 )
b - b1
=1
2 (a - a1 )
ÞÞ b - b1 = 2 ( a - a1 ) ...
...(ii)
Now p (– 2) = 2
Þ f (– 2) – g (– 2) = 2
Þ 4a – 2b + c – 4a1 + 2b1 – c1 = 2
Þ 4 (a – a1) – 2 (b – b1) + (c – c1) = 2 ...(iii)
From equations, (i), (ii) and (iii)
a - a1 = c - c1 =
1
( b - b1 ) = 2
2
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M-51
Complex Numbers and Quadratic Equations
Now, p ( 2) = f ( 2) - g (2)
Þ 12 b2 (3 c2 – 2 c2+ y ) ³ 0 [Q b2 ³ 0]
Þ c2 + y ³ 0
Þ y ³ – c2
But from eqn. (i), c2 < 4ab or – c2 > – 4ab
= 4 ( a - a1 ) + 2 ( b - b1 ) + ( c - c1 )
= 8 + 8 + 2 = 18
114. (a) Let the correct equation be
\
Þ y > – 4 ab
ax 2 + bx + c = 0
Now, Sachin’s equation
118. (c) Let a and b are roots of the equation
x2 + ax + 1 = 0
ax 2 + bx + c ' = 0
Given that, roots found by Sachin’s are 4 and 3
Þ -
b
=7
a
a + b = – a and ab = 1
....(i)
Given that | a - b | < 5
Þ
Rahul’s equation, ax 2 + b ' x + c = 0
Given that roots found by Rahul’s are 3 and 2
(a + b)2 - 4ab < 5
(Q
c
=6
Þ
...(ii)
a
From (i) and (ii), roots of the correct equation
x 2 - 7 x + 6 = 0 are 6 and 1.
115. (c) Since both the roots of given quadratic equation lie
in the line Re z = 1 i.e., x = 1, hence real part of both the
roots are 1.
Let both roots be 1 + ia and 1 – ia
Product of the roots, 1 + a2 = b
(a - b) 2 = (a + b ) 2 - 4ab
)
Þ a 2 - 4 < 5 Þ a2 – 4 < 5
Þ a2 – 9 < 0 Þ a2 < 9 Þ – 3 < a < 3
Þ a Î (–3, 3)
119. (c) Given equation is x 2 - 2mx + m 2 - 1 = 0
Þ ( x - m) 2 - 1 = 0
Þ ( x - m + 1)( x - m - 1) = 0
Þ x = m - 1, m + 1
Q a +1 ³ 1
m – 1 > –2 and m + 1 < 4
2
Þ m > - 1 and m < 3 Þ -1 < m < 3
\b ³ 1 Þ Q b Î (1, ¥)
116. (b)
we get y ³ – c2 > – 4ab
x2 - x + 1 = 0 Þ x =
x=
1± 3 i
2
a=
1
3
+i
= -w2
2
2
120. (b) Given that x 2 + px + q = 0
1± 1- 4
2
Sum of roots = tan30 + tan15 = – p
Product of roots = tan30 . tan15 = q
tan 45° =
Þ – p = 1- q Þ q - p = 1
\ 2+ q - p = 3
1 i 3
b= = -w
2
2
121. (d)
a2009 + b2009 = (-w2 )2009 + (-w)2009
= -w2 - w = 1
117. (b) Given that roots of the equation
bx2 + cx + a = 0 are imaginary
\ c2 – 4ab < 0
Let y = 3b2x2 + 6 bc x + 2c2
Þ 3b2x2 + 6 bc x + 2c2 – y = 0
As x is real, D ³ 0
Þ 36 b2c2 – 12 b2 (2c2 – y ) ³ 0
-p
tan30° + tan15°
Þ
=1
1 - tan30°.tan15° 1 - q
z2 + z + 1 = 0 Þ
So, z +
[Q w3 = 1]
....(i)
= w or w 2
1
= w + w 2 = -1
z
é 1
ù
2
2
êëQ z = w and 1 + w + w = 0 úû
z2 +
z3 +
1
z2
1
z
3
= w 2 + w = -1,
= w3 + w3 = 2
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[Q w3 = 1]
EBD_8344
M-52
Mathematics
z4 +
1
z
4
= -1, z 5 +
1
z
5
Þ ( x - 2)( x - 1) = 0
= -1
Þ x = 1, 2 Þ x = ±1, ±2
1
=2
z6
\ The given sum = 1+1 + 4 + 1 + 1 + 4 = 12
and z 6 +
æ Pö
æ Qö
122. (b) tan ç ÷ , tan ç ÷ are the roots of ax 2 + bx + c = 0
è 2ø
è 2ø
b
æ Pö
æ Qö
tan ç ÷ + tan ç ÷ = è 2ø
è 2ø
a
\ No.of solution = 4
127. (b) Let one roots of given equation be a
\ Second roots be 2a then
a + 2a = 3a =
Þ a=
æ Pö
æ Qö c
tan ç ÷ × tan ç ÷ =
è 2ø
è 2ø a
(
1 - 3a
a - 5a + 3
2
1 - 3a
)
2
3 a - 5a + 3
....(i)
2
and a.2a = 2a 2 = 2
a - 5a + 3
æ Pö
æ Qö
tan ç ÷ + tan ç ÷
è 2ø
è 2ø
æ P Qö
= tan ç + ÷ = 1
è 2 2ø
æ Pö
æ Qö
1 - tan ç ÷ tan ç ÷
è 2ø
è 2ø
é 1 (1 - 3a)2 ù
2
ú= 2
\ 2ê
2
2
9
ëê (a - 5a + 3) ûú a - 5a + 3
[from (i)]
pù
é
êëQ P + Q = 2 úû
b
b a-c
a =1
Þ
Þ - =
c
a
a
1a
(1 - 3a) 2
(a 2 - 5a + 3)
=9
Þ 9a 2 - 6a + 1 = 9 a 2 - 45a + 27
-
Þ 39 a = 26 Þ a =
Þ –b=a–c Þ c=a+b
123. (d) Let a , a + 1 be roots
Then a + a + 1 = b = sum of roots
a (a + 1) = c = product of roots
\ b 2 - 4c = (2a + 1)2 - 4a (a + 1) = 1 .
2
124. (d) Given that 4 is a root of x + px + 12 = 0
2
3
128. (d) Given that Z 2 + aZ + b = 0 ;
Z1 + Z 2 = - a & Z1Z 2 = b
0, Z1, Z 2 form an equilateral triangle
\ 02 + Z12 + Z 2 2 = 0.Z1 + Z1.Z 2 + Z 2 .0
(for an equilateral triangle,
Þ 16 + 4 p + 12 = 0 Þ p = -7
Z12 + Z 22 + Z32 = Z1Z 2 + Z 2 Z 3 + Z3Z1 )
Now, the equation x 2 + px + q = 0
Þ Z12 + Z 2 2 = Z1Z 2
has equal roots.
\D =0
Þ ( Z1 + Z 2 ) 2 = 3Z1Z 2
p 2 49
=
4
4
125. (c) Let the second root be a.
2
Þ p - 4q = 0 Þ q =
Then a + (1 - p ) = - p Þ a = -1
Also a.(1 - p) =1 - p
Þ (a - 1)(1 - p) = 0 Þ p =1 [Q a = -1]
\ Roots are a = -1 and 1 - p = 0
126. (c) Given that
\ a 2 = 3b
129. (a) p + q = – p Þ q = 2p
and pq = q Þ q (p – 1) = 0
Þ q = 0 or p = 1.
If q = 0, then p = 0.
or p = 1, then q = –2.
c 9
= > 0, " t Î R
a t2
\ Product of real roots is always positive.
130. (a) Product of real roots =
x 2 - 3 x + 2 = 0 Þ| x |2 -3 | x | +2 = 0
Downloaded from @Freebooksforjeeneet
M-53
Complex Numbers and Quadratic Equations
131. (a) Let a and b are r oots of th e equation
x2 + ax + b = 0 and g and d be the roots of the equation x2
+ bx + a = 0 respectively.
\ a + b = –a, ab = b and g + d = –b, g d = a.
Given |a – b| = |g – d| Þ (a – b)2 = (g – d)2
Þ (a+ b)2 – 4ab = (g + d)2 – 4gd
Þ a2 – 4b = b2 – 4a
Þ (a2 – b2) + 4(a – b)= 0
Þ a+ b+4=0
(Q a ¹ b)
132. (a) Given that a2 = 5a – 3 and b2 = 5b – 3;
Þ a & b are roots of equation, x2 = 5x – 3
or x2 – 5x + 3 = 0
\ a + b = 5 and ab = 3
Thus, the equation whose roots are
a
b
and
is
b
a
For x2 – 2bx – 10 = 0
a + b = 2b
and ab = – 10
...(ii)
...(iii)
b
is also root of x2 – 2bx – 10 = 0
a
Þ b2 – 2ab2 – 10a2 = 0
By eqn. (i) Þ 5a – 10a2 – 10a2 = 0
Þ 20a2 = 5a
a=
1
5
a = and b2 =
4
4
a2 = 20 and b2 = 5
Now, a2 + b2 = 5 + 20 = 25
135. (b) Let, the roots of the equation, x2 + (2 – l) x + (10 – l)
= 0 are a and b.
Also roots of the given equation are
Þ
æ a b ö ab
=0
x2 - x ç + ÷ +
è b aø ab
l – 2 ± 4 – 4 l + l 2 – 40 + 4l l – 2 ± l 2 – 36
=
2
2
æ a 2 + b2 ö
Þ x2 - x ç
÷ +1 = 0
è ab ø
The magnitude of the difference of the roots is
or 3x2 – 19x +3 = 0
So, a3 + b3 =
P
5r2
133. (c)
Q
(l – 2)3 3 (l – 2) (l 2 – 36)
+
4
4
(l – 2) (4l 2 – 4l – 104)
= (l – 2) (l 2 – l – 26) = f (l )
4
As f (l) attains its minimum value at l = 4.
Therefore, the magintude of the difference of the roots is
=
5r
R
5
DPQR is possible if
5 + 5r > 5r2
|i
Þ
1 + r > r2
Þ
r2 – r – 1 < 0
Þ
æ
1
5öæ
1
5ö
çr - 2 + 2 ÷ çr - 2 - 2 ÷ < 0
è
øè
ø
Þ
æ - 5 + 1 5 + 1ö
r Îç
,
2 ø÷
è 2
20 |= 2 5
136. (b) | z – (3 – 2i) | £ 4 represents a circle whose centre is
(3, – 2) and radius = 4.
| z | = | z – 0 | represents the distance of point ‘z’ from origin
(0, 0)
y
R
4
x
O
7 æ - 5 + 1 5 + 1ö
7
Ï
,
\ r¹
4 çè 2
2 ø÷
4
Q
G
2b
a
Þ a=
Þ
b2 = 5a
S
b
a
2
and product of roots = a =
(a ¹ 0)
(3 , –2 )
4
134. (a) ax2 – 2bx + 5 = 0,
If a and a are roots of equations, then sum of roots
2a =
l 2 – 36
2
5
5
b
Þ 2=
a
a
a
...(i)
Suppose RS is the normal of the circle passing through
origin ‘O’ and G is its center (3, – 2).
Here, OR is the least distance
and OS is the greatest distance
OR = RG – OG and OS = OG + GS
...(i)
Downloaded from @Freebooksforjeeneet
EBD_8344
M-54
Mathematics
As, RG = GS = 4
OG =
139. (b) Let a, b be the common roots of both the
equations.
32 + ( - 2) 2 = 9 + 4 = 13
For first equation ax 2 + bx + c = 0 ,
we have
From (i), OR = 4 - 13 and OS = 4 + 13
(
) (
So, required difference = 4 + 13 - 4 - 13
)
= 13 + 13 = 2 13
a.b =
137. (c)
x 2 + bx - 1 = 0 common root
x2 + x + b = 0
-
- x=
b +1
b -1
b +1
Put x =
in equation
b -1
-b
a
a+b=
c
a
...(ii)
For second equation 2 x 2 + 3 x + 4 = 0 ,
we have
-3
2
a+b=
...(iii)
2
...(iv)
1
Now, from (i) & (iii) & from (ii) & (iv)
a.b=
æ b + 1 ö æ b +1 ö
ç
÷ +ç
÷ +b =0
è b - 1 ø è b -1 ø
-b -3
=
a
2
(b + 1)2 + (b + 1) (b – 1) + b (b – 1)2 = 0
b2 + 1 + 2b + b2 – 1 + b (b2 – 2b + 1) = 0
2b2 + 2b + b3– 2b2 + b = 0
b3 + 3b = 0
b(b2 + 3) = 0
b2 = – 3
b
3/ 2
=
a
1
2
...(i)
c
2
=
a
1
3
&c=2
2
putting these values in first equation, we get
Therefore on comparing we get a = 1, b =
3
x + 2 = 0 or 2 x 2 + 3x + 4 = 0
2
from this, we get a = 2, b = 3; c = 4
or a : b : c = 2 : 3 : 4
140. (a) Given equations are
x2 + 2x + 3 = 0
…(i)
ax2 + bx + c = 0
…(ii)
Roots of equation (i) are imaginary roots in order pair.
According to the question (ii) will also have both roots
same as (i). Thus
x2 +
b = ± 3i
|b| = 3
138. (d) We have
2
2
f (x) = x + 2bx + 2c
and g(x) = - x 2 - 2cx + b2 , ( x Î R )
Þ f (x) = ( x + b)2 + 2c2 - b2
2
2
2
and g(x) = -( x + c) + b + c
Now, fmin = 2c2 – b2 and gmax = b2 + c2
Given : min f (x) > max g(x)
Þ 2c 2 - b 2 > b 2 + c 2
Þ c 2 > 2b 2
Þ |c| > | b | 2
a b c
= = = l (say)
1 2 3
Þ a = l, b = 2l, c = 3l
Hence, required ratio is 1 : 2 : 3
Þ
|c|
c
> 2Þ > 2
|b|
b
Þ
c
Î ( 2, ¥) .
b
x-5
> 0 Þ x2 + 5x – 14 < x – 5
x + 5x - 14
Þ x2 + 4x – 9 < 0
Þ a = – 5, – 4, – 3, – 2, – 1, 0, 1
a = – 5 does not satisfy any of the options
a = – 4 satisfy the option (a) a2 + 3a – 4 = 0
141. (a)
2
Downloaded from @Freebooksforjeeneet
M-55
Complex Numbers and Quadratic Equations
142. (c) x2 – (a + 1) x + a2 + a – 8 = 0
Since roots are different, therefore D > 0
Þ (a + 1)2 – 4(a2 + a – 8) > 0
Þ (a – 3) (3a + 1) < 0
There are two cases arises.
Case I. a – 3 > 0 and 3a + 1 < 0
z
=
£
z
Þ
z
Þ
æ
2 + 20 ö
çç z –
÷÷
2
è
ø
Þ
(z
2
(-
Þ
z
)
5 +1 £
max
+
5) (1 + 5)
(
Q x is real
x=5
)
z £
D³0
0
– (1 - 5) £ 0
–
(1 -
3x 2 ( y - 1) + 9 x( y - 1) + 7 y - 17 = 0
4
4
+
z
z
æ
2 - 20 ö
çç z –
÷÷ £ 0
2
è
ø
)(z
+
–¥
Þ
z-
–2 z -4£ 0
– (1 + 5)
3 x2 + 9 x + 7
Y axis
4
z
£ 2+
Þ
3 x 2 + 9 x + 17
Þ ( y - 1)( y - 41) £ 0 Þ 1 £ y £ 41
\ Max value of y is 41
–
+
+
8
–8
1
41
146. (c) Given that both roots of quadratic equation are less
than 5 then (i)
4
=2
z
4 4
+
z z
z-
y=
81( y - 1) 2 - 4 ´ 3( y - 1)(7 y - 17) ³ 0
11
3
Hence, no solution in this case
Case II : a – 3 < 0 and 3a + 11 > 0
11
Þ a < 3 and a > 3
11
\
<a<3 Þ –2<a<3
3
Þ a > 3 and a < -
143. (a) Given that z -
145. (b)
¥
)
5 +1
= 5 +1
144. (d) Let the roots of equation x2 – 6x + a = 0 be a and 4
b and that of the equation
x2 –cx + 6 = 0 be a and 3 b .Then
a + 4b = 6 ...(i) 4a b = a ...(ii)
and a + 3b = c ...(iii)
3a b = 6 ...(iv)
Þ a = 8 (from (ii) and (iv))
\ The equation becomes x2 – 6x + 8 = 0
Þ (x –2) (x – 4) = 0
Þ roots are 2 and 4
Þ a = 2, b = 1 \ Common root is 2.
X axis
Discriminant ³ 0
4k2– 4(k2 + k – 5) ³ 0
4k2 – 4k2 – 4k + 20 ³ 0
4k £ 20 Þ k £ 5
(ii) p(5) > 0
Þ f(5) > 0 ; 25 – 10 k + k2 + k – 5 > 0
Þ k2 – 9k + 20 > 0
Þ k (k – 4) –5(k – 4) > 0
Þ (k – 5) (k – 4) > 0
–
+
+
¥
–¥
4
5
Þ k Î ( – ¥ , 4 ) U (5, ¥ )
(iii)
Sum of roots
<5
2
2k
b
=
<5
2a
2
Þ k<5
The intersection of (i), (ii) & (iii) gives
k Î ( – ¥ , 4 ).
Þ–
147. (a) Given equation is x 2 - (a - 2) x - a - 1 = 0
Þ a + b = a - 2 ; a b = -(a + 1)
a 2 + b2 = (a + b) 2 - 2 a b
= a2 - 2a + 6 = (a - 1)2 + 5
For min. value of a2 + b2, a – 1 = 0
Þ a = 1.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-56
Mathematics
6
Linear Inequalities
TOPIC Ć
1.
Solution of Linear Inequality and
System of Linear Inequalities,
Representation of Solution of
Linear Inequality in One Variable
on a Number Line, Representation
of Solution of a Linear Inequality
and System of Linear Inequalities
in a Cartesian Plane, Equations and
Inequations Involving Absolute
Value Functions, Greatest Integer
Functions, Logarithmic Functions
The region represented by {z = x + iy Î C : |z| – Re(z) £ 1} is
also given by the inequality:
[Sep. 06, 2020 (I)]
(a) y 2 ³ 2( x + 1)
1
2
Consider the two sets :
(c) y 2 £ x +
2.
(a)
(b)
(c)
(d)
5.
6.
7.
3.
4.
[Jan. 9, 2020 (II)]
(a) A Ç B = (–2, –1)
(b) B – A = R – (–2, 5)
(c) A È B = R – (2, 5)
(d) A – B = [–1, 2)
Let S be the set of all real roots of the equation,
3x(3x – 1) + 2 = |3x – 1| + | 3x – 2|. Then S:[Jan. 8, 2020 (II)]
[April 10, 2019 (I)]
(b) 2 sin x = sin y
(d) sin x = | sin y |
The number of integral values of m for which the equation
(1 + m2)x2 – 2(1 + 3m)x + (1 + 8m) = 0 has no real root is :
[April 08, 2019 (II)]
(a) 1
(b) 2
(c) infinitely many
(d) 3
The number of integral values of m for which the quadratic
expression, (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), xÎR, is
always positive, is :
[Jan. 12, 2019 (II)]
(a) 3
(b) 8
(c) 7
(d) 6
x
8.
(b) A Ç B = {-3}
(c) B – A = (–3, 5)
(d) A È B = R
If A = {x Î R : |x| < 2} and
B = {x Î R : |x – 2| ³ 3}; then :
1
£ 1 also satisfy the equation:
4sin 2 y
(a) 2|sin x| = 3sin y
(c) sin x = 2 sin y
A = {m Î R : both the roots of x2 – (m + 1)x + m + 4 = 0
are real} and B = [– 3, 5).
Which of the following is not true? [Sep. 03, 2020 (I)]
(a) A - B = (-¥, - 3) È (5, ¥)
All the pairs (x,y) that satisfy the inequality
2 sin 2 x - 2sin x + 5.
1ö
æ
(b) y 2 £ 2 ç x + ÷
è
2ø
2
(d) y ³ x + 1
contains exactly two elements.
is a singleton.
is an empty set.
contains at least four elements.
9.
x
æ 3ö æ 4 ö
If f ( x ) = ç ÷ + ç ÷ - 1 , x Î R , then the equation
è5ø è5ø
f(x) = 0 has :
[Online April 9, 2014]
(a) no solution
(b) one solution
(c) two solutions
(d) more than two solutions
If a, b, c are distinct +ve real numbers and a2 + b2 + c2 = 1
then ab + bc + ca is
[2002]
(a) less than 1
(b) equal to 1
(c) greater than 1
(d) any real no.
Downloaded from @Freebooksforjeeneet
M-57
Linear Inequality
1.
(b) Q | z | - Re ( z ) £ 1
(Q z = x + iy )
y2 + y – 1 = 0
Þ x 2 + y 2 - x £ 1 Þ x2 + y 2 £ 1 + x
\y =
-1 + 5
2
=
-1 - 5
2
Þ x2 + y2 £ 1 + x 2 + 2x
1ö
æ
Þ y £ 1 + 2x Þ y £ 2 ç x + ÷
è
2ø
2
2.
2
(a) A = {m Î R : x 2 - (m + 1) x + m + 4 = 0
roots}
has
\Only one –1 +
real
5.
D³0
Þ m 2 - 2m - 15 ³ 0
–
–3
5
satisfy equation
2
(d) Given inequality is,
2
2 sin 2 x - 2sin x + 5 £ 2 2sin y
Þ (m + 1) 2 - 4(m + 4) ³ 0
+
[\ Equation not Satisfy]
+
Þ
sin 2 x - 2sin x + 5 £ 2sin 2 y
Þ
(sin x - 1)2 + 4 £ 2sin 2 y
5
It is true if sin x = 1 and |sin y| = 1
A= {(-¥, - 3] È [5, ¥)}
B= [ -3, 5) Þ A - B= ( -¥, - 3) È [5, ¥)
3.
Therefore, sin x = |sin y|
6.
(b) A = {x : x Î (–2, 2)}
(1 + m2 ) x 2 - 2(1 + 3m) x + (1 + 8m) = 0
B = {x : x Î (– ¥, –1] È [5, ¥)}
Q equation has no real solution
A Ç B = {x : x Î (–2, –1]}
4.
(c) Given equation is
\ D<0
A È B = {x : x Î (–¥, 2) È [5, ¥)}
Þ 4 (1 + 3m)2 < 4 (1 + m2) (1 + 8m)
A – B = {x : x Î (–1, 2)}
Þ 1 + 9m2 + 6m < 1 + 8m + m2 + 8m3
B – A = {x : x Î (–¥, –2] È [5, ¥)}
Þ 8m3 – 8m2 + 2m > 0
x
(b) Let 3 = y
\
Þ 2m (4m2 – 4m + 1) > 0 Þ 2m (2m – 1)2 > 0
y(y – 1) + 2 = |y – 1| + |y – 2|
Þ m > 0 and m ¹
Case 1: when y > 2
y2 – y + 2 = y – 1 + y – 2
D < 0 [ \ Equation not satisfy.]
Case 2: when 1 £ y £ 2
y2 – y2 + 2 = y – 1 – y + 2
7.
(3) Let the given quadratic expression
(1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), is positive for all x Î R,
then
1 + 2m > 0
y2 – y + 1 = 0
Q
é 1
ù
êëQ 2 is not an integer úû
Þ number of integral values of m are infinitely many.
y2 – 3y + 5 = 0
Q
1
2
D < 0 [ \ Equation not satisfy.]
Case 3: when y £ 1
y2 – y + 2 = – y + 1 – y + 2
...(i)
D<0
Þ 4(1 + 3m)2 – 4(1 + 2m)4 (1 + m) < 0
Þ 1 + 9m2 + 6m – 4[1 + 2m2 + 3m] < 0
Downloaded from @Freebooksforjeeneet
EBD_8344
M-58
Mathematics
Þ m2 – 6m – 3 < 0
x
Þ m Î (3 - 2 3,3 + 2 3)
æ 3ö
æ 4ö
çè ÷ø + çè ÷ø
5
5
From (i)
Þ
3x + 4x = 5x
\
m> -
=1
...(i)
For x = 1
1
2
31 + 41 > 51
For x = 3
m Î ( 3 - 2 3, 3 + 2 3 )
33 + 43 = 91 < 53
Then, integral values of m = {0, 1, 2, 3, 4, 5, 6}
Only for x = 2, equation (i) Satisfy
Hence, number of integral values of m = 7
So, only one solution (x = 2)
x
8.
x
Þ
(b)
x
æ 3ö
æ 4ö
f (x) = ç ÷ + ç ÷ - 1
è 5ø
è 5ø
Þ
(a) Q (a – b)2 + (b – c)2 + (c – a)2 > 0
Þ 2(a2 + b2 + c2 – ab – bc – ca) > 0
[Q a2 + b2 + c2 = 1]
Put f (x) = 0
x
9.
Þ 2 > 2(ab + bc + ca) Þ ab + bc + ca < 1
x
æ 3ö
æ 4ö
0 = ç ÷ + ç ÷ -1
è 5ø
è 5ø
Downloaded from @Freebooksforjeeneet
M-59
Permutations and Combinations
7
Permutations and
Combinations
TOPIC Ć
1.
2.
3.
4.
5.
Fundamental Principle of Counting,
Factorials, Permutations, Counting
Formula for Permutations,
Permutations in Which Things may
be Repeated, Permutations in
Which all Things are Different,
Number of Permutations Under
Certain Restricted Conditions,
Circular Permutations
Two families with three members each and one family with
four members are to be seated in a row. In how many ways
can they be seated so that the same family members are
not separated?
[Sep. 06, 2020 (I)]
(a) 2! 3! 4!
(b) (3!)3 × (4!)
(c) (3!)2 × (4!)
(d) 3! (4!)3
The value of (2 × 1P0 - 3 × 2 P1 + 4 × 3 P2 - ... up to 51th term)
+ (1! – 2! + 3! – ... up to 51th term) is equal to :
[Sep. 03, 2020 (I)]
(a) 1 – 51(51)!
(b) 1 + (51)!
(c) 1 + (52)!
(d) 1
If the letters of the word 'MOTHER' be permuted and all
the words so formed (with or without meaning) be listed as
in a dictionary, then the position of the word 'MOTHER' is
________.
[NA Sep. 02, 2020 (I)]
If the number of five digit numbers with distinct digits
and 2 at the 10th place is 336 k, then k is equal to:
[Jan. 9, 2020 (I)]
(a) 4
(b) 6
(c) 7
(d) 8
Total number of 6-digit numbers in which only and all the
five digits 1, 3, 5, 7 and 9 appear, is: [Jan. 7, 2020 (I)]
1
(6!)
(a)
2
(b) 6!
(c) 56
(d)
5
(6!)
2
6.
The number of 6 digit numbers that can be formed using
the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no
digit is repeated, is:
[April 10, 2019 (I)]
(a) 72
(b) 60
(c) 48
(d) 36
7.
The number of four-digit numbers strictly greater than
4321 that can be formed using the digits 0, 1, 2, 3, 4, 5
(repetition of digits is allowed) is: [April 08, 2019 (II)]
(a) 288
(b) 360
(c) 306
(d) 310
Consider three boxes, each containing 10 balls labelled
1, 2,..., 10. Suppose one ball is randomly drawn from each
of the boxes. Denote by n i, the label of the ball drawn from
the ith box, (i = 1, 2, 3). Then, the number of ways in which
the balls can be chosen such that n1 < n2 < n3 is :
[Jan. 12, 2019 (I)]
(a) 120
(b) 82
(c) 240
(d) 164
The number of natural numbers less than 7,000 which
can be formed by using the digits 0,1,3,7,9 (repetition of
digits allowed) is equal to:
[Jan. 09, 2019 (II)]
(a) 374
(b) 372
(c) 375
(d) 250
Let S be the set of all triangles in the xy-plane, each
having one vertex at the origin and the other two
vertices lie on coordinate axes with integral coordinates.
If each triangle in S has area 50 sq. units, then the
number of elements in the set S is:[Jan. 09, 2019 (II)]
(a) 9
(b) 18
(c) 36
(d) 32
The number of numbers between 2,000 and 5,000 that can
be formed with the digits 0, 1, 2, 3, 4, (repetition of digits is
not allowed) and are multiple of 3 is?
[Online April 16, 2018]
(a) 30
(b) 48
8.
9.
10.
11.
(c) 24
(d) 36
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EBD_8344
M-60
12.
Mathematics
n – digit numbers are formed using only three digits 2, 5
and 7. The smallest value of n for which 900 such distinct
numbers can be formed, is
[Online April 15, 2018]
(a) 6
(b) 8
(c) 9
13.
14.
15.
21.
(d) 7
The number of ways in which 5 boys and 3 girls can be
seated on a round table if a particular boy B1 and a particular
girl G1 never sit adacent to each other, is :
[Online April 9, 2017]
(a) 5 × 6!
(b) 6 × 6!
(c) 7!
(d) 5 × 7!
If all the words, with or without meaning, are written using
the letters of the word QUEEN and are arranged as in
English dictionary, then the position of the word QUEEN
is :
[Online April 8, 2017]
(a) 44th
(b) 45th
(c) 46th
(d) 47th
If all the words (with or without meaning) having five
letters, formed using the letters of the word SMALL and
arranged as in a dictionary; then the position of the word
SMALL is :
[2016]
(a) 52nd
(b) 58th
(c) 46th
(d) 59th
22.
23.
24.
10
16.
17.
The sum
å (r 2 + 1) × (r!) is equal to :
r =1
[Online April 10, 2016]
(a) 11 × (11!)
(b) 10 × (11!)
(c) (11!)
(d) 101 × (10!)
If the four letter words (need not be meaningful) are to be
formed using the letters from the word
"MEDITERRANEAN" such that the first letter is R and
the fourth letter is E, then the total number of all such
words is :
[Online April 9, 2016]
(a) 110
(b) 59
25.
26.
11!
(c)
18.
19.
(2!)3
The number of points, having both co-ordinates as
integers, that lie in the interior of the triangle with vertices
(0, 0), (0, 41) and (41, 0) is :
[2015]
(a) 820
(b) 780
(c) 901
(d) 861
The number of integers greater than 6,000 that can be
formed, using the digits 3, 5, 6, 7 and 8, without repetition,
is :
[2015]
(a) 120
(b) 72
(c) 216
20.
(d) 56
(d) 192
The number of ways of selecting 15 teams from 15 men
and 15 women, such that each team consists of a man and
a woman, is:
[Online April 10, 2015]
27.
28.
(a) 1120
(b) 1880
(c) 1960
(d) 1240
Two women and some men participated in a chess
tournament in which every participant played two games
with each of the other participants. If the number of games
that the men played between themselves exceeds the
number of games that the men played with the women by
66, then the number of men who participated in the
tournament lies in the interval: [Online April 19, 2014]
(a) [8, 9]
(b) [10, 12)
(c) (11, 13]
(d) (14, 17)
8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4,
4. The number of such numbers in which the odd digits do
no occupy odd places, is:
[Online April 12, 2014]
(a) 160
(b) 120
(c) 60
(d) 48
An eight digit number divisible by 9 is to be formed using
digits from 0 to 9 without repeating the digits. The number
of ways in which this can be done is:
[Online April 11, 2014]
(a) 72 (7!)
(b) 18 (7!)
(c) 40 (7!)
(d) 36 (7!)
The sum of the digits in the unit’s place of all the 4-digit
numbers formed by using the numbers 3, 4, 5 and 6, without
repetition, is:
[Online April 9, 2014]
(a) 432
(b) 108
(c) 36
(d) 18
5 - digit numbers are to be formed using 2, 3, 5, 7, 9 without
repeating the digits. If p be the number of such numbers
that exceed 20000 and q be the number of those that lie
between 30000 and 90000, then p : q is :
[Online April 25, 2013]
(a) 6 : 5
(b) 3 : 2
(c) 4 : 3
(d) 5 : 3
Assuming the balls to be identical except for difference in
colours, the number of ways in which one or more balls
can be selected from 10 white, 9 green and 7 black balls is:
[2012]
(a) 880
(b) 629
(c) 630
(d) 879
If seven women and seven men are to be seated around a
circular table such that there is a man on either side of
every woman, then the number of seating arrangements is
[Online May 26, 2012]
(a) 6! 7!
(b) (6!)2
(c) (7!)2
(d) 7!
If the letters of the word SACHIN are arranged in all possible
ways and these words are written out as in dictionary,
then the word SACHIN appears at serial number [2005]
(a) 601
(b) 600
(c) 603
(d) 602
Downloaded from @Freebooksforjeeneet
M-61
Permutations and Combinations
29.
30.
31.
32.
33.
34.
How many ways are there to arrange the letters in the
word GARDEN with vowels in alphabetical order[2004]
(a) 480
(b) 240
(c) 360
(d) 120
The range of the function f ( x) =7 - x Px -3 is
[2004]
(a) {1, 2, 3, 4, 5}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3, 4,}
(d) {1, 2, 3,}
The number of ways in which 6 men and 5 women can dine
at a round table if no two women are to sit together is
given by
[2003]
(a) 6! × 5!
(b) 6 × 5
(c) 30
(d) 5 × 4
The sum of integers from 1 to 100 that are divisible by 2 or
5 is
[2002]
(a) 3000
(b) 3050
(c) 3600
(d) 3250
Number greater than 1000 but less than 4000 is formed
using the digits 0, 1, 2, 3, 4 (repetition allowed). Their
number is
[2002]
(a) 125
(b) 105
(c) 374
(d) 625
Total number of four digit odd numbers that can be formed
using 0, 1, 2, 3, 5, 7 (using repetition allowed) are [2002]
(a) 216
(b) 375
(c) 400
(d) 720
TOPIC n
35.
36.
37.
38.
Combinations, Counting Formula
for Combinations, Division and
Distribution of Objects,
Dearrangement Theorem, Sum
of Numbers, Important Result
About Point
The number of words (with or without meaning) that can
be formed from all the letters of the word “LETTER” in
which vowels never come together is ______.
[NA Sep. 06, 2020 (II)]
The number of words, with or without meaning, that can
be formed by taking 4 letters at a time from the letters of the
word ‘SYLLABUS’ such that two letters are distinct and
two letters are alike, is ______. [NA Sep. 05, 2020 (I)]
There are 3 sections in a question paper and each section
contains 5 questions. A candidate has to answer a total of
5 questions, choosing at least one question from each
section. Then the number of ways, in which the candidate
can choose the questions, is :
[Sep. 05, 2020 (II)]
(a) 3000
(b) 1500
(c) 2255
(d) 2250
A test consists of 6 multiple choice questions, each having
4 alternative answers of which only one is correct. The
number of ways, in which a candidate answers all six
questions such that exactly four of the answers are correct,
is __________.
[NA Sep. 04, 2020 (II)]
39.
40.
41.
42.
43.
The total number of 3-digit numbers, whose sum of digits
is 10, is _________.
[NA Sep. 03, 2020 (II)]
Let n > 2 be an integer. Suppose that there are n Metro
stations in a city located along a circular path. Each pair of
stations is connected by a straight track only. Further, each
pair of nearest stations is connected by blue line, whereas
all remaining pairs of stations are connected by red line. If
the number of red lines is 99 times the number of blue lines,
then the value of n is :
[Sep. 02, 2020 (II)]
(a) 201
(b) 200
(c) 101
(d) 199
If Cr º 25Cr and C0 + 5×C1 + 9×C2 + ... + (101)×C25 = 225×k, then
k is equal to ________.
[NA Jan. 9, 2020 (II)]
An urn contains 5 red marbles, 4 black marbles and 3 white
marbles. Then the number of ways in which 4 marbles can
be drawn so that at the most three of them are red is
_________.
[NA Jan. 8, 2020 (I)]
If a, b and c are the greatest values of 19Cp, 20Cq and 21Cr
respectively, then:
[Jan. 8, 2020 (I)]
(a)
a
b
c
=
=
11 22 21
(b)
a
b
c
= =
10 11 21
a
b
c
a
b
c
=
=
= =
(d)
11 22 42
10 11 42
The number of 4 letter words (with or without meaning)
that can be formed from the eleven letters of the word
‘EXAMINATION’ is _________. [NA Jan. 8, 2020 (II)]
The number of ordered pairs (r, k) for which 6. 35Cr
= (k2 – 3).36Cr + 1, where k is an integer, is:
[Jan. 7, 2020 (II)]
(a) 3
(b) 2
(c) 6
(d) 4
(c)
44.
45.
46.
The number of ways of choosing 10 obects out of 31
obects of which 10 are identical and the remaining 21 are
distinct is:
[April 12, 2019 (I)]
(a) 220 – 1
(b) 221
(c) 220
(d) 220+1
47.
A group of students comprises of 5 boys and n girls. If
the number of ways, in which a team of 3 students can
randomly be selected from this group such that there is at
least one boy and at least one girl in each team, is 1750,
then n is equal to :
[April 12, 2019 (II)]
(a) 28
(b) 27
(c) 25
(d) 24
48.
Suppose that 20 pillars of the same height have been
erected along the boundary of a circular stadium. If the
top of each pillar has been connected by beams with the
top of all its non-adacent pillars, then the total number of
beams is :
[April 10, 2019 (II)]
(a) 170
(b) 180
(c) 210
(d) 190
Downloaded from @Freebooksforjeeneet
EBD_8344
M-62
49.
50.
51.
Mathematics
A committee of 11 members is to be formed from 8 males
and 5 females. If m is the number of ways the committee is
formed with at least 6 males and n is the number of ways
the committee is formed with at least 3 females, then:
[April 9, 2019 (I)]
(a) m + n = 68
(b) m = n = 78
(c) n = m – 8
(d) m = n = 68
All possible numbers are formed using the digits 1, 1, 2, 2,
2, 2, 3, 4, 4 taken all at a time. The number of such numbers
in which the odd digits occupy even places is :
[April 8, 2019 (I)]
(a) 180
(b) 175
(c) 160
(d) 162
There are m men and two women participating in a chess
tournament. Each participant plays two games with every
other participant. If the number of games played by the
men between themselves exceeds the number of games
played between the men and the women by 84, then the
value of m is
[Jan. 12, 2019 (II)]
(a) 12
(b) 11
(c) 9
(d) 7
20
52.
53.
54.
If
å
i =1
56.
n+2
57.
58.
59.
60.
61.
62.
63.
(d) 270
From 6 different novels and 3 different dictionaries, 4
novels and 1 dictionary are to be selected and arranged in
a row on a shelf so that the dictionary is always in the
middle. The number of such arrangements is :
[2018]
(a) less than 500
(b) at least 500 but less than 750
(c) at least 750 but less than 1000
(d) at least 1000
A man X has 7 friends, 4 of them are ladies and 3 are
men. His wife Y also has 7 friends, 3 of them are ladies
and 4 are men. Assume X and Y have no common
friends. Then the total number of ways in which X and
n-2
C6
P2
= 11, then n satisfies the equation :
[Online April 10, 2016]
(b) n2 + 2n – 80 = 0
(d) n2 + 5n – 84 = 0
æ 15 C r ö
r2 ç
÷
ç 15 C ÷ is equal to :
r =1
è
r -1 ø
[Online April 9, 2016]
(a) 1240
(b) 560
(c) 1085
(d) 680
Let A and B be two sets containing four and two elements
respectively. Then the number of subsets of the set A × B,
each having at least three elements is :
[2015]
(a) 275
(b) 510
(c) 219
(d) 256
If in a regular polygon the number of diagonals is 54,
then the number of sides of this polygon is
[Online April 11, 2015]
(a) 12
(b) 6
(c) 10
(d) 9
Let A and B two sets containing 2 elements and 4 elements
respectively. The number of subsets of A × B having 3 or
more elements is
[2013]
(a) 256
(b) 220
(c) 219
(d) 211
Let Tn be the number of all possible triangles formed by
oining vertices of an n-sided regular polygon. If
Tn+1 – Tn = 10, then the value of n is :
[2013]
(a) 7
(b) 5
(c) 10
(d) 8
On the sides AB, BC, CA of a DABC, 3, 4, 5 distinct points
(excluding vertices A, B, C) are respectively chosen. The
number of triangles that can be constructed using these
chosen points as vertices are : [Online April 23, 2013]
(a) 210
(b) 205
(c) 215
(d) 220
The number of ways in which an examiner can assign 30
marks to 8 questions, giving not less than 2 marks to any
question, is :
[Online April 22, 2013]
15
3
[Jan. 10, 2019 (I)]
(a) 400
(b) 50
(c) 200
(d) 100
Consider a class of 5 girls and 7 boys. The number of
different teams consisting of 2 girls and 3 boys that can
be formed from this class, if there are two specific boys A
and B, who refuse to be the members of the same team, is:
[Jan. 9, 2019 (I)]
(a) 500
(b) 200
(c) 300
(d) 350
The number of four letter words that can be formed using
the letters of the word BARRACK is
[Online April 15, 2018]
(a) 144
(b) 120
If
(a) n2 + n – 110 = 0
(c) n2 + 3n – 108 = 0
20
æ
ö
Ci - 1
k
ç 20
÷ = , then k equals:
20
ç Ci + Ci - 1 ÷
21
è
ø
(c) 264
55.
Y together can throw a party inviting 3 ladies and 3 men,
so that 3 friends of each of X and Y are in this party, is :
[2017]
(a) 484
(b) 485
(c) 468
(d) 469
64.
The value of å
(a)
C7
(b)
21
C8
(d) 30 C8
C7
A committee of 4 per sons is to be formed from
2 ladies, 2 old men and 4 young men such that it includes
at least 1 lady, at least 1 old man and at most 2 young men.
Then the total number of ways in which this committee
can be formed is :
[Online April 9, 2013]
(a) 40
(b) 41
(c) 16
(d) 32
(c)
65.
30
21
Downloaded from @Freebooksforjeeneet
M-63
Permutations and Combinations
66.
67.
68.
69.
The number of arrangements that can be formed from the
letters a, b, c, d, e, f taken 3 at a time without repetition and
each arrangement containing at least one vowel, is
[Online May 19, 2012]
(a) 96
(b) 128
(c) 24
(d) 72
If n = mC2, then the value of nC2 is given by
[Online May 19, 2012]
(a) 3(m + 1C4)
(b) m – 1 C4
(c) m + 1C4
(d) 2(m + 2C4)
Statement 1: If A and B be two sets having p and q elements
respectively, where q > p. Then the total number of functions
from set A to set B is qp.
[Online May 12, 2012]
Statement 2: The total number of selections of p different
obects out of q obects is qCp.
(a) Statement 1 is true, Statement 2 is false.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation of Statement 1.
(c) Statement 1 is false, Statement 2 is true
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation of Statement 1.
If the number of 5-elemen t subsets of the set
A= {a1, a2, ...., a20} of 20 distinct elements is k times the
number of 5-element subsets containing a4, then k is
[Online May 7, 2012]
(a) 5
70.
71.
72.
73.
(b)
74.
75.
Thus A È B È C = S, A Ç B = B Ç C = A Ç C = f. The
number of ways to partition S is
[2007]
12!
(a)
(4!)
76.
77.
78.
80.
(4!) 4
12!
(d)
3
The value of
50
C4 +
6
å 56 - r C3
is
[2005]
r =1
(a)
55
C4
(b)
(c)
56
C3
(d)
55
C3
56
C4
The number of ways of distributing 8 identical balls in 3
distinct boxes so that none of the boxes is empty is [2004]
8
C3
(b) 21
(c) 38
(d) 5
A student is to answer 10 out of 13 questions in an
examination such that he must choose at least 4 from the
first five questions. The number of choices available to
him is
[2003]
(a) 346
(b) 140
(c) 196
(d) 280
If nCr denotes the number of combination of n things
taken r at a time, then the expression
n
Cr +1 + nC r -1 + 2´n Cr equals
(a)
n +1
Cr +1
n+2
81.
12!
3!(4!) 4
3!(4!)
At an election, a voter may vote for any number of
candidates, not greater than the number to be elected.
There are 10 candidates and 4 are of be selected, if a voter
votes for at least one candidate, then the number of ways
in which he can vote is
[2006]
(a) 5040
(b) 6210
(c) 385
(d) 1110
(a)
79.
(b)
3
12!
(c)
20
7
10
(c) 4
(d)
3
There are 10 points in a plane, out of these 6 are collinear.
If N is the number of triangles formed by oining these
points. Then :
[2011RS]
(a) N £ 100
(b) 100 < N £ 140
(c) 140 < N £ 190
(d) N > 190
Statement-1: The number of ways of distributing 10
identical balls in 4 distinct boxes such that no box is empty
is 9C3 .
Statement-2: The number of ways of choosing any 3 places
from 9 different places is 9C3.
[2011]
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation for Statement-1.
There are two urns. Urn A has 3 distinct red balls and urn
B has 9 distinct blue balls. From each urn two balls are
taken out at random and then transferred to the other. The
number of ways in which this can be done is
[2010]
(a) 36
(b) 66
(c) 108
(d) 3
From 6 different novels and 3 different dictionaries,4 novels
and 1 dictionary are to be selected and arranged in a row
on a shelf so that the dictionary is always in the middle.
Then the number of such arrangement is:
[2009]
(a) at least 500 but less than 750
(b) at least 750 but less than 1000
(c) at least 1000
(d) less than 500
How many different words can be formed by umbling the
letters in the word MISSISSIPPI in which no two S are
adacent?
[2008]
(a) 8. 6C4. 7C4
(b) 6.7. 8C4
(c) 6. 8. 7C4.
(d) 7. 6C4. 8C4
The set S = {1, 2, 3, ......., 12} is to be partitioned into three
sets A, B, C of equal si e.
(b)
[2003]
n+ 2
n +1
Cr
(d)
(c)
Cr
Cr +1
Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4,
6 and 7 without repetition. Total number of such numbers
are
[2002]
(a) 312
(b) 3125
(c) 120
(d) 216
Downloaded from @Freebooksforjeeneet
EBD_8344
M-64
1.
Mathematics
(b) Number of arrangement
Number of four-digit numbers starting with 44 is,
= (3! ´ 3! ´ 4!) ´ 3! = (3!)3 4!
2.
4
r!
(c) We know, (r + 1) × r Pr -1 = ( r + 1) × = ( r + 1)!
1!
= 6 × 6 = 36
6 6
Number of four-digit numbers starting with 43 and greater
than 4321 is,
So, (2 × 1P0 - 3 × 2 P1 + .....51 terms) +
3.
(1! – 2! + 3! – ... upto 51 terms)
= [2! – 3! + 4! – ... + 52!] + [1! – 2! + 3! – ... + 51!]
= 52! + 1! = 52! + 1
(309)
4
Þ 2 ´ 5! + 2 ´ 4! + 3 ´ 3! + 2! + 1
4.
5.
6.
(d) Five digits numbers be 1, 3, 5, 7, 9
For selection of one digit, we have 5C1 choice.
6!
And six digits can be arrange in
ways.
2!
5.6! 5
= .6!
Hence, total such numbers =
2! 2
(b) Given digit 0, 1, 2, 5, 7, 9
a1 a 2
7.
a3 a 4
a5
5
= 6 × 6 × 6 = 216
6
6
6
Number of four-digit numbers starting with 45 is,
4
5
= 6 × 6 = 36
6
4
3
2
=4
8.
9.
a6
(a1 + a3 + a5) – (a2 + a4 + a6 ) = 11 K
Therefore, (1, 2, 9) (0, 5, 7)
Number of ways to arranging them
= 3! × 3! + 3! × 2 × 2 = 6 × 6 + 6 × 4 = 6 × 10 = 60
(d) 0, 1, 2, 3, 4, 5
Number of four-digit number starting with 5 is,
3
= 3 × 6 = 18
(5/4/3) 3 6
Number of four-digit numbers starting with 432 and greater
than 4321 is,
M O T H E R
3 4 6 2 1 5
= 240 + 48 + 18 + 2 + 1 = 309
(d) Number of five digit numbers with 2 at 10th place
= 8 × 8 × 7 × 6 = 2688
Q It is given that, number of five digit number with 2 at
10th place = 336k
\ 336 k = 2688 Þ k = 8
4
10.
4
Hence, required numbers = 216 + 36 + 36 + 18 + 4 = 310.
(a) Collecting different labels of balls drawn = 10 × 9 × 8
Q arrangement is not required.
\ the number of ways in which the balls can be chosen is,
10 ´ 9 ´ 8
= 120
3!
(a) Number of numbers with one digit = 4 = 4
Number of numbers with two digits = 4 ´ 5 = 20
Number of numbers with three digits = 4 ´ 5 ´ 5
= 100
Number of numbers with four digits = 2 ´ 5 ´ 5 ´ 5
= 250
\ Total number of numbers = 4 + 20 + 100 + 250
= 374
(c) One of the possible DOAB is A(a, 0) and B(0, b).
Area of DOAB =
1
|ab| .
2
\
|ab| = 100
|a| |b| = 100
But 100 = 1 ´ 100, 2 ´ 50, 4 ´ 25, 5 ´ 20 or 10 ´ 10
\ For 1 ´ 100, a = 1 or –1 and b = 100 or – 100
\ Total possible pairs are 8.
Total possible pairs for 1 ´ 100, 2 ´ 50, 4 ´ 25 or 5 ´ 20 are 4 ´ 8.
And for 10 ´ 10 total possible pairs are 4.
\ Total number of possible triangles with integral
coordinates are 4 ´ 8 + 4 = 36.
6
Downloaded from @Freebooksforjeeneet
M-65
Permutations and Combinations
11.
(a) The thousands place can only be filled with 2, 3 or 4,
since the number is greater than 2000.
For the remaining 3 places, we have pick out digits such
that the resultant number is divisible by 3.
It the sum of digits of the number is divisible by 3, then the
number itself is divisible by 3.
Case 1: If we take 2 at thousands place.
15.
(b) ALLMS
No. of words starting with
_____
A: A
4!
= 12
2!
_ _ _ _ _ 4! = 24
L : L
_____
M: M
The remaining digits can be filled as:
4!
= 12
2!
0, 1 and 3 as 2 + 1 + 0 + 3 = 6 is divisible by 3.
_A
_____
S : S
0, 3 and 4 as 2+ 3 + 0 + 4 = 9 is divisible by 3.
In both the above combinations the remaining three digits
can be arranged in 3! ways.
: S_ L_ _ _ _ 3! = 6
\ Total number of numbers in this case = 2 × 3! = 12.
Case 2: If we take 3 at thousands place. The remaining
digits can be filled as:
0, 1 and 2 as 3 + 1 + 0 + 2 = 6 is divisible by 3.
SMALL ® 58 th word
10
16.
0, 2 and 4 as 3 + 2 + 0 + 4 = 9 is divisible by 3.
R -1
T1 = r r + r – (r – 1) r
Case 3: If we take 4 at thousands place.
T2 = 2 3 – 1 2
T1 = 1 2 – 0
T3 = 3 4 – 2 3
T10 = 10 11 – 9 10
In the above combination, the remaining three digits can
be arranged in 3! ways.
10
å (r 2 + 1) r = 10 11
\ Total number of numbers in this case = 3! = 6.
\ Total number of numbers between 2000 and 5000 divisible
by 3 are 12 + 12 + 6 = 30.
R -1
17.
(d) Required n digit numbers is 3n as each place can be
filled by 2, 5, 7.
So smallest value of n such that 3n > 900. Therefore n = 7.
(a) 4 boys and 2 girls in circle
Þ 5! ´
14.
å (r 2 + 1) r
T1 = (r2 + 1 + r – r) r = (r2 + r) r – (r – r) r
0, 2 and 3 as 4 + 2 + 0 + 3 = 9 is divisible by 3.
13.
(b)
In both the above combinations, the remaining three digits
can be arranged in 3! ways. Total number of numbers in
this case = 2 × 3! = 12.
The remaining digits can be filled as:
12.
3!
=3
2!
6!
´ 2!
4!2!
18.
Þ 5 ´ 6!
(c) E, E, N, Q, U
(i) E ................. = 4! = 24
(b) M, EEE, D. I, T, RR, AA, NN
R––E
Two empty places can be filled with identical letters [EE,
AA, NN] Þ 3 ways
Two empty places, can be filled with distinct letters [M, E,
D, I, T, R, A, N] Þ 8P2
\ Number of words 3 + 8P2 = 59
(b) Total number of integral points inside the square
OABC = 40 × 40 = 1600
No. of integral points on AC
(0, 41) C
(41, 41) B
4!
= 12
2
(iii) Q E ............... = 3! = 6
(ii) N ................. =
(iv) Q N ............... =
3!
=3
2!
(v) Q U E E N = 1
\ Required rank
= (24) + (12) + (6) + (c) + (a) = 46th
O
(0, 0)
A
(41, 0)
= No. of integral points on OB
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EBD_8344
M-66
Mathematics
= 40 [namely (1, 1), (2, 2) ... (40, 40)]
\
=
23.
No. of integral points inside the DOAC
1600 – 40
= 780
2
(d) Four digits number can be arranged in 3 × 4! ways.
Five digits number can be arranged in 5! ways.
Number of integers = 3 × 4! + 5! = 192.
20. (d) Number of ways of selecting a man and a woman for
a team from 15 men and 15 women
= 15 × 15 = (15)2
Number of ways of selecting a man and a woman for next
team out of the remaining 14 men and 14 women.
= 14 × 14 = (14)2
Similarly for other teams
Hence required number of ways
19.
= (15)2 + (14)2 + .... + (1)2 =
21.
22.
Digits which are not used Number of 8 digits
to form 8 digits number
numbers which are
divisible by 9
divisible by 9
0 and 9
8 ´ 7!
15 ´16 ´ 31
= 1240
6
(b) Let no. of men = n
No. of women = 2
Total participants = n + 2
No. of games that M1 plays with all other men
= 2(n – 1)
These games are played by all men
M2, M3, ......, Mn.
So, total no. of games among men = n.2(n – 1).
However, we must divide it by ‘2’, since each game is
counted twice (for both players).
So, total no. of games among all men
= n(n – 1)
....... (i)
Now, no. of games M1 plays with W1 and W2 = 4
(2 games with each)
Total no. of games that M1, M2, ....., Mn play with W1
and W2 = 4n
....... (ii)
Given : n(n – 1) – 4n = 66
Þ n = 11, –6
As the number of men can't be negative.
So, n = 11
(b) In 8 digits numbers, 4 places are odd places.
Also, in the given 8 digits, there are three odd digits 1,
1 and 3.
No. of ways three odd digits arranged at four even
24.
25.
=
4! 5!
´
= 120
2! 3!2!
7 ´ 7!
2 and 7
3 and 6
7 ´ 7!
7 ´ 7!
4 and 5
7 ´ 7!
Hence total number of 8 digits numbers which are
divisible by 9
= 8 × (7!) + 7 × (7!) + 7 × (7!) + 7 × (7!) + 7 × (7!)
= 36 × (7!)
(b) With 3 at unit place,
total possible four digit number (without repetition) will
be 3 ! = 6
With 4 at unit place,
total possible four digit numbers will be 3! = 6
With 5 at unit place,
total possible four digit numbers will be 3! = 6
With 6 at unit place,
total possible four digit numbers will be 3! = 6
Sum of unit digits of all possible numbers
=6×3+6×4+6×5+6×6
= 6 [3 + 4 + 5 + 6]
= 6 [18]= 108
(d)
q:
4 P3 4!
=
2! 2!
No. of ways the remaining five digits 2, 2, 2, 4 and 4
5!
3!2!
Hence, required number of 8 digits number
1 and 8
p : 0 0 0 0 0 place
5 4 3 2 1 ways
Total no. of ways = 5 ! = 120
Since all numbers are > 20,000
\ all numbers 2, 3, 5, 7, 9 can come at first place.
places =
arranged at remaining five places =
(d) We know that any number is divisible by 9 if sum
of the digits of the number is divisible by 9.
Now sum of the digits from 0 to 9
= 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
Hence to form 8 digits numbers which are divisible by
9, a pair of digits either 0 and 9, 1 and 8, 2 and 7, 3 and
6 or 4 and 5 are not used.
0 0 0 0 0 place
3 4 3 2 1 ways
Total no. of ways = 3 × 4 ! = 72
(Q 2 and 9 can not be put at first place)
So, p : q = 120 : 72 = 5 : 3
26.
(d) Given that number of white balls = 10
Number of green balls = 9
and Number of black balls = 7
\ Required probability
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M-67
Permutations and Combinations
= (10 + 1) (9 + 1) (7 + 1) – 1
= 11.10.8 –1 = 879
[Q The total number of ways of selecting one or more
items from p identical items of one kind, q identical items of
second kind; r identical items of third kind is
(p + 1) (q + 1) (r + 1) –1 ]
27. (a) 7 women can be arranged around a circular table in
6! ways.
Among these 7 men can sit in 7! ways.
Hence, number of seating arrangement
= 7! × 6!
28. (a) Alphabetical order is
A, C, H, I, N, S
No. of words starting with A = 5! = 120
No. of words starting with C = 5! = 120
No. of words starting with H = 5! = 120
No. of words starting with I = 5! = 120
No. of words starting with N = 5! = 120
SACHIN – 1
\ Sachin appears at serial no. 601
29. (c) Total number of arrangements of letters in the word
GARDEN = 6 ! = 720 there are two vowels A and E, in half of
the arrangements A preceeds E and other half A follows E.
So, numbers of word with vowels in alphabetical order in
32.
33.
34.
35.
7- x
4!
2!
Then put both E in 5 gaps formed in 5 C2 ways.
\ No. of ways =
36.
4! 5
× C2 = 120
2!
(240)
S ® 2, L ® 2, A, B, Y ,U .
2
5
\ Required number of ways = C1 ´ C2 ´
37.
4!
= 240.
2!
(d) Since, each section has 5 questions.
\ Total number of selection of 5 questions
= 3 ´ 5C1 ´ 5C1 ´ 5C3 + 3 ´ 5C1 ´ 5C2 ´ 5C2
\ x = 3, 4, 5
= 3 × 5 × 5 × 10 + 3 × 5 × 10 × 10
= 750 + 1500 = 2250.
(135)
Select any 4 correct questions in 6C4 ways.
Number of ways of answering wrong question = 3
\ f (3) =7 -3 P3-3 =4 P0 = 1
\ Required number of ways = 6C4 (1)4 ´ 32 = 135.
(d)
Px -3 is defined if
7 - x ³ 0, x - 3 ³ 0 and 7 - x ³ x - 3
Þ 3 £ x £ 5 and x ÎI
38.
39.
\ f (4) =7 - 4 P4 -3 =3 P1 = 3
31.
(120.00)
For vowels not together
Number of ways to arrange L, T, T, R =
1
´ 720 = 360
2
30.
(b) Required sum
= (2 + 4 + 6 + ... + 100) + (5 + 10 + 15 + ... + 100)
– (10 + 20 + ... + 100)
= 2(1 + 2 + 3 ...+ 50) + 5(1 + 2 + 3 + ...+50)
–10(1 + 2 + 3 + ....+10)
= 2550 + 1050 – 530 = 3050.
(c) Total number of numbers
= 3 ´ 5 ´ 5 ´ 5 –1 = 374
(d) Total number of numbers formed using 0, 1, 2, 3, 5, 7
= 5 ´ 6 ´ 6 ´ 4 = 36 ´ 20 = 720.
(54)
Let xyz be the three digit number
\ f (5) =7 -5 P5 -3 =2 P2 = 2
x + y + z = 10, x £ 1, y ³ 0, z ³ 0
Hence range = {1, 2, 3}
x -1 = t Þ x = 1+ t
x - 1 ³ 0, t ³ 0
(a) No. of ways in which 6 men can be arranged at a
round table = (6 – 1)! = 5!
t + y + z = 10 - 1 = 9
0 £ t , z, z £ 9
\ Total number of non-negative integral solution
M
M
M
M
M
M
40.
6
Now women can be arranged in P5
= 6! ways.
Total Number of ways = 6! × 5!
11×10
= 55
2
But for t = 9, x = 10, so required number of integers
= 55 – 1 = 54.
(a) Number of two consecutive stations (Blue lines) = n
Number of two non-consecutive stations (Red lines)
= 9 + 3-1C3-1 = 11C2 =
= n C2 - n
Now, according to the question, n C2 - n = 99n
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EBD_8344
M-68
Mathematics
Þ
n(n - 1)
- 100n = 0 Þ n(n - 1 - 200) = 0
2
46.
Þ n - 1 - 200 = 0 Þ n = 201
25
25
25
r =0
r =0
r =0
å (4r + 1) 25Cr = 4 å r. 25Cr + å 25 Cr
41. (51)
25
= 4å r ´
r =1
25
25 24
Cr -1 + 225 = 100 å 24 Cr -1 + 2 25
r
r =1
= 100.224 + 225 = 225(50 + 1) = 51.225
Hence, by comparison k = 51
42. (490)
0 Red, 1 Red, 2 Red, 3 Red
Number of ways of selecting atmost three red balls
= 7C4 + 5C1 · 7C3 + 5C2 · 7C2 + 5C3 · 7C1
= 35 + 175 + 210 + 70 = 490
43. (c) We know nCr is greatest at middle term.
So, a = (19Cp)max = 19C10 = 19C9
b = (20Cq)max = 20C10
c = (21C6)max = 21C10 = 21C11
a
Now,
19
=
C9
221
= 220
2
(c) Number of ways of selecting three persons such that
there is atleast one boy and atleast one girl in the selected
persons
= n + 5C3 – nC3 – 5C3 = 1750
= 1 + 21C1 + 21C2 + ... + 21C10 =
47.
Þ
(n + 5)!
5!
n!
3!(n + 2)! 3!(n - 3)! 3! 2! = 1750
(n + 5) (n + 4)(n + 3) n( n - 1)( n - 2)
= 1760
6
6
Þ n2 + 3n – 700 = 0 Þ
n = 25
[n = – 28 reected]
20
(a) Total number of beams = C2 – 20 = 190 – 20 = 170
(b) Since, m = number of ways the committee is formed
with at least 6 males
= 8C6 . 5C5 + 8C7 . 5C4 + 8C8 . 5C3 = 78
and n = number of ways the committee is formed with at
least 3 females
= 5C3 . 8C8 + 5C4 . 8C7 + 5C5 . 8C6 = 78
Hence, m = n = 78
(a) Q There are total 9 digits and out of which only 3
digits are odd.
Þ
48.
49.
b
c
=
20 19
21 20 19
C9
. C9
.
10
11 10
a b
c
a
b
c
= =
=
=
\
1 2 42 / 11
11 22 42
44. (2454) EXAMINATION
2N, 2A, 2I, E, X, M, T, O
Case I : If all are different, then
(c) Number of ways of selecting 10 obects
= (10I, 0D) or (9I, 1D) or (8I, 1D) or ... (0I, 10D)
Here, D signifies distinct obect and I indicates identical
obect
Þ
50.
8!
= 8.7.6.5 = 1680
4!
Case II : If two are same and two are different, then
8
p4 =
\
4!
= 3.21.12 = 756
2!
Case III : If two are same and other two are same, then
Hence, total number of 9 digit numbers
3
C1 . 7C2 .
4!
= 3.6 = 18
2!2!
Total cases = 1680 + 756 + 18 = 2454
3! ö 6!
æ4
= ç C3 . 2! ÷ . 2! 4! = 180
è
ø
3
C2 .
\
45. (1)
Þ
r +1
6
r +1
6
r can be 5, 35 for k Î I
r = 5, k = ± 2
r = 35, k = ± 3
Hence, number of ordered pairs = 4.
Þ
51.
36
× 35Cr(k2 – 3) = 35Cr.6
r +1
k2 - 3 =
4
Number of ways to arrange odd digits first = C3.
k2 = 3 +
52.
(a) mC2 ´ 2 = mC1 × 2C1 ´ 2 + 84
m(m – 1) = 4m + 84
m2 – 5m – 84 = 0
m2 – 12m – 7m – 84 = 0
m(m – 12) + 7 (m – 12) = 0
m = 12, m = – 7
Q m>0
m = 12
(d) Consider the expression,
20
20
Ci -1
20
Ci + Ci -1
20
=
Ci -1
21
C1
Downloaded from @Freebooksforjeeneet
3!
2!
M-69
Permutations and Combinations
=
(2) 2 ladies, 1 man
(3) 1 lady, 2 men
(4) 0 ladies, 3 men
Possible cases for Y are
20!
i !(21 - i )! i
´
=
(i - 1)!(21 - i )!
21!
21
3
20
æ
ö 20 i 3
C
(1) 20
å çç 20C + 20i -C1 ÷÷ = å æç 21 ö÷ = (21)3 å i3
ø
i =1 è
i
i -1 ø
i =1 è
i=1
20
\
(1) 0 ladies, 3 men
(2) 1 lady, 2 men
2
=
1
æ 20 ´ 21 ö 100
´ç
÷ =
21
(21)3 è 2 ø
æ
å çç
i =1 è
20
Q
(3) 2 ladies, 1 man
(4) 3 ladies, 0 man
No. of ways = 4C3 . 4C3 + (4C2 . 3C1)2 + (4C1 . 3C2)2
3
Ci-1 ö
k
÷ =
20
Ci + 20 Ci-1 ÷ø
21
20
+ (3C3)2
= 16 + 324 + 144 + 1 = 485
\ k = 100
53. (c) Since, the number of ways to select 2 girls is 5C2.
Now, 3 boys can be selected in 3 ways.
(a) Selection of A and selection of any 2 other boys
(except B) in 5C2 ways
(b) Selection of B and selection of any 2 two other boys
(except A) in 5C2 ways
(c) Selection of 3 boys (except A and B) in 5C3 ways
Hence, required number of different teams
= 5C2 (5C2 + 5C2 + 5C3) = 300
54. (d) If all four letters are different then the number of words
5C × 4! = 120
4
If two letters are R and other two different letters are chosen
from B, A, C, K then the number of words
= 4 C2 ´
n +2
57.
15
58.
15Cr
15Cr -1
4!
=6
2! 2!
15
1
15 - r r
15 - r r - 1
15
1
=
=
r - 1| 15 - r .(16 - r )
r - 1 16 - r
=
16 - r
r
=
å r 2 çè
15
Therefore, the number of four-letter words that can be
formed = 120 + 72 + 72 + 6 = 270
æ 15C ö
r
÷
è Cr -1 ø
å r 2 ç 15
(d)
r =1
4!
= 72
2!
r =1
æ 16 - r ö
÷ å r (16 - r )
r ø = r =1
15
15
15
r =1
r =1
= 16å r - å r
(d) \ Required number of ways = C4 ´ C1 ´ 4!
6
= 11
P2
(n + 2)(n + 1) n (n - 1)(n - 2)(n - 3)
6.5.4.3.2.1
= 11
(n - 2)(n - 3)
2.1
4!
= 72
2!
of words =
C6
Þ (n + 2) (n + 1) n (n – 1) = 11 . 10. 9. 4
Þ n=9
n2 + 3n – 108 = (9)2 + 3(9) – 108
= 81 + 27 – 108
= 108 – 108 = 0
If word is formed using two R’s and two A’s then the number
55.
n-2
Þ
If two letters are A and other two different letters are chosen
from B, R, C, K then the number of words
= 4 C2 ´
(c)
3
= 15 × 3 × 24 = 1080
2
16 ´ 15 ´ 16 15 ´ 31 ´16
2
6
= 8 × 15 × 16 – 5 × 8 × 31 = 1920 – 1240 = 680
(c) Given
n(A) = 4, n(B) = 2, n(A × B) = 8
=
4 ladies
56.
(b)
3 ladies
Y
X
3 men
Possible cases for X are
(1) 3 ladies, 0 man
59.
4 men
Required number of subsets
= 8C3 + 8C4 +.... + 8C8 = 28 – 8C0 – 8C1 – 8C2
= 256 – 1 – 8 – 28 = 219
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EBD_8344
M-70
60.
61.
62.
63.
64.
Mathematics
(a) Number of diagonal = 54
n(n - 3)
Þ
= 54
2
2
Þ n – 3n – 108 = 0 Þ n2 – 12 n + 9n – 108 = 0
Þ n (n – 12)+ 9 (n – 12) = 0
Þ n = 12, –9 Þ n = 12 (Q n ¹ –9)
(c) Given
n(A) = 2, n(B) = 4, n(A × B) = 8
Required number of subsets =
8C + 8C +.... + 8C = 28 – 8C – 8C – 8C
3
4
8
0
1
2
= 256 – 1 – 8 – 28 = 219
(b) We know,
Tn = nC3, Tn+1 = n+1C3
ATQ, Tn+1 – Tn = n+1C3 – nC3 = 10
Þ nC2 = 10
Þ n = 5.
(b) Required number of triangles
= 12C3 – (3C3 + 4C3 + 5C3) = 205
(c) 30 marks to be alloted to 8 questions. Each question
has to be given ³ 2 marks
Let questions be a, b, c, d, e, f, g, h
and a + b + c + d + e + f + g + h = 30
Let a = a1 + 2 so, a1 ³ 0
b = a2 + 2 so, a2 ³ 0,......, a8 ³ 0
= 2 C2 ´ 4C1 ´ 3! = 1´ 4 ´ 6 = 24
Hence, total number of arrangements
= 72 + 24 = 96
67.
m(m - 1) is also a natural number..
2
Now
Þ a1 + a2 + ...... + a8 = 30 – 16 = 14
So, this is a problem of distributing 14 articles in 8 groups.
Number of ways = 14+8–1C8–1 = 21C7
65.
(b)
L
2
³1
O
2
³1
Y
4
2£
Þ
O
1
2
1
2
Y
2
1
1
0
68.
Required number of ways
= 2C1 × 2C1 × 2C2 + 2C1 × 2C2 × 4C1 + 2C2 × 2C1 × 4C1 + 2C2
× 2C2 × 4C0
4´3
+2×1×4+1×2×4+1×1×1
2
= 24 + 8 + 8 + 1 = 41
(a) There are 2 vowels and 4 consonants in the letters a,
b, c, d, e, f.
If we select one vowel, then number of arrangements
=2×2×
66.
4´3
= 2C1 ´ 4 C2 ´ 3! = 2 ´
´ 3 ´ 2 = 72
2
If we select two vowels, then number of arrangements
m2 - m
m(m - 1)
=
2
2
æ m 2 - m öæ m2 - m ö
- 1÷
ç
֍
ç 2 ÷ç 2
÷
m(m - 1)
è
øè
ø
C2 =
\
2
2
So, a1 + a2 + ...... + a8 ü
= 30
+ 2 + 2 +...... + 2 ýþ
L
1
1
2
2
m(m - 1)
2
Since m and (m – 1) are two consecutive natural numbers,
therefore their product is an even natural number. So
(a) n = mC2 =
69.
=
m(m - 1) (m2 - m - 2)
8
=
m(m - 1)[m2 - 2m + m - 2]
8
=
m(m - 1)[m(m - 2) + 1(m - 2)]
8
=
m(m - 1)(m - 2)(m + 1)
8
=
3 ´ ( m + 1) m (m - 1)(m - 2)
=3
4 ´ 3 ´ 2 ´1
(
m +1
C4
)
(d) Statement - 1 : n(A) = p, n(B) = q, q > p
Total number of functions from A ® B = qp
It is a true statement.
Statement - 2 : The total number of selections of p different
obects out of q obects is qCp.
It is also a true statement and it is a correct explanation for
statement - 1 also.
(c) Set A = {a1, a2, ....., a20} has 20 distinct elements.
We have to select 5-element subset.
\ Number of 5-element subsets = 20C5
According to question
(
)
20 C = 19 C .k
5
4
Þ
æ 19! ö
20!
= k. ç
5! 15!
è 4! 15!÷ø
Þ
20
= k Þk=4
5
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M-71
Permutations and Combinations
70. (a) Number of required triangles =
10
= (50 C4 +50 C3 ) +51 C3 +52 C3 + 53 C3 +54 C3 +55 C3
C3 -6 C3
10 ´ 9 ´ 8 6 ´ 5 ´ 4
= 120 - 20 = 100
6
6
(a) The number of ways of distributing 10 identical balls
in 4 distinct boxes
= (51 C4 + 51C3 ) + 52 C3 + 53 C3 + 54 C3 + 55 C3
=
71.
= 10 -1 C4 -1 = 9 C3
72.
Proceeding in the same way, we get
=
78.
(c) Two balls are taken from each urn
3
74.
Cr -1
9´8
= 3 ´ 36 = 108
2
(c) 4 novels, out of 6 novels and 1 dictionary out of 3 can
be selected in 6 C4 ´3C1 ways
7!
7´ 6
=
= 21
2!5!
2 ´1
(c) According to given question two cases are possible.
(i) Selecting 4 out of first five question and 6 out of
remaining question
Then 4 novels with one dictionary in the middle can be
arranged in 4! ways.
\ Total ways of arrangement
= 5C4 ´8 C6 = 140 ways
(ii) Selecting 5 out of first five question and 5 out of remaining
= 6 C4 ´3C1 ´ 4! = 1080
(d) First let us arrange M, I, I, I, I, P, P
8 questions = 5C5 ´8 C5 = 56 ways
Therefore, total number of choices
=140 + 56 = 196.
7! 8
7! 8
C4 =
C4 = 7 ´ 6C4 ´ 8C4
4!2!
4!2!
(a) Set S = {1, 2, 3, ...... 12}
AÈ B È C = S, AÇ B = B ÇC = A ÇC = f
\ Each sets contain 4 elements.
\ The number of ways to partition
= 12C4 × 8C4 × 4C4
=
=
12!
12!
8!
4!
=
´
´
4!8! 4!4! 4!0! (4!)3
(c) The number of ways can vote
= 10 C1 + 10 C2 + 10 C3 + 10 C4
= 10 + 45 + 120 + 210 = 385
77.
n -1
\ The required number of ways
7!
ways
4!2!
*M*I*I*I*I*P*P*
Now 4 S can be kept at any of the * places in 8C4 ways so
that no two S are adacent.
Total required ways
76.
(b) We know that the number of ways of distributing n
identical items among r persons, when each one of them
Total number of ways = C2 ´ C2
Which can be done in
75.
C4 + 55 C3 = 56 C4 .
receives at least one item is
9
=3×
73.
55
50
(d)
=
50
C4 +
6
å 56- r C3
r =1
é 55 C3 + 54 C3 + 53C3 + 52 C3 ù
ú
C4 + ê
êë
+ 51C3 + 50 C3 úû
=
79.
80.
8 -1
(c)
C3-1 = 7C2 =
n
Cr +1 + n Cr -1 + 2 n Cr
= n Cr -1 + n Cr + n Cr + n Cr +1
éëQ nCr + nCr -1 = n+1Cr ùû
81.
= n +1Cr + n +1Cr +1 = n + 2 Cr +1
(d) We know that a number is divisible by 3 only when
the sum of the digits is divisible by 3. The given digits are
0, 1, 2, 3, 4, 5.
Herethe possible number of combinations of 5 digits out
of 6 are 5C4 = 5, which are as follows–
1 + 2 + 3 + 4 + 5 = 15 = 3 × 5 (divisible by 3)
0 + 2 + 3 + 4 + 5 = 14 (not divisible by 3)
0 + 1 + 3 + 4 + 5 = 13 (not divisible by 3)
0 + 1 + 2 + 4 + 5 = 12 = 3 × 4 (divisible by 3)
0 + 1 + 2 + 3 + 5 = 11 (not divisible by 3)
0 + 1 + 2 + 3 + 4 = 10 ( not divisible by 3)
Thus the number should contain the digits 1, 2, 3, 4, 5 or
the digits 0, 1, 2, 4, 5.
Taking 1, 2, 3, 4, 5, the 5 digit numbers are
= 5! = 120
Taking 0, 1, 2, 4, 5, the 5 digit numbers are
= 5! – 4! = 96
\ Total number of numbers = 120 + 96 = 216
n
n
n +1
Cr ù
We know éë Cr + Cr -1 =
û
Downloaded from @Freebooksforjeeneet
EBD_8344
M-72
Mathematics
8
Binomial Theorem
7.
TOPIC Ć
1.
Binomial Theorem for a Positive
Integral Index ‘x’, Expansion of
Binomial, General
Term, Coefficient of any Power of ‘x’
2.
1
[Sep. 06, 2020 (I)]
(a)
5
8
(b)
7
8
(c)
3
8
(d)
1
8
The natural number m, for which the coefficient of x in the
æ
1ö
binomial expansion of ç x m + 2 ÷
è
x ø
3.
6
5.
10.
æ
in the expansion of ç x +
è
1) is equal to :
(a) 29
(c) 26
If the fourth term in
11.
1 ö
æ
1
ç 1 + log x + x 12 ÷ is equal to 200, and x > 1, then the
10
ç x
÷
è
ø
value of x is:
[April 08, 2019 (II)]
(a) 100
(b) 10
(c) 103
(d) 104
Let (x + 10)50 + (x – 10)50 = a0 + a1x + a2x2 + .... + a50x50,
8.
is 1540, is ______.
[NA Sep. 05, 2020 (I)]
The coefficient of x4 in the expansion of (1 + x + x2 + x3)6
in powers of x, is ____________.[NA Sep. 05, 2020 (II)]
a7
is equal to
a13
r =0
___________.
[NA Sep. 04, 2020 (I)]
If a and b be the coefficients of x4 and x2 respectively in
the expansion of
Let (2 x 2 + 3 x + 4)10 = å ar x r . Then
(x +
6.
9.
[April 9, 2019 (I)]
(a) 83
(b) 82
(c) 8
(d) 8–2
If some three consecutive coefficients in the binomial
expansion of (x + 1)n in powers of x are in the ratio 2:15:70,
then the average of these three coefficients is:
[April 09, 2019 (II)]
(a) 964
(b) 232
(c) 227
(d) 625
The sum of the co-efficients of all even degree terms in x
22
20
4.
æ2
log x ö
ç + x 8 ÷ (x > 0) is 20 × 87, then a value of x is:
èx
ø
If {p} denotes the fractional part of the number p, then
200
ïì 3 ïü
í
ý , is equal to :
îï 8 ïþ
) (
6
If the fourth term in the Binomial expansion of
)
6
for all x Î R; then
n
æ 2 1 ö
of x in the expansion of ç x + 3 ÷ is nC23 , is :
x ø
è
[April 10, 2019 (II)]
(a) 38
(b) 58
(c) 23
(d) 35
6
6
x 2 - 1 + x - x 2 - 1 , then: [Jan. 8, 2020 (II)]
(a) a + b = 60
(b) a + b = –30
(c) a – b = 60
(d) a – b = –132
The smallest natural number n, such that the coefficient
6
x 3 - 1 ö÷ + æç x - x3 - 1 ö÷ , (x >
ø è
ø
[April 8, 2019 (I)]
(b) 32
(d) 24
the binomial expansion of
(a) 12.50
(c) 12.25
12.
a2
is equal to : [Jan. 11, 2019 (II)]
a0
(b) 12.00
(d) 12.75
(
If the third term in the binomial expansion of 1 + x log 2 x
)
5
equals 2560, then a possible value of x is:
[Jan. 10, 2019 (I)]
Downloaded from @Freebooksforjeeneet
M-73
Binomial Theorem
21.
1
(a)
4
(b) 4 2
18
æ 1
ö
ç 3
1 ÷
m
x
+
ç
1 ÷ , (x > 0), are m and n respectively, then
n
ç
÷
2x 3 ø
è
is equal to :
[Online April 10, 2016]
(a) 27
(b) 182
1
(d) 2 2
8
The positive value of l for which the co-efficient of x2
(c)
13.
10
l ö
æ
in the expression x 2 ç x + 2 ÷
x ø
è
is 720, is:
(c)
22.
(b) 2 2
(d) 3
5
k
2403
is
, then k
15
15
[Jan. 9, 2019 (I)]
14.
If the fractional part of the number
15.
is equal to:
(a) 6
(b) 8
(c) 4
(d) 14
The coefficient of x10 in the expansion of (1 + x)2 (1 + x2)3
(1 + x3)4 is equal to
[Online April 15, 2018]
(a) 52
(b) 44
(c) 50
(d) 56
If n is the degree of the polynomial,
16.
8
17.
18.
23.
(1 + ax + bx ) (1 - 2 x )
(a, b) is equal to:
24.
19.
æ
x +1
x -1
ç
1
1
ç 23
è x - x3 +1 x - x 2
20.
æ 272 ö
(b) ç16,
÷
3 ø
è
æ 251 ö
(c) ç 16,
÷
3 ø
è
æ 251 ö
(d) ç 14,
÷
3 ø
è
{
}
If X = 4n - 3n - 1 : n Î N and
5
25.
If 1 + x4 + x5 =
å a i (1 + x )
i =0
i
[2014]
, for all x in R, then a2 is:
[Online April 12, 2014]
(a) – 4
(c) – 8
26.
where x ¹ 0, 1, is :
[Online April 9, 2017]
(a) 1
(b) 4
(c) – 4
(d) – 1
If (27)999 is divided by 7, then the remainder is :
[Online April 8, 2017]
(a) 1
(b) 2
(c) 3
(d) 6
æ 272 ö
(a) ç14,
÷
3 ø
è
numbers, then X È Y is equal to:
(a) X
(b) Y
(c) N
(d) Y – X
10
ö
÷
÷
ø
in powers of x are both ero, then
[2014]
Y = {9 ( n - 1) : n Î N } , where N is the set of natural
(x + x3 - 1)5 + (x - x3 - 1)5 ,(x > 1) is :
(a) 0
(b) 1
(c) 2
(d) – 1
The coefficient of x –5 in the binomial expansion of
18
2
8
é
ù
é
ù
1
1
ê
ú +ê
ú and
ê 5 x 3 + 1 – 5 x3 – 1 ú
ê 5 x 3 + 1 + 5 x3 – 1 ú
ë
û
ë
û
m is the coefficient of xn in it, then the ordered pair (n, m) is
equal to
[Online April 15, 2018]
(a) (12 , (20)4)
(b) (8, 5 (10)4)
(c) (24 , (10)8)
(d) (12, 8 (10)4)
2
The coefficient of x in the expansion of the product
(2 – x2). ((1 + 2x + 3x2)6 + (1 – 4x2)6) is
[Online April 16, 2018]
(a) 106
(b) 107
(c) 155
(d) 108
The sum of the co-efficients of all odd degree terms in the
expansion of
[2018]
5
4
(d)
4
5
If the coefficients of the three successive terms in the
binomial expansion of (1 + x)n are in the ratio 1 : 7 : 42, then
the first of these terms in the expansion is:
[Online April 10, 2015]
(a) 8th
(b) 6th
(c) 7th
(d) 9th
If the coefficents of x 3 and x 4 in the expansion of
(c)
[Jan. 10, 2019 (II)]
(a) 4
If the coefficients of x–2 and x–4 in the expansion of
27.
xö
æ
If ç 2 + ÷
è
3ø
(b) 6
(d) 10
55
is expanded in the ascending powers of x and
the coefficients of powers of x in two consecutive terms of
the expansion are equal, then these terms are:
[Online April 12, 2014]
(a) 7th and 8th
(b) 8th and 9th
(c) 28th and 29th
(d) 27th and 28th
The number of terms in the expansion of
(1 + x)101 (1 + x2 – x)100 in powers of x is:
[Online April 9, 2014]
(a) 302
(b) 301
(c) 202
(d) 101
Downloaded from @Freebooksforjeeneet
EBD_8344
M-74
Mathematics
29.
If for positive integers r > 1, n > 2, the coefficients of the
(3r)th and (r + 2)th powers of x in the expansion of (1 + x)2n
are equal, then n is equal to : [Online April 25, 2013]
(a) 2r + 1
(b) 2r –1
(c) 3r
(d) r + 1
The sum of the rational terms in the binomial expansion of
30.
1ö
æ 1
ç 2 2 + 35 ÷ is :
[Online April 23, 2013]
ç
÷
è
ø
(a) 25
(b) 32
(c) 9
(d) 41
If the 7th term in the binomial expansion of
28.
35.
36.
10
31.
(
3 +1
)
2n
-
(
3 -1
)
(d) 144
The remainder left out when 82n – (62)2n+1 is divided by 9
is:
[2009]
37.
2n
38.
(b) an odd positive integer
(c) an even positive integer
32.
33.
34.
(
The number of terms in the expansion of y
+x
)
39.
(b) Statement-1 is true, Statement-2 is true; Statement-2
is NOT a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
r =0
= (1 + x)n + nx(1 + x )n –1.
n -5
6
(b)
n-4
5
(c)
5
n-4
(d)
6
.
n -5
For natural numbers m, n if (1 - y ) m (1 + y ) n
= 1 + a1 y + a2 y 2 + ....... and a1 = a2 = 10, then (m, n) is
(a) (20, 45)
(c) (45, 35)
(b) (35, 20)
(d) (35, 45)
[2006]
11
40.
7
If the coefficient of x in éê ax 2 + æç 1 ö÷ ùú
è bx ø û
ë
equals the
11
é
-7
1 öù
coefficient of x in ê ax - æç
÷ ú , then a and b satisfy
è
bx 2 ø û
ë
the relation
[2005]
(a) a – b = 1
(b) a + b = 1
Statement - 2 : For each natural number n, n 7 - n is
divisible by 7.
[2011 RS]
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation for Statement-1.
å (r + 1) nCr x r
(a)
,
in which powers of x and y are free from radical signs are
[Online May 12, 2012]
(a) six
(b) twelve
(c) seven
(d) five
If f(y) = 1 – (y – 1) + (y – 1)2 – (y – 1)3
+ ... – (y – 1)17,
then the coefficient of y2 in it is [Online May 7, 2012]
(a) 17 C2
(b) 17 C3
(c) 18 C2
(d) 18 C3
Statement - 1 : For each natural number n, (n + 1)7–1 is
divisible by 7.
r =0
[2008]
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2
is a correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2
is not a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
In the binomial expansion of (a – b)n, n ³ 5, the sum of 5th
and 6th terms is ero, then a/b equals
[2007]
(d) a rational number other than positive integers
1/10 55
å (r + 1) nCr = (n + 2)2n –1.
n
(a) an irrational number
1/5
(d) 0
Statement -1 :
Statement-2:
is :
[2012]
(b) 7
n
(d) 2e
If n is a positive integer, then
(b) –144
(c) 132
(c) 8
9
e
2
(a) –132
(a) 2
æ 3
ö
+ 3 ln x ÷ , x > 0, is equal to 729, then x can be :
ç3
è 84
ø
[Online April 22, 2013]
(a) e 2
(b) e
(c)
The coefficient of x7 in the expansion of (1– x – x2 + x3 )6
is
[2011]
(c)
41.
a
=1
b
(d) ab = 1
The coefficient of xn in expansion of (1 + x )(1 - x )n is
[2004]
(a) ( -1)n -1 n
(b) ( -1)n (1 - n)
(c) ( -1)n -1 (n - 1)2
(d) (n - 1)
Downloaded from @Freebooksforjeeneet
M-75
Binomial Theorem
42.
43.
44.
The number of integral terms in the expansion of
( 3 + 8 5)256 is
[2003]
(a) 35
(b) 32
(c) 33
(d) 34
r and n are positive integers r > 1, n > 2 and coefficient of
(r+2)th term and 3rth term in the expansion of (1 + x)2n are
equal, then n equals
[2002]
(a) 3r
(b) 3r + 1
(c) 2r
(d) 2r + 1
The coefficients of xp and xq in the expansion of (1+ x )p+q
are
[2002]
(a) equal
(b) equal with opposite signs
(c) reciprocals of each other
(d) none of these
Middle Term, Greatest Term,
Independent Term, Particular Term
from end in Binomial Expansion,
Greatest Binomial Coefficients
TOPIC n
n
50.
1
For a positive integer n, æç1 + ö÷ is expanded in increasing
è xø
powers of x. If three consecutive coefficients in this
expansion are in the ratio, 2 : 5 : 12, then n is equal to
__________.
[NA Sep. 02, 2020 (II)]
16
51.
1 ö
æ x
+
In the expansion of ç
÷ , if l1 is the least
è cos q x sin q ø
p
p
£ q £ and
8
4
l2 is the least value of the term independent of x when
value of the term independent of x when
p
p
£ q £ , then the ratio l2 : l1 is equal to :
16
8
[Jan. 9, 2020 (II)]
(a) 1 : 8
(b) 16 : 1
52.
(c) 8 : 1
(d) 1 : 16
The total number is irrational terms in the binomial
53.
1 ö
æ 1
ç 7 5 - 310 ÷
expansion of ç
[Jan. 12, 2019 (II)]
÷ is :
è
ø
(a) 55
(b) 49
(c) 48
(d) 54
A ratio of the 5th term from the begining to the 5th term
60
45.
If the constant term in the binomial expansion of
10
æ
k ö
ç x 2÷
x ø
è
46.
47.
is 405, then |k| equals:
[Sep. 06, 2020 (II)]
(a) 9
(b) 1
(c) 3
(d) 2
If for some positive integer n, the coefficients of three
consecutive terms in the binomial expansion of (1 + x)n +5
are in the ratio 5 : 10 : 14, then the largest coefficient in this
expansion is :
[Sep. 04, 2020 (II)]
(a) 462
(b) 330
(c) 792
(d) 252
If the number of integral terms in the expansion of
10
æ 1
ö
ç
÷
1
23 +
1 ÷
from the end in the binomial expansion of ç
çç
÷
2(3) 3 ÷ø
è
is:
(a)
(31 2 + 51 8 )n is exactly 33, then the least value of n is :
48.
49.
Let a > 0, b > 0 be such that a + b = 4. If the maximum
value of the term independent of x in the binomial
3
expansion of
(a) 336
(c) 84
1
(a x 9
1
+ b x 6 )10
2
is 10k, then k is equal to :
[Sep. 02, 2020 (I)]
(b) 352
(d) 176
1
(c) 4(36) 3 :1
54.
(d) 2(36) 3 :1
The term independent of x in the binomial expansion of
8
æ 1
5 öæ 2 1 ö
ç 1 - + 3 x ÷ç 2 x - ÷ is : [Online April 11, 2015]
xø
è x
øè
9
æ3 2 1 ö
ç x - ÷ is k, then 18k is equal to :
3x ø
è2
[Sep. 03, 2020 (II)]
(a) 5
(b) 9
(c) 7
(d) 11
1
(b) 1: 4(16) 3
1
[Sep. 03, 2020 (I)]
(a) 264
(b) 128
(c) 256
(d) 248
If the term independent of x in the expansion of
[Jan. 12, 2019 (I)]
1
1 : 2(6) 3
55.
(a) 496
(b) –496
(c) 400
(d) –400
The term independent of x in expansion of
x +1
x -1 ö
æ
çè 2/ 3 1/ 3
÷
x - x + 1 x - x1/ 2 ø
(a) 4
(c) 210
10
is
(b) 120
(d) 310
Downloaded from @Freebooksforjeeneet
[2013]
EBD_8344
M-76
56.
Telegram @unacademyplusdiscounts
The ratio of the coefficient of x15 to the term independent
2ö
æ
of x in the expansion of ç x 2 + ÷
è
xø
57.
58.
67.
68.
æ x3 3 ö
the binomial expansion of çç + ÷÷ equals 5670 is :
è 3 xø
[Jan. 11, 2019 (I)]
(a) 0
(b) 6
(c) 4
(d) 8
The value of r for which
is :
æ 1ö
The middle term in the expansion of ç1 - ÷
è
xø
powers of x is
(a) – 2nCn–1
(c) 2nCn – 1
If 20C1 + (22) 20C2 +(32) 20C3+ ………. + (202) 20C20 = A(2b), then
the ordered pair (A, b) is equal to : [April 12, 2019 (II)]
(a) (420, 19)
(b) (420, 18)
(c) (380, 18)
(d) (380, 19)
18
The coefficient of x in the product (1+x)(1–x) 10
(1+x+x2)9 is :
[April 12, 2019 (I)]
(a) 84
(b) –126
(c) –84
(d) 126
If the coefficients of x2 and x3 are both ero, in the expansion
of the expression (1 + ax + bx2) (1–3x)15 in powers of x, then
the ordered pair (a, b) is equal to: [April 10, 2019 (I)]
(a) (28, 861)
(b) (–54, 315)
(c) (28, 315)
(d) (–21, 714)
The sum of the series
2·20C0 + 5·20C1 + 8·20C2 + 11·20C3 + … + 62·20C20
�矄 ⼄盐 ᨨ [April 8, 2019 (I)]
(a) 226
(b) 225
(c) 223
(d) 224
The sum of the real values of x for which the middle term in
63.
15
[Online April 9, 2013]
(b) 7 : 64
(d) 1 : 32
(a) 7 : 16
(c) 1 : 4
n
(1 - x )n
in
64.
65.
[Online May 26, 2012]
(b) – 2nCn
(d) 2n Cn
The coefficient of the middle term in the binomial expansion
4
66.
6
in powers of x of (1 + ax ) and of (1 - ax ) is the same if
a equals
(a)
[2004]
3
5
(b)
10
3
Mathematics
8
(c)
-3
10
TOPIC Đ
(d)
-5
3
Properties of Binomial Coefficients,
Number of Terms in the Expansion
of (x+y+z)n, Binomial theorem for
any Index, Multinomial theorem,
Infinite Series
20
Cr 20C0 + 20 Cr -120C1 + 20 Cr - 220C2 + ... + 20 C020 Cr
is maximum, is :
[Jan. 11, 2019 (I)]
(a) 15
(b) 20
(c) 11
(d) 10
20
59.
The value of
å 50-r C6
is equal to : [Sep. 04, 2020 (I)]
r=0
(a)
(c)
60.
51C
7
50C
6
–
–
30C
7
30C
6
(b)
(d)
–
+
30C
7
30C
7
If
to:
(a) (25)2
(c) 224
[NA Jan. 9, 2020 (I)]
[Jan. 10, 2019 (II)]
(b) 225 – 1
(d) 225
3
If the sum of the coefficients of all even powers of x in the
æ 1 - t6 ö
÷÷
è 1- t ø
[Jan. 09, 2019 (II)]
70.
The coefficient of t4 in the expansion of ç
ç
71.
(a) 14
(b) 15
(c) 10
(d) 12
The value of
(21C1 – 10C1) + (21C2 – 10C2) + (21C3 – 10C3) + (21C4 – 10C4)
product
(1 + x + x2 + ... + x2n) (1 – x + x2 – x3 + ... + x2n) is 61, then n
is equal to ¾¾¾.
62.
[NA Jan. 7, 2020 (I)]
The term independent of x in the expansion of
æ 1 x8 ö æ 2 3 ö 6
çç - ÷÷ . ç 2 x - 2 ÷ is equal to :
x ø
è 60 81 ø è
[NA April 12, 2019 (II)]
(a) –72
(b) 36
(c) –36
(d) –108
then K is equal
r=0
The coefficient of x4 in the expansion of (1 + x + x2)10 is
_______.
61.
50C
7
51C
7
å { 50 Cr × 50 - r C25 - r } = K ( 50 C25 ),
25
69.
+ .... + (21C10 – 10C10) is :
[2017]
(a) 220 – 210
(b) 221 – 211
(c) 221 – 210
(d) 220 – 29
Downloaded from @Freebooksforjeeneet
M-77
Binomial Theorem
(a) Statement -1 is false, Statement-2 is true
n
72.
73.
æ 2 4ö
If the number of terms in the expansion of ç1 - + 2 ÷ ,
è x x ø
x ¹ 0, is 28, then the sum of the coefficients of all the terms
in this expansion, is :
[2016]
(a) 243
(b) 729
(c) 64
(d) 2187
The sum of coefficients of integral power of x in the
(
binomial expansion 1 - 2 x
74.
75.
)
is :
(b)
1 50
2 +1
2
(
)
(d)
1 50
3
2
(c)
1 50
3 +1
2
(d) Statement -1 is true, Statement-2 is false
77.
(
)
( )
The coefficient of x 1012 in the expansion of
(1 + xn + x253)10, (where n £ 22 is any positive integer), is
[Online April 19, 2014]
(a) 1
(b) 10 C4
(c) 4n
(d) 253 C 4
10
10
j =1
j =1
78.
C0 -
20
C1 +
20
and S3 = å j
j =1
2 10
Cj.
C2 -
[2007]
20
C3 + ..... -..... +
(a) 0
(b)
20
C10
(c)
(d)
1
2
20
- 20 C10
20
(1 +
neglected,
C10
then
3
x) 2
æ 1 ö
- ç 1 + x÷
è 2 ø
3
1
(1 - x ) 2
approximated as
[2010]
3
(a) 1 - x 2
8
Statement -1 : S3 = 55 × 29.
79.
(c) Statement -1 is false, Statement -2 is true .
80.
In a shop there are five types of ice-creams available. A
child buys six ice-creams.
Statement-1 : The number of different ways the child can
buy the six ice-creams is 10C5.
Statement -2 : The number of different ways the child can
buy the six ice-creams is equal to the number of different
ways of arranging 6 A’s and 4 B’s in a row.
[2008]
be
[2005]
(b)
3
3 x + x2
8
x 3 2
3
(d)
- x
- x2
2 8
8
If x is positive, the first negative term in the expansion of
(1 + x)27 5 is
(b) Statement -1 is true, Statement -2 is false.
(d) Statement - 1 is true, Statement 2 is true ; Statement -2
is a correct explanation for Statement -1.
may
(c)
Statement - 2: S1 = 90 × 28 and S2 = 10 × 28 .
(a) Statement -1 is true, Statement -2 is true ; Statement 2 is not a correct explanation for Statement -1.
C10 is
3
If x is so small that x and higher powers of x may be
Let S1 = å j ( j - 1)10C J , S2 = å j10C j
10
76.
The sum of the series
20
[2015]
)
1 50
3 -1
2
(c) Statement -1 is true, Statement-2 is true; Statement 2 is not a correct explanation for Statement-1
50
(
(a)
(b) Statement -1 is true, Statement-2 is true; Statement -2
is a correct explanation for Statement-1
81.
[2003]
(a) 6th term
(b) 7th term
(c) 5th term
(d) 8th term
The positive integer ust greater than (1 + 0.0001)10000 is
[2002]
(a) 4
(b) 5
(c) 2
(d) 3
If the sum of the coefficients in the expansion of (a + b)n is
4096, then the greatest coefficient in the expansion is
[2002]
(a) 1594
(b) 792
(c) 924
(d) 2924
Downloaded from @Freebooksforjeeneet
EBD_8344
M-78
1.
Mathematics
(d)
3200 1 100
= (9 )
8
8
r1 r2
r3
n(n + 1) 2
1
1é
ù
= (1 + 8)100 = ê1 + n × 8 +
× 8 + ....ú
8
8ë
2
û
0
7
3
1
5
4
2
3
5
1
= + Integer
8
3
1
6
ìï 3200 üï ì 1
ü 1
\í
ý = í + integer ý =
þ 8
ïî 8 ïþ î 8
2.
a7 =
10!37 43 10!(2)(3)5 (4) 4
+
7!3!
5!4!
10!(2) 2 (3)3 (4)5 10!(2) 3 (3)(4) 6
+
2!3!5!
3!6!
a13 = Coeff. of x13
2r1 + r2 = 13 and r1 + r2 + r3 = 10
Possibilities are
(13)
+
Tr +1 =
22
Tr +1 =
22
æ 1ö
Cr × ( xm )22- r × ç 2 ÷
èx ø
r
Cr × x 22 m - mr - 2r
Q 22m - mr - 2r = 1
3.
r1 r2
22 m - 1
3×3×5
Þr=
Þ r = 22 m+ 2
m+2
So, possible value of m = 1, 3, 7, 13, 43
But 22Cr = 1540
\ Only possible value of m = 13.
(120.00)
æ 1- x
Coefficient of x 4 in çç
è 1- x
4
a13 =
6
ö
÷÷ = coefficient of x 4 in
ø
(1 - 6 x4 )(1 - x)-6
= 9C4 - 6 ×1 = 126 - 6 = 120.
4.
\
5.
(8.00)
r =0
General term =
10!
(2 x 2 )r1 (3 x) r2 (4) r3
r1 !r2 !r3 !
Since, a7 = Coeff. of x7
2r1 + r2 = 7 and r1 + r2 + r3 = 10
Possibilities are
0
4
5
1
5
3
2
6
1
3
10!(25 )(33 )(42 ) 10!(26 )(3)(43 )
+
5! 3! 2!
6!1! 3!
a7
= 23 = 8
a 13
(d) Using Binomial expansion
(x + a)n + (x – a)n = 2(T1 + T3 + T5 + T7...)
20
The given expression is (2 x 2 + 3x + 4)10 = å ar x r
7
10!(23 )(37 ) 10!(24 )(35 )(4)
+
3!7!
4!5!
+
4
6
7
2
= coefficient of x4 in (1 - 6 x ) éë1 + C1 x + C2 x + ....ùû
r3
3
6
6
\ æç x + x 2 - 1 ö÷ + æç x - x 2 - 1 ö÷ = 2 (T1 + T3 + T5 + T7 )
è
ø è
ø
2[6C0 x5 + 6C2 x4 (x2 – 1) + 6C4 x2 (x2 – 1)2 + 6C6 (x2 – 1)3]
= 2[x6 + 15(x6 – x4) + 15x2(x4 – 2x2 + 1) + (–1 + 3x2 – 3x4 + x6)]
=2(32x6 – 48x4 + 18x2 – 1)
a = – 96 and b = 36
\
a – b = – 132
Downloaded from @Freebooksforjeeneet
M-79
Binomial Theorem
6.
(a)
æ 2 1 ö
çx + 2 ÷
x ø
è
n
Þ
1
n
( )
General term Tr +1 = Cr x
7.
2 n-r æ
According to the question x > 1, \ x = 10.
3
æ xlog8 x ö
log8 x
7
Þ 8 ´ 20 ´ ç
÷ = 20 ´ 8 Þ x
= 64
ç x ÷
x
è
ø
11.
= a0 + a1x + a2x2 + ¼ + a50x50
\
\
1
or x = 82
8
(b) Given nCr–1 : nCr : nCr+1 = 2 : 15 : 70
\
Cr -1
=
Cr
=
2
r
r +1 3
=
and
=
n - r + 1 15
n - r 14
Þ 17r = 2n + 2 and 17r = 3n – 14
i.e., 2n + 2 = 3n – 14 Þ n = 16 & r = 2
\ Average =
=
16
C1 + 16C2 +
3
16
C3
50
C2 ´1048
50
C50 1050
50 ´ 49 49
=
= 12.25
2 ´100 4
(a) Third term of (1 + xlog2 x )5 = 5C2 ( xlog2 x )5-3
Given, 5C2 ( xlog2 x )2 = 2560
Þ
( xlog2 x )2 = 256 = (±16)2
Þ
x log 2 x = 16 or x log 2 x = –16 (reected)
Þ
x log 2 x = 16 Þ log2 xlog2 x = log2 16 = 4
Þ log2 x = ±2 Þ x = 22 or 2–2
( x + x 3 - 1)6 + ( x - x 3 - 1)6
+ 6C6(x3 – 1)3]
6
7
4
8
5
2
= 2[x + 15x – 15x + 15x – 30x + 15x + x9 – 3x6
+ 3x3 – 1]
Hence, the sum of coefficients of even powers of
x = 2[1 – 15 + 15 + 15 – 3 – 1] = 24
(b) \ fourth term is equal to 200.
æ æ 1 öö
÷÷
ç ç
T4 = C ç x è 1+ log10 x ø ÷
3
ç
÷
ç
÷
è
ø
6
a2
a0 =
=5C2 ( xlog2 x )2
16 + 120 + 560
=
3
= 2[6C0x6 + 6C2x4 (x3 – 1) + 6C4x2 (x3 – 1)2
10.
12.
696
= 232
3
(d)
a0 = 2.50C501050
a2 = 2.50C2.1048
n
2
15
Cr
=
and
n
15
Cr +1 70
Þ
a0 + a1x + a2x2 + ¼ + a50x50
= 2(50C0x50 + 50C2x48 . 102 + 50C4x46 . 104 + ¼)
Þx=
9.
(c) (x + 10)50 + (x – 10)50
Now, take log8 on both sides, then
(log8x)2 – (log8x) = 2
Þ log8x = –1
or log8x =2
n
3
Þ t2 + 4t – t – 4 = 0 Þ t (t + 4) – 1 (t + 4) = 0
Þ t = 1 or t = – 4
log10 x = 1 Þ x = 10
or log10 x = – 4 Þ x = 10–4
3
n
+
æ1
3 ö
ç +
÷ t = 1 Þ t2 + 3t – 4 = 0
4
2(1
+ t) ø
è
To find coefficient of x, 2n – 5r = 1
Given nCr = nC23 Þ r = 23 or n – r = 23
\ n = 58 or n = 38
Minimum value is n = 38
(b) Q T4 = 20 ´ 87
Þ
1
x 4 2(1+ log10 x ) = 10
Taking log10 on both sides and putting log10 x = t
r
1 ö n
2n – 5r
ç 3 ÷ = Cr × x
èx ø
æ 2ö
Þ 6C3 ç ÷ ´ ( xlog8 x )3 = 20 ´ 87
è xø
8.
3
20 x 2(1+ log10 x ) .x 4 = 200
3
3
æ 1 ö
ç x12 ÷ = 200
ç
÷
è
ø
1
4
(a) Since, coefficient of x 2 in the expression x 2
Þ x = 4 or
13.
lö
æ
çè x + 2 ÷ø is a constant term, then
x
lö
æ
Coefficient of x2 in x2 ç x + 2 ÷
è
x ø
10
lö
æ
= co-efficient of constant term in ç x + 2 ÷
è
x ø
lö
æ
General term in ç x + 2 ÷
è
x ø
10
=
10
Cr
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( x)
10
10- r
æ lö
çè 2 ÷ø
x
r
EBD_8344
M-80
Mathematics
= 10C ( x )
r
10 - r
- 2r
2
× l2
720
= 16 Þ l = 4d
Þ
5´ 9
Hence, required value of l is 4.
(b) 2403 = 2400 × 23
= 24 ×100 × 23
= (24)100 × 8
= 8(24)100 = 8(16)100
= 8 (1 + 15)100
= 8 + 15 m
When 2403 is divided by 15, then remainder is 8.
l2 =
Hence, fractional part of the number is
15.
é
ù
é
ù
1
1
ê
ú +ê
ú
ê 5 x 3 + 1 – 5 x3 – 1 ú
ê 5 x 3 + 1 + 5 x3 – 1 ú
ë
û
ë
û
17.
=
1 é
ê
28 ë
(
5 x3 + 1 + 5 x3 – 1
) +(
)
28
24
= 54 × 24 ×
2
2
(a) Let a = ((1 + 2x + 3x2)6 + (1 – 4x2)6)
\
Coefficient of x2 in the expansion of the product
(2 – x2) ((1 + 2x + 3x2)6 + (1 – 4x2)6)
= 2 (Coefficient of x2 in a) – 1 (Constant of expansion)
In the expansion of ((1 + 2x + 3x2)6 + (1 – 4x2)6).
Constant = 1 + 1 = 2
Coefficient of x2 = [Coefficient of x2 in (6C0 (1 + 2x)6 (3x2)0)]
+ [Cofficient of x2 in (6C1 (1 + 2x)5 (3x2)1)]
8
– [6C1 (4x2)]
= 60 + 6 × 3 – 24 = 54
\
8
The coefficient of x2 in (2 – x2) ((1 + 2x + 3x2)6 +
(1 – 4x2)6)
= 2 × 54 – 1 (2) = 108 – 2 = 106
18.
8
é 5 x 3 + 1 + 5 x3 – 1 ù
é 5 x3 + 1 – 5 x 3 – 1 ù
ú +ê
ú
=ê
ê (5 x 3 + 1) – (5 x3 – 1) ú
ê (5 x3 + 1) – (5 x 3 – 1) ú
ë
û
ë
û
8
(
= 104 × 23 = 8 (10)4
After rationalise the polynomial we get
8
)
)
ù
ú
ú
ú
ú
ú
8
ú
ú
û
+ 8C854 ]
= 54 ×
é
5 x 3 + 1 – 5 x3 – 1 ù
1
ú
+ê
´
ê 5 x3 + 1 + 5 x3 – 1
5 x 3 + 1 – 5 x 3 – 1 úû
ë
)
Now, coefficient of x12 = [8C054 + 8C254 + 8C454 + 8C654
8
15
é
5 x3 + 1 + 5 x 3 – 1 ù
1
ê
ú
´
3
3
ê 5 x3 + 1 – 5x3 – 1
ú
+
+
5
x
1
5
x
–
1
ë
û
)(
So, the degree of polynomial is 12,
Therefore value of k is 8
(a) Q (1 + x)2 = 1 + 2x + x2,
(1 + x2)3 = 1 + 3x2 + 3x4 + x6,
and (1 + x3)4 = 1 + 4x3 + 6x6 + 4x9 + x12
So, the possible combinations for x10 are:
x . x9, x . x6 . x3, x2 . x2 . x6, x4 . x6
Corresponding coefficients are 2 × 4, 2 × 1 × 4, 1 × 3 × 6, 3
× 6 or 8, 8, 18, 18.
\ Sum of the coefficient is 8 + 8 + 18 + 18 = 52
Therefore, the coefficient of x 10 in the expansion of
(1 + x)2 (1 + x2)3 (1 + x3)4 is 52.
(d)
(
é 8C0 (5 x 3 + 1)4 + 8C2 (5 x 3 + 1)3 (5 x 3 – 1) + 8C ù
4
ú
1 ê 3
2
3
2
ê
ú
= 8 (5 x + 1) (5 x – 1)
2 ê 8
ú
3
3
3
3
4
ê + C6 (5 x + 1) (5x – 1) + 8C8 (5 x – 1)
ú
ë
û
8
16.
)(
)(
(
(
10 - r
- 2r = 0 Þ r = 2
2
Co-efficient is x2 in expression = 10C2l2 = 720
14.
)
(
8
6
2
é8
3
8
3
5 x3 – 1
ê C0 5 x + 1 + C2 5 x + 1
ê
4
4
1 ê
= 8 ê + 8 C4 5 x 3 + 1
5 x3 – 1
2 ê
2
6
ê 8
3
5 x3 – 1 + 8C8 5 x3 - 1
ê + C6 5 x + 1
ë
Then, for constant term,
(c) Since we know that,
(x + a)5 + (x – a)5
= 2[5C0 x5 + 5C2 x3× a2 + 5C4x×a4]
\
8
(x +
x3 - 1
) + (x 5
x3 - 1
)
5
= 2[5C0 x5 + 5C2 x3(x3 – 1) + 5C4x(x3 – 1)2]
)
8ù
5 x3 + 1 - 5 x3 – 1 ú
û
Þ 2[x5 + 10x6 – 10x3 + 5x7 – 10x4 + 5x]
\
Sum of coefficients of odd degree terms = 2.
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M-81
Binomial Theorem
19.
(a)
(
)(
) (
é x1 3 + 1 x 2 3 - x1 3 + 1
ê
23
13
ê
+
x
x
1
ëê
(
)
(
= x1 3 + 1 - 1 - 1 x1 2
Tr + 1 = 10Cr x
for r = 10
T11 =
10
C10 x
) = (x
10
13
)(
)
10
x -1 x +1 ù
ú
x x -1 ú
ûú
(
- 1 x1 2
)
24.
(b) 4n – 3n – 1 = (1 + 3)n – 3n –1
= [nC0 + nC1.3 + nC2.32 +......+ nCn3n] – 3n – 1
= 9 [nC2 +nC3.3+....+nCn.3n–2]
\ 4n – 3n – 1 is a multiple of 9 for all n.
\ X = {x : x is a multiple of 9}
Also, Y = {9 (n – 1) : n ÎN}
= {All multiples of 9}
Clearly X Ì Y. \ X È Y = Y
25.
(a)
)
10
20-5r
6
5
-5
4
5
1+ x + x =
Coefficient of x–5 = 10C10 (1) (–1)10 = 1
20.
(d)
( 28 - 1)
Þ
\
21.
7
(b) Tr + 1 =
18
Cr
çè
÷ø
Þ 1 + x 4 + x5
= a0 + a1 (1 + x) + a2 (1 + 2 x + x 2 ) + a3 (1 + 3x + 3x 2 + x3 )
+ a4 (1 + 4 x + 6 x 2 + 4 x3 + x 4 ) + a5 (1 + 5x + 10 x 2 + 10 x3 + 5x 4 + x5 )
r
æ
ö
ç 1 ÷ =
ç 1÷
çè 3 ÷ø
2x
18
Cr x
6-
2r
3
coefficient of x -2
coefficient of x -4
n
22. (c)
18
=
Þ 1 + x 4 + x5
1
2
r
2
= a0 + a1 + a1 x + a2 + 2a2 x + a2 x + a3 + 3a3 x
+3a3 x 2 + a3 x3 + a4 + 4a4 x + 6a4 x 2 + 4a4 x3 + a4 x 4 + a5
2r
ì
ü
ïï 6 - 3 = -2 Þ r = 12 ïï
í
ý
ï & 6 - 2r = -4 Þ r = 15 ï
îï
þï
3
Þ
+5a5 x + 10a5 x 2 + 10a5 x3 + 5a5 x 4 + a5 x5
Þ 1 + x 4 + x5
1
= (a0 + a1 + a2 + a3 + a4 + a5 ) + x(a1 + 2a2 + 3a3 + 4a4 + 5a5 )
C12 12
2 = 182
1
18
C15 15
2
+ x 2 (a2 + 3a3 + 6a4 + 10a5 ) + x3 (a3 + 4 a4 + 10a5 )
+ x 4 (a4 + 5a5 ) + x5 (a5 )
On comparing the like coefficients, we get
a5 = 1
C r n Cr +1 n C r + 2
=
=
1
7
42
18 ´ 17 ´ 16
4 a + 18 ´ 17
.8 +
- 36b = 0
6
2
= –51 × 16 × 8 + a × 36 × 17 – 36b = 0
= –34 × 16 + 51a – 3b = 0
= 51a – 3b = 34 × 16 = 544
= 51a – 3b = 544
....(i)
Only option number (b) satisfies the equation number (i)
...(i)
;
a4 + 5a=
5 1 ...(ii);
a3 + 4a4 + 10 a=
5 0 ...(iii)
By solving we get r = 6
so, it is 7th term.
23. (b) Consider (1 + ax + bx2) (1 – 2x)18
= (1 + ax + bx2) [18C0 – 18C1 (2x)
+ 18C2(2x)2– 18C3(2x)3 + 18C4(2x)4 –.......]
3
18
Coeff. of x = C3 (–2)3 + a. (–2)2.18C2 + b (–2).18C1 = 0
Coeff. of x3 = – 18C3.8 + a × 4. 18C2 – 2b × 18 = 0
=-
ai (1 + x)i
+ a4 (1 + x )4 + a5 (1 + x )5
28l - 7 + 7 - 1 7 ( 4l - 1) + 6
=
7
7
Remainder = 6
18 - r
æ 1ö
ç x3 ÷
i= 0
1
2
3
= a0 + a1 (1 + x) + a2 (1 + x ) + a3 (1 + x )
28l - 1
=
7
999
å
and a2 + 3a3 + 6 a4 + 10a5 = 0 ...(iv)
from (i) & (ii), we get
a4 = -4 ...(v) from (i), (iii) & (v), we get
a3 = +6 ...(vi)
26.
Now, from (i), (v) and (vi), we get
a2 = – 4
(a) Let rth and (r + 1)th term has equal coefficient
xö
æ
çè 2 + ÷ø
3
55
xö
55 æ
= 2 ç1 + ÷
è 6ø
æ xö
rth term = 255 55Cr ç ÷
è 6ø
55
r
Downloaded from @Freebooksforjeeneet
EBD_8344
M-82
Mathematics
1
Coefficient of xr is 255 55Cr r
6
æ xö
(r +1)th term = 255 55Cr +1 ç ÷
è 6ø
30.
9
r +1
æ 3
ö
+ 3 ln x ÷ , x > 0
ç3
84
è
ø
1
55 55
Cr +1. r +1
Coefficient of xr+1 is 2
6
Both coefficients are equal
We have
Tr+1 = nCr (x)n–r ar for (x + a)n.
\ According to the question
1
1
255 55Cr r = 255 55Cr +1 r +1
6
6
729 =
1
1
1
.
=
r 55 - r r + 1 54 - r 6
27.
6 (r + 1) = 55 – r
6r + 6 = 55 – r
7r = 49
r=7
(r + 1) = 8
Coefficient of 7th and 8th terms are equal.
(c) Given expansion is
(1 + x)101 (1 – x + x2)100
= (1 + x) (1 + x)100 (1 – x + x2)100
= (1 + x) [(1 –
x3)100]
Expansion (1 –
x3)100
(2n)!
(2n)!
=
Þ
(r + 2)! (2n - r - 2)! (3r )! (2 n - 3r )!
Þ (3r) ! (2n – 3r) ! = (r + 2) ! (2n – r – 2) ! ...(1)
Now, put value of n from the given choices.
= 2 é 2n C1
êë
(
( 3)
3 +1
2 n -1
)
2n
-
+ 2 n C3
(
( 3)
)
2n
2 n -3
( 3)
2n-5
+ ....ù
úû
Q (a + b)n – (a – b)n
= 2[n C1a n -1b + nC3 a n -3b3 ...]
= which is an irrational number.
55
32.
1 ö
æ 1
(a) Given expansion is çç y 5 + x 10 ÷÷
è
ø
The general term is
Tr +1 =
55
æ 1ö
Cr ç y 5 ÷
ç ÷
è ø
55- r
æ 1 ö
.ç x10 ÷
ç
÷
è
ø
r
Tr + 1 would free from radical sign if powers of y and x are
integers.
r
55 - r
and
are integer.
5
10
LHS : (3r) ! (4r + 2 – 3r) ! = (3r) ! (r + 2) !
i.e.
RHS : (r + 2) ! (3r) !
Þ r is multiple of 10.
Þ LHS = RHS
Hence, r = 0, 10, 20, 30, 40, 50
(d) (21/2 + 31/5)10 = 10C0(21/2)10
It is an A.P.
+ 10C1(21/2)9 (31/5) + ...... + 10C10(31/5)10
3 -1
+ 2 n C5
Choice (a) put n = 2r + 1 in (1)
29.
33 3
´ 3 ´ (6 ln x)
84
31. (a) Consider
will have 100 + 1 = 101 terms
As given 2nCr+2 = 2nC3r
3
æ 3 ö
6
C6 ç
÷ .( 3 ln x )
3
è 84 ø
Þ (ln x)6 = 1 Þ (ln x)6 = (ln e)6
Þ x=e
So, (1 + x) (1 – x3)100 will have 2 × 101 = 202 terms
(a) Expansion of (1 + x)2n is 1 + 2nC1x + 2nC2 x2 + ...... +
2nC xr + 2nC
r+1 + ...... + 2nC x2n
r
r+1 x
2n
9
Þ 36 = 84 ´
= (1 + x) [(1 + x) (1 – x + x2)]100
28.
(b) Let r + 1 = 7 Þ r = 6
Given expansion is
Thus, 50 = 0 + (k – 1) 10
There are only two rational terms – first term and last term.
50 = 10k – 10 Þ k = 6
Now sum of two rational terms
Thus, the six terms of the given expansion in which x and
y are free from radical signs.
= (2)5 + (3)2 = 32 + 9 = 41
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M-83
Binomial Theorem
33.
(d) Given function is
36.
f(y) = 1 – (y – 1) + (y – 1)2 – (y – 1)3 + ........ – (y – 1)17
In the expansion of (y – 1)n
(a) (8)2n – (62) 2n + 1
= (64) n – (62)2n + 1
= (63 + 1)n – (63 – 1)2n + 1
= é nC0 (63) n + n C1 (63) n -1 + nC2 (63)n - 2
ë
Tr + 1 = nCr yn – r (– 1)r
coeff of y2 in (y – 1)2 = 2C0
coeff of y2 in (y – 1)3 = – 3C1
coeff of y2 in (y – 1)4 = 4C2
So, coeff of termwise is
+ ........+
Cn-1 (63) + nCn ù
û
- é 2n+1C0 (63)2n+1 - 2n+1C1 (63)2n
ë
+ 2 n +1C2 (63)2 n -1 – ........+ (–1)2n+1 2n+1C2n+1 ù
û
2C + 3C + 4C + 5C + .......... + 17C
0
1
2
3
15
= 1 + 3C1 + 4C2 + 5C3 + .......... + 17C15
= (3C0 + 3C1) + 4C2+ 5C3 + .......... + 17C15
= 4C1 + 4C2 + 5C3 + .......... + 17C15
= 5C2 + 5C3 + .......... + 17C15
= 18C15 = 18C3
n
n-1 n
n- 2 n
+ C2 (63) n-3
= 63 × ëé C0 (63) + C1 (63)
+........ + nCn -1 ùû + 1 – 63 ×
é 2n+1C0 (63) 2n - 2n+1C1 (63)2 n-1 + ....... + 2 n+1 C2n ù + 1
ë
û
34. (a) Statement 2 :
P ( n ) : n 7 - n is divisible by 7
37.
Put n = 1, 1 – 1 = 0 is divisible by 7, which is true
Let n = k, P (k) : k7 – k is divisible by 7, true
Put n = k + 1
\
n
( n + 1) 7 - n7 - 1 is divisible by 7
Hence both Statements 1 and 2 are correct and Statement
2 is the correct explanation of Statement -1.
35. (b) (1 – x – x2 + x3)6 = [(1– x) – x2 (1 – x)]6
= (1– x)6 (1 – x2)6
= (1 – 6x + 15x2 – 20x3 + 15x4 – 6x5 + x6) × (1 – 6x2 + 15x4
– 20x6 + 15x8 – 6x10 + x12)
7
Coefficient of x = (– 6) (– 20) + (– 20)(15) + (– 6) (–6)
= – 144
r =0
r =1
n –1
Cr –1 x r + (1 + x)n
n –1
Cr –1 x r –1 + (1 + x )n
= nx (1+ x) n–1 + (1+ x) n = RHS
\ Statement 2 is correct.
Putting x = 1, we get
n
å (r + 1)n Cr = n × 2n –1 + 2n = (n + 2) × 2n –1.
r =0
Statement 1 : n - n is divisible by 7
Þ
r =0
n
7
is divisible by 7
r =0
n
= nx å
\ P(k + 1) is divisible by 7
Hence P(n) : n7 – n is divisible by 7
( n + 1) 7 - n7 - 1 + ( n7 - n)
n
r =1
So 7 C1 , 7 C 2 , ......7 C 6 are all divisible by 7
Þ
n
n
=år× r
P(k + 1) : k7 + 7C1k6 + 7C2k2 +......+ 7C6k + 1 – k – 1, is div.
by 7.
P(k + 1) : (k7 – k) + (7C1k6 + 7C2k5 +........+ 7C6k) is div. by
7.
Since 7 is coprime with 1, 2, 3, 4, 5, 6.
( n + 1) 7 - ( n + 1) is divisible by 7
n
n
r
n
r
å (r + 1) nCr x r = å r. Cr x + å Cr x
7
P (k + 1) : (k + 1) – ( k + 1) is div. by 7
Þ
= 63 × some integral value + 2
Hence, when divided by 9 leaves 2 as the remainder.
(b) From statement 2:
38.
\ Statement 1 is also true and statement 2 is a correct
explanation for statement 1.
(b) Tr + 1 = (–1)r. nCr (a)n – r. (b)r is an expansion
of (a – b)n
\ 5th term = t5 = t4+1
= (–1)4. nC4 (a)n–4.(b)4 = nC4 . an–4 . b4
6th term = t6 = t5+1 = (–1)5 nC5 (a)n–5 (b)5
Given t5 + t6 = 0
\ nC4 . an–4 . b4 + (– nC5 . an–5 . b5) = 0
Þ
n!
an
n!
a n b5
. .b4 .
=0
4!(n - 4)! a 4
5!(n - 5)! a5
Þ
é 1
b ù
ê
ú = 0 [Q a ¹ 0, b ¹ 0]
4!( n - 5)!.a 4 ë (n - 4) 5.a û
Þ
b
1
a n-4
=0 Þ
=
n - 4 5a
5
b
n !.a n b4
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EBD_8344
M-84
39.
Mathematics
(d)
(1 - y ) m (1 + y ) n
42.
= [1 - m C1 y + m C2 y 2 - ......] [1 + n C1 y + n C 2 y 2 + .....]
=
ì m(m - 1) n(n - 1)
ü
+
- mn ý y2 + .....
= 1 + (n - m) y + í
2
2
î
þ
\ a1 = n - m = 10
and a2 =
...(i)
é 2 1 ù
ê ax + bx ú
ë
û
43.
(c) tr + 2 = 2nCr + 1 xr + 1;t3r = 2nC3r – 1 x3r – 1
Given that, 2nCr + 1 = 2nC3r – 1 ;
Þ r + 1 + 3r – 1 = 2n
Þ 2n = 4r Þ n = 2r
[from (i)]
44.
(a) We know that tp + 1 = p + qCp xp and
tq + 1 =
2 11 - r
11
= Cr (ax )
æ 1ö
çè ÷ø
bx
256 - r
2 (5) r / 8
It is possible if r is an integral multiple of 8 and 0 £ r £ 256
Tr +1 in the expansion
11
Cr (3)
\ r = 33
m2 + n2 - m - n - 2mn
= 10
2
...(ii)
Þ m + n = 80
Solving (i) and (ii), we get
\ m = 35, n = 45
(d)
256
Cr ( 3)256 - r (8 5)r
256
Terms will be integral if 256 - r & r both are +ve integer..
2
8
(m - n) 2 - (m + n) = 20
40.
Tr +1 =
(c)
Q
r
45.
p+qC xq
q
p + qC = p + qC .[ Remember nC = nC
p
q
r
n–r
(c) General term = Tr +1 =
= 11 Cr (a)11 - r (b) - r ( x)22 - 2r - r
æ kö
Cr ( x )10- r × ç - 2 ÷
è x ø
For the Coefficient of x7, we have
22 – 3r = 7 Þ r = 5
=
10
Cr ( - k ) r × x
\ Coefficient of x 7
=
10
Cr ( - k ) r × x
= 11C5 (a)6 (b)- 5
11
1 ù
é
ê ax - 2 ú
bx û
ë
-r
r
r
11
Þ k2 =
- 2 r +11- r
46.
-7
...(ii)
2
nn
Þ
= (-1) n + ( -1) n -1.n
= ( -1)n (1 - n)
Cr
Cr +1
=
n+ 5
Cr ,
n+ 5
Cr +1 ,
n+ 5
Cr + 2
1
2
(Given)
r +1
1
= Þ 3r = n + 3
n+ 5- r 2
n +5
n
(1 + x )(1 - C1 x + C2 x - .... + (-1) Cn x )
= (-1) nn Cn + ( -1) n -1n Cn -1
n +5
Q n+ 5
\ 11 C5 (a)6 (b) - 5 = 11C6 a5 ´ (b) - 6
Þ ab = 1.
(b) Coeff. of xn in ( 1+x) (1 – x)n
= coeff of xn in
n
(a) Consider the three consecutive coefficients of
(1 + x) n+ 5 be
= C6 a ´ 1 ´ (b)
Q Coefficient of x7 = Coefficient of x–7
From (i) and (ii),
n
405 ´ 2 81
= =9
10 ´ 9
9
\| k | = 3
-6
5
10 - 5 r
2
\10C2 (- k )2 = 405
(-1) ´ (b) ( x)
= Cr (a)
For the Coefficient of x–7, we have
Now 11 – 3r = – 7 Þ 3r = 18 Þ r = 6
\ Coefficient of x
10 - r
- 2r
2
10 - 5r
=0Þr =2
2
æ 1 ö
= 11Cr (ax 2 )11 - r ç è bx 2 ÷ø
11- r
11
r
Since, it is constant term, then
...(i)
Again T r + 1 in the expansion
41.
10
]
and
n+ 5
Cr +1
Cr + 2
=
...(i)
5
7
r+2
5
= Þ 12r = 5n + 6
...(ii)
n+4-r 7
Solving (i) and (ii) we get r = 4 and n = 6
\ Largest coefficient in the expansion is 11C6 = 462.
Þ
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M-85
Binomial Theorem
47.
n- r
(3) 2
Tr +1 = Cr
n
50.
n
1ö
æ 1
(c) Here, ç 3 2 + 5 8 ÷
è
ø
n
r
(5) 8
n-r
r
and are integer
2
8
So, r must be 0, 8, 16, 24 ......
n
n
Cr
æ 3x2 ö
9
(c) General term = Tr +1 = Cr ç
÷
è 2 ø
9- r
æ 1ö
çè - ÷ø
3x
r
Cr
=
12
n - r 12
Þ
=
5
r +1 5
16- r
r
T9 = 16C8
18 - 3r = 0 Þ r = 6
6
æ 3 ö æ 1ö
æ 1ö
\ T7 = C6 ç ÷ ç - ÷ = 9C3 ç ÷
è 2 ø è 3ø
è 6ø
T9 = 16C8
3
9
7
= 7.
18
(a) General term of
\18k = 18 ´
1
-1
=
Cr (ax 9 )10 - r (bx 6 ) r
10
Cr
10
Then, by AM-GM inequality
1
a 3 + b2
³ ( a 3b 2 ) 2
2
Þ (2)2 ³ a3b2 Þ a 6b4 £ 16
Q The maximum value of the term independent of
x = 10k
10
C4 ×16 Þ k = 336.
l
2
Now, l =
1
C4 a 6b4
Since a 3 + b2 = 4
\10k =
(sin 2q)8
28
8
æ 1 ö
ç
÷
è 2ø
= 16C8.28.24
[Q min. value of l2 at q = p/8]
10 - r r
- = 0 Þ r = 4.
9
6
\ Term independent of x =
28
l2 = 16C8 =
10 - r r
a10 - r b r ( x ) 9 6
Term independent of x if
1
sin8 q cos8 q
é p pù
When qÎ ê , ú , th en least value of the term
ë16 8 û
independent of x,
-1
1
10
r
p p
When qÎ éê , ùú , then least value of the term
ë8 4û
independent of x,
l 1 = 16C8 28
[Q min. value of l1 at q = p/4]
3
9 ´ 8 ´ 7 æ 1ö
æ 7ö
=
ç ÷ = çè ÷ø .
3 ´ 2 ´ 1 è 6ø
18
(ax 9 + bx 6 )10 =
...(i)
æ x ö
æ 1 ö
Tr +1 =16Cr ç
÷
ç x cos q ÷
è sin q ø
è
ø
For r = 8 term is free from ‘x’
The term is independent of x, then
49.
5
n - r +1 5
Þ
=
2
r
2
...(ii)
Þ 5n - 17r - 12 = 0
Solving eqns. (i) and (ii),
n = 118, r = 34
51. (b) General term of the given expansion
æ 1 ö 18-3r
çè - ÷ø x
3
3
=
Cr -1
Cr +1
n
Þ n = 256
9-r
n
Þ 2n - 7r + 2 = 0
Now n = t33 = a + (n - 1)d = 0 + 32 ´ 8 = 256
æ 3ö
= 9Cr ç ÷
è 2ø
Cr -1 : nCr : n Cr +1 = 2 : 5 :12
Þ
Q
48.
(118)
According to the question,
52.
16
C8 .28.24
16
C8.28
= 16
(d) Let the general term of the expansion
Tr+1 =
=
60
60
æ 1ö
Cr ç 7 5 ÷
è ø
12-
Cr × (7)
60- r
r
5 ( -1) r
æ 1ö
ç -310 ÷
è
ø
r
r
× (3)10
Then, for getting rational terms, r should be multiple of
L.C.M. of (5, 10)
Then, r can be 0, 10, 20, 30, 40, 50, 60.
Since, total number of terms = 61
Hence, total irrational terms = 61 – 7 = 54
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EBD_8344
M-86
Mathematics
55.
10
53.
æ 1
ö
0
æ 1 ö æ 1 ö 10
ç 23 + 1 ÷
10
3
C
2
+
÷ ç
1÷ =
0ç
ç
è 2(3)1/ 3 ÷ø
è
ø
çè
÷
3
2(3) ø
(c)
(c) Given expression can be written as
10
é 1/3 3 3
ù
( x )2 - 12 ú
ê ( x ) +1
ê 2/3 1/3
x ( x - 1) úû
ë x - x +1
10
æ 1ö æ 1 ö0
L + C10 ç 2 3 ÷ ç
1/ 3 ÷
è ø è 2(3) ø
10
æ
æ x +1 ö ö
= ç ( x1/3 + 1) - çç
÷÷
ç
x ÷ø ÷ø
è
è
10
6
æ 1ö
1
ç 23 ÷
1 4
5th term from beginning T5 =10C4 è ø æ 3 ö
ç 2.3 ÷
è
ø
and 5th term from end T11–5+1 =
æ 1ö
3
T5 : T7 = C4 ç 2 ÷
è ø
10
\
2
æ 1ö æ 1 ö
3
ç
÷
2
= ç ÷ : ç 1÷
è ø è 3ø
2.3
2
2
10
4
æ 1ö æ
23 ç
C6 ç
è
ö
1 ÷
÷
1
ø çè 3 ø÷
2.3
4
6æ
= (x1/3 – x–1/2)10
General term = Tr+1
= 10Cr (x1/3)10–r(–x–1/2)r =
6
4
ö
ö
æ 1ö æ
ç 1 ÷ : 10C ç 2 3 ÷ ç 1 ÷
6
1
ç 1÷
è ø çè 3 ø÷
è 2.33 ø
2.3
= 10 Cr ( -1) r · x
6
56.
1
8
æ 2 1ö
(c) General term of ç 2x - ÷ is
è
xø
8
15
Required ratio =
r
8
-r
2
10 - r r
- =0
3
2
C10 ´ (2)10
15 !
10 ! 5 !
=
15 !
´ (2)5
5 !10 !
= 1 : 32
57. (d) Given expansion can be written as
r
æ 1ö
+3x 5 . 8 Cr (2x 2 )8- r ç - ÷
è xø
=
15
C5 ´ (2)5
r
æ 1ö 1
æ 1ö
Cr (2x 2 )8- r ç - ÷ - 8Cr (2x 2 )8-r ç - ÷
è xø x
è xø
· ( -1) r · x
(d) Tr+1 = 15Cr(x2)15–r . (2x–1)r = 15Cr × (2)r × x30–3r
For independent term, 30 – 3r = 0 Þ r = 10
Hence the term independent of x,
T11 = 15C10 × (2)10
For term involve x15, 30 – 3r = 15 Þ r = 5
Hence coefficient of x15 = 15C5 × (2)5
r
æ -1 ö
Cr (2x 2 )8- r ç ÷
è x ø
\ Given expression is equal to
=
10 - r
3
10- r - r
3 2
8
æ 1
5ö 8
2 8- r æ 1 ö
çè1 - + 3x ÷ø Cr (2x ) çè - ÷ø
x
x
Cr x
Þ r=4
So, required term = T5 = 10C4 = 210
2
3
3
= 2 × 2 × 3 = 4(6) 3 :1 = 4.(36) 3 :1
1
54.
10
Term will be independent of x when
2
2
10
æ
1 ö
= ç x1/ 3 + 1 - 1 ÷
xø
è
n
r
Cr 28- r (-1)r x16 -3r – 8 C r 28- r ( -1)r x15-3r
r
æ 1ö
+3. 8Cr 2(8- r) ç - ÷ (-1)r x 21-3r
è xø
For the term independent of x, we should have
16 – 3r = 0, 15 – 3r = 0, 21 – 3r = 0
From the simplification we get r = 5 and r = 7
\ – 8C5 (23) (–1)5 – 3. 8C7.2
é 8!
ù
é 8!
ù
+ê
´ 8ú - 3 ´ ê
´ 2ú
ë 5!3! û
ë 7!1! û
æ x -1 ö
n
n –n
2n
ç x ÷ .(1 - x ) = (–1) x (1 – x)
è
ø
Total number of terms will be 2n + 1 which is odd (Q
2n is always even)
\ Middle term =
2n + 1 + 1
= (n + 1) th
2
Now, Tr +1 = nCr (1)r x n - r
2n
So,
Cn × x 2 n - n
n
n
x .( -1)
= 2 n Cn .( -1)n
Middle term is an odd term. So, n + 1 will be odd.
So, n will be even.
\ Required answer is 2nCn.
= (56 × 8) – 48
= 448 – 6 × 8 = 448 – 48 = 400
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M-87
Binomial Theorem
58.
(c) The middle term in the expansion of
6
1 æ 2 3 ö
3 ö
x8 æ
2x - 2 ÷ - ç 2x2 - 2 ÷
ç
60 è
x ø 81 è
x ø
Term independent of x,
(1 + ax) 4 = T3 =4 C2 (ax )2 = 6a 2 x 2
The middle term in the expansion of
(1 - ax)6 = T4 =6 C3 ( -ax )3 = -20a 3 x3
According to the question
6a 2 = -20a 3 Þ a = -
6
1 æ 2 3 ö
1
2x - 2 ÷ - .
= Coefficient of x in
60 çè
81
x ø
3
10
æ 2 3 ö
coefficient of x–8 in ç 2x - 2 ÷
x ø
è
20
59.
(a) The given series,
å 50- r C6
=
= 50C6 + 49C6 + 48C6 + 47 C6 + ... + 32C6 + 31C6 + 30C6
= (30 C7 + 30C6 ) + 31C6 + 32C6 + .... + 49C6 + 50C6 - 30C7
63.
= (31 C7 + 31C6 ) + 32C6 + .... + 49C6 + 50C6 - 30C7
= (32 C7 + 32C6 ) + .... + 49C6 + 50C6 - 30C7
=
10! b+ 2 g
x
a !b!g !
Now, compare it with R.H.S., A = 420 and b = 18
64.
Case-2 : When g = 1, b = 2, a = 7
10!
= 360
7! 2!1!
(a) Given expression,
(1 – x)10 (1 + x + x2)9 (1 + x) = (1 – x3)9 (1 – x2)
= (1 – x3)9 – x2(1 – x3)9
Þ Coefficient of x18 in (1 – x3)9 – coeff. of x16 in (1 – x3)9
9! 7 ´ 8 ´ 9
=
= 84
6!3!
6
(c) Given expression is (1 + ax + bx2) (1 – 3x)15
Co-efficient of x2 = 0
Þ 15C2 (– 3)2 + a . 15C1(– 3) + b . 15C0 = 0
9
= C6 - 0 =
Case-3 : When g = 2, b = 0, a = 8
10!
= 45
8!0! 2!
æ 1 x8 ö æ 2 3 ö6
ç - ÷ ç 2x - 2 ÷
ç 60 81 ÷ è
x ø
è
ø
r =1
= 420 × 218
10! b+ 2 g
x
a !b! g !
Hence, total = 615
61. (30) Let (1– x + x2 ..... x2n) (1 + x + x2 ..... x2n)
= a0 + a1x + a2x2 + .....
put x = 1
1(2n + 1) = a0 + a1 + a2 + ..... a2n
put x = – 1
(2n + 1) × 1 = a0 – a1 + a2 + ...... a2n
Adding (i) and (ii), we get,
4n + 2 = 2(a0 + a2 + .....) = 2 × 61
Þ 2n + 1 = 61 Þ n = 30.
62. (d) Given expression is,
r =1
20
19
= 20 å r. Cr -1
é 20
ù
18
19
= 20 êê19 å Cr - 2 + 2 úú = 20[19 . 218 + 219]
ë r=2
û
For coefficient of x4; b + 2g = 4
Here, three cases arise
Case-1 : When g = 0, b = 4, a = 6
Þ
å r 2 . 20 Cr
20
é 20
ù
19
19
ê
ú
+
20
(
r
1)
C
C
å
å
1
1
r
r
= ê
úû
r =1
ër = 1
= 51C7 - 30C7
Þ
(b) Given, 20C1+ 22.20C2 + 32.20C3 + ... + 202.20C20
= A(2b)
Taking L.H.S.,
20
.................................................................
.................................................................
.................................................................
Þ
6
-1 6
1 6
C3 (2)3 (3)3 +
C (2) (3)5
60
81 5
= – 72 + 36 = – 36
r=0
60. (615) General term of the expansion =
6
=
65.
15 ´14
´ 9 - 15 ´ 3a + b = 0
2
Þ 945 – 45a + b = 0
...(i)
Now, co-efficient of x3 = 0
Þ 15C3 (– 3)3 + a . 15C2(– 3)2 + b .15C1 (– 3) = 0
Þ
...(i)
...(ii)
15 ´14 ´13
15 ´14 ´ 9
´ (-3 ´ 3 ´ 3) + a ´
– b × 3 × 15 = 0
3´ 2
2
Þ 15 × 3 [– 3 × 7 × 13 + a × 7 × 3 – b] = 0
Þ 21a – b = 273
...(ii)
From (i) and (ii), we get,
a = 28, b = 315 Þ (a, b) º (28, 31,5)
Þ
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EBD_8344
M-88
66.
Mathematics
(b) 2.20C0 + 5.20C1 + 8.20C2 + ...... + 62.20C20
20
=
å (3r + 2) 20Cr
20
20
r =0
r =0
+
20
20
= 3 å r. Cr + 2 å Cr
r =0
20
20
r =1
r=0
Hence, the coefficient of t4 = 1 ×
19
20
= 60å Cn -1 + 2 å Cr
71.
3´ 4´ 5´ 6
= 15
4 ´ 3 ´ 2 ´1
21
(a) We have ( C1 + 21C2 ...... + 21C10)
– (10C1 + 10C2 ..... 10C10)
th
æn ö
(a) Middle Term, ç + 1÷ term in the binomial
è2 ø
=
8
æ x3 3 ö
expansion of ç + ÷ is,
è 3 xø
3ö4
=
69.
4
(d)
25
r =0
r =0
æ
å ( 50Cr ×50-r C25- r ) = å çè 50 - r
50
æ 50 1 æ
25 ö ö
= å ç 25 ´ 25 ´ çè 25 - r r ø÷ ÷
ø
r =0 è
25
= C25 å Cr =
25
50
25
C25 (2 )
r =0
70.
.... + 10C10 = 210 – 1)
1 21
[2 - 2] - (210 - 1)
2
72.
(b) Total number of terms = n+2C2 = 28
(n + 2) (n + 1) = 56; n = 6
6
æ 2 4 ö
\ Put x = 1 in expansion ç 1 - + 2 ÷ ,
è x x ø
we get sum of coefficient = (1 – 2 + 4)6
= 36 = 729.
73.
(c) We know that (a + b)n + (a – b)n
= 2[n C0 a n b0 + n C2 a n - 2b 2 + n C4 a n -4b 4 ...]
(1 - 2
Then, by comparison, K = 225
(b) Consider the expression
3
æ 1 - t6 ö
ç 1 - t ÷ = (1 – t6)3(1 – t)–3
è
ø
3× 4 2
æ
t
= (1 – 3t6 + 3t12 – t18) çè 1 + 3t +
2!
x)
50
+ (1 + 2 x )
50
2 éë 50 C0 + 50C2 (2 x )2 + 50C4 (2 x ) 4 ...ùû
= 2 éë 50 C0 + 50C2 22 x +
50 - r ö
r 25 25 - r ÷ø
50
C4 24 x 2 + ...ùû
Putting x = 1, we get,
50
25
50
10C + 10C +
1
2
= (220 – 1) – (210 – 1) = 220 – 210
8 ´ 7 ´ 6 ´ 5 12- 4
´x
Þ
= 5670
4´3´2
Þ x8 = 81
Þ x8 – 81 = 0
\ sum of all values of x = sum of roots of equation (x8 – 81 =
0).
(b) Consider the expression
20C 20C + 20C
20
20
20
20
20
r
0
r – 1 C1 + Cr – 2 C2 + ... + C0 × Cr
For maximum value of above expression r should be
equal to 20.
as 20C20 × 20C0 + 20C19 × 20C1 + ... + 20C20 × 20C0
= (20C0)2 + (20C1)2 + ... + (20C20)2 = 40C20.
Which is the maximum value of the expression,
So, r = 20.
25
1 21
[( C1 + .... + 21C10 ) + (21C11 + .... 21C20)] – (210 – 1)
2
(Q
æx
æ 3ö
8
T4 + 1 = C4 ç ÷ çè ÷ø = 5670
3
è ø x
68.
3× 4 × 5 × 6
4!
=
= 60 × 219 + 2 × 220 = 221 [15 + 1] = 225
67.
3× 4×5 3 3× 4×5×6 4
ö
t +
t + K ฀÷
ø
3!
4!
74.
(b)
C0 + 50C2 22 + 50C4 24... =
350 + 1
2
Given expansion (1 + x n + x 253 )10
Let x1012 = (1)a (xn)b. (x253)c
Here a, b, c, n are all +ve integers and a £ 10,
b £ 10, c £ 4, n £ 22, a + b + c = 10
Now bn + 253c = 1012
Þ bn = 253 (4 – c)
For c < 4 and n £ 22; b > 10, which is not possible.
\ c = 4, b = 0, a =6
\ x1012 = (1)6. (xn)0 . (x253)4
Hence the coefficient of x1012 =
Downloaded from @Freebooksforjeeneet
10!
=
6!0!4!
10C
4
M-89
Binomial Theorem
10
75.
S2 = å j
(b)
10
j =1
10
C j = å 10 9C j -1
= (1 - x )
j =1
n n -1
é n
ù
Cr -1 ú
êëQ Cr = r
û
76.
= 10 é 9 C0 + 9C1 + 9 C2 + .... + 9 C9 ù = 10.29
ë
û
(a) The given situation in statement 1 is equivalent to
find the non negative integral solutions of the equation
x1 + x2 + x3 + x4 + x5 = 6
which is coeff. of x6 in the expansion of
(1 + x + x2 + x3 + .....¥)5= coeff. of x6 in (1– x)–5
= coeff. of x6 in 1 + 5x +
é
ê x
= ê1 + +
ë 2
79.
5.6 2
x ..........
2!
5 × 6 × 7 × 8 × 9 ×10 10! 10
=
= C6
6!
6!4!
\ Statement 1 is wrong.
Number of ways of arranging 6A’s and 4B’s in a row
1 3 ù
.
2 2 x 2 ú é -3 x 2 ù = -3 x 2
úê
ú 8
2!
ûë8
û
Tr +1 =
27 ö
æ
32
\ r = 7 . çèQ n = 5 ÷ø
5
Therefore, first negative term is T8 .
n
80.
10! 10
= C 6 which is same as the number of ways the
6!4!
child can buy six icecreams.
\ Statement 2 is true.
(d) We know that, (1 + x)20 = 20C0 + 20C1x + 20C2 x2
(d)
æ 1ö
(1 + 0.0001)10000 = ç1 + ÷ , n = 10000
è nø
1 n ( n - 1) 1 n ( n - 1)( n - 2) 1
1
= 1 + n. +
+
+ ... + n
2! n 2
3!
n
n3
n
=
20C x10 + ..... 20C x20
10
20
x = –1, (0) = 20C0 – 20C1 + 20C2 – 20C3 + ......
+ 20C10 – 20C11 .... + 20C20
0 = 2[20C0 – 20C1 + 20C2 – 20C3
+ ..... – 20C9] + 20C10
ù
3 1 ö
.
2
2 2 x 2 ÷ - æç1 + 3 x + 3.2 x ö÷ ú
ú
÷ ç
2!
2
2! 4 ÷ø ûú
ø è
n(n - 1)(n - 2).........( n - r + 1) r
( x)
r!
For first negative term,
n - r +1 < 0 Þ r > n +1
(d)
Þr >
=
77.
éæ
êç 3
êç 1 + x +
ëêè 2
-1
2
=1+1 +
1 æ 1 ö 1 æ 1 öæ 2 ö
1
1- + 11 - ÷ + ... + n
2! çè n ÷ø 3! çè n ÷ç
øè n ø
n
+ ......
Put
Þ
Þ 20C
10
= 2[20C0 – 20C1 + 20C2 – 20C3
+ ...... – 20C9 + 20C10]
Þ 20C – 20C + 20C – 20C + .... + 20C
0
1
2
3
10
=
78.
1 20
C10
2
< 1+
1 1 1
1
+ + + ........+
1! 2! 3!
(9999)!
1 1
+ + .......¥ = e < 3
1! 2!
(c) Take a = 1 and b = 1 in (a + b)n ×
=1+
81.
2n = 4096 = 212 Þ n = 12;
The greatest coeff = coeff of middle term.
So middle term = t7.
Þ t7 = t6 + 1 = 12C6a6b6
Þ Coeff of t7 = 12C6 =
12!
= 924.
6!6!
(c) Q x3 and higher powers of x may be neglected
3
\
(1 + x) 2 - æçè1 +
(
1-
1
x2
)
xö
÷
2ø
3
Downloaded from @Freebooksforjeeneet
EBD_8344
M-90
Mathematics
9
Sequences and Series
7.
TOPIC Ć Arithmetic Progression
1.
2.
3.
The common difference of the A.P. b1, b2, ..., bm is 2 more
than the common difference of A.P. a 1, a 2, ..., a n. If
a40 = – 159, a100 = – 399 and b100 = a70, then b1 is equal to:
[Sep. 06, 2020 (II)]
(a) 81
(b) – 127
(c) – 81
(d) 127
2sin2a–1
4–2sin2a
If 3
, 14 and 3
are the first three terms of an
A.P. for some a, then the sixth term of this A.P is:
[Sep. 05, 2020 (I)]
(a) 66
(b) 81
(c) 65
(d) 78
If the sum of the first 20 terms of the series
(a1 ¹ 0), then the sum of the A.P., a1 , a3 , a5 , ..., a23 is
ka1, where k is equal to :
[Sep. 02, 2020 (II)]
(a) 8.
9.
log (71/ 2 ) x + log (71/ 3 ) x + log (71/ 4 ) x + ... is 460, then x is
equal to :
(a) 72
4.
(b) 71/2
(c) e 2
[Sep. 05, 2020 (II)]
(d) 746/21
10.
difference is an integer and Sn = a1 + a2 + .... + an . If
( Sn - 4 , an - 4 ) is equal to :
5.
(a)
6.
1
6
(b)
1
5
(c)
1
4
(d)
(d) nth term is -4
2
5
121
10
(c)
72
5
(d) -
72
5
The number of terms common to the two A.P.’s 3, 7, 11, ...,
407 and 2, 9, 16, ..., 709 is _____.
[NA Jan. 9, 2020 (II)]
1
1
and its 20th term is ,
20
10
then the sum of its first 200 terms is: [Jan. 8, 2020 (II)]
If the 10th term of an A.P. is
(b) 50
1
4
(c) 100
(d) 100
1
2
Let f : R ® R be such that for all x Î R, (21+x + 21–x), f (x)
and (3x + 3–x) are in A.P., then the minimum value of f (x) is:
[Jan. 8, 2020 (I)]
(a) 2
(b) 3
(c) 0
(d) 4
Five numbers are in A.P., whose sum is 25 and product is
(a) 27
12.
(b) 7
(c)
21
2
(d) 16
Let Sn denote the sum of the first n terms of an A.P. If S4 = 16
and S6 = –48, then S10 is equal to : [April 12, 2019 (I)]
(a) –260
13.
3
1
4
20 + 19 + 19 + 18 + ... upto nth term is 488 and then
5
5
5
nth term is negative, then :
[Sep. 03, 2020 (II)]
(a) n = 60
(b) nth term is –4
(b)
1
2520. If one of these five numbers is - , then the greatest
2
number amongst them is:
[Jan. 7, 2020 (I)]
1
7
In the sum of the series
(c) n = 41
11.
[Sep. 04, 2020 (II)]
(a) (2490, 249)
(b) (2480, 249)
(c) (2480, 248)
(d) (2490, 248)
If the first term of an A.P. is 3 and the sum of its first 25
terms is equal to the sum of its next 15 terms, then the
common difference of this A.P. is :
[Sep. 03, 2020 (I)]
121
10
(a) 50
Let a1 , a2 ,....., an be a given A.P. whose common
a1 = 1, an = 300 and 15 £ n £ 50, then the ordered pair
If the sum of first 11 terms of an A.P., a1 , a2 , a3 , ... is 0
(b) –410
(c) –320
(d) –380
If a1, a2, a3, …… are in A.P. such that a1 + a7 + a16 = 40,
then the sum of the first 15 terms of this A.P. is :
[April 12, 2019 (II)]
(a) 200
14.
(b) 280
(c) 120
(d) 150
If a1, a2, a3, ..... an are in A.P. and a1 + a4 + a7 + ….+ a16 = 114,
then a1 + a6 + a11 + a16 is equal to : [April 10, 2019 (I)]
(a) 98
(b) 76
(c) 38
Downloaded from @Freebooksforjeeneet
(d) 64
M-91
Sequences and Series
15.
Let the sum of the first n terms of a non-constant A.P.,
n(n - 7)
a1, a2, a3, ………….. be 50n +
A, where A is a
2
constant. If d is the common difference of this A.P., then
the ordered pair (d, a50) is equal to: [April 09, 2019 (I)]
(a) (50, 50 + 46A)
(b) (50, 50 + 45A)
(c) (A, 50 + 45A)
(d) (A, 50 + 46A)
24.
Let
25.
12
k =0
k =1
17.
18.
19.
20.
21.
22.
Let a 1 , a 2 , ....., a 30 be an A.P., S =
T=
Let a1,a 2 ,a 3 ,..., a 49 be in A.P. such that
å a 4k +1 = 416 and a 9 + a 43 = 66. If
å f (a + k) = 16(210 - 1) , where the function f satisfies
f(x + y) = f(x) f(y) for all natural numbers x, y and f(a) = 2.
Then the natural number ‘a’ is:
[ April 09, 2019 (I)]
(a) 2
(b) 16
(c) 4
(d) 3
If the sum and product of the first three terms in an A.P.
are 33 and 1155, respectively, then a value of its 11th term
is:
[April 09, 2019 (II)]
(a) –35
(b) 25
(c) –36
(d) –25
The sum of all natural numbers ‘n’ such that 100 < n < 200
and H.C.F. (91, n) > 1 is :
[April 08, 2019 (I)]
(a) 3203
(b) 3303
(c) 3221
(d) 3121
If nC4, nC5 and nC6 are in A.P., then n can be :
[Jan. 12, 2019 (II)]
(a) 9
(b) 14
(c) 11
(d) 12
If 19th term of a non- ero A.P. is ero, then its (49th term) :
(29th term) is :
[Jan. 11, 2019 (II)]
(a) 4 : 1
(b) 1 : 3
(c) 3 : 1
(d) 2 : 1
The sum of all two digit positive numbers which when
divided by 7 yield 2 or 5 as remainder is:
[Jan. 10, 2019 (I)]
(a) 1256
(b) 1465
(c) 1365
(d) 1356
1 1
1
, 2 ,.....
are two A.P's such that
h1 h
hn
x3 = h2 = 8 and x8 = h7 = 20, then x5. h10 equals.
[Online April 15, 2018]
(a) 2560
(b) 2650
(c) 3200
(d) 1600
10
16.
If x1 , x2, ....., xn and
2
a12 + a 22 + ... + a17
= 140m , then m is equal to :
26.
27.
(a) 68
(b) 34
(c) 33
(d) 66
For any three positive real numbers a, b and c,
9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c). Then :
[2017]
(a) a, b and c are in G.P.
(b) b, c and a are in G.P.
(c) b, c and a are in A.P.
(d) a, b and c are in A.P.
If three positive numbers a, b and c are in A.P. such that
abc = 8, then the minimum possible value of b is :
[Online April 9, 2017]
1
(a) 2
28.
29.
30
is:
i=1
(a)
15
å a (2i - 1) .
i= 1
30.
If a5 = 27 and S – 2T = 75, then a10 is equal to:
[Jan. 09, 2019 (I)]
(a) 52
(b) 57
(c) 47
(d) 42
1 1 1
, , ,....., (xi ¹ 0 for i = 1, 2, ...., n) be in A.P..
23. Let
x1 x2 x3
2
(b) 4 3
(c) 4 3
(d) 4
Let a1, a2, a3, ...., an, be in A.P. If a3 + a7 + a11 + a15 = 72,
then the sum of its first 17 terms is equal to :
[Online April 10, 2016]
(a) 306
(b) 204
(c) 153
(d) 612
Let a and b be the roots of equation px2 + qx + r = 0, p ¹ 0.
If p, q, r are in A.P and
å a i and
34
9
(b)
2 13
9
1 1
+ = 4, then the value of | a – b|
a b
[2014]
(c)
61
9
æ1ö
÷ is equal to.
i =1 è i ø
[Online April 16, 2018]
(a) 3
(b)
13
8
(c)
13
4
(d)
1
8
2 17
9
[Online April 11, 2014]
(a) 4000
31.
(b) 4020
(c) 4200
(d) 4220
Given an A.P. whose terms are all positive integers. The
sum of its first nine terms is greater than 200 and less than
220. If the second term in it is 12, then its 4th term is:
[Online April 9, 2014]
n
åç x
(d)
The sum of the first 20 terms common between the series 3
+ 7 + 11 + 15 + ......... and 1 + 6 + 11 + 16 + ......, is
such that x1 = 4 and x21 = 20. If n is the least positive
integer for which xn > 50, then
[2018]
(a) 8
32.
(b) 16
(c) 20
(d) 24
If a1, a2, a3,...., an, .... are in A.P. such that a4 – a7 + a10 = m,
then the sum of first 13 terms of this A.P., is :
[Online April 23, 2013]
(a) 10 m
(b) 12 m
(c) 13 m
Downloaded from @Freebooksforjeeneet
(d) 15 m
EBD_8344
M-92
33.
34.
Mathematics
Given sum of the first n terms of an A.P. is 2n + 3n2. Another
A.P. is formed with the same first term and double of the
common difference, the sum of n terms of the new A.P. is :
[Online April 22, 2013]
(a) n + 4n2 (b) 6n2 – n (c) n2 + 4n (d) 3n + 2n2
Let a1, a2, a3,... be an A.P, such that
a1 + a2 + ... + a p
a1 + a2 + a3 + ... + aq
=
p3
q3
(a) 34 minutes
(c) 135 minutes
41.
35.
36.
37.
41
11
(b)
31
121
a1 + a2 + ........... + aq
11
41
(d)
121
1861
If 100 times the 100th term of an AP with non ero common
difference equals the 50 times its 50th term, then the 150th
term of this AP is :
[2012]
th
(a) – 150
(b) 150 times its 50 term
(c) 150
(d) Zero
If the A.M. between pth and qth terms of an A.P. is equal
to the A.M. between rth and sth terms of the same A.P.,
then p + q is equal to
[Online May 26, 2012]
(a) r + s – 1 (b) r + s – 2 (c) r + s + 1 (d) r + s
Suppose q and f (¹ 0) are such that sec (q + f), sec q and
(a) ± 2
42.
(c) ±
1
2
(d) ± 2
Let an be the n th term of an A.P. If
41
11
(b)
2
(c) m2 – m (4r + 1) + 4 r 2 – 2 = 0
43.
2
2
(d) m – m (4r – 1) + 4 r + 2 = 0
Let Tr be the rth term of an A.P. whose first term is a and
common difference is d. If for some positive integers
m, n, m ¹ n, Tm =
1
1
and Tn = , then a – d equals
n
m
[2004]
1 1
1
+
(b) 1
(c)
(d) 0
m n
mn
If 1, log9 (31–x + 2), log3 (4.3x – 1) are in A.P. then x equals
[2002]
(a) log3 4
(b) 1 – log3 4
(c) 1 – log4 3
(d) log4 3
(a)
å a2r = a and
r=1
å a2r –1 = β, then the common difference of the A.P. is
a6
equals
a21
(b) m 2 – m (4r + 1) + 4 r + 2 = 0
44.
100
, p ¹ q , then
2
(a) m 2 – m (4r – 1) + 4 r – 2 = 0
100
38.
q2
the binomial expansion of (1 + y )m are in A.P., then m and
r satisfy the equation
[2005]
[Online May 19, 2012]
(b) ± 1
p2
2
7
11
(c)
(d)
7
2
41
If the coefficients of rth, (r + 1)th, and (r + 2)th terms in the
(a)
æ fö
sec (q – f) are in A.P. If cos q = k cos ç ÷ for some k, then
è 2ø
k is equal to
=
[2006]
; p ¹ q . Then a6 is equal to:
a21
(c)
Let a1 , a2 , a3 ............ be terms on A.P. If
a1 + a2 + ...........a p
[Online April 9, 2013]
(a)
(b) 125 minutes
(d) 24 minutes
TOPIC n Geometric Progression
r=1
[2011]
(a) a - b
39.
40.
(b)
a -b
100
(c) b - a
(d)
α–β
200
A man saves ` 200 in each of the first three months of his
service. In each of the subsequent months his saving
increases by ` 40 more than the saving of immediately
previous month. His total saving from the start of service
[2011]
will be ` 11040 after
(a) 19 months
(b) 20 months
(c) 21 months
(d) 18 months
A person is to count 4500 currency notes. Let an denote
the number of notes he counts in the nth minute. If a1 = a2
= ... = a10 = 150 and a10, a11, ... are in an AP with common
difference –2, then the time taken by him to count all notes
is
[2010]
¥
45.
If f (x + y) = f (x) f (y) and
å f ( x) = 2 , x, y Î N, where N is
x =1
the set of all natural numbers, then the value of
f (4)
is :
f (2)
[Sep. 06, 2020 (I)]
(a)
46.
2
3
(b)
1
9
(c)
1
3
(d)
4
9
Let a, b, c, d and p be any non ero distinct real numbers
such that (a2 + b2 + c2)p2 – 2 (ab + bc + cd)p + (b2 + c2 +
d2) = 0. Then :
[Sep. 06, 2020 (I)]
(a) a, c, p are in A.P.
(b) a, c, p are in G.P.
(c) a, b, c, d are in G.P.
(d) a, b, c, d are in A.P.
Downloaded from @Freebooksforjeeneet
M-93
Sequences and Series
47.
Suppose that a function f : R®R satisfies f (x + y) = f(x)f(y)
for all x, y Î R and f (a) = 3. If
S=
n
å f (i) = 363 , then n is
x10 - x + 45a( x - 1)
, then k is equal to :
x -1
i =1
equal to ______.
48.
(a) – 5
[NA Sep. 06, 2020 (II)]
If 210 + 29×31 + 28×32 + . . . . + 2×39 + 310 = S – 211 then S is
equal to:
[Sep. 05, 2020 (I)]
(a) 311 – 212
1 49
(3 - 1)
26
(b)
2 50
(3 - 1)
(c)
13
50.
52.
53.
56.
100
(a) (-¥, - 9] È [3, ¥)
(b) [-3, ¥)
(c) (-¥, - 3] È [9, ¥)
(d) (-¥, 9]
to :
(a) 300
58.
59.
x + y - xy
(1 + x)(1 + y )
(b)
2
2
3
x + y + xy
(1 + x)(1 + y )
x + y + xy
(d)
(1 - x )(1 - y )
{x + ka} + {x 2 + (k + 2)a} + {x 3 + (k + 4)a}
+{x4 + (k + 6)a} + ... where
a¹0
and
x ¹ 1. If
n =1
n =1
(c) 175
¥
¥
n= 0
n=0
is equal
[Jan. 9, 2020 (II)]
(d) 150
p
,
4
then :
[Jan. 9, 2020 (II)]
(a) x(1 + y) = 1
(b) y(1 – x) = 1
(c) y(1 + x) = 1
(d) x(1 – y) = 1
The greatest positive integer k, for which 49k + 1 is a
factor of the sum 49125 + 49124 + ... + 492 + 49 + l, is:
[Jan. 7, 2020 (I)]
(a) 32
(b) 63
(c) 60
(d) 65
Let a1, a2, a3, ... be a G. P. such that a1 < 0, a1 + a2 = 4 and
å
i =1
ai = 4l, then l is equal to:
[Jan. 7, 2020 (II)]
(a) –513
(b) –171
(c) 171
(d)
511
3
The coefficient of x7 in the expression
(1 + x)10 + x(l + x)9 + x2(l + x)8 + ... + x10 is:
[Jan. 7, 2020 (II)]
(a) 210
(b) 330
(c) 120
(d) 420
61. If a, b and g are three consecutive terms of a nonconstant G.P. such that the equations ax 2 + 2bx + g = 0
and x 2 + x – 1 = 0 have a common root, then a (b+ g) is
equal to :
[April 12, 2019 (II)]
62.
Let S be the sum of the first 9 terms of the series :
200
å a2n = 100, then å an
2n
n
If x = å (-1) tan2n q and y = å cos q, for 0 < q <
(a) 0
x + y - xy
(c)
(1 - x )(1 - y )
100
9
60.
( x + y ) + ( x + xy + y ) + ( x + x y + xy + y ) + .... is :
3
and
(b) 225
a3 + a4 = 16. If
If | x |< 1, | y |< 1 and x ¹ y, then the sum to infinity of the
following series
[Sep. 02, 2020 (I)]
2
å a2n+1 = 200
n =1
57.
æ1 1 1
ö
Th e value of (0.16) log 2.5 ç + 2 + 3 + ... to ¥ ÷ is
3
3
3
è
ø
equal to _________ .
[NA Sep. 03, 2020 (I)]
The sum of the first three terms of a G.P. is S and their
product is 27. Then all such S lie in : [Sep. 02, 2020 (I)]
(a) 2 2
(b)
(c) 1
(d) 2
Let an be the nth term of a G.P. of positive terms.
If
1 50
(3 - 1)
26
1 50
(3 - 1)
(d)
13
2
54.
1
1
24
1
Let a and b be the roots of x2 - 3x + p = 0 and g and d be
(a)
1
The product 2 4 . 416 . 8 48 . 16128 ... to ¥ is equal to:
[Jan. 9, 2020 (I)]
the roots of x2 - 6 x + q = 0. If a, b, g, d form a geometric
progression. Then ratio (2q + p) : (2q – p) is :
[Sep. 04, 2020 (I)]
(a) 3 : 1
(b) 9 : 7
(c) 5 : 3
(d) 33 : 31
51.
1
[Sep. 02, 2020 (II)]
(d) 3
(b) 311
311 10
+2
(d) 2×311
2
If the sum of the second, third and fourth terms of a positive
term G.P. is 3 and the sum of its sixth, seventh and eighth
terms is 243, then the sum of the first 50 terms of this G.P. is:
[Sep. 05, 2020 (II)]
(a)
(c) – 3
1
55.
(c)
49.
(b) 1
(b) ab
(c) ag
(d) bg
Let a, b and c be in G.P. with common ratio r, where a ¹ 0
1
. If 3a, 7b and 15c are the first three terms of
2
an A.P., then the 4th term of this A.P. is :
[April 10, 2019 (II)]
and 0 < r £
(a)
2
a
3
(b) 5 a
(c)
7
a
3
Downloaded from @Freebooksforjeeneet
(d) a
EBD_8344
M-94
63.
Mathematics
If three distinct numbers a, b, c are in G.P. and the equations
ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a common root,
then which one of the following statements is correct?
consecutive terms of a G.P., then
[April 08, 2019 (II)]
(a)
d e f
, , are in A.P..
a b c
d e f
, , are in G.P..
a b c
The product of three consecutive terms of a G.P. is 512. If
4 is added to each of the first and the second of these
terms, the three terms now form an A.P. Then the sum of
the original three terms of the given G.P. is :
[Jan. 12, 2019 (I)]
(a) 36
(b) 32
(c) 24
(d) 28
Let a and b be the roots of the quadratic equation
x2 sinq – x (sinq cosq + 1) + cosq = 0 (0 < q < 45 ), and
(c) d, e, f are in G.P.
64.
65.
(b) d, e, f are in A.P.
¥
a < b. Then
æ
å çç an +
n =0 è
(d)
1
7
(c)
(d) 4
2
13
70. If a, b and c be three distinct real numbers in G.P. and
a + b + c = xb, then x cannot be:
[Jan. 09, 2019 (I)]
(a) – 2
(b) – 3
(c) 4
(d) 2
71. If b is the first term of an infinite G. P whose sum is five,
then b lies in the interval.
[Online April 15, 2018]
(a) (– ¥, – 10)
(b) (10, ¥)
(c) (0, 10)
(d) (– 10, 0)
(a) 2
73.
[Jan. 11, 2019 (II)]
(a)
1
1
1 - cos q 1 + sin q
1
1
+
(c)
1 - cos q 1 + sin q
66.
(b)
a < b < c and a + b + c =
1
1
(d)
1 + cos q 1 - sin q
74.
[Jan. 11, 2019 (I)]
67.
(b) 4(52)
(c) 53
(d) 2(52)
27
. Then the
19
[Jan. 11, 2019 (I)]
is 3 and the sum of the cubes of its terms is
common ratio of this series is :
(a)
68.
1
3
(b)
2
3
(c)
2
9
(d)
æ q + 1ö æ q + 1ö
æ q + 1ö
Tn = 1 + ç
÷ +ç
÷ + ... + ç
÷
è 2 ø è 2 ø
è 2 ø
69.
4
9
Let Sn = 1 + q + q2 + .... + qn and
2
75.
76.
n
where q is a real number and q ¹ 1. If
101C + 101C .S + .... + 101C
.
101 S100 = aT100, then a is
1
2 1
equal to :
[Jan. 11, 2019 (II)]
99
(a) 2
(b) 202
(c) 200
(d) 2100
th
th
th
Let a, b and c be the 7 , 11 and 13 terms respectively
of a non-constant A.P. If these are also the three
3
, then the value of a is
4
[Online April 15, 2018]
(a)
1
1
4 3 2
(b)
1
1
4 4 2
(c)
1
1
4
2
(d)
1
1
4 2 2
If the 2nd, 5th and 9th terms of a non-constant A.P. are in
G.P., then the common ratio of this G.P. is :
[2016]
8
4
7
(c)
(d)
5
3
4
Let = 1 + ai be a complex number, a > 0, such that 3 is
areal number. Then the sum 1 + + 2 + .... + 11 is equal to:
(Online April 10, 2016)
(a) 1
The sum of an infinite geometric series with positive terms
n
3
æ3ö æ3ö æ 3ö
æ 3ö
Let A n = ç ÷ - ç ÷ + ç ÷ - ... + (-1)n - 1 ç ÷ and
è4ø è4ø è 4ø
è 4ø
Bn = 1 – An. Then, the least odd natural number p, so that
Bn > An, for all n ³ p is
[Online April 15, 2018]
(a) 5
(b) 7
(c) 11
(d) 9
If a, b, c are in A.P. and a2, b2, c2 are in G.P. such that
1
1
+
1 + cos q 1 - sin q
a3
a9
Let a1, a2, ..., a10 be a G.P. If a = 25, then a equals :
5
1
(a) 54
(b)
2
72.
( -1)n ö÷
is equal to :
bn ÷ø
a
is equal to:
c
[Jan. 09, 2019 (II)]
(b)
(a) 1365 3i
(b) -1365 3i
(c) -1250 3i
(d) 1250 3i
If m is the A.M. of two distinct real numbers l and n(l, n >
1) and G1, G2 and G3 are three geometric means between l
and n, then G14 + 2G 42 + G 34 equals.
(a) 4 lmn2 (b) 4 l2m2n2 (c) 4 l2 mn
77.
[2015]
(d) 4 lm2n
The sum of the 3rd and the 4th terms of a G.P. is 60 and
the product of its first three terms is 1000. If the first term
of this G.P. is positive, then its 7th term is :
[Online April 11, 2015]
(a) 7290
(b) 640
(c) 2430
Downloaded from @Freebooksforjeeneet
(d) 320
M-95
Sequences and Series
78.
Three positive numbers form an increasing G. P. If the middle
term in this G.P. is doubled, the new numbers are in A.P.
then the common ratio of the G.P. is:
[2014]
(a)
(b) 2 + 3
(a) 2 - 3
(c)
(d) 3 + 2
2+ 3
The least positive integer n such that
5
(
1
1- 5
2
(c)
79.
10
87.
2 2
2
1
1 - - 2 - .... - n -1 <
, is: [Online April 12, 2014]
3 3
100
3
(a) 4
80.
81.
(b) 5
(c) 6
(1000 )!
(c)
82.
84.
85.
86.
(b)
(1001)!
( 51)!( 950 )!
(d)
( 49 )!( 951)!
(b) 8
(c) 4
(a2 + b2 + c2) p2 – 2p (ab + bc + cd) + (b2 + c2 + d2) £ 0,
then
[Online May 12, 2012]
(a) a, b, c, d are in A.P. (b) ab = cd
(c) ac = bd
(d) a, b, c, d are in G.P.
The difference between the fourth term and the first term
of a Geometrical Progresssion is 52. If the sum of its first
three terms is 26, then the sum of the first six terms of the
progression is
[Online May 7, 2012]
(a) 63
(b) 189
(c) 728
(d) 364
The first two terms of a geometric progression add up to
12. the sum of the third and the fourth terms is 48. If the
terms of the geometric progression are alternately positive
and negative, then the first term is
[2008]
(a) –4
(b) –12
(c) 12
(d) 4
In a geometric progression consisting of positive terms,
each term equals the sum of the next two terms. Then the
common ratio of its progression is equals
[2007]
1
2
(d)
1
5
2
)
5 -1
2k p
2k p ö
+ i cos
÷ is
11
11 ø
[2006]
(a) i
(b) 1
(c) – 1
(d) – i
If the expan sion in powers of x of the fun ction
a0 + a1 x + a2 x 2 + a3 x 3 ......
then an is
[2006]
89.
n
n
(a) b - a
b-a
n
n
(b) a - b
b-a
n +1
- bn +1
(c) a
b-a
(d)
bn +1 - a n +1
b-a
Let two numbers have arithmetic mean 9 and geometric
mean 4. Then these numbers are the roots of the quadratic
equation
[2004]
(a) x 2 - 18 x - 16 = 0
(b) x 2 - 18 x + 16 = 0
2
(c) x + 18 x - 16 = 0
90.
(d) x 2 + 18 x + 16 = 0
Sum of infinite number of terms of GP is 20 and sum of their
square is 100. The common ratio of GP is
[2002]
(a) 5
(d) 2
If a, b, c, d and p are distinct real numbers such that
æ
å çè sin
1
is
(1 - ax)(1 - bx)
(1001)!
(50 )!( 951)!
Given a sequence of 4 numbers, first three of which are in
G.P. and the last three are in A.P. with common difference
six. If first and last terms of this sequence are equal, then
the last term is :
[Online April 25, 2013]
(a) 16
83.
88.
(d) 7
(1000 )!
(50 )!( 950 )!
The value of
k =1
In a geometric progression, if the ratio of the sum of first 5
terms to the sum of their reciprocals is 49, and the sum of
the first and the third term is 35. Then the first term of this
geometric progression is:
[Online April 11, 2014]
(a) 7
(b) 21
(c) 28
(d) 42
50
The coefficient of x in the binomial expansion of
(1 + x) 1000 + x (1 + x) 999 + x 2 (1 + x) 99 8 + ....
+ x1000 is:
[Online April 11, 2014]
(a)
)
(
(b)
91.
92.
(b) 3/5
(c) 8/5
(d) 1/5
Fifth term of a GP is 2, then the product of its 9 terms is
[2002]
(a) 256
(b) 512
(c) 1024
(d) none of these
æ pö
The sum of all values of qÎ ç 0, ÷ satisfying
è 2ø
sin22q + cos4 2q =
(a) p
(b)
3
is:
4
5p
4
[Jan. 10, 2019 (I)]
(c)
p
2
(d)
3p
8
Harmonic Progression, Relation
TOPIC Đ Between A. M., G. M. and H.M. of
two Positive Numbers
93.
If m arithmetic means (A.Ms) and three geometric means
(G.Ms) are inserted between 3 and 243 such that 4th A.M.
is equal to 2nd G.M., then m is equal to ___________.
[Sep. 03, 2020 (II)]
Downloaded from @Freebooksforjeeneet
EBD_8344
M-96
94.
Mathematics
If the arithmetic mean of two numbers a and b, a > b > 0, is
a+b
five times their geometric mean, then
is equal to :
a-b
[Online April 8, 2017]
(a)
95.
6
2
(b)
3 2
4
If A > 0, B > 0 and A + B =
tanA + tanB is :
96.
(a)
3- 2
(c)
2
3
(c)
7 3
12
(d)
5 6
12
p
, then the minimum value of
6
[Online April 10, 2016]
(b) 4 - 2 3
97.
f :R ® R
be a function which satisfies
f ( x + y ) = f ( x) + f ( y ), " x, y ÎR. If f (a) = 2 and
å f (k ), n Î N , then the value of n, for which
g(n) = 20, is :
(a) 5
(b) 20
(c) 4
1
1
1
:G
and . If
a
b
M
[Online April 12, 2014]
(a) 1 : 4
(b) 1 : 2
(c) 2 : 3
(d) 3 : 4
If a1, a2, .........., an are in H.P., then the expression
a1a2 + a2a3 + .......... + an–1an is equal to
[2006]
(a) n(a1 - an )
(b) (n - 1)(a1 - an )
(c) na1an
(d) (n - 1)a1an
¥
If x =
¥
n(n + 1)(2 n + 1)
is equal to __________.
4
n =1
å
[Jan. 8, 2020 (II)]
n =0
n=0
n=0
where a, b, c are in
A.P and |a | < 1, | b | < 1, | c | < 1 then x, y, z are in [2005]
(a) G. P.
(b) A.P.
(c) Arithmetic - Geometric Progression
(d) H.P.
100. If the sum of the roots of the quadratic equation
ax 2 + bx + c = 0 is equal to the sum of the squares of
a b
c
their reciprocals, then , and are in
c a
b
(a) Arithmetic - Geometric Progression
(b) Arithmetic Progression
(c) Geometric Progression
(d) Harmonic Progression.
20
104. The sum
å (1 + 2 + 3 + ... + k )
is _______.
k =1
[Jan. 8, 2020 (I)]
105. If the sum of the first 40 terms of the series, 3 + 4 + 8 + 9
+ 13 + 14 + 18 +19 + ... is (102)m, then m is equal to:
[Jan. 7, 2020 (II)]
(a) 20
(b) 25
(c) 5
(d) 10
106. For x Î R, let [x] denote the greatest integer < x, then the
sum of the series
¥
å a n , y = å bn , z = å c n
[Sep. 02, 2020 (II)]
(d) 9
7
103. The sum,
Let G be the geometric mean of two positive numbers a
is 4 : 5, then a : b can be:
99.
102. Let
(d) 270
and b, and M be the arithmetic mean of
98.
[Sep. 04, 2020 (I)]
(b) (11, 103)
(d) (11, 97)
(a) (10, 97)
(c) (10, 103)
k =1
[Online April 9, 2016]
(c) 258
19) = a - 220b, then an ordered pair (a, b) is equal to :
( n -1)
(d) 2 - 3
(b) 216
101. If 1 + (1 – 22 × 1) + (1 – 42 × 3) + (1 – 62 × 5) + . . . . . . . + (1 – 202 ×
g (n) =
Let x, y, be positive real numbers such that x + y + = 12
and x3y4 5 = (0.1) (600)3. Then x3 + y3 + 3 is equal to :
(a) 342
Arithmetic-Geometric Sequence
TOPIC Ė (A.G.S.), Some Special Sequences
[2003]
é 1ù é 1 1 ù é 1 2 ù
é 1 99 ù
êë - 3 úû + êë - 3 - 100 úû + êë- 3 - 100 úû + L + êë - 3 - 100 úû is
[April 12, 2019 (I)]
(a) –153
(b) –133
(c) –131
(d) –135
3 ´ 13 5 ´ (13 + 23 ) 7 ´ (13 + 23 + 33 )
+
+
) + …...
12
12 + 22
12 + 22 + 32
upto 10th term, is :
[April 10, 2019 (I)]
107. The sum
(a) 680
108. The sum 1 +
(b) 600
(c) 660
(d) 620
13 + 23 13 + 23 + 33
+
+ ¼¼ +
1+ 2
1+ 2 + 3
13 + 23 + 33 + ... + 153 1
- (1 + 2 + 3 + ... + 15 is equal to :
1 + 2 + 3 + ... + 15
2
(a) 620
(b) 1240
(c) 1860
Downloaded from @Freebooksforjeeneet
[April 10, 2019 (II)]
(d) 660
M-97
Sequences and Series
109. The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 + ….. upto
11th term is:
[April 09, 2019 (II)]
(a) 915
(b) 946
(c) 945
(d) 916
110. Some identical balls are arranged in rows to form an
equilateral triangle. The first row consists of one ball, the
second row consists of two balls and so on. If 99 more
identical balls are added to the total number of balls used
in forming the equilateral triangle, then all these balls can
be arranged in a square whose each side contains exactly
2 balls less than the number of balls each side of the triangle
contains. Then the number of balls used to form the
equilateral triangle is:
[April 09, 2019 (II)]
(a) 157
2) 262
(c) 225
(d) 190
20
111. The sum
(a) 2 –
1
å k 2k
is equal to :
k =1
3
11
(b) 1 – 20
17
2
2
112. Let Sk =
[April 08, 2019 (II)]
(c) 2 –
11
219
(d) 2 –
21
220
1 + 2 + 3 + ..... + k
.
2
2
If S12 + S22 + ..... + S10
=
5
A. Then A is equal to
12
[Jan. 12, 2019 (I)]
(a) 283
(b) 301
(c) 303
(d) 156
113. If the sum of the first 15 terms of the series
3
3
3
3
æ3ö æ 1ö æ 1ö
3 æ 3ö
ç ÷ + ç1 ÷ + ç 2 ÷ + 3 + ç 3 ÷ + ....... is equal to
è4ø è 2ø è 4ø
è 4ø
225 k then k is equal to :
[Jan. 12, 2019 (II)]
(a) 108
(b) 27
(c) 54
(d) 9
114. The sum of the following series [Jan. 09, 2019 (II)]
1+ 6 +
9 (1 + 2 + 3 ) 12 (1 + 2 + 3 + 4 )
+
7
9
2
2
2
2
2
2
15 (12 + 22 + .... + 52 )
+... up to 15 terms, is:
11
(a) 7520
(b) 7510
(c) 7830
(d) 7820
115. The sum of the first 20 terms of the series
(a) 38 +
1
220
[Online April 16, 2018]
(b) 39 +
If B - 2A = 100 l , then l is equal to :
[2018]
(a) 248
(b) 464
(c) 496
(d) 232
117. Let a, b, c Î R. If f(x) = ax2 + bx + c is such that a + b + c = 3
and f(x + y) = f(x) + f(y) + xy, " x, y Î R, then
1
219
1
1
(c) 39 + 20
(d) 38 + 19
2
2
116. Let A be the sum of the first 20 terms and B be the sum of
the first 40 terms of the series
10
å f ( n ) is equal
n =1
to :
(a) 255
[2017]
(b) 330
(c) 165
(d) 190
1
1+ 2
1+ 2 + 3
118. Let Sn = 3 + 3 3 + 3 3 3 + .....
1 1 +2 1 +2 +3
1 + 2 + ...... + n
, If 100 Sn = n, then n is equal to :
+ 3 3
1 + 2 + ....... + n3
[Online April 9, 2017]
(a) 199
(b) 99
(c) 200
(d) 19
119. If the sum of the first n terms of the series
3 + 75 + 243 + 507 + ...... is 435 3 , then n equals :
[Online April 8, 2017]
(a) 18
(b) 15
(c) 13
(d) 29
120. If the sum of the first ten terms of the series
2
2
2
2
16
æ 3ö æ 2ö æ 1ö
2 æ 4ö
ç 1 5 ÷ + ç 2 5 ÷ + ç 3 5 ÷ + 4 + ç 4 5 ÷ + ......., is 5 m,
è ø è ø è ø
è ø
then m is equal to :
(a) 100
(b) 99
[2016]
(c) 102
(d) 101
121. For x Î R, x ¹ -1 , if (1 + x)2016 + x (1 + x)2015 + x2
2016
(1 + x)2014 + .... + x2016 =
å ai xi , then a17 is equal to :
i =0
[Online April 9, 2016]
2
+
3 7 15 31
1 + + + + + ... is?
2 4 8 16
12 + 2 × 22 + 32 + 2.4 2 + 52 + 2.62 + ...
(a)
2016!
(b) 17! 1999!
2017!
17!2000!
2016!
2017!
(d)
16!
2000!
122. The sum of first 9 terms of the series.
(c)
13 13 + 23 13 + 23 + 33
+
+
+ ....
1 1+ 3
1+ 3 + 5
(a) 142
(b) 192
(c) 71
5
123. If
1
å n(n + 1)(n + 2)( n + 3) =
n =1
[2015]
(d) 96
k
, then k is equal to
3
[Online April 11, 2015]
(a)
1
6
(b)
17
105
(c)
55
336
Downloaded from @Freebooksforjeeneet
(d)
19
112
EBD_8344
M-98
Mathematics
30
124. The value of
å ( r + 2) ( r - 3)
r =16
(a) 7770
[Online April 10, 2015]
(c) 7775
(d) 7780
(b) 7785
( ) + 3(11)
125. If (10 ) + 2 (11) 10
9
1
8
2
(10 )
7
9
9
(a) 100
(b) 110
121
10
(c)
k =1
[2014]
(d)
441
100
126. The number of terms in an A.P. is even; the sum of the odd
terms in it is 24 and that the even terms is 30. If the last term
exceeds the first term by 10
the A.P. is:
(a) 4
(b) 8
127. If the sum
3
12
+
5
12 + 22
+
7
12 + 22 + 32
1
, then the number of terms in
2
[Online April 19, 2014]
(c) 12
(d) 16
+ ...... + up to 20 terms is equal
k
to
, then k is equal to:
[Online April 9, 2014]
21
(a) 120
(b) 180
(c) 240
(d) 60
128. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,.....,
is
[2013]
number n.
[2012]
(a) Statement-1 is false, Statement-2 is true.
(b) Statement-1 is true, statement-2 is true; statement-2 is
a correct explanation for Statement-1.
(c) Statement-1 is true, statement-2 is true; statement-2 is
not a correct explanation for Statement-1.
(d) Statement-1 is true, statement-2 is false.
134. If the sum of the series 12 + 2.22 + 32 + 2.42 + 52 + ... 2.62 +...
2
of the series, when n is odd, is
(a) n2(n + 1)
135. The sum of the series 1 +
(a)
(c)
7
(179 + 10 -20 )
81
(d)
7
(99 + 10 -20 )
9
(c) n +
1
+
5
2
1 +2
2
is:
7
11
(b)
2
4
132. The sum of the series :
(a)
+
7
+ .... upto 11-terms
1 + 22 + 32
[Online April 22, 2013]
2
(c)
11
60
(d)
2
11
[Online April 9, 2013]
1
1
1+
+
+ ....... upto 10 terms, is :
1+ 2 1+ 2 + 3
4 10 28
+ +
+ ... upto n terms is
3 9 27
[Online May 19, 2012]
7
(99 - 10-20 )
9
2
n 2 ( n - 1)
2
(d) n2(n – 1)
2
(b)
3
(b)
n2 ( n + 1)
(c)
7
(179 - 10-20 )
81
129. The value of l2 + 32 + 52 + .......................+ 252 is :
[Online April 25, 2013]
(a) 2925
(b) 1469
(c) 1728
(d) 1456
130. The sum of the series :
(b)2 + 2(d)2 + 3(6)2 + ... upto 10 terms is :
[Online April 23, 2013]
(a) 11300 (b) 11200
(c) 12100
(d) 12300
n ( n + 1)
, then the sum
2
[Online May 26, 2012]
upto n terms, when n is even, is
(a)
131. The sum
å ( k 3 - ( k - 1)3 ) = n3, for any natural
n
Statement-2:
+ .....
+10 (11) = k (10 ) , then k is equal to:
16
20
18
22
(b)
(c)
(d)
9
11
11
13
133. Statement-1: The sum of the series 1 + (1 + 2 + 4) +
(4 + 6 + 9) + (9 + 12 + 16) + .... + (361 + 380 + 400) is 8000.
(a)
is equal to :
7
1
2
n + - n -1
6
6 3.2
(b)
1
1
2 2.3 n
5
7
1
n - + n -1
3
6 2.3
1
1
(d) n - - n -1
3 3.2
136. The sum of the series
1
1+ 2
+
1
2+ 3
+
1
3+ 4
+ ...
upto 15 terms is
[Online May 12, 2012]
(a) 1
(b) 2
(c) 3
(d) 4
137. The sum of the series
12 + 2.22 + 32 + 2.42 + 52 + 2.62 + .... + 2(2m)2 is
[Online May 7, 2012]
(a) m(2m + 1)2
(b) m2(m + 2)
(c) m2(2m + 1)
(d) m(m + 2)2
138. The sum to infinite term of the series
1+
2 6 10 14 ... is
+
+ +
+
3 32 33 34
(a) 3
(b) 4
[2009]
(c) 6
Downloaded from @Freebooksforjeeneet
(d) 2
M-99
Sequences and Series
1 1 1
- + - ....... upto infinity is
2! 3! 4!
[2007]
139. The sum of series
(a)
-
1
2
+
(b)
1
2
e
e
140. The sum of the series
(c) e –2
(d) e –1
1
1
1
1+
+
+
+ ....................ad inf. is
4.2! 16.4! 64.6 !
é n(n + 1) ù
(a) ê
ë 2 úû
(c)
[2005]
2
n(n + 1)2
4
143. If Sn =
n
1
å nC
r=0
and t n =
r
(b)
n2 (n + 1)
2
(d)
3n( n + 1)
2
n
r
å nC
r =0
r
t
, then n is equal to
Sn
[2004]
(a)
e -1
e
(b)
e +1
e
141. The sum of series
(a)
(e2 - 2)
e
2
(c)
e -1
2 e
1 1 1
+ + + ..... is
2! 4! 6!
(b)
(d)
e +1
2 e
(c) n – 1
(d)
1
n
2
[2003]
[2004]
(e - 1)2
2e
2
(e - 1)
(e - 1)
(d)
2e
2
142. The sum of the first n terms of the series
(c)
12 + 2.2 2 + 3 2 + 2.4 2 + 5 2 + 2.6 2 + ...
is
2n – 1
1
n -1
(b)
2
2
144. The sum of the series
(a)
n(n + 1)2
when n is even. When n is odd the sum is
2
[2004]
1
1
1
+
.......... .. up to ¥ is equal to
1.2 2.3 3.4
æ 4ö
èeø
(a) log e ç ÷
(b) 2 log e 2
(d) log e 2
(c) log e 2 - 1
3
3
3
3
3
145. 1 – 2 + 3 – 4 +...+9 =
(a) 425
(b) –425
(c) 475
146. The value of 21/4. 41/8. 81/16 ... ¥ is
(a) 1
(b) 2
(c) 3/2
Downloaded from @Freebooksforjeeneet
[2002]
(d) –475
[2002]
(d) 4
EBD_8344
M-100
1.
Mathematics
(c) Let common difference of series
4.
\ 300 = 1 + (n - 1)d
a1 , a2 , a3 ,......., an be d.
Q a40 = a1 + 39d = -159
...(i)
and a100 = a1 + 99d = -399
...(ii)
Þd =
From equations (i) and (ii),
d = – 4 and a1 = – 3
Since, the common difference of b1, b2, ......, bn is 2 more
than common difference of a1, a2, ......, an.
\ Common difference of b1, b2, b3, ..... is (– 2).
\ n - 1 = 13 or 23
Þ n = 24 and d = 13
a20 = 1 + 19 ´ 13 = 248
s20 =
Þ b1 = 198 - 279 Þ b1 = -81
5.
32sin 2a
81
+ 2sin 2 a = 28
3
3
(a) Given a = 3 and S25 = S40 – S25
Þ 2´
Þ 5(2 + 8d ) = 4(2 + 13d )
x 81
+ = 28
3 x
Þ 10 + 40 d = 8 + 52d
Þ x - 84 x + 243 = 0 Þ x = 81, x = 3
2
Þd =
When x = 81 Þ sin 2a = 2 (Not possible)
When x = 3 Þ a =
p
12
6.
\ a = 30 = 1, d = 14 - 1 = 13
a6 = a + 5d = 1 + 65 = 66.
3.
Q S = 460
Þ log 7 ( x × x × x × ......x ) = 460
3
4
21
Þ log7 x(2+3+ 4......21) = 460
Þ (2 + 3 + 4 + ..... + 21) log 7 x = 460
20
(2 + 21) log 7 x = 460
Þ
2
460
Þ log 7 x =
= 2 Þ x = 7 2 = 49
230
1
6
(b) Sn = 20 + 19 3 + 19 1 + 18 4 + ....
5
5
5
Q Sn = 488
488 =
n é æ 100 ö
æ 2ö ù
2ç
÷ + ( n - 1) çè - ÷ø ú
2 êë è 5 ø
5 û
488 =
n
(101 - n) Þ n2 - 101n + 2440 = 0
2
(a) S = log 7 x 2 + log 7 x 3 + log 7 x 4 + ...20 terms
2
25
40
[6 + 24 d ] = [6 + 39 d ]
2
2
Þ 25[6 + 24d ] = 20[6 + 39d ]
Let 32sin 2 a = x
Þ
20
(2 + 19 ´ 13) = 2490.
2
Þ 2S25 = S40
So, 32sin 2a-1 + 34 - 2sin 2a = 28
Þ
(Q15 £ n £ 50)
Þ n = 14 or 24
Þ b1 + 99(-2) = (-3) + 69( -4)
(a) Given that 32sin 2a -1 , 14, 34 - 2sin 2a are in A.P..
299
23 ´ 13
,
=
(n - 1) (n - 1)
Q d is an integer
Q b100 = a70
2.
(d) Given that a1 = 1 and an = 300 and d ÎZ
Þ n = 61 or 40
For n = 40 Þ Tn > 0
For n = 61 Þ Tn < 0
nth term = T61 =
100
æ 2ö
+ (61 - 1) ç - ÷ = -4
è 5ø
5
Downloaded from @Freebooksforjeeneet
M-101
Sequences and Series
7.
(d) Let common difference be d.
Q S11 = 0 \
11.
11
{2a1 + 10 × d } = 0
2
Þ a1 + 5d = 0 Þ d = -
a1
5
...(i)
Now, S = a1 + a3 + a5 + ... + a23
= a1 + (a1 + 2d ) + (a1 + 4d ) + .... + (a1 + 22d )
= 12 a1 + 2d
11 ´ 12
2
é
æ a öù
= 12 êa1 + 11 × ç - 1 ÷ ú
è 5 øû
ë
(From (i))
72
æ 6ö
= 12 ´ ç - ÷ a1 = - a1
è 5ø
5
8.
(14)First common term of both the series is 23 and common
difference is 7 × 4 = 28
Q Last term £ 407
Þ 23 + (n – 1) × 28 £ 407
Þ (n – 1) × 28 £ 384
384
+1
28
Þ n £ 14.71
Hence, n = 14
12.
Þ n£
9.
(d) T10 =
T20 =
(d) Let 5 terms of A.P. be
a – 2d, a – d, a, a + d, a + 2d.
Sum = 25 Þ 5a = 25 Þ a = 5
Product = 2520
(5 – 2d) (5 – d) 5(5 + d) (5 + 2d) = 2520
Þ (25 – 4d 2) (25 – d 2) = 504
Þ 625 – 100d 2 – 25d 2 + 4d 4 = 504
Þ 4d 4 – 125d 2 + 625 – 504 = 0
Þ 4d 4 – 125d 2 + 121 = 0
Þ 4d 4 – 121d 2 – 4d 2 + 121 = 0
Þ (d 2 – 1) (4d 2 – 121) = 0
11
Þ d = ± 1, d = ±
2
-1
11
d = ± 1 and d = –
, does not give
as a term
2
2
11
\ d=
2
\ Largest term = 5 + 2d = 5 + 11 = 16
(c) Given, S4 = 16 and S6 = – 48
Þ 2(2a + 3d) = 16 Þ 2a + 3d = 8
...(i)
And 3[2a + 5d] = – 48 Þ 2a + 5d = – 16
Þ 2d = – 24
[using equation (i)]
Þ d = – 12 and a = 22
10
= (44 + 9(– 12)) = – 320
2
(a) Let the common difference of the A.P. is ‘d’.
Given, a1 + a7 + a16 = 40
Þ a1 + a1 + 6d + a1 + 15d = 40
Þ 3a1 + 21d = 40
\
13.
1
= a + 9d
20
...(i)
1
= a + 19d
10
40
...(i)
3
Now, sum of first 15 terms of this A.P. is,
Þ a1 + 7d =
...(ii)
Solving equations (i) and (ii), we get
S15 =
1
1
a=
,d =
200
200
Þ S200
10.
(b) If 2
200 é 2 199 ù 201
1
=
+
=
= 100
2 êë 200 200 úû
2
2
1– x
+2
1+ x
x
14.
15.
Using AM ³ GM
f(x) ³ 3
(b) a1 + a4 + a7 + .... + a16 = 114
Þ a1 + a16 = 38
Now, a1 + a6 + a11 + a16 = 2(a1 + a16 ) = 2 × 38 = 76
ö
÷
÷
ø
ö æ x 1 ö
÷ +ç3 + x ÷
3 ø
ø è
[Using (i)]
Þ 3(a1 + a 16 ) = 114
, f(x), 3 + 3 are in A.P., then
1
æ
2 f ( x) = 2 ç 2 x + x
2
è
15
[2a1 + 14d] = 15 (a1 + 7d)
2
æ 40 ö
= 15 ç ÷ = 200
è 3 ø
–x
æ 21+ x + 21- x + 3x + 3- x
f ( x) = ç
ç
2
è
S10
A
7Aö
æ
n + n2 ´
(d) Q Sn = ç 50 Þ a1 = 50 – 3S
2 ÷ø
2
è
\ d = a2 – a1 = Sn 2 - Sn1 - Sn1 Þ d =
Now, a50 = a1 + 49 × d
= (50 – 3A) + 49 A = 50 + 46 A
So, (d, a50) = (A, 50 + 46 A)
Downloaded from @Freebooksforjeeneet
A
´2 = A
2
EBD_8344
M-102
16.
Mathematics
(d) Q f(x + y) = f(x) . f(y)
Þ Let f(x) = tx
Q f(1) = 2
\ t' = 2
Þ f(x) = 2x
tn = a + (n – 1)d
\ t19 = a + 18d = 0
\ a = –18d
t49
a + 48d
t29 = a + 28d
\
10
Since,
å f (a + k ) = 16(210 - 1)
k =1
=
10
a +k
= 16(210 - 1)
Then, å 2
k =1
21.
10
Þ 2a å 2k = 16(210 - 1)
k =1
a
Þ2 ´
17.
18.
19.
((210 ) - 1) ´ 2
= 16 ´ (210 - 1)2.2a = 16
(2 - 1)
Þa=3
(d) Let three terms of A.P. are a – d, a, a + d
Sum of terms is, a – d + a + a + d = 33 Þ a = 11
Product of terms is, (a – d) a (a + d) = 11(121 – d2) = 1155
Þ 121 – d2 = 105 Þ d = ± 4
if d = 4
T11 = T1 + 10d = 7 + 10(4) = 47
if d = – 4
T11 = T1 + 10d = 15 + 10 (– 4) = – 25
(d) Q 91 = 13 × 7
Then, the required numbers are either divisible by 7 or 13.
\ Sum of such numbers = Sum of no. divisible by 7 +
sum of the no. divisible by 13 – Sum of the numbers
divisible by 91
= (105 + 112 + ... + 196) + (104 + 117 + ... +195) – 182
= 2107 + 1196 – 182 = 3121
(b) Since nC4, nC5 and nC6 are in A.P.
2nC5 = nC4 + nC6
n
2=
n
C4
C5
+
n
C6
n
C5
5
n-5
+
5
n-4
Þ 12(n – 4) = 30 + n2 – 9n + 20
Þ n2 – 21n + 98 = 0
(n – 7) (n – 14) = 0
(n – 7) (n – 14) = 0
n = 7, n = 14
20.
–18d + 48d 30d
=
=3
–18d + 28d 10d
t49 : t29 = 3 : 1
(d) Two digit positive numbers which when divided by 7
yield 2 as remainder are 12 terms i.e,16, 23, 30, ..., 93
Two digit positive numbers which when divided by 7 yield
5 as remainder are 13 terms i.e,12, 19, 26, ..., 96
By using AP sum of 16, 23, ..., 93, we get
S1 = 16 + 23 + 30 + ... + 93 = 654
By using AP sum of 12, 19, 26, ..., 96, we get
S1 = 12 + 19 + 26 + ... + 96 = 702
\ required Sum = S1 + S2 = 654 + 702 = 1356
30
22. (a) S = å ai =
30
[2a1 + 29d]
2
15
15
[2a1 + 28d ]
2
i =1
T = å a(2i -1) =
i =1
Since, S – 2T = 75
Þ 30 a1 + 435d – 30a1 – 420d = 75
Þd=5
Also, a5 = 27 Þ a1 + 4d = 27 Þ a1 = 7,
Hence, a10 = a1 + 9d = 7 + 9 ´ 5 = 52
23.
(c) Q
(c) Let first term and common difference of AP be a and d
respectively, then
1 1 1
1
, , ,.....,
are in A.P
x1 x2 x3
xn
x1 = 4 and x21 = 20
Let 'd ' be the common difference of this A.P
\ its 21st term =
Þd=
2=
¼(i)
1
1
= + [(21 – 1) ´ d ]
x21 x1
1 æ 1 1ö
´ç
– ÷
20 è 20 4 ø
Þ d=–
Also xn > 50 (given).
\
1
1
= + [(n – 1) ´ d ]
xn x1
Þ xn =
\
x1
1 + (n – 1) ´ d ´ x1
x1
> 50
1 + ( n – 1) ´ d ´ x1
Downloaded from @Freebooksforjeeneet
1
100
M-103
Sequences and Series
Þ
26.
4
> 50
æ 1 ö
1 + (n – 1) ´ ç –
÷´4
è 100 ø
4
æ 1 ö
Þ 1 + ( n – 1) ´ ç –
÷´4<
50
è 100 ø
1
[(15 a – 3b)2 + (3b – 5c)2 + (5c – 15a)2 ] = 0
2
It is possible when 15a – 3b = 0, 3b – 5 c = 0 and
5c – 15a = 0
Þ 15a = 3b Þ b = 5a
1
23
(n – 1) < –
100
100
Þ n – 1 > 23 Þ n > 24
Therefore, n = 25.
Þ –
Þ
25
1
åx
i =1 i
24.
=
Þ b=
25 éæ
1ö
æ 1 ö ù 13
ç 2 ´ ÷ + (25 – 1) ´ ç –
÷ú =
ê
2 ëè
4ø
è 100 ø û 4
Þ x5 = x3 + 2d1 = 8 + 2 ×
12
= 2.4
5
27.
12
= 12.8
5
Suppose d 2 is the common difference of the A.P
1 1
1
, ,.....
h1 h2
hn then
5d2 =
28.
1 1 –3
–3
– =
Þ d2 =
20 8 40
200
29.
Þ x5 . h10 = 12.8 × 200 = 2560
12
Þ a1 + 24d = 32
Now, a9 + a43 = 66 Þ 2a1 + 50d = 66
From eq. (i) & (ii) we get; d = 1 and a1 = 8
Also,
17
17
r =1
r =1
å a 2r = å [8 + (r - 1)1]2 = 140 m
17
Þ
å (r + 7)
Þ
å (r 2 + 14r + 49) = 140 m
2
= 140 m
r =1
17
r =1
æ 17 ´18 ´ 35 ö
æ 17 ´18 ö
÷ + 14 ç
÷ + (49 ´17) = 140
Þ ç
è
ø
è 2 ø
6
Þ m = 34
17
[a + a ] = 17 × 18 = 306
2 1 17
(b) Let p, q, r are in AP
Þ 2q = p + r
...(i)
...(ii)
Þ
...(i)
1 1
+ =4
a b
Given
13
(b) Q å a 4k +1 = 416 Þ [2a1 + 48d] = 416
2
k =0
c 5c 6c
+
=
3 3
3
Þ a + b = 2c
Þ b, c, a are in A.P.
(a) By Arithmetic Mean:
a + c = 2b
Consider a = b = c = 2
Þ abc = 8
Þ a + b = 2b
\ minimum possible value of b = 2
(a) a3 + a7 + a11 + a15 = 72
(a3 + a15) + (a7 + a11) = 72
a3 + a15 + a7 + a11 = 2 (a1+ a17)
a1 + a17 = 36
S17 =
1
1
1
Q
=
+ 3d 2 =
Þ h10 = 200
h10 h7
200
25.
5c
c
,a =
3
3
Þ a+b=
(a) Suppose d1 is the common difference of the A.P.
x1, x2, .... xn then
Q x8 – x3 = 5d1 = 12 Þ d1 =
(c) We have
9(25a2 + b2) + 25 (c2 – 3ac) = 15b (3 a + c)
Þ 225a2 + 9b2 + 25c2 – 75ac = 45ab + 15bc
Þ (15a)2 + (3b)2 + (5c)2 – 75ac – 45ab– 15 bc = 0
a +b
=4
ab
We have a + b = – q/p and ab =
q
p
= 4 Þ q = - 4r
r
p
r
p
-
Þ
....(ii)
From (i), we have
2( – 4r) = p + r
p = –9r
....(iii)
Now, | a - b | = (a + b)2 - 4ab
2
æ -q ö
4r
= ç ÷ =
è pø
p
q 2 - 4 pr
| p|
Downloaded from @Freebooksforjeeneet
EBD_8344
M-104
Mathematics
From (ii) and (iii)
16r 2 + 36r 2
2 13
=
| -9 r |
9
(b) Given n = 20; S20 = ?
Series (1) ® 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47,
51, 55, 59...
Series (2) ® 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56,
61, 66, 71.
The common terms between both the series are
11, 31, 51, 71...
Above series forms an Arithmetic progression (A.P).
Therefore, first term (a) = 11 and
common difference (d) = 20
n
Now, Sn = [2a + (n - 1)d ]
2
20
S20 =
[2 × 11 + (20 – 1) 20]
2
S20 = 10 [22 + 19 × 20]
S20 = 10 × 402 = 4020
\ S20 = 4020
(c) Let a be the first term and d be the common
difference of given A.P.
Second term, a+ d = 12
...(i)
Sum of first nine terms,
33.
(b) Given Sn = 2n + 3n2
Now, first term = 2 + 3 = 5
second term = 2(2) + 3(4) = 16
third term = 2(3) + 3 (9) = 33
Now, sum given in option (b) only has the same first term
and difference between 2nd and 1st term is double also.
34.
(b)
a1 + a2 + a3 + ...... + a p p3
=
a1 + a2 + a3 + ...... + aq q 3
Þ
a1 + a2 8
=
Þ a1 + (a1 + d) = 8a1
1
a1
=
30.
31.
9
(2a + 8d ) = 9( a + 4d )
2
Given that S9 is more than 200 and less than 220
Þ 200 < S9 < 220
Þ 200 < 9 (a + 4d) < 220
Þ 200 < 9 (a + d + 3d) < 220
Putting value of (a + d) from equation (i)
200 < 9 (12 + 3d) < 220
Þ 200 < 108 + 27d < 220
Þ 200 – 108 < 108 + 27d – 108 < 220 – 108
Þ 92 < 27d < 112
Possible value of d is 4
27 × 4 = 108
Thus, 92 < 108 < 112
Putting value of d in equation (i)
a + d = 12
a = 12 – 4 = 8
4th term = a + 3d = 8 + 3 × 4 = 20
(c) If d be the common difference, then
Þ d = 6a1
=
m = a4 – a7 + a10 = a4 – a7 + a7 + 3d = a7
S13 =
13
13
[a1 + a13 ] = [ a1 + a7 + 6d ]
2
2
13
[2a7 ] = 13a7 = 13 m
=
2
a1 + 5 ´ 6a1
1 + 30
31
=
=
a1 + 20 ´ 6a1 1 + 120 121
35.
(d) Let 'a' is the first term and 'd ' is the common difference
of an A.P.
Now, According to the question
100a100 = 50a50
100 (a + 99d) = 50 (a + 49d)
Þ 2a + 198 d = a + 49d Þ a + 149 d = 0
Hence, T150 = a + 149 d = 0
36.
(d) Given :
37.
(a) Since, sec (q – f), secq and sec (q + f) are in A.P.,
S9 =
32.
a + 5d
a6
= 1
a21 a1 + 20d
Now
a p + aq
a + as
= r
2
2
Þ a + (p – 1) d +a + (q – 1)d
= a + (r – 1)d + a + (s – 1)d
Þ 2a + (p + q)d – 2d = 2a + (r + s) d – 2d
Þ (p + q)d = (r + s)d Þ p + q = r + s.
\ 2 secq = sec (q – f) + sec (q + f)
Þ
cos ( q + f) + cos ( q - f )
2
=
cos q
cos ( q - f ) cos ( q + f )
(
)
Þ 2 cos 2 q - sin 2 f = cos q [ 2 cos q cos f]
Þ cos 2 q (1 - cos f ) = sin 2 f = 1 - cos 2 f
2
2
Þ cos q = 1 + cos f = 2 cos
f
2
f
k
But given cosq = cos
2
\ k= 2
f
2
\ cos q = 2 cos
Downloaded from @Freebooksforjeeneet
M-105
Sequences and Series
38.
(b) Let A.P. be a, a + d , a + 2d ,.........
a2 + a4 + ........... + a200 = a
100
é2 ( a + d ) + (100 - 1) 2d ùû = a ...(i)
Þ
2 ë
and a1 + a3 + a5 + ......... + a199 =b
100
Þ
[ 2a + (100 – 1) 2d] = β
2
Subtracting (ii) from (i), we get
39.
40.
43.
(m - n)d =
...(ii)
44.
Þ 31– x + 2 = 3 (4.3x – 1)
Þ 3.3–x + 2 = 12.3x – 3.
Put 3x = t
3
+ 2 = 12t - 3 Þ 12t2 – 5t – 3 = 0;
Þ
t
1 3
Hence t = - ,
3 4
3
Þ 3x = (as 3 x ¹ - ve )
4
æ 3ö
è 4ø
Þ x = log3 ç ÷ or x = log3 3 – log3 4
45.
Cr ,
m
¥
å f ( x) = 2 Þ k + k 2 + k 3 + .....¥ = 2
m
Cr
m
Cr +1
m
Cr
2
4
f (4) k 4
=
= k2 = .
9
f (2) k 2
(c) Rearrange given equation, we get
Now,
46.
(a 2 p 2 - 2abp + b2 ) + (b2 p 2 - 2bcp + c 2 )
+(c2 p 2 - 2cdp + d 2 ) = 0
Þ (ap - b)2 + (bp - c)2 + (cp - d )2 = 0
=
r
m-r
+
m - r +1 r +1
Þ m - m(4r + 1) + 4r - 2 = 0 .
2
2
k
=2Þk =
1- k
3
Cr +1 are in A.P..
2m Cr =m Cr -1 + m Cr +1
+
Þ x = 1 – log3 4
(d) Let f (1) = k, then f (2) = f (1 + 1) = k2
f (3) = f (2 + 1) = k3
Þ
(c) Coeficient of rth, (r + 1)th and (r + 2)th terms is mCr–1,
mC and mC
r
r+1 resp.
Cr -1
p
log b a
q
x =1
a
11
a1 + 5d 11
Þ 6 =
=
a21 41
a1 + 20d 41
m
ap =
Þ log3 (31– x + 2) = log3 [3(4 × 3x – 1)]
Put p = 11 and q = 41
Þ 2=
bq
Þ log3 (31 – x + 2) = log33 + log3 (4.3x – 1)
2a1 + ( p - 1)d p
=
2a1 + (q - 1)d q
m
1
Þa-d =0
mn
(b) 1, log9 (31 – x + 2), log3 (4.3x – 1) are in A.P.
Q a, b, c are in A.P then b = a + c
Q log
But n = 125 is not possible
\ Total time = 24 + 10 = 34 minutes.
(d) Given that
Given that m Cr -1 ,
1 1
1
- Þd =
n m
mn
Þ 2 log9 (31– x+2) = 1 + log3 (4.3x – 1)
Þ n - 3 [ 2 ´ 240 + (n - 4) ´ 40] = 11040 - 600
2
Þ (n - 3)[240 + 20n - 80] = 10440
Þ (n - 3)(20n + 160) = 10440
Þ (n - 3)(n + 8) = 522
Þ n2 + 5n - 546 = 0
Þ (n + 26) (n – 21) = 0
\ n = 21
(a) Till 10th minute number of counted notes = 1500
Remaining notes = 4500 – 1500 = 3000
n
3000 = [ 2 ´ 148 + ( n - 1)( -2) ] = n [148 - n + 1]
2
p
S p 2 [2a1 + ( p - 1)d ] p 2
=
=
2
q
Sq
[2a1 + (q - 1)d ] q
2
.....(ii)
From (i) a =
Þ n = 125, 24
42.
.....(i)
1
m
Subtracting (ii) from (i), we get
a -b
d=
100
(c) Let number of months = n
\ 200 × 3 + (240 + 280 + 320 + ... + (n – 3)th term) = 11040
Þ
1
n
Tn = a + (n - 1) d =
n 2 - 149 n + 3000 = 0
41.
(d) Tm = a+ (m – 1) d =
\ ap - b = bp - c = cp - d = 0
Þ
b c d
= =
a b c
\ a, b, c, d are in G.P.
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EBD_8344
M-106
47.
Mathematics
(5.00)
51.
Q f ( x + y ) = f ( x) × f ( y )
(4)
"x Î R and f(1) = 3
(0.16)
Þ f ( x) = 3x Þ f (i ) = 3i
æ1 1 1
ö
log 2.5 ç + 2 + 3 + .....¥÷
è3 3 3
ø
æ 1 ö
ç
÷
log 2.5 ç 3 ÷
1
ç 1- ÷
è 3ø
n
Þ å f (i) = 363
i =1
= 0.16
Þ 3 + 3 + 3 + .... + 3 = 363
2
Þ
3
n
= 0.16
é
a (r - 1) ù
êQ S n =
ú
(r - 1) û
ë
n
3(3n - 1)
= 363
3 -1
52.
Þ 3 = 243 = 3 Þ n = 5
(b) Given sequence are in G.P. and common ratio
æ æ 3ö
210 ç ç ÷
è è 2ø
ö
- 1÷
ø
11
(c) Let terms of G.P. be
æ3 ö
çè - 1÷ø
2
-2
= 4.
a
, a, ar
r
. . . (i)
and a3 = 27
Þa=3
Put a = 3 in eqn. (1), we get
= S - 211
If f ( x) = x +
1
, then f ( x) Î ( -¥, - 2] È [2, ¥)
x
Þ 311 - 211 = S - 211 Þ S = 311
(b) Let the first term be 'a' and common ratio be 'r'.
Þ 3 f ( x) Î ( -¥, - 6] È [6, ¥)
Q ar (1 + r + r ) = 3
...(i)
Then, it concludes that
and ar 5 (1 + r + r 2 ) = 243
...(ii)
\ S50 =
Þ 3 + 3 f ( x) Î ( -¥, - 3] È [9, ¥)
S Î (-¥, - 3] È [9, ¥)
53.
From (i) and (ii),
r 4 = 81 Þ r = 3 and a =
1
13
a(r 50 - 1) 350 - 1
=
r -1
26
. . . (ii)
1ö
æ
S = 3 + 3ç r + ÷
è
rø
2
50.
æ 1ö
=ç ÷
è 2ø
æ1
ö
\ a ç + 1 + r÷ = S
èr
ø
3
2
æ 311 - 211 ö
ç
÷
è 211 ø
Þ 210
= S - 211
1
2
49.
æ 1ö
-2log 2.5 ç ÷
è 2ø
5
n
\
æ 1ö
log 2.5 ç ÷
è 2ø
= (2.5)
363 ´ 2
Þ 3n - 1 =
= 242
3
48.
a ù
é
êëQ S¥ = 1 - r úû
é
a(r n - 1) ù
êQ Sn =
ú
(r - 1) û
ë
(c) S = ( x + y) + ( x 2 + y 2 + xy) + ( x 3 + x 2 y + xy 2 + y 3 ) + ....¥
=
1
é( x 2 - y 2 ) + ( x 3 - y 3 ) + ( x 4 - y 4 + ....¥ù
û
x- y ë
=
1 é x2
y 2 ù ( x - y )( x + y - xy )
ê
ú=
x - y ë1 - x 1 - y û ( x - y )(1 - x )(1 - y )
(b) Let a, b, g , d be in G.P., then ad = bg
Þ
Þ
a ù
é
êëQ S¥ = 1 - r úû
a g
a -b
g -d
= Þ
=
b d
a+b
g +d
9- 4p
36 - 4q
=
3
6
=
54.
x + y - xy
(1 - x)(1 - y)
(c) S = ( x + x2 + x 3 + ....9 terms)
Þ 36 - 16 p = 36 - 4q Þ q = 4 p
\
2q + p 8 p + p 9 p 9
=
=
=
2q - p 8 p - p 7 p 7
+ a[k + (k + 2) + + (k + 4) + ...9 terms]
ÞS=
x ( x 9 - 1) 9
+ [2ak + 8 ´ (2a)]
x -1
2
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M-107
Sequences and Series
x10 - x 9a(k + 8) x10 - x + 45a( x - 1)
(Given)
+
=
x -1
x -1
1
ÞS =
Þ
æ 4 ö
ç
÷
è 1+ r ø
(a)
1
24
+
2 3
+ +.....¥
16 48
a
a1 (r q - 1) (-4)((-2)9 - 1)
=
r -1
-2 - 1
å ai =
100
4
= (-513) = 4l Þ l = – 171
3
(b) The given series is in G.P. then
Þ
i =1
60.
ar 2 ( r 200 - 1)
2
r -1
= 200
...(i)
Sn =
100
å a2n = a2 + a4 + ..... + a200 = 100
ar (r 200 - 1)
= 100
r2 -1
From equations (i) and (ii), r = 2 and
a2 + a3 + ..... + a200 + a201 = 300
Þ r(a1 + ..... + a200) = 300
Þ
Þ
200
å an =
n =1
...(ii)
(1 + x)10 é(1 + x )11 - x11 ù
ë
û
Þ
= (1 + x)11 – x11
1
11
(1 + x) ´
(1 + x )
300
= 150
r
(b) y = 1 + cos q + cos q + .....
2
61.
4
1
1
Þ
= sin 2 q
y
1 - cos 2 q
x = 1 – tan 2q + tan4q + .....
Þ y=
x=
1
2
1 - (- tan q)
y=
1
sin 2 q
\ y(1 – x) = 1
=
1
2
sec q
Þ y=
1
1- x
n
é
ù
(49)126 - 1 ((49)63 + 1)(4963 - 1) êQS = a(r -1) ú
n
=
(b)
r -1 úû
êë
48
48
\ K = 63
From eqn. (i), a1 =
62.
...(i)
...(ii)
4
and substituting the value of a1,
1+ r
\ Coefficient of x7 is 11C7 = 11C11–7 = 11C4 = 330
(d) Q a, b, g are three consecutive terms of a nonconstant G.P.
\ b2 = ag
So roots of the equation ax2 + 2bx + g = 0 are
-2b ± 2 b2 - ag b
=
2a
a
2
Q ax + 2bx + g = 0 and x2 + x – 1 = 0 have a common
root.
\ this root satisfy the equation x2 + x – 1 = 0
b2 – ab – a2 = 0
Þ ag – ab – a2 = 0 Þ a + b = g
Now, a(b + g) = ab + ag
Þ x = cos2q
(b) Since, a1 + a2 = 4 Þ a1 + a1r = 4
a3 + a4 = 16
Þ a1r2 + a1r3 = 16
a(1 - r n )
1- r
é æ x ö11 ù
(1 + x)10 ê1 - ç
÷ ú
êë è 1 + x ø úû
x ö
æ
ç1 ÷
è 1+ x ø
n =1
59.
4
3
=
= 2
(d) Let G.P. be a, ar, ar2 .....
n =1
58.
= 16
r = – 2, a1(1 – 2) = 4 Þ a1 = – 4
å a2n+1 = a3 + a5 + ..... + a201 = 200
57.
r3
r = 2, a1(1 + 2) = 4 Þ a1 =
1 1 1
+ + +.....¥
2 4 8 16
56.
æ 4 ö
+ç
÷
è 1+ r ø
Þ 4r2 (1 + r) = 16(1 + r)
Þ r2 = 4
\ r=±2
x10 - x + 9a(k + 8)( x - 1) x10 - x + 45a( x - 1)
=
x -1
x -1
Þ 9a(k + 8) = 45a Þ k + 8 = 5 Þ k = -3.
55.
r2
= ab + b2 = (a + b)b = bg
(d) Q a, b, c are in G.P. Þ b = ar, c = ar2
Q 3a, 7b, 15c are in A.P. Þ 3a, 7ar, 15ar2 are in A.P.
\ 14ar = 3a + 15ar2
1 3
Þ 15r2 – 14r + 3 = 0 Þ r = or
3 5
in eqn (ii),
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EBD_8344
M-108
Mathematics
\ x = cos q, cosec q, qÎ (0, 45°)
1
3
\ r = reected
2
5
Fourth term = 15ar2 + 7ar – 3a
Qr <
\ a = cos q, b = cosec q
฀
å an
æ 15 7 ö
= a (15r2 + 7r – 3) = a ç + - 3 ÷ = a
è9 3 ø
63.
n= 0
฀
(–1) n
1
1
1
= 1– cosec q +
+ ¼฀
n
cosec 2 q cosec3 q
n=0 b
= 1 – sin q + sin 2 q – sin3 q + ¼ ฀.
Þ ax2 + 2 acx + c = 0 Þ ( a x + c )2 = 0
1
1+ sin q
=
c
a
Also, given that ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have
a common root.
Þx= -
=
66.
Þ r=
1
or 2
2
1
1
+
.
1– cos q 1+ sin q
(a) Let a1 = a, a2 = ar, a3 = ar2 ... a10 = ar9
where r = common ratio of given G.P.
ar 2
= 25
a
Þ r = ±5
67.
a9 ar 8
4
4
4
Now, a = 4 = r = (±5) = 5
5
ar
(b) Let the terms of infinite series are a, ar, ar2, ar3, ...
So,
a
=3
1- r
Since, sum of cubes of its terms is
27
19
a3
27
So,
=
19
1 - r3
a3r3, ... ฀ is
Þ
27
a
a2
´
=
1 - r (1 + r 2 + r ) 19
9(1 + r 2 - 2r ) ´ 3 27
=
19
1 + r2 + r
Þ 6r2 – 13r + 6 = 0
Þ (3r – 2)(2r – 3) = 0
Þ
8
+ 8 + 16 = 28
2
(c) x2 sin q – x (sin q . cos q + 1) + cos q = 0.
Therefore, sum of three terms =
65.
n =0
(–1)n ö ฀ n ฀ ( –1)n
÷ = åa + å n
bn ø n = 0
n=0 b
Þ
Þ
64.
å çè an +
a3
Given, a = 25
1
d
f
2e
d 2e f
+
=0Þ - + =0
a
ac c
a b c
d e f
2e d f
= + Þ , , are in A.P..
b a c
a b c
a
(d) Let three terms of a G.P. be , a, ar
r
a
× a × ar = 512
r
a3 = 512 Þ a = 8
4 is added to each of the first and the second of three
8
terms then three terms are, + 4, 8 + 4, 8r.
r
8
Q + 4,12,8r form an A.P.
r
8
\ 2 × 12 = + 8r + 4
r
Þ 2r2 – 5r + 2 = 0
Þ (2r – 1) (r – 2) = 0
æ
฀
\
c
must satisfy dx2 + 2ex + f = 0
a
æ cö
c
Þ d . a + 2e çç - a ÷÷ + f = 0
è
ø
x2 sin q – x sin q . cos q – x + cos q = 0.
x sin q (x – cos q) – 1 (x – cos q) = 0.
(x – cos q) (x sin q – 1) = 0.
1
1– cosq
å
(a) Since a, b, c are in G.P.
\ b2 = ac
Given equation is, ax2 + 2bx + c = 0
Þx= -
= 1 + cos q + cos2 q + ¼ ฀ =
2
3
, or
3
2
As | r | < 1
Þ r=
So, r =
2
3
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27
that is sum of a3,
19
M-109
Sequences and Series
n +1
68.
æ q + 1ö
1– ç
è 2 ÷ø
æ 1– q n+1 ö
(d) Sn = ç
,
T
=
÷
æ q + 1ö
è 1– q ø n
1– ç
è 2 ÷ø
\
70.
(d) Q a, b, c, are in G.P.
71.
Þ b2 = ac
Since, a + b + c = xb
Þ a + c = (x – 1)b
Take square on both sides, we get
a2 + c2 + 2ac = (x –1)2 b2
Þ a2 + c2 = (x – 1)2 ac – 2ac [Q b2 = ac]
Þ a2 + c2 = ac[(x – 1)2 – 2]
Þ a2 + c2 = ac[x2 – 2x – 1]
Q a2 + c2 are positive and b2 = ac which is also positive.
Then, x2 – 2x – 1 would be positive but for x = 2, x2 – 2x – 1
is negative.
Hence, x cannot be taken as 2.
(c) First term = b and common ratio = r
101
æ q +1ö
1– ç
è 2 ÷ø
Þ T100 =
æ q + 1ö
1– ç
è 2 ÷ø
Sn =
2101 – (q +1)101
q n +1
1
–
, T100 =
1– q 1– q
2100 (1– q)
Now, 101C1 +
101C
2
S1 +
101C
3
S2 + ¼ + 101C101 S100
æ 1 ö
=ç
(101C2 + ¼ + 101C101)
è 1– q ÷ø
1
–
(101C2 q2 + 101C3 q3 + ¼ + 101C101 q101) + 101
1– q
=
For infinite series, Sum =
æ 1 ö
1
(2101 – 1 – 101) – ç
((1 + q)101 – 1
1– q
è 1– q ÷ø
1
=
[2101 – 101 + 101q – (1 + q)101] + 101
1– q
æ 1 ö
[2101 – (1 + q)101] = 2100 T100
=ç
è 1– q ÷ø
Hence, by comparison a = 2100
69.
(d) Let first term and common difference be A and D
respectively.
\ a = A + 6D, b = A + 10D
and c = A + 12D
Since, a, b, c are in G.P.
Hence, relation between a, b and c is,
b
=5
1– r
Þ b = 5 (1 – r)
So, interval of b = (0, 10) as, – 1< r < 1 for infinite G.P.
– 101C1 q) + 101
1
=
[2101 – 102 – (1 + q)101 + 1 + 101q] + 101
1– q
a -8 D
=
=4
c -2 D
2
72.
3
æ3ö æ3ö æ 3ö
æ 3ö
(b) An = ç ÷ - ç ÷ + ç ÷ - ... + (-1)n - 1 ç ÷
è4ø è4ø è 4ø
è 4ø
Which is a G.P. with a =
3
-3
,r =
and number of
4
4
terms = n
\ An =
n
3 æ æ -3 ö ö
´ ç1 - ç
÷ ÷
4 çè è 4 ø ÷ø
æ -3 ö
1- ç ÷
è 4 ø
=
n
3 æ æ -3 ö ö
´ ç1 - ç
÷ ÷
4 çè è 4 ø ÷ø
7
4
é æ - 3 ön ù
...(i)
ê1 - ç
÷ ú
êë è 4 ø úû
As, Bn = 1 – An
For least odd natural number p, such that Bn > An
Þ An =
3
7
Þ 1 – An > An
Þ 1 > 2 × An Þ An <
From eqn. (i), we get
\ b2 = a.c.
\ (A + 10D)2 = (A + 6D) (A + 12D)
\ 14D + A = 0
\ A = – 14D
n
n
3 é æ -3ö ù 1
æ - 3ö 7
´ ê1 - ç
÷ ú < Þ 1- ç 4 ÷ < 6
7 ëê è 4 ø ûú 2
è
ø
Þ 1-
\ a = –8D, b = –4D and c = –2D
n
7 æ - 3ö
<ç
÷
6 è 4 ø
n
Þ
-1 æ - 3ö
<ç
÷
6 è 4 ø
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n
1
2
EBD_8344
M-110
Mathematics
n
3
æ - 3ö
As n is odd, then ç
÷ =4
è 4 ø
n
=
(1 + 3i)12 - 1
3i
12
-1
æ3ö
So
<-ç ÷
6
è4ø
73.
n
1 æ3ö
Þ >ç ÷
6 è4ø
n
12
(1 + 3i )
12
Þ
3
4
...(i)
76.
1
=
Þ ac = ±
16
From (i), (ii) and (iii), we get;
a2c2
1
1
=
16a 2
Case I: 16a2 – 8a + 1 = 0
Þa=
...(iii)
77.
1
(not possible as a < b)
4
74.
1
1
4 2 2
8 ± 128
32
78.
(Q a < b)
ar 2 - ar (A + 8d) - (A + 4d)
=
ar - a
(A + 4d) - (A + d)
4
3
(b) z = 1 + ai
z2 = 1 – a2 + 2ai
z2. z ={(1 – a2) + 2ai}
{1 + ai}
= (1 – a2) + 2ai + (1 – a2)
ai – 2a2
Q z3 is real Þ 2a + (1 – a2) a = 0
a (3 – a2) = 0 Þ a =
1 + z + z2 .......... z11 =
4095
3i = – 1365 3i
3
l+n
n
and common ratio of G.P. = r = æç ö÷ 4
(d) m =
2
èlø
= ln (l + n)2 = ln × (2m)2 = 4lm2n
(d) Let a, ar and ar2 be the first three terms of G.P
According to the question
a (ar) (ar2) = 1000 Þ (ar)3 = 1000 Þ ar = 10
and ar2 + ar3 = 60 Þ ar (r + r2) = 60
Þ r2 + r – 6 = 0
Þ r = 2, –3
10
(reect)
3
Hence, T7 = ar6 = 5(2)6 = 5 × 64 = 320.
(b) Let a, ar, ar2 are in G.P.
According to the question
a, 2ar, ar2 are in A.P.
Þ 2 × 2ar = a + ar2
Þ 4r = 1 + r2 Þ r2 – 4r + 1 = 0
4 ± 16 - 4
= 2± 3
2
Since r > 1
\ r = 2 - 3 is reected
Hence, r = 2 + 3
r=
75.
3i
=–
r=
(d) Let the GP be a, ar and ar2 then a = A + d; ar= A + 4d;
ar2 = A + 8d
Þ
4095
a = 5, a = –
1
1
Þ a= ±
4 2 2
\ a=
=
3
2 2
3
G14 + 2G 42 + G34 = l n + 2l n + ln
Þ 16a2 – 8a ± 1 = 0
Case II: 16a2 – 8a – 1 = 0 Þ a =
3i
\ G1 = l3/4n1/4, G2 = l1/2n1/2, G3 = l1/4 n3/4
3
1
Þ b=
...(ii)
4
4
Again it is given that, a2, b2, c2 are in G.P. then
a±
212 - 1
1
Þ 2b + b =
(b2)2
æ1
3 ö
i÷
çç +
÷
2
2
è
ø
p
pö
æ
= 212 ç cos + i sin ÷ = 212 (cos 4p + i sin 4p) = 212
3
3ø
è
æ1ö
æ3ö
log ç ÷ = n log ç ÷ Þ 6.228 < n
è4ø
è6ø
Hence, n should be 7.
(d) Q a, b, c are in A.P. then
a + c = 2b
also it is given that,
a+b+c=
= 212
3 (a > 0)
12
z12 - 1 (1 + 3i) - 1
=
1 + 3i - 1
z -1
79.
(b) 1 -
2 2
2
1
- ....
<
3 32 3n-1 100
2 é1 1
1
1 ù
1
Þ 1 - ê + + + ...
<
ú
2
3
n1
3 ë3 3
3
3 û 100
é1 æ 1
öù
1 - 2 ê ç n - 1÷ ú
øû
1
ë3 è 3
Þ
<
1
100
-1
3
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M-111
Sequences and Series
80.
é 3n - 1ù
1
Þ 1- 2 ê
ú<
n
êë 2.3 úû 100
é 3n - 1ù
1
Þ 1- ê
ú<
n
êë 3 úû 100
1
1
Þ 1 -1 + n <
Þ 100 < 3n
100
3
Thus, least value of n is 5
(c) According to Question
S
Þ 5' = 49 {here, S5 = Sum of first 5 terms
S5
and S5 = Sum of their reciprocals)
Þ
a(r 5 - 1)
(r - 1)
a -1 (r -5 - 1)
(r
Þ
or
-1
(1 + x)1001 é(1 + x )1001 - x1001 ù
ë
û
=
(1 + x )1001
= [(1 + x)1001 - x1001 ]
Now coeff of x 50 in above expansion is equal
to coeff of x50 in (1 + x)1001 which is 1001C50
=
82.
(b) Let a, b, c, d be four numbers of the sequence.
Now, according to the question b2 = ac and c – b = 6 and a
–c=6
Also, given a = d
= 49
éa +bù
\ b2 = ac Þ b 2 = a ê
ë 2 úû
- 1)
a ( r 5 - 1) ´ ( r -1 - 1)
a -1 ( r -5 - 1) ´ ( r - 1)
(Q 2c = a + b)
Þ a2 – 2b2 + ab = 0
Now, c – b = 6 and a – c = 6,
gives a – b = 12
= 49
a 2 (1 - r 5 ) ´ (1 - r ) ´ r 5
(1 - r 5 ) ´ (1 - r ) ´ r
Þ b = a – 12
= 49
\ a2 – 2b2 + ab = 0
Þ a2 – 2(a – 12)2 + a(a – 12) = 0
Þ a2r4 = 49 Þ a2r4 = 72
Þ ar 2 = 7
...(i)
Also, given, S1 + S3 = 35
a + ar2 = 35
...(ii)
Now substituting the value of eq. (i) in eq. (ii)
a + 7 = 35
81.
(1001)!
50!(951)!
Þ a2 – 2a2 – 288 + 48a + a2 – 12a = 0
Þ 36a = 288 Þ a = 8
83.
a = 28
(d) Let given expansion be
S = (1 + x)1000 + x (1 + x)999 + x2 (1 + x)998 + ...
+ ... + x1000
Put 1 + x = t
S = t1000 + xt999 + x2 (t)998 + ... + x1000
x
This is a G.P with common ratio
t
é æ x ö 1001 ù
t1000 ê1 - ç ÷
ú
ètø
ëê
ûú
S=
x
1t
é æ x ö 1001 ù
(1 + x)1000 ê1 - ç
ú
÷
êë è 1 + x ø
úû
=
x
11+ x
Hence, last term is d = a = 8.
(d) The given relation can be written as
(a2p2 – 2abp + b2) + (b2p2 + c2 – 2bpc) +
(c2p2 + d2 – 2pcd) £ 0
or (ap – b)2 + (bp – c)2 + (cp – d)2 £ 0
...(i)
Since a, b, c, d and p are all real, the inequality (i) is possible
only when each of factor is ero.
i.e., ap – b = 0, bp – c = 0 and cp – d = 0
or
84.
p=
b c d
= =
a b c
or a, b, c, d are in G.P.
(c) Let a, ar, ar2, ar3, ar4, ar5 be six terms of a G.P. where
‘a’ is first term and r is common ratio.
According to given conditions, we have
ar3 – a = 5 Þ a(r3 – 1) = 52
...(i)
and a + ar + ar2 = 26
Þ a (1 + r + r2) = 26
...(ii)
To find: a (1 + r + r2 + r3 + r4 + r5)
Consider
a[1 + r + r2 + r3 + r4 + r5]
= a [1 + r + r2 + r3 (1 + r + r2)]
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EBD_8344
M-112
Mathematics
= a[1 + r + r2] [1 + r3]
Divide (i) by (ii), we get
r3 -1
2
85.
x n = b n + ab n -1 + a 2 b n - 2 + ....... + a n -1b + a n
=2,
{which is a G.P. with r =
1+ r + r
we know r3 – 1 = (r – 1) (1 + r + r2)
\ r – 1 = 2 Þ r = 3 and a = 2
\ a (1+ r + r2 + r3 + r4 + r5) = a (1 + r + r2) (1 + r3)
= 2(1 + 3 + 9) (1 + 27) = 26 × 28 = 728
(b) ATQ,
a + ar = 12
…(i)
ar2 + ar3 = 48
…(ii)
ar 2 (1 + r ) 48
= Þ r2 = 4, Þ r = –2
a(1 + r )
12
Þ
86.
\ Coefficient of xn
...(iii)
(Q terms are alternately + ve and –ve)
Þ a = –12
(b) Let the series a, ar, ar 2 , ..... are in geometric
progression.
Given that, a = ar + ar2
Þ 1= r + r2
Þ r2 + r – 1 = 0
n +1 ù
é
b n ê1 - æç a ö÷ ú
ë è bø û
\ Its sum is =
}
a
1b
=
89.
90.
x 2 - (a + b ) x + ab = 0 Þ x 2 - 18 x + 16 = 0
(b) Let a = first term of G.P. and r = common ratio of G.P.;
Then G.P. is a, ar, ar2
Þ
k =1
Þ
10
k =1
10
= iå e
-
2k p
i
11
k =1
=
2p
é
- i
ê
i 1 + e 11
ê
ë
[Qeiq = cosq + i sinq]
91.
ì 10 - 2k p i ü
ï
ï
= i í å e 11 - 1ý
ïî k =0
ïþ
4p
- i
+ e 11
( )
é
2 p 11 ù
é 1 - e - 2p i ù
ê1 - - 11 ú
e
ú
= iê
i
ú
= iê
2p - i
2p
- iú
ê
ê
- i ú
ë1 - e 11 û
ëê 1 - e 11 ûú
[Q e
=i×0–i
a
= 20
1- r
a2
1- r
2
= 100 Þ
[20(1 - r)]2
1- r2
= 100
... (i)
[from (i)]
400(1 - r)2
= 100 Þ 4(1 - r) = 1 + r
(1 - r)(1 + r)
Þ 1 + r = 4 – 4r Þ 5r = 3 Þ r = 3/5.
(b) Q a4 = 2 Þ ar4 = 2
Now, a ´ ar ´ ar 2 ´ ar 3 ´ ar 4 ´ar 5 ´ ar 6 ´ ar 7 ´ ar 8
= a9 r36 = (ar4)9 = 29 = 512
ù
+ ....11 terms ú - i
ú
û
-2 pi
a+b
=9
2
Þ a = 20(1 – r)
Also a2 + a2r2 + a2r4 + ... to ¥ =100
2k p
2k p ö
æ
+ i cos
÷
(d) å çè sin
11
11 ø
2k p
2k p ö
æ
= i å ç cos
- i sin
÷
è
11
11 ø
bn +1 - a n +1
b-a
(b) Let two numbers be a and b then
10
87.
\ an =
Given S ¥ = 20 Þ
5 -1
[Q terms of G.P. are positive
2
\ r should be positive]
Þ r=
bn+1 - an+1
b-a
Þ a + b = 18 and ab = 4 Þ ab = 16
\ Equation with roots a and b is
-1 ± 1 - 4 ´ -1 -1 ± 5
=
2
2
Þ r=
a
b
= 1]
92.
(c) sin2 2q + cos4 2q =
3
4
Þ 1 – cos2 2q + cos4 2q =
3
4
Þ cos2 2q(1 – cos2 2q) =
1
4
...(i)
Q G.M. £ A.M.
æ cos 2 2q + (1 - cos 2 2q ö
\ (cos2 2q)(1 – cos2 2q) £ ç
÷
2
è
ø
=–i
88.
(d) (1 - ax) -1 (1 - bx) -1
= (1 + ax + a 2 x 2 + ...)(1 + bx + b 2 x 2 + ...)
1
4
So, from equation (i) and (ii), we get.
=
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...(ii)
2
M-113
Sequences and Series
G.M. = A.M.
It is possible only if,
cos2 2q = 1 – cos2 2q
Þ cos2 2q =
93.
Þ y ³ 2 1- 3y
Þ y2 ³ 4 - 4 3y
1
1
Þ cos 2q = ±
2
2
p 3p
p 3p p
=
\ Sum = +
Þ q= ,
8 8
8 8
2
(39)
Þ y2 + 4 3y - 4 ³ 0
Þ y £ -2 3 - 4 or y ³ -2 3 + 4
( y £ -2 3 - 4 is not possible as tan A tan B > 0)
Let m arithmetic mean be A1, A2 ... Am and G1, G2, G3 be
(b) x + y + z = 12
AM > GM
geometric mean.
The A.P. formed by arithmetic mean is,
3, A1, A2, A3, .......Am, 243
æ xö æ yö æ zö
3ç ÷ + 4 ç ÷ + 5 ç ÷
è3ø è 4ø è5ø
12
243 - 3 240
=
m +1
m +1
The G.P. formed by geometric mean
3, G1, G2, G3, 243
x3 y 4 z 5
96.
\d=
33 4 4 55
1
Q A4 = G2
960
= 27 Þ m + 1 = 40 Þ m = 39.
m +1
(d) A.T.Q.,
A.M. = 5 G.M.
97.
(a) G =
a 10 + 96 10 + 4 6
=
=
b 10 - 96 10 - 4 6
Use Componendo and Dividendo
a +b
20
5
5 6
=
=
=
a -b 8 6 2 6
12
tan A + tan B
1 - tan A tan B
1
y
Þ
where y = tan A + tan B
=
3 1 - tan A tan B
Þ tan A tan B = 1 – 3y
M=
Also AM > GM
tan A + tan B
³ tan A tan B
2
a+b
2ab
1
:G = 4 :5
M
Given that
2ab
(a + b) ab
(b) tan (A + B) =
Þ
ab
1 1
+
M= a b
2
a+b
= 10
ab
95.
<1
x
y z
= = (= k)
3 4 5
x = 3k; y = 4k; z = 5k
x + y + z = 12
3k + 4k + 5k = 12
k=1
\ x = 3; y = 4; z = 5
\ x3 + y3 + z3 = 216
Þ 3+
\
5
\
æ 240 ö
Þ 3+ 4ç
= 3(3)2
è m + 1÷ø
a+b
= 5 ab
2
4
x3 y4 z5 < 33 .44 .55
x3 y4 z5 < (0.1) (600)3
But, given x3 y4 z5 = (0.1) (600)3
\ all the number are equal
æ 243 ö 3+1
= (81)1/ 4 = 3
r =ç
è 3 ÷ø
94.
3
æxö æ yö æ zö
> 12 ç ÷ ç ÷ ç ÷
è3ø è 4ø è5ø
Þ
Þ
a+b
2 ab
4
5
=
=
5
4
a + b + 2 ab
a + b - 2 ab
=
5+ 4
5- 4
{Using Componendo & Dividendo}
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EBD_8344
M-114
Mathematics
¥
( a )2 + ( b ) 2 + 2 ab
Þ
=
( a ) + ( b ) - 2 ab
2
2
9
1
99.
å an
(d) x =
n=0
¥
æ b + aö
Þç
÷
è b - aø
2
Þ
y=
9
b+ a 3
=
= Þ
1
b- a 1
b+ a+ b- a
b+ a- b+ a
=
3 +1
3 -1
¥
z=
{Using
b
4
= =2
a
2
b
4
=
a
1
a
1
=
Þa:b=1:4
b
4
(d) Q a1, a2, a3.....an are in H.P.
Þ b = 1-
1
y
1
Þ c = 1-
1
z
å cn = 1 - c
n =0
a, b, c are in A.P. Þ
æ 1ö
1
1
2 ç1 - ÷ = 1 - + 1 è
yø
x
z
2b = a + c
2 1 1
Þ x, y, z are in H.P..
= +
y x z
c
-b
100. (d) ax 2 + bx + c = 0, a + b =
, ab =
a
a
1
1
ATQ, a + b =
+
a 2 b2
b2
2
a +b
a+b =
a 2 b2
1 1 1
1
, , ....
are in A.P.
a1 a2 a3 an
\
1
å bn = 1 - b
n =0
Componendo & Dividendo}
98.
1
1
Þ a = 1x
1- a
=
2
2c
b a2 a
Þ- =
a
c2
a2
On simplification 2a 2 c = ab 2 + bc 2
1
1
1
1
1
1
= d (say)
\ a - a = a - a =..........=
an an -1
2
1
3
2
Then a1a2 =
a2 - a3
a1 - a2
,
, a2 a3 =
d
d
an -1 - an
..........., an -1an =
d
Adding all equations, we get
\ a1a2 + a2 a3 + ......... + an-1an
=
a -a
a1 - a2 a2 - a3
+
+ .... + n -1 n
d
d
d
a -a
1
= [a1 - a2 + a2 - a3 + .... + an -1 - an ] = 1 n
d
d
Þ
2a c b
= +
b
a c
[Divide both side by abc]
c a b
Þ , , are in A.P..
a b c
a b
c
\ , , & are in H.P..
c a
b
101. (b) The given series is
1 + (1 - 22 ×1) + (1 - 42 × 3) + (1 - 62 × 5) + ...(1 - 202 ×19)
10
S = 1 + å [1 - (2r ) 2 (2r - 1)]
r =1
10
10
r =1
r =1
= 1 + å (1 - 8r 3 + 4r 2 ) = 1 + 10 - å (8r 3 - 4r 2 )
2
Also,
1
1
= + (n - 1)d
an a1
a -a
Þ 1 n = (n - 1)d
a1an
a1 - an
= (n - 1)a1an
d
Which is the required result.
Þ
æ 10 ´ 11ö
æ 10 ´ 11 ´ 21ö
= 11 - 8 ç
+ 4´ç
÷ø
è 2 ÷ø
è
6
= 11 - 2 ´ (110)2 + 4 ´ 55 ´ 7
= 11 - 220(110 - 7)
= 11 - 220 ´ 103 = a - 220b
Þ a = 11, b = 103
\ (a, b) = (11, 103)
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M-115
Sequences and Series
102. (a) Given : f ( x + y ) = f ( x) + f ( y), "x, y Î R, f (1) = 2
Þ f (2) = f (1) + f (1) = 2 + 2 = 4
f (3) = f (1) + f (2) = 2 + 4 = 6
f (n - 1) = 2(n - 1)
n -1
Now, g (n) = å f (k )
20
[2 ´ 7 + (19)10] = 10[14 + 190]
2
= 10[2040] = (102)(20)
Þ m = 20
S40 =
2ù é
n -1ù
é 1ù é
= [nx]
106. (b) Q [ x ] + ê x + ú + ê x + ú .... ê x +
nû ë
nû ë
n úû
ë
and [x] + [– x] = – 1 (x Î z)
k =1
= f (1) + f (2) + f (3) + .... f (n - 1)
= 2 + 4 + 6 + ..... + 2 (n – 1)
= 2[1 + 2 + 3 + ..... + (n - 1)]
=2´
(n - 1)(n)
= n2 - n
2
Q g (n) = 20 (given)
So, n2 - n = 20
Þ (n - 5)(n + 4) = 0
Þ n = 5 or n = – 4 (not possible)
é 7 n(n + 1)(2n + 1) ù 1 é 7
ù
3
2
ú ê å (2n + 3n + n) ú
103. (504) ê å
4
ëê n =1
ûú 4 êë n=1
ûú
2
1 é æ 7.8 ö
æ 7.8.15 ö 7.8 ù
= 4 ê 2 ç 2 ÷ + 3ç 6 ÷ + 2 ú
ø
è
ø
ëê è
ûú
=
ìé 1 ù é 1
1 ù
é 1 99 ù ü
+ ... ê +
ý
= – 100 – í ê ú + ê +
ú
3
3
100
û
ë 3 100 úû þ
îë û ë
é100 ù
= – 100 – ê
= – 133
ë 3 úû
107. (c) rth term of the series,
Tr =
Þ n2 - n - 20 = 0
1
[2 ´ 49 ´16 + 28 ´15 + 28]
4
Þ
é 1ù é 1
ù
é 1 99 ù
êë - 3 úû + êë- 3 - 100 úû + ... + êë - 3 - 100 úû
\
1
[1568 + 420 + 28] = 504
4
104. (1540) Given series can be written as
k (k + 1) 1 20 2
= å (k + k )
2
2 k =1
k =1
20
å
1 é 20(21)(41) 20(21) ù
= ê
+
2ë
6
2 úû
1 é 420 ´ 41 20 ´ 21ù 1
= ê
+
= [2870 + 210] = 1540
2ë 6
2 úû
2
14 + 18
+ 4 + 8{
+ 9 + 13
+3
+ 19.....40 terms
105. (a) S = 3{
12
123
S = 7 + 17 + 27 + 37 + 47 + ..... 20 terms
(2 r + 1)(13 + 23 + 33 + ... + r 3 )
12 + 22 + 32 + ... + r 2
2
3r (r + 1)
6
æ r (r + 1) ö
Tr = (2r + 1) ç
÷ ´ r (r + 1)(2r + 1) =
2
2
è
ø
10
\ sum of 10 terms is = S =
3 10
å Tr = 2 å (r 2 + r )
r =1
r =1
=
3 ì10 ´ (10 + 1)(2 ´10 + 1) 10 ´11ü
+
í
ý
2î
6
2 þ
=
3
´ 5 ´ 11´ 8 = 660
2
108. (a) Let, S = 1 +
13 + 23 13 + 23 + 33
+
+ ...15 terms
1+ 2
1+ 2 + 3
2
æ n(n + 1) ö
ç
÷
n(n + 1)
1 + 2 + ...n
2 ø
Tn =
=è
=
n(n + 1)
1 + 2 + ...n
2
2
3
3
3
1 æ 15 2 15 ö 1 æ 15(16)(31) 15(16) ö
S
=
+
ç å n + å n ÷÷ = ç
÷
Now,
2 çè n=1
6
2 ø
n =1 ø 2 è
= 680
1 15(16)
= 680 - 60 = 620
2 2
109. (b) 1 + 2.3 + 3.5 + 4.7 + ........
Let, S = (2.3 + 3.5 + 4.7 + .......)
\ required sum is, 680 -
10
10
n =1
n =1
(
)
2
Now, S10 = å ( n + 1)( 2n + 1) = å 2n + 3n + 1
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EBD_8344
M-116
Mathematics
2 n ( n + 1)( 2n + 1) 3n ( n + 1)
+
+n
6
2
Put n = 10
=
2.10.11.21 3.10.11
=
+
+ 10 = 945
6
2
Hence required sum of the series = 1 + 945 = 946
110. (d) Number of balls used in equilateral triangle
n (n + 1)
=
2
Q side of equilateral triangle has n-balls
\ no. of balls in each side of square is = (n – 2)
According to the question,
1 2
[1 + 22 + ... + 112 - 1]
4
=
1 é11(11 + 1)(2 ´ 11 + 1) ù
- 1ú
4 êë
6
û
1 é11 ´12 ´ 23 ù
- 1ú
4 êë
6
û
=
1
[505]
4
A=
505 12
´ = 303
4
5
3
3
3
æ 3 ö æ 3ö æ 9 ö
3
113. (b) S = ç ÷ + ç ÷ + ç ÷ + (3) + ...
è 4 ø è 2ø è 4 ø
n ( n + 1)
2
+ 99 = ( n - 2 )
2
Þ n2 + n + 198 = 2n2 – 8n + 8
Þ n2 – 9n – 190 = 0 Þ (n –19) (n + 10) = 0
Þ n = 19
Number of balls used to form triangle
=
=
3
3
3
3
æ 3 ö æ 6 ö æ 9 ö æ 12 ö
S = ç ÷ + ç ÷ + ç ÷ + ç ÷ + ...
è 4 ø è 4 ø è 4ø è 4 ø
Let the general term of S be
3
æ 3r ö
Tr = ç ÷ , then
è 4ø
n ( n + 1) 19 ´ 20
=
= 190
2
2
15
æ 3ö
3 15
å k . 2k
255K =
å Tr = çè 4 ø÷ å r 3
1
1
1
1
S = + 2. 2 + 3. 3 + .... + 20. 20
2
2
2
2
255K =
27 æ 15 ´ 16 ö
´ç
÷
64 è 2 ø
20
111. (c) Let, S =
1
k =1
1
1
1
1
1
S = 2 + 2. 3 + .... + 19 20 + 20 21
2
2
2
2
2
On subtracting equations (ii) by (i),
...(i)
r =1
r =1
2
Þ K = 27
...(ii)
114. (d) S = 1 + 6 +
1
1 ö
1
S æ1 1
= +
+
+ .... + 20 ÷ - 20 21
2 çè 2 22 23
2 ø
2
9(12 + 22 + 32 ) 12(12 + 22 + 32 + 42 )
+
7
9
+
15(12 + 22 + 32 + 42 + 52 )
+K
11
3 × (1)2 6 × (12 + 22 ) 9 × (12 + 22 + 32 )
+
+
+
3
5
7
1æ
1 ö
11
1
1
2 çè 220 ÷ø
- 20. 21 = 1 - 20 - 10. 20
=
1
2
2
2
12
S=
1
1
11
S
= 1 - 11. 20 Þ S = 2 - 11. 19 = 2 - 19
2
2
2
2
Now, nth term of the series,
112. (c) Q 1 + 2 + 3 + ... + k =
\ Sk =
k (k + 1)
2
k (k + 1) k + 1
=
2k
2
12 × (12 + 22 + 32 + 42 )
+K
9
tn =
3n × (12 + 2 2 + K + n 2 )
(2n + 1)
Þ tn =
3n × n ( n + 1)(2n + 1) n3 + n 2
=
6(2 n + 1)
2
1 ìïæ n(n + 1) ö
n(n + 1)(2n + 1) üï
ý
\ Sn = Stn = 2 íçè 2 ø÷ +
6
ïî
ïþ
2
5
1
A = [2 2 + 33 + ... + 112 ]
Þ
12
4
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M-117
Sequences and Series
=
f(2) = 7
Þ f(3) = 12
Sn = 3 + 7 + 12 + .........+ tn
Sn = 3 + 7 + 12 + .........+ tn–1 + tn
–
– – –
–
–
0 = 3 + 4 + 5 .........to n term – tn
tn = 3 + 4 + 5 + .... upto n terms
n(n + 1) æ n(n + 1) 2n + 1ö
+
ç
÷
4 è 2
3 ø
Hence, sum of the series upto 15 terms is,
S15 =
15 ´ 16 ì15 ×16 31 ü
+ ý
í
4 î 2
3þ
31
= 60 ´ 120 + 60 ´
3
= 7200 + 620 = 7820
115. (d) The general term of the given series =
where r ³ 0
\ req. sum = 1 +
2 ´ 2 -1
r =1
2r
å
æ 2 ´ 2 -1ö
÷=
2r
r =1 è
ø
åç
19
æ
1ö
å çè 2 - 2r ÷ø
19
19
20
n =1
n =1
40
n =21
n(n + 1)(n + 8)
6
10 ´11 ´18
= 330
6
n ( n + 1)
2
118. (a) Tn =
æ n ( n + 1) ö
ç
÷
2
è
ø
n =1
B – 2A = (212 + 2.222 + 232 + 2.242 + ..... + 402)
– (12 + 2.22 + 32 + 2.42 ..... + 202)
= 20 [22 + 2.24 + 26 + 2.28 + ..... + 60]
é
+ 24 + 26....60) + (24 + 28 + .... + 60) ù
= 20 ê(22
14442444
3 144
42444
3ú
ë
û
20 terms
10 terms
10
é 20
ù
20 ê (22 + 60) + (24 + 60) ú
ë2
û
2
= 10[20.82 + 10.84]
= 100[164 + 84] = 100.248
117. (b) f(x) = ax2 + bx + c
f(1) = a + b + c = 3 Þ f (1) = 3
Now f(x + y) = f(x) + f(y) + xy ...(i)
Put x = y = 1 in eqn (i)
f(2) = f(1) + f(1) + 1 = 2f(1) + 1
2
Þ Tn =
2
n ( n + 1)
Þ Sn =
å Tn = 2 å çè n - n + 1 ÷ø = 2 ìíî1 - n + 1üýþ
20
å a n - 2å an
(n 2 + 5n)
2
Sn= 1 é n(n + 1)(2n + 1) 5n(n + 1) ù
+
úû
2 êë
6
2
19
æ1ö
æ1ö
\ req. sum = 1 + 37 + ç ÷ = 38 + ç ÷
è2ø
è2ø
116. (a) Here, B – 2A
40
å tn = å
r =1
æ1ö
æ1ö
= 38 + ç ÷ - 1 = 37 + ç ÷
è2ø
è2ø
= å an - 2å a n =
Sn =
S10 =
19
19
1 æ æ1ö ö
æ1ö
ç1 - ç ÷ ÷
ç ÷ -1
2 ç è 2 ø ÷ø
2
= 38 + è ø
= 2(19) - è
1
1
12
19
(n 2 + 5n)
2
=
r
r
19
Now,
19
2 ´ 2r - 1
,
2r
tn =
Þ
Sn =
n
æ1
1 ö
n =1
2n
n +1
Q 100 Sn = n
2n
=n
n +1
Þ n + 1 = 200
Þ n = 199
119. (b) Q
Þ 100 ´
3 éë1 + 25 + 81 + 69 + ......ùû = 435 3
Þ
3 [1 + 5 + 9 + 13 + ..... + Tn ] = 435 3
Þ
n
3 ´ éë 2 + ( n - 1)4 ùû = 435 3
2
Þ 2n + 4n2 – 4n = 870
Þ 4n2 – 2n – 870 = 0
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1
EBD_8344
M-118
Mathematics
Þ 2n2 – n – 435 = 0
5
å
1 ± 1 + 4 ´ 2 ´ 435 1 ± 59
n =
=
4
4
\ n=
r =1
1 + 59
1 - 59
= 15; or n =
= 14.5
4
4
2
2
2
Þ
2
2
æ 8 ö æ 12 ö æ 16 ö æ 20 ö
æ 44 ö
120. (d) ç ÷ + ç ÷ + ç ÷ + ç ÷ ... + ç ÷
è5ø è 5 ø è 5 ø è 5 ø
è 5 ø
(
16 2 2 2
S=
2 + 3 + 4 + ... + 112
25
=
16 æ 11(11 + 1)(22 + 1) ö
- 1÷
25 çè
6
ø
Þ
Tr =
1 é1
1 ù k
=
3 êë 6 6.7.8 úû 3
1 1æ
1ö k
1 1 55 k
´ ç1 - ÷ = Þ ´ ´
=
3 6 è 56 ø 3
3 6 56 3
Þ k=
55
336
å ( r 2 - r - 6 ) = 7780
20
)
124. (d)
=
16
16
´ 505 = ´101
25
5
16
16
m = ´101
5
5
Þ m = 101.
121. (a) S = (1 + x)2016 + x (1 + x)2015 + x2(1 + x)2014
+.....+ x2015 (1 + x) + x2016 ...(i)
æ x ö
ç
÷ S = x (1 + x)2015 + x2 (1 + x)2014 + ..... +
è 1+ x ø
x
1+ x
S
x 2017
= (1 + x)2016 –
1+ x
1+ x
\ S = (1 + x)2017 – x2017
2017!
17!2000!
2
é n(n + 1) ù
êë 2 úû
1
= (n + 1) 2
122. (d) nth term of series =
2
4
n
1
1
2
2
Sum of n term = S (n + 1) = éë Sn + 2Sn + n ùû
4
4
1 é n(n + 1)(2n + 1) 2n(n + 1)
ù
+
+ nú
4 êë
6
2
û
Sum of 9 terms
=
1 é 9 ´ 10 ´ 19 18 ´ 10 ù 384
+
+ 9ú =
= 96
4 êë
6
2
û
4
123. (c) General term of given expression can be written as
=
Multiplied by
11
on both the sides
10
11
x = 11.108 + 2.(11)2.(10)7 + ...+ 9(11)9 + 1110 ...(ii)
10
Subtract (ii) from (i), we get
...(ii)
Subtracting (i) from (ii)
a17 = coefficient of x17 = 2017C17 =
125. (a) Given that 109 + 2. (11)(10)8 + 3(11)2 (10)7 + ... + 10(11)9
= k(10)9
9
8
2
7
9
Let x = 10 + 2.(11)(10) + 3(11) (10) + ... + 10(11)
...(i)
æ 11ö
x ç1 - ÷ = 109 + 11 (10)8 + 112 × (10)7 + ... + 119 – 1110
è 10 ø
2017
x2016 +
r =16
é æ 11 ö10 ù
ê ç ÷ - 1ú
x
9 ê è 10 ø
ú - 1110
Þ - = 10 ê
ú
11
10
-1 ú
ê
10
ëê
ûú
x
10
10
10
10
Þ - = (11 - 10 ) - 11 = -10
10
Þ x = 1011 = k.109 Given
Þ k = 100
126. (b) Let a, d and 2n be the first term, common difference
and total number of terms of an A.P. respectively i.e. a
+ (a + d) + (a + 2d) + ... + (a + (2n – 1)d )
No. of even terms = n, No. of odd terms = n
Sum of odd terms :
n
[2a + (n - 1)(2d )] = 24
2
Þ n [a + (n – 1) d ] = 24
Sum of even terms :
So =
n
[ 2(a + d ) + (n - 1)2d ] = 30
2
Þ n [a + d + (n – 1) d] = 30
Subtracting equation (i) from (ii), we get
nd = 6
...(i)
Se =
...(ii)
...(iii)
ù
1é
1
1
Tr = ê
3 ë n(n + 1)(n + 2) (n + 1)(n + 2)(n + 3) úû
Also, given that last term exceeds the first term by
on taking summation both the side, we get
a + (2n – 1) d = a +
21
2
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21
2
M-119
Sequences and Series
21
2
21
Þ 2´6=d
(Q nd = 6)
2
3
d=
2
Putting value of d in equation (3)
20
é
1 æ æ 1ö öù
ê
ç 1 - çè ÷ø ÷ ú
10 è
10 ø ú
7ê
= ê 20 ú
1
9ê
ú
110
ê
ú
ë
û
2nd – d =
=
6´2
n=
=4
3
Total no. of terms = 2n = 2 × 4 = 8
127. (a) nth term of given series is
129. (a) Consider 12 + 32 + 52 + ...... + 252
nth term Tn = (2n – 1)2, n = 1,...... 13
13
6
2n + 1
=
n ( n + 1)(2 n + 1)
n(n + 1)
6
Now, Sn =
1 ù
é1
Let
term, an = 6 ê + 1úû
n
n
ë
Sum of 20 terms, S20 = a1 + a2 + a3 + .... + a20
nth
S20
æ 1 1ö
æ 1 1ö
æ 1 1ö
= 6 çè - ø÷ + 6 çè - ø÷ + 6 çè - ÷ø + ...
1 2
2 3
3 4
1ö
æ 1 1ö
æ 1 1ö
æ 1
+6 ç - ÷ + 6 ç - ÷ + 6 ç - ÷
è 18 19 ø
è 19 20 ø
è 20 21ø
éæ
1ö æ 1 1ö æ 1 1ö
S20 = ê çè1 - 2 ÷ø + çè 2 - 3 ÷ø + çè 3 - 4 ÷ø + ...
êë
æ 1
1 öù
1 ö æ 1
1 ö æ 1
- ÷ú
+ç
+ç
+ç
÷
÷
18
19
19
20
20
21
è
ø è
ø è
ø ûú
S20
1 ö 120
æ
= 6 ç1 - ÷ =
è 21ø 21
k
21
On comparing (i) and (ii), we get
k = 120
Given that S20 =
.....(i)
.....(ii)
7 77 777
128. (c) Let S =
+
+
+ ...¼+ up to 20 terms
10 100 103
é 1 11 111
ù
= 7ê +
+ 3 +¼... + up to 20 termsú
10
100
10
ë
û
Multiply and divide by 9
=
=
7 é 9 99 999
ù
+
+
+ ...¼up to 20 termsú
9 êë10 100 1000
û
éæ
1ö æ
1 ö æ
1 öù
7 êç1 - ÷ + ç1 - 2 ÷ + ç1 - 3 ÷ ú
10
è
ø
è
ø
è
øú
10
10
9ê
êë
+¼...up to 20 terms úû
20
7 é179 1 æ 1 ö ù
7
+ ç ÷ ú =
[179 + (10)–20]
ê
9 ëê 9 9 è 10 ø ûú 81
13
=
å
n =1
å
n =1
13
Tn = å (2n - 1) 2
n =1
13
13
n =1
n =1
4n 2 + å 1 - å 4n = 4 å n 2 + 13 - 4 å n
n( n + 1)
é n (n + 1) (2 n + 1) ù
= 4ê
úû + 13 - 4
6
2
ë
Put n = 13, we get
Sn = 26 × 14 × 9 + 13 – 26 × 14
= 3276 + 13 – 364 = 2925.
130. (c) 22 + 2(4)2 + 3(6)2 + ...... upto 10 terms
= 22 [13 + 23 + 33 + ...... upto 10 terms]
2
æ 10 ´ 11 ö
= 4.ç
÷ = 12100
è 2 ø
131. (c) Given sum is
3
5
7
+
+
+ .....
12 12 + 22 12 + 22 + 32
nth term = Tn
=
2n + 1
6
=
n ( n + 1) (2n + 1) n( n + 1)
6
1 ù
é1
or Tn = 6 ê ë n n + 1 úû
\ Sn =
å
Tn = 6 å
1
1
6n
6
-6å
=
n +1 n n +1
n
6
6n
=
n +1 n +1
So, sum upto 11 terms means
=6-
S11 =
6 ´ 11 66 33 11
=
=
=
11 + 1 12 6
2
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EBD_8344
M-120
Mathematics
132. (c) Tr =
1
2
=
1 + 2 + 3 + ... + r r ( r + 1)
15 + 16
Thus, given series upto 15 terms is
10
1
r ù
é r +1
=2å ê
(
1)
(
1)
(
r
r
r
r
r
r
+
+
+ 1) úû
ë
r =1
r =1
10
S10 = 2 å
1
1 ö
æ1
= 2å ç ÷
r =1 è r r + 1 ø
éæ 1 1 ö æ 1 1 ö æ 1 1 ö
æ 1 1 öù
= 2 êç - ÷ + ç - ÷ + ç - ÷ + ... + ç - ÷ ú
è 10 11 ø û
ëè 1 2 ø è 2 3 ø è 3 4 ø
10 20
é 1ù
= 2 ê1 - ú = 2 ´ =
ë 11 û
11 11
133. (b) nth term of the given series
((n - 1)3 - n3 )
= n3 - (n - 1)3
(n - 1) - n
k =1
Þ n = 20 which is a natural number.
Hence, both the given statements are true.
and statement 2 is correct explanation for statement 1.
134. (c) If n is odd, the required sum is
12 + 2.22 + 32 + 2.42 + ..... + 2 (n – 1)2 + n2
( n - 1)( n - 1 + 1) 2 + n2 (Q n - 1 is even )
æ n - 1 ö 2 n (n + 1)
+ 1÷ n =
=ç
è 2
ø
2
2
4 10 28
+ +
+ .... n terms
3 9 27
1 ö
æ 1ö æ 1ö æ
= 1 + ç1 + ÷ + ç1 + ÷ + ç1 + ÷ + .... n terms
è 3 ø è 9 ø è 27 ø
= (1 + 1 + 1 + .... + n terms)
æ1 1 1
ö
+ç + +
+ ....n terms ÷
3
9
27
è
ø
1æ
1ö
13 çè 3n ÷ø
1 3
= n+
= n + ´ [1 - 3- n ]
1
3 2
13
1
n + n +1
2m ( 2m + 1)
2
= m ( 2m + 1)
2
2 6 10 14
138. (a) Let S = 1 + + + + + .......¥
3 32 33 34
1
Multiplying both sides by , we get
3
1
1 2 6 10
S = + 2 + 3 + 4 + ........¥
3
3 3
3 3
Subtracting eqn. (ii) from eqn. (i), we get
....(i)
....(i)
2
4 4
4 4
S = + 2 + 3 + 4 + ........¥
3
3 3
3 3
4
2
4 3
Þ
S = 3 = ´ Þ S =3
1 3 2
3
13
1
2+ 3
139. (d) We know that ex = 1 + x +
x 2 x3
+
+ ........¥
2! 3!
Put x = – 1
\ e–1 = 1 - 1 + 1 - 1 + 1 ........¥
2! 3! 4!
1 1 1 1
- + - ........¥
2! 3! 4! 5!
140. (d) We know that
\ e–1 =
1
1
1
-n
= n + [1 - 3 ] = n + 2 2.3n
2
nth term =
15 + 16
Þ
135. (c) Given series is 1 +
+
1
2
1 4 4
4
S = 1 + + 2 + 3 + 4 + ........¥
3
3 3
3 3
2
1+ 2
3+ 4
+ ...... +
2
n
136. (c) Given series is
1
1- 2
2- 3
3- 4
15 - 16
+
+
+ ...... +
-1
-1
-1
-1
(By rationali ation)
+ 2.62 + ........ + 2(2m)2 is
å éëk 3 - ( k - 1)3 ùû Þ 8000 = n3
1
+
- 15 + 16
= -1 + 16 = -1 + 4 = 3
Hence, the required sum = 3
137. (a) The sum of the given series 12 + 2.22 + 32 + 2.42 + 52
2
=
1
= -1 + 2 - 2 + 3 - 3 + 4 + ..... - 14 + 15
= Tn = ( n - 1) + ( n - 1) n + n 2
Þ Sn =
+
1+ 2
2+ 3
This can be re-written as
10
=
1
\ 15th term =
+
1
3+ 4
+ .....
e x + e- x
x 2 x 4 x6
= 1+
+
+ .............
2
2! 4! 6!
1
Putting x = , we get
2
Downloaded from @Freebooksforjeeneet
M-121
Sequences and Series
-1
1
Q n Cr = n Cn - r
1
1
1
e2 + e 2
+
+
+ ......¥ =
1+
4.2! 16.4! 64.6!
2
1
e+
e = e +1
=
2
2 e
\
1
1
1
+
...............¥
1.2 2.3 3.4
1
1 ö
æ1
Tn =
=ç ÷
n(n + 1) è n n + 1 ø
144. (a) Let S =
141. (b) We know that
x x 2 x3
+
+ .....¥
1! 2! 3!
1 1 1
\ e = 1 + + + + .......
1! 2! 3!
ex = 1 +
-1
and e = 1 -
æ 1 1 ö æ 1 1 ö æ 1 1 ö æ 1 1ö
\ S = ç - ÷ - ç - ÷ + ç - ÷ - ç - ÷ .....
è 1 2ø è 2 3ø è 3 4 ø è 4 5ø
1 1 1
+ - + .......
1! 2! 3!
é1 1 1 1
ù
= 1 - 2 ê - + - ................¥ú
ë2 3 4 5
û
é 1 1
ù
\ e + e -1 = 2 ê1 + + + ....ú
ë 2! 4!
û
\
tn n
=
Sn 2
é
ù
x 2 x3 x 4
+
+ .......¥ ú
êQ log(1 + x) = x ë
û
2
3
4
1 1 1
e + e -1
+ + + ...... =
-1
2! 4! 6!
2
æ4ö
= 1 - 2[- log(1 + 1) + 1] = 2log 2 - 1 = log ç ÷ .
èeø
3
3
3
3
3
145. (a) 1 – 2 + 3 – 4 + ........+ 9
= 13 + 23 + 33 +.......+ 93 – 2(23 + 43 + 63 + 83)
e 2 + 1 - 2e (e - 1)2
=
2e
2e
142. (b) If n is odd, the required sum is
=
2
é
æ n(n + 1) ö ù
êQ Sn3 = ç
÷ ú
è 2 ø û
ë
12 + 2.22 + 32 + 2.4 2 + ...... + 2.( n - 1)2 + n 2
(n - 1)(n - 1 + 1) 2
+ n2
2
[Q (n–1) is even
\ using given formula for the sum of
(n–1) terms.]
=
2
tn =
tn =
0
n
C0
n
1
n
+
C0
+
1
n
1
n
+
C1
n -1
+
C1
2
n
1
n
2
146. (b) Let P = 21/4.2 2/8.23/16.........¥
+ .... +
C2
+ .... +
C2
n-2
= 21 / 4 + 2 / 8 + 3 / 16+..........¥
1
n
Cn
n
n
]
é4´5ù
= (45) 2 - 16.ê
ú = 2025 – 1600 = 425
ë 2 û
2
æ n - 1 ö 2 n (n + 1)
=ç
+ 1÷ n =
è 2
ø
2
143. (d) Sn =
[
é 9 ´10 ù
3 3
3
3
3
=ê
ú - 2.2 1 + 2 + 3 + 4
2
ë
û
...(i)
Now, let S = 1 + 2 + 3 + ........¥
4 8 16
...(ii)
1
1 2
S = + + ........¥
2
8 16
Subtracting (ii) from (i)
Cn
0
+
+
+ .... + n
Cn n Cn-1 n Cn- 2
C0
Adding (i) and (ii), we get,
n
é 1
1
1 ù
2tn = (n) ê n
+ n + .... n ú = nSn
C1
Cn ûú
ëê C0
Þ
1
1 1 1
S = + + + ........¥
2
4 8 16
or
1
a
1/ 4
1
=
= Þ S =1
S=
2
1 - r 1 - 1/ 2 2
\
P = 2S = 2
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......(i)
......(ii)
EBD_8344
M-122
Mathematics
10
Straight Lines and
Pair of Straight Lines
5.
TOPIC Ć
Distance Formula, Section
Formula, Results of Triangle,
Locus, Equation of Locus, Slope of
a Straight Line, Slope of a line
joining two points, Parallel and
Perpendicular Lines
6.
1.
A triangle ABC lying in the first quadrant has two vertices as
A(1, 2) and B(3, 1). If ÐBAC = 90°, and ar (DABC ) = 5 5 sq.
units, then the abscissa of the vertex C is:
[Sep. 04, 2020 (I)]
2.
(a) – 2
3.
(b) – 4
(c)
(d)
14
15
8.
If a DABC has vertices A(–1, 7), B(–7, 1) and C(5, –5), then
its orthocentre has coordinates :
[Sep. 03, 2020 (II)]
æ 3 3ö
(a) ç - , ÷
è 5 5ø
æ3 3ö
(c) ç , - ÷
è5 5ø
4.
7.
(a) 1 + 5 (b) 1 + 2 5 (c) 2 + 5 (d) 2 5 - 1
If the perpendicular bisector of the line segment oining
the points P (1, 4) and Q (k, 3) has y-intercept equal to – 4,
then a value of k is :
[Sep. 04, 2020 (II)]
9.
(d) (3, –3)
æ3 ö
Let A(l, 0), B(6, 2) and C ç , 6÷ be the vertices of a triangle
è2 ø
æ 7 1ö
PQ, where Q is the point çè - , - ÷ø , is ¾¾¾.
6 3
[NA Jan. 7, 2020 (I)]
æ 7ö
æ1 ö
æ1 ö
æ1 5ö
(a) ç1, ÷ (b) ç , 2 ÷ (c) ç ,1÷ (d) ç , ÷
è 3ø
è3 ø
è3 ø
è 3 3ø
Let O(0, 0) and A(0, 1) be two fixed points. Then the locus
of a point P such that the perimeter of DAOP is 4, is :
[April 8, 2019 (I)]
(a) 8x2 – 9y2 + 9y = 18
(b) 9x2 – 8y2 + 8y = 16
(c) 9x2 + 8y2 – 8y = 16
(d) 8x2 + 9y2 – 9y = 18
Two vertices of a triangle are (0, 2) and (4, 3). If its
orthocentre is at the origin, then its third vertex lies in
which quadrant?
[Jan. 10, 2019 (II)]
(a) third
(b) second
(c) first
(d) fourth
Let the orthocentre and centroid of a triangle be A(–3, 5)
and B(3, 3) respectively. If C is the circumcentre of this
triangle, then the radius of the circle having line segment
AC as diameter, is :
[2018]
(a) 2 10
(b) (–3, 3)
ABC. If P is a point inside the triangle ABC such that the
triangles APC, APB and BPC have equal areas, then the
length of the line segment
A triangle has a vertex at (1, 2) and the mid points of the
two sides through it are (–1, 1) and (2, 3). Then the centroid
of this triangle is :
[April 12, 2019 (II)]
10.
(b) 3
5
2
(c)
3 5
2
(d)
10
A square, of each side 2, lies above the x-axis and has one
vertex at the origin. If one of the sides passing through the
origin makes an angle 30 with the positive direction of the
x-axis, then the sum of the x-coordinates of the vertices of
the square is :
[Online April 9, 2017]
(a) 2 3 - 1 (b) 2 3 - 2 (c) 3 - 2 (d) 3 - 1
A ray of light is incident along a line which meets another
line, 7x – y + 1 = 0, at the point (0, 1). The ray is then
reflected from this point along the line, y + 2x = 1. Then the
equation of the line of incidence of the ray of light is :
[Online April 10, 2016]
(a) 41x – 25y + 25 = 0
(b) 41x + 25y – 25 = 0
(c) 41x – 38y + 38 = 0
(d) 41x + 38y – 38 = 0
Downloaded from @Freebooksforjeeneet
M-123
Straight Lines and Pair of Straight Lines
11.
12.
13.
14.
15.
16.
17.
Let L be the line passing through the point P(1, 2) such
that its intercepted segment between the co-ordinate axes
is bisected at P. If L1 is the line perpendicular to L and
passing through the point (–2, 1), then the point of
intersection of L and L1 is :
[Online April 10, 2015]
æ 4 12 ö
(a) çè , ÷ø
5 5
æ 3 23ö
(b) çè , ÷ø
5 10
æ 11 29 ö
(c) çè , ÷ø
20 10
æ 3 17 ö
(d) çè , ÷ø
10 5
(a) (3x + 1) 2 + (3 y ) 2 = a 2 - b 2
(b) (3x - 1) 2 + (3 y ) 2 = a 2 - b 2
(c) (3x - 1) 2 + (3 y ) 2 = a 2 + b 2
(d) (3x + 1) 2 + (3 y ) 2 = a 2 + b 2
19.
æ 8ö
The points çè 0, ÷ø , (1, 3) and (82, 30) :
3
[Online April 10, 2015]
(a) form an acute angled triangle.
(b) form a right angled triangle.
(c) lie on a straight line.
(d) form an obtuse angled triangle.
The x-coordinate of the incentre of the triangle that has
the coordinates of mid points of its sides as (0, 1) (1, 1) and
(1, 0) is :
[2013]
(a) 2 + 2 (b) 2 - 2 (c) 1 + 2 (d) 1 - 2
A light ray emerging from the point source placed at P( l, 3)
is reflected at a point Q in the axis of x. If the reflected ray
passes through the point R (6, 7), then the abscissa of Q is:
[Online April 9, 2013]
7
5
(a) 1
(b) 3
(c)
(d)
2
2
Let A (h, k), B(1, 1) and C (2, 1) be the vertices of a right
angled triangle with AC as its hypotenuse. If the area of
the triangle is 1square unit, then the set of values which 'k'
can take is given by
[2007]
(a) {–1, 3} (b) {–3, –2} (c) {1, 3}
(d) {0, 2}
If a vertex of a triangle is (1, 1) and the mid points of two
sides through this vertex are (–1, 2) and (3, 2) then the
centroid of the triangle is
[2005]
7ö
æ
æ -1 7 ö
æ 7ö
1 7
(a) ç - 1, ÷ (b) ç , ÷ (c) ç1, ÷ (d) æç , ö÷
3ø
3
3
3
è
ø
è
ø
è
è3 3ø
If the equation of the locus of a point equidistant from the
(a)
a12 + b12 - a 2 2 - b2 2
(c) a12 - a2 2 + b12 - b2 2
18.
(b)
1 2
a2 + b2 2 - a12 - b12
2
(d)
1 2
( a1 + a2 2 + b12 + b2 2 )
2
Locus of centroid of the triangle whose vertices are
(a cos t , a sin t ), (b sin t , - b cos t ) and (1, 0), where t is a
parameter, is
[2003]
[2002]
Various Forms of Equation of a
TOPIC n Line
20.
Let 墰: R ® R be defined as
ì 5
æ 1ö
2
ï x sin çè x ÷ø + 5 x , x < 0
ïï
f ( x) = í
x=0
0,
ï
1
æ
ö
ï x 5 cos ç ÷ + lx 2 , x > 0
è xø
ïî
21.
22.
The value of l for which 墰" (0) exists, is ______.
[NA Sep. 06, 2020 (I)]
Let C be the centroid of the triangle with vertices (3, –1),
(1, 3) and (2, 4). Let P be the point of intersection of the
lines x + 3y – 1 = 0 and 3x – y + 1 = 0. Then the line passing
through the points C and P also passes through the point:
[Jan. 9, 2020 (I)]
(a) (–9, –6) (b) (9, 7)
(c) (7, 6)
(d) (–9, –7)
Slope of a line passing through P(2, 3) and intersecting
the line x + y = 7 at a distance of 4 units from P, is:
[April 9, 2019 (I)]
1- 5
7 -1
5 -1
(d)
(b) 1 - 7 (c)
1+ 5
7 +1
5 +1
1+ 7
A point on the straight line, 3x + 5y = 15 which is
equidistant from the coordinate axes will lie only in :
[April 8, 2019 (I)]
(a) 4th quadrant
(b) 1st quadrant
(c) 1st and 2nd quadrants (d) 1st, 2nd and 4th quadrants
Two vertical poles of heights, 20 m and 80 m stand apart
on a hori ontal plane. The height (in meters) of the point
of intersection of the lines oining the top of each pole to
the foot of the other, from this hori ontal plane is :
[April 08, 2019 (II)]
(a) 15
(b) 18
(c) 12
(d) 16
If a straight line passing through the point P(–3, 4) is such
that its intercepted portion between the coordinate axes is
bisected at P, then its equation is :
[Jan. 12, 2019 (II)]
(a) 3x – 4y + 25 = 0
(b) 4x – 3y + 24 = 0
(c) x – y + 7 = 0
(d) 4x + 3y = 0
(a)
23.
point (a1, b1 ) and (a2, b2 ) is
(a1 - b2 ) x + (a1 - b2 ) y + c = 0 , then the value of c is
[2003]
A triangle with vertices (4, 0), (–1, –1), (3, 5) is
(a) isosceles and right angled
(b) isosceles but not right angled
(c) right angled but not isosceles
(d) neither right angled nor isosceles
24.
25.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-124
26.
27.
Mathematics
If in a parallelogram ABDC, the coordinates of A, B and C
are respectively (1, 2), (3, 4) and (2, 5), then the equation of
the diagonal AD is :
[Jan. 11, 2019 (II)]
(a) 5x – 3y + 1 = 0
(b) 5x + 3y – 11 = 0
(c) 3x – 5y + 7 = 0
(d) 3x + 5y – 13 = 0
A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and
R(3, –2) are fixed points, then the locus of the centroid of
DPQR is a line:
[Jan. 10, 2019 (I)]
(a) with slope
3
2
34.
by 2 3 units. If the new points Q lies in the third quadrant,
then the equation of the line passing through Q and
perpendicular to L is :
[Online April 9, 2016]
35.
(b) parallel to x-axis
2
(d) parallel to y-axis
3
If the line 3x + 4y – 24 = 0 intersects the x-axis at the point
A and the y-axis at the point B, then the incentre of the
triangle OAB, where O is the origin, is: [Jan. 10, 2019 (I)]
(a) (3, 4) (b) (2, 2)
(c) (4, 3)
(d) (4, 4)
A straight line through a fixed point (2, 3) intersects the
coordinate axes at distinct points P and Q. If O is the origin
and the rectangle OPRQ is completed, then the locus of R
is :
[2018]
29.
(a) 2x + 3y = xy
30.
31.
32.
33.
æ1 8ö
æ 10 7 ö
(a) ç , - ÷
(b) ç - , - ÷
è 3 3ø
è 3 3ø
(c) (–3, –9)
(d) (–3, –8)
A straight line through origin O meets the lines 3y = 10 – 4x
and 8x + 6y + 5 = 0 at points A and B respectively. Then O
divides the segment AB in the ratio :
[Online April 10, 2016]
(a) 2 : 3
(b) 1 : 2
(c) 4 : 1
(d) 3 : 4
If a variable line drawn through the intersection of the
lines
(c) x + y = 3 - 3 6
(d) x + y = 3 - 2 6
A straight line L through the point (3, – 2) is inclined at
(b)
36.
37.
38.
39.
3x+2–3 3 = 0
3y+x–3+2 3 =0
(c) y –
x y
x y
+ = 1 and + = 1 , meets the coordinate axes
3 4
4 3
at A and B, (A ¹ B) , then the locus of the midpoint of AB
is :
[Online April 9, 2016]
(a) 7xy = 6 (x + y)
(b) 4 (x + y)2 – 28 (x + y) + 49 = 0
(c) 6xy = 7 (x + y)
(d) 14 (x + y)2 – 97 (x + y) + 168 = 0
(b) 2x + 2y = 1 - 6
(a) y +
(b) 3x + 2y = xy
(c) 3x + 2y = 6xy
(d) 3x + 2y = 6
In a triangle ABC, coordianates of A are (1, 2) and the
equations of the medians through B and C are x + y = 5 and
x = 4 respectively. Then area of D ABC (in sq. units) is
[Online April 15, 2018]
(a) 5
(b) 9
(c) 12
(d) 4
Two sides of a rhombus are along the lines, x – y + 1 = 0
and 7x – y – 5 = 0. If its diagonals intersect at (–1, –2), then
which one of the following is a vertex of this rhombus?
[2016]
(a) x + y = 2 - 6
an angle of 60 to the line 3 x + y = 1. If L also
intersects the x-axis, then the equation of L is :
[Online April 11, 2015]
(c) with slope
28.
The point (2, 1) is translated parallel to the line L : x – y = 4
3x+2+ 3 3 = 0
(d) 3 y – x + 3 + 2 3 = 0
The circumcentre of a triangle lies at the origin and its
centroid is the mid point of the line segment oining the
points (a2 + 1, a2 + 1) and (2a, – 2a), a ¹ 0. Then for any a,
the orthocentre of this triangle lies on the line:
[Online April 19, 2014]
(a) y – 2ax = 0
(b) y – (a2 + 1)x = 0
(c) y + x = 0
(d) (a – 1)2x – (a + 1)2y = 0
If a line intercepted between the coordinate axes is trisected
at a point A(4, 3), which is nearer to x-axis, then its equation
is:
[Online April 12, 2014]
(a) 4x – 3y = 7
(b) 3x + 2y = 18
(c) 3x + 8y = 36
(d) x + 3y = 13
Given three points P, Q, R with P(5, 3) and R lies on the
x-axis. If equation of RQ is x – 2y = 2 and PQ is parallel to
the x-axis, then the centroid of DPQR lies on the line:
[Online April 9, 2014]
(a) 2x + y – 9 = 0
(b) x – 2y + 1 = 0
(c) 5x – 2y = 0
(d) 2x – 5y = 0
A ray of light along x + 3 y = 3 gets reflected upon
reaching x-axis, the equation of the reflected ray is [2013]
(a) y = x +
(c) y =
40.
3
(b)
3y = x –
3x - 3
(d)
3y = x -1
3
Let A (–3, 2) and B (–2, 1) be the vertices of a triangle ABC.
If the centroid of this triangle lies on the line 3x + 4y + 2 = 0,
then the vertex C lies on the line :
[Online April 25, 2013]
(a) 4x + 3y + 5 = 0
(b) 3x + 4y + 3 = 0
(c) 4x + 3y + 3 = 0
(d) 3x + 4y + 5 = 0
Downloaded from @Freebooksforjeeneet
M-125
Straight Lines and Pair of Straight Lines
41.
If the extremities of the base of an isosceles triangle are the
points (2a, 0) and (0, a) and the equation of one of the
sides is x = 2a, then the area of the triangle, in square units,
is :
[Online April 23, 2013]
(a)
42.
43.
5 2
a
4
(b)
5 2
a
2
(c)
25a
4
49.
2
(d) 5a2
50.
If the x-intercept of some line L is double as that of the line,
3x + 4y = 12 and the y-intercept of L is half as that of the
same line, then the slope of L is : [Online April 22, 2013]
(a) – 3
(b) – 3/8
(c) – 3/2
(d) – 3/16
If the line 2x + y = k passes through the point which divides
the line segment oining the points (1,1) and (2,4) in the
ratio 3 :2, then k equals :
[2012]
11
29
(b) 5
(c) 6
(d)
5
5
The line parallel to x-axis and passing through the point of
intersection of lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0,
where (a, b) ¹ (0, 0) is
[Online May 26, 2012]
(a) above x-axis at a distance 2/3 from it
(b) above x-axis at a distance 3/2 from it
(c) below x-axis at a distance 3/2 from it
(d) below x-axis at a distance 2/3 from it
If the point (1, a) lies between the straight lines
x + y = 1 and 2(x + y) = 3 then a lies in interval
[Online May 12, 2012]
A straight line through the point A (3, 4) is such that
its intercept between the axes is bisected at A. Its equation
is
[2006]
(a) x + y = 7
(b) 3x - 4 y + 7 = 0
(c) 4 x + 3 y = 24
(d) 3 x + 4 y = 25
The line parallel to the x- axis and passing through the
intersection of the lines ax + 2by + 3b = 0 and
bx – 2ay – 3a = 0, where (a, b) ¹ (0, 0) is
[2005]
(a) below the x - axis at a distance of
3
from it
2
(b) below the x - axis at a distance of
2
from it
3
(c) above the x - axis at a distance of
3
from it
2
(d) above the x - axis at a distance of
2
from it
3
(a)
44.
45.
æ3 ö
æ 3ö
(a) ç , ¥ ÷ (b) ç1, ÷
è2 ø
è 2ø
46.
47.
(c)
( -¥, 0 )
51.
æ 1ö
(d) ç 0, ÷
è 2ø
If the straight lines x + 3y = 4, 3x + y = 4 and x + y = 0 form
a triangle, then the triangle is
[Online May 7, 2012]
(a) scalene
(b) equilateral triangle
(c) isosceles
(d) right angled isosceles
If A (2, – 3) and B (– 2, 1) are two vertices of a triangle and
52.
48.
(b) 2 x + 3 y = 1
(c) 2 x + 3 y = 3
(d) 2 x - 3 y = 1
x > 0 and y = 3x , x > 0 , then a belong to
1
(a) æç 0, ö÷ (b) (3, ¥)
è 2ø
æ1 ö
(c) ç , 3÷
è 2 ø
[2006]
1ö
æ
(d) ç - 3, - ÷
2ø
è
x y
x y
- = 1 and
+ =1
2 3
-2 1
(b)
x y
x y
- = -1 and
+ = -1
2 3
-2 1
(c)
x y
x y
+ = 1 and + = 1
2 3
2 1
(d)
x y
x y
+ = -1 and
+ = -1
2 3
-2 1
Let A(2, - 3) and B ( -2, 3) be vertices of a triangle ABC.
[2004]
53.
x
If (a, a ) falls inside the angle made by the lines y = ,
2
2
(a)
If the centroid of this triangle moves on the line
2 x + 3 y = 1, then the locus of the vertex C is the line
third vertex moves on the line 2 x + 3 y = 9, then the locus
of the centroid of the triangle is :
[2011RS]
(a) x - y = 1
The equation of the straight line passing through the point
(4, 3) and making intercepts on the co-ordinate axes whose
sum is –1 is
[2004]
(a) 3x - 2 y = 3
(b) 2 x - 3 y = 7
(c) 3x + 2 y = 5
(d) 2 x + 3 y = 9
Locus of mid point of the portion between the axes of
x cos a + y sina = p whre p is constant is
[2002]
(a) x2 + y2 =
(c)
1
x
2
+
1
y
2
4
(b) x2 + y2 = 4p2
p2
=
2
p
2
(d)
1
x
2
Downloaded from @Freebooksforjeeneet
+
1
y
2
=
4
p2
EBD_8344
M-126
Mathematics
59.
TOPIC Đ
54.
55.
56.
57.
Distance Between two Lines, Angle
Between two Lines and Bisector of
the Angle Between the two Lines,
Perpendicular Distance of a Point
from a Line, Foot of the
Perpendicular, Position of a Point
with Respect to a Line, Pedal
Points, Condition for Concurrency
of Three Lines
æ 29 8 ö
(b) ç , ÷
è 5 5ø
æ 8 29 ö
(c) ç , ÷
è5 5 ø
æ 29 11 ö
(d) ç , ÷
è 5 5ø
If the line, 2x – y + 3 = 0 is at a distance
1
60.
61.
62.
and
2
from
5
5
the lines 4x – 2y + a = 0 and 6x – 3y + b = 0, respectively,
then the sum of all possible value of a and b is ______.
[NA Sep. 05, 2020 (I)]
The locus of the mid-points of the perpendiculars drawn
from points on the line, x = 2y to the line x = y is:
[Jan. 7, 2020 (II)]
(a) 2x – 3y = 0
(b) 5x – 7y = 0
(c) 3x – 2y = 0
(d) 7x – 5y = 0
A straight line L at a distance of 4 units from the origin
makes positive intercepts on the coordinate axes and the
perpendicular from the origin to this line makes an angle
of 60o with the line x + y = 0. Then an equation of the line
L is:
[April 12, 2019 (II)]
(a) x + 3 y = 8
63.
64.
65.
(c) 3x + y = 8
(d) None of these
Lines are drawn parallel to the line 4x – 3y + 2 = 0, at a
3
from the origin. Then which one of the
5
following points lies on any of these lines ?
[April 10, 2019 (II)]
distance
æ 1 2ö
(a) ç - , ÷
è 4 3ø
æ 1 1ö
(b) ç , - ÷
è 4 3ø
æ1 1ö
(c) ç , ÷
è 4 3ø
æ 1 2ö
(d) ç - , - ÷
è 4 3ø
35
35
(d) 5
(b) –5
(c) 3
3
Two sides of a parallelogram are along the lines, x + y = 3
and x – y + 3 = 0. If its diagonals intersect at (2, 4), then
one of its vertex is:
[Jan. 10, 2019 (II)]
(a) (3, 5) (b) (2, 1)
(c) (2, 6)
(d) (3, 6)
Consider the set of all lines px + qy + r = 0 such that
3p + 2q + 4r = 0. Which one of the following statements is
true?
[Jan. 9, 2019 (I)]
(a)
(b) ( 3 + 1) x + ( 3 - 1) y = 8 2
58.
2
2
2
2
(b)
(c)
(d)
5
5
5
5
A rectangle is inscribed in a circle with a diameter lying
along the line 3y = x + 7. If the two adacent vertices of the
rectangle are (–8, 5) and (6, 5), then the area of the rectangle
(in sq. units) is:
[April 09, 2019 (II)]
(a) 84
(b) 98
(c) 72
(d) 56
Suppose that the points (h, k), (1, 2) and (– 3, 4) lie on the
line L1. If a line L2 passing through the points (h, k) and
(4, 3) is perpendicular on L1, then equals :
[April 08, 2019 (II)]
1
1
(a) 3
(b) 0
(c) 3
(d) –
7
If the straight line, 2x – 3y + 17 = 0 is perpendicular to the
line passing through the points (7, 17) and (15, b), then b
equals :
[Jan. 12, 2019 (I)]
(a)
Let L denote the line in the xy-plane with x and y intercepts
as 3 and 1 respectively. Then the image of the point (–1, – 4)
in this line is:
[Sep. 06, 2020 (II)]
æ 11 28 ö
(a) ç , ÷
è5 5 ø
If the two lines x + (a–1) y = 1 and 2x + a2y =1 (a Î R – {0,1})
are perpendicular, then the distance of their point of
intersection from the origin is:
[April 09, 2019 (II)]
66.
67.
æ3 1ö
(a) The lines are concurrent at the point ç , ÷ .
è4 2ø
(b) Each line passes through the origin.
(c) The lines are all parallel.
(d) The lines are not concurrent.
Let the equations of two sides of a triangle be
3x – 2y + 6 = 0 and 4x + 5y – 20 = 0. If the orthocentre
of this triangle is at (1, 1), then the equation of its third
side is:
[Jan. 09, 2019 (II)]
(a) 122y – 26x – 1675 = 0
(b) 122y + 26x + 1675 = 0
(c) 26x + 61y + 1675 = 0
(d) 26x – 122y – 1675 = 0
The foot of the perpendicular drawn from the origin, on
the line, 3x + y = l(l ¹ 0) is P. If the line meets x-axis at A and
y-axis at B, then the ratio BP : PA is
[Online April 15, 2018]
(a) 9 : 1
(b) 1 : 3
(c) 1 : 9
(d) 3 : 1
The sides of a rhombus ABCD are parallel to the lines,
x – y + 2 = 0 and 7x – y + 3 = 0. If the diagonals of the
rhombus intersect at P(1, 2) and the vertex A (different
from the origin) is on the y-axis, then the ordinate of A is
[Online April 15, 2018]
(a) 2
(b)
7
4
(c)
7
2
Downloaded from @Freebooksforjeeneet
(d)
5
2
M-127
Straight Lines and Pair of Straight Lines
68.
69.
70.
Let a, b, c and d be non- ero numbers. If the point of
intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d
=0 lies in the fourth quadrant and is equidistant from the
two axes then
[2014]
(a) 3bc – 2ad = 0
(b) 3bc + 2ad = 0
(c) 2bc – 3ad = 0
(d) 2bc + 3ad = 0
Let PS be the median of the triangle vertices
P(2, 2), Q(6, –1) and R(7, 3). The equation of the line
passing through (1, –1) and parallel to PS is:
[2014]
(a) 4x + 7y + 3 = 0
(b) 2x – 9y – 11 = 0
(c) 4x – 7y – 11 = 0
(d) 2x + 9y + 7 = 0
If a line L is perpendicular to the line 5x – y = 1, and the area
of the triangle formed by the line L and the coordinate axes
is 5, then the distance of line L from the line x + 5y = 0 is:
[Online April 19, 2014]
(a)
71.
72.
7
(b)
5
(c)
7
(d)
76.
2 3
4 3
4 3
2 3
(b)
(c)
(d)
15
15
5
5
If the image of point P(2, 3) in a line L is Q(4, 5), then the
image of point R(0, 0) in the same line is:
[Online April 25, 2013]
(a) (2, 2) (b) (4, 5)
(c) (3, 4)
(d) (7, 7)
Let q1 be the angle between two lines 2x + 3y + c1 = 0 and
– x + 5y + c2 = 0 and q2 be the angle between two lines
2x + 3y + c1 = 0 and – x + 5y + c3 = 0, where c1, c2, c3 are any
real numbers :
Statement-1: If c2 and c3 are proportional, then q1 = q2.
Statement-2: q1 = q2 for all c2 and c3.
[Online April 23, 2013]
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation of Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation of Statement-1.
(c) Statement-1 is false; Statement-2 is true.
(d) Statement-1 is true; Statement-2 is false.
If the three lines x – 3y = p, ax + 2y = q and
ax + y = r form a right-angled triangle then :
[Online April 9, 2013]
(a) a2 – 9a + 18 = 0
(b) a2 – 6a – 12 = 0
(c) a2 – 6a – 18 = 0
(d) a2 – 9a + 12 = 0
(b) L1 ^ L2 , L1 P L3 , L1 intersect L2.
(c) L1 ^ L2 , L2 P L3 ,L1 intersect L4.
(d) L1 ^ L2 , L1 ^ L3 , L2 intersect L4.
77.
78.
74.
75.
If a, b, c Î R and 1 is a root of equation ax2 + bx + c = 0,
then the curve y = 4ax2 + 3bx + 2c, a ¹ 0 intersect x-axis at
[Online May 26, 2012]
(a) two distinct points whose coordinates are always
rational numbers
(b) no point
(c) exactly two distinct points
(d) exactly one point
Let L be the line y = 2x, in the two dimensional plane.
[Online May 19, 2012]
Statement 1: The image of the point (0, 1) in L is the point
æ 4 3ö.
ç , ÷
è5 5ø
(a)
73.
[Online May 26, 2012]
(a) L1 P L4 , L2 P L3 , L1 intersect L4.
5
5
13
13
7
If the three distinct lines x + 2ay + a = 0, x + 3by + b = 0 and
x + 4ay + a = 0 are concurrent, then the point (a, b) lies on
a:
[Online April 12, 2014]
(a) circle
(b) hyperbola
(c) straight line
(d) parabola
The base of an equilateral triangle is along the line given
by 3x + 4y = 9. If a vertex of the triangle is (1, 2), then the
length of a side of the triangle is: [Online April 11, 2014]
Consider the straight lines
L1 : x – y = 1
L2 : x + y = 1
L3 : 2x + 2y = 5
L4 : 2x – 2y = 7
The correct statement is
79.
80.
81.
æ 4 3ö
Statement 2: The points (0, 1) and ç , ÷ lie on opposite
è5 5ø
sides of the line L and are at equal distance from it.
(a) Statement 1 is true, Statement 2 is false.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation for Statement 1.
(c) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation for Statement 1.
(d) Statement 1 is false, Statement 2 is true.
If two vertices of a triangle are (5, –1) and (–2, 3) and its
orthocentre is at (0, 0), then the third vertex is
[Online May 12, 2012]
(a) (4, – 7) (b) (– 4, – 7) (c) (– 4, 7) (d) (4, 7)
If two vertical poles 20 m and 80 m high stand apart on a
hori ontal plane, then the height (in m) of the point of
intersection of the lines oining the top of each pole to the
foot of other is
[Online May 7, 2012]
(a) 16
(b) 18
(c) 50
(d) 15
The point of intersection of the lines
(a3 + 3)x + ay + a – 3 = 0 and
(a5 + 2)x + (a + 2)y + 2a + 3 = 0 (a real) lies on the y-axis for
[Online May 7, 2012]
(a) no value of a
(b) more than two values of a
(c) exactly one value of a (d) exactly two values of a
Downloaded from @Freebooksforjeeneet
EBD_8344
M-128
82.
The lines x + y = a and ax – y = 1 intersect each other in
the first quadrant. Then the set of all possible values of a
in the interval :
[2011RS]
(a)
83.
84.
85.
86.
87.
88.
Mathematics
( 0, ¥ )
(b) [1, ¥)
(c)
( -1, ¥)
(d)
89.
æ
è
equation of its diagonal not passing through the origin is
[2003]
(a) y (cos a + sin a ) + x(cos a - sin a ) = a
The lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line
L3 : y + 2 = 0 at P and Q respectively. The bisector of the
acute angle between L1 and L2 intersects L3 at R. [2011]
2 3
3 2
3
3 2
(a)
(b)
(c)
(d)
8
5
4
8
The perpendicular bisector of the line segment oining
P (1, 4) and Q(k, 3) has y-intercept –4. Then a possible
value of k is
[2008]
(a) 1
(b) 2
(c) –2
(d) – 4
(b) y (cos a - sin a ) - x(sin a - cos a ) = a
(c) y (cos a + sin a ) + x(sin a - cos a ) = a
(d) y (cos a + sin a ) + x (sin a + cos a ) = a
TOPIC Ė Pair of Straight Lines
90.
91.
92.
3
x+ y = 0
2
(b) x + 3 y = 0
(c)
3x + y = 0
(d) x +
93.
94.
3
y=0
2
If x1, x2 , x3 and y1 , y2 , y3 are both in G.P. with the same
( x3 , y3 )
(a) are vertices of a triangle
(b) lie on a straight line
(c) lie on an ellipse
(d) lie on a circle.
[2003]
(a) –3
(b) 1
(c) 3
(d) 1
If the sum of the slopes of the lin es given by
x 2 - 2cxy - 7 y 2 = 0 is four times their product c has the
value
[2004]
(a) –2
(b) –1
(c) 2
(d) 1
95.
common ratio, then the points ( x1, y1 ), ( x 2 , y 2 ) and
The equation y = sin x sin (x + 2) – sin 2 (x + 1) represents
a straight line lying in :
[April 12, 2019 (I)]
(a) second and third quadrants only
(b) first, second and fourth quadrant
(c) first, third and fourth quadrants
(d) third and fourth quadrants only
If one of the lines of my2 + (1– m2) xy – mx2= 0 is a bisector
of the angle between the lines xy = 0, then m is [2007]
(a) 1
(b) 2
(c) –1/2
(d) –2
If one of the lines given by
6 x 2 - xy + 4cy 2 = 0 is 3x + 4y = 0, then c equals [2004]
Let P = (–1, 0), Q = (0, 0) and R = (3, 3 3 ) be three point.
The equation of the bisector of the angle PQR is [2007]
(a)
pö
4ø
angle aç 0 < a < ÷ with the positive direction of x-axis. The
( -1,1)
Statement-1: The ratio PR : RQ equals 2 2 : 5
Statement-2: In any triangle, bisector of an angle divides
the triangle into two similar triangles.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation for Statement-1.
The lines p(p2 +1)x – y + q = 0 and
(p2 + 1)2x + (p2 + 1)y + 2q = 0 are perpendicular to a common
line for :
[2009]
(a) exactly one values of p
(b) exactly two values of p
(c) more than two values of p
(d) no value of p
The shortest distance between the line y – x = 1 and the
curve x = y2 is :
[2009]
A square of side a lies above the x-axis and has one vertex
at the origin. The side passing through the origin makes an
96.
If the pair of straight lines x 2 - 2 pxy - y 2 = 0 and
x 2 - 2qxy - y 2 = 0 be such that each pair bisects the angle
between the other pair, then
[2003]
(a) pq = –1 (b) p = q
(c) p = –q (d) pq = 1.
The pair of lines represented by
3ax2 + 5xy + (a2 – 2)y2 = 0
are perpendicular to each other for
[2002]
(a) two values of a
(b) " a
(c) for one value of a
(d) for no values of a
If the pair of lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
intersect on the y-axis then
[2002]
(a) 2fgh = bg2 + ch2
(b) bg2 ¹ ch2
(c) abc = 2fgh
(d) none of these
Downloaded from @Freebooksforjeeneet
M-129
Straight Lines and Pair of Straight Lines
1.
(b) Let DABC be in the first quadrant
Slope of line AB = -
1
2
3.
A(–1, 7)
(b)
Slope of line AC = 2
Length of AB = 5
Q
P
R
Y
C
B(–7, 1)
A (1
,2)
C(5, –5)
S
90
mBC =
B (3,1)
X
6
1
=2
-12
\ Equation of AS is y - 7 = 2( x + 1)
It is given that ar(DABC) = 5 5
y = 2x + 9
1
\ AB × AC = 5 5 Þ AC = 10
2
mAC =
\ Coordinate of vertex C = (1 + 10cos q, 2 + 10sin q)
\ Equation of BP is y - 1 =
Q tan q = 2 Þ cos q =
1
5
, sin q =
x+9
2
Þ 4 x + 18 = x + 9
æ k + 1 7ö
(b) Mid point of line segment PQ be ç
.
,
è 2 2 ÷ø
\
15
1
´
= -1
1+ k 1- k
...(ii)
2x + 9 =
\ Abscissa of vertex C is 1 + 2 5.
7
+4
15
æ k + 1 7ö
=
, ÷ = 2
(0, –4) and ç
k +1
è 2 2ø
k +1
-0
2
4-3
1
=
Slope of PQ =
1- k 1- k
1
( x + 7)
2
y=
5
\ Slope of perpendicular line passin g th rough
12
= -2
-6
x 9
+
2 2
From equs. (i) and (ii),
2
\ Coordinate of C = (1 + 2 5, 2 + 4 5)
2.
...(i)
Þ 3x = 9 Þ x = -3
\y =3
4.
(5) P will be centroid of DABC
æ 17 8 ö
P ç , ÷ Þ PQ = (4)2 + (3) 2 = 5
è 6 3ø
5.
(b) From the mid-point formula co-ordinates of vertex B
and C are B (– 3, 0) and C (3, 4).
Now, centroid of the triangle
æ 3- 3+1 0 + 4 + 2 ö
æ1 ö
G ºç
,
÷ Þ G º ç ,2÷
3
è 3
ø
è3 ø
1 - k 2 = -15 Þ k = ±4.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-130
Mathematics
8.
A (1 2)
(2 3)
(–1 1)
6.
B
(c) Let point P (h, k)
Q OA = 1
So, OP + AP = 3
Þ
(b) Since Orthocentre of the triangle is A(–3, 5) and
centriod of the triangle is B(3, 3), then
AB = 40 = 2 10
A
B C
Centroid divides orthocentre and circumcentre of the
triangle in ratio 2 : 1
\ AB : BC = 2 : 1
ᔴ
2
Now, AB = AC
3
3
3
Þ AC = AB = (2 10) Þ AC = 3 10
2
2
\ Radius of circle with AC as diametre
h2 + k 2 + h 2 + (k - 1)2 = 3
=
Y
9.
AC 3
5
=
10 = 3
2
2
2
(b)
A (0 1)
B
P (h k)
1
2
C
3 - 1, 3 + 1
3
–1,
X
O (0 0)
2
X
Y
A
o
Þ
h2
+ (k –
1)2
=9+
h2 +
k2 –
6 h2 + k 2
45
o
60
O
Þ 6 h2 + k 2 = 2k + 8
3, 1
o
30
Þ 9h2 + 8k2 – 8k – 16 = 0
For A;
Hence, locus of point P is
7.
x
y
9x2 + 8y2 – 8y – 16 = 0
(b) Since, mQR ´ mPH = –1
cos30
1
Þ mQR = - m
Þ x = 3 and y = 1
For C,
PH
Þ mQR =
y -3
=0
x-4
=
x
cos120o
Þ y=3
mPQ ´ mRH = –1
sin 30o
=
=2
y
sin120o
=2
Þ x = -1, y = 3
For B,
1 y
´ = -1
4 x
Þ y = –4x
Þ
Þ x=-
o
x
cos 75o
3
4
=
y
sin 75o
=2 2
Þ x = 3 -1
and y = 3 + 1
æ -3 ö
Vertex R is çè , 3÷ø
4
Hence, vertex R lies in second quadrant.
10.
\ Sum = 2 3 - 2
(c) Let slope of incident ray be m.
\ angle of incidence = angle of reflection
Downloaded from @Freebooksforjeeneet
M-131
a
q
q
y+
2x
=1
Straight Lines and Pair of Straight Lines
a
7x–y+1=0
\
9
m-7
-2 - 7
=
=
1 + 7m
1 - 14 13
Þ
9
9
m-7
m-7
=
=or
1 + 7 m 13
1 + 7m
13
Now, x-co-ordinate of incentre is given as
ax1 + bx2 + cx3
a+b+c
Þ x-coordinate of incentre
Þ 13m – 91 = 9 + 63m or 13m – 91 = – 9 – 63m
Þ 50m = –100 or 76m = 82
Þ m=–
1
2
Þ y– 1= –
11.
or
1
(x – 0)
2
i.e x + 2y – 2 = 0
Þ 41x – 38y + 38 = 0
(a) Equation of line L
x y
+ =1
2 4
2x + y = 4
m=
41
38
or y – 1 =
=
41
(x – 0)
38
14.
2 ´ 0 + 2 2.0 + 2.2
2
=
2+ 2+ 2 2
2+ 2
(d) Let abcissa of Q = x
= 2- 2
Y
Q (6, 7)
or 38y – 38 – 41x = 0
P(1, 3)
(180 – q)
q
...(i)
O
X
Q
(x, 0)
(1, 2)
\ Q = (x, 0)
tan q =
For line
x – 2y = – 4
...(ii)
solving equation (i) and (ii); we get point of intersection
12.
Now, tan (180 – q) = – tan q
\
12 ö
æ
ç 4 / 5, ÷
5ø
è
15.
æ 8ö
(c) A ç 0, ÷ B (1, 3) C (89,30 )
è 3ø
0-7
0 -3
, tan (180° - q) =
x-6
x -1
-3
-7
5
=
Þ x =
x -1 x - 6
2
(a) Given : A(l, k), B(1, 1) and C(2, 1) are vertices of a right
angled triangle and area of DABC = 1 square unit
Y
A (1, k)
1
Slope of AB =
3
Slope of BC =
13.
1
3
B (1, 1)
So, lies on same line
(b) From the figure, we have
a = 2, b = 2 2, c = 2
x1 = 0, x2 = 0, x3 = 2
C (2, 1)
O
X
We know that, area of right angled triangle
1
1
= × BC × AB = 1 = (a) | (k – 1)|
2
2
Þ ± (k - 1) = 2 Þ k = – 1, 3
Downloaded from @Freebooksforjeeneet
EBD_8344
M-132
16.
Mathematics
(c) Vertex of triangle is (1, 1) and midpoint of sides through
this vertex is (– 1, 2) and (3, 2)
A (1, 1)
(-1, 2)
20.
ì 4
æ 1ö
æ 1ö
3
ï5 x × sin çè x ø÷ - x cos èç x ø÷ + 10 x, x < 0
ïï
0,
f '( x ) = í
x=0
ï
1
1
æ
ö
æ
ö
ï 5 x 4 cos ç ÷ + x 3 sin ç ÷ + 2 lx, x > 0
è xø
è xø
îï
(3, 2)
C
(x2, y2)
B
(x1, y1)
ì
æ 1ö
æ 1ö
3
2
ï (20 x - x )sin çè x ÷ø - 8 x cos çè x ÷ø + 10,
ïï
0,
f ''( x ) = í
ï
1
1
ï(20 x 3 - x) cos æç ö÷ + 8 x 2 sin æç ö÷ + 2l,
è
ø
è
x
xø
îï
1 + x1
1 + y1
=-1,
=2
2
2
Þ B (–3, 3)
1 + x2
1 + y2
= 3,
=2
2
2
Þ C (–5, 3)
\ Centroid is 1 - 3 + 5 , 1 + 3 + 3
3
3
æ 7ö
Þ ç1, ÷
è 3ø
17.
18.
2
2
2
(3 x - 1) + (3 y ) = a + b
19.
(a) AB =
BC =
2
21.
CA = ( 4 - 3) 2 + ( 0 - 5) 2 = 26 ;
Q AB = CA
\ Isosceles triangle
Q ( 26 )2 + ( 26 )2 = 52
x=0
x>0
(a) Coordinates of centroides
æ x + x + x y + y + y3 ö
C =ç 1 2 3, 1 2
÷
3
3
è
ø
æ 3 + 1 + 2 -1 + 3 + 4 ö
=ç
,
÷ = (2, 2)
3
è 3
ø
The given equation of lines are
x + 3y – 1 = 0
3x – y + 1 = 0
Then, from (i) and (ii)
22.
...(i)
...(ii)
æ 1 2ö
point of intersection P ç - , ÷
è 5 5ø
equation of line DP
8x – 11y + 6 = 0
(b)
Y
䰘
4
?
2
3
( 4 + 1) 2 + (0 + 1) 2 = 26 ;
(3 + 1) 2 + (5 + 1) 2 = 52
x<0
Now, f ''(0+ ) = f ''(0- ) Þ 2l = 10 Þ l = 5
(b) ( x - a1 ) 2 + ( y - b1 ) 2 = ( x - a2 )2 + ( y - b2 )2
1
( a1 - a2 ) x + (b1 - b2 ) y + (a2 2 + b2 2 - a12 - b12 ) = 0
2
Comparing with given eqn. we get
1
c = ( a2 2 + b2 2 - a12 - b12 )
2
(c) We know that centroid
æ x + x + x y + y2 + y3 ö
( x, y) = ç 1 2 3 , 1
÷ø
è
3
3
a cos t + b sin t + 1
x=
3
Þ a cos t + b sin t = 3 x - 1
a sin t - b cos t
y=
3
Þ a sin t - b cos t = 3 y
Squaring and adding,
BC2 = AB2 + AC2
\ right angled triangle,
So, the given triangle is isosceles right angled .
(5)
䰘
X
Since point at 4 units from P (2, 3) will be
A (4 cosq + 2, 4 sinq + 3) and this point will satisfy the
equation of line x + y = 7
Þ cos q + sin q =
1
2
Downloaded from @Freebooksforjeeneet
M-133
Straight Lines and Pair of Straight Lines
\ substitute the value of x from equation (i) to equation
(ii), we get
On squaring
Þ sin 2q -
3
2 tan q
3
Þ
=2
4 1 + tan q
4
y
y
+
=1
80 20
Þ y + 4y = 80 Þ y = 16 m
Þ 3tan2 q + 8tan q + 3 = 0
Þ tan q =
Þ tan q =
23.
-8 ± 2 7
6
(ignoring –ve sign)
25.
Hence, height of intersection point is 16 m.
(b) Since, P is mid point of MN
-8 ± 2 7 1 - 7
=
6
1+ 7
N(0, y)
P( 3, 4)
(c) A point which is equidistant from both the axes lies
on either y = x and y = – x.
Since, point lies on the line 3x + 5y = 15
Then the required point
M (x, 0)
3x + 5 y = 15
x + y = 0
15
x= 2
0+ x
= –3
2
Þ x = –3 ´ 2 Þ x = –6
Then,
y+0
=4Þy+0=2´4Þy=8
2
Hence required equation of straight line MN is
æ 15 15 ö
15
y=
Þ (x, y) = ç - , ÷ {2nd quadrant}
2
è 2 2ø
and
3x + 5 y = 15
x–y =0
or
15
x=
8
x y
+ = 1 Þ 4x – 3y + 24 = 0
-6 8
y=
26.
(a) Since, in parallelogram mid points of both diagonals
coinsides.
æ 15 15 ö
15
Þ (x, y) = ç , ÷ {1st quadrant}
8
è8 8ø
\ mid-point of AD = mid-point of BC
Hence, the required point lies in 1st and 2nd quadrant.
24.
y
(d)
B (x, 80)
C (0, 20)
20
O
80
D
A
(x1, 0)
x
B
æ x1 +1 y1 + 2 ö æ 3 + 2 4 + 5 ö
,
,
÷
çè
÷ =ç
2
2 ø è 2
2 ø
\ (x1, y1) = (4, 7)
Then, equation of AD is,
Equations of lines OB and AC are respectively
80
x
y=
x1
x
y
+
x1 20 = 1
y–7=
2–7
(x – 4)
1– 4
y–7=
5
(x – 4)
3
...(i)
...(ii)
Q equations (i) and (ii) intersect each other
3y – 21 = 5x – 20
5x – 3y + 1 = 0
Downloaded from @Freebooksforjeeneet
EBD_8344
M-134
27.
Mathematics
P(h, k)
(c)
2 3
+ = 1 or 3x + 2y = xy..
x y
(b) Median through C is x = 4
So the x coordianate of C is 4. let C º (4, y), then the
midpoint of A (1, 2) and C (4, y) is D which lies on the median
through B.
Then, the locus of R is
30.
C(a, b)
A
R(3, –2)
Q(1, 4)
Let centroid C be (a, b)
D
1+ 3 + h
we have a =
Þ h = 3a – 4
3
G
4- 2+ k
Þ k = 3b – 2
3
but P(h, k) lies on 2x – 3y + 4 = 0
Þ 2(3a – 4) – 3(3b – 2) + 4 = 0
Þ 6a – 9b – 8 + 6 + 4 = 0
Þ 6a – 9b + 2 = 0
Locus: 6x – 9y + 2 = 0
B
b=
æ1+ 4 2 + y ö
,
\ Dº ç
2 ÷ø
è 2
1+ 4 + 2 + y
= 5 Þ y = 3.
2
So, C º (4, 3).
The centroid of the triangle is the intersection of the
mesians. Here the medians x = 4 and x + 4 and x + y = 5
intersect at G (4, 1).
The area of triangle D ABC = 3 × D AGC
Now ,
æ 8 ´ 0 + 6 ´ 8 + 10 ´ 0 8 ´ 6 + 6 ´ 0 + 10 ´ 0 ö
,
I ºç
÷ º (2, 2).
è
8 + 6 + 10
8 + 6 + 10
ø
= 3´
D
A
29.
(b) Equation of PQ is
x y
+ =1
h k
...(i)
R(h, k)
(0, k)Q
C
5=0
O (–1,–2)
x–y+l=0
B
Let other two sides of rhombus are
x– y+ l = 0
and 7x –y + m = 0
then O is equidistant from AB and DC and from AD and
BC
\ -1 + 2 + 1 = -1 + 2 + l Þ l = – 3
and -7 + 2 - 5 = -7 + 2 + m Þ m = 15
(2, 3)
O
x – y +1= 0
0
(a)
+m=
31.
1
[1 (1 – 3) + 4 (3 – 2) + 4(2 – 1)] = 9.
2
7x – y
x y
or + = 1
8 6
\ coordinates of A, B & O are (8, 0), (0, 6) & (0, 0)
respectively.
Þ OA = 8, OB = 6 & AB = 10.
\ Incentre of DOAB is given as:
7x – y
–
6
2
6 2
x + \ its slope = =
9
9
9 3
(b) Equation of the line is:
3x + 4y = 24
Þ y=
28.
C
P(h, 0)
Since, (i) passes through the fixed point (2, 3) Then,
2 3
+ =1
h k
\ Other two sides are x – y – 3 = 0 and
7x – y + 15 = 0
\ On solving the eqns of sides pairwise, we get
the vertices as
æ 1 -8 ö
æ -7 -4 ö
ç 3 , 3 ÷ , (1, 2), ç 3 , 3 ÷ , ( -3, -6 )
è
ø
è
ø
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M-135
Straight Lines and Pair of Straight Lines
32.
(c) Length of ^ to 4x + 3y = 10 from origin (0, 0)
35.
5 1
=
10 2
Q Lines are parallel to each other Þ ratio will be 4 : 1 or 1 : 4
(a) L1 : 4x + 3y – 12 = 0
L2 : 3x + 4y – 12 = 0
L1 + lL2 = 0
(4x + 3y – 12) + l (3x + 4y – 12) = 0
x (4 + 3l) + y (3 + 4l) – 12 (1 + l) = 0
P2 =
æ 12(1 + l) ö
, 0÷
Point A ç
è 4 + 3l
ø
\ tan 60 =
mid point Þ h =
m1 - (- 3)
1 + ( - 3m1 )
Þ m1 = 0, m2 = 3
Since line L is passing through (3, –2)
\ y – (–2) = +
Þ y+ 2 =
36.
æ 12(1 + l) ö
Point B ç 0,
÷
è 3 + 4l ø
6 (1 + l )
4 + 3l
3(x - 3)
y – 3x + 2 + 3 3 = 0
(d) Circumcentre = (0, 0)
We know the circumcentre (O),
Centroid (G) and orthocentre (H) of a triangle lie on the
line oining the O and G.
..... (i)
6 (1 + l )
3 + 4l
Eliminate l from (i) and (ii), then
6(h + k) = 7hk
6(x + y) = 7xy
(d) x – y = 4
To find equation of R
slope of L = 0 is 1
Þ slope of QR = – 1
Let QR is y = mx + c
y=–x+c
x+y–c=0
..... (ii)
Also,
HG
2
=
GO
1
3(a + 1) 2 3( a - 1) 2
,
2
2
Now, these coordinates satisfies eqn given in option (d)
Hence, required eqn of line is
(a – 1)2 x – (a + 1)2 y = 0
(b)
Þ Coordinate of orthocentre =
37.
C(0, b)
2
distance of QR from (2, 1) is 2 3
2 3 =
3(x - 3)
æ (a + 1)2 (a - 1)2 ö
Centroid = ç 2 , 2 ÷
è
ø
k=
34.
3x - 1 = 0
Þ (slope) m2 = - 3
Let the other slope be m1
Length of ^ to 8x + 6y + 5 = 0 from origin (0, 0)
33.
(c) Given eqn of line is y +
Þ y = – 3x + 1
10
P1 =
=2
5
| 2 +1- c |
A(4, 3)
2
1
O (0, 0)
P(2,1)
L=0
(4,0)
2 3
x–y=4
Q
R
2 6 = |3 – c|
c – 3 = ±2 6 c = 3 ± 2 6
Line can be x + y = 3 ± 2 6
x+y=3– 2 6
B(a, 0)
A divides CB in 2 : 1
æ 1 ´ 0 + 2 ´ a ö 2a
Þ 4 = çè
÷=
1+ 2 ø
3
Þ a = 6 Þ coordinate of B is B (6, 0)
æ 1´ b + 2 ´ 0ö b
3 = çè
÷=
1+ 2 ø 3
Þ b = 9 and C (0, 9)
Slope of line passing through (6, 0), (0, 9)
slope, m =
9
3
=2
-6
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EBD_8344
M-136
Mathematics
Equation of line y – 0 =
38.
-3
( x - 6)
2
æ x - 5 y1 + 3 ö
Centroid, E = ç 1
,
÷
3 ø
è 3
2y = –3x + 18
3x + 2y = 18
(d)
Since centroid lies on the line
3x + 4y + 2 = 0
æ x -5ö
æ y1 + 3 ö
\ 3ç 1
÷ + 4ç
÷+2 = 0
è 3 ø
è 3 ø
Þ 3x1 + 4y1 + 3 = 0
Hence vertex (x1, y1) lies on the line
3x + 4y + 3 = 0
P (5, 3)
Q (8, 3)
x – 2y = 2
R (2, 0)
41.
(b) Let y-coordinate of C = b
\ C = (2a, b)
Y
C (2a, b)
Equation of RQ is x – 2y = 2 ...(i)
at y = 0, x = 2 [R (2, 0)]
as PQ is parallel to x, y-coordinates of Q is also 3
Putting value of y in equation (i), we get
Q (8, 3)
39.
x = 2a
B
(0, a)
X
A
(2a, 0)
O
æ 8 + 5 + 2 3 + 3ö
,
Centroid of DPQR = ç
÷ = (5, 2)
è
3
3 ø
Only (2x – 5y = 0) satisfy the given co-ordinates.
(b) Suppose B(0, 1) be any point on given line and
AB =
co-ordinate of A is ( 3, 0). So, equation of
Now, AC = BC Þ b = 4a 2 + (b - a)2
4 a 2 + a 2 = 5a
Þ b2 = 4a2 + b2 + a2 – 2ab
B
(0, 1)
Þ 2ab = 5a2 Þ b =
A
5a ö
æ
\ C = ç 2a , ÷
è
2 ø
Hence area of the triangle
3, 0
B' (0, –1)
Reflected ray is
40.
-1 - 0
0- 3
=
5a
2
=
y -0
1
2
2a
0
2a
x- 3
Þ
3y = x - 3
(b) Let C = (x1, y1)
0 1
2a
1
0
a 1 =
2
5a
1
0
2
0
a
5a
2
1
1
0
1
5a 2
æ 5a ö
= ´ 2a ç - ÷ = è 2 ø
2
2
Since area is always +ve, hence area
A(– 3, 2)
5a 2
sq. unit
2
(d) Given line 3x + 4y = 12 can be rewritten as
=
2
42.
E
1
B
(– 2, 1)
D
æ x1 - 2 y1 + 1 ö
,
÷
ç
2 ø
è 2
C
(x1, y1)
3x 4 y
x y
+
=1 Þ + =1
12 12
4 3
Þ x-intercept = 4 and y-intercept = 3
Let the required line be
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M-137
Straight Lines and Pair of Straight Lines
x y
+ = 1 where
a b
a = x-intercept and b = y-intercept
According to the question
a = 4 × 2 = 8 and b = 3/2
L:
43.
2(1 + y) = 3 Þ y =
x 2y
=1
\ Required line is +
8 3
Þ 3x + 16y = 24
24
-3
x+
Þ y=
16
16
-3
.
Hence, required slope =
16
(c) Let the points be A (1,1) and B (2,4).
Let point C divides line AB in the ratio 3 : 2.
So, by section formula we have
æ 3 ´ 2 + 2 ´ 1 3 ´ 4 + 2 ´1 ö æ 8 14 ö
C =ç
,
= ,
3 + 2 ÷ø çè 5 5 ÷ø
è 3+2
46.
3
1
-1 =
2
2
æ 1ö
Thus ‘a’ lies in ç 0, ÷
è 2ø
(c) Let equation of AB : x + 3y = 4
Let equation of BC : 3x + y = 4
Let equation of CA : x + y = 0
Now, By solving these equations we get
A = (– 2, 2), B = (1, 1) and C = (2, – 2)
Now, AB = 9 + 1 = 10 ,
BC = 1 + 9 = 10
and CA = 16 + 16 = 32
Since, length of AB and BC are same therefore triangle is
isosceles.
47.
A(2, –3)
(b)
æ 8 14 ö
Since Line 2x + y = k passes through C ç , ÷
è5 5 ø
G
(h, k)
2 ´ 8 14
+ =k Þk=6
5
5
(c) Given lines are
ax + 2by + 3b = 0 and bx – 2ay – 3a = 0
Since, required line is || to x-axis
\x=0
We put x = 0 in given equation, we get
Þ
44.
æ 2 - 2 + a -3 + 1 + b ö
,
÷
Centroid (h,k) = ç
è
ø
3
3
\ a = 3h
b - 2 = 3k
3
2
This shows that the required line is below x-axis at a
2by = – 3b Þ y = -
distance of
45.
b = 3k + 2
Third vertex (α, β ) lies on the line
3
from it.
2
(d)
C (a, b)
B(–2, 1)
2x + 3 y = 9
2a + 3b = 9
Y
2 ( 3h) + 3 ( 3k + 2) = 9
2h + 3k = 1
3
x+
y=
48.
2
1
(0, 23 )
2x + 3 y =1
(c) Clearly for point P,
y
y = 3x
1
2
掸 P(a, a )
( 3 , 0)
2
O
2
x+
3
3
2
y=
1
(1, 0)
Since, (1, a) lies between x + y = 1
and 2(x + y) = 3
\ Put x = 1 in 2 (x + y) = 3.
We get the range of y. Thus,
X
y=
O
x
2
a 2 - 3a < 0 and a -
+
-¥
x
2
–
0
a
>0
2
+
3
¥
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EBD_8344
M-138
+
-¥
Þ
49.
Mathematics
–
1
2
0
52.
+
x + x + x y + y + y3 ö
centroid of DABC is æç 1 2 3 , 1 2
÷
è
3
3
ø
æ 2 - 2 + h -3 + 1 + k ö
=ç
,
÷
3
3
è
ø
¥
1
<a<3
2
(c)
y
P(0, b)
A(3, 4)
Q(a, 0)
O
50.
53.
x
Q A is the mid point of PQ,
a+0
0+b
= 3,
= 4 Þ a = 6, b = 8
\
2
2
x y
\ Equation of line is + = 1
6 8
or 4x + 3y = 24
(a) The eqn. of line passing through the intersection of
lines ax + 2by + 3b = 0 and
bx - 2ay - 3a = 0 is
ax + 2by + 3b + l (bx – 2ay – 3a) = 0
Þ (a + b l ) x + (2b – 2a l )y + 3b – 3 l a = 0
Required line is parallel to x-axis.
\ a + b l = 0 Þ l = – a/b
a
Þ ax + 2by + 3b – (bx – 2ay – 3a) = 0
b
2a 2
3a 2
=0
y+
Þ ax + 2by + 3b – ax +
b
b
2ö
54.
2
51.
A
Q cos2 a + sin2 a = 1
1
1
4
Locus of (x1, y1) is
.
+
=
2
2
x
y
p2
(a) The line in xy-plane is,
a +1 b + y
2( -1 - 12 - 3)
=
=1
3
10
-3(a 2 + b 2 )
a 2 - 4 = 0 Þ a = ±2 Þ b = -3 or 1
\ Equations of straight lines are
x y
x
y
+
= 1 or
+ =1
2 -3
-2 1
M (x1, y1)
x
+ y = 1 Þ x + 3y - 3 = 0
3
Let image of the point (–1, –4) be (a, b), then
æ 2b 2 + 2a 2 ö
æ 3b2 + 3a 2 ö
yç
÷ = -ç
÷
b
b
è
ø
è
ø
-3
=
2
2(b 2 + a 2 )
So it is 3/2 units below x-axis.
x y
(a) Let the required line be + = 1
a b
then a + b = –1 Þ b = –a –1
4 3
(i) passes through (4, 3) , Þ + = 1
a b
Þ 4b + 3a = ab
Putting value of b from (ii) in (iii), we get
æ h -2 + k ö
=ç ,
÷ . It lies on 2x + 3y = 1
è3
3 ø
2h
Þ
- 2 + k = 1 Þ 2h + 3k = 9
3
Þ Locus of C is 2x + 3y = 9
Y
(d) Equation of AB is
x cos a + y sin a = p;
B
x cos a y sin a
Þ
+
= 1;
p
p
x
y
Þ
+
=1
p / cos a p / sin a
O
So, co-ordinates of A and B are
p ö
æ p
ö
æ
, 0÷ and ç 0,
;
çè
è sin a ÷ø
cos a ø
So, coordinates of midpoint of AB are
p ö
æ p
M ( x1, y1) = ç
,
è 2cos a 2sin a ÷ø
p
p
x1 =
& y1 =
;
2 cos a
2 sin a
Þ cos a = p/2x1 and sin a = p/2y1 ;
æ
2a
3a
y ç 2b +
=0
÷ + 3b +
b ø
b
è
y=
(d) Let the vertex C be (h, k), then the
Þ a +1 =
....(i)
....(ii)
Þa=
55.
b + 4 16
=
3
5
11
28
,b=
5
5
(30)
L1 : 2 x - y + 3 = 0
....(iii)
L1 : 4 x - 2 y + a = 0 Þ 2 x - y +
a
=0
2
L1 : 6 x - 3 y + b = 0 Þ 2 x - y +
b
=0
3
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X
M-139
Straight Lines and Pair of Straight Lines
Distance between L1 and L2 ;
a-6
2 5
=
1
5
( 3 – 1) x + ( 3 + 1) y = 8 2
Þ| a - 6 |= 2
58.
Þ a = 4, 8
Q distance from origin =
Distance between L1 and L3 :
b-9
3 5
=
2
5
3
5
3 a
\ =
Þa=±3
5 5
Þ| b - 9 |= 6
Hence, line is 4x – 3y + 3 = 0 or 4x – 3y – 3 = 0
Þ b = 15, 3
æ 1 2ö
Clearly ç - , ÷ satisfies 4x – 3y + 3 = 0
è 4 3ø
Sum of all values = 4 + 8 + 15 + 3 = 30.
56.
or ( 3 + 1) x + ( 3 – 1) y = 8 2
(a) Let straight line be 4x – 3y + a = 0
(b)
59.
(a) Q two lines are perpendicular Þ m1m2 = – 1
æ -1 ö æ -2 ö
Þ ç - ÷ ç 2 ÷ = -1
è a 1øè a ø
Þ 2 = a2(1 – a) Þ a3 – a2 + 2 = 0
Þ (a + 1) (a2 + 2a + 2) = 0 Þ a = –1
Hence equations of lines are x – 2y = 1 and 2x + y = 1
Since, slope of PQ =
k -a
=–1
h - 2a
æ 3 -1 ö
\ intersection point is ç , ÷
è5 5 ø
Þ k – a = – h + 2a
h+k
3
Also, 2h = 2a + b and
2k = a + b
Þ 2h = a + 2k
Þ a = 2h – 2k
From (i) and (ii), we have
Þ a=
57.
Now, distance from origin =
...(i)
60.
(a) Given situation
x 㦨 –1
...(ii)
B(6 5)
h+k
= 2(h - k )
3
So, locus is 6x – 6y = x + y
Þ 5x = 7y Þ 5x – 7y = 0
(b) Q perpendicular makes an angle of 60 with the line
x + y = 0.
\ the perpendicular makes an angle of 15 or 75 with
x-axis.
x졔y㦨0
Hence, the equation of line will be
x cos 75 + y sin 75 = 4
or x cos 15 + y sin 15 = 4
(–1 2)
Q perpendicular bisector of AB will pass from centre.
\ equation of perpendicular bisector x = – 1
Hence centre of the circle is (– 1, 2)
Let co-ordinate of D is (a, b)
60°
O
9
1
10
2
+
=
=
25 25
25
5
ᔴ
a+6
b+5
= -1 and
=2
2
2
Þ a = – 8 and b = – 1, \ D º (– 8, – 1)
| AD | = 6 and | AB | = 14
Area = 6 × 14 = 84
Þ
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EBD_8344
M-140
61.
Mathematics
(a) Q (h, k), (1, 2) and (– 3, 4) are collinear
\
h
k 1
1
2 1
3p + 2q + 4r = 0
3
2
p+ q+r =0
4
4
From (i) & (ii) we get :
Þ
= 0 Þ – 2h – 4k + 10 = 0
-3 4 1
Þ h + 2k = 5
3
1
,y=
4
2
Hence the set of lines are concurrent and passing through
\ x=
...(i)
4-2
1
= - Þ mL2 = 2
2
-3 - 1
By the given points (h, k) and (4, 3),
Now, mL1 =
[Q L1 ^ L2 ]
k -3
k -3
= 2 Þ k - 3 = 2h - 8
Þ
h-4
h-4
2h – k = 5
From (i) and (ii)
...(ii)
æ 3 1ö
the fixed point çè , ÷ø
4 2
mL2 =
...(ii)
65.
A x2 ,
(d)
20 4 x2
5
k 1
=
h 3
(d) Q Equation of straight line can be rewritten as,
h = 3, k = 1 Þ
62.
y=
2
17
x+ .
3
3
2
3
Slope of line passing through the points (7, 17) and (15, b)
\ Slope of straight line =
b - 17
b - 17
=
=
15 - 7
8
Since, lines are perpendicular to each other.
Hence, m1m2 = –1
æ 2 ö æ b - 17 ö
Þ çè ÷ø çè
÷ = –1 Þ
3
8 ø
63.
(d)
b=5
S
R
æ 3 x1 + 6 ö
çè x1 ,
÷
2 ø
Since, AH is perpendicular to BC
Hence, mAH × mBC = –1
æ 20 - 4 x2 ö
–1
ç
÷ 3
5
´ = –1
ç
x2 - 1 ÷ 2
ç
÷
è
ø
15 - 4 x2
2
5( x2 - 1) = 3
45 – 12x2 = –10x2 + 10
x–y+3=0
M(2, 4)
2x2 = 35 Þ x2 =
P (0, 3) x + y = 3
64.
Q
Since, x – y + 3 = 0 and x + y = 3 are perpendicular lines and
intersection point of x – y + 3 = 0 and x + y = 3 is P(0, 3).
Þ M is mid-point of PR Þ R(4, 5)
Let S(x1, x1 + 3) and Q(x2, 3 – x2)
M is mid-point of SQ
Þ x1 + x2 = 4, x1 + 3 + 3 – x2 = 8
Þ x1 = 3, x2 = 1
Then, the vertex D is (3, 6).
(a) The given equations of the set of all lines
px + qy + r = 0
...(i)
and given condition is :
35
2
æ 35
ö
Þ A çè , - 10÷ø
2
Since, BH is perpendicular to CA.
Hence, mBH × mCA = –1
æ 3 x1
ö
+ 3 -1
ç 2
÷ æ 4ö
ç x - 1 ÷ çè - 5 ø÷ = –1
1
èç
ø÷
(3x1 + 4)
´4 =5
2( x1 - 1)
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M-141
Straight Lines and Pair of Straight Lines
\ Parallel equations of the diagonals are 2x + 4y – 7 = 0
and 12x – 6y + 13 = 0
-33ö
æ
Þ 6x1 + 8 = 5x1 – 5 Þ x1 = –13 Þ çè -13,
÷
2 ø
-1
and 2.
2
Now, slope of the diagonal from A(0, c) and passing through
P(1, 2) is (2 – c)
\ 2 – c = 2 Þ c = 0 (not possible)
Þ Equation of line AB is
\ slopes of diagonals are
æ 33
ö
- + 10
35 ö
ç 2
־
y + 10 = ç
ç x - ø÷
2
-13 - 35 ÷ è
çè
ø÷
\2–c=
455
2
Þ –122y – 1220 = –26x + 455
Þ 26x – 122y – 1675 = 0
(d) Equation of the line, which is perpendicular to the
line, 3x + y = l(l ¹ 0) and passing through origin, is given
by
Þ –61y – 610 = –13x +
66.
\ ordinate of A is
68.
- ( (3´ 0) + (1´ 0) - l )
32 + 12
=
l
10
5bc - 4ad 4 ad - 5bc
=
-2ab
2 ab
Q Point of intersection is in fourth quadrant so x is
positive and y is negative.
Also distance from axes is same
So x = – y
(Q distance from x-axis is –y as y is negative)
æl ö
Given the line meets X-axis at A = ç , 0 ÷ and meets
è3 ø
Y-axis at B = (0, l)
2
9l 2 81l 2
æ 3l ö æ l
ö
+
ç ÷ + ç - l ÷ Þ BP =
100 100
è 10 ø è 10
ø
Þ BP =
2
2(ad - bc) bc - ad
=
-2ab
ab
Þ y=
æ 3l l ö
So, foot of perpendicular P = ç , ÷
è 10 10 ø
So, BP =
(a) Given lines are
4ax + 2ay + c = 0
5bx + 2by + d = 0
The point of intersection will be
Þ x=
2
5
.
2
x
-y
1
=
=
2ad - 2bc 4ad - 5bc
8ab - 10ab
x -0 y -0
=
=r
3
1
For foot of perpendicular
r=
-1
5
Þc=
2
2
lö
æ l 3l ö æ
Now, PA = ç - ÷ + ç 0 - ÷
10 ø
è 3 10 ø è
69.
bc - ad 5bc - 4ad
=
Þ 3bc – 2ad = 0
ab
2ab
(d) Let P, Q, R, be the vertices of DPQR
P (2, 2)
90l 2
100
2
Q (6, – 1)
l
l
10l
+
Þ PA =
900 100
900
Therefore BP : PA = 3 : 1
(d) Let the coordinate A be (0, c)
Equations of the given lines are
x – y + 2 = 0 and
7x – y + 3 = 0
We know that the diagonals of the rhombus will be parallel
to the an gle bisectors of the two given lines;
y = x + 2 and y = 7x + 3
\ equation of angle bisectors is given as:
Þ PA
67.
2
2
7x - y + 3
x- y+2
=±
2
5 2
5x – 5y + 10 = ± (7x – y + 3)
2
S
R (7, 3)
Since PS is the median
S is mid-point of QR
æ 7 + 6 3 - 1ö æ 13 ö
So, S = ç
,
÷ = ç ,1÷
è 2
2 ø è2 ø
2 -1
2
=13
9
22
Since, required line is parallel to PS therefore slope of
required line = slope of PS
Now, eqn. of line passing through (1, –1) and having
2
slope - is
9
Now, slope of PS =
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EBD_8344
M-142
70.
Mathematics
2
y - (-1) = - ( x - 1)
9
9y + 9 = –2x + 2 Þ 2x + 9y + 7 = 0
(b) Let equation of line L, perpendicular to 5x – y = 1
be x + 5y = c
72.
(b)
P (1, 2)
A
x + 5y = c
æ cö
B ç 0, ÷
è 5ø
C
A(c, 0)
ax1 + by1 - c
a2 + b2
Now shortest distance of P (1, 2) from 3x + 4y = 9 is
3(1) + 4(2) - 9
PC = d =
Given that area of DAOB is 5.
We know
1 é æ cö ù
cç ÷
2 êë è 5 ø úû
2
Þ c = ± 50
\ Equation of line L is x + 5y = ± 50
Distance between L and line x + 5y = 0 is
± 50 - 0
71.
12 + 52
2
5
a2
æ 2ö
a2 = ç ÷ +
è 5ø
4
æ
æ cö ö
çQ ( x1, y1 ) = (10,0), ( x3 , y3 ) = çè 0, 5 ÷ø ÷
ç
÷
è ( x2 , y2 ) = (c, 0)
ø
d=
32 + 4 2
=
Given that DAPB is an equilateral triangle
Let 'a' be its side
a
then PB = a, CB =
2
Now, In DPCB, (PB)2 = (PC)2 + (CB)2
(By Pythagoras theoresm)
1
ì
ü
íarea, A = [ x1 ( y2 - y3 ) + x2 ( y3 - y1) + x3 ( y1 - y2 )]ý
2
î
þ
Þ5=
3x + 4y = 9
Shortest distance of a point (x1, y1) from line
ax + by = c is d =
O (0, 0)
B
=
50
=
26
5
13
(c) x + 2ay + a = 0
...(i)
x + 3by + b = 0
...(ii)
x + 4ay + a = 0
...(iii)
Subtracting equation (iii) from (i)
–2ay = 0
ay = 0 c y = 0
Putting value of y in equation (i), we get
x+0+a=0
x=–a
Putting value of x and y in equation (ii), we get
–a+b=0 Þ a=b
Thus, (a, b) lies on a straight line
a2 -
a4
4
3a 2
4
Þ
=
=
4
25
4
25
a2 =
16
Þa=
75
16
4
3 4 3
=
´
=
75 5 3
15
3
4 3
15
(d) Mid-point of P(2, 3) and Q(4, 5) = (3, 4)
Slope of PQ = 1
Slope of the line L = – 1
Mid-point (3, 4) lies on the line L.
Equation of line L,
y – 4 = – 1(x – 3) Þ x + y – 7 = 0
Let image of point R(0, 0) be S(x1, y1)
\ Length of Equilateral triangle (a) =
73.
...(i)
æ x1 y1 ö
Mid-point of RS = ç , ÷
è 2 2ø
æx y ö
Mid-point ç 1 , 1 ÷ lies on the line (i)
è 2 2ø
\ x1 + y1 = 14
Slope of RS =
y1
x1
Since RS ^ line L
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...(ii)
M-143
Straight Lines and Pair of Straight Lines
y1
´ (-1) = -1
x1
\ x1 = y1
...(iii)
From (ii) and (iii),
x1 = y1 = 7
Hence the image of R = (7, 7)
(a) Two lines – x + 5y + c2 = 0 and – x + 5y + c3 = 0 are
parallel to each other. Hence statement-1 is true, statement2 is true and statement-2 is the correct explanation of
statement-1.
(a) Since three lines x – 3y = p,
ax + 2y = q and ax + y = r
form a right angled triangle
\ product of slopes of any two lines = –1
\
74.
75.
78.
(c) Statement - 1
Let P¢ (x1, y1) be the image of (0, 1) with respect to the
line 2x – y = 0 then
x1 y1 - 1 -4(0) + 2(1)
=
=
-1
2
5
4
3
Þ x1 = , y1 =
5
5
Thus, statement-1 is true.
Also, statement-2 is true and correct explanation for
statement-1.
79.
(b)
A(5, – 1)
Suppose ax + 2y = q and x – 3y = p are ^ to each other..
\ - a ´ 1 = -1 Þ a = 6
2 3
H
(0, 0)
Now, consider option one by one
a = 6 satisfies only option (a)
76.
\ Required answer is a 2 - 9a + 18 = 0
(d) Consider the lines
L1 : x – y = 1
L2 : x + y = 1
L3 : 2x + 2y = 5
L4 : 2x – 2y = 7
L1 ^ L2 is correct statement
(Q Product of their slopes = – 1)
L1 ^ L3 is also correct statement
(Q Product of their slopes = – 1)
Now, L2 : x + y = 1
L4 : 2x – 2y = 7
Þ 2x – 2 (1 – x) = 7
Þ 2x – 2 + 2x = 7
9
-5
Þx=
and y =
4
4
Let the third vertex of DABC be (a, b).
Orthocentre = H(0, 0)
Let A (5, – 1) and B (– 2, 3) be other two vertices of DABC.
Now, (Slope of AH) × (Slope of BC) = – 1
æ -1 - 0 ö æ b - 3 ö
Þ çè
֍
÷ = -1
5 - 0 ø è a + 2ø
Þ b – 3 = 5 (a + 2)
Similarly,
(Slope of BH) × (Slope of AC) = – 1
(d) Given ax2 + bx + c = 0
Þ ax2 = –bx – c
Now, consider
y = 4ax2 + 3bx + 2c
= 4 [–bx – c] + 3bx + 2c
= – 4bx – 4c + 3bx + 2c = – bx – 2c
Since, this curve intersects x-axis
\ put y = 0, we get
–bx – 2c = 0 Þ – bx = 2c
-2c
b
Thus, given curve intersects x-axis at exactly one point.
Þx=
...(i)
æ 3ö æ b + 1ö
Þ - çè ÷ø ´ çè
÷ = -1
2
a - 5ø
Þ 3b + 3 = 2a – 10
Þ 3b – 2a + 13 = 0
...(ii)
On solving equations (i) and (ii) we get
a = – 4, b = – 7
Hence, third vertex is (– 4, – 7).
Hence, L2 intersects L4.
77.
C (a, b)
B (– 2, 3)
80.
(a)
C(a, 80)
A (0, 20)
M
O(0, 0)
N
We put one pole at origin.
Downloaded from @Freebooksforjeeneet
B (a, 0)
EBD_8344
M-144
Mathematics
Case II : If a < 0
x + y = -a
BC = 80 m, OA = 20 m
Line OC and AB intersect at M.
To find: Length of MN.
ax - y = 1
On adding equations (iii) and (iv), we get
æ 80 - 0 ö
x
Eqn of OC: y = ç
è a - 0 ÷ø
Þ y=
80
x
a
1- a
a -1
>0Þ
<0
1+ a
a +1
Since a – 1 < 0
\ a+1>0
Þ a > –1
x=
æ 20 - 0 ö
Eqn of AB: y = çè
÷ ( x - a)
0-a ø
81.
Þ
80
-20
x=
( x - a)
a
a
Þ
a
80
-20
x=
x + 20 Þ x =
a
a
5
...(ii)
y = -a -
Since a2+ 1 > 0
\ a+1<0
Þ a < –1
.... (vi)
From (v) and (vi), a Î f
Hence, Case-II is not possible.
So, correct answer is a Î[1, ¥)
–1
L3
X
83.
(b)
y–
x=
0
R(–1, –2)
O
(0, 0)
2x
+
y=
0
.... (i)
.... (ii)
x (1 + a ) = 1 + a Þ x = 1
y=a–1
Since given that intersection point lies in first quadrant
So, a – 1 ³ 0
Y'
Q y+2=0
X'
(b) Given that x + y = a
L1
P(–2, –2)
Y
on the y-axis for no value of ‘a’.
Þ a ³1
Þ a Î[1, ¥)
1- a
-a - a2 - 1 + a
>0 =
>0
1+ a
1+ a
æ a 2 + 1ö
a2 + 1
>
0
Þ
<0
Þ -ç
÷
a +1
è a +1 ø
80 a
\ y = ´ = 16
a 5
(a) Given equation of lines are
(a3 + 3)x + ay + a – 3 = 0 and
(a5 + 2)x + (a + 2)y + 2a + 3 = 0 (a real)
Since point of intersection of lines lies on y-axis.
\ Put x = 0 in each equation, we get
ay + a – 3 = 0 and
(a + 2)y + 2a + 3 = 0
On solving these we get
(a + 2) (a – 3) – a (2a + 3) = 0
Þ a2 – a – 6 – 2a2 – 3a = 0
Þ – a2 – 4a – 6 = 0 Þ a2 + 4a + 6 = 0
and ax - y = 1
Case I : If a > 0
x+ y = a
ax - y = 1
On adding equations (i) and (ii), we get
.... (v)
–1
-4 ± 16 - 24 -4 ± -8
=
Þ a=
2
2
(not real)
This shows that the point of intersection of the lines lies
82.
.... (iv)
x (1 + a) = 1 - a
...(i)
-20
( x - a)
Þ y=
a
At M: (i) = (ii)
.... (iii)
L2
L1 : y – x = 0
L2 : 2x + y = 0
L3 : y + 2 = 0
On solving the equation of line L1 and L2 we get their point
of intersection (0, 0) i.e., origin O.
On solving the equation of line L1 and L3,
we get P = (– 2, – 2).
Similarly, solving equation of line L2 and L3, we get
Q = (– 1, – 2)
We know that bisector of an angle of a triangle, divide the
opposite side the triangle in the ratio of the sides including
the angle [Angle Bisector Theorem of a Triangle]
\
(-2)2 + (-2) 2 2 2
PR OP
=
=
=
RQ OQ
5
( -1)2 + ( -2) 2
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M-145
Straight Lines and Pair of Straight Lines
84.
\ Statement 1 is true but ÐOPR ¹ ÐOQR
So DOPR and DOQR not similar
\ Statement 2 is false.
(a) Given that the lines p (p2 + 1) x – y + q = 0
and (p2 + 1)2 x + (p2 + 1) y +2q = 0
are perpendicular to a common line then these lines must
be parallel to each other,
\ m1 = m2
Þ -
y2 - y1 3 3
=
3
x2 - x1
p
Þ tan q = 3 Þ q =
3
p
Þ ÐRQX =
3
p 2p
\ ÐRQP = p - =
3
3
Let QM bisects the ÐPQR ,
Slope of QR =
( p 2 + 1) 2
p ( p 2 + 1)
=-1
p2 + 1
p
2p
Þ ÐMQX =
3
3
2p
\ Slope of the line QM = tan
=– 3
3
\ Equation of line QM is (y – 0) = – 3 (x – 0)
\ ÐMQR =
(p+1)=0
Þ
Þ p=–1
\ p can have exactly one value.
(p2 + 1)2
85.
(d) Let (a2, a) be the point of shortest distance on x = y2
Then distance between (a2, a) and line
x – y + 1 = 0 is given by
D=
ax1 + by1 + c
2
a +b
=
2
It is min when a =
Dmin =
86.
3
4 2
=
a2 - a + 1
2
=
1
1
3
(a - ) 2 +
2
4
2
1
and
2
3 2
8
3–4
–1
=
k –1 k –1
\ Slope of perpendicular bisector of
PQ = ( k –1)
(d) Slope of PQ =
æ k +1 7ö
, .
Also, mid point of PQ ç
è 2 2 ÷ø
\ Equation of perpendicular bisector of PQ is
7
k + 1ö
æ
= (k –1) ç x –
÷
è
2
2 ø
Þ 2y – 7 = 2(k –1) x –(k2 –1)
Þ 2(k – 1)x – 2y + ( 8 – k2) = 0
Given that y-intercept
y–
8 – k2
=–4
2
2
Þ 8 – k = –8 or k2 = 16 Þ k = ± 4
(c) Given : The coordinates of points P, Q, R are (–1, 0),
(0, 0), (3,3 3) respectively..
Y
R (3, 3 3 )
M
X'
2p / 3
P (-1, 0)
p/3
Q (0, 0)
89.
Þ y= – 3 x Þ 3 x + y= 0
(b) Taking co-ordinates as
æ x yö
A ç , ÷ ; B( x, y ) andC ( xr , yr ) .
èr rø
Then slope of line oining
æ 1ö
y ç1 - ÷
x
y
rø y
æ
ö
A ç , ÷ , B ( x, y ) = è
=
1
r
r
æ
ö x
è
ø
x ç1 - ÷
è rø
and slope of line oining B(x, y) and C(xr, yr)
y ( r - 1) y
=
=
x ( r - 1) x
\ m1 = m2
Q Slope of AB and BC are same and one point B common.
Þ Points lie on the straight line.
(a) Co-ordinates of A = (a cos α, a sin α)
B
Equation of OB,
Y
æp
ö
C
y = tan ç + a ÷ x
4
è
ø
A
CA ^ r to OB
p
4
=
87.
88.
X
a
O
æp
ö
X
\ Slope of CA = - cot ç + a ÷
è4
ø
Equation of CA
æp
ö
y - a sin a = - cot ç + a ÷ ( x - a cos a)
è4
ø
æ
æp
öö
Þ ( y - a sin a ) ç tan ç + a ÷ ÷ = ( a cos a - x )
è4
øø
è
p
æ
ö
ç tan 4 + tan a ÷
Þ ( y - a sin a) ç
÷ = (a cos a - x )
ç 1 - tan p tan a ÷
è
4
ø
Þ ( y - a sin a) (1 + tan a) = (a cos a - x)(1 - tan a)
Þ ( y - a sin a ) (cos a + sin a)
= (a cos a - x )(cos a - sin a )
Y'
Downloaded from @Freebooksforjeeneet
EBD_8344
M-146
Mathematics
Þ y (cos + sin a ) - a sin a cos a - a sin 2 a
92.
= a cos a - a cos a sin a - x (cos a - sin a)
Þ y (cos a + sin a ) + x(cos a - sin a ) = a
y (sin a + cos a ) + x (cos a - sin a ) = a.
(d) Consider the equation,
y = sin x. sin (x + 2) – sin2(x + 1)
2
90.
=
=
1
cos(2 x + 2) é1 - cos(2 x + 2) ù
cos(-2) -ê
úû
2
2
2
ë
93.
(cos 2) - 1
= - sin 2 1
2
y
94.
(0 0)
91
(a) 3 x + 4 y = 0 is one of the line of the pair equations.
of lines
3
Put y = - x ,
6 x 2 - xy + 4cy 2 = 0 ,
4
2
3 2
3 ö
æ
2
we get, 6 x + x + 4c ç - x÷ = 0
è 4 ø
4
3 9c
Þ 6+ +
= 0 Þ c = -3
4 4
(c) Let the lines be y = m1x and y = m2x then
2c
1
m1 + m2 = - and m1m2 = 7
7
Given that m1 + m2 = 4 m1m2
2c
4
Þ - =- Þc=2
7
7
(a) Equation of bisectors of second pair of straight lines
is,
x
qx 2 + 2 xy - qy 2 = 0
....(i)
It must be identical to the first pair
y = - sin 2 1
x 2 - 2 pxy - y 2 = 0
from (i) and (ii)
q
2
-q
=
=
Þ pq = -1 .
1 - 2 p -1
By the graph y lies in III and IV quadrant.
(a) From figure equation of bisectors of lines, xy = 0 are
y= ± x
y
95.
....(ii)
(a) We know that pair of straighty lines
ax2 + 2hxy + by2 = 0 are perpendicular when a + b = 0
3a + a2 – 2 = 0 Þ a2 + 3a – 2 = 0.;
y = –x
(0, 0)
\ Put y = ± x in the given equation
my2 + (1 – m2)xy – mx2 = 0
\ mx2 ± (1 – m2)x2 – mx2 = 0
Þ 1 – m2 = 0 Þ m = ± 1
- 3 ± 9 + 8 - 3 ± 17
=
2
2
96. (a) Put x = 0 in the given equation
Þ by2 + 2 fy + c = 0.
For unique point of intersection, f 2 – bc = 0
Þ af 2 – abc = 0.
We know that for pair of straight line
abc + 2fgh – af 2 –bg2 – ch2 = 0
Þ 2fgh – bg2 – ch2 = 0
Þa =
y=x
x
Downloaded from @Freebooksforjeeneet
11
Conic Sections
TOPIC Ć
1.
8.
Circles
If the length of the chord of the circle, x2 + y 2 = r 2 (r > 0)
along the line, y – 2x = 3 is r, then r2 is equal to :
[Sep. 05, 2020 (II)]
9
24
12
(b) 12
(c)
(d)
5
5
5
The circle passing through the intersection of the circles,
(a)
2.
9.
x2 + y 2 - 6 x = 0 and x2 + y 2 - 4 y = 0, having its centre
3.
4.
5.
on the line, 2 x - 3 y + 12 = 0, also passes through the
point:
[Sep. 04, 2020 (II)]
(a) (–1, 3) (b) (–3, 6) (c) (–3, 1) (d) (1, –3)
Let PQ be a diameter of the circle x2 + y2 = 9. If a and b are
the lengths of the perpendiculars from P and Q on the
straight line, x + y = 2 respectively, then the maximum
value of ab is ______________. [NA Sep. 04, 2020 (II)]
The diameter of the circle, whose centre lies on the line
x + y = 2 in the first quadrant and which touches both the
lines x = 3 and y = 2, is __________.
[NA Sep. 03, 2020 (I)]
The number of integral values of k for which the line,
3x + 4y = k intersects the circle, x2 + y 2 - 2 x - 4 y + 4 = 0
at two distinct points is ________. [NA Sep. 02, 2020 (I)]
6.
7.
A circle touches the y-axis at the point (0, 4) and passes
through the point (2, 0). Which of the following lines is
not a tangent to this circle?
[Jan. 9, 2020 (I)]
(a) 4x – 3y + 17 = 0
(b) 3x – 4y – 24 = 0
(c) 3x + 4y – 6 = 0
(d) 4x + 3y – 8 = 0
If the curves, x2 – 6x + y2 + 8 = 0 and x2 – 8y + y2 + 16 – k = 0,
(k > 0) touch each other at a point, then the largest value
of k is ______.
[NA Jan. 9, 2020 (II)]
If a line, y = mx + c is a tangent to the circle, (x – 3)2 + y2 = 1
and it is perpendicular to a line L1, where L1 is the tangent
æ 1 1 ö
to the circle, x2 + y2 = 1 at the point ç
,
÷ ; then:
2ø
è 2
[Jan. 8, 2020 (II)]
(a) c2 – 7c + 6 = 0
(b) c2 + 7c + 6 = 0
(c) c2 + 6c + 7 = 0
(d) c2 – 6c + 7 = 0
Let the tangents drawn from the origin to the circle,
x2 + y2 – 8x – 4y + 16 = 0 touch it at the points A and B. The
(AB)2 is equal to:
[Jan. 7, 2020 (II)]
(a)
10.
52
5
(b)
56
5
(c)
64
5
(d)
32
5
If the angle of intersection at a point where the two circles
with radii 5 cm and 12 cm intersect is 90o, then the length
(in cm) of their common chord is : [April 12, 2019 (I)]
13
120
60
13
(b)
(c)
(d)
5
13
13
2
A circle touching the x-axis at (3, 0) and making an intercept
of length 8 on the y-axis passes through the point :
[April 12, 2019 (II)]
(a) (3, 10)
(b) (3, 5)
(c) (2, 3)
(d) (1, 5)
If the circles x2 + y2 + 5Kx + 2y + K = 0 and 2 (x2 + y2) +
2Kx + 3y – 1= 0, (KÎR), intersect at the points P and Q,
then the line 4x + 5y – K = 0 passes through P and Q, for:
[April 10, 2019 (I)]
(a) infinitely many values of K
(b) no value of K.
(c) exactly two values of K
(d) exactly one value of K
The line x = y touches a circle at the point (1, 1). If the
circle also passes through the point (1, –3), then its radius
is:
[April 10, 2019 (I)]
(a)
11.
12.
13.
(a) 3
(b) 2 2
(c) 2
Downloaded from @Freebooksforjeeneet
(d) 3 2
EBD_8344
M-148
14.
15.
Mathematics
The locus of the centres of the circles, which touch the
circle, x2 + y2 = 1 externally, also touch the y-axis and lie in
the first quadrant, is:
[April 10, 2019 (II)]
(a) x = 1 + 4 y , y ³ 0
(b) y = 1 + 2 x , x ³ 0
(c) y = 1 + 4 x , x ³ 0
(d) x = 1 + 2 y , y ³ 0
21.
If a variable line, 3x + 4y – l = 0 is such that the two circles
x2 + y2 – 2x – 2y + 1 = 0 and x2 + y2 – 18x – 2y + 78 = 0 are
on its opposite sides, then the set of all values of l is the
interval :
[Jan. 12, 2019 (I)]
(a) (2, 17)
(b) [13, 23]
(c) [12, 21]
(d) (23, 31)
22.
A square is inscribed in the circle x 2 + y 2 - 6 x + 8 y - 103 = 0
with its sides parallel to the coordinate axes. Then the
distance of the vertex of this square which is nearest to the
origin is :
[Jan. 11, 2019 (I)]
ìa + i
ü
: µ Î R ý (i = -1)
All the points in the set S = í
îa -1
þ
lie on a:
[April 09, 2019 (I)]
(a) straight line whose slope is 1.
(b) circle whose radius is 1.
23.
(c) circle whose radius is
16.
17.
18.
19.
2.
(d) straight line whose slope is –1.
If a tangent to the circle x2 + y2 = 1intersects the coordinate
axes at distinct points P and Q, then the locus of the midpoint of PQ is:
[April 09, 2019 (I)]
(a) x2 + y2 – 4x2y2 = 0
(b) x2 + y2 – 2xy = 0
(c) x2 + y2 – 16x2y2 = 0
(d) x2 + y2 – 2x2y2 = 0
The common tangent to the circles x2 + y2 = 4 and x2 + y2 +
6x + 8y – 24 = 0 also passes through the point:
[April 09, 2019 (II)]
(a) (4, –2) (b) (– 6, 4) (c) (6, –2) (d) (– 4, 6)
The sum of the squares of the lengths of the chords
intercepted on the circle, x2 + y2 = 16, by the lines, x + y = n,
n Î N, where N is the set of all natural numbers, is :
[April 08, 2019 (I)]
(a) 320
(b) 105
(c) 160
(d) 210
If a circle of radius R passes through the origin O and
intersects the coordinate axes at A and B, then the locus of
the foot of perpendicular from O on AB is :
[Jan. 12, 2019 (II)]
(a)
(x
(b)
( x2 + y 2 )
(c)
( x2 + y 2 )
(d)
( x2 + y2 ) ( x + y ) = R2 xy
2
+y
)
2 2
3
20.
2
24.
25.
26.
27.
(a) 6
(b) 137
(c)
(d) 13
41
Two circles with equal radii are intersecting at the points
(0, 1) and (0, –1). The tangent at the point (0, 1) to one of
the circles passes through the centre of the other circle.
Then the distance between the centres of these circles is :
[Jan. 11, 2019 (I)]
(a) 1
(b) 2
(c) 2 2
(d)
2
A circle cuts a chord of length 4a on the x-axis and passes
through a point on the y-axis, distant 2b from the origin.
Then the locus of the centre of this circle, is :
[Jan. 11, 2019 (II)]
(a) a hyperbola
(b) an ellipse
(c) a straight line
(d) a parabola
If a circle C passing through the point (4, 0) touches the
circle x2 + y2 + 4x – 6y = 12 externally at the point
(1, – 1), then the radius of C is:
[Jan 10, 2019 (I)]
(a) 2 5
(b) 4
(c) 5
(d) 57
If the area of an equilateral triangle inscribed in the
circle, x2 + y2 + 10x + 12y + c = 0 is 27 3 sq. units then
c is equal to:
[Jan. 10, 2019 (II)]
(a) 13
(b) 20
(c) – 25
(d) 25
Three circles of radii a, b, c (a < b < c) touch each other
externally. If they have x-axis as a common tangent, then:
[Jan 09, 2019 (I)]
1
1
1
=
+
a
b
c
= 4R x y
(a)
= 4R2 x 2 y2
(c) a, b, c are in A.P
2 2 2
= 4 Rx 2 y 2
Let C1 and C2 be the centres of the circles x2 + y2 – 2x –2y – 2 = 0
and x2 + y2 – 6x –6y + 14 = 0 respectively. If P and Q are
the points of intersection of these circles then, the area
(in sq. units) of the quadrilateral PC1QC2 is :
[Jan. 12, 2019 (I)]
(a) 8
(b) 6
(c) 9
(d) 4
(b)
1
1
1
=
+
b
a
c
(d)
a , b, c are in A.P..
28. If the circles x + y – 16x – 20y + 164 = r2 and
(x – 4)2 + (y – 7)2 = 36 intersect at two distinct points,
then:
[Jan. 09, 2019 (II)]
(a) r > 11
(b) 0 < r < 1
(c) r = 11
(d) 1 < r < 11
29. The straight line x + 2y = 1 meets the coordinate axes at A
and B. A circle is drawn through A, B and the origin. Then
the sum of perpendicular distances from A and B on the
tangent to the circle at the origin is : [Jan. 11, 2019 (I)]
2
(a)
5
2
(b) 2 5
2
(c)
5
4
Downloaded from @Freebooksforjeeneet
(d) 4 5
M-149
Conic Sections
30.
If the tangent at (1, 7) to the curve x 2 = y - 6 touches the
31.
circle x 2 + y2 + 16x + 12y + c = 0 then the value of c is :
[2018]
(a) 185
(b) 85
(c) 95
(d) 195
If a circle C, whose radius is 3, touches externally the circle,
x2 + y2 + 2x – 4y – 4 = 0 at the point (2, 2), then the length
of the intercept cut by this circle c, on the x-axis is equal to
[Online April 16, 2018]
32.
33.
34.
35.
(a) 5
(b) 2 3
(c) 3 2
(d) 2 5
A circle passes through the points (2, 3) and (4, 5). If its
centre lies on the line, y – 4x + 3 = 0, then its radius is equal
to
[Online April 15, 2018]
(a) 5
2
(b) 1
(c)
(d) 2
Two parabolas with a common vertex and with axes along
x-axis and y-axis, respectively, intersect each other in the
first quadrant. if the length of the latus rectum of each
parabola is 3, then the equation of the common tangent to
the two parabolas is?
[Online April 15, 2018]
(a) 3 (x + y) + 4 = 0
(b) 8 (2x + y) + 3 = 0
(c) 4 (x + y) + 3 = 0
(d) x + 2y + 3 = 0
The tangent to the circle C1 : x2 + y2 – 2x – 1 = 0 at the point
(2, 1) cuts off a chord of length 4 from a circle C2 whose
centre is (3, – 2). The radius of C2 is
[Online April 15, 2018]
(a) 6
2
(b) 2
(c)
(d) 3
The radius of a circle, having minimum area, which touches
the curve y = 4 – x2 and the lines, y = |x| is :
[2017]
(
2(
(a) 4
(c)
36.
)
2 - 1)
2 +1
(
4(
(b) 2
(d)
)
2 - 1)
3
1
(d)
4
2
A line drawn through the point P(4, 7) cuts the circle
x2 + y2 = 9 at the points A and B. Then PA·PB is equal to :
[Online April 9, 2017]
(a) 53
(b) 56
(c) 74
(d) 65
The two adacent sides of a cyclic quadrilateral are 2 and
5 and the angle between them is 60 . If the area of the
38.
39.
(b) 1
16
.
3
16
9
If a point P has co–ordinates (0, –2) and Q is any point on
the circle, x2 + y2 – 5x – y + 5 = 0, then the maximum value
of (PQ)2 is :
[Online April 8, 2017]
(d) an ellipse with length of minor axis
40.
(a)
25 + 6
2
(b) 14 + 5 3
47 + 10 6
(d) 8 + 5 3
2
If two parallel chords of a circle, having diameter 4 units,
lie on the opposite sides of the centre and subtend angles
(c)
41.
æ1ö
cos -1 ç ÷ and sec–1 (7) at the centre respectively, then
è7ø
the distance between these chords, is :
[Online April 8, 2017]
42.
43.
44.
45.
4
8
8
16
(d)
7
7
7
7
If one of the diameters of the circle, given by the equation,
x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose
centre is at (–3, 2), then the radius of S is:
[2016]
(a)
(c)
quadrilateral is 4 3 , then the perimeter of the quadrilateral is :
[Online April 9, 2017]
(a) 12.5
(b) 13.2
(c) 12
(d) 13
Let Î C, the set of complex numbers. Then the equation,
2| +3i| – | – i| = 0 represents :
[Online April 8, 2017]
10
.
3
(c) an ellipse with length of maor axis
The equation
(a) 2
8
.
3
(b) a circle with diameter
2 +1
æ iz - 2 ö
Im ç
÷ + 1 = 0, z Î C, z ¹ i represents a part of a circle
è z -i ø
having radius equal to :
[Online April 9, 2017]
37.
(a) a circle with radius
(b)
(a) 5
(b) 10
(c) 5 2
(d) 5 3
Equation of the tangent to the circle, at the point (1, –1)
whose centre is the point of intersection of the straight
lines x – y = 1 and 2x + y = 3 is : [Online April 10, 2016]
(a) x + 4y + 3 = 0
(b) 3x – y – 4 = 0
(c) x – 3y – 4 = 0
(d) 4x + y – 3 = 0
A circle passes through (–2, 4) and touches the y-axis at
(0, 2). Which one of the following equations can represent
a diameter of this circle ?
[Online April 9, 2016]
(a) 2x – 3y + 10 = 0
(b) 3x + 4y – 3 = 0
(c) 4x + 5y – 6 = 0
(d) 5x + 2y + 4 = 0
Locus of the image of the point (2, 3) in the line
(2x – 3y + 4) + k (x – 2y + 3) = 0, k Î R, is a :
[2015]
(a) circle of radius
46.
(c)
2.
(b) circle of radius 3.
(c) straight line parallel to x-axis
(d) straight line parallel to y-axis
The n umber of common tangents to the cir cles
x2 + y2 – 4x – 6x – 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0, is:
[2015]
(a) 3
(b) 4
(c) 1
(d) 2
Downloaded from @Freebooksforjeeneet
EBD_8344
M-150
47.
48.
If the incentre of an equilateral triangle is (1, 1) and the
equation of its one side is 3x + 4y + 3 = 0, then the
equation of the circumcircle of this triangle is :
[Online April 11, 2015]
(a) x2 + y2 – 2x – 2y – 14 = 0
(b) x2 + y2 – 2x – 2y – 2 = 0
(c) x2 + y2 – 2x – 2y + 2 = 0
(d) x2 + y2 – 2x – 2y – 7 = 0
If a circle passing through the point (–1, 0) touches yaxis at (0, 2), then the length of the chord of the circle
along the x-axis is :
[Online April 11, 2015]
(a)
49.
Mathematics
3
2
51.
(c)
5
2
(b) 3 3
(c) 3 2
57.
The set of all real values of l for which exactly two common
tangents can be drawn to the circles
x2 + y2 – 4x – 4y + 6 = 0 and
x2 + y2 – 10x – 10y + l = 0 is the interval:
[Online April 11, 2014]
(a) (12, 32)
(b) (18, 42)
(c) (12, 24)
(d) (18, 48)
If the point (1, 4) lies inside the circle
x2 + y2 – 6x – 10y + P = 0 and the circle does not touch or
intersect the coordinate axes, then the set of all possible
values of P is the interval:
[Online April 9, 2014]
(a) (0, 25)
(b) (25, 39)
(c) (9, 25)
(d) (25, 29)
1
1 1
+
= .
a 2 b2 4
Then, the foot of perpendicular from the origin on the
Let a and b be any two numbers satisfying
variable line,
(d) 4 3
If y + 3x = 0 is the equation of a chord of the circle,
x2 + y2 – 30x = 0, then the equation of the circle with this
chord as diameter is :
[Online April 10, 2015]
(a) x2 + y2 + 3x + 9y = 0 (b) x2 + y2 + 3x – 9y = 0
(c) x2 + y2 – 3x – 9y = 0 (d) x2 + y2 – 3x + 9y = 0
The largest value of r for which the region represented by
the set {w Î C|w – 4 – i| £r} is contained in the region
represented by the set ( z Î c / | z - 1 |£| z + i | ) , is equal
to:
56.
(d) 5
Let the tangents drawn to the circle, x2 + y2 = 16 from the
point P(0, h) meet the x-axis at point A and B. If the area of
DAPB is minimum, then h is equal to :
[Online April 10, 2015]
(a) 4 2
50.
(b) 3
55.
(a) a hyperbola with each semi-axis = 2
(b) a hyperbola with each semi-axis = 2
(c) a circle of radius = 2
(d) a circle of radius =
58.
59.
[Online April 10, 2015]
5
3
2 (b) 2 2
2 (d)
(c)
17
2
2
Let C be the circle with centre at (1, 1) and radius = 1. If T is
the circle centred at (0, y), passing through origin and
touching the circle C externally, then the radius of T is
equal to
[2014]
x y
+ = 1 , lies on: [Online April 9, 2014]
a b
2
The circle passing through (1, –2) and touching the axis of
x at (3, 0) also passes through the point
[2013]
(a) (–5, 2) (b) (2, –5) (c) (5, –2)
(d) (–2, 5)
If a circle of unit radius is divided into two parts by an arc
of another circle subtending an angle 60 on the
circumference of the first circle, then the radius of the arc
is:
[Online April 25, 2013]
(a)
52.
(a)
53.
54.
1
2
(b)
1
4
(c)
3
2
(d)
(a)
60.
3
2
The equation of circle described on the chord
3x + y + 5 = 0 of the circle x2 + y2 = 16 as diameter is:
[Online April 19, 2014]
(a) x2 + y2 + 3x + y – 11 = 0
(b) x2 + y2 + 3x + y + 1 = 0
(c) x2 + y2 + 3x + y – 2 = 0
(d) x2 + y2 + 3x + y – 22 = 0
For the two circles x2 + y2 = 16 and
x2 + y2 – 2y = 0, there is/are
[Online April 12, 2014]
(a) one pair of common tangents
(b) two pair of common tangents
(c) three pair of common tangents
(d) no common tangent
61.
62.
3
(b)
1
2
(c) 1
(d)
2
Statement 1: The only circle having radius 10 and a
diameter along line 2x + y = 5 is x2 + y2 – 6x + 2y = 0.
Statement 2 : 2x + y = 5 is a normal to the circle
x2 + y2 – 6x + 2y = 0.
[Online April 25, 2013]
(a) Statement 1 is false; Statement 2 is true.
(b) Statement 1 is true; Statement 2 is true, Statement 2 is
a correct explanation for Statement 1.
(c) Statement 1 is true; Statement 2 is false.
(d) Statement 1 is true; Statement 2 is true; Statement 2 is
not a correct explanation for Statement 1.
If the circle x2 + y2 – 6x – 8y + (25 – a2) = 0 touches the axis
of x, then a equals.
[Online April 23, 2013]
(a) 0
(b) ± 4
(c) ± 2
(d) ± 3
If a circle C passing through (4, 0) touches the circle
x2 + y2 + 4x – 6y – 12 = 0 externally at a point (1, –1), then
the radius of the circle C is :
[Online April 22, 2013]
(a) 5
(b) 2 5
(c) 4
Downloaded from @Freebooksforjeeneet
(d)
57
M-151
Conic Sections
63.
If two vertices of an equilateral triangle are
A (– a, 0) and B (a, 0), a > 0, and the third vertex C lies
above x-axis then the equation of the circumcircle of DABC
is :
[Online April 22, 2013]
70.
(a) x2 + y 2 - 2 x - 2 y + 1 = 0
(a) 3x2 + 3y2 – 2 3ay = 3a 2
(b) 3 x 2 + 3 y 2 - 2ay = 3a 2
71.
(c) x 2 + y 2 - 2ay = a 2
(d) x 2 + y 2 - 3ay = a 2
64.
If each of the lines 5x + 8y = 13 and 4x – y = 3 contains a
diameter of the circle
x2 + y2 – 2(a2 – 7a + 11) x – 2 (a2 – 6a + 6) y + b3 + 1 = 0,
then
[Online April 9, 2013]
72.
(a) a = 5 and b Ï (-1,1)
73.
(b) a = 1 and b Ï (-1,1)
The length of the diameter of the circle which touches the
x-axis at the point (1,0) and passes through the point (2,3) is:
[2012]
10
3
6
5
(b)
(c)
(d)
3
5
5
3
The number of common tangents of the circles given by
x2 + y2 – 8x – 2y + 1 = 0 and x2 + y2 + 6x + 8y = 0 is
[Online May 26, 2012]
(a) one
(b) four
(c) two
(d) three
If the line y = mx + 1 meets the circle x2 + y2 + 3x = 0 in two
points equidistant from and on opposite sides of x-axis,
then
[Online May 19, 2012]
(a) 3m + 2 = 0
(b) 3m – 2 = 0
(c) 2m + 3 = 0
(d) 2m – 3 = 0
If three distinct points A, B, C are given in the 2-dimensional
coordinate plane such that the ratio of the distance of
each one of them from the point (1, 0) to the distance from
1
(– 1, 0) is equal to , then the circumcentre of the triangle
2
ABC is at the point
[Online May 19, 2012]
(a)
66.
67.
68.
æ5 ö
(a) ç , 0 ÷
è3 ø
69.
(b) (0, 0)
æ1 ö
(c) ç , 0 ÷
(d) (3, 0)
è3 ø
The equation of the circle passing through the point (1, 2)
and through the points of intersection of
x2 + y2 – 4x – 6y – 21 = 0 and 3x + 4y + 5 = 0 is given by
[Online May 7, 2012]
(a) x2 + y2 + 2x + 2y + 11 = 0
(b) x2 + y2 – 2x + 2y – 7 = 0
(c) x2 + y2 + 2x – 2y – 3 = 0
(d) x2 + y2 + 2x + 2y – 11 = 0
(b) x2 + y2 – x – y = 0
(c) x2 + y2 + 2x + 2y – 7= 0
(d) x2 + y2 + x + y – 2 = 0
The two circles x2 + y2 = ax and x2 + y2 = c2 (c > 0) touch
each other if
[2011]
(a) | a | = c
(b) a = 2c
(c) | a | = 2c
(d) 2 | a | = c
The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m
at two distinct points if
[2010]
(a) – 35 < m < 15
(b) 15 < m < 65
(c) 35 < m < 85
(d) – 85 < m < – 35
If P and Q are the points of intersection of the circles
x 2 + y2 + 3 x + 7 y + 2 p - 5 = 0 and
(c) a = 2 and b Ï (-¥,1) (d) a = 5 and b Î (-¥,1)
65.
The equation of the circle passing through the point (1, 0)
and (0, 1) and having the smallest radius is - [2011 RS]
74.
x2 + y2 + 2x + 2y – p2=0 then there is a circle passing
through P, Q and (1, 1) for:
[2009]
(a) all except one value of p
(b) all except two values of p
(c) exactly one value of p
(d) all values of p
Three distinct points A, B and C are given in the
2-dimensional coordinates plane such that the ratio of the
distance of any one of them from the point (1, 0) to the
1
. Then the
3
circumcentre of the triangle ABC is at the point: [2009]
distance from the point (–1, 0) is equal to
75.
76.
æ5 ö
(a) çè , 0÷ø
4
æ5 ö
(b) çè , 0÷ø
2
æ5 ö
(c) çè , 0÷ø
3
(d) (0, 0)
The point diametrically opposite to the point
P(1, 0) on the circle x2 + y2 + 2x + 4y – 3 = 0 is
[2008]
(a) (3, – 4)
(b) (–3, 4)
(c) (–3, –4)
(d) (3, 4)
Consider a family of circles which are passing through the
point (– 1, 1) and are tangent to x-axis. If (h, k) are the
coordinate of the centre of the circles, then the set of values
of k is given by the interval
[2007]
(a) - 1 £ k £ 1
2
2
(c) 0 £ k £
1
2
(b) k £
1
2
(d) k ³
1
2
Downloaded from @Freebooksforjeeneet
EBD_8344
M-152
77.
Mathematics
Let C be the circle with centre (0, 0) and radius 3 units. The
equation of the locus of the mid points of the chords of the
82.
circle x 2 + y 2 = 4 orthogonally, then the locus of its centre
is
[2004]
circle C that subtend an angle of 2p at its center is
3
(b) x 2 + y 2 = 1
(a) x 2 + y 2 = 3
2
(c) x 2 + y 2 =
78.
(d) x 2 + y 2 =
27
4
(a) 2ax - 2by - (a 2 + b 2 + 4) = 0
[2006]
(b) 2ax + 2by - (a 2 + b 2 + 4) = 0
(c) 2ax - 2by + (a 2 + b 2 + 4) = 0
9
4
(d) 2ax + 2by + (a 2 + b 2 + 4) = 0
If the lines 3x - 4 y - 7 = 0 and 2 x - 3 y - 5 = 0 are two
diameters of a circle of area 49p square units, the equation of the circle is
[2006]
83.
(a) x 2 + y 2 + 2 x - 2 y - 47 = 0
(b) x 2 + y 2 + 2 x - 2 y - 62 = 0
2
2
(c) x + y - 2 x + 2 y - 62 = 0
84.
(d) x 2 + y 2 - 2 x + 2 y - 47 = 0
79.
If a circle passes through the point (a, b) and cuts the
2
2
2
2
2
2
2
(d) 2ax + 2by – ( a + b + p ) = 0
If the pair of lines ax 2 + 2 (a + b)xy + by 2 = 0 lie along
diameters of a circle and divide the circle into four sectors
such that the area of one of the sectors is thrice the area of
another sector then
[2005]
2
(a) 3a - 10ab + 3b = 0
(d) ( x - p)2 = 4qy
If the lines 2 x + 3 y + 1 = 0 and 3x - y - 4 = 0 lie along
diameter of a circle of circumference 10p, then the equation
of the circle is
[2004]
(c) x 2 + y 2 + 2 x + 2 y - 23 = 0
(d) x 2 + y 2 - 2 x + 2 y - 23 = 0
85.
86.
(c) 3a 2 + 10ab + 3b 2 = 0
(d) 3a 2 + 2ab + 3b 2 = 0
If the circles x 2 + y 2 + 2ax + cy + a = 0 and
x 2 + y 2 – 3ax + dy – 1 = 0 intersect in two distinct points
P and Q then the line 5x + by – a = 0 passes through P and
Q for
[2005]
(a) exactly one value of a
(b) no value of a
(c) infinitely many values of a
(d) exactly two values of a
Intercept on the line y = x by the circle x 2 + y 2 - 2 x = 0 is
AB. Equation of the circle on AB as a diameter is [2004]
(a) x 2 + y 2 + x - y = 0
(b) x 2 + y 2 - x + y = 0
(c) x 2 + y 2 + x + y = 0
(d) x 2 + y 2 - x - y = 0
If the two circles ( x - 1) 2 + ( y - 3) 2 = r 2 and
x 2 + y 2 - 8 x + 2 y + 8 = 0 intersect in two distinct point,
then
[2003]
(a) r > 2
(b) 2 < r < 8
(c) r < 2
(d) r = 2.
(b) 3a 2 - 2ab + 3b 2 = 0
81.
(c) ( y - p )2 = 4qx
2
2
2
(c) x + y – 2ax – 3by + ( a2 - b2 - p 2 ) = 0
2
(b) ( x - q)2 = 4 py
(b) x 2 + y 2 - 2 x - 2 y - 23 = 0
(b) 2ax + 2by – ( a - b + p ) = 0
80.
(a) ( y - q)2 = 4 px
2
(a) x + y – 3ax – 4by + ( a + b - p ) = 0
2
A variable circle passes through the fixed point A( p, q ) and
touches x-axis . The locus of the other end of the diameter
through A is
[2004]
(a) x 2 + y 2 + 2 x - 2 y - 23 = 0
circle x 2 + y 2 = p 2 orthogonally, then the equation of
the locus of its centre is
[2005]
2
If a circle passes through the point (a, b) and cuts the
87.
The lines 2 x - 3 y = 5 and 3x - 4 y = 7 are diameters of a
circle having area as 154 sq.units.Then the equation of
the circle is
[2003]
(a) x 2 + y 2 - 2 x + 2 y = 62
(b) x 2 + y 2 + 2 x - 2 y = 62
(c) x 2 + y 2 + 2 x - 2 y = 47
(d) x 2 + y 2 - 2 x + 2 y = 47 .
Downloaded from @Freebooksforjeeneet
M-153
Conic Sections
88.
If the chord y = mx + 1 of the circle x2+y2=1 subtends an
angle of measure 45 at the maor segment of the circle
then value of m is
[2002]
89.
(c) –1 ± 2
(d) none of these
The centres of a set of circles, each of radius 3, lie on the
circle x2+y2=25. The locus of any point in the set is[2002]
(a) 4 £
90.
x2 + y2
(b)
æ1
æ3 1ö
ö
Parabola
92.
Let L1 be a tangent to the parabola y2 = 4(x + 1) and L2 be
a tangent to the parabola y2 = 8(x + 2) such that L1 and L2
intersect at right angles. Then L1 and L2 meet on the
straight line :
[Sep. 06, 2020 (I)]
(a) x + 3 = 0
(b) 2x + 1 = 0
(c) x + 2 = 0
(d) x + 2y = 0
93.
The centre of the circle passing through the point (0, 1)
and touching the parabola y = x2 at the point (2,4) is:
[Sep. 06, 2020 (II)]
94.
æ -53 16 ö
(a) ç
, ÷
è 10 5 ø
æ 6 53 ö
(b) ç , ÷
è 5 10 ø
æ 3 16 ö
(c) ç , ÷
è 10 5 ø
æ -16 53 ö
, ÷
(d) ç
è 5 10 ø
If the common tangent to the parabolas, y2 = 4x and x2 = 4y
also touches the circle, x2 + y2 = c2, then c is equal to:
[Sep. 05, 2020 (I)]
(a)
95.
1
1
1
(c)
4
1
(d)
2
(b)
2
2 2
Let P be a point on the parabola, y2 = 12x and N be the foot
of the perpendicular drawn from P on the axis of the
parabola. A line is now drawn through the mid-point M of
PN, parallel to its axis which meets the parabola at Q. If the
y-intercept of the line NQ is
1
(d) PN = 3
4
Let the latus ractum of the parabola y2 = 4x be the common
chord to the circles C1 and C2 each of them having radius
2 5. Then, the distance between the centres of the circles
C1 and C2 is :
[Sep. 03, 2020 (II)]
97.
æ1 3ö
(a) ç 2 , 2 ÷ (b) ç 2 , - 2 ÷ (c) ç 2 , 2 ÷ (d) ç 2 , 2 ÷
è
è
è
è
ø
ø
ø
ø
The equation of a circle with origin as a centre and passing
through equilateral triangle whose median is of length 3a
is
[2002]
(a) x2 + y2 = 9a2
(b) x2 + y2 = 16a2
(c) x2 + y2 = 4a2
(d) x2 + y2 = a2
TOPIC n
96.
£ 25
(c) x2 + y2 ³ 25
(d) 3 £ x2 + y2 £ 9
The centre of the circle passing through (0, 0) and (1, 0)
and touching the circle x2 + y2 = 9 is
[2002]
æ1 1ö
91.
£ 64
x2 + y2
1
3
(c) MQ =
(b) –2 ± 2
(a) 2 ± 2
(b) MQ =
(a) PN = 4
4
, then : [Sep. 03, 2020 (I)]
3
(b) 8
(c) 12
(d) 4 5
(a) 8 5
The area (in sq. units) of an equilateral triangle inscribed in
the parabola y2 = 8x, with one of its vertices on the vertex
of this parabola, is :
[Sep. 02, 2020 (II)]
(a) 64 3
98.
(b) 256 3
(c) 192 3
(d) 128 3
If one end of a focal chord AB of the parabola y2 = 8x is at
æ 1
ö
Aç
, - 2 ÷ , then the equation of the tangent to it at B
è 2
ø
is:
[Jan. 9, 2020 (II)]
(a) 2x + y – 24 = 0
(b) x – 2y + 8 = 0
(c) x + 2y + 8 = 0
(d) 2x – y – 24 = 0
99. The locus of a point which divides the line segment oining
the point (0, –1) and a point on the parabola, x2 = 4y,
internally in the ratio 1 : 2, is:
[Jan. 8, 2020 (I)]
(a) 9x2 – 12y = 8
(b) 9x2 – 3y = 2
(c) x2 – 3y = 2
(d) 4x2 – 3y = 2
100. Let a line y = mx (m > 0) intersect the parabola, y2 = x at a
point P, other than the origin. Let the tangent to it at P
meet the x-axis at the point Q, If area (DOPQ) = 4 sq. units,
then m is equal to _______.
[NA Jan. 8, 2020 (II)]
101. If y = mx + 4 is a tangent to both the parabolas, y2 =4x and
x2 = 2by, then b is equal to:
[Jan. 7, 2020 (I)]
(a) –32
(b) –64
(c) –128
(d) 128
102. The tangents to the curve y = (x – 2)2 –1 at its points of
intersection with the line x – y = 3, intersect at the point :
[April 12, 2019 (II)]
æ5 ö
æ 5
ö
æ5
ö
æ 5 ö
(a) ç ,1÷ (b) ç - , -1÷ (c) ç , -1÷ (d) ç - ,1÷
è2 ø
è 2
ø
è2
ø
è 2 ø
103. If the line ax + y = c, touches both the curves x2 + y2 = 1
and y2 = 4 2x , then |c| is equal to [April 10, 2019 (II)]
1
2
1
(d)
2
2
104. The area (in sq. units) of the smaller of the two circles that
touch the parabola, y2 = 4x at the point (1, 2) and the x-axis
is:
[April 09, 2019 (II)]
(a) 2
(b)
(a) 8p (2 –
2)
(c) 4p (3 + 2 )
(c)
(b) 4p (2 –
(d) 8p (3 – 2
Downloaded from @Freebooksforjeeneet
2)
2)
EBD_8344
M-154
Mathematics
105. If one end of a focal chord of the parabola, y2 = 16x is at
(1, 4), then the length of this focal chord is:
[April 09, 2019 (I)]
(a) 25
(b) 22
(c) 24
(d) 20
106. The shortest distance between the line y = x and the curve
y2 = x – 2 is :
[April 08, 2019 (I)]
(a) 2
7
(b)
8
(c)
7
4 2
11
(d)
4 2
107. If the tangents on the ellipse 4x2 + y2 = 8 at the points (1, 2)
and (a, b) are perpendicular to each other, then a2 is equal
to :
[April 08, 2019 (I)]
128
64
4
2
(b)
(c)
(d)
17
17
17
17
108. The tangent to the parabola y2 = 4x at the point where it
intersects the circle x2 + y2 = 5 in the first quadrant, passes
through the point :
[April 08, 2019 (II)]
(a)
æ 1 4ö
(a) ç - , ÷
è 3 3ø
110. Equation of a common tangent to the parabola y2 = 4x and
the hyperbola xy = 2 is :
[Jan. 11, 2019 (I)]
(a) x + y + 1 = 0
(b) x – 2y + 4 = 0
(c) x + 2y + 4 = 0
(d) 4x + 2y + 1 = 0
111. If the area of the triangle whose one vertex is at the vertex
of the parabola, y2 + 4 (x – a2) = 0 and the other two
vertices are the points of intersection of the parabola and
y-axis, is 250 sq. units, then a value of ‘a’ is :
[Jan. 11, 2019 (II)]
(a) 5 5
(b) 5(21/3)
(c) (10)2/3 (d) 5
2
112. If the parabolas y = 4b(x – c) and y2 = 8ax have a common
normal, then which one of the following is a valid choice
for the ordered triad (a, b, c)?
[Jan 10, 2019 (I)]
(b) (1, 1, 3)
æ1
ö
(c) ç , 2, 0 ÷
(d) (1, 1, 0)
è2
ø
113. The length of the chord of the parabola x2 = 4y having
(a) 3 2
(b) 2 11
[Jan. 10, 2019 (II)]
(c) 8 2
(b) (8, 6)
(c) (6, 4 2 )
(d) (4, – 4)
115. Equation of a common tangent to the circle, x2 + y2 – 6x = 0
and the parabola, y2 = 4x, is :
[Jan 09, 2019 (I)]
(a) 2 3 y = 12x + 1
(b)
3 y=x+3
(c) 2 3 y = – x – 12
(d) 3 y = 3x + 1
116. Let A (4, – 4) and B (9, 6) be points on the parabola,
y2 = 4x. Let C be chosen on the arc AOB of the parabola,
where O is the origin, such that the area of DACB is
maximum. Then, the area (in sq. units) of DACB, is:
[Jan. 09, 2019 (II)]
1
1
3
(b) 30
(c) 32
(d) 31
4
2
4
117. Tangent and normal are drawn at P(16, 16) on the parabola
æ3 7ö
æ 1 1ö
(c) ç , ÷
(d) ç - , ÷
è4 4ø
è 4 2ø
109. The equation of a tangent to the parabola, x2 = 8y, which
makes an angle q with the positive direction of x-axis, is :
[Jan. 12, 2019 (II)]
(a) y = x tanq + 2 cotq
(b) y = x tanq – 2 cotq
(c) x = y cotq + 2 tanq
(d) x = y cotq – 2 tanq
equation x - 2 y + 4 2 = 0 is:
(a) (5, 2 6 )
(a) 31
æ1 3ö
(b) ç , ÷
è4 4ø
æ1
ö
(a) ç , 2, 3 ÷
è2
ø
114. Axis of a parabola lies along x-axis. If its vertex and focus
are at distance 2 and 4 respectively from the origin, on the
positive x-axis then which of the following points does
not lie on it?
[Jan 09, 2019 (I)]
(d) 6 3
y2 = 16x , which intersect the axis of the parabola at A and
B, respectively. If C is the centre of the circle through the
points P, A and B and ÐCPB = q , then a value of tan q is:
[2018]
4
1
(d)
3
2
118. Tangents drawn from the point (– 8, 0) to the parabola
y2 = 8x touch the parabola at P and Q. If F is the focus of
the parabola, then the area of the triangle PFQ (in sq. units)
is equal to
[Online April 15, 2018]
(a) 48
(b) 32
(c) 24
(d) 64
119. If y = mx + c is the normal at a point on the parabola y2 = 8x
whose focal distance is 8 units, then |c| is equal to :
[Online April 9, 2017]
(a) 2
(b) 3
(c)
(b) 8 3
(c) 10 3 (d) 16 3
(a) 2 3
120. If the common tangents to the parabola, x2 = 4y and the
circle, x2 + y2 = 4 intersect at the point P, then the distance
of P from the origin, is :
[Online April 8, 2017]
(a)
2 +1
(b) 2(3 + 2 2)
(c) 2( 2 + 1)
(d) 3 + 2 2
121. Let P be the point on the parabola, y2 = 8x which is at a
minimum distance from the centre C of the circle,
x2 + (y + 6)2 = 1. Then the equation of the circle, passing
through C and having its centre at P is:
[2016]
x
+ 2y - 24 = 0
4
(b) x2 + y2 – 4x + 9y + 18 = 0
(c) x2 + y2 – 4x + 8y + 12 = 0
(d) x2 + y2 – x + 4y – 12 = 0
2
2
(a) x + y -
Downloaded from @Freebooksforjeeneet
M-155
Conic Sections
122. P and Q are two distinct points on the parabola, y2 = 4x,
with parameters t and t1 respectively. If the normal at P
passes through Q, then the minimum value of t12 is :
[Online April 10, 2016]
(a) 8
(b) 4
(c) 6
(d) 2
123. Let O be the vertex and Q be any point on the parabola,
x2 = 8y. If the point P divides the line segment OQ internally
in the ratio 1 : 3, then locus of P is :
[2015]
(a) y2 = 2x
(b) x2 = 2y
(c) x2 = y
(d) y2 = x
124. Let PQ be a double ordinate of the parabola, y2 = – 4x,
where P lies in the second quadrant. If R divides PQ in
the ratio 2 : 1 then the locus of R is :
[Online April 11, 2015]
(a) 3y2 = – 2x
(b) 3y2 = 2x
(c) 9y2 = 4x
(d) 9y2 = – 4x
125. The slope of the line touching both the parabolas y 2 = 4 x
and x 2 = -32 y is
[2014]
1
2
1
3
(b)
(c)
(d)
8
3
2
2
126. A chord is drawn through the focus of the parabola y2 = 6x
such that its distance from the vertex of this parabola is
(a)
5
, then its slope can be:
2
[Online April 19, 2014]
2
2
5
3
(b)
(c)
(d)
5
3
2
2
127. Two tangents are drawn from a point (– 2, – 1) to the curve,
y2 = 4x. If a is the angle between them, then |tan a| is equal
to:
[Online April 12, 2014]
(a)
1
1
(b)
(c) 3
(d) 3
3
3
128. Let L1 be the length of the common chord of the curves
x2 + y2 = 9 and y2 = 8x, and L2 be the length of the latus
rectum of y2 = 8x, then:
[Online April 11, 2014]
(a) L1 > L2
(b) L1 = L2
(a)
L1
(d) L = 2
2
(c) L1 < L2
129. Given : A circle, 2x2 + 2y2 = 5 and a parabola, y2 = 4 5 x.
Statement-1 : An equation of a common tangent to these
curves is y = x + 5 .
5
(m ¹ 0) is their
m
common tangent, then m satisfies m4 – 3m2 + 2 = 0.[2013]
(a) Statement-1 is true; Statement-2 is true; Statement-2
is a correct explanation for Statement-1.
Statement-2 : If the line, y = mx +
(b) Statement-1 is true; Statement-2 is true; Statement-2
is not a correct explanation for Statement-1.
(c) Statement-1 is true; Statement-2 is false.
(d) Statement-1 is false; Statement-2 is true.
130. The point of intersection of the normals to
the parabola y 2 = 4x at the ends of its latus rectum is :
[Online April 23, 2013]
(a) (0, 2) (b) (3, 0)
(c) (0, 3)
(d) (2, 0)
131. Statement-1: The line x – 2y = 2 meets the parabola,
y2 + 2x = 0 only at the point (– 2, – 2).
Statement-2: The line y = mx -
1
( m ¹ 0) is tangent to
2m
æ
1
1ö
the parabola, y2 = – 2x at the point ç ,- ÷ .
2
mø
è 2m
[Online April 22, 2013]
(a) Statement-1 is true; Statement-2 is false.
(b) Statement-1 is true; Statement-2 is true; Statement-2
is a correct explanation for statement-1.
(c) Statement-1 is false; Statement-2 is true.
(d) Statement-1 a true; Statement-2 is true; Statement-2 is
not a correct explanation for statement-1.
æ 3ö
x2 y 2
132. The normal at ç 2, ÷ to the ellipse,
+
= 1 touches
è 2ø
16 3
a parabola, whose equation is [Online May 26, 2012]
(a) y2 = – 104 x
(b) y2 = 14 x
(c) y2 = 26x
(d) y2 = – 14x
133. The chord PQ of the parabola y2 = x, where one end P of
the chord is at point (4, – 2), is perpendicular to the axis of
the parabola. Then the slope of the normal at Q is
[Online May 26, 2012]
1
1
(a) – 4
(b) (c) 4
(d)
4
4
1
is always a tangent to the
m
parabola, y2 = – 4x for all non- ero values of m.
Statement 2: Every tangent to the parabola, y2 = – 4x will
meet its axis at a point whose abscissa is non-negative.
[Online May 7, 2012]
(a) Statement 1 is true, Statement 2 is true; Statement 2 is
a correct explanation of Statement 1.
(b) Statement 1 is false, Statement 2 is true.
(c) Statement 1 is true, Statement 2 is false.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation of Statement 1.
135. The shortest distance between line y – x =1 and curve
x = y2 is
[2011]
134. Statement 1: y = mx -
(a)
3 2
8
(b)
8
3 2
(c)
4
3
Downloaded from @Freebooksforjeeneet
(d)
3
4
EBD_8344
M-156
Mathematics
136. If two tangents drawn from a point P to the parabola
y2 = 4x are at right angles, then the locus of P is [2010]
(a) 2x + 1 = 0
(b) x = – 1
(c) 2x – 1 = 0
(d) x = 1
137. A parabola has the origin as its focus and the line x = 2 as
the directrix. Then the vertex of the parabola is at [2008]
(a) (0, 2)
(b) (1, 0)
(c) (0, 1)
(d) (2, 0)
138. The equation of a tangent to the parabola
y2 = 8x is y = x + 2. The point on this line from which the
other tangent to the parabola is perpendicular to the given
tangent is
[2007]
(a) (2, 4) (b) (–2, 0) (c) (–1, 1) (d) (0, 2)
139. The locus of the vertices of the family of parabolas
y=
a3 x2 a2 x
+
- 2a
3
2
(a) xy =
is
105
64
[2006]
(b) xy =
y 2 = 8 x . The locus of mid point of PQ is
[2005]
(b) y 2 + 4x + 2 = 0
2
(d) x 2 – 4y + 2 = 0
(c) x + 4y + 2 = 0
141. A circle touches the x- axis and also touches the circle with
centre at (0,3 ) and radius 2. The locus of the centre of the
circle is
[2005]
(a) an ellipse
(b) a circle
(c) a hyperbola
(d) a parabola
142. If a ¹ 0 and the line 2bx + 3cy + 4d = 0 passes through
the points of intersection of the parabolas
y 2 = 4ax and x 2 = 4ay, then
[2004]
(a) d 2 + (3b - 2c ) 2 = 0
(b) d 2 + (3b + 2c)2 = 0
(c) d 2 + (2b - 3c ) 2 = 0
(d) d 2 + (2b + 3c ) 2 = 0
143. The normal at the point (bt12 , 2bt1 ) on a parabola meets
the parabola again in the point (bt 2 2 , 2bt 2 ) , then [2003]
(a) t 2 = t1 +
2
t1
(c) t 2 = -t1 +
2
t1
(b) t 2 = -t1 (d) t 2 = t1 -
2
t1
2
t1
144. Two common tangents to the circle x2 + y2 = 2a2 and
parabola y2 = 8ax are
[2002]
(a) x = ± ( y + 2 a )
(b) y = ± ( x + 2 a )
(c) x = ± ( y + a )
Ellipse
145. Which of the following points lies on the locus of the foot
of perpendicular drawn upon any tangent to the ellipse,
x2 y 2
+
= 1 from any of its foci?
4
2
[Sep. 06, 2020 (I)]
(a) (-2, 3)
(b) (-1, 2)
(c) (-1, 3)
(d) (1, 2)
146. If the normal at an end of a latus rectum of an ellipse
passes through an extermity of the minor axis, then the
eccentricity e of the ellipse satisfies: [Sep. 06, 2020 (II)]
(a) e4 + 2e2 – 1 = 0
(b) e2 + e – 1 = 0
4
2
(c) e + e – 1 = 0
(d) e2 + 2e – 1 = 0
147. If the co-ordinates of two points A and B are ( 7, 0) and
3
4
35
(c) xy =
(d) xy = 64
16
105
140. Let P be the point ( 1, 0 ) and Q a point on the locus
(a) y 2 – 4x + 2 = 0
TOPIC Đ
(d) y = ± ( x + a )
(- 7, 0) respectively and P is any point on the conic,
9x2 + 16y2 = 144, then PA + PB is equal to :
[Sep. 05, 2020 (I)]
(a) 16
(b) 8
(c) 6
(d) 9
148. If the point P on the curve, 4x2 + 5y2 = 20 is farthest from
the point Q(0, – 4), then PQ2 is equals to :
[Sep. 05, 2020 (I)]
(a) 36
(b) 48
(c) 21
(d) 29
x2
+
y2
= 1 (a > b) be a given ellipse, length of whose
a 2 b2
latus rectum is 10. If its eccentricity is the maximum value
149. Let
5
+ t - t 2 , then a2 + b2 is equal to:
12
[Sep. 04, 2020 (I)]
(a) 145
(b) 116
(c) 126
(d) 135
150. Let x = 4 be a directrix to an ellipse whose centre is at the
of the function, f(t ) =
1
. If P(1, b), b > 0 is a point on
2
this ellipse, then the equation of the normal to it at P is :
[Sep. 04, 2020 (II)]
(a) 4x – 3y = 2
(b) 8x – 2y = 5
(c) 7x – 4y = 1
(d) 4x – 2y = 1
origin and its eccentricity is
151. A hyperbola having the transverse axis of length 2 has
the same foci as that of the ellipse 3x2 + 4y2 = 12, then this
hyperbola does not pass through which of the following
points?
[Sep. 03, 2020 (I)]
æ 1
ö
, 0÷
(a) ç
è 2 ø
æ 3 ö
(b) çç - 2 , 1÷÷
è
ø
1 ö
æ
(c) ç 1, ÷
2ø
è
æ 3 1 ö
÷
(d) çç 2 ,
2 ÷ø
è
Downloaded from @Freebooksforjeeneet
M-157
Conic Sections
|x| | y|
+
= 1 and
2
3
152. Area (in sq. units) of the region outside
x2 y 2
+
= 1 is :
4
9
inside the ellipse
[Sep. 02, 2020 (I)]
(a) 6(p - 2)
(b) 3(p - 2)
(c) 3(4 - p)
(d) 6(4 - p)
159. An ellipse, with foci at (0, 2) and (0, –2) and minor axis of
length 4, passes through which of the following points ?
[April 12, 2019 (II)]
(a) ( 2, 2)
(b) (2, 2)
(c) (2, 2 2)
(d) (1, 2 2)
160. If the line x – 2y = 12 is tangent to the ellipse
153. If e1 and e2 are the eccentricities of the ellipse,
x2 y 2
+
=1
18 4
2
2
and the hyperbola, x - y = 1 respectively and (e1, e2) is
9
4
a point on the ellipse, 15x2 + 3y2 = k, then k is equal to
[Jan. 9, 2020 (I)]
(a) 16
(b) 17
(c) 15
(d) 14
154. The length of the minor axis (along y-axis) of an ellipse in
4
. If this ellipse touches the line, x
3
+ 6y = 8; then its eccentricity is:
[Jan. 9, 2020 (II)]
the standard form is
1 11
5
1 5
1 11
(b)
(c)
(d)
2 3
6
2 3
3 3
155. Let the line y = mx and the ellipse 2x2 + y2 = 1 intersect at a
point P in the first quadrant. If the normal to this ellipse at
(a)
æ 1
ö
,0 and (0, b),
P meets the co-ordinate axes at ç è 3 2 ÷ø
then b is equal to:
[Jan. 8, 2020 (I)]
2
2
2 2
2
(b)
(c)
(d)
3
3
3
3
156. If the distance between the foci of an ellipse is 6 and the
distance between its directrices is 12, then the length of
its latus rectum is:
[Jan. 7, 2020 (I)]
(a)
(a)
3
(b) 3 2
(c)
3
2
(d) 2 3
x2 y 2
+
=1
a2 9
for some a Î R, then the distance between the foci of the
ellipse is:
[Jan. 7, 2020 (II)]
157. If 3x + 4y = 12 2 is a tangent to the ellipse
(a) 2 7
(b) 4
(c) 2 5
(d) 2 2
158. If the normal to the ellipse 3x2 + 4y2=12 at a point P on it is
parallel to the line, 2x + y = 4 and the tangent to the ellipse
at P passes through Q (4,4) then PQ is equal to :
[April 12, 2019 (I)]
(a)
5 5
2
(b)
61
2
(c)
221
2
(d)
157
2
x2 y 2
+
=1
a 2 b2
æ -9 ö
at the point ç 3, ÷ , then the length of the latus rectum
è 2 ø
of the ellipse is :
[April 10, 2019 (I)]
(a) 9
(b) 12 2
(c) 5
(d) 8 3
2
161. The tangent and normal to the ellipse 3x + 5y2 = 32 at the
point P(2, 2) meet the x-axis at Q and R, respectively. Then
the area (in sq. units) of the triangle PQR is :
[April 10, 2019 (II)]
(a)
34
15
(b)
14
3
(c)
16
3
(d)
68
15
162. If the tangent to the parabola y2 = x at a point (a, b), (b Ō
Ā) is also a tangent to the ellipse, x2 + 2y2 = 1, then a is
equal to:
[April 09, 2019 (II)]
2 -1
(a)
(b) 2 2 - 1
(c) 2 2 + 1
(d)
2 +1
163. In an ellipse, with centre at the origin, if the difference of
the lengths of maor axis and minor axis is 10 and one of
the foci is at (0, 5 3 ), then the length of its latus rectum
is:
[April 08, 2019 (II)]
(a) 10
(b) 5
(c) 8
(d) 6
164. Let S and S¢ be the foci of an ellipse and B be any one of
the extremities of its minor axis. If DS¢BS is a right angled
triangle with right angle at B and area (DS¢BS) = 8 sq. units,
hen the length of a latus rectum of the ellipse is :
[Jan. 12, 2019 (II)]
(a) 4
(b) 2 2
(c) 4 2
(d) 2
165. If tangents are drawn to the ellipse x 2 + 2 y 2 = 2 at all
points on the ellipse other than its four vertices then the
mid points of the tangents intercepted between the
coordinate axes lie on the curve :
[Jan. 11, 2019 (I)]
1
(a)
4x2
1
(c)
2x
2
+
+
1
2 y2
1
4y
2
=1
(b)
x2 y 2
+
=1
4
2
=1
(d)
x2 y 2
+
=1
2
4
Downloaded from @Freebooksforjeeneet
EBD_8344
M-158
Mathematics
172. If the distance between the foci of an ellipse is half the
length of its latus rectum, then the eccentricity of the
ellipse is:
[Online April 11, 2015]
166. Two sets A and B are as under :
A = {(a, b) Î R ´ R :| a - 5 | < 1 and | b - 5 | < 1};
2
2
B = {(a,b) Î R ´ R : 4(a - 6) + 9(b - 5) £ 36}. Then :
[2018]
(a) A Ì B
(b) A Ç B = f (an empty set)
(d) B Ì A
167. If the length of the latus rectum of an ellipse is 4 units and
the distance between a focus and its nearest vertex on the
3
units, then its eccentricity is?
2
2
(b)
3
1
(c)
9
1
(d)
3
168. The eccentricity of an ellipse having centre at the origin,
axes along the co-ordinate axes and passing through the
points (4, –1) and (–2, 2) is :
[Online April 9, 2017]
(a)
1
2
(b)
2
5
3
2
(c)
3
4
(d)
169. Consider an ellipse, whose centre is at the origin and its
maor axis is along the x–axis. If its eccentricity is
3
and
5
the distance between its foci is 6, then the area (in sq.
units) of the quadrilateral inscribed in the ellipse, with the
vertices as the vertices of the ellipse, is :
[Online April 8, 2017]
(a) 8
(b) 32
(c) 80
(d) 40
2
170. If the tangent at a point on the ellipse
2
x
y
+
= 1 meets
27 3
the coordinate axes at A and B, and O is the origin, then the
minimum area (in sq. units) of the triangle OAB is :
[Online April 9, 2016]
(a) 3 3
(b)
9
2
(c) 9
(d)
9
x 2 y2
+
= 1, is :
9
5
27
2
3
[2015]
(b) 27
(c)
2 –1
( x2 + y2 )
2
(a)
= 6 x2 + 2 y 2
( x2 + y2 )
2
(b)
= 6 x2 - 2 y 2
( x2 - y2 )
2
(c)
= 6x2 + 2 y2
( x2 - y 2 )
2
(d)
= 6 x2 - 2 y 2
27
4
(d) 18
174. A stair-case of length l rests against a vertical wall and a
floor of a room. Let P be a point on the stair-case, nearer to
its end on the wall, that divides its length in the ratio 1 : 2.
If the stair-case begins to slide on the floor, then the locus
of P is:
[Online April 11, 2014]
(a) an ellipse of eccentricity
(b) an ellipse of eccentricity
(c) a circle of radius
1
2
3
2
l
2
3
l
2
175. If OB is the semi-minor axis of an ellipse, F1 and F2 are its
foci and the angle between F1B and F2B is a right angle,
then the square of the eccentricity of the ellipse is:
[Online April 9, 2014]
(d) a circle of radius
1
1
1
1
(b)
(c)
(d)
2
2 2
2
4
176. The equation of the circle passing through the foci of the
(a)
171. The area (in sq. units) of the quadrilateral formed by the
tangents at the end points of the latera recta to the ellipse
(a)
(b)
1
2 –1
(d)
2
2
173. The locus of the foot of perpendicular drawn from the
centre of the ellipse x2 + 3y2 = 6 on any tangent to it is
[2014]
[Online April 16, 2018]
1
(a)
2
2 2 –1
2
(c)
(c) neither A Ì B nor B Ì A
maor axis is
(a)
x2 y 2
+
= 1, and having centre at (0, 3) is [2013]
16 9
x2 + y2 – 6y – 7 = 0
x2 + y2 – 6y + 7 = 0
x2 + y2 – 6y – 5 = 0
x2 + y2 – 6y + 5 = 0
ellipse
(a)
(b)
(c)
(d)
Downloaded from @Freebooksforjeeneet
M-159
Conic Sections
177. A point on the ellipse, 4x2 + 9y2 = 36, where the normal is
parallel to the line, 4x –2y – 5 = 0, is :
[Online April 25, 2013]
æ 9 8ö
(a) ç , ÷
è 5 5ø
æ8 9ö
(b) ç , - ÷
è5 5ø
æ 9 8ö
æ8 9ö
(c) ç - , ÷
(d) ç , ÷
è 5 5ø
è 5 5ø
178. Let the equations of two ellipses be
E1 :
x2 y 2
x2 y 2
+
= 1 and E2 :
+
=1,
3
2
16 b 2
1
, then the length
2
of the minor axis of ellipse E2 is :[Online April 22, 2013]
(a) 8
(b) 9
(c) 4
(d) 2
179. Equation of the line passing through the points of
intersection of the parabola x 2 = 8y and the ellipse
If the product of their eccentricities is
x2
+ y 2 = 1 is :
3
[Online April 9, 2013]
(a) y – 3 = 0
(c) 3y + 1 = 0
(b) y + 3 = 0
(d) 3y – 1 = 0
185. An ellipse has OB as semi minor axis, F and F ' its focii
and the angle FBF ' is a right angle. Then the eccentricity
of the ellipse is
[2005]
1
1
1
(a) 1
(b)
(c)
(d)
2
3
4
2
186. The eccentricity of an ellipse, with its centre at the origin,
1
is . If one of the directrices is x = 4, then the equation of
2
the ellipse is:
[2004]
(a) 4 x 2 + 3 y 2 = 1
(b) 3 x 2 + 4 y 2 = 12
(c) 4 x 2 + 3 y 2 = 12
(d) 3 x 2 + 4 y 2 = 1
TOPIC Ė
Hyperbola
187. If the line y = mx + c is a common tangent to the hyperbola
2
x
+ y 2 = 1 at
4
which the tangents are parallel to the chord oining the
points (0, 1) and (2, 0), then the distance between P1 and
P2 is
[Online May 12, 2012]
180. If P1 and P2 are two points on the ellipse
(a) 2 2
(b) 5
(c) 2 3
(d) 10
181. Equation of the ellipse whose axes are the axes of
coordinates and which passes through the point (–3, 1)
2
is
[2011]
5
2
2
2
2
(a) 5x + 3y – 48 = 0
(b) 3x + 5y – 15 = 0
(c) 5x2 + 3y2 – 32 = 0
(d) 3x2 + 5y2 – 32 = 0
and has eccentricity
x 2 + 4 y 2 = 4 is inscribed in a rectangle
aligned with the coordinate axes, which in turn is inscribed
in another ellipse that passes through the point (4, 0). Then
the equation of the ellipse is :
[2009]
182. The ellipse
(a) x 2 + 12 y 2 = 16
8
2
4
5
(b)
(c)
(d)
3
3
3
3
184. In an ellipse, the distance between its foci is 6 and minor
axis is 8. Then its eccentricity is
[2006]
(a) 3
(b) 1
(c) 4
(d) 1
5
5
2
5
(a)
(b) 4 x 2 + 48 y 2 = 48
(c) 4 x 2 + 64 y 2 = 48
(d) x 2 + 16 y 2 = 16
183. A focus of an ellipse is at the origin. The directrix is the line
1
x = 4 and the eccentricity is . Then the length of the
2
semi-maor axis is
[2008]
x2 y 2
= 1 and the circle x2 + y2 = 36, then which one of
100 64
the following is true?
[Sep. 05, 2020 (II)]
(a) c2 = 369
(b) 5m = 4
(c) 4c2 = 369
(d) 8m + 5 = 0
x2
-
y2
= 1. If the
a2 b2
normal to it at P intersects the x-axis at (9, 0) and e is its
eccentricity, then the ordered pair (a2, e2) is equal to :
[Sep. 04, 2020 (I)]
188. Let P(3, 3) be a point on the hyperbola,
æ9 ö
æ9 ö
æ3 ö
(a) ç , 3 ÷ (b) ç , 2 ÷ (c) ç , 2 ÷ (d) (9, 3)
è2 ø
è2 ø
è2 ø
189. Let e 1 and e 2 be the eccentricities of the ellipse,
x2 y 2
x2 y 2
+ 2 = 1 (b < 5) and the hyperbola,
=1
25 b
16 b2
respectively satisfying e1e2 = 1. If a and b are the distances
between the foci of the ellipse and the foci of the hyperbola
respectively, then the ordered pair (a, b) is equal to :
[Sep. 03, 2020 (II)]
(a) (8, 12)
æ 20
ö
(b) ç , 12 ÷
3
è
ø
æ 24
ö
(c) ç , 10 ÷
è 5
ø
(d) (8, 10)
Downloaded from @Freebooksforjeeneet
EBD_8344
M-160
Mathematics
190. A line parallel to the straight line 2x – y = 0 is tangent to the
2
2
x
y
= 1 at the point (x1, y1). Then x12 + 5 y12
4
2
is equal to :
[Sep. 02, 2020 (I)]
(a) 6
(b) 8
(c) 10
(d) 5
hyperbola
p
191. For some q Î æç 0, ö÷ , if the eccentricity of the hyperbola,
è 2ø
x2 - y 2 sec2 q = 10 is
5 times the eccentricity of the
ellipse, x2 sec 2 q + y 2 = 5, then the length of the latus
rectum of the ellipse, is :
[Sep. 02, 2020 (II)]
(a) 2 6
(b)
30
2 5
3
(d)
4 5
3
(c)
through the point (4, -2 3) is 5x = 4 5 and its
eccentricity is e, then :
[April 10, 2019 (I)]
(a) 4e4 – 24e2 + 27 = 0
(b) 4e4 – 12e2 – 27 = 0
(c) 4e4 – 24e2 + 35 = 0
(d) 4e4 + 8e2 – 35 = 0
196. If 5x + 9 = 0 is the directrix of the hyperbola
16x2 – 9y2 = 144, then its corresponding focus is :
[April 10, 2019 (II)]
æ 5 ö
æ5 ö
(b) ç - ,0 ÷ (c) ç , 0 ÷ (d) (–5, 0)
è 3 ø
è3 ø
197. If the line y = mx + 7 3 is normal to the hyperbola
x 2 y2
= 1, then a value of m is : [April 09, 2019 (I)]
24 18
(a)
5
2
(b)
15
2
(c)
2
5
(a)
(c)
( -6, 2 10 )
( 4, 15 )
(b)
(d)
( 2 6, 5)
( 6,5 2 )
200. If a hyperbola has length of its conugate axis equal to 5
and the distance between its foci is 13, then the eccentricity
of the hyperbola is :
[Jan. 11, 2019 (II)]
13
13
13
(b) 2
(c)
(d)
6
8
12
201. Let the length of the latus rectum of an ellipse with its
maor axis along x-axis and centre at the origin, be 8. If the
distance between the foci of this ellipse is equal to the
length of its minor axis, then which one of the following
points lies on it?
[Jan. 11, 2019 (II)]
(a)
192. If a hyperbola passes through the point P(10,16) and it
has vertices at (± 6,0), then the equation of the normal to
it at P is:
[Jan. 8, 2020 (II)]
(a) 3x + 4y = 94
(b) 2x + 5y = 100
(c) x + 2y = 42
(d) x + 3y = 58
193. Let P be the point of intersection of the common tangents
to the parabola y2 = 12x and hyperbola 8x2 – y2 = 8. If S and
S¢ denote the foci of the hyperbola where S lies on the
positive x-axis then P divides SS¢ in a ratio :
[April 12, 2019 (I)]
(a) 13 : 11 (b) 14 : 13 (c) 5 : 4
(d) 2 : 1
194. The equation of a common tangent to the curves, y2 = 16x
and xy = – 4, is :
[April 12, 2019 (II)]
(a) x – y + 4 = 0
(b) x + y + 4 = 0
(c) x – 2y + 16 = 0
(d) 2x – y + 2 = 0
195. If a directrix of a hyperbola centred at the origin and passing
(a) (5, 0)
198. If the eccentricity of the standard hyperbola passing
through the point (4, 6) is 2, then the equation of the
tangent to the hyperbola at (4, 6) is : [April. 08, 2019 (II)]
(a) x – 2y + 8 = 0
(b) 2x – 3y + 10 = 0
(c) 2x – y – 2 = 0
(d) 3x – 2y = 0
199. If the vertices of a hyperbola be at (–2, 0) and (2, 0) and one
of its foci be at (–3, 0), then which one of the following
points does not lie on this hyperbola?[Jan. 12, 2019 (I)]
(d)
3
5
(a)
(c)
(4
(4
)
3)
2, 2 2
(b)
3, 2
(d)
(4
(4
)
3)
3, 2 2
2, 2
202. The equation of a tangent to the hyperbola 4x2 – 5y2 = 20
parallel to the line x – y = 2 is:
[Jan 10, 2019 (I)]
(a) x – y + 1 = 0
(b) x – y + 7 = 0
(c) x – y + 9 = 0
(d) x – y – 3 = 0
203. Let
ìï
üï
y2
x2
S = í( x, y ) Î R 2 :
= 1ý ,
1 + r 1 - r ïþ
ïî
where r ¹ ± 1 Then S represents: [Jan. 10, 2019 (II)]
(a) a hyperbola whose eccentricity is
2
, when 0 < r < 1
1- r
(b) an ellipse whose eccentricity is
2
, when r > 1
r +1
(c) a hyperbola whose eccentricity is
2
, when 0 < r < 1
r +1
(d) an ellipse whose eccentricity is
1
, when r > 1
r +1
Downloaded from @Freebooksforjeeneet
M-161
Conic Sections
204. Let 0 < q <
210. The eccentricity of the hyperbola whose length of the latus
rectum is equal to 8 and the length of its conugate axis is
equal to half of the distance between its foci, is : [2016]
p
. If the eccentricity of the hyperbola
2
x2
y2
= 1 is greater than 2, then the length of
2
cos q sin 2 q
its latus rectum lies in the interval:
[Jan 09, 2019 (I)]
(a) (3, ¥) (b) (3/2, 2] (c) (2, 3]
(d) (1, 3/2]
205. A hyperbola has its centre at the origin, passes through
the point (4, 2) and has transverse axis of length 4 along
the x-axis. Then the eccentricity of the hyperbola is :
[Jan. 09, 2019 (II)]
(a)
3
2
(b)
3
(c) 2
(d)
2
3
206. Tangents are drawn to the hyperbola 4x 2 - y2 = 36 at
the points P and Q. If these tangents intersect at the point
T(0, 3) then the area (in sq. units) of DPTQ is :
[2018]
(a) 54 3 (b) 60 3
(c) 36 5 (d) 45 5
207. The locus of the point of intersection of the lines,
2 x – y + 4 2k = 0 and
2kx + ky – 4 2 = 0 (k is any
non- ero real parameter) is.
[Online April 16, 2018]
(a) A hyperbola with length of its transverse axis 8 2
3
(
2, 3
)
and has foci at (± 2, 0). Then the tangent to this hyperbola
at P also passes through the point :
[2017]
(a)
(c)
((2
)
3)
2, - 3
(b)
2,3
(d)
(3 2, 2 3 )
( 3, 2 )
209. The locus of the point of intersection of the straight lines,
tx – 2y – 3t = 0
x – 2ty + 3 = 0 (t ÎR), is :
[Online April 8, 2017]
(a) an ellipse with eccentricity
of the conic,
3
(c)
x 2 y2
+
= 4 and has vertices at the foci of
3
4
3
, then
2
which of the following points does NOT lie on it ?
[Online April 10, 2016]
this conic. If the eccentricity of the hyperbola is
(a) ( 5, 2 2)
(b) (0, 2)
(c) (5, 2 3)
(d) ( 10, 2 3)
212. Let a aand b respectively be the semitransverse and semiconugate axes of a hyperbola whose eccentricity satisfies
the equation 9e2 – 18e + 5 = 0. If S(5, 0) is a focus and
5x = 9 is the corresponding directrix of this hyperbola, then
a2 – b2 is equal to :
[Online April 9, 2016]
(a) –7
(b) –5
(c) 5
(d) 7
113. An ellipse passes through the foci of the hyperbola,
9x2 – 4y2 = 36 and its maor and minor axes lie along the
tr ansverse and conugate axes of th e hyperbola
respectively. If the product of eccentricities of the two
æ 39
ö
, 3÷
è 2
ø
æ 13
ö
, 6÷
(a) ç
è 2
ø
3
208. A hyperbola passes through the point P
(b)
conics is
1
(d) A hyperbola whose eccentricity is
2
1
, then which of the following points does not
2
lie on the ellipse?
[Online April 10, 2015]
(b) An ellipse with length of its maor axis 8 2
(c) An ellipse whose eccentricity is
4
4
(d)
3
3
3
211. A hyperbola whose transverse axis is along the maor axis
(a)
2
5
(b) an ellipse with the length of maor axis 6
(c) a hyperbola with eccentricity 5
(d) a hyperbola with the length of conugate axis 3
(b) ç
æ1
3ö
(c) çè 13,
(d) ( 13, 0 )
÷
2
2 ø
114. The tangent at an extremity (in the first quadrant) of latus
x 2 y2
= 1 , meet x-axis and
4
5
y-axis at A and B respectively. Then (OA)2 – (OB)2, where
O is the origin, equals:
[Online April 19, 2014]
4
20
16
(b)
(a) (c) 4
(d) 9
9
3
115. Let P (3 sec q, 2 tan q) and Q (3 sec f, 2 tan f) where
rectum of the hyperbola
q+f =
p
, be two distinct points on the hyperbola
2
x 2 y2
= 1 . Then the ordinate of the point of intersection
9
4
of the normals at P and Q is:
[Online April 11, 2014]
(a)
11
3
(b) -
11
3
(c)
13
2
Downloaded from @Freebooksforjeeneet
(d) -
13
2
EBD_8344
M-162
Mathematics
216. A common tangent to the conics x 2 = 6y and
2x2 – 4y2 = 9 is:
[Online April 25, 2013]
(a) x - y =
3
2
(b) x + y = 1
(c) x + y =
9
2
(d) x – y = 1
x2 y 2
= 1 meets x-axis at P
4
2
and y-axis at Q. Lines PR and QR are drawn such that
OPRQ is a rectangle (where O is the origin). Then R lies on:
[Online April 23, 2013]
4
x
(c)
2
2
x2
+
2
y
+
2
4
y2
=1
(b)
2
x
=1
(d)
218. If the foci of the ellipse
2
4
x2
-
4
y
-
2
2
y2
=1
=1
x2 y 2
+
= 1 coincide with the foci
16 b 2
2
2
of the hyperbola x - y = 1 , then b2 is equal to
144 81 25
[Online May 19, 2012]
(a) 8
(b) 10
(c) 7
(d) 9
219. If the eccentricity of a hyperbola
passes through (k, 2), is
(a) 18
(b) 8
x2
-
y2
= 1 , which of the
cos 2 a sin 2 a
following remains constant when a varies = ?
[2007]
(a) abscissae of vertices (b) abscissae of foci
(c) eccentricity
(d) directrix.
221. For the Hyperbola
217. A tangent to the hyperbola
(a)
220. The equation of the hyperbola whose foci are
(– 2, 0) and (2, 0) and eccentricity is 2 is given by :
[2011RS]
(a) x2 – 3y2 = 3
(b) 3x2 – y2 = 3
(c) – x2 + 3y2 = 3
(d) – 3x2 + y2 = 3
222. The locus of a point P (a, b) moving under the condition
that the line y = ax + b is a tangent to the hyperbola
x2
-
y2
= 1 is
a
b2
(a) an ellipse
(c) a parabola
2
223. The foci of the ellipse
[2005]
(b) a circle
(d) a hyperbola
x2 y 2
+
= 1 and the hyperbola
16 b2
1
x2 y 2
coincide. Then the value of b 2 is[2003]
=
144 81 25
(a) 9
(b) 1
(c) 5
(d) 7
x2 y 2
= 1 , which
9 b2
13
, then the value of k2 is
3
[Online May 7, 2012]
(c) 1
(d) 2
Downloaded from @Freebooksforjeeneet
M-163
Conic Sections
1.
(d) In right DRSQ , sin 60° =
RS
r
\ ab |max =
4.
P
R(0, 0)
r
60
S r/2
5+ 9
= 7.
2
(3)
Q
r2
3
3r
=
2
2
Now equation of PQ is y – 2x – 3 = 0
Þ
2.
(0, 0)
3r | 0 + 0 - 3 |
=
2
5
x=3
Þ Radius (r1) = 3 – k
Q Centre lies on x + y = 2
Let x = k
3r
3
2 3
12
=
Þr=
Þ r2 =
2
5
5
5
(b) We know family of circle be S1 + lS2 = 0
\y = 2-k
x2 + y 2 - 6 x + l ( x2 + y 2 - 4 y) = 0
Þ (1 + l) x 2 + (1 + l ) y 2 - 6 x - 4ly = 0
Þ Centre = (k, 2 – k)
Also, radius (r2) = 2 – (2 – k)
...(i)
\ 3 - k = 2 - (2 - k )
2l ö
æ 3
Centre (– g, – f ) = ç
,
è 1 + l l + 1÷ø
Þk=
Centre lies on 2x – 3y + 12 = 0, then
6
6l
+ 12 = 0 Þ l = -3
l +1 l +1
Equation of circle (i),
5.
Þ x 2 + y 2 + 3x - 6 y = 0
...(ii)
Only (– 3, 6) satisfy equation (ii).
(7)
Let P (3cos q, 3sin q), Q( -3cos q, - 3sin q)
a=
3cos q + 3sin q - 2
ab =
2
, b=
3
2
3 3
=
2 2
Hence, diameter = 3.
(9)
r = 3-
-2 x 2 - 2 y 2 - 6 x + 12 y = 0
3.
r1
(k, 2 – k)
Þ RS = r ´
\
y=2
-3cos q - 3sin q - 2
2
(3cos q + 3sin q) 2 - 4
5 + 9sin 2q
=
2
2
The given circle is x2 + y 2 - 2 x - 4 y + 4 = 0
\ Centre of circle (1, 2), r = 1.
If line cuts circle then p < r, where p =
Þ
3+8- k
< 1 Þ k Î (6, 16)
5
k = 7, 8, 9, 10, 11, 12, 13, 14, 15
ab is max. when sin2q = 1
Downloaded from @Freebooksforjeeneet
ax1 + by1 + c
a 2 + b2
EBD_8344
M-164
Mathematics
x + y 2 = 0, which is perpendicular to x – y + c = 0
6.
æ 1 1 ö
,
At ç
÷ which is tangent of (x – 3)2 + y2 = 1
è 2 2ø
So, m = 1 Þ y = x + c
Now, distance of (3, 0) from y = x + c is
(d)
c+3
Equation of family of circle
(x – 0)2 + (y – 4)2 + lx = 0
Passes through the point (2, 0) then
4 + 16 + 2l = 0 Þ l = – 10
Hence, the equation of circle
x2 + y2 – 10x – 8y + 16 = 0
Þ (x – 5)2 + (y – 4)2 = 25
Centre (5, 4).
Þ
Þ
Þ
\
9.
1
1
coeff. of x + coeff. of y - constant
R=
2
2
(c) L = S1 = 16 = 4
=
Perpendicular distance of 4x + 3y – 8 = 0 from the centre of
circle
7.
20 + 16 - 8
16 + 9
=
28
¹5
5
( k)
2
2 LR
2
L +R
2
=
2 ´ 4 ´ 2 16
=
16 + 4
20
of chord of contact =
Hence, 4x + 3y – 8 = 0 can not be tangent to the circle.
(36)The given equation of circle
x2 – 6x + y2 + 8 = 0
(x – 3)2 + y2 = 1
...(i)
So, centre of circle (i) is C1(3, 0) and radius r1 = 1.
And the second equation of circle
x2 – 8y + y2 + 16 – k = 0 (k > 0)
x 2 + ( y - 4)2 =
c = -3 ± 2
(c + 3)2 = 2
c2 + 6c + 9 = 2
c2 + 6c + 7 = 0
R = 16 + 4 - 16 = 2
Length of chord of contact
= 25 + 16 - 16 = 5
=
=1
2
10.
So, centre of circle (ii) is C2(0, 4) and radius r2 =
Two circles touches each other when
P
(b)
12
ᔴ1
k
Distance between C2(3, 0) and C1(0, 4) is
Þ
k + 1 or
k +1= 5
k - 1 (C1C2 = 5)
or
k -1 = 5
Þ k = 16 or k = 36
Hence, maximum value of k is 36
The given equation of circles
x2 – 6x + y2 + 8 = 0
Þ (x – 3)2 + y2 = 1
8.
æ 1 1 ö
,
(c) Slope of tangent of x2 + y2 = 1 at ç
÷
è 2 2ø
1
1
x+
y -1 = 0
2
2
5
M
ᔴ2
Q
According to the diagram,
In DPC1C2, tan a =
5
5
Þ sin a =
12
13
In DPC1M, sin a =
PM
5
PM
60
Þ
=
Þ PM =
12
13
12
13
C1C2 = |r1 ± r2| Þ 5 = 1 ± k
either
64
5
a
...(ii)
Square of length
120
13
(a) Let centre of circle is C and circle cuts the y-axis at B
and A. Let mid-point of chord BA is M.
Hence, length of common chord (PQ) =
11.
A
M
4
3
ᔴ
B
(3 0)
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M-165
Conic Sections
CB =
14.
MC 2 + MB 2
(b) Let centre of required circle is (h, k).
\ OO’ = r + r’
[By the diagram]
32 + 42 = 5 = radius of circle
\ equation of circle is,
(x – 3)2 + (y – 5)2 = 52
(3, 10) satisfies this equation.
Although there will be another circle satisfying the same
conditions that will lie below the x-axis having equation
(x – 3)2 + (y – 5)2 = 52
12.
OἘ
rἘ
r
O (0 0)
(b) S1 º x2 + y2 + 5Kx + 2y + K = 0
3
1
y- =0
2
2
Equation of common chord is S1 – S2 = 0
S2 º x2 + y2 + Kx +
1
y
+ K + =0
...(i)
2
2
Equation of the line passing through the intersection
points P & Q is,
4x + 5y – K = 0
...(ii)
Comparing (i) and (ii),
Þ h2 + k 2 = 1 + h
Þ h2 + k2 = 1 + h2 + 2h
Þ k2 = 1 + 2h
Þ 4Kx +
4K 1 2K + 1
= =
4 10 -2 K
Þ K=
\ locus is y = 1 + 2 x , x ³ 0
15.
a +i
a-i
Since, is a complex number and let = x + iy
(b) Let ÎS then z =
Then, x + iy =
...(iii)
1
and – 2K = 20K + 10
10
Þ 22K = –10 Þ K =
13.
(h k)
Þ x + iy =
-5
11
a2 + 1
(by rationalisation)
( a 2 - 1)
i (2a)
+ 2
a +1 a + 1
2
Then compare both sides
a2 - 1
x= 2
a +1
1 -5
Q K = or
is not satisfying equation (3)
10 11
\ No value of K exists.
(b) Equation of circle which touches the line y = x at
(1, 1) is, (x – 1)2 + (y – 1)2 + l (y – x) = 0
This circle passes through (1, – 3)
y=
2a
16.
\ 0 + 16 + l (– 3 – 1) = 0
Þ 16 + l (– 4) = 0 Þ l = 4
Hence, equation of circle will be,
(x – 1)2 + (y – 1)2 + 4y – 4x = 0
Þ x2 + y2 – 6x + 2y + 2 = 0
y㦨x
...(ii)
Now squaring and adding equations (i) and (ii)
(1 –3)
(1 1)
...(i)
a2 + 1
Þ x2 + y2 =
\ Radius =
(a + i ) 2
(a 2 - 1)2
(a 2 + 1) 2
+
4a 2
(a 2 + 1) 2
=1
(a) Let any tangent to circle x2 + y2 = 1 is
x cosq + y sinq = 1
Since, P and Q are the point of intersection on the coordinate axes.
1 ö
æ 1
ö
æ
,0 ÷ & Q º ç 0,
Then P º ç
÷
è cos q ø
è sin q ø
1 ö
æ 1
,
\ mid-point of PQ be M º ç
÷ º ( h, k )
è 2cos q 2sin q ø
9 +1 - 2 = 2 2
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EBD_8344
M-166
Mathematics
Þ cos q =
1
2h
Hence, the sum of squares of lengths of chords
...(i)
1
sin q =
...(ii)
2k
Now squaring and adding equation (i) and (ii)
=
19.
5
æ
n =1
è
å 4 çç16 -
n2 ö
5 ´ 6 ´11
÷
= 210
2 ÷ø = 64 × 5 – 2.
6
(b) As ÐAOB = 90
1
1
+
=4
h2 k 2
\ locus of M is : x2 + y2 = 4x2y2
Þ h2 + k2 = 4h2k2
17.
(h, k)
(c) By the diagram, dc1c2 = r1 - r2
2
Let AB diameter and M(h, k) be foot of perpendicular, then
2
x +y =4
-h
k
Then, equation of AB
MAB =
2
2
x + y + 6x + 8y – 24 = 0
-h
( x - h)
k
Þ hx + ky = h2 + k2
(y – k) =
18.
Equation of common tangent is,
S1 – S2 = 0
6x + 8y – 20 = 0 Þ 3x + 4y – 10 = 0
Hence (6, – 2) lies on it.
(d) Let the chord x + y = n cuts the circle x2 + y2 = 16 at P
and Q Length of perpendicular from O on PQ
0+ 0-n
=
2
2
1 +1
=
æ h2 + k 2 ö
æ h2 + k 2 ö
Then, A ç h , 0÷ and B ç 0, k ÷
è
ø
è
ø
Q AB is the diameter, then
AB = 2R
Þ AB2 = 4R2
n
2
2
æ h 2 + k 2 ö æ h2 + k 2 ö
Þ ç
÷ +ç
÷ = 4R2
è h ø è k ø
Hence, required locus is (x2 + y2)3 = 4R2 x2 y2
P
20.
P
(d)
4
O
C1
Q
C2
x+y=n
Q
2
n2
2 æ n ö
Then, length of chord PQ = 2 4 - ç
÷ = 2 16 2
è 2ø
Thus only possible values of n are 1, 2, 3, 4, 5.
2g1g2 + 2f1f2 = 2(–1)(–3) + 2(–1)(–3) = 12
c1 + c2= 14 – 2 = 12
Since, 2g1g2 + 2f1f2= c1 + c2
Hence, circles intersect orthogonally
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M-167
Conic Sections
\ Area of the quadrilateral PC1QC1
23.
æ1
ö
= 2 çè (C1 P)(C2 P )÷ø
2
21.
(b) Q Two circles of equal radii intersect each other
orthogonally. Then R is mid point of PQ.
(0, 1)
P
1
= 2 ´ r1r2 = (2)(2) = 4 sq, units
2
(c) Condition 1: The centre of the two circles are (1, 1)
and (9, 1). The circles are on opposite sides of the line
3x + 4y – l = 0.
Put x = 1, y = 1 in the equation of line,
3(1) + 4(1) – l = 0 Þ 7 – l = 0
Now, put x = 9, y = 1 in the equation of line,
3(9) + 4(1) – l = 0
Then, (7 – l) (27 + 4 – l) < 0
Þ (l – 7) (l – 31) < 0
l Î (7, 31)
…(i)
Condition 2: Perpendicular distance from centre on line ³
radius of circle.
For x2 + y2 – 2x – 2y = 1,
90
O1
O2
R
Q
(0, –1)
and PR = O1R = O2R
1
(0 - 0) 2 + (1 + 1) 2 = 1
2
\ Distance between centres = 1 + 1 = 2.
PR =
24.
(d)
C (h, k )
|3 + 4 - l |
³1
5
Þ |l – 7| ³ 5
Þ l ³ 12 or l Þ 2
...(ii)
For x2 + y2 – 18x – 2y + 78 = 0
Þ
Let centre be C(h, k)
CQ = CP = r
Þ CQ2 = CP2
|27 + 4 - l |
³2
5
Þ l ³ 41 or l £ 21
...(iii)
Intersection of (1), (2) and (3) gives l Î [12, 21].
(h – 0)2 + (k ± 0)2 = CM2 + MP2
h2 + (k ± 2b)2 = k2 + 4a2
h2 + k2 + 4b2 ± 4bk = k2 + 4a2
Then, the locus of centre C(h, k)
22.
(c) The equation of circle is,
x2 + y2 – 6x + 8y – 103 = 0
x2 + 4b2 ± 4by = 4a2
Hence, the above locus of the centre of circle is a
Þ (x – 3)2 + (y + 4)2 = (8 2)2
parabola.
C(3, – 4), r = 8 2
Þ Length of side of square =
25.
2r = 16
S
P
2r
(–5, –4)
(c) The equation of circle x2 + y2 + 4x – 6y = 12 can be
written as (x + 2)2 + (y – 3)2 = 25
Q
Q(4, 0)
(11, –4)
R
Þ P(–5, 4), Q(–5, –12)
R(11, –12), S(11, 4)
Þ Required distance = OP
(–2, 3)
P
= (-5 - 0) 2 + (-4 - 0) 2 = 25 + 16 = 41
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EBD_8344
M-168
Mathematics
AM2 = AC2 – MC2
= (a + c)2 – (a – c)2 = 4ac
Þ AM2 = XY2 = 4ac
Let P = (1, –1) & Q = (4, 0)
Equation of tangent at P (1, –1) to the given circle :
x(1) + y(–1) + 2(x + 1) – 3(y – 1) –12 = 0
3x – 4y – 7 = 0
...(i)
The required circle is tangent to (1) at (1, –1).
\ (x – 1)2 + (y + 1)2 + l (3x – 4y – 7) = 0
...(ii)
Equation (ii) passes through Q(4, 0)
Þ 32 + 12 + l(12 – 7) = 0 Þ 5l + 10 = 0 Þ l = –2
Equation (2) becomes x2 + y2 – 8x + 10y + 16 = 0
radius =
26.
Þ XY = 2 ac
Similarly, YZ = 2 ba and XZ = 2 bc
Then, XZ = XY + YZ
Þ 2 bc = 2 ac + 2 ba
(-4)2 + (5)2 - 16 = 5
Þ
28.
(d)
30 r
a
2
(–5,–6)
1
1
1
+
=
a
b
c
(d) Consider the equation of circles as,
x2 + y2 – 16x – 20y + 164 = r2
i.e. (x – 8)2 + (y – 10)2 = r2
...(i)
and (x – 4)2 + (y – 7)2 = 36
...(ii)
Both the circles intersect each other at two distinct points.
Distance between centres
= (8 - 4)2 + (10 - 7)2 = 5
\ |r – 6| < 5 < |r + 6|
\ If |r – 6| < 5 Þ r Î (1, 11)
and |r + 6| > 5 Þ r Î (–฀ , –11) È (–1, ฀ )
From (iii) and (iv),
r Î (1, 11)
Let the sides of equilateral D inscribed in the circle be a,
a
2r
then cos 30 =
3 a
=
Þ a = 3r
2 2r
29.
Then, area of the equilateral triangle =
=
3
4
( 3r )
2
=
3 2
a
4
(a)
æ 1ö
çè 0, ÷ø
2
B
3 3 2
r
4
But it is given that area of equilateral triangle = 27 3
Let equation of circle be x2 + y2 + 2gx + 2fy = 0
As length of intercept on x axis is 1 = 2 g 2 - c
3 3 2
r
Then, 27 3 =
4
r2 = 36 Þ r = 6
Þ |g| =
2
æ 1
ö æ 1
ö
But ç - coeff. of x÷ + ç - coeff. of y÷
è 2
ø è 2
ø
(a)
C
B
M
1
2
2
– constant term = r2
(–5)2 + (–6)2 – c = 36 Þ c = 25
27.
...(iii)
...(iv)
length of intercept on y-axis =
1
= 2 f 2 -c
2
1
4
Equation of circle that passes through given points is
Þ |f|=
y
=0
2
Tangent at (0, 0) is,
x2 + y2 – x –
A
X
Y
Z
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M-169
Conic Sections
( x - 0)
(x2 – x1 )2 + (y2 – y1)2 = (2 – 3)2 + (3 – 4)2 = 2
æ dy ö
× (x - 0)
(y – 0) = çè dx ø÷
(0,0)
Þ 2x + y = 0
Perpendicular distance from B(1, 0) on the tangent to the
1
circle = 2
5
æ 1ö
Perpendicular distance from B ç 0, ÷ on the tangent to
è 2ø
the circle =
30.
33.
3
4
\ Two parabolas are y2 = 3x and x2 = 3y
Suppose y = mx + c is the common tangent.
\ y2 = 3x Þ (mx + c)2 = 3x Þ m2x2 + (2mc – 3) x + c2 = 0
As, the tangent touches at one point only
So, b2 – 4ac = 0
Þ (2mc – 3)2 – 4m2c2 = 0
Þ 4m2c2 + 9 – 12mc – 4m2c2 = 0
\ 4a = 4b = 3 Þ a = b =
2
5
1
+2
5
2
=
.
Sum of perpendicular distance =
2
5
2
(c) Equation of tangent at (1, 7) to x = y – 6 is
2x – y + 5 = 0.
(–8, –6)
O
2x–y+5=0
Þc=
P
31.
-16 + 6 + 5
= 64 + 36 - C
5
Þ m3 = – 1 Þ m = – 1 Þ c =
Hence, y = mx + c = – x –
\
32.
2
= 2 g - c = 2 25 - 20 = 2 5
(c) Equation of the line passing through the points (2, 3)
and (4, 5) is
æ5–3ö
y–3= ç
÷ x – 2 Þ x – y + 1 = 0 ..... (i)
è4– 2ø
Equation of the perpendicular line passing through the
midpoint (3, 4) is x + y – 7 = 0
..... (ii)
Lines (1) and (2) intersect at the center of the circle. So, the
center of the circle is (3, 4)
Therefore, the radius is
.... (i)
34.
..... (ii)
– 4c – 4 æ 3 ö
=
ç
÷
3
3 è 4m ø
m2 =
Þ 5 = 100 - c Þ c = 95
(d) Given circle is:
x2 + y2 + 2x – 4y – 4 = 0
\ its centre is (– 1, 2) and radius is 3 units.
Let A = (x, y) be the centre of the circle C
x -1
y+2
= 2 Þ x = 5 and
=2 Þ y=2
2
2
So the centre of C is (5, 2) and its radius is 3
\ equation of centre C is:
x2 + y2 – 10x – 4y + 20 = 0
\ The length of the intercept it cuts on the x-axis
9
3
=
12 m 4 m
\ x2 = 3y Þ x2 = 3 (mx + c) Þ x2 – 3mx – 3c = 0
Again, b2 – 4ac = 0
Þ 9m2 – 4 (1) (– 3c) = 0
Þ 9m2 = – 12c
Form (i) and (ii)
Now, perpendicular from centre O(–8, –6) to
2x – y + 5 = 0 should be equal to radius of the circle
\
units.
(c) As origin is the only common point to x-axis and
y-axis, so, origin is the common vertex
Let the equation of two of parabolas be y 2 = 4ax and
x2 = 4by
Now latus rectum of both parabolas = 3
–3
4
3
4
Þ 4 (x + y) + 3 = 0
(a) Here, equation of tangent on C1 at (2, 1) is:
2x + y – (x + 2) – 1 = 0
Or x + y = 3
If it cuts off the chord of the circle C2 then the equation of
the chord is:
x+y=3
\ distance of the chord from (3, – 2) is :
3- 2-3
= 2
2
Also, length of the chord is l = 4
d=
2
\ radius of C2 = r =
=
(2)2 + ( 2) 2 =
æl ö
2
ç ÷ +d
è2ø
6
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EBD_8344
M-170
35.
Mathematics
2x2 + 2y2 – y – 1 = 0
Þ x2 + y2 – 1/2y – 1/2 = 0
(None)
(Let the equation of circle be
x2 + (y – k)2 = r2
It touches x – y = 0
2
1ö
9
2 æ
Þ x +çy- ÷ =
4ø
16
è
3
4
(b) P (4, 7). Here, x = 4, y = 7
x – y = –3
\ PA × PB = PT2
r=
Þ
37.
Also; PT =
x 2 + y2 - ( x - y )
2
Þ PT = 16 + 49 - 9 = 56
Þ PT2 = 56 \ PA × PB = 56
0
38.
(c)
5
Þ
0-k
2
60o
=r
Þ k= r 2
\ Equation of circle becomes
k2
...(i)
2
It touches y = 4 – x2 as well
Solving the two equations
a
c
120o
2
b
x2 + (y – k)2 =
\
k2
Þ 4 – y + (y – k) =
2
Here; cos q =
a 2 + b 2 - c2
2ab
and q = 60o
2
Þ 1y2 – y(2k + 1) +
k2
+ 4= 0
2
It will give equal roots \ D = 0
æ k2
ö
Þ (2k + 1) = 4 ç + 4 ÷
2
è
ø
4 + 25 - c2
2.2.5
Þ 10 = 29 – c2
Þ c2 = 19
Þ cos 60o =
Þ c = 19
2
Þ 2k2 + 4k – 15 = 0
Þ k=
1
a 2 + b2 - 19
=
2
2ab
2
2
Þ a + b – 19 = –ab
Þ a2 + b2 + ab = 19
Þ-
\ Area =
éæ ix - y - 2 öæ x - ( y - 1) i ö ù
Im êçç
֍
÷÷ ú + 1 = 0
֍
ëêè x + ( y - 1) i øè x - ( y - 1) i ø ûú
Þ
r=
a 2 + b 2 - c2
2ab
and q = 120o
k
-2 + 34
=
2
2 2
Which is not matching with any of the option given here.
(c) Let = x + y
\
36.
-2 + 34
2
also; cos q =
1
1
´ 2 ´ 5 sin 60 + ab sin120o = 4 3
2
2
5 3 ab 3
+
= 4 3
2
4
On solving, we get:
Downloaded from @Freebooksforjeeneet
M-171
Conic Sections
Þ 2 cos2 q = 8/7
Þ cos2 q = 4/7
4
Þ cos2 q =
7
ab
5 3
= 4- =
Þ
4
2 2
Þ
ab = 6
\ a + b2 = 13
Þ a = 2, b = 3
Perimeter = Sum of all sides = 2 + 5 + 2 + 3 = 12
(a) Let = x + iy
Þ 2 |x + i (y + 3) = |x + i (y – 1)|
2
39.
Þ
2 x2 + ( y + 3) = x2 + ( y -1)
2
Þ cos2 f =
Þ r=
= cos2 f – 1 =
2
40.
26
35
y+
=0
3
3
42.
2 cos 2 f - 1
(d)
=7
1
8
2
= 2 cos2 f = = cos f =
7
7
7
(–3, 2)
4
4
8
+
=
7
7
7
S
O
B
64 8
=
9 3
5
5 2
A
(2, –3)
(b) Given that x 2 + y 2 - 5 x - y + 5 = 0
Given, centre of S is O (–3, 2) and centre of given circle is
A(2, –3) and radius is 5.
25
2
++ (yy –- 1/2)2 –-1/4 =+ 0 =
4
Þ (x – 5/2)2 + (y – 1/2)2 = 3/2
2
x -– 5/2)2 –Þ (x
é
æ
1
OA = 5 2
Also AB = 5 (Q AB = radius of the given circle)
Using pythagoras theorem in DOAB
öù
on circle êQ º ç 5 / 2 + 3 / 2 cos Q, + 3 / 2 sin Q ÷ ú
2
è
øû
ë
2
æ5
ö æ5
ö
Þ PQ2 = ç + 3 / 2 cos Q ÷ + ç + 3 / 2 sin Q ÷
2
2
è
ø è
ø
Þ PQ2 =
25 3
+ + 5 3 / 2 ( cos Q + sin Q )
2 2
( cos Q + sin Q )
= 14 + 5 3 / 2
\ Maximum value of PQ2
= 14 + 5 3 / 2 ´ 2 = 14 + 5 3
41.
1
P1P2 = r cos q + r cos f =
169 35
0+
9
3
Þ r=
7
Also, sec2 f = 7 =
Þ 4x2 + 4(y + 3)2 = x2 + (y – 1)2
Þ 3x2 = y2 – 2y + 1 – 4y2 – 24y – 36
Þ 3x2 + 3y2 + 26y + 35 = 0 (which is a circle)
Þ x2 + y2 +
2
(b)
P1
2
f
2
43.
r=5 3
(a) Point of intersection of lines
æ 4 1ö
x – y = 1 and 2x + y = 3 is ç , ÷
è 3 3ø
1
4
+1
3
= 3 =4
Slope of OP =
4
1
-1
3
3
1
Slope of tangent = 4
Equation of tangent
2
O
2
1
(x – 1)
4
4y + 4 = – x + 1
x + 4y + 3 = 0
y+1=–
2
q
P2
(1,–1) P
Since cos 2q = 1/7 Þ 2 cos2 Q – 1 = 1/7
O æ 4 1ö
ç
÷
è 3, 3 ø
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EBD_8344
M-172
44.
Mathematics
(a – 1)2 + (b – 2)2 = (2 – 1)2 + (3 – 2)2
(a – 1)2 + (b –2)2 = 1 + 1 = 2
Y
(a)
A
(x – 1)2 + (y – 2)2 = ( 2)2
Compare with
(x – a)2 + (y – b)2 = r2
Therefore, given locus is a circle with centre (1, 2) and
–2,4
E
D
G
X¢
C
radius
46.
B (0, 2)
F
O
X
Y¢
45.
(a) x + y2 – 4x – 6y – 12 = 0
Centre, C1 = (2, 3)
Radius, r1 = 5 units
x2 + y2 + 6x + 18y + 26 = 0
Centre, C2 = (–3, –9)
Radius, r2 = 8 units
C1C2 =
...(ii)
(2 + 3)2 + (3 + 9)2 = 13 units
4-2
=–1
-2 - 0
Q EF ^ AB
\ Slope of EF = 1
Equation of EF, y – 3 = 1(x +1)
Þ y=x+4
...(i)
Equation of BG
y=2
...(ii)
From equations (i) and (ii)
x = – 2, y = 2
since C be the point of intersection of EF and BG, therefore
centre, C = (–2, 2)
Now coordinates of centre C satisfy the equation
2x – 3y + 10 = 0
Hence 2x – 3y + 10 = 0 is the equation of the diameter
(a) Intersection point of 2x – 3y + 4 = 0 and
x – 2y + 3 = 0 is (1, 2)
...(i)
r1 + r2 = 5 + 8 = 13
\ C1 C2 = r1 + r2
EF = perpendicular bisector of chord AB
BG = perpendicular to y-axis
Here C = centre of the circle
mid-point of chord AB, D = (– 1, 3)
slope of AB =
2.
2
C1
C2
47.
48.
Therefore there are three common tangents.
(a) Let radius of circumcircle be
According to the question,
r 10
=
Þr=4
2
5
So equation of required circle is
(x – 1)2 + (y – 1)2 = 16
Þ x2 + y2 – 2x – 2y – 14 = 0
(b) Let ‘h’ be the radius of the circle and since circle
touches y–axis at (0, 2) therefore centre = (h, 2)
A(2, 3)
(h, 2) h (0, 2)
P
(1, 2)
A
B(a, b)
Let image of A(2, 3) is B(a, b).
Since, P is the fixed point for given family of lines
So, PB = PA
m
B
(–1, 0)
Now, eqn of circle is
(h + 1)2 + 22 = h2
Þ 2h + 5 = 0
h=–
5
2
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M-173
Conic Sections
From the figure, it is clear that AB is the chord along
x –axis
\ AB = 2 (AM) = 2
52.
(b)
C
æ 3ö
25
- 4 = 2ç ÷ = 3
è 2ø
4
(0, y)
(1, 1)
T
49.
(a)
OP =
P (o, h)
4
sin q
Equation of circle
4
OB =
cos q
4
A
C º ( x - 1)2 + ( y - 1)2 = 1
q
0
Radius of T = | y |
T touches C externally therefore,
Distance between the centres = sum of their radii
B
Þ
16
32
=
sin q cos q sin 2q
least value sin2q = 1; q = 45
Þ (0 – 1)2 + (y –1)2 = (1 + |y|)2
Area = OP × OB =
Þ 1 + y2 + 1 – 2y = 1 + y2 + 2| y |
4
=4 2
sin 45°
(d) Given that y + 3x = 0 is the equation of a chord of the
circle then
y = – 3x
....(i)
2
2
(x ) + (– 3x) – 30x = 0
10x2 – 30x = 0
10x(x – 3) = 0
x = 0, y = 0
so the equation of the circle is
(x – 3) (x – 0) + (y + 9) (y – 0) = 0
x2 – 3x + y2 + 9y = 0
x2 + y2 – 3x + 9y = 0
(a) Radius
2 | y | = 1 – 2y
So, h =
50.
51.
CP =
4 +1
2
=
5
2
=
(0 - 1)2 + ( y - 1)2 = 1+ | y |
5
2
2
1
4
If y < 0 then –2y = 1 – 2y Þ 0 = 1 (not possible)
1
\ y=
4
(a) Given circle is x2 + y2 – 16 = 0
Eqn of chord say AB of given circle is
3x + y + 5 = 0.
Equation of required circle is
If y > 0 then 2y = 1 – 2y Þ y =
53.
x 2 + y 2 - 16 + l (3x + y + 5) = 0
Þ x 2 + y 2 + (3l ) x + (l ) y + 5l - 16 = 0 ...(1)
æ -3l -l ö
, ÷.
Centre C = çè
2
2 ø
If line AB is the diameter of circle (1), then
æ -3l -l ö
,
Cç
will lie on line AB.
è 2 2 ÷ø
y
æ -3l ö æ -l ö
i.e. 3 çè
÷ +ç ÷ +5 = 0
2 ø è 2 ø
C(4, 1)
(0, 0)
(0, – 1)
P
9l - l
+5=0Þl =1
2
Hence, required eqn of circle is
Þ -
x
(1, 0)
x+
y=
x 2 + y 2 + 3x + y + 5 - 16 = 0
Þ x 2 + y 2 + 3x + y - 11 = 0
0
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EBD_8344
M-174
54.
55.
Mathematics
(d) Let, x2 + y2 = 16 or x2 + y2 = 42
radius of circle r1 = 4, centre C1 (0, 0)
we have, x2 + y2 – 2y = 0
Þ x2 + (y2 – 2y + 1) – 1 = 0 or x2 + (y – 1)2 = 12
Radius 1, centre C2 (0, 1)
|C1C2 | = 1
| r2 – r1| = |4 – 1| = 3
| C1C2| < |r2 – r1|
Þ 34 – P < 25
ÞP>9
...(ii)
Also if the circle does not touch or intersect x-axis the
radius r < x-coordinate of centre C.
or 34 - P < 3 Þ 34 - P < 9 Þ P > 25
... (iii)
If the point (1, 4) is inside the circle, then its distance
from centre C < r.
(b) The equations of the circles are
Þ 5 < 34 – K
Þ P < 29
... (iv)
Now all the conditions (ii), (iii) and (iv) are satisfied if 25
< P < 29 which is required value of P.
(c) Let the foot of the perpendicular from (0, 0) on the
x y
variable line + = 1 is ( x1 > y1 )
a b
Hence, perpendicular distance of the variable line
x y
+ = 1 from the point O (0, 0) = OA
a b
O (0, 0)
2
2
x + y - 10 x - 10 y + l = 0
2
or
...(1)
2
and x + y - 4 x - 4 y + 6 = 0 ...(2)
C1 = centre of (1) = (5, 5)
57.
C2 = centre of (2) = (2, 2)
d = distance between centres
= C1C2 =
9 + 9 = 18
50 - l , r2 =
r1 =
2
[(3 - 1)2 + (5 - 4)2 ] < 34 - P
For exactly two common tangents we have
Þ
50 - l - 2 < 3 2 < 50 - l + 2
Þ
50 - l - 2 < 3 2 or 3 2 < 50 - l + 2
Þ
Þ
50 - l < 4 2 or 2 2 < 50 - l
x12 + y12
b2
1
1
=
1
+ 2
a
b
= x12 + y12
A
(x1, y1, 0)
Y
58.
C (3, 5)
r
4 = x12 + y12
X
The equation of circle is
x 2 + y 2 - 6 x - 10 y + P = 0 ...(i)
A (3, 0)
( x - 3)2 + ( y - 5)2 = ( 34 - P)2
Centre (3, 5) and radius 'r' = 34 - P
If circle does not touch or intersect the x-axis then radius
x < y - coordiante of centre C
or
x x
+ = 1
a b
1 1ù
é 1
êQ 2 + 2 = 4 ú ,
b
ë a
û
which is equation of a circle with radius 2.
Hence (x1, y1) i.e., the foot of the perpendicular from the
x x
point (0, 0) to the variable line + = 1 is lies on a
a b
circle with radius = 2
(c) Since circle touches x-axis at (3, 0)
\ The equation of circle be
(x – 3)2 + (y – 0)2 + ly = 0
Þ
Required interval is (18, 42)
r
+
1
2
Þ l > 18 or l < 42
(d)
1
a2
Þ 50 - l < 32 or 8 < 50 - l
56.
-1
Þ
r1 - r2 < C1C2 < r1 + r2
A
(1, –2)
34 - P < 5
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M-175
Conic Sections
59.
60.
As it passes through (1, –2)
\ Put x = 1, y = –2
Þ (1 – 3)2 + (–2)2 + l(–2) = 0
Þ l=4
\ equation of circle is
(x – 3)2 + y2 – 8 = 0
Now, from the options (5, –2) satisfies equation of circle.
(*) Given information is incomplete in the question.
(a) Circle : x2 + y2 – 6x + 2y = 0
...(i)
Line : 2x + y = 5
...(ii)
Centre = (3, – 1)
Now, 2 × 3 – 1 = 5, hence centre lies on the given line.
Therefore line passes through the centre. The given line is
normal to the circle.
Thus statement-2 is true, but statement-1 is not true as
there are infinite circle according to the given conditions.
x 2 + y2 + 4x - 6y -12 = 0
\ A = (– 2, 3) and B = (g, f)
Now, from the figure, we have
-2 + g
3+ f
= 1 and
= -1 (By mid point formula)
2
2
Þ g = 4 and f = – 5
63.
(a) Let C = (x, y)
Now, CA2 = CB2 = AB2
Þ (x + a)2 + y2 = (x – a)2 + y2 = (2a)2
Þ x2 + 2ax + a2 + y2 = 4a2
...(i)
and x2 – 2ax + a2 + y2 = 4a2
...(ii)
From (i) and (ii), x = 0 and y = ± 3a
Since point C(x, y) lies above the x-axis and a > 0, hence y
10
10
10
=
2x + y = 5
3a
\ C = (0, 3a)
Let the equation of circumcircle be
61.
(b)
x2 + y2 + 2gx + 2fy + C = 0
Y
Since points A(– a, 0), B(a, 0) and C(0,
circle, therefore
a2 – 2ga + C = 0
...(iii)
2
(3, 4)
4
3a ) lie on the
a + 2ga + C = 0
...(iv)
and 3a2 + 2 3af + C = 0
...(v)
X
From (iii), (iv), and (v)
3
O
g = 0, c = – a2, f = -
3
Hence equation of the circumcircle is
x2 + y2 – 6x – 8y + (25 – a2) = 0
Radius = 4 = 9 + 16 + (25 - a 2 )
62.
x2 + y2 -
Þ a=±4
(a) Let A be the centre of given circle and B be the centre
of circle C.
B
3
y - a2 = 0
2 3ay
- a2 = 0
3
Þ 3x2 + 3y2 – 2 3ay = 3a2
64.
O
(1, –1)
2a
Þ x2 + y2 –
D(4, 0)
A
a
(d) Point of intersection of two given lines is (1, 1). Since
each of the two given lines contains a diameter of the
given circle, therefore the point of intersection of the two
given lines is the centre of the given circle.
Hence centre = (1, 1)
C
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EBD_8344
M-176
Mathematics
\ a2 – 7a + 11 = 1 Þ a = 2, 5
...(i)
and a – 6a + 6 = 1 Þ a = 1, 5
...(ii)
2
Radius =
3
units.
2
From both (i) and (ii), a = 5
Y
2
Now on replacing each of (a – 7a + 11) and
(a2 – 6a + 6) by 1, the equation of the given circle is
x2 + y2 – 2x – 2y + b3 + 1 = 0
A(0, 1)
Þ (x – 1)2 + (y – 1)2 + b3 = 1
Þ b3 = 1 – [(x – 1)2 + (y – 1)2]
65.
\ b Î (– ¥, 1)
(a) Since, circle touches , the x-axis at (1, 0). So, let centre
of the circle be (1, h)
B
æ 3 ö
çè – , 0÷ø
2
X¢
O
X
Y
Y¢
(1, h) (2, 3)
C
B
Line : y = mx + 1
y-intercept of the line = 1
\ A = (0, 1)
X
A (1, 0)
Slope of line, m = tan q =
Given that circle passes through the point B (2,3)
\ CA = CB
(radius)
Þ CA2 = CB2
Þ (1 – 1)2 + (h – 0)2 = (1 – 2)2 + ( h – 3)2
Þ h2 = 1 + h2 + 9 – 6h
66.
Þ m=
68.
h=
\
Length of the diameter =
10
3
x2 + y2 – 8x – 2y + 1 = 0
and x2 + y2 + 6x + 8y = 0
Their centres and radius are
25 = 5
Now, C1C2 = 49 + 25 = 74
r1 – r2 = – 1, r1 + r2 = 9
Since, r1 – r2 < C1C2 < r1 + r2
\ Number of common tangents = 2
(b) Circle : x2 + y2 + 3x = 0
æ 3 ö
Centre, B = ç – , 0÷
è 2 ø
10
x +1=0
...(1)
3
\ A lies on the circle given by (1). As B and C also follow
the same condition.
\ Centre of circumcircle of DABC = centre of circle given
æ5 ö
by (1) = ç ,0÷ .
è3 ø
(d) Point (1, 2) lies on the circle x2 + y2 + 2x + 2y – 11 = 0,
because coordinates of point (1, 2) satisfy the equation
x2 + y2 + 2x + 2y – 11 = 0
Now, x2 + y2 – 4x – 6y – 21 = 0 ...(i)
x2 + y2 + 2x + 2y – 11 = 0
...(ii)
3x + 4y + 5 = 0
...(iii)
Þ x2 + y2 –
C1 (4, 1), r1 = 16 = 4
67.
Þ 3m – 2 = 0
(a) Let P(1, 0) and Q(–1, 0), A(x, y)
Given:
(c) Given circles are
C2 (– 3, – 4), r2 =
1 2
=
3 3
2
AP BP CP 1
=
=
=
AQ BQ CQ 2
Þ 2AP = AQ
Þ 4(AP)2 = AQ2
Þ 4[(x – 1)2 + y2] = (x + 1)2 + y2
Þ 4(x2 + 1 – 2x) + 4y2 = x2 + 1 + 2x + y2
Þ 3x2 + 3y2 – 8x – 2x + 4 – 1 = 0
Þ 3x2 + 3y2 – 10x + 3 = 0
10 5
=
6 3
Þ
OA
OB
69.
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M-177
Conic Sections
73.
From (i) and (iii),
2
æ 3x + 5 ö
æ 3x + 5 ö
x2 + ç –
÷ – 4 x – 6 èç –
÷ – 21 = 0
è
4 ø
4 ø
Þ 16x2 + 9x2 + 30x + 25 – 64x + 72x + 120 – 336 = 0
Þ 25x2 + 38x –191 = 0
...(iv)
From (ii) and (iii),
Þ L º x + 5 y + p2 + 2 p - 5 = 0
Þ Equation of circle passing through P and Q is
S1 + l L = 0
Þ (x2 + y2 + 3x + 7y + 2p – 5)
+ l (x + 5y + p2 +2p – 5) = 0
Given that it passes through (1, 1), therefore
(7 + 2p ) + l (2p + p2 + 1) = 0
2
æ 3x + 5 ö
æ 3x + 5 ö
x2 + ç –
÷ + 2 x + 2 èç –
÷ –11 = 0
è
4 ø
4 ø
Þ 16x2 + 9x2 + 30x + 25 + 32x – 24x – 40 – 176 = 0
Þ 25x2 + 38x –191 = 0
...(v)
Thus we get the same equation from (ii) and (iii) as we get
from equation (i) and (iii). Hence the point of intersections
of (ii) and (iii) will be same as the point of intersections of
(i) and (iii). Therefore the circle (ii) passing through the
point of intersection of circle(i) and point (1, 2) also as
shown in the figure.
Þ l =–
74.
x2 + y2 – 4x – 6y – 21 = 0
71.
72.
x2 + y 2 - 4 x - 8 y - 5 = 0
4 + 16 + 5 = 5
Given circle is intersecting the line 3 x - 4 y = m, at two
distinct points.
Þ length of perpendicular from centre to the line < radius
Centre = (2, 4), Radius =
6 - 16 - m
< 5 Þ 10 + m < 25
5
Þ –25 < m + 10 < 25 Þ – 35 < m < 15
Þ
AP BP CP 1
=
=
=
AQ BQ CQ 3
Þ 3AP = AQ
Let A = (x, y) then
3AP = AQ Þ 9 AP2= AQ2
Þ 9 (x – 1)2 + 9y2 = (x + 1)2 + y2
Þ 9 x2 – 18x + 9 + 9y2 = x2 +2x +1 + y2
Þ 8x2 – 20x + 8y2 + 8 = 0
x2 + y2 + 2x+ 2y – 11 = 0
70.
2p + 7
( p + 1) 2
which does not exist for p = – 1
(a) Given that P (1, 0), Q (– 1, 0)
and
(1, 2)
Hence equation(ii) i.e.
x2 + y2 + 2x + 2y – 11 = 0 is the equation of required circle.
(b) Given circle whose diametric end points are (1,0) and
(0,1) will be of smallest radius. Equation of this smallest
circle is
(x – 1) (x – 0) + (y – 0) (y – 1) = 0
Þ x2 + y2 – x – y = 0
(a) If the two circles touch each other and centre (0, 0) of
x2 + y2 = c2 is lies on circle x2 + y2 = ax then they must
touch each other internally.
a
a
So,
Þ a =c
= c2
2
(a) Given equation of circle is
(a) The given circles are
S1 º x2 + y2 + 3x + 7y + 2p – 5 = 0....(1)
S 2 º x2 + y2 + 2x + 2y – p2 = 0 ....(2)
\ Equation of common chord PQ is
S1 – S2 = 0 [From (i) and (ii)]
Þ x2 + y2 –
5
x +1 = 0
3
....(1)
\ A lies on the circle given by eq (1). As B and C also
follow the same condition, they must lie on the same circle.
\ Centre of circumcircle of D ABC
æ5 ö
= Centre of circle given by (1) = ç , 0÷
è4 ø
75.
(c) The given circle is x2 + y2 + 2x + 4y –3 = 0
P(1,0)
Q(a,b)
C(–1, –2)
Centre (–g, –f ) = (–1, –2)
Let Q ( h, k) be the point diametrically opposite to the
point P(1, 0),
0+k
1+ h
= –2
= –1 and
then
2
2
Þ h = –3, k = – 4
So, Q is (–3, –4)
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EBD_8344
M-178
76.
Mathematics
\ Using 2 g1 g2 + 2 f1 f 2 = c1 + c2 , we get
(d) Equation of circle whose centre is (h, k) and touch
the x-axis
i.e (x – h)2 + (y – k)2 = k2
2(-a ) ´ 0 + 2(- b) ´ 0 = c1 - p2
Þ c1 = p 2
Let equation of circle is x 2 + y 2 - 2ax - 2by + p 2 = 0
Q It passes through (a, b)
(h, k)
Þ a 2 + b 2 - 2aa - 2bb + p 2 = 0
\ Locus of (a, b) is
(-1,1)
X'
X
\ 2ax + 2by - (a 2 + b 2 + p 2 ) = 0 .
80.
(radius of circle = k because circle is tangent to x-axis)
Q Equation of circle passing through (–1, 1)
2
2
2
\ (–1 –h) + (1 – k) = k
2
2
Þ 1 + h + 2h + 1 + k – 2k = k2
Þ h2 + 2h – 2k + 2 = 0
D³0
2
\ (2) – 4 × 1.(–2k + 2) ³ 0
77.
(d)
q
3q
Given that area of one sector
= 3 × area of another sector
Þ Angle at centre by one sector = 3 ´ angle at centre
by another sector
Let one angle be q then other = 3q
1
Þ 4 – 4(–2k + 2) ³ 0 Þ 1 + 2k – 2 ³ 0 Þ k ³
2
(d) Given that centre of circle be (0, 0) and radius is 3 unit
Let M(h, k) be the mid point of chord AB where
2p
ÐAOB =
3
Clearly q + 3q = 180 Þ q = 45o
(Linear pair)
\ Angle between the diameters represented by pair of
equation
ax 2 + 2 ( a + b ) xy + by 2 = 0 is 45o
O (0, 0)
3
p/3
A
M(h, k) B
\ Using tan q =
p 3
p
\ ÐAOM = . Also OM = 3cos =
3 2
3
Þ
h2 + k 2 =
we get, tan 45o =
3
9
Þ h2 + k 2 =
2
4
Þ1=
9
4
(d) On solving we get point of intersection of
3x - 4 y - 7 = 0 and 2 x - 3 y - 5 = 0 is (1, - 1) which is
the centre of the circle
Area of circle = pr2 = 49p
\ radius = 7
79.
2
( a + b) 2 - ab
a+b
2 a 2 + b2 + ab
a+b
(
Þ ( a + b) = 4 a 2 + b2 + ab
2
2
\ Locus of (h, k) is x + y =
78.
2 h2 - ab
a+b
2
)
Þ a 2 + b 2 + 2 ab = 4a 2 + 4b 2 + 4ab
81.
Þ 3a 2 + 3b 2 + 2 ab = 0
(b) Given that
s1 = x 2 + y 2 + 2ax + cy + a = 0
\ Equation is ( x - 1)2 + ( y + 1) 2 = 49
s2 = x 2 + y 2 - 3ax + dy - 1 = 0
Þ x 2 + y 2 - 2 x + 2 y - 47 = 0
Equation of common chord PQ of circles s1 and s2 is
(d) Let the centre variable circle be (a, b)
given by s1 - s2 = 0
2
2
2
Q It cuts the circle x + y = p orthogonally
Þ 5ax + (c - d ) y + a + 1 = 0
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M-179
Conic Sections
82.
Given that 5x + by – a = 0 passes through P and Q
\ The two equations should represent the same line
a
b
c
Þ 1 = 1 = 1
a2 b2 c2
a c - d a +1
Þ a + 1 = -a2
Þ =
=
1
b
-a
[Q D = –3]
a2 + a + 1 = 0
\ No real value of a.
(b) Let the equation of circle is
Circumference of circle = 2pr = 10p
\ r = 5.
Required circle is, ( x - 1)2 + ( y + 1)2 = 52
Þ x 2 + y 2 - 2 x + 2 y - 23 = 0
85.
x 2 + y 2 - 2 x = 0, we get
x = 0, y = 0 and x = 1, y = 1
x 2 + y 2 + 2 gx + 2 fy + c = 0 ......(1)
It passes through (a, b)
\ Extremities of diameter of the required circle are A (0, 0)
and B (1, 1). Hence, the equation of circle is
\ a 2 + b2 + 2 ga + 2 fb + c = 0
( x - 0)( x - 1) + ( y - 0)( y - 1) = 0
......(2)
Circle (1) cuts x 2 + y 2 = 4 orthogonally
Two circles intersect orthogonally if 2g1g2 + 2f1f2 = c1 + c2
86.
Þ r -3 < 5 Þ 0 < r < 8
\ from (2) a 2 + b2 + 2 ga + 2 fb + 4 = 0
\ Locus of centre (–g, –f ) is
87.
2
or 2ax + 2by = a + b + 4
(d) Let the variable circle be
....(2)
88.
....(3)
Let the other end of diameter through (p, q) be (h, k),
then
æ h + pö
æ k + qö æ h + pö
p + q + 2pç÷ + 2q ç ÷ +ç
÷ =0
è
è
2 ø
2 ø è 2 ø
±2 m
=
;
1 - m2
1 - m2
Þ 1 – m2 = ± 2m Þ m2 ± 2m – 1 = 0
2
2
Þ h + p - 2hp - 4kq = 0
\ locus of (h, k) is
2
2 m2 - 0
-2 ± 4 + 4
-2 ± 2 2
=
= -1 ± 2 .
2
2
(a) Q The centre C of circle of radius 3 lies on circle of
radius 5. Let P(x, y) in the smaller circle.
Þm=
2
2
x 2 + y 2 - 2 x + 2 y = 47
(c) Given equation of circle x2 + y2 = 1 = (1)2
Þ x2 + y2 = (y – mx)2
Þ x2 = m2x2 – 2 mxy;
Þ x2 (1 – m2) + 2mxy = 0. Which represents the pair of
lines between which the angle is 45o.
\ tan 45 = ±
h+ p
k+q
= - g and
=-f
2
2
Putting value of g and f in (3), we get
2
(d) Area of circle = pr 2 = 154 Þ r = 7
For centre, solving equation
Equation of circle, ( x - 1)2 + ( y + 1)2 = 72
\ g 2 - c = 0 Þ c = g 2 . From (2)
p 2 + q 2 + 2 gp + 2 fq + g 2 = 0
89.
Y
2
x + p - 2 xp - 4 yq = 0
A C
Þ ( x - p )2 = 4qy
84.
B
P
O
(d) Two diameters are along
2 x + 3 y + 1 = 0 and 3x - y - 4 = 0
On solving we get centre (1, –1)
.... (2)
2 x - 3 y = 5& 3x - 4 y = 7 we get, x = 1, y = -1
\ centre = (1, –1 )
x 2 + y 2 + 2 gx + 2 fy + c = 0 ....(1)
Since it passes through (p, q)
\ p 2 + q 2 + 2 gp + 2 fq + c = 0
Circle (1) touches x-axis,
.... (1)
and r1 + r2 > C1C 2 , r + 3 > 5 Þ r > 2
From (1) and (2), 2 < r < 8.
a 2 + b2 - 2ax - 2by + 4 = 0
2
Þ x2 + y 2 - x - y = 0
(b) Q Given two circles intersect at two points
\ r1 - r2 < C1C2
\ 2( g ´ 0 + f ´ 0) = c - 4 Þ c = 4
83.
(d) Solving y = x and the circle
we should have
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X
EBD_8344
M-180
90.
Mathematics
OA £ OP £ OB
[From (ii)]
Þ (5 - 3) £ x 2 + y 2 £ 5 + 3
Þ 2m12 + ym1 + ( x + 2) = 0
Þ 4 £ x 2 + y 2 £ 64
(b) Let the required circle be
x2 + y2 + 2gx + 2fy + c = 0
Since it passes through (0, 0) and (1, 0)
On putting these values, we get
From (i) and (iii),
x +1 - y
1
=
=
Þ x+3=0
y
x+2
2
93.
1
2
Points (0, 0) and (1, 0) lie inside the circle x2 + y2 = 9, so
two circles touch internally
Þ c1c2 = r1 – r2
3
\ g 2 + f 2 = 3 - g 2 + f 2 Þ g2 + f 2 =
2
Squaring both side, we get
Þ c = 0 and g = -
5
2
= - ( x - 1) Þ 4 x + 6 y = 19
...(i)
2
3
Equation of normal to the parabola at (2, 4) is,
1
y - 4 = - ( x - 2) Þ x + 4 y = 18
4
\ From (i) and (ii), x = -
ö
ö æ1
æ1
ç , 2 ÷ or ç ,- 2 ÷
ø
ø è2
è2
(c) Let ABC be an equilateral triangle, whose median is
AD.
In equilateral triangle median is also altitude
So, AD ^ BC
A
Given AD = 3a.
Let AB = BC = AC = x.
O
In DABD, AB2 = AD2 + BD2 ;
Þ x2 = 9a2 + (x2/4)
B
D
94.
1
...(i)
m
Q Equation of tangent to x2 = 4y with slope m be :
1
= - m 2 Þ m = -1
m
\ Equation tangent : x + y + 1 = 0
It is tangent to circle x2 + y2 = c2
C
Þc=
95.
2
m2
[Tangent to y2 = 8(x + 2)]
m12 ( x + 1) - ym1 + 1 = 0
...(i)
m22 ( x + 2) - ym2 + 2 = 0
...(ii)
1
m1
(Q L1 ^ L2 )
(c)
1
2
P
R
M
Q
1
[Tangent to y2 = 4(x + 1)]
m1
Q m2 = -
...(ii)
From eq. (i) and (ii),
L1 : y = m1 ( x + 1) +
L2 : y = m2 ( x + 2) +
(b) Equation tangent to parabola y2 = 4x with slope m be:
y = mx +
Þ r2 = (3a – r)2 +
(a)
16
53
, y=
5
10
y = mx - am 2
x2
4
Þ r2 = 9a2 – 6ar + r2 + 3a2
Þ 6ar = 12a2
Þ r = 2a
So equation of circle is x2 + y2 = 4a2
...(ii)
æ 16 53ö
\ Centre of the circle is ç - , ÷
è 5 10 ø
3 2
x = 9a2 Þ x2 = 12a2.
4
In D OBD, OB2 = OD2 + BD2
92.
(d) Circle passes through A(0, 1) and B(2, 4). So its centre
is the point of intersection of perpendicular bisector of AB
and normal to the parabola at (2, 4).
Perpendicular bisector of AB;
y-
9 1
Þ f2 = - =2
\f =± 2.
4 4
Hence, the centres of required circle are
91.
...(iii)
N
Q y 2 = 12 x
\a = 3
Let P(at 2 , 2at )
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M-181
Conic Sections
Þ N (at 2 , 0) Þ M (at 2 , at )
Q Equation of QM is y = at
So, y 2 = 4ax Þ x =
tan 30° =
at 2
4
Area =
æ at 2
ö
Þ Qç
, at ÷
è 4
ø
98.
Þ Equation of QN is y =
-4
( x - at 2 )
3t
96.
3 2 1
at = and PN = 2at = 2
4
4
2
(b)
y = 4x
L
1
3
=
4t
2t 2
Þt = 2 3
1
× 8(2 3) × 2 × 24 = 192 3.
2
-1
2
Parameter of other end of
focal chord is 2
So, coordinates of
B is (8, 8)
Þ Equation of tangent at B
is 8y – 4(x + 8) = 0
Þ 2y – x = 8
Þ x – 2y + 8 = 0
4
4
1
= - ( - at 2 ) Þ at = 1 Þ t =
3
3t
3
Now, MQ =
Þ
æ1
ö
(b) Let parabola y2 = 8x at point ç , -2 ÷ is (2t 2, 4t)
è2
ø
Þ t=
æ 4ö
Q QN passes through ç 0, ÷ , then
è 3ø
4t
2t 2
99.
(a) Let point P be (2t, t2) and Q be (h, k)
Using section formula,
2 5
h=
C1 S(1,0)
-2 + t 2
2t
,k =
3
3
C2
L'
æ 3h ö
Hence, locus is 3k + 2 = ç ÷
è 2 ø
2
Þ 9x2 = 12y + 8
100. (0.5) Let the coordinates of P = P(t2, t)
Distance between the centres
= C1C2 = 2C1S = 2 20 - 4 = 8.
97.
2
2
(c) Let A = (2t , 4t ) and B = (2t , - 4t )
2
A (2 t , 4t)
O
(0, 0)
30
30
M
2
B (2 t , –4t)
For equilateral triangle (ÐAOM = 30°)
Tangent at P(t2, t) is ty =
x + t2
2
Þ 2ty = x + t2
Q(–t2, 0), O(0, 0)
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EBD_8344
M-182
Mathematics
0
1 2
t
\ Area of DOPQ = 2
-t 2
0 1
0 1
Þ |t|3 = 8
t = ± 2 (t > 0)
\ 4y = x + 4 is a tangent
\ P is (4, 2)
101. (c) y = mx + 4
Tangent of y2 = 4x is
Þ y = mx +
1
m
B
ᔴ
(h k)
Þ h=
x–y㦨3
5
,k=–1
2
103. (d) Equation of tangent on y 2 = 4 2 x is yt = x + 2t 2
1
\m=
2
Now, y = mx
A
t 1 =4
This is also tangent on circle
...(i)
...(ii)
2t 2
1+ t2
= 1 Þ 2t4 = 1 + t2 Þ t2 = 1
Hence, equation is ± y = x + 2 Þ | c | =
2
104. (d) The circle and parabola will have common tangent at
P (1, 2).
P(
1
2)
a
[Q Equation of tangent of y2 = 4 ax is y = mx + ]
m
\
From (i) and (ii)
4=
1
1
Þ m=
m
4
So, line y =
1
x + 4 is also tangent to parabola
4
x2 = 2by, so solve both equations.
æ x + 16 ö
x 2 = 2b ç
÷
è 4 ø
Þ 2x2 – bx – 16b = 0
Þ D=0
[For tangent]
Þ b2 – 4 × 2 × (–16b) = 0
Þ b2 + 32 × 4b = 0
b = – 128, b = 0 (not possible)
102. (c) Tangent to the curve y = (x – 2)2 – 1 at any point (h, k)
is,
Þ
1
( y + k ) = ( x - 2)(h - 2) - 1
2
y+k
= xh - 2 x - 2h + 3
Þ
2
Þ (2h – 4) x – y – 4h + 6 – k = 0
Given line, x – y – 3 = 0
Þ
2h - 4 4h - 6 + k
=
=1
1
3
So, equation of tangent to parabola is,
4 ( x + 1)
Þ 2 y = 2x + 2 Þ y = x + 1
2
Let equation of circle (by family of circles) is
(x – x1)2 + (y – y1)2 + lT = 0
Þ c º (x – 1)2 + (y – 2)2 + l(x – y + 1) = 0
Q circles touches x-axis.
\ y-coordinate of centre = radius
Þ c = x2 + y2 + (l – 2)x + (– l – 4)y + (l + 5) = 0
y ´ ( 2) =
2
2
l+4
æ l - 2 ö æ -l - 4 ö
= ç
÷ +ç
÷ - ( l + 5)
2
è 2 ø è 2 ø
Þ
l 2 - 4l + 4
= l + 5 Þ l2 – 4l + 4 = 4l + 20
4
Þ l2 – 8l – 16 = 0 Þ l = 4 ± 4 2
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M-183
Conic Sections
Þ l = 4 - 4 2 (Q l = 4 + 4 2 forms bigger circle)
(
)
Hence, centre of circle 2 2 - 2, 4 - 2 2 and radius
=4-2 2
(
\ area = p 4 - 2 2
)
2
(
= 8p 3 - 2 2
)
-4a 8
+ .
b
b
Since, tangents are perpendicular to each other.
105. (a) Q
Þa=4
One end of focal chord of the parabola is at (1, 4)
y – coordinate of focal chord is 2at
\ 2 at = 4
y2 = 16x
4ax + by = 8 Þ y =
-4a -1
=
Þ b = 8a
b
2
from (1) & (2) we get:
Þ
1
2
Hence, the required length of focal chord
Þt =
2
2
106. (c) The shortest distance between line y = x and parabola
= the distance LM between line y = x and tangent of
parabola having slope 1.
Y
y㦨x
2
2
Þ a2 =
34
17
108. (c) To find intersection point of x2 + y2 = 5 and y2 = 4x,
substitute y2 = 4x in x2 + y2 = 5, we get
x2 + 4x – 5 = 0 Þ x2 + 5x – x – 5 = 0
Þ x (x + 5) – 1 (x + 5) = 0
\ x = 1, – 5
Intersection point in 1st quadrant be (1, 2).
Now, equation of tangent to y2 = 4x at (1, 2) is
y × 2 = 2 (x + 1) Þ y = x + 1
Þx–y+1=0
...(i)
æ3 7ö
Hence, ç , ÷ lies on (i)
è4 4ø
L
y 2= x – 2
X
(2 0)
...(ii)
Þa=±
1ö
æ 1ö
æ
= a ç t + ÷ = 4 ´ ç 2 + ÷ = 25
2ø
è tø
è
M
107. (d) Since (a, b) touches the given ellipse 4x 2 + y 2 = 8
\ 4a2 + b2 = 8
...(i)
Equation of tangent on the ellipse at the point A (1, 2)
is:
4x + 2y = 8 Þ 2x + y = 4 Þ y = –2x + 4
But, also equation of tangent at P (a, b) is:
109. (c) x2 = 8y
P(2at, at2)
q
Let equation of tangent of parabola having slope 1 is,
y = m (x – 2) +
a
m
Here m = 1 and a =
1
4
\ equation of tangent is: y = x –
7
4
Distance between the line y – x = 0 and y – x +
7
-0
4
=
2
2
1 +1
=
7
4 2
7
=0
4
Then, equation of tangent at P
tx = y + at2
Þ y = tx – at2
Then, slope t = tan q
Now, y = tan qx – 2 tan2 q
Þ cot qy = x – 2 tan q
x = y cot q + 2tan q
110. (c) Equation of a tangent to parabola y2 = 4x is:
y = mx +
1
m
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EBD_8344
M-184
Mathematics
This line is a tangent to xy = 2
1ö
æ
1
2
\ x çè mx + ÷ø = 2 Þ mx + x - 2 = 0
m
m
Q Tangent is common for parabola and hyperbola.
2
x- 2
x2
+4 2 =0
4
2 x 2 - 4 x - 16 2 = 0
æ 1ö
\ D = ç ÷ - 4 × m × (-2) = 0
è mø
x1 + x2 = 2 2, x1x2 = –16, (x1 – x2)2 = 8 + 64 = 72
Since, points P and Q both satisfy the equations (ii), then
1
+ 8m = 0
m2
1 + 8 m3 = 0
x1 - 2 y1 + 4 2 = 0
m3 = -
x1 - 2 y2 + 4 2 = 0
(x2 – x1) =
1
1
Þm= 8
2
1
\ Equation of common tangent: y = - x - 2
2
Þ 2y = –x – 4 Þ x + 2y + 4 = 0
111. (d) y2 = –4(x – a2)
Þ PQ = ( x2 - x1 )2 + ( y2 - y1 )2
= ( x2 - x1 )2 +
= | x2 - x1 | ×
Q
(0, 2a)
P
(a2, 0)
R
2( y2 - y1 ) Þ (x2 – x1)2 = 2(y2 – y1)2
( x2 - y1) 2
2
3
3
=6 2´
=6 3
2
2
Hence, length of chord = 6 3.
114. (b) Since, vertex and focus of given parabola is (2, 0) and
(4, 0) respectively
y
(0, –2a)
Area =
1
(4a)(a2) = 2a3
2
x¢
Since 2a3 = 250 Þ a = 5
112. (a, b, c, d)
Normal to y2 = 8ax is
y = mx – 4am – 2am3
...(i)
and normal to y2 = 4b (x – c) with slope m is
y = m(x – c) – 2bm – bm3 ...(ii)
Since, both parabolas have a common normal.
\ 4am + 2am3 = cm + 2bm + bm3
Þ 4a + 2am2 = c + 2b + bm2 or m = 0
Þ (4a – c – 2b) = (b – 2a) m2
or (X-axis is common normal always)
Since, x-axis is a common normal. Hence all the options
are correct for m = 0.
113. (d) Let intersection points be P(x1, y1) and Q(x2, y2)
The given equations
x2 = 4y
...(i)
x - 2y + 4 2 = 0
Use eqn (i) in eqn (ii)
...(ii)
O
(4, 0)
(2, 0)
x
y¢
Then, equation of parabola is
(y – 0)2 = 4 ´ 2(x – 2)
Þ y2 = 8x – 16
Hence, the point (8, 6) does not lie on given parabola.
115. (b) Since, the equation of tangent to parabola y2 = 4x is
y = mx +
1
m
...(i)
The line (i) is also the tangent to circle
x2 + y2 – 6x = 0
Then centre of circle = (3, 0)
radius of circle = 3
The perpendicular distance from centre to tangent is equal
to the radius of circle
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M-185
Conic Sections
\
1
m =3Þ
1 + m2
3m +
,
16
P(
1
3
nt
ge
n
Ta
A
(–16, 0)
X'
1
x± 3
Then, from equation (i): y = ±
3
Hence,
)
16
al
rm
No
Þ m= ±
Y
2
1ö
æ
2
ç 3m + ÷ = 9 (1 + m )
è
mø
X
B(24, 0)
3 y = x + 3 is one of the required common
tangent.
Y'
116. (a)
4
3
Slope of PB (m2) = –2
Slope of PC (m1) =
4
+2
m1 - m 2
Hence, tan q =
= 3
1 + m1 × m 2 1 - 4 × 2
3
Let the coordinates of C is (t2, 2t).
Since, area of DACB
t2
1
9
=2
4
2t 1
6 1
-4 1
1
= |t2(6 + 4) – 2t(9 – 4) + 1(–36 – 24)|
2
Þ tan q = 2
118. (a) Equation of the chord of contact PQ is given by:
T=0
or T º yy1 – 4(x + x1), where (x1, y1) º (– 8, 0)
\ Equation becomes: x = 8
& Chord of contact is x = 8
\ Coordinates of point P and Q are (8, 8) and (8, – 8)
and focus of the parabola is F (2, 0)
1
\ Area of triangle PQF = × (8 – 2) × (8 + 8) = 48 sq. units
2
119. (c) c = –29m – 9m3
a= 2
Given (at2 – a)2 + 4a2t2 = 64
Þ (a(t2 + 1)) = 8
Þ t2 + 1 = 4
Þ t2 = 3
Þ t=
1
= |10t2 – 10t – 60|
2
= 5|t2 – t – 6|
\
3
c = 2at (2 + t2) = 2 3 (5)
c = 10 3
2
25
æ 1ö
= 5 çè t - 2 ÷ø - 4
[Here, t Î (0, 3)]
For maximum area, t =
1
2
125
1
= 31 sq. units
Hence, maximum area =
4
4
117. (a) Equation of tnagent at P(16, 16) is given as:
x – 2y + 16 = 0
120. (c)
y
x’
x
P
y’
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EBD_8344
M-186
Mathematics
Tangent to x2 + y2 = 4 is
123. (b) Let P(h, k) divides
OQ in the ratio 1 : 3
Let any point Q on x2 = 8y is (4t, 2t2).
y = mx + 2 1 + m2
Also, x2 = 4y
x2 = 4mx + 8 1 + m 2 or x2 = 4mx – 8 1 + m 2
For D = 0
1
2
we have; 16 m2 + 4.8 1 + m = 0
P3
Q (4t, 2t2)
O
Þ m2 + 2 1 + m2 = 0
2
Þ m2 = -2 1 + m
4
2
Þ m = 4 + 4m
Þ m4 – 4m2 – 4 = 0
Then by section formula
Þ m2 =
4 ± 16 + 16
2
Þ m2 =
4± 4 2
2
Þ
2
P (2t , 4t)
C
Centre of new circle = P(2t2, 4t) = P(2, – 4)
Radius = PC = (2 – 0)2 + (–4 + 6)2 = 2 2
\ Equation of circle is :
( )
2
(x –2)2 + (y + 4) = 2 2
Þ x2 + y2 – 4x + 8y + 12 = 0
t12 = t 2 +
t2 +
4
t2
4
t2
2
t
124. (d) Let P ( -at12 , 2at1 ), Q( -at12 , -2at1 ) and R (h, k)
By using section formula, we have
h = -at12 , k =
-2at1
3
2at1
3
Þ 3k = –2at1
Þ 9k2 = 4a2 t12 = 4a (–h)
Þ 9k2 = –4ah
Þ 9k2 = –4h Þ 9y2 = –4x
125. (c) Given parabolas are
y2 = 4x
x2 = –32y
Let m be slope of common tangent
Equation of tangent of parabola (1)
1
y = mx +
m
Equation of tangent of parabola (2)
y = mx + 8m2
(i) and (ii) are identical
k= -
1
1
1
= 8m2 Þ m3 = Þ m =
2
m
8
ALTERNATIVE METHOD:
Þ
1
m
Since this is also tangent to x2 = – 32y
Let tangent to y2 = 4x be y = mx +
+4
³ 2 t 2.
t2
and h = t
2
Þ 2k = h 2
Required locus of P is x2 = 2y
Þ m2 = 2 + 2 2
121. (c) Minimum distance Þ perpendicular distance
Eqn of normal at p(2t2, 4t)
y = –tx + 4t + 2t3
It passes through C(0, –6)
t3 + 2t + 3 = 0 Þ t = – 1
122. (a) t1 = – t –
k=
\
4
t2
=4
Minimum value of t12 = 8
1ö
æ
x 2 = -32 ç mx + ÷
è
mø
Þ x2 + 32mx +
32
=0
m
Now, D = 0
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...(1)
...(2)
...(i)
...(ii)
M-187
Conic Sections
æ 32 ö
(32) 2 - 4 ç ÷ = 0
è mø
4
1
3
Þm =
Þ m=
32
2
126. (a) Equation of parabola, y2 = 6x
3
Þ y2 = 4 ´ x
2
128. (c)
A
x2 + y 2 = 32
B
æ3 ö
\ Focus = çè , 0÷ø
2
Let equation of chord passing through focus be
ax + by + c = 0 ...(1)
æ3 ö
Since chord is passing through ç , 0÷
è2 ø
3
\ Put x = , y = 0 in eqn (1), we get
2
3
a+c = 0
2
3
Þ c = - a ...(2)
2
distance of chord from origin is
a (0) + b(0) + c
=
2
2
=
5 5
2 2
c
a +b
a + b2
Squaring both sides
c2
5
= 2
4
a + b2
2
4 2
c
5
Putting value of c from (2), we get
4 9 2
a2 + b 2 = ´ a
5 4
9
4 2
2
2
b2 = a - a = a
5
5
a2
5 a
5
2 = 4,b = ± 2
b
Þ a2 + b2 =
æ ± 5ö
dy
5
a
= - = -ç
÷=m 2
dx
b
è 2 ø
127. (d) The locus of the point of intersection of tangents
to the parabola y2 = 4 ax inclined at an angle a to each
other is
tan2a (x + a)2 = y2 – 4ax
Given equation of Parabola y2 = 4x {a = 1}
Point of intersection (–2, –1)
tan2a (–2 + 1)2 = (–1)2 – 4 × 1 × (–2)
Þ tan2a = 9
Þ tan a = ± 3
Þ |tan a| = 3
Slope of chord,
y2 = 8x
We have
x2 + (8x) = 9
x2 + 9x – x – 9 = 0
x (x + 9) – 1 (x + 9) = 0
(x + 9) (x – 1) = 0
x = –9, 1
for x = 1, y = ± 2 2 x = ± 2 2
(2 2 + 2 2) 2 + (1 - 1) 2 = 4 2
L2 = Length of latus rectum = 4a = 4 × 2 = 8
L1 < L2
129. (b) Let common tangent be
L1 = Length of AB =
5
m
Since, perpendicular distance from centre of the circle to
the common tangent is equal to radius of the circle, therefore
y = mx +
5
m
=
5
2
1+ m
On squaring both the side, we get
m2 (1 + m2) = 2
Þ m4 + m2 – 2 = 0
Þ (m2 + 2)(m2 – 1) = 0
Þ m = ± 1 (Q m2 ¹ –2)
2
(
)
y = ± x + 5 , both statements are correct as m = ±1
satisfies the given equation of statement-2.
130. (b) We know that point of intersection of the normal to
the parabola y 2 = 4ax at the ends of its latus rectum is
(3a, 0)
Hence required point of intersection = (3, 0)
131. (b) Both statements are true and statement-2 is the correct
explanation of statement-1
a
\ The straight line y = mx +
is always a tangent to the
m
2
parabola y = 4ax for any value of m.
æ a 2a ö
The co-ordinates of point of contact ç 2 ,
÷
èm m ø
Now, required radius = OB =
9 + 16 = 25 = 5
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EBD_8344
M-188
Mathematics
x2 y 2
+
=1
16 3
Now, equation of normal at (2, 3/2) is
135. (a)
132. (a) Ellipse is
y
(t2, t)
16 x 3 y
= 16 - 3
2
3/ 2
Þ 8x – 2y = 13
Þ
y = 4x -
Let y = 4 x -
13
2
13
touches a parabola
2
y2 = 4ax.
We know, a straight line y = mx + c touches a parabola y2
= 4ax if a – mc = 0
\
x
O
æ 13ö
a - ( 4) ç - ÷ = 0 Þ a = – 26
è 2ø
\ Distance =
=
Hence, required equation of parabola is
y2 = 4 (– 26)x = – 104 x
133. (a) Point P is (4, –2) and PQ ^ x-axis
So, Q = (4, 2)
Y
Let (t2, t) be point on parabola from that line have shortest
distance.
t 2 - t +1
2
2
1 éæ 1 ö
3ù
êç t - ÷ + ú
4ú
2 êëè 2 ø
û
Distance is minimum when t -
1
=0
2
1 é 3ù
3 2
0+
=
8
2 êë 4 úû
136. (b) We know that the locus of perpendicular tangents is
directrix i.e., x = - a; x = -1
137. (b) We know that vertex of a parabola is the mid point
of focus and the point
\ Shortest distance =
nt
ange
T
Q
Normal
X
P
(4, – 2)
Y
X
OA
B
Y¢
x= 2
Equation of tangent at (4, 2) is
yy1 =
Þ 2y =
1
(x + x1)
2
1
(x + 2) Þ 4y = x + 2
2
x 1
Þy= +
4 2
where directrix meets the axis of the parabola.
Given that focus is O(0, 0) and directrix meets the axis at
B(2, 0)
æ 0+ 2 ö
\ Vertex of the parabola is ç
, 0÷ = (1, 0)
è 2
ø
138. (b) Given that parabola y2 = 8x
1
4
\ Slope of normal = – 4
134. (d) Both the given statements are true.
Statement - 2 is not the correct explanation for
statement - 1.
So, slope of tangent =
y 2 = 8x
Y
(2,0)
X'
x+2=0
F
Y'
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X
M-189
Conic Sections
We know that the locus of point of intersection of two
perpendicular tangents to a parabola is its directrix.
Point must be on the directrix of parabola
Q Equation of directrix x + 2 = 0
Þ x = –2
Hence the point is (–2, 0)
139. (a) Given that family of parabolas is
y=
a3 x 2 a 2 x
+
- 2a
3
2
9 ö 3a
a3 æ 2 3
x+
- 2a
Þ y = çè x +
÷2a
3
16a2 ø 16
æ -3 -35a ö
÷
\ Vertex of parabola is çè ,
4a 16 ø
To find locus of this vertex,
Þ 64xy = 105
105
which is the required equation of locus.
64
140. (a) Given P = (1, 0), let Q = (h, k)
Since Q lies on y2 = 8x
Þ xy =
\ K = 8h
2
...(i)
Let (a, b) be the midpoint of PQ
k+0
h +1
\a=
, b=
2
2
2 a -1 = h
2 b = k.
Putting value of h and k in (i)
(2b) 2 = 8(2a - 1) Þ b 2 = 4a - 2
Þ y2 - 4x + 2 = 0 .
141. (d) Equation of circle with centre (0, 3) and radius 2 is
x 2 + ( y - 3)2 = 4
Let locus of the centre of the variable circle is (a , b )
Q It touches x - axis.
\
Circle touch externally Þ c1c2 = r1 + r2
\ a 2 + (b - 3) 2 = 2 + b
Þ
a 2 = 10(b - 1 / 2)
\
1ö
æ
Locus is x 2 = 10 ç y - ÷
è
2ø
y 2 = 4ax and x 2 = 4ay , we get (0, 0) and ( 4a, 4a)
Putting in the given equation of line
2bx + 3cy + 4d = 0, we get
d = 0 and 2b +3c = 0
-3
16 y
and a = 4x
35
-3 -16 y
=
4x
35
r2 c2
(a, b)
Which is equation of parabola.
142. (d) Solving equations of parabolas
-3
-35a
x=
and y =
4a
16
Þ
r1
a 2 + (b - 3) 2 = b 2 + 4 + 4b
a2 + b2 – 6b + 9 = b2 + 4 + 4b
2
35a a3 æ
3ö
= çx+ ÷
Þy+
16
3 è
4a ø
Þ a=
c1
It¢s equation is ( x - a )2 + ( y – b)2 = b2
Þ d 2 + (2b + 3c )2 = 0
143. (b) Equation of the normal to a parabola y2 = 4bx at point
(bt
2
1 , 2bt1
) is
3
y = – t1 x + 2bt1 + bt1
(
)
2
Given that, it also passes through bt 2 , 2bt 2 then
2bt2 = – t1 bt22 + 2 bt1 + bt13
(
2
2
Þ 2t2 – 2t1 = – t1 t2 – t1
)
Þ 2(t2 – t1) = –t1(t2 + t1) (t2 – t1)
2
Þ 2 = – t1(t2 + t1) Þ t2 + t1 = – t
1
2
t1
144. (b) The equation of any tangent to the parabola y2 = 8ax is
Þ t2 = – t1 –
2a
...(i)
m
2
If (i) is also a tangent to the circle, x + y2 = 2a2 then,
y = mx+
2a = ±
2a
m m2 + 1
Þ
Þ ( m2 + 2)(m2 – 1) = 0 Þ m = ± 1.
Putting the value of m in eqn (i), we get
y = ± (x + 2a).
m2(1 + m2) = 2
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EBD_8344
M-190
Mathematics
145. (c) We know that the locus of the feet of the perpendicular
draw from foci to any tangent of the ellipse
the auxiliary circle x + y = a
2
2
2
x
2
a2
+
y
2
b2
= 1 is
\ Auxiliary circle : x2 + y 2 = 4
149. (c) The given ellipse is
x2
a
2
+
y2
b2
= 1, (a > b)
Length of latus rectum =
\ (-1, 3) satisfies the given equation.
æ
b2 ö
146. (c) Normal to the ellipse 2 + 2 = 1 at ç ae, ÷ is
aø
è
a
b
x2
y2
Þ x - ey =
2
...(i)
Q (0, - b) lies on equation (i), then
be =
e( a - b )
a
1
2
f ''(t ) = -2 < 0 Þ maximum
Þ f (t ) max =
2
5 1 1 8 2
+ - =
=
12 2 4 12 3
Since, f(t )max. = eccentricity
Þ ab = a 2 e 2 Þ b = ae 2 Þ
b2
a2
= e4
\1 - e 2 = e 4 Þ e 4 + e 2 - 1 = 0
x2 y 2
147. (b) Ellipse :
+
= 1,
16 9
a = 4, b = 3, c = 16 - 9 = 7
Þe=
on it; PA + PB = 2a
Þ PA + PB = 2(4) = 8.
x2 y 2
+
=1
5
4
2
3
Now, b 2 = a 2 (1 - e2 )
5a 2
æ 4ö
5a = a 2 ç 1 - ÷ Þ 5a =
Þ a 2 - 9a = 0
è 9ø
9
2
Þ a = 9 Þ a 2 = 81 and b = 45
\ a 2 + b2 = 81 + 45 = 126
\ (± 7, 0) are the foci of given ellipse. So for any point P
148. (a) Ellipse º
...(i)
5
+ t - t2
12
f '(t ) = 1 - 2t = 0 Þ t =
e( a - b )
a
2
2b 2
= 10 Þ b2 = 5a
a
Now f(t ) =
a 2 x b2 y
= a 2 - b2
ae b2 / a
2
Þ
2b 2
a
150. (d)
1
a
= 4Þ a =4´ =2
e
2
Now, b 2 = a 2 (1 - e2 )
3
æ 1ö
Þ b2 = 4 ç1 - ÷ = 4 ´ = 3
è 4ø
4
Let a point on ellipse be ( 5 cos q, 2sin q)
So, equation
\ PQ 2 = ( 5 cos q)2 + (-4 - 2sin q)2
= 5cos 2 q + 4sin 2 q + 16 + 16sin q
x2 y 2
+
=1
4
3
Þ 3x2 + 4 y 2 = 12
Now, P (1, b) lies on it
= 21 + 16 sin q - sin 2 q
= 21 + 64 - (sin q - 8) 2 = 85 - (sin q - 8) 2
PQ2 to be maximum when sin q = 1
2
\ PQmax
= 85 - 49 = 36.
...(i)
Þ 3 + 4b 2 = 12 Þ b =
3
2
æ 3ö
So, equation of normal at P ç1, ÷
è 2ø
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M-191
Conic Sections
a 2 x b2 y
= a 2 - b2 Þ 4 x - 2 y = 1
1
3/ 2
151. (d) The given ellipse :
Þ
x2 y 2
+
=1
4
3
2
Given : 2a = 2 Þ a =
Q c 2 = a 2 + b2 Þ 1 =
154. (a) Let
1
2
2b =
1
1
+ b2 Þ b =
2
2
x2
y2
+
= 1; a > b
a 2 b2
4
2
Þ b=
3
3
Equation of tangent º y = mx ±
So, equation of hyperbola is
Comparing with º y =
x2 y 2
=1
1
1
2
2
m=
Þ 2x2 - 2 y2 = 1
So, option (d) does not satisfy it.
(0, 3) Q
152. (a)
(–2, 0)
|x| | y|
+
=
1
2
3
O P
x2 y 2
+
=
1
4
9
(2, 0)
Q Area of ellipse = pab = p ´ 2 ´ 3 = 6p
\ Required area = Area of ellipse
– 4 (Area of triangle OPQ)
æ1
ö
= 6p - 4 ç ´ 2 ´ 3÷
è2
ø
= 6p - 12 = 6(p - 2) sq. units
153. (a) Eccentricity of ellipse
a 2 m2 + b2
-x 4
+
6 3
-1
16
and a 2m2 + b 2 =
6
9
a 2 4 16
a 2 16 4 4
+ =
= - =
Þ
36 3 9
36 9 3 9
Þ a2 = 16 Þ a = ± 4
Þ
Now, eccentricity of ellipse (e) = 1 -
Þ e = 1-
b2
a2
4
11 1 11
=
=
3 ´16
12 2 3
155. (d) Let P be (x1, y1).
So, equation of normal at P is
1
x
y
- =2 x1 y1
2
(0, –3)
4
7
7
e1 = 1 - =
=
18
9
3
Eccentricity of hyperbola
Since, the point (e1, e2) is
7
13
Þ k = 15 æç ö÷ + 3 æç ö÷
è9ø è 9 ø
Þ k = 16
Qc = a - b = 4 - 3 = 1
\ Foci = (±1, 0)
Now for hyperbola :
2
4
13
13
=
=
9
9
3
on the ellipse
15x2 + 3y2 = k.
Then, 15e12 + 3e22 = k
e2 = 1 +
1
æ
ö
, 0÷
It passes through ç è 3 2 ø
Þ
-1
6 2x1
So, y1 =
=-
1
1
x =
2 Þ 1 3 2
2 2
(as P lies in Ist quadrant)
3
y
2
So, b = 1 =
2
3
156. (b) 2ae = 6 and
2a
= 12
e
Þ ae = 3
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...(i)
EBD_8344
M-192
Mathematics
a
a
...(ii)
=6 Þ e=
6
e
Þ a2 = 18
[From (i) and (ii)]
Þ b2 = a2 – a2e2 = 18 – 9 = 9
Case-2: q =
and
\ Latus rectum =
5p
, then tangent does not pass through
3
Q (4, 4).
159. (a) Let the equation of ellipse :
2b 2 2 ´ 9
=
=3 2
a
3 2
x2
y2
+
=1
a 2 b2
157. (a) 3 x + 4 y = 12 2
Given that length of minor axis is 4 i.e. a = 4.
Also given be = 2
Þ 4 y = -3x + 12 2
Q a2 = b2 (1 – e2) Þ 4 = b2 – 4 Þ b = 2 2
3
Þ y =- x+3 2
4
Hence, equation of ellipse will be
Now, condition of tangency, c2 = a2m2 + b2
2
\ 18 = a .
Q ( 2 , 2) satisfies this equation.
9
9
+ 9 Þ a2. = 9
16
16
\ ellipse passes through ( 2 , 2).
Þ a2 = 16 Þ a = 4
160. (a) Equation of tangent to
Eccentricity e = 1 -
b2
a
2
= 1-
9
7
=
16
4
3x
a
\ ae =
-
2
9y
2b 2
\
\ Focus are (± 7,0)
3
a
\ Distance between foci of ellipse = 2 7
158. (a) Slope of tangent on the line 2x + y = 4 at point P is
Given ellipse is,
=1
2
-9
=
2
2b .(-2)
=
1
12
Þ a2 = 3 × 12 and b2 =
1
.
2
(On comparing)
9 ´12
4
Þ a = 6 and b = 3 3
Therefore, latus rectum =
x2
22
+
9ö
æ
y2
+
= 1 at ç 3, - ÷ is,
2
2
2ø
è
a
b
x2
But given equation of tangent is, x – 2y = 12
7
.4 = 7
4
3x2 +4y2 = 12 Þ
y2
( 3) 2
=1
161. (d) 3x2 + 5y2 = 32 Þ
3 sin q)
\ equation of tangent on the ellipse, at P is,
Let point P(2cos q,
2b2 2 ´ 27
=
=9
a
6
3x 2 5 y 2
+
=1
32
32
P(2 2)
x
y
cos q +
sin q = 1
2
3
Þ mT = –
R
3
cot q
2
Q both the tangents are parallel Þ Þ tan q = –
Case-1: q =
x2 y 2
+
=1
4
8
3
1
cot q =
2
2
p
p
3 Þ q = p – 3 or q = 2p – 3
2p
, then point P
3
3ö
æ
5 5
ç -1, 2 ÷ and PQ =
è
ø
2
Tangent on the ellipse at P is
3(2) x 5(2) y
3x 5 y
+
=1Þ
+
=1
32
32
16 16
æ 16 ö
\ co-ordinates of Q will be ç ,0 ÷
è 3 ø
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Q
M-193
Conic Sections
Now, normal at P is
32 32 y 32 32
= 3(2) 5(2) 3
5
æ4 ö
\ co-ordinates of R will be ç , 0 ÷
è5 ø
(-ae, 0)
S(ae, 0)
1
Hence, area of DPQR = ( PQ)( PR)
2
=
68
1 136 136
=
.
15
2 9
25
1 ö
æ 1
162. (d) Let tangent to parabola at point ç 2 , - m ÷ is
2 ø
è 4m
2
and tangent to ellipse is, y = mx ± m +
1
× 2ae × b = 8 Þ b2 = 8
2
From eqn (i)
a2e2 = 8
1
2
Also, e2 = 1 -
Now, condition for common tangency,
2
Þ 16m4 + 8m2 – 1 = 0 Þ m =
4m
2
=
b2
a2
Hence, required length of latus rectum =
-8 ± 64 + 64
2 (16 )
x2
+ y2 = 1
( 2)2
1
= 2 +1
2 -1
4
4
( 2 cos q, sin q)
163. (b) Given that focus is (0,5 3) Þ| b | > | a |
( 2, 0)
Let b > a > 0 and foci is (0, ± be)
Q a2 = b2 – b2e2 Þ b2e2 = b2 – a2
be =
2
2
b2 - a 2 Þ b – a = 75
...(i)
Q 2b – 2a = 10 Þ b – a = 5
From (i) and (ii)
b + a = 15
On solving (ii) and (iii), we get
Þ b = 10, a = 5
...(ii)
...(iii)
2 cos q x
+ y sin q = 1
2
æ 2
ö
1 ö
æ
Pç
, 0÷ and Q ç 0,
÷
è
sin q ø
è cos q ø
Let mid point be (h, k)
2
Now, length of latus rectum =
2b 2 2(8)
=
4
a
= 4 units
165. (c) Given the equation of ellipse,
-8 ± 8 2
2 -1
=
2 (16 )
4
1
...(ii)
Þ a2e2 = a2 – b2 Þ 8 = a2 – 8 Þ a2 = 16
1
1 Þ 1 = m2 + 1
= ± m2 +
2
16m 2
4m
2
a=
...(i)
Þ
1
y = mx +
4m
=
b
b
´
= –1 Þ b2 = a2e2
- ae ae
Since, area of DS¢BS = 8
2a
50
= =5
b
10
164. (a) Q DS¢BS is right angled triangle, then
(Slope of BS) ´ (Slope of BS¢) = –1
1
1
,k=
2 sin q
2 cos q
As cos2q + sin 2q = 1
Þ h=
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EBD_8344
M-194
\
Mathematics
1
1
+ 2 =1
2
2h
4k
Locus is
Þ a2 +
9
4
\ from eq. (iii) we get;
1
1
+
=1
2 x2 4 y 2
Þ a=
166. (a) A = {(a,b) Î R ´ R :| a - 5 |< 1,| b - 5 |< 1}
e2 = 1 -
Let a – 5, = x, b – 5 = y
Set A contains all points inside | x |< 1,| y |< 1
2
1
3
168. (c) Centre at (0, 0)
x2
y2
+ 2 =1
a
b
at point (4, –1)
(x - 1)2 y 2
+
=1
9
4
2
Y
16
1
+
=1
a 2 b2
Þ 16b2 + a2 = a2b2
at point (–2, 2)
(0, 1)
(0, 1)
X¢
(1, 0)
(1, –1)
(–1, –1) (0, –1)
(–1, 0)
X
4
+ 2 =1
a
b
Þ 4b2 + 4a2 = a2b2
2
\ (±1, ±1) lies inside the ellipse.
Hence, A Ì B .
167. (d) Let for ellipse coordinates of focus and vertex are
(ae, 0) and (a, 0) respectively.
3
Þ a - = ae
2
9
- 3a = a 2e 2
4
Length of latus rectum =
Þ b2 = 2a
e2 = 1 -
b2
a2
Þ e2 = 1 -
2a
a2
3
2
(given)
2
Þ 3a 2 = 12b 2
b2 = a2 (1– e2)
2
Þ
2
a 2 = 4b2
3
3
Þ e=
2
4
169. (d) e = 3/5 & 2ae = 6 Þ a = 5
Þ e2 =
2
2
2
Q b = a (1 – e )
2
Þ b = 25 (1 – 9/25)
...(i)
2b 2
=4
a
...(ii)
(from (ii))
2
... (iii)
a
Substituting the value of e2 in eq. (i) we get;
Þ e2 = 1 -
...(ii)
Þ 16b + a = 4a + 4b
From equations (i) and (ii)
2
\ Distance between focus and vertex = a(1 – e) =
...(i)
4
Y¢
Þ a2 +
2
8 1
=1 - =
a
9 9
Þ e=
2
B = {(a,b) ÎR ´ R: 4(a - 6) + 9(B- 5) £ 36}
Set B contains all points inside or on
(–1, 1)
9
2ö
æ
- 3a = a 2 ç1 - ÷
4
a
è
ø
Þ b= 4
\ area of required quadrilateral
= 4 (1/2 ab) = 2ab = 40
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M-195
Conic Sections
Þ 2a2e = b2 = a2 (1 – e2)
170. (c) Equation of tangent to ellipse
Þ 2e = 1 – e2 Þ (e + 1)2 = 2 Þ e =
2 -1
173. (a) Given equation of ellipse can be written as
x
y
cos q +
sin q = 1
27
3
Area bounded by line and co-ordinate axis
D=
1 27
3
9
.
.
=
2 cos q sin q sin 2q
Þ a2 = 6, b2 = 2
Now, equation of any variable tangent is
D = will be minimum when sin 2q = 1
Dmin = 9
y = mx ± a 2 m2 + b2
171. (b) The end point of latus rectum of ellipse
æ
b2 ö
x
y
+
= 1 in first quadrant is ç ae, ÷ and the tangent
aø
è
a 2 b2
2
2
æa ö
at this point intersects x-axis at ç , 0÷ and y-axis at (0, a).
èe ø
x2 y 2
The given ellipse is
+
=1
9
5
Then
...(i)
where m is slope of the tangent
So, equation of perpendicular line drawn from centre to
tangent is
y=
-x
m
...(ii)
Eliminating m, we get
( x4 + y 4 + 2 x2 y 2 ) = a 2 x2 + b2 y 2
a2 = 9, b2 = 5
Þ ( x 2 + y 2 )2 = a 2 x 2 + b2 y 2
5 2
Þ e = 1– =
9 3
\ End point of latus rectum in first quadrant is L (2, 5/3)
2x y
Equation of tangent at L is
+ =1
9 3
[Q It meets x-axis at A (9/2, 0) and y-axis at B (0, 3)]
\ Area of DOAB =
x2 y 2
+
=1
6
2
Þ ( x2 + y 2 )2 = 6 x2 + 2 y 2
174. (b) Let point A (a, 0) is on x-axis and B (0, b) is on
y-axis.
Y
1 9
27
´ ´3=
2 2
4
B (0, b)
P (h, k)
Y
l
B
(0, 3)
L(2,5/3)
C
O
S
A
X
(9/2, 0)
D
By symmetry area of quadrilateral
= 4 × (Area DOAB) = 4 ´
27
= 27 sq. units.
4
172. (b) Focus of an ellipse is given as (± ae, 0)
Distance between them = 2ae
According to the question, 2ae =
b2
a
A(a, 0)
X
Let P (h, k) divides AB in the ratio 1 : 2.
So, by section formula
h=
2(0) + 1( a )
a
=
1+ 2
3
k=
2(b) + 1(0) 2b
=
3
3
Þ a = 3h and b =
3k
2
Now, a 2 + b 2 = l2
Þ 9 h2 +
9k 2
= l2
4
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EBD_8344
M-196
Þ
Mathematics
h2
æ lö
çè ÷ø
3
+
2
k2
æ 2l ö
çè ÷ø
3
2
=1
æ l2
9 ö
1- ç ´ 2 ÷ =
è 9 4l ø
Now e =
1-
1
3
=
4
2
Thus, required locus of P is an ellipse with eccentricity
3
.
2
x2
Þ Focii = (± 7, 0)
y2
+
= 1 be the equation of ellipse.
a 2 b2
Given that F1B and F2B are perpendicular to each other.
Slope of F1B × slope of F2B = – 1
175. (a) Let
7
4
Now, radius of this circle = a2 = 16
Þ e=
æ 0-b ö æ 0-b ö
çè
÷ ´ç
÷ =–1
- ae - 0 ø è ae - 0 ø
x2 y 2
+
=1
9
4
Normal at the point is parallel to the line
Þ
B (0, b)
F1
(–a, 0) (–ae, 0) F2(ae, 0)
Now equation of circle is
(x – 0)2 + (y – 3)2 = 16
x2 + y2 – 6y – 7 = 0
177. (c) Given ellipse is 4x2 + 9y2 =36
4x – 2y – 5 = 0
Slope of normal = 2
(a, 0)
Slope of tangent =
-1
2
Point of contact to ellipse
(0, – b)
æ b ö æ -b ö
çè ÷ø ´ çè ÷ø = – 1
ae
ae
e =
1-
b
b2 ïü
ïì
2
íQ e = 1 - 2 ý
a ïþ
ïî
b2
a
2
2
a2
=
b2
a2
b2
b2 1
1= 2 2 Þ 2 =
2
a
a
1 1
b2
e2 = 1 - 2 = 1 - =
2 2
a
y2
b2
=1
æ
a2m
b
and line is ç
,
ç 2 2
2
2 2
a m + b2
è a m +b
ö
÷
÷
ø
1
2
No common tangents for these two circles.
176. (a) From the given equation of ellipse, we have
æ -9 8 ö
\ Point = ç , ÷
è 5 5ø
178. (c) Given equations of ellipses
x2 y 2
+
=1
3
2
2
1
Þ e1 = 1 - =
3
3
E1 :
and E2 :
e2 =
9
a = 4, b = 3, e = 1 16
a2
+
Now, a2 = 9, b2 = 4
b2 = a2e2
2
x2
Þ e2 =
x2 y 2
+
=1
16 b 2
1 - b2
16 - b2
=
16
4
Also, given e1 × e2 =
Þ
1
3
´
1
2
16 - b 2 1
= Þ 16 - b 2 = 12
4
2
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M-197
Conic Sections
Þ b2 = 4
\ Length of minor axis of
E2 = 2b = 2 × 2 = 4
Þ x = 2 or – 2
179. (d) x2 = 8y
...(ii)
2
When y = – 3, then x2 = – 24, which is not possible.
2
æ 1ö
c = 4 ç – ÷ + 12 = 2
è 2ø
We put y = –
2
x2
a2
x
a
2
+
y
2
or x 2 – 2 2 x + 2 = 0
y2
b2
=1
b
2
= 1 is given by
...(ii)
32 2 32
,b =
3
5
So, the equation of the ellipse is
Solving (i) and (ii) we get a2 =
3x 2 + 5 y 2 = 32
182. (a) The given equation of ellipse is
x2
y2
+
=1
4
1
So, A = (2, 0) and B= (0, 1)
If PQRS is the rectangle in which it is inscribed, then
P = (2, 1).
x2
+
y2
Q
B (0,1)
O
2
Þ x2 + 2 2 x + 2 = 0
+
Þ 5b2 = 3a 2
2
1
x+ 2
2
x
æ x
ö
+ ç – + 2÷ = 1
ø
4 è 2
ö
x 2 æ x2
æ xö
+ç
– 2 ç ÷ 2 + 2÷ = 1
è
ø
4 è 4
2
ø
\
2
= 2 + 8 = 10
= 1 be the ellipse
a 2 b2
the rectangle PQRS.
x2 y 2
+
=1
4
1
)}
b 2 = a 2 (1 - e 2 ) = a 2 (1 - 2 / 5)
Let
1
So, y = – x ± 2
2
For ellipse :
2
(
2– – 2
Given it passes through (–3, 1) so
9
1
+ 2 = 1 ...(i)
2
a
b
Also, we know that
1
= 0 Þ 3y – 1 = 0
3
1– 0
1
=–
0–2
2
)
{
181. (d) Let the equation of ellipse be
æ 2 6 1ö
æ 2 6 1ö
, ÷ and ç , ÷
ç
3
3
3 3ø
è
ø
è
Required equation of the line,
y = mx ± a 2 m2 + b2
Here a = 2, b = 1
(
æ 2 ö
= ç
+ 2 2
è 2 ÷ø
2 6
1
When y = , then x = ±
3
3
180. (d) Any tangent on an ellipse
2
1 ö æ
1 ö
æ
\ Points are ç 2,
, ç – 2, –
÷
÷
è
2ø è
2ø
2
Point of intersection are
1
and x = – 2, y = –
ì 1 æ 1 öü
\ P1 P2 = í
–ç–
÷ý +
2øþ
î 2 è
8y
1
+ y 2 = 1 Þ y = – 3,
3
3
m=
2
...(i)
x2
+ y2 = 1
3
From (i) and (ii),
y-
1
If x = 2, y =
R
Then it passed through P (2,1 )
4
1
\
+
= 1 ....(i)
2
a
b2
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circumscribing
P (2, 1)
A
(2,0) (4,0)
S
EBD_8344
M-198
Mathematics
Also, given that, it passes through (4, 0)
\ 16 + 0 = 1 Þ a 2 = 16
a2
Þ b2 = 4/3 [putting a2 = 16 in eqn (i)]
\ The required equation of ellipse is
[from (i)]
Þ 2e2 = 1, e =
x2
y2
+
=1
16 4 / 3
or x2 + 12y2 =16
a
183. (a) Perpendicular distance of directrix x = ± from
e
focus ( ± ae, 0)
X
O S
(ae, 0)
a
x= a/e
Y´
– ae = 4
e
1ö
æ
Þaç2 – ÷ = 4
è
2ø
8
Þa =
3
\ Semi maor axis = 8/3
184. (a) Given that distance between foci is
2ae = 6 Þ ae = 3 and length of minor axis is 2b = 8 Þ
=
b=4
we know that b 2 = a 2 (1 - e 2 )
Þ 16 = a 2 - a 2 e 2 Þ a 2 = 16 + 9 = 25 Þ a = 5
1
a
186. (b) Given that e = . Directrix , x = = 4
2
e
1
1
= 2 \b = 2 1- = 3
2
4
Equation of ellipse is
\a = 4´
y = mx ± 6 1 + m2
For common tangent,
36(1 + m 2 ) = 100m 2 - 64
100
64
´ 36 369
100
164
æ
ö
\ c 2 = 36 ç1 +
=
÷=
64 ø
64
4
è
Þ 100 = 64m 2 Þ m 2 =
Þ 4c 2 = 369
188. (a) Q The equation of hyperbola is
x2
y2
=1
a 2 b2
Q Equation of hyperbola passes through (3, 3)
1
2
-
-
1
2
=
1
9
Þ FB 2 + F ' B 2 = FF ' 2
a 2e2 + b2
2
) (
2
+
a 2e2 + b2
2 2
)
Þ 2(a e + b ) = 4a e Þ e =
2
2
= (2ae) 2
b2
a2
...(i)
B (0, b)
Q It passes through (9, 0)
6
-3
=
1
1
- 2
a2
b
1
1
=
b 2 2a 2
From equations (i) and (ii),
\
F' ( -ae, 0)
O
...(i)
x -3
y -3
=
1
1
×3 - 2 ×3
a2
b
185. (a) Given that ÐFBF ' = 90°
2 2
.
a
b
Equation of normal at point (3, 3) is :
3 3
\ e= =
a 5
(
2
y = mx ± 100m2 - 64
and the general tangent to the circle in slope form is
æa
ö
çè - ae÷ø
e
\
1
x2 y 2
+
= 1 Þ 3x 2 + 4 y 2 = 12
4
3
187. (c) General tangent to hyperbola in slope form is
Y
X´
We know that e 2 = 1 - b 2 / a 2 = 1 - e2
F (ae, 0)
9
a2 = , b2 = 9
2
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...(ii)
M-199
Conic Sections
Q Eccentricity = e, then e 2 = 1 +
b2
a
2
=3
æ9 ö
\ (a 2 , e 2 ) = ç , 3÷
è2 ø
x2 y 2
+
=1
25 b2
\ x12 + 5 y12 =
b2
25
x2 y 2
+
=1
16 b2
x2
5cos q
2
+
y2
=1
5
According to the question, e1 = 5e2
2
3
Now length of latus rectum of ellipse
2
Þ 1 + cos 2 q = 5sin 2 q Þ cos q =
æ b öæ b ö
Þ (e1e2 ) 2 = 1 Þ çç 1 - ÷ç
÷ç 1 + ÷÷ = 1
è 25 øè 16 ø
2
2
=
b2 b 2
b4
Þ 1+ - =1
16 25 25 ´ 16
9
b4
b2 Þ
= 0 Þ b2 = 9
16 × 25
25 ×16
2a 2 10 cos 2 q
20
4 5
=
=
=
3
b
5
3 5
192. (b) Let the hyperbola is
x2
y2
=1
a 2 b2
If a hyperbola passes through vertices at (± 6, 0), then
\ a=6
As hyperbola passes through the point P(10, 16)
9
4
\ e1 = 1 =
25 5
\
9
5
=
16 4
Distance between focii of ellipse
100 256
= 1 Þ b2 = 144
36 b 2
\ Required hyperbola is
= a = 2ae1 = 2(5)(e1 ) = 8
Distance between focii of hyperbola
x2 y 2
=1
36 144
36 x 144 y
+
= 36 + 144
10
16
\ At P(10, 16) normal is
Equation of normal is
= b = 2ae2 = 2(4)(e2 ) = 10
\ (a, b) = (8, 10)
190. (a) The tangent to the hyperbola at the point (x1, y1) is,
xx1 - 2 yy1 - 4 = 0
The given equation of tangent is
2x – y = 0
Þ
and Ellipse :
x2
y2
2
= 1 Þ e1 = 1 + cos q
10 10cos 2 q
Þ e2 = 1 - cos 2 q = sin q
b2
16
e1e2 = 1
And, e2 = 1 +
32
2
+5´ = 6
7
7
191. (d) Hyperbola :
The equation of hyperbola,
Then, e2 = 1 +
...(ii)
2
32
y12 = , x12 =
7
7
189. (d) Equation of ellipse is
Then, e1 = 1 -
x12 y12
-1 = 0
4
2
On solving eqs. (i) and (ii)
\
36 x 144 y
+
= 36 + 144
10
16
\ 2x + 5y = 100.
193. (c) Equation of tangent to y2 = 12x is y = mx +
x1
=2
2 y1
3
m
Equation of tangent to
Þ x1 = 4 y1
...(i)
Since, point (x1, y1) lie on hyperbola.
x2 y 2
= 1 is y = mx ± m2 - 8
1
8
Q parabola and hyperbola have common tangent.
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EBD_8344
M-200
Mathematics
9
3
= ± m2 - 8 Þ 2 = m2 – 8
m
m
\
Put m2 = u
u2 – 8u – 9 = 0 Þ
u2 – 9u + u – 9 = 0
Þ (u + 1) (u – 9) = 0
Q u = m2 ³ 0 Þ u = m2 = 9 Þ m = ± 3
\ equation of tangent is y = 3x + 1
or y = – 3x – 1
æ 1 ö
\ intersection point is P ç - ,0 ÷ .
è 3 ø
e = 1+
b2
a2
Þ e = 1+
8
Þe=3
1
\ foci (± 3, 0)
\
16
a2
S
æ 1 ö
ç - 3 ,0÷ P
è
ø
12
- 2 2
=1
a e - a2
Þ
æ a2 ö
4 é4
3 ù
Þ 4e 2 - 4 - 3 = (e2 - 1) ç ÷
=
1
ê
ú
ç 4 ÷
a 2 ë 1 e2 - 1û
è ø
æ 4e ö
Þ 4(4e2 - 7) = (e2 - 1) ç
÷
è 5ø
(3, 0)
x2 y 2
=1
9 16
Then focus is S’ (– ae, 0)
S'
(–3, 0)
4
m
Q This is common tangent.
y = mx +
So, put y = mx +
x=
4
in xy = – 4.
m
= 16m Þ m3 = 1Þ m = 1
\ equation of common tangent is y = x + 4
195. (c) Q directrix of a hyperbola is,
5x = 4 5 Þ x =
4
a
4
Þ =
e
5
5
(3, 0)
–9
5
2
a = 3, b = 4 Þ e = 1 +
4ö
æ
4
x ç mx + ÷ + 4 = 0 Þ mx 2 + x + 4 = 0
mø
m
è
16
2
196. (d) 16x2 – 9y2 = 144 Þ
194. (a) Given curves, y2 = 16x and xy = – 4
Equation of tangent to the given parabola;
m2
y2
- 2 = 1 passes throug (4, -2 3)
a
b
2
é 2
ù
b2
2 2
2
2
êQ e = 1 + 2 Þ a e - a = b ú
a
êë
úû
1
3+
SP
3 = 10 = 5
=
1
¢
SP 3 8 4
3
D=0Þ
x2
Þ 4e 4 - 24e 2 + 35 = 0
S'
(–3, 0)
Now, hyperbola
é
2 ù
16 25 êQ e = 1 + b ú
=
a2 úû
9
9 êë
5 ö
æ
\ the focus S ¢ º ç 3 - ´ ,0 ÷ º (-5,0)
3 ø
è
197. (c) Since, lx + my + n = 0 is a normal to
then
x2
y2
= 1,
a 2 b2
a2
b 2 (a 2 + b2 )2
=
l 2 m2
n2
but it is given that mx - y + 7 3 is normal to hyperbola
x2 y 2
=1
24 18
24
æa ö
ç e 0÷
è
ø
(4 -2 3)
then
m
2
-
18
(-1)
2
=
(24 + 18) 2
(7 3) 2
Þm=
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2
5
M-201
Conic Sections
Þ a = 8, b2 = 32
198. (c) Let equation of hyperbola be
Then, the equation of the ellipse
x2
y2
=1
a2 b2
Qe = 1+
b2
a2
...(i)
16
2
-
36
=1
a
b2
On solving (i) and (ii), we get
a2 = 4, b2 = 12
\
(
Þ b2 = a2 (e2 – 1)
e = 2 Þ b2 = 3a2
Equation (i) passes through (4, 6),
...(ii)
...(iii)
x2 y 2
=1
4 12
Now equation of tangent to the hyperbola at (4, 6) is
4x 6 y
y
= 1 Þ x - = 1 Þ 2x – y = 2
4 12
2
199. (d) Let the points are,
A(2, 0), A¢(–2, 0) and S(–3, 0)
Þ Centre of hyperbola is O(0, 0)
A A¢ = 2a Þ 4 = 2a Þ a = 2
Q Distance between the centre and foci is ae.
200. (a) \ Conugate axis = 5
\ 2b = 5
Distance between foci =13
x2 y 2
= 1 Þ a2 = 5, b2 = 4
5
4
The equation of tangent to the hyperbola is,
y = mx ± a 2 m 2 - b2
= x ± 5-4 Þy = x ±1
203. (b) Since, r ¹ ±1, then there are two cases, when r > 1
Þ e2 = 1 Þ e=
...(i)
(1 – r) = (1 + r) (e2 – 1) Þ e2 = 1 +
13
13
Þe=
2
12
201. (b) Let the ellipse be
Then,
(r - 1)
2r
=
(r + 1) ( r + 1)
2r
r +1
204. (a) Q a2 = cos2 q, b2 = sin 2 q
and e > 2 Þ e2 > 4 Þ 1 + b2/a2 > 4
Þ 1 + tan2 q > 4
\ a=6
ae =
2
(r + 1)
x2
y2
= –1 (Hyperbola)
1- r 1+ r
Then,
Þ e=
Þ a2 = 36
(r - 1)
2
=
(r + 1) (r + 1)
When 0 < r < 1, then
2ae = 13
Then, b2 = a2 (e2 – 1)
( r - 1)
( r + 1)
(r – 1) = (r + 1) (1 – e2) Þ 1 – e2 =
3
2
Þ b2 = a2 (e2 – 1) = a2e2 – a2 = 9 – 4 = 5
\ (6, 5 2) does not lie on this hyperbola.
202. (a) Given, the equation of line,
x–y=2Þy=x–2
\ its slope = m = 1
Equation of hyperbola is:
x2
y2
+
= 1 (Ellipse)
r -1 r +1
Then,
\ OS = ae Þ 3 = 2e Þ e =
x2 y 2
=1
4
5
Q (6, 552) does not satisfy eq (i).
)
Hence, the point 4 3,2 2 lies on the ellipse.
Now equation of hyperbola is
Þ Equation of hyperbola is
x2 y2
+
=1
64 32
x2 y 2
+
=1
a2 b2
2b 2
= 8, 2ae = b2 and b2 = a2(1 – e2)
a
æ p pö
Þ sec2 q > 4 Þ q Î çè , ÷ø
3 2
Latus rectum,
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EBD_8344
M-202
Mathematics
2b2 2sin 2 q
=
= 2(sec q – cos q)
a
cos q
æ p pö
d ( LR )
Þ
= 2(sec q tan q + sin q) > 0 "q Î çè , ÷ø
3 2
dq
p
p
1
æ
ö
æ
ö
\ min (LR) = 2 çè sec - cos ÷ø = 2 çè 2 - ÷ø = 3
3
3
2
LR =
p
2
Hence, length of latus rectum lies in the interval (3, ¥)
max (LR) tends to infinity as q ®
1
\ Area of DPQT = ´ TR ´ PQ
2
Q P º (3 5, -12) \ TR = 3 + 12 = 15,
1
\ Area of DPQT = ´15 ´ 6 5 = 45 5 sq. units
2
207. (a) Here, lines are:
2 x – y + 4 2k = 0
2 x + 4 2k = y
Þ
2kx + ky - 4 2 = 0
and
205. (d)
...(i)
...(ii)
Put the value of y from (i) in (ii) we get;
(
)
Þ 2 2 kx + 4 2 k 2 – 1 = 0
Þ x=
2 (1 – k 2 )
2 2(1 + k 2 )
,y=
k
k
Consider equation of hyperbola
2
2
æ y ö
æxö
\ç
÷ – ç ÷ =1
è4ø
è4 2ø
x2 y 2
=1
22 b 2
Q (4, 2) lies on hyperbola
\ length of transverse axis
16 4
=1
4 b2
4
\ b2 =
3
2a = 2 × 4 2 = 8 2
Hence, the locus is a hyperbola with length of its transverse
\
axis equal to 8 2
b2
Since, eccentricity = 1 + 2
a
4
Hence, eccentricity = 1 + 3 = 1 + 1 = 2
4
3
3
206. (d) Here equation of hyperbola is
x 2 y2
=1
9 36
Now, PQ is the chord of contant
\ Equation of PQ is :
Þ y = –12
x(0) y(3)
=1
9
36
x2
-
y2
=1
a 2 b2
foci is (±2, 0) Þ ae = ±2 Þ a2e2 = 4
Since
b2 = a2 (e2 – 1)
b2 = a2 e2 – a2 \ a2 + b2 = 4
208. (c) Equation of hyperbola is
Hyperbola passes through ( 2, 3 )
2
3
\ 2 - 2 =1
a
b
...(ii)
2
3
- 2 =1
4 -b b
[from (i)]
2
Y
x 2 y2
=1
9 36
(0, 3)
X
X'
Q
R
Y'
...(i)
Þ b4 + b2 – 12 = 0
Þ (b2 – 3) (b2 + 4) = 0
Þ b2 = 3
b2 = – 4
For b2 = 3
(Not possible)
2
2
Þ a2 = 1 \ x - y = 1
1
3
P
Equation of tangent is
2x
3y
=1
1
3
Clearly (2 2,3 3) satisfies it.
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M-203
Conic Sections
209. (d) Here, tx – 2y – 3t = 0 & x – 2ty + 3 = 0
On solving, we get;
6t
3t 2 + 3
3t
=
& x=
2t 2 - 2 t 2 - 1
t2 -1
Put t = tan q
\ x = – 3 sec 2q & 2y = 3 (–tan 2q)
2
2
Q sec 2q – tan 2q = 1
y=
(
2b 2
=8
a
13
2
If e, be the eccentricity of the ellipse, then
94
1
5
=1+ Þ e=
9
4
2
x2
a2
+
2
Now
x2 y 2
+
=1
211. (c)
12 16
æ9 ö
Þ a2 = 4 ç - 1 ÷
è4 ø
Þ a2 = 5
a 2 - b 2 =ae1
The point (5, 2 3 ) does not satisfy the above equation.
212. (a) S(5, 0) is focus Þ ae = 5 (focus)
Þ b2 = 12
Hence, equation of ellipse is
x 2 y2
+
=1
13 12
æ 13 3 ö
Now putting the coordinate of the point ç 2 , 2 ÷ in
è
ø
the equation of the ellipse, we get
13
3
+
=1
4 ´ 13 4 ´ 12
x2 y 2
\ It's equation is
=–1
5
4
5
(a) Þ (e) =
3
b2 = a2 (e2 – 1) Þ b2 = 16
=1
\ 13 – b2 = 1
1
12
e = 1=
2
16
Foci (0, 2) & (0, – 2)
So, transverse axis of hyperbola = 2b = 4 Þ b = 2
& a2 = 12 (e2 – 1)
(a) & (b) Þ a2 = 9
b2
a2 = 13
3
a
a
9
Þ = (directrix)
5
e
5
y2
Since ellipse passes through the foci (+ 13 , 0) of the
hyperbola, therefore
Þ 4b2 = a 2 e2 Þ 4a 2 (e 2 - 1) = a 2 e 2
x=
1
13 1
=
Þ e1 =
2
2
13
Equation of ellipse is
e1 ´
1
and 2b = (2ae)
2
Þ 3e2 = 4 Þ e =
)
e=
x2 y 2
=1
9 94
which represents a hyperbola
\ a2 = 9 & b2 = 9/4
210. (a)
x 2 y2
=1
4
9
Its Foci = ± 13, 0
Þ
l(T.A.) = 6; e2 = 1 +
213. (c) Equation of hyperbola is
––––– (a)
––––– (b)
Þ
1 1
+ = 1 , which is not true,
4 16
æ 13 3 ö
Hence the point ç 2 , 2 ÷ does not lie on the ellipse.
è
ø
214. (a)
2ö
æ
b
ç - ae, ÷
aø
è
a2 – b2 = 9 – 16 = – 7
Downloaded from @Freebooksforjeeneet
æ
b2 ö
ç ae, ÷
aø
è
L
(ae, 0)
EBD_8344
M-204
Mathematics
x2
4
Þ a2 = 4,
Given
a 2 + b2
e=
æp
ö
æp ö
3 x cos ç - q÷ + 2 y cot ç - q÷ = 32 + 22
è2
ø
è2 ø
y2
=1
5
b2 = 5
a2
=
Þ
4+5 3
=
4
2
3h sin q + 2k tan q = 32 + 2 2
...(3)
and 3h cos q + 2k cot q = 32 + 22
...(4)
3 5ö
æ
æ 5ö
L = çè 2 ´ , ÷ø = çè 3, ÷ø
2 2
2
Comparing equation (3) & (4), we get
Equation of tangent at (x1, y1) is
x x1 y y1
- 2 =1
a2
b
5
Here x1 = 3, y1 =
2
3h cos q - 3h sin q = 2k tan q - 2k cot q
Þ
3h cos q + 2 k cot q = 3h sin q + 2 k tan q
3h(cos q - sin q) = 2k (tan q - cot q)
3h(cos q - sin q) = 2k
3x y
x
y
+
=1
=1Þ
4 -2
4 2
3
or, 3h =
20
16
-4 = 9
9
215. (d) Let the coordinate at point of intersection of
normals at P and Q be (h, k)
Since, equation of normals to the hyperbola
a
-
y2
b
2
= 1 At point (x1, y1) is
a 2 x b2 y
+
= a2 + b2
x1
y1
therefore equation of normal to the hyperbola
2
x
2
-
3
y
2
22
= 1 at point P (3 secq , 2 tanq) is
Þ 3 x cos q + 2 y cot q = 3 + 2
–2k = 13 Þ k =
-13
2
Hence, ordinate of point of intersection of normals at P
and Q is
-13
2
216. (a) x2 – 6y = 0
...(i)
2
2x – 4y = 9
...(ii)
Consider the line,
2
...(1)
at point Q (3 sec f, 2 tanf) is
x2
32
-
y2
22
32 x
22 y
+
= 32 + 22
3sec f 2 tan f
Þ 3 x cos f + 2 y cot f = 32 + 2 2
3
2
On solving (i) and (iii), we get only
x- y =
Similarly, Equation of normal to the hyperbola
Given q + f =
Þ 2k tan q - 2 k + 2k tan q = 13
2
32 x
22 y
+
= 32 + 22
3sec q 2 tan q
2
...(5)
-2k (sin q + cos q)sin q
+ 2k tan q = 32 + 22
sin q cos q
OA2 – OB2 =
2
-2 k (sin q + cos q)
sin q cos q
Now, putting the value of equation (5) in eq. (3)
4
3
y-intercept of the tangent, OB = –2
x-intercept of the tangent, OA =
x2
(sin q - cos q)(sin q + cos q)
sin q cos q
x = 3, y =
3
2
æ
Hence ç 3,
è
...(2)
p
p
Þ f = - q and these passes through
2
2
...(iii)
3ö
÷ is the point of contact of conic (i), and
2ø
line (iii)
On solving (ii) and (iii), we get only x = 3, y =
(h, k)
\ From eq. (2)
Downloaded from @Freebooksforjeeneet
3
2
M-205
Conic Sections
æ 3ö
Hence ç 3, ÷ is also the point of contact of conic (ii) and
è 2ø
b2
16
eccentricity = e = 1 -
line (iii).
Hence line (iii) is the common tangent to both the given
conics.
foci: ± ae = ± 4 1 -
b2
16
217. (d) Equation of the tangent at the point ‘q’ is
Equation of hyperbola is
x sec q y tan q
=1
a
b
Þ
Þ P = (a cos q, 0) and Q = (0, – b cot q)
Let R be (h, k) Þ h = a cos q, k = –b cot q
Þ
-b
k
-bh
h
=
Þ sin q =
and cos q =
ak
a
h a sin q
x2
y2
=1
144 81
25 25
2 2
a k
+
h2
a2
=
=1
81 25
81
´
= 1+
25 144
144
eccentricity = e = 1 +
By squaring and adding,
b2 h2
x2 y 2
1
=
144 81 25
foci: ± ae = ±
225 15
=
144 12
12 15
´ = ±3
5 12
Since, foci of ellipse and hyperbola coincide
Y
\ ± 4 1R
Q
O
P
b2
= ± 3 Þ b2 = 7
16
K2 4
- =1
9 4
X
(Qb = ± 2)
Þ K2 = 18
219. (a) Given hyperbola is
Þ
b2
k2
+1 =
a2
Þ
h2
a2
h2
-
b2
k2
Now, given eqn of hyperbola is
x2 y 2
=1
9 b2
=1
Since this passes through (K, 2), therefore
x2 y 2
=1
4
2
K2 4
=1
9 b2
...(1)
Þ a 2 = 4, b 2 = 2
\ R lies on
a
2
x2
-
b
2
y2
= 1 i.e.,
4
x2
218. (c) Given equation of ellipse is
x2 y 2
+
=1
16 b 2
-
2
y2
=1
Also, given e = 1 +
Þ
1+
b2
a
2
=
13
3
13
b2
=
Þ 9 + b2 = 13
9
3
Þ b=±2
Downloaded from @Freebooksforjeeneet
EBD_8344
M-206
Mathematics
Now, from eqn (1), we have
K2 4
- =1
9 4
Co-ordinates of foci are (± ae, 0)
(Qb = ± 2)
Þ K2 = 18
220. (b) Given that ae = 2 and e = 2
\ a =1
(
222. (d) We know that tangent to the hyperbola
y = mx ±
b 2 = 1( 4 - 1)
\ Equation of hyperbola,
x2
a2
-
y2
b2
=1
cos 2 a
2 2
2
2
Þ m = a and a m - b = b
\
a 2a 2 - b 2 = b 2
Locus is a 2 x 2 - y 2 = b2 which is hyperbola.
3 x2 - y 2 = 3
221. (b) Given, equation of hyperbola is
-
y2
sin 2 a
=1
223. (d)
a=
1
x2
y2
=
144 81 25
144 12
= ,b =
25
5
81 9
= ,
25 5
Compare with equation of hyperbola
x2
a2
-
y2
b2
e = 1+
= 1, we get a 2 = cos 2 a and
81 15 5
=
=
144 12 4
\ Foci = ( ± ae, 0) = ( ±3 , 0)
\ foci of ellipse = foci of hyperbola
\ for ellipse ae = 3 but a = 4,
b 2 = sin 2 a
We know that, b 2 = a 2 (e2 - 1)
Þ sin 2 a = cos 2 a (e 2 - 1)
\ e=
3
4
Þ sin 2 a + cos2 a = cos 2 a.e2
Then, b 2 = a 2 (1 - e 2 )
Þ e 2 = sec2 a
Þ b 2 = 16æç1 - 9 ö÷ = 7
è 16 ø
Þ e = sec a
\ ae = cos a .
y2
b2
a 2 m2 - b2
2
x
y
=1
1
3
x2
a2
-
Given that y = a x + b is the tangent of hyperbola.
b2 = 3
Þ
x2
is
)
We know, b 2 = a 2 e 2 - 1
2
i.e. ( ± 1, 0)
Hence, abscissae of foci remain constant when a varies.
1
=1
cos a
Downloaded from @Freebooksforjeeneet
=1
12
Limits and Derivatives
TOPIC Ć
1.
x ®a +
(a)
3.
lim
3
2
( x - 1)( x 2 - 5 x + 6)
is equal to :
x ®-ab
x2 - 6 x + 8
then lim
x (e (
x®0
1 - cos(p( x ))
is equal to: [Sep. 05, 2020 (I)]
x+a-4
3
(b)
(c)
1
2
2
1+ x 2 + x 4 -1) / x
- 1)
(d)
(a) 1/2
7.
1
2
[Sep. 05, 2020 (II)]
1 + x2 + x4 - 1
x ®0
1 - x+ | x |
= L, then L is equal to :
l - x + [ x]
8.
9.
4.
5.
(b) 2
(c)
1
2
(d) 0
ìï 1 æ
x2
x2
x2
x 2 ö üï
If lim í 8 çç1 - cos - cos + cos cos ÷÷ ý = 2 - k ,
x ®0 ï x
2
4
2
4 ø ïþ
î è
then the value of k is __________. [NA Sep. 03, 2020 (I)]
3x + 33- x - 12
lim - x /2 1- x is equal to ¾¾¾.
x ®2 3
-3
[NA Jan. 7, 2020 (I)]
x + 2sin x + 1 - sin 2 x - x + 1
2
(b) 2
(c) 3
is
[April 12, 2019 (II)]
(d) 1
x4 - 1
x3 - k 3
= lim 2
, then k is:[April 10, 2019 (I)]
x ®1 x - 1
x ®k x - k 2
8
3
(b)
3
8
(c)
3
2
(d)
4
3
x 2 - ax + b
= 5 , then a + b is equal to :
x ®1
x -1
If lim
(b) 5
(c) –7
sin 2 x
equals :
x ®0 2 - 1 + cos x
lim
(a) 4 2
11.
[April 12, 2019 (II)]
(d) 3/2
If lim
(a) –4
10.
(c) –1/2
x + 2 sin x
lim
x ®0
(a)
[Sep. 03, 2020 (I)]
(a) 1
(b) –3/2
(a) 6
(a) is equal to e
(b) is equal to 1
(c) is equal to 0
(d) does not exist
Let [t] denote the greatest integer £ t. If for some
l Î R - {0, 1}, lim
Let f(x) = 5 - x - 2 and g ( x) = x + 1 , xÎ R. If f(x) attains
maximum value at a and g(x) attains minimum value at b,
If a is the positive root of the equation, p(x) = x2 – x – 2 = 0,
then lim
2.
6.
Limit of a Function, Left Hand &
Right Hand limits, Existance of
Limits, Sandwitch Theorem,
Evaluation of Limits when X® ¥,
Limits by Factorisation,
Substitution & Rationalisation
(b)
2
(c) 2 2
cot3 x - tanx
is:
p
pö
æ
x ® cos x +
ç
4
4 ÷ø
è
(b) 4 2
[April 8, 2019 (I)]
(d) 4
[Jan. 12, 2019 (I)]
lim
(a) 4
[April 10, 2019 (II)]
(d) 1
(c) 8 2
Downloaded from @Freebooksforjeeneet
(d) 8
EBD_8344
M-208
12.
Mathematics
x ®1-
lim
(
) (
tan p sin x + | x | - sin ( x [ x ])
2
x2
)
x cot ( 4 x )
lim
x ®0 sin 2
x cot 2 ( 2 x )
is equal to :
21.
3x - 3
lim
2x - 4 - 2
(a)
3
lim
x ®0
(b) exists and equals 2 2
23.
(b)
e x - cos x
lim
sin 2 x
x® 0
25.
2 2
(d) does not exist
For each xÎR, let [x] be greatest integer less than or equal
to x. Then
[Jan. 09, 2019 (II)]
x ([ x ] + | x |) sin [ x]
lim
is equal to:
x ®0
x
(a) – sin 1 (b) 1
(c) sin 1
lim
x ®0
(
lim
x®0
x
(b) -
1
2
(c)
1
4
(d)
2
1
8
1
2 2
[2015]
(d) 3
(c)
lim
x®0
27.
[Online April 10, 2015]
3
2
(d)
5
4
) is equal to:
(c)
[2014]
p
2
(d) 1
{
2
x - 4x + 4
x ®2
} = 5,
[Online April 11, 2014]
(c) 2
(d) 3
(1 - cos 2 x)(3 + cos x)
is equal to
x tan 4 x
1
4
lim
x ®0
(a) – p
(d) 0
1
2
(d)
(c) 4
is equal to :
(b) p
(b)
(
sin p cos 2 x
(d)
3
2
(c)
tan ( x - 2 ) x 2 + ( k - 2 ) x - 2k
If lim
(a) -
26.
x tan 2 x - 2 x tan x
equals. [Online April 15, 2018]
(1 - cos 2 x) 2
(a) 1
1
2
then k is equal to:
(a) 0
(b) 1
)
2 +1
1
(c) exists and equals
2
(b) 3
(a) -p
24.
(
1
16
is equal to : [Online April 8, 2017]
1
(b)
sin p cos 2 x
4 2
1
(c)
(1 - cos 2x)(3 + cosx)
is equal to :
x tan 4x
1
(a) exists and equals
18.
[2017]
2
22.
[Jan. 9, 2019 (I)]
y4
1
24
x ®3
(a) 2
(b) equals 0
(d) does not exist
1 + 1 + y4 - 2
y®0
(b)
[Jan. 11, 2019 (II)]
(a) 0
(b) 2
(c) 4
(d) 1
For each t Î R, let [t] be the greatest integer less than or
equal to t. Then,
[Jan. 10, 2019 (I)]
lim
1
4
(a) 2
(a) equals 1
(c) equals – 1
17.
equals :
( p - 2x )3
p
x®
2
:
æp
ö
(1 - | x | + sin |1 - x |) sin ç [1 - x] ÷
2
è
ø
lim
x ®1+
|1 - x | [1 - x]
16.
cot x - cos x
lim
(a)
20.
2
[Jan. 11, 2019 (I)]
(b) equals p
(d) equals 0
(a) does not exist
(c) equals p + 1
15.
19.
2
p
(b)
(c)
(d) p
2p
p
2
Let [x] denote the greatest integer less than or equal to x.
Then :
x ®0
14.
[Jan. 12, 2019 (II)]
1
(a)
13.
p - 2 sin -1 x
is equal to:
1- x
lim
x
2
1
2
) equals
(b) 1
(c) 1
(d) 2
[Online May 26, 2012]
(c) – 1
æ x - sin x ö
æ 1ö
lim ç
÷ sin çè ÷ø
x ø
x
(d) p
[Online May 7, 2012]
x®0 è
(a) equals 1
(c) does not exist
[2013]
(b) equals 0
(d) equals – 1
Downloaded from @Freebooksforjeeneet
M-209
Limits and Derivatives
28.
Let f : R ® [0, ¥) be such that lim f(x) exists and
x®5
( f ( x ))
lim
2
-9
x -5
x®5
35.
=0
lim
x ®0
1 - cos 2 x
x®5
29.
(b) 1
(c) equals
30.
1¥, Limits by Expansion Method
[2011]
1
36.
2
(a)
2
3
4
1
[2010]
æ 2 öæ 2 ö 3
(b) ç ÷ç ÷
è 3 øè 9 ø
(c) 3
1 - cos(ax 2 + bx + c)
( x - a) 2
is equal to
1
æ 2 ö3
(c) ç ÷
è9ø
(d) 1
[2005]
37.
æ 2 öæ 2 ö 3
(d) ç ÷ç ÷
è 9 øè 3 ø
x + x 2 + x3 + ... + x n - n
= 820, (n Î N) then the
x ®1
x -1
If lim
value of n is equal to ________.
2
32.
(a)
a
(a - b ) 2
2
(b) 0
(c)
-a 2
(a - b ) 2
2
(d)
33.
34.
38.
1
(a - b) 2
2
(b)
1
8
39.
(a)
(c) 0
(d)
1
32
40.
log x n - [ x]
, n Î N , ([x] denotes greatest integer less
[ x]
x ®0
than or equal to x)
[2002]
(a) has value -1
(b) has value 0
(c) has value 1
(d) does not exist
(a)
e4
(b) e2
1/ x
is equal to :
(b) 2
æ 3 x2 + 2 ö
lim ç 2
÷
x ®0 è 7 x + 2 ø
[NA Sep. 02, 2020 (I)]
(d) e 2
(c) 1
1/ x 2
is equal to:
[Jan. 8, 2020 (I)]
1
e
lim ò
(b)
x t sin(10t ) dt
x® 0 0
x
(c) e2
is equal to:
[Jan. 8, 2020 (II)]
(b)
n
n
r =1
r =1
å a r + nlim
å br is equal to :
then nlim
®¥
®¥
[April 12, 2019 (I)]
[2002]
e3
(a)
(d) 1
(d) e
1
1
1
(c) (d) 10
5
10
If a and b are the roots of the equation 375x2–25x–2=0,
(a) 0
41.
1
e2
x
(c)
[Sep. 02, 2020 (II)]
[2003]
lim
æ x2 + 5 x + 3 ö
lim ç 2
÷
x ®¥ çè x + x + 2 ÷ø
æ
æp
öö
lim ç tan ç + x÷ ÷
è4
øø
x®0 è
(a) e
é
æ xö ù
ê1 - tan çè 2 ÷ø ú [1 - sin x]
ë
û
lim
is
p é
x ® 1 + tan æ x ö ù [ p - 2 x ]3
2 ê
èç 2 ø÷ úû
ë
(a) ¥
( a ¹ 0) is equal to :
æ 2 ö3
(a) ç ÷
è3ø
Let a and b be the distinct roots of ax 2 + bx + c = 0 ,
x®a
1
- (4 x ) 3
4
3
2
(b)
then lim
(3a +
1
x) 3
[Sep. 03, 2020 (II)]
f (3 x)
f (2 x)
= 1 then lim
=
f ( x)
x ®¥ f ( x)
lim
lim
x ®a
1
(a + 2 x) 3 - (3 x ) 3
(d) does not exist
2
Let f : R ® R be a positive increasing function with
x ®¥
31.
1
TOPIC n Evaluation of Limits of the form
(d) 3
(b) equals –
2
(b) –1
(d) does not exist
Limits Using L-hospital's Rule,
(c) 2
æ 1 - cos{2( x - 2)} ö
lim ç
÷
x®2 è
x-2
ø
(a) equals
[2002]
(a) 1
(c) ero
[2011RS]
Then lim f(x) equals :
(a) 0
is
2x
21
346
(b)
29
358
(c)
1
12
Downloaded from @Freebooksforjeeneet
(d)
7
116
EBD_8344
M-210
42.
Mathematics
Let f : R ® R be a differentiable function satisfying
1
öx
43.
æ 1 + f ( 3 + x ) – f ( 3)
f ¢(3) + f ¢(2) = 0. Then lim ç
÷ is equal
x ®0 è 1 + f ( 2 – x ) – f ( 2 ) ø
to :
[April 08, 2019 (II)]
(a) 1
(b) e–1
(c) e
(d) e2
For each t Î R , let [t] be the greatest integer less than or
equal to t. Then
[2018]
48.
æ a b ö
If lim çç1+ + ÷÷
x ®¥ è x x 2 ø
49.
(a) a = 1 and b = 2
(b) a = 1, b Î R
(c) a Î R, b = 2
(d) a Î R, b Î R
x ®2
is given by
(a) 2
(b) –2
1
lim
x®0
2
45.
[Online April 16, 2018]
9 - (27 + x ) 3
(a) -
1
3
1
6
(b)
(c) -
(
Let p = lim 1 + tan 2 x
x ®0+
)
1
2x
1
6
(d)
50.
1
3
Let f (x) be a polynomial of degree 4 having extreme values
æ f ( x) ö
at x = 1 and x = 2. If lim ç 2 + 1÷ = 3 then f (– 1) is equal
x ®0 è x
ø
to
[Online April 15, 2018]
then log p is equal to :
(a)
[2016]
(a)
46.
1
2
51.
1
4
(c) 2
(1 - cos 2x) 2
is :
x ®0 2x tan x - x tan 2x
lim
(a) 2
47.
(b)
(b) -
1
2
æ a 4ö
If lim ç1 + - 2 ÷
x x ø
x ®¥ è
(c) –2
(d) 1
[Online April 10, 2016]
(d)
1
2
52.
2x
= e3, then 'a' is equal to :
53.
(a) 2
1
(c)
2
2
(d)
3
1
2
(b)
3
2
(c)
5
2
(d)
9
2
Let f(x) be a polynomial of degree four having extreme
é f (x) ù
1 + 2 ú = 3, then f(2) is
values at x = 1 and x = 2. If xlim
®0 ê
x û
ë
equal to :
[2015]
(a) 0
(b) 4
(c) – 8
(d) – 4
Let f (1) = –2 and f ¢ (x) ³ 4.2 for 1 £ x £ 6 . The possible
value of f (6) lies in the interval :
[April 25, 2013]
(a) [15, 19)
(b) (– ¥ , 12)
(c) [12, 15)
(d) [19, ¥ )
If f (x) = 3x10 – 7x8 + 5x6 – 21x3 + 3x2 – 7, then
[Online April 9, 2016]
3
(b)
2
(c) – 4
Derivatives of Polynomial &
Trigonometric Functions,
Derivative of Sum, Difference,
Product & Quotient of two
functions
TOPIC Đ
equals.
xf (2) - 2 f ( x)
x-2
[2002]
(d) 3
Let f (x) = 4 and f ¢ (x) = 4. Then lim
(a) is equal to 15.
(b) is equal to 120.
(c) does not exist (in R). (d) is equal to 0.
44.
= e 2 , then the values of a and b, are
[2004]
æé 1 ù é 2 ù
é15 ù ö
lim x ç ê ú + ê ú + ... + ê ú ÷
è x
x
ë x ûø
x ®0+ ë û ë û
(27 + x ) 3 - 3
2x
lim
f (1 - a ) - f (1)
a®0
(a) -
a3 + 3a
53
3
(b)
53
3
is
[Online May 19, 2012]
(c) -
55
3
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(d)
55
3
M-211
Limits and Derivatives
1.
(b) x2 - x - 2 = 0 Þ ( x - 2)( x + 1) = 0
R.H.L. = lim
h® 0
Þ x = 2, - 1 Þ a = 2
1- h + h
1
=
l+h+0
l
Given that limit exists. Hence L.H.L. = R.H.L.
1 - cos( x - x - 2)
x-2
2
\ lim
x ® 2+
Þ | l - 1| = | l |
1
1
=2
and L =
l
2
Þl=
æ x - x - 2ö
2 sin ç
÷
2
è
ø
2
= lim
x ® 2+
( x 2 - x - 2)
( x 2 - x - 2)
2
´
2( x - 2)
æ x2 - x - 2ö
ç
÷
2
è
ø
lim
2 sin
2.
1
2
Put
x4
Þ
1+ x 2 + x 4 -1
x
e
5.
1 + x2 + x4 - 1
x
(36)Let 3x = t2
t2 +
lim
t ®3
-1
1 + x2 + x4 - 1
= t when x ® 0 Þ t ® 0
x
27
t2
- 12
1 3
t t2
t 4 - 12t 2 + 27
t -3
t ®3
= lim
(t 2 - 3)(t + 3)(t - 3)
t -3
t ®3
2
= (3 – 3) (3 + 3) = 36.
(a) f (x) = 5 – | x – 2 |
Graph of y = f (x)
y
= lim
6.
et - 1
=1
t ®0 t
(b) Given lim
x®0
1- x + | x |
=L
l - x + [ x]
Here, L.H.L. = lim
h® 0
sin q ù
é
lim
= 1ú
êëQ q®
0 q
û
4
= 2-8 = 2- k
16 ´ 64
\ L = lim
3.
= 2-k
\k = 8
æ 1+ x2 + x 4 -1 ö
x
xçe
- 1÷
ç
÷
ç
÷
ø
(b) Let L = lim è
2
4
x®0
1 + x + x -1
x®0
x4
x2
x2
2sin 2
8 = 2- k
Þ lim 4 4 ´ 4
x®0 x
x
´ 16
´ 64
16
64
2
= lim
æ
x2 ö
1
cos
ç
4 ÷ø
è
2sin 2
3
´ 1´ 3 =
æ
x2 ö
1
cos
ç
2 ÷ø
è
x®0
æ
æ x2 - x - 2ö ö
ç sin ç
÷÷
2
è
ø÷
1
( x - 2)( x + 1)
=
´ lim
lim ç
( x - 2)
x 2 - x - 2 ÷ x ® 2+
2 x ® 2+ ç
ç
÷
2
è
ø
=
(8)
x-2
x ® 2+
= lim
4.
1+ h + h
1
=
l + h -1 l -1
(2 5)
O
2
By the graph f (x) is maximum at x = 2
\ a = 2 g (x) = | x + 1 |
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x
EBD_8344
M-212
Mathematics
Graph of y = g (x)
y
9.
x 2 - ax + b
=5
x -1
x ®1
(c) lim
Q limit is finite. \ 1 – a + b = 0
Þ lim
x ®1
–1
O
2
x
By the graph g (x) is minimum at x = – 1
\ b=–1
Now, xlim
®2
= lim
x ®2
7.
x® 0
10.
( x - 1)( x - 2)( x - 3)
( x - 2)( x - 4)
= lim
x ®0
x + 2sin x
x 2 + 2sin x + 1 - sin 2 x - x + 1
8.
11.
4 x3
=4
x ®1 1
x® K
x3 - k 3
x2 - k 2
tan x ö
æ
cot 3 x ç1 è cos3 x ø÷
cos ( x + p /4 )
(1 - tan 4 x)
3
p
x ® tan x cos( x + p / 4)
4
(1 + tan 2 x)(1 - tan x)(1 + tan x)
p
æ cos x - sin x ö
x®
tan 3 x ç
4
è
ø÷
2
= lim
æ0
ö
ç 0 form ÷
è
ø
(1 + tan 2 x )(1 + tan x )(cos x - sin x)
sin 3 x æ cos x - sin x ö
x®
4
ç
ø÷
cos 2 x è
2
[Using L Hospital’s Rule]
=4
3x 2
=4
x ® K 2x
Þ lim
Þ
cot 3 x - tan x
(d) lim
= lim
p
p
x ® cos æ x + p ö
x®
4
çè
÷ø
4
4
= limp
Lt
\ lim
sin 2 x
sin 2 x
= lim
x ù x®0 2 2 sin 2 x
é
2 ê1 - cos ú
4
2û
ë
= lim
æ x3 - k 3 ö
x4 - 1
= lim ç 2
÷
(a) Given, xlim
x ® K çè x - k 2 ÷ø
®1 x - 1
x4 -1
x ®1 x - 1
é0ù
êë 0 úû
2
é
sin ù
= 1ú
êQ xlim
x
®
0
ë
û
Taking L.H.S. lim
x
2 - 2cos2
2
æ sin x ö
ç
÷ .16
16
è x ø
=4 2
= lim
=
2
2 2
x ®0
xö
æ
ç sin 4 ÷
2 2ç
÷
ç x ÷
è 4 ø
( x + 2sin x) ëé x 2 + 2sin x + 1 + sin 2 x - x + 1 ûù
( x 2 - sin 2 x ) + ( x + 2sin x)
3´ 2
=2
3
sin 2 x
x ®0
( x - 1)( x - 3) 1
=
2
x-4
é
æ sin x ö ù é 2
2
ù
ê1 + 2 ç x ÷ ú ë x + 2sin x + 1 + sin x - x + 1 û
è
øû
ë
lim
= x® 0
æ
sin 2 x ö æ
æ sin x ö ö
çx÷ + 1+ 2ç
÷÷
x ø çè
è
è x øø
=
sin 2 x
x ®0 2 - 1 + cos x
(a) lim
= lim
On rationalising,
= lim
x ®0
æ0
ö
ç 0 form ÷ (By L Hospital’s rule)
è
ø
Þ 2 – a = 5 Þ a = – 3 and b = – 4
Then a + b = – 3 – 4 = – 7
(b) Given limit is,
lim
2x - a
=5
1
=8
( 2 )( 2 )
[Using L Hospital’s Rule]
12.
3
8
k =4 Þk=
2
3
(2)(2)
1
=
(b)
lim-
x ®1
= lim
h®0
p - 2sin -1 x
= lim f (1 - h )
h®0
1- x
p - 2sin -1 (1 - h)
1 - (1 - h)
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M-213
Limits and Derivatives
p - 2sin -1 (1 - h)
= lim
h® 0
-
= lim
16.
h
1
-1
2 2sin (1 - h)
h®0
1
-1
2sin (1 - h )
1
2 h
= lim
h® 0
=2´
1
´2´
1 - (1 - h)2
2
1
1
´
=
p
p
2
y ®0 4 æ
y ç
è
= lim
y® 0
æ tan(p sin x ö
+ 1÷ = 1 + p
= lim+ ç
x ®0 è
x2
ø
x ®0
tan(p sin x) + ( - x + sin x)
x2
=
2
17.
tan(p sin x ) + x + sin x - 2 x sin x
x ®0
x2
= p+1+1–2=p
Since, LHL ¹ RHL
Hence, limit does not exist.
2
2
(a)
2
(d) lim
x ®0
lim
x ® 0-
2
15.
2
æ 4x ö 4
.ç
.
=1
è tan 4 x ø÷ 22
æp
ö
(1 - | x | + sin (|1 - x |))sin ç [1 - x]÷
è2
ø
(b) lim
x ®1+
|1 - x | [1 - x]
æp
ö
(1 - |1 + h | + sin (|1 - 1 - h |))sin ç [1 - 1 - h]÷
è2
ø
= lim
h®0
|1 - 1 - h | [1 - 1 - h]
æp
ö
(1 - 1 - h + sinh)sin ç ( -1)÷
è2
ø
= lim
h®0
h([0 - h])
= lim
(- h)( -1 + h)sin( -1)
h
= lim (1 - h)sin(-1) = –sin1
h®0
18.
(d) Let, L = lim
x ®0
( x tan 2 x - 2 x tan x)
= lim K (say)
x ®0
(1 - cos 2 x)2
é 2 tan x ù
xê
ú - 2 x tan x
1 - (tan x )2 û
ë
ÞK=
(1 - (1 - 2sin 2 x ))2
=
2 x tan x - [2 x tan x - 2 x tan 3 x ]
4 sin 4 x ´ (1 - tan 2 x )
=
2 x tan 3 x
=
4 sin x ´ (1 - tan 2 x)
4
2 x tan 3 x
æ cos 2 x - sin 2 x ö
4 sin 4 x ´ çç
÷÷
cos 2 x
è
ø
sin 3 x
cos3 x
=
æ
cos 2 x - sin 2 x ö
4 sin 4 x ´ çç
÷÷
cos 2 x
è
ø
2x
æ pö
(- h + sin h) sin ç - ÷
è 2ø
= lim
=0
h® 0
h (-1)
)
x ([ x ] + | x |) × sin[ x]
| x|
(0 - h)([0 - h]+ |0 - h |) × sin[0 - h]
|0 - h |
h®0
2
æ x ö æ tan 2 x ö
.ç
= lim ç
÷
x ®0 è sin x ÷
ø è 2x ø
(
= lim
h®0
x cot 4 x
x .tan 2 2 x
= lim
2
x®0 sin 2 x .tan 4 x
sin x .cot 2 x
æ
y4 ç
è
æ 1 + y4 –1öæ 1 + y4 + 1 ö
ç
֍
÷
è
øè
ø
ö
1 + 1 + y 4 + 2 ÷ æç 1 + y4 + 1ö÷
ø
øè
1+ y4 -1
ö
1+ 1+ y4 + 2 ÷ 1+ y4 +1
ø
1
1
=
2 2´2 4 2
= lim-
14.
æ
ö
y4 ç 1+ 1+ y4 + 2÷
è
ø
= lim
2
2
And LHL is, lim-
1+ 1+ y4 - 2
y® 0
2
y4
y® 0
= lim
tan(p sin x ) + ( x - 0)
(a) RHL is, lim+
x ®0
x2
1+ 1+ y4 - 2
æ
öæ
ö
4
4
çè 1 + 1 + y - 2 ÷ø çè 1 + 1 + y + 2 ÷ø
= lim
y® 0
æ
ö
y4 ç 1+ 1+ y4 + 2÷
è
ø
1
2 h
1
h(2 - h )
2
13.
(-1)
(a) L = lim
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EBD_8344
M-214
Mathematics
x
2 sin x ´ (cos x - sin 2 x) cos x
ÞK=
1
x
´ lim
2 sin x x ® 0 cos x (cos2 x - sin 2 x)
\ L = lim
x ®0
= lim
x®0
19.
(c)
23.
2
cot x(1 - sin x)
p
x®
2
pö
æ
-8 ç x - ÷
2ø
è
3
= limp
x®
æp
ö
8ç - x ÷
è2
ø
x2
= lim sin
= lim sin
x ®0
tan t(1 - cos t)
t®0
8t
1 1 1
= .1. =
8 2 16
20.
3
= lim
t ®0
24.
tan t 1 - cos t
.
8t
t2
3 ( x - 3)
x ®3 2
21.
( x - 3)
´
2x - 4 + 2
(
3x + 3
)
or k + 2 = 5
25.
3 2 2
1
´
=
2
6
2
=
x
2sin x 3 + cos x
·
·
= lim
x®0
1
tan 4 x
x2
= 2 lim
= 2.4
x
· lim 3 + cos x · lim
x®0
x ®0
= 2.4.
x
tan 4 x
1
4x
1
lim
= 2.4. = 2
4 x®0 tan 4 x
4
(c) lim
x ®0
sin 2 x
x ®0
x2
22.
Þ k=3
(d) Multiply and divide by x in the given expression, we
get
x
(1 - cos 2 x ) (3 + cos x)
lim
·
x®0
1
tan 4 x
x2
é
2 xù
êëQ1 - cos 2 x = 2sin 2 úû
x
2sin 2 x 3 + cos x
·
·
2
x®0
1
tan
4x
x
2
2
=5
tan h ü
ì
= 1ý
íQ lim
î h®0 h
þ
(k + x ) = 5
Þ 1 ´ xlim
®2
= lim
x®0
( x - 2) 2
tan( x - 2){x ( x - 2) + k ( x - 2)}
=5
( x - 2) ´ ( x - 2)
x® 2
x
(1 - cos 2 x ) (3 + cos x)
lim
·
x®0
1
tan 4 x
x2
sin 2 x
tan( x - 2){x 2 + kx - 2 x - 2 k}
=5
Þ lim
(a) Multiply and divide by x in the given expression, we
get
= 2 lim
x2 - 4 x + 4
x ®2
æ (k + x )( x - 2 ) ö
æ tan( x - 2) ö
´ lim ç
Þ xlim
=5
ç
÷
è
(
x
2)
ø
®2
x® 2 è
( x - 2 ) ÷ø
( 3x - 9 ) ´ ( 2x - 4 + 2 )
A = lim
x ®3 {( 2x - 4 ) - 2} ´ ( 3x + 3 )
= lim
tan( x - 2){x 2 + (k - 2) x - 2k}
x®2
Rationalise
Þ
(d) lim
Þ lim
2x - 4 - 2
x ®3
(p sin 2 x) p sin 2 x
´
x2
p sin 2 x
2
3x - 3
(b) Let A = lim
[Q sin (p – q) = sin q]
x2
æ sin x ö
= lim 1´ p ç
÷ =p
x ®0
è x ø
Put
= lim
( p - p sin 2 x)
x ®0
3
p
p
- x = t Þ as x ® Þ t ® 0
2
2
æp öæ
æ p öö
cot ç - t ÷ ç 1 - sin ç - t÷ ÷
è2 øè
è 2 øø
= lim
3
t® 0
8t
sin éëp(1 - sin 2 x) ùû
x ®0
cot x(1 - sin x)
2
x2
x®0
= lim
1
x
1
´ lim
=
2
2
x
0
®
2 sin x
cos 0 (cos 0 - sin 0) 2
lim
sin( p cos 2 x)
(b) lim
2xe + sin x
2sin x cos x
1 3
æ x x2 1 ö 1
= 1+ =
lim ç
e + ÷
x ®0 è sin x
2 ø cos x
2 2
x2
· lim ( 3 + cos x ) · lim
x® 0
x® 0
1
=2
4
(
4x
1
´
tan 4 x 4
sin p cos 2 x
26.
(d) Consider, lim
x®0
(
sin p - p sin 2 x
= lim
x®0
x
2
(
sin p sin 2 x
= lim
x®0
p sin 2 x
x
)
)
2
éëQ sin ( p - q ) = sin q ùû
) ´ ( p sin x) = p
2
x2
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M-215
Limits and Derivatives
27.
æ ( x - a )( x - b) ö
2sin 2 ç a
÷ø
è
2
= lim
x® a
( x - a)2
æ x - sin x ö
æ 1ö
(b) Consider lim çè
÷ sin çè ÷ø
x ø
x
x ®0
é æ sin x ö ù
ê x çè1 - x ÷ø ú
1
= lim ê
ú ´ lim sin æç ö÷
x
x ®0 ë
û x® 0 è x ø
2
= lim
x ® a ( x - a )2
´
æ ( x - a ) ( x - b) ö
sin 2 ç a
÷ø
è
2
é sin x ù
æ 1ö
´ lim sin ç ÷
= lim ê1 è xø
x úû x®0
x® 0 ë
28.
sin x ù
é
æ 1ö
´ lim sin ç ÷
= ê1 - lim
ú
è
xø
ë x® 0 x û x ® 0
=
æ 1ö
= 0 ´ lim sin ç ÷ = 0
è xø
x ®0
æp xö
tan ç - ÷ .(1 - sin x )
è4 2ø
(d) lim
p
( p - 2 x )3
x®
(d) Given that lim
x®5
32.
( f ( x) ) 2 - 9
x®5
Þ lim
x-5
2
Let x =
=0
[(f (x) )2 – 9] = 0
= lim
2
Þ é lim f ( x) ù = 9 Þ lim f (x) = 3
ë x®5
û
x®5
29.
a 2 (a - b)2
.
2
(d) lim
x®2
1 - cos{2( x - 2)}
x-2
é
2 vù
êëQ1 - cos q = 2sin 2 úû
x-2
x®2
L.H.L = - lim
x® 2
(at x = 2)
R.H.L = lim
x®2
(at x = 2)
2 sin( x - 2)
=- 2
( x - 2)
2 sin( x - 2)
= 2
( x - 2)
Thus L.H.L ¹ R.H.L
(at x = 2)
(at x = 2)
1 - cos{2( x - 2)}
does not exist.
x-2
(d) Given that f(x) is a positive increasing function.
\ 0 < f ( x ) < f (2 x ) < f (3 x )
Divided by f (x)
33.
( -2 y )3
y
y
- tan 2sin 2
2
2
é
= lim
2 qù
y ®0
y3
êëQ1 - cos q = 2sin 2 úû
( -8). .8
8
y
2
tan é
1
2 . sin y / 2 ù = 1
= lim
ê
ú
32
y ®0 32 æ y ö ë y / 2 û
ç ÷
è2ø
sin q
tan q ù
é
= lim
= 1ú
êëQ lim
q® 0
q®
0
q
q
û
(d) Since, lim [ x ] = -1 ¹ lim [ x] = 0. So lim [ x] does
x ®0-
x ®¥
x ®¥
f (2 x )
f (3 x )
£ lim
f ( x) x ®¥ f ( x)
By Sandwich Theorem.
f (2 x)
=1
f ( x)
(a) Given that
ax2 + bx + c = a(x – a)(x – b)
1 - cos a ( x - a )( x - b )
lim
x ®a
( x - a )2
Þ lim
34.
(a)
x ®0
æ x 2 + 5 x + 3ö
æ
4x +1 ö
lim ç 2
çè1 + 2
÷
÷ = xlim
®¥
x
+ x + 2ø
x ®¥ è x + x + 2 ø
( 4 x +1) x
é
x2 + x + 2 ù x2 + x + 2
êæ
4 x + 1 ö 4 x+1 ú
ú
= lim êç1 + 2
÷
x ®¥ ê è
x + x + 2ø
ú
ëê
ûú
lim
4x2 + x
= e x®¥ x
x ®¥
31.
x ®0 +
x
x® 2
Þ lim 1 £ lim
æ yö
tan ç - ÷ .(1 - cos y )
è 2ø
not exist, hence the required limit does not exist.
Hence, lim
f (2 x ) f (3 x)
<
Þ 0<1<
f ( x)
f ( x)
p
+ y; y ® 0
2
y®0
2 sin( x - 2)
= lim
30.
a 2 ( x - a ) 2 ( x - b) 2
4
a 2 ( x - a )2 ( x - b)2
´
4
2
+ x+ 2
4+
lim
=e
1
é
lù
êQ lim (1 + lx ) x = e ú
ë x®¥
û
1
x
x ®¥ 1+ 1 + 2
x 2
x
=e
4
é 1
ù
êëQ ¥ = 0úû
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x
EBD_8344
M-216
35.
Mathematics
1 - cos 2 x
(d) lim
x ®0
2x
40.
(a) Using L’ Hospital rule,
;
lim
x ®0
lim
x ®0
sin q ù
é
= 1ú
êëQ lim
q® 0
q
û
2sin 2 x
sin x
Þ lim
x®0 x
2x
41.
The limit of above does not exist as
x sin(10 x)
=0
1
(c) Given equation is, 375x2 – 25x – 2 = 0
Sum and product of the roots are,
a+b=
LHS = –1 ¹ RHL = 1
25
-2
and ab =
375
375
n
36.
37.
(b) lim
x ®a
1
( a + 2 x) 3
1
- (3 x ) 3
1
x) 3
1
- (4 x ) 3
(3a +
å (ar + br )
n ®¥
lim
é0
ù
ê 0 case ú
ë
û
r =1
= (a + a 2 + a3 + ...¥ (b + b 2 + b3... + ¥ )
a
b
a + b - 2ab
+
=
1 - a 1 - b 1 - (a + b) + ab
Apply L'Hospital rule
=
1
1
(a + 2 x)-2 / 3 × 2 - × (3x )-2 /3 × 3
3
3
lim
1
x ®a 1
-2/ 3
(3a + x )
× - (4 x) -2 / 3 × 4
3
3
25
4
+
29
1
29
375 375 =
=
=
=
25
2
375 - 25 - 2 348 12
1375 375
1
1/ 3
(3a) -2 / 3 × (2 - 3)
24 / 3 1 2 æ 2 ö
3-2/ 3 1
3
=
= -2 /3 × = 1/3 × = × ç ÷
1
3 3 è9ø
3
9
(4a)-2 / 3 × (1 - 4) 4
3
(40)
æ 1 + f (3 + x) - f (3) ö x
¥
(a) I = lim ç
÷ [1 form]
x ®0 è 1 + f (2 - x ) - f (2) ø
1
42.
Þ I = el1, where
æ æ 1 + f (3 + x ) - f (3) ö ö æ 1 ö
- 1÷ ÷ ç ÷
I1 = lim ç ç
x ®0 è è 1 + f (2 - x) - f (2)
øø è x ø
x + x 2 + x3 + .... + x n - n
æ0
ö
= 820 ç case÷
è0
ø
x ®1
x -1
lim
1 + 2 x + 3x 2 + .... + nx n -1
= 820
x ®1
1
(Using L' Hospital rule)
æ 1 ö æ f (3 + x) - f (3) - f (2 - x) + f (2) ö
= lim ç ÷ ç
÷
1 + f (2 - x) - f (2)
x ®0 è x ø è
ø
lim
Þ 1 + 2 + 3 + .... + n = 820
(
n(n + 1)
= 820 Þ n2 + n - 1640 = 0
2
Þ
By L. Hospital Rule,
æ
ö
1
æ f ¢(3 + x) + f ¢(2 - x) ö
I1 = lim ç
÷
÷ lim ç
1
x ®0 è
ø x®0 è 1 + f (2 - x) - f (2) ø
Þ n = 40, n Î N
38.
æ 1 + tan x ö
(d) lim ç
÷
x®0 è 1 - tan x ø
= f¢ (3) + f¢ (2) = 0
1/ x
I
0
Þ I = e 1 = e =1
1é æ p ö ù
lim ê tan ç + x÷ -1ú
è4 ø û
Þe
1 æ 1+ tan x ö
lim ç
-1
è 1- tan x ÷ø
x®0 x ë
Þ
æ 2 tan x ö 1
lim ç
÷
e x ®0è 1- tan x ø x
43.
Þ e x®0 x
æ tan x ö æ 2 ö
lim ç
֍
÷
x ø è 1- tan x ø
= e x ® 0è
1
39.
1 ìï 3 x 2 + 2
=e
-4
=e 2
1
= e -2 = 2
e
æ 1
2
15 ö
(b) Since, lim x ç é ù + é ù + ... + é ù ÷
êë x úû êë x úû
êë x úû ø
+
è
x®0
æ 1 + 2 + 3 + ... + 15 ö
= lim x ç
÷ø x
x ® 0+ è
= e2
üï
lim í
- 1ý
æ 3x 2 + 2 ö x 2
x ®0 x 2 ï 7 x 2 + 2
î
þï
(b) Let R = lim ç 2
÷ =e
x ®0 çè 7 x + 2 ÷ø
1 ìï -4 x 2 üï
lim í
ý
x ®0 x 2 ï 7 x 2 + 2 ï
î
þ
0
form)
0
ìrü
Q 0 £ í ý <1
îxþ
Þ
æì 1 ü ì 2 ü
ì15 ü ö
çè í x ý + í x ý + ... + í x ý ÷ø
î þ î þ
î þ
ìrü
0 £ xí ý < x
îxþ
æ 1 + 2 + 3 + ... + 15 ö 15 ´ 16
= 120
\ lim+ x çè
÷ø =
x
2
x ®0
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M-217
Limits and Derivatives
44.
(c) Let L = lim
x®0
1
1
x) 3
-3
9 - (27 +
2
x) 3
(27 +
48.
Here ‘L’ is in the indeterminate form i.e.,
\ using the L’Hospital rule we get:
-2
æ 1 öù
é
êæ a b ö ç a + b ÷ ú
Þ lim ê ç1 + + ÷ çè x 2 ÷ø ú
x
x x2 ø
x®¥ è
ê
ú
ë
û
-2
1
1
(27 + x ) 3
´ (27) 3
1
3
3
=
=L = lim
-1
-1
x®0
6
2
-2
- (27 + x ) 3
´ 27 3
3
3
(
1
lim ln sec x
+ x
x ®0
Þ
49.
(1 - cos 2 x )
3
2
4
5
4
æ x 2 x4
ö
4 ç1 +
- ..... ÷
ç
÷
3! 5!
ø
= lim è
x ®0
2 æ 60 ö
- 2 + x ç - ÷ + .....
è 15 ø
(dividing numerator & denominator by x4)
= –2
(b)
(c) Given that f (2) = 4 and f(2) = 4
x® 2
æ
ö
x
x
4ç x - +
- ..... ÷
ç
÷
3! 5!
è
ø
= lim
4
64 ö
x ®0 4 æ 2 8 ö
æ
x ç - ÷ + x6 ç - ÷
3
5
15
15
è
ø
è
ø
47.
= e 2 Þ e2 a = e 2
= lim f (2) - 2 f ¢( x ) = f (2) - 2 f ¢(2)
5
3
= e2
xf (2) - 2 f ( x) æ 0 ö
, ç ÷
è 0ø
x-2
Applying L-Hospital's rule, we get
)
(2 sin 2 x ) 2
x® 0
æ
ö
æ
ö
2x
23 x 3
25 x5
x
2xç x +
+
+ ..... ÷ - x ç 2 x +
+2
+ ..... ÷
ç
÷
ç
÷
3
15
3
15
è
ø
è
ø
lim
æa b ö
2xç + ÷
è x x2 ø
x®2
x ®0 2 x tan x - x tan 2 x
=
= e2
We have, lim
sec x tan x
tan x 1
= lim
= lim
=
+
+
sec
x
2
x
2
×
x®0
x ®0 2 x
(c) lim
é bù
lim 2 ê a + ú
xû
e x®¥ ë
2x
Þ a = 1 and b Î R
Applying L hospital's rule :
46.
x ®¥
æ a bö
Given that lim ç1 + + ÷
x x2 ø
x ®¥ è
0
0
1
ln(1 + tan 2 x)
45. (a) ln p = lim+
2x
x ®0
(b) We know that lim (1 + x ) x = e
æ a 4 ö
lim ç 1 + - 2 ÷
x x ø
x ®¥ è
2x
é
æ a 4
ö ù
= e ê lim ç 1 + - 2 - 1÷ 2 x ú
è
ø û
x
®¥
x
x
ë
8ö
æ
= e lim ç 2a - ÷ = e2a
x ®¥ è
xø
\ 2a = 3 Þ a = 3/2
(1¥ form)
= 4 – 2 ´ 4 = –4.
50.
(d) Q f (x) has extremum values at x = 1 and x = 2
Q f ¢(1) = 0 and f ¢(2) = 0
As, f (x) is a polynomial of degree 4.
Suppose f (x) = Ax4 + Bx3 + Cx2 + Dx + E
Q
æ f ( x) ö
lim ç 2 + 1÷ = 3
x ®0 è x
ø
æ Ax 4 + Bx 3 + Cx 2 + Dx + E
ö
+ 1 ÷÷ = 3
Þ lim çç
2
x ®0
x
è
ø
D E
æ
ö
Þ lim ç Ax 2 + Bx + C + + 2 + 1÷ = 3
x ®0 è
x x
ø
As limit has finite value, so D = 0 and E = 0
Now A(0)2 + B(0) + C + 0 + 0 + 1 = 3
Þ c +1=3 Þc=2
f ¢(x) = 4Ax3 + 3Bx2 + 2Cx + D
f ¢(1) = 0 Þ 4A(1) + 3B(1) + 2C(1) + D = 0
Þ 4A + 3B = – 4
...(i)
f ¢(2) = 0 Þ 4A(8) + 3B(4) + 2C(2) + D = 0
Þ 8A + 3B = –2
...(ii)
From equations (i) and (ii), we get
1
A = and B = – 2
2
x4
– 2x3 + 2x2
2
(- 1) 4
– 2(– 1)3 + 2(– 1)2
Therefore, f (– 1) =
2
1
9
9
= + 2 + 2 = . Hence f (– 1) =
2
2
2
So, f (x) =
Downloaded from @Freebooksforjeeneet
EBD_8344
M-218
51.
Mathematics
é f (x) ù
(a) lim ê1 + 2 ú = 3
x ®0 ë
x û
Þ lim
x ®0
f (x)
x
2
=2
So, f(x) contain terms in x2, x3 and x4.
52.
x ®0
f (x)
x
2
1 4
x
2
f(2) = 2 × 4 – 2 × 8 +
1
× 16 = 0
2
(d) Given f (1) = – 2 and f ¢(x) ³ 4.2 for 1 £ x £ 6
Consider f ¢(x) =
Let f(x) = a1x2 + a2x3 + a3x4
Since lim
Þ f(x) = 2x2 – 2x3 +
Þ f (x + h) – f (x) = f ¢(x) . h ³ (4.2)h
= 2 Þ a1 = 2
So, f (x + h) ³ f (x) + (4.2) h
put x = 1 and h = 5, we get
Hence, f(x) = 2x2 + a2x3 + a3x4
f (6) ³ f (1) + 5(4.2) Þ f (6) ³ 19
f ¢(x) = 4x + 3a2x2 + 4a3x3
As given : f ¢ (1) = 0 adn f ¢(2) = 0
Hence, 4 + 3a2 + 4a3 = 0
...(i)
and 8 + 12a2 + 32a3 = 0
...(ii)
By 4x(i) – (ii), we get
53.
Hence f (6) lies in [19, ¥)
(b) Let f (x) = 3x10 – 7x8 + 5x6 – 21x3 + 3x2 – 7
f ¢ (x) = 30x9 – 56x7 + 30x5 – 63x2 + 6x
f ¢ (1) = 30 – 56 + 30 – 63 + 6
= 66 – 63 – 56 = – 53
f (1 – a ) – f (1)
a®0
a3 +3a
Consider lim
16 + 12a2 + 16a3 – (8 + 12a2 + 32a3) = 0
Þ 8 – 16a3 = 0 Þ a3 = 1/2
f ( x + h) - f ( x )
h
=
and by eqn. (i), 4 + 3a2 + 4/2 = 0 Þ a2 = –2
lim
a®0
f ¢ (1– a )(–1)–0
(By using L’hospital rule)
3a 2 +3
f ¢(1 – 0)(–1)
=
3(0)2 + 3
=
– f ¢ (1) 53
=
3
3
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13
Mathematical
Reasoning
TOPIC Ć
1.
Statement, Truth value of a statement,
Logical Connectives, Truth Table,
Logical Equivalance, Tautology &
Contradiction, Duality
The negation of the Boolean expression p Ú (: p Ù q) is
equivalent to :
[Sep. 06, 2020 (I)]
(a) p Ù : q
(b) : p Ù : q
(c) : p Ú : q
2.
7.
[Jan. 8, 2020 (I)]
8.
9.
(a) ( x Ù y ) Ú ( : x Ù : y ) (b) ( x Ù y ) Ù ( : x Ú : y )
Given the following two statements :
( S1 ) : (q Ú p) ® ( p « ~ q) is a tautology..
4.
(a) q
(b) (~ p ) Ú q
(c) (~ p ) Ù q
(d) (~ p) Ú (~ q)
Let p, q, r be three statements such that the truth value of
If p ® (p Ù ~q) is false, then the truth values of p and q are
(a) F, F
[Jan. 9, 2020 (II)]
(b) T, F
(c) T, T
(b) ~(p Ù ~ q) ® p Ú q
(c) ~(p Ú ~ q) ® p Ù q
(d) ~(p Ú ~ q) ® p Ú q
The logical statement
(b) q
(c) ~p
[Jan. 7, 2020 (I)]
(d) ~q
11.
The Boolean expression ~ ( p Þ (: q)) is equivalent to :
(a) p Ù q
The proposition p ®~ ( p Ù ~ q) is equivalent to :
respectively:
(a) p Ú (~q) ® p Ù q
[April 12, 2019 (II)]
( p Ù q) ® (~ q Ú r ) is F. Then the truth values of p, q, r
are respectively :
[Sep. 03, 2020 (II)]
(a) T, F, T (b) T, T, T (c) F, T, F
(d) T, T, F
6.
Which of the following statements is a tautology?
If the truth value of the statement p ® (~q Ú r) is false (F),
then the truth values of the statements p, q, r are
respectively.
[April 12, 2019 (I)]
(a) T, T, F (b) T, F, F
(c) T, F, T
(d ) F, T, T
(b) q Þ: p (c) p Ú q
(d) (: p) Þ q
12.
Which one of the following Boolean expressions is a
tautology ?
[April 10, 2019 (I)]
(a) (p Ù q) Ú (p Ù ~ q)
(b) (p Ú q) Ú (p Ú ~ q)
(c) (p Ú q) Ù (p Ú ~ q)
(d) (p Ú q) Ù (~p Ú ~ q)
13.
If p Þ(q Ú r) is false, then the truth values of p, q, r are
respectively:
[April 09, 2019 (II)]
(a) F, T, T (b) T, F, F
(c) T, T, F (d) F, F, F
14.
Which one of the following statements is not a tautology?
[Sep. 03, 2020 (I)]
5.
(d) p Ú (p Ù q)
10.
[Sep. 04, 2020 (I)]
both (S1) and (S2) are correct
only (S1) is correct
only (S2) is correct
both (S1) and (S2) are not correct
(c) p Ù (p Ú q)
(a) p
( S2 ) : ~ q Ù (~ p « q) is a fallacy. Then :
(a)
(b)
(c)
(d)
(b) q ® (p Ù (p ® q))
( p Þ q ) ^ ( q Þ : p ) is equivalent to:
(c) ( x Ù : y ) Ú ( : x Ù y ) (d) ( : x Ù y ) Ú ( : x Ù : y )
3.
(a) (p Ù (p ® q)) ® q
[Jan. 8, 2020 (II)]
(d) : p Ú q
The negation of the Boolean expression x «: y is
equivalent to:
[Sep. 05, 2020 (I)]
Which one of the following is a tautology?
[April 08, 2019 (II)]
(a) (p Ú q) ® (p Ú (~ q)) (b) (p Ù q) ® (~ p) Ú q
(c) p ® (p Ú q)
(d) (p Ù q) ® p
(d) F, T
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EBD_8344
M-220
Mathematics
15 . The Boolean expression
25.
The following statement
(p ® q) ® [(~p ® q) ® q] is :
[2017]
(a) a fallacy
(b) a tautology
(c) equivalent to ~ p ® q (d) equivalent to p ® ~q
26.
The proposition (~p) Ú (p Ù ~ q)
[Online April 8, 2017]
( ( p Ù q ) Ú ( p Ú : q ) ) Ù ( : p Ù : q ) is equivalent to :
[Jan. 12, 2019 (I)]
16.
17.
(a) p Ù q
(b) p Ù ( : q )
(c ) ( : p ) Ù ( : q )
(d) p Ú ( : q )
The expression : ( : p ® q) is logically equivalent to :
[Jan. 12, 2019 (II)]
(a) : p Ù : q
(b) p Ù : q
(d) p Ù q
(c) : p Ù q
If q is false and p ^ q « r is true, then which one of the
following statements is a tautology? [Jan. 11, 2019 (I)]
(a)
18.
( p Ú r) ® (p Ù r)
( p Ù r) ® (p Ú r)
(c) p Ù r
(d) p Ú r
Consider the following three statements:
P : 5 is a prime number.
Q : 7 is a factor of 192.
R : L.C.M. of 5 and 7 is 35.
Then the truth value of which one of the following
statements is true?
[Jan. 10, 2019 (II)]
(a) (~ P) Ú (Q Ù R)
(b) (P Ù Q) Ú (~ R)
(c) (~ P) Ù (~ Q Ù R)
19.
(b)
(a)
20.
21.
22.
( Ú , Ù)
(b)
(Ú , Ú)
( Ù , Ú)
(d)
The negation of ~ s Ú (~ r Ù s) is equivalent to :
29.
If (pÙ ~ q) Ù (p Ù r) ® ~ p Ú q is false, then the truth values
of p, q and r are respectively
[Online April 15, 2018]
(a) F, T, F (b) T, F, T (c) F, F, F
(d) T, T, T
24.
Which of the following is a tautology?
[2017]
(a) (~ p) Ù (p Ú q) ® q
(b) (q ® p)Ú ~ (p ® q)
(c) (~ q) Ú (p Ù q) ® q
(d) (p ® q) Ù (q ® p)
(b) s Ù r
(c) s Ù ~ r
(d) s Ù (r Ù ~ s)
The statement : ( p « : q ) is:
[2014]
(b) a fallacy
(c) eqivalent to p « q
(d) equivalent to : p « q
30.
Let p, q, r denote arbitrary statements. Then the logically
equivalent of the statement p Þ ( q Ú r ) is:
[Online April 12, 2014]
31.
32.
~ (pÚ q) Ú (~ p Ù q) is equivalent to :
[2018]
(a) p
(b) q
(c) ~q
(d) ~p
If p ® (~ p Ú ~ q) is false, then the truth values of p and q
are respectively.
[Online April 16, 2018]
(a) T, F
(b) F, F
(c) F, T
(d) T, T
23.
(a) s Ú (r Ú ~ s)
[2015]
(a) a tautology
( Ù , Ù)
The logical statement
[~ (~ p Ú q) Ú (p Ù r)] Ù (~ p Ù r)
is equivalent to:
[Jan. 09, 2019 (II)]
(a) (~ p Ù ~ q) Ù r
(b) ~ p Ú r
(c) (p Ù r) Ù ~ q
(d) (p Ù ~ q) Ú r
The Boolean expression
(d) p Ú ( ~ q )
28.
[Jan. 09, 2019 (I)]
(c)
(c) q ® p
The Boolean Expression (p Ù : q) Ú qÚ (: p Ù q) is
equivalent to:
[2016]
(a) p Ú q (b) p Ú : q (c) : p Ù q (d) p Ù q
equivalent to p Ù q , where Å, e Î {Ù , Ú} then the
ordered pair ( Å, e ) is:
(b) p Ù ( ~ q )
27.
(d) P Ú (~ Q Ù R)
If the Boolean expression (p Å q) Ù ( : p e q ) is
(a) p ® ~ q
33.
(a)
( p Ú q) Þ r
(b)
( p Þ q) Ú ( p Þ r )
(c)
( p Þ~ q ) Ù ( p Þ r )
(d)
( p Þ q) Ù ( p Þ ~ r)
The proposition : ( p Ú : q ) Ú : ( p Ú q ) is logically
equivalent to:
[Online April 11, 2014]
(a) p
(b) q
(c) : p
(d) : q
Consider
Statement-1 : (p ^ ~ q) ^ (~ p ^ q) is a fallacy.
Statement-2 : (p ® q) « (~ q ® ~ p) is a tautology.
[2013]
(a) Statement-1 is true; Statement-2 is true;
Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true; Statement-2 is true; Statement-2
is not a correct explanation for Statement-1.
(c) Statement-1 is true; Statement-2 is false.
(d) Statement-1 is false; Statement-2 is true.
Let p and q be any two logical statements and
r : p ® (: p Ú q) . If r has a truth value F, then the truth
values of p and q are respectively :
[Online April 25, 2013]
(a) F, F
(b) T, T
(c) T, F
(d) F, T
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M-221
Mathematical Reasoning
34.
35.
36.
For integers m and n, both greater than 1, consider the
following three statements :
P : m divides n
Q : m divides n2
R : m is prime,
then
[Online April 23, 2013]
(a) Q Ù R ® P
(b) P Ù Q ® R
(c) Q ® R
(d) Q ® P
42.
The statement p ® (q ® p) is equivalent to :
[Online April 22, 2013]
(a) p ® q
(b) p ® ( p Ú q)
(c) p ® ( p ® q)
(d) p ® ( p Ù q)
Statement-1: The statement A ® ( B ® A) is equivalent
TOPIC n
to A ® ( A Ú B ) .
37.
38.
39.
Statement-2: The statement ~ [(A Ù B) ® (~ A Ú B)] is a
Tautology.
[Online April 9, 2013]
(a) Statement-1 is false; Statement-2 is true.
(b) Statement-1 is true; Statement-2 is true; Statement- 2 is
not correct explanation for Statement-1.
(c) Statement-1 is true; Statement-2 is false.
(d) Statement-1 is true; Statement-2 is true; Statement- 2 is
the correct explanation for Statement-1.
Let p and q be two Statements. Amongst the following,
the Statement that is equivalent to p ® q is
[Online May 19, 2012]
(a) pÙ ~ q (b) ~ p Ú q (c) ~ p Ù q (d) pÚ ~ q
The logically equivalent preposition of p Û q is
[Online May 12, 2012]
(a)
( p Þ q) Ù ( q Þ p)
(b) p Ù q
(c)
( p Ù q) Ú ( q Þ p)
(d)
( p Ù q) Þ ( q Ú p)
Consider the statement: “For an integer n, if n3 – 1 is even,
then n is odd.” The contrapositive statement of this statement is:
[Sep. 06, 2020 (II)]
(a) For an integer n, if n is even, then n 3 – 1 is odd.
(b) For an intetger n, if n3 – 1 is not even, then n is not
odd.
(c) For an integer n, if n is even, then n 3 – 1 is even.
(d) For an integer n, if n is odd, then n 3 – 1 is even.
44.
The statement ( p ® (q ® p)) ® ( p ® ( p Ú q)) is :
[Sep. 05, 2020 (II)]
(a) equivalent to ( p Ù q) Ú (~ q)
(b) a contradiction
45.
(b) A Ú (A Ù B)
(c) [A Ù (A ® B)] ® B (d) B ® [A Ù (A ® B)]
40.
Statement-1 : ~ ( p «~ q) is equivalent to p « q .
Statement-2 : ~ ( p «~ q) is a tantology
41.
[2009]
(a) Statement-1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is true,
Statement-2 is a correct explanation for statement -1
The statement p ® (q®p) is equivalent to
[2008]
(a) p ® (p® q)
(b) p ® (p Ú q)
(c) p ® (p Ù q)
(d) p ® (p «q)
Converse, Inverse & Contrapositive
of the Conditional Statement,
Negative of a Compound Statement,
Algebra of Statement
43.
The only statement among the following that is a tautology
is
[2011RS]
(a) A Ù (A Ú B)
Let p be the statement “x is an irrational number”, q be the
statement “y is a transcendental number”, and r be the
statement “ x is a rational number iff y is a transcendental
number”.
[2008]
Statement-1 : r is equivalent to either q or p
Statement-2 : r is equivalent to ~(p«~q).
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2
is a correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2
is not a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
46.
(c) equivalent to ( p Ú q) Ù (~ p)
(d) a tautology
Contrapositive of the statement :
'If a function f is differentiable at a, then it is also
continuous at a', is :
[Sep. 04, 2020 (II)]
(a) If a function f is continuous at a, then it is not
differentiable at a.
(b) If a function f is not continuous at a, then it is not
differentiable at a.
(c) If a function f is not continuous at a, then it is
differentiable at a
(d) If a function f is continuous at a, then it is differentiable
at a.
The contrapositive of the statement "If I reach the station
in time, then I will catch the train" is : [Sep. 02, 2020 (I)]
(a) If I do not reach the station in time, then I will catch
the train.
(b) If I do not reach the station in time, then I will not
catch the train.
(c) If I will catch the train, then I reach the station in time.
(d) If I will not catch the train, then I do not reach the
station in time.
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EBD_8344
M-222
47.
Mathematics
53.
Negation of the statement:
5 is an integer of 5 is irrational is:
48.
[Jan. 9, 2020 (I)]
(a)
5 is not an integer or 5 is not irrational
(b)
5 is not an integer and 5 is not irrational
(c)
5 is irrational or 5 is an integer..
(d)
5 is an integer and 5 is irrational
the equation 2 sin
contrapositive statement of “If A Í B and B Í D, then
[Jan. 7, 2020 (II)]
(a) If A Í C, then A Í B and B Í D
54.
(b) If A Í C, then B Ì A or D Ì B
(c) If A Í C, then A Í B and B Í D
(d) If A Í C, then A Í B or B Í D
49.
50.
51.
52.
The negation of the Boolean expression ~ s Ú (~ r Ù s) is
equivalent to :
[April 10, 2019 (II)]
(a) ~ s Ù ~ r
(b) r
(c) s Ú r
(d) s Ù r
q
= 1 + sin q - 1 - sin q.
2
Statement q:
The angles A, B, C and D of any quadrilateral ABCD satisfy
Let A, B, C and D be four non-empty sets. The
A Í C ” is:
Consider the following two statements.
Statement p:
The value of sin 120 can be divided by taking q = 240 in
For any two statements p and q, the negation of the
expression p Ú (~ p Ù q) is:
[April 9, 2019 (I)]
(a) ~ p Ù ~ q
(b) p Ù q
(c) p « q
(d) ~ p Ú ~ q
The contrapositive of the statement "If you are born in
India, then you are a citi en of India", is :
[April 8, 2019 (I)]
(a) If you are not a citi en of India, then you are not born
in India.
(b) If you are a citi en of India, then you are born in
India.
(c) If you are born in India, then you are not a citi en of
India.
(d) If you are not born in India, then you are not a citi en
of India.
Contrapositive of the statement “If two numbers are not
equal, then their squares are not equal”. is :
[Jan. 11, 2019 (II)]
(a) If the squares of two numbers are not equal, then the
numbers are equal.
(b) If the squares of two numbers are equal, then the
numbers are not equal.
(c) If the squares of two numbers are equal, then the
numbers are equal.
(d) If the squares of two numbers are not equal, then the
numbers are not equal.
55.
56.
57.
æ1
ö
æ1
ö
the equation cos ç ( A + C ) ÷ + cos ç ( B + D) ÷ = 0
è2
ø
è2
ø
Then the truth values of p and q are respectively.
[Online April 15, 2018]
(a) F, T
(b) T, T
(c) F, F
(d) T, F
Contrapositive of the statement
‘If two numbers are not equal, then their squares are not
equal’, is :
[Online April 9, 2017]
(a) If the squares of two numbers are equal, then the
numbers are equal.
(b) If the squares of two numbers are equal, then the
numbers are not equal.
(c) If the squares of two numbers are not equal, then the
numbers are not equal.
(d) If the squares of two numbers are not equal, then the
numbers are equal.
The contrapositive of the following statement,
"If the side of a square doubles, then its area increases
four times", is :
[Online April 10, 2016]
(a) If the area of a square increases four times, then its
side is not doubled.
(b) If the area of a square increases four times, then its
side is doubled.
(c) If the area of a square does not increases four times,
then its side is not doubled.
(d) If the side of a square is not doubled, then its area
does not increase four times.
Consider the following two statements :
P : If 7 is an odd number, then 7 is divisible by 2.
Q : If 7 is a prime number, then 7 is an odd number.
If V1 is the truth value of the contrapositive of P and V 2 is
the truth value of contrapositive of Q, then the ordered
pair (V1, V2) equals:
[Online April 9, 2016]
(a) (F, F) (b) (F, T)
(c) (T, F)
(d) (T, T)
Consider the following statements :
P : Suman is brilliant
Q : Suman is rich.
R : Suman is honest
the negation of the statement
“Suman is brilliant and dishonest if and only if suman is
rich” can be equivalently expressed as :
[Online April 11, 2015]
(a) ~ Q « ~ P Ú R
(b) ~ Q « ~ P Ù R
(c) ~ Q « P Ú ~ R
(d) ~ Q « P Ù ~ R
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M-223
Mathematical Reasoning
58.
59.
60.
61.
The contrapositive of the statement “If it is raining, then
I will not come”, is :
[Online April 10, 2015]
(a) If I will not come, then it is raining.
(b) If I will not come, then it is not raining.
(c) If I will come, then it is raining.
(d) If I will come, then it is not raining.
The contrapositive of the statement “if I am not feeling
well, then I will go to the doctor” is
[Online April 19, 2014]
(a) If I am feeling well, then I will not go to the doctor
(b) If I will go to the doctor, then I am feeling well
(c) If I will not go to the doctor, then I am feeling well
(d) If I will go to the doctor, then I am not feeling well.
The contrapositive of the statement “I go to school if it
does not rain” is
[Online April 9, 2014]
(a) If it rains, I do not go to school.
(b) If I do not go to school, it rains.
(c) If it rains, I go to school.
(d) If I go to school, it rains.
The negation of the statement
"If I become a teacher, then I will open a school", is :
[2012]
(a) I will become a teacher and I will not open a school.
(b) Either I will not become a teacher or I will not open a
school.
(c) Neither I will become a teacher nor I will open a school.
(d) I will not become a teacher or I will open a school.
62.
63.
64.
Let p and q denote the following statements
p : The sun is shining
q: I shall play tennis in the afternoon
The negation of the statement “If the sun is shining then I
shall play tennis in the afternoon”, is
[Online May 26, 2012]
(a) q Þ: p
(b) qÙ : p
(c) p Ù : q
(d) : q Þ: p
The Statement that is TRUE among the following is
[Online May 7, 2012]
(a) The contrapositive of 3x + 2 = 8 Þ x = 2 is x ¹ 2
Þ 3x + 2 ¹ 8.
(b) The converse of tanx = 0 Þ x = 0 is x ¹ 0 Þ tan x = 0.
(c) p Þ q is equivalent to pÚ : q .
(d) p Ú q and p Ù q have the same truth table.
Let S be a non-empty subset of R. Consider the following
statement :
P : There is a rational number x Î S such that
x > 0.
Which of the following statements is the negation of the
statement P ?
[2010]
(a) There is no rational number x Î S such than x < 0.
(b) Every rational number x Î S satisfies x < 0.
(c) x Î S and x < 0 Þ x is not rational.
(d) There is a rational number x Î S such that x < 0.
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EBD_8344
M-224
Mathematics
(b) Negation of given statement = ~ ( p Ú (~ p Ù q))
1.
2.
(b)
= ~ p Ù ~ (~ p Ù q) = ~ p Ù ( p Ú ~ q)
p
q
~q
p^ ~ q
~p
= (~ p Ù q) Ú (~ p Ù ~ q)
T
T
F
F
F
T
T
= F Ú (~ pÙ ~ q) = ~ p Ù ~ q
T
F
T
T
F
F
F
(a) p : x « ~ y = ( x ®~ y ) Ù (~ y ® x )
F
T
F
F
T
T
T
= (~ x Ú ~ y ) Ù ( y Ú x )
F
F
T
F
T
T
T
=~ ( x Ù y ) Ù ( x Ú y )
Negation of p is
(Q~ ( x Ù y ) = ~ xÚ ~ y)
(d) ( p Ù q) ® (~ q Ú r )
= ~ ( p Ù q ) Ú (~ q Ú r )
p
q
~p ~q q Ú p p« ~q (S 1 ) ~p« q (S 2 )
T
T
F
F
T
F
F
F
F
T
F
F
T
T
T
T
T
T
F
T
T
F
T
T
T
T
F
F
F
T
T
F
F
T
F
F
= (~ p Ú ~ q ) Ú (~ q Ú r )
= (~ p Ú ~ q Ú r )
Q (~ pÚ ~ q Ú r ) is false, then ~p, ~q and r all these must
be false.
Þ p is true, q is true and r is false.
6.
(c)
\ S1 is not tautology and
S2 is not fallacy.
Hence, both the statements (S1) and (S2) are not correct.
7.
F
9.
p
q
~q
p Ù ~q
T
T
F
F
p ® (p Ù ~ q)
F
T
F
T
T
T
F
T
F
F
T
F
F
T
F
T
(a)
p®q
T
F
T
p q
T T
T F
F T
8.
p ® ~ (p^ ~ q) ~ p q
\ p ®~ ( p Ù ~ q) is equivalent to ~ p Ú q
5.
~ p = ( x Ù y )Ú ~ ( x Ú y ) = ( x Ù y ) Ú (~ x Ù ~ y )
(d) The truth table of both the statements is
3.
4.
F
p Ù ( p ® q ) ( p Ù ( p ® q )) ® q q ® p Ù ( p ® q )
T
T
T
F
T
T
F
T
F
T
F
T
(d) (~ p Ù q) ® (p Ú q)
Þ ~ {(~ p Ù q) Ù (~ p Ù ~ q)}
Þ ~ {~ p Ù f }
(c)
p
T
T
F
q
T
F
T
F
F
10.
p Þ q ~ p q Þ ~ p ( p Þ q) Ù ( p Þ ~ q)
T
F
F
F
F
F
T
F
T
T
T
T
T
T
T
T
T
Clearly (p Þ q) Ù (q Þ ~p) is equivalent to ~p
11.
12.
pÙq
T
F
F
p Ú ( p Ù q)
T
T
F
pÚq
T
T
T
p Ù ( p Ú q)
T
T
F
F
F
F
F
(a) Given statement p ® (~ q Ú r) is False.
Þ p is True and ~ q Ú r is False
Þ p is True and ~ q is False and r is False
\ truth values of p, q r are T, T, F respectively.
(a) Given Boolean expression is,
~ (p Þ (~ q))
{Q p Þ q is same as ~ p Ú q}
º ~ ((~ p) Ú (~ q)) º p Ù q
(b)
(p Ú q) Ú (pÚ ~ q) = p Ú (q Ú p)Ú ~ q
= (p Ú p) Ú (qÚ ~ q) = p Ú T = T
Hence first statement is tautology.
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M-225
Mathematical Reasoning
13.
14.
(b) For p Þ q Ú r to be F.
r should be F and p Þ q should be F
for p Þ q to be F, p Þ T and q Þ F
p, q, r º T, F, F
(a) By truth table :
p q
T T
T F
~q
F
T
F T
F F
F
T
pv ~ q ~ p
T
F
T
F
F
T
pÙ ~ q
F
T
pvq
T
T
p ® pvq
T
T
F
F
T
F
T
T
T
T
18.
p Ù q ( p Ù q ) ® p ~ pvq
T
T
T
F
T
F
F
F
T
T
T
T
( p Ù q ) ® (~ p) Ú q (p Ú q) ® (p Ú (~ q))
T
T
15.
16.
17.
23.
T
T
T
F
T
T
19.
(c) Consider the Boolean expression
((p Ù q) Ú (pv ~ q)) Ù (~ pÙ ~ q)
= (pÚ ~ q) Ù (~ pÙ ~ q)
= ((pÚ ~ q)Ù ~ p) Ù ((pÚ ~ q)Ù ~ q)
= ((pÙ ~ p) Ú (~ qÙ ~ p))Ù ~ q
= (~ pÙ ~q)Ù ~ q = (~ pÙ ~q)
(a) ~ (~ p ® Q) º ~ (p Ú q) º ~ p Ù ~ q
(b) q is false and [(p Ù q) « r] is true
As (p Ù q) is false
[False « r] is true
Hence r is false
Option (a): says p Ú r,
Since r is false
Hence (p Ú r) can either be true or false
Option (b): says (p Ù r) ® (p Ú r)
(p Ù r) is false
Since, F ® T is true and
F ® F is also true
Hence, it is a tautology
Option (c): (p Ú r) ® (p Ù r)
i.e. (p Ú r) ® F
20.
21.
It can either be true or false
Option (d): (p Ù r),
Since, r is false
Hence, (p Ù r) is false.
(d) P is True, Q is False and R is True
(a) (~ P) Ú (Q Ù R) º F Ú (F Ù T) º F Ú F = F
(b) (P Ù Q) Ú (~R) º (T Ù F) Ú (F) º F Ú F = F
(c) (~ P) Ù (~ Q Ù R) º F Ù (T Ù T) º F Ù T = F
(d) P Ú (~Q Ù R) º T Ú (T Ù T) º T Ú T = T
(c) Check each option
(a) (p Ú q) Ù (~p Ù q) = (~p Ù q)
(b) (p Ú q) Ù (~p Ú q) = q
(c) (p Ù q) Ù (~p Ú q) = p Ù q
(d) (p Ù q) Ù (~p Ù q) = F
(c) Logical statement,
= [~ (~ p Ú q) Ú (p Ù r)] Ù (~ q Ù r)
= [(p Ù ~ q) Ú (p Ù r)] Ù (~ q Ù r)
= [(p Ù ~ q) Ù (~ q Ù r)] Ú [(p Ù r) Ù (~ q Ù r)]
= [p Ù ~ q Ù r] Ú [p Ù r Ù ~ q]
= (p Ù ~ q ) Ù r
= (p Ù r) Ù ~ q
(d)
: (pÚ q) Ú (: p Ù q)
Þ ( : p Ù : q) Ú (: p Ù q)
Þ : pÙ (: q Ú q)
22.
Þ : pÙ t º : p
(d)
p
q
~p ~q ~pÚ~q
p ® (~ p Ú ~ q)
T
T
F
F
F
F
T
F
F
T
T
T
F
T
T
F
T
T
F
F
T
T
T
T
From the truth table,
p ® (~ p Ú ~ q) is false only when p and q both are true.
(b) As the truth table for the (pÙ ~ q) Ù (p Ù r) ® ~ p Ú q is false, then only possible values of (p, q, r) is (T, F, T)
p
q
r
~q
p Ù ~q
pÙ r
~p
~p Ú q
(p Ù ~q) Ù (p Ù r)
(p Ù ~q) Ù (p Ù r) ® ~p Ú q
T
T
T
F
F
T
F
T
F
T
T
F
T
T
T
T
F
F
T
F
T
T
F
F
F
F
F
T
F
T
F
T
T
F
F
F
T
T
F
T
F
F
T
T
F
F
T
T
F
T
F
T
F
F
F
F
T
T
F
T
T
F
F
T
T
F
F
F
F
T
F
F
F
T
F
F
T
T
F
T
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EBD_8344
M-226
Mathematics
(a) Truth table
24.
30.
p Þ (q Ú r ) which is equivalent to
p
q
~p
T
T
F
T
F
T
T
F
F
T
F
T
º (~ p Ú q) Ú (~ pÚ ~ q )
F
T
T
T
T
T
F
F
T
F
F
T
º ~ p Ú (qÚ ~ q)
º~ p
(b) Statement-2 : (p ® q) « (~q ® ~p)
º (p ® q) « (p ® q)
which is always true.
So, statement 2 is true
Statement-1: (p ^ ~q) ^ (~p ^ q)
= p ^ ~q ^ ~p ^ q
= p ^ ~p ^ ~q ^ q
= f^f=f
So statement-1 is true
(c) p ® (~ p Ú q) has truth value F.
It means p ® (~ p Ú q) is false.
It means p is true and ~ p Ú q is false.
Þ p is true and both ~ p and q are false.
(~p) ^ (p q) (~p) ^ (p q) ® q
p
( p Þ q) Ú ( p Þ r )
31.
32.
\ (a) ~ p Ù (p Ú q) ® q be a tautology
Other options are not tautology.
25.
(b) We have
p q ~ p p ® q ~ p ® q (~ p ® q) ® q (p ® q) ® (( ~p ®q)® q)
T F F
F
T
F
T
T T F
F F T
T
T
T
F
T
T
T
T
F T T
T
T
T
T
\ It is tautology.
26.
33.
(b) (~p) Ú (p Ù ~ q)
p q ~ p ~ q pÙ ~ q
T T F
F
F
T F F
T
T
F T T
F
F
F
27.
(b) Given statement is
F
T
T
( ~ p ) Ú (p Ù ~ q )
F
T
F
F
34.
F
(c) Given ~ ( p Ú ~ q)Ú ~ ( p Ú q)
Þ p is true and q is false.
(a)
(b)
(a) (pÙ : q) Ú q Ú (: p Ù q)
Hence P Ù Q ® R, false
Þ {(p Ú q) Ù (: q Ú q)} Ú (: p Ù q)
Þ {(p Ú q) Ù T} Ú (: p Ù q)
28.
29.
8
64
= 2,
= 16 ; but 4 is not prime.
4
4
(6)2 36
=
= 3 ; but 12 is not prime
12
12
Þ (p Ú q) Ú (: p Ù q)
(c)
Þ {(p Ú q) Ú : p} Ù (p Ú q Ú q)
Þ T Ù (p Ú q)
Hence Q ® R, false
Þ pÚq
(b) :[:sÚ(:r Ù s)]
= sÙ:(:r Ù s)
= sÙ(r Ú:s)
= (s Ù r) Ú (s Ù: s)
= (s Ù r) Ú f
=sÙr
(d)
35.
(i)
(c)
p q
F F
F T
T F
T
T
~q
T
F
T
F
p « ~ q ~ ( p « ~ q)
F
T
T
F
T
F
F
T
(ii)
p«q
T
F
F
T
4
(4) 2 16
=
= 2 ; is not an integer
8
8
8
Hence Q ® P, false
(b)
q p q® p
T T
T
T F
F
F T
T
F
F
T
p ® (q ® p)
T
T
T
pÚq
T
T
T
p ® ( p Ú q)
T
T
T
T
F
T
Since truth value of p ® (q ® p) and
p ® (p Ú q) are same, hence p ® (q ® p) is equivalent
to p ® (p Ú q).
From column (i) and (ii) are equivalent.
Clearly equivalent to p « q
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M-227
Mathematical Reasoning
36.
(c)
A B ~ A AÙB ~AÚB
T
T
F
F
39.
T
F
T
F
F
F
T
T
T
F
F
F
T
F
T
T
( A Ù B) ® ~ [( A Ù B) ®
(~ A Ú B) (~ A Ú B )]
T
F
T
F
T
F
T
F
37. (b) Let p and q be two statements.
p ® q is equivalent to : p Ú q .
38. (a)
( p Þ q ) Ù ( q Þ p) means
pÛq
(c) Truth table of all options is as follows.
A
B A Ú B A Ù B A Ù (A Ú B) A Ú (A Ù B) A ® B A Ù (A ® B) [A Ù (A ® B) ®B]
T
F
T
F
T
T
F
F
T
[B ® [A Ù (A ®B)]
T
F
T
T
F
F
F
T
F
T
F
T
T
T
T
T
T
T
T
T
T
F
F
F
F
F
F
T
F
T
T
\ It is tautology.
40.
(b) The truth table for the logical statements, involved in
statement 1, is as follows :
(i)
(ii)
p
q
:q
T
T
F
F
T
F
T
F
F
T
F
T
p «: q : ( p « : q )
F
T
T
F
43.
(a) Contrapositive statement will be
"For an integer n, if n is not odd then n3 – 1 is not even".
or
"For an integer n, if n is even then n3 – 1 is odd".
44.
(d) The truth table of ( p ® (q ® p)) ® ( p ® ( p Ú q))
is
p«q
T
F
F
T
T
F
F
T
p
We observe the columns (i) and (ii) are identical, therefore
~(p « ~q) is equivalent to p « q
But ~ (p « ~q) is not a tautology as all entries in its column
are not T.
\ Statement-1 is true but statement-2 is false.
41. (b) The truth table for the given statements, as follows :
p
T
T
F
F
q pÚ q q® p
T T
T
F T
T
T T
F
F F
T
p ® (q® p)
T
T
T
T
p ® (pÚ q)
T
T
T
T
From table we observe that
p ® (q®p) is equivalent to p®(pÚq)
42. (None)
Given that
p : x is an irrational number
q : y is a transcendental number
r : x is a rational number iff y is a transcendental number.
clearly r : : p « q
Truth table to check the equivalence of ‘r’ and ‘q or p’; ‘r’
and : ( p «: q)
p
(i)
(ii)
(iii)
q ~p ~ q ~p « q q or p p« ~ q ~ (p« ~ q)
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
F
T
T
F
T
T
T
F
F
T
T
F
T
F
F
T
From columns (i), (ii) and (iii), we observe, that none of the
these statements are equivalent to each other.
\ Statement 1as well as statement 2 both are false.
\ None of the options is correct.
q pÚ q p® (pÚ q) q® p p® (q® p)
(p® (q® p))
® (p® (pÚ q))
T T
O
T
T
T
T
T
F
T
T
T
T
T
F T
T
T
F
T
T
F
F
T
T
T
T
F
45.
Hence, the statement is tautology.
(b) Contrapositive statement will be "If a function is not
continuous at 'a', then it is not differentiable at 'a'.
46.
(d) Contrapositive of p ® q is ~ q ® ~ p
i.e. contrapositive of 'if p then q' is 'if not q then not p'.
47. (b) Let p and q the statements such that p = 5 is an
integer q = 5 is an irrational number.
Then, negation of the given statement
5 is not an integer and 5 is not an irrational Number
~ (p Ú q) = ~ p Ù ~ q
48. (d) Let P = A Í B, Q = B Í D, R = A Í C
Contrapositive of (P Ù Q) ® R is ~ R ® ~ (P Ù Q)
~R®~ PÚ~Q
49. (d) ~ s Ú (~ r Ù s) º (~ s Ú ~ r) Ù (~ s Ú s)
º (~ s Ú ~ r)
(Q ~ s Ú s) is tautology)
º ~ (s Ù r)
Hence, its negation is s Ù r.
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EBD_8344
M-228
50.
51.
52.
53.
Mathematics
(d) ~ (p Ú (~p Ù q)) = ~ (~p Ù q) Ù ~p
= (~q Ú p) Ù ~p
= ~p Ù (p Ú ~q)
= (~q Ù ~p) Ú (p Ù ~p)
= (~p Ù ~q)
(a) S: ‘’If you are born in India, then you are a citi en of
India.’’
Contrapositive of p ® q is ~ q ® ~ p
So contrapositive of statement S will be :
‘’If you are not a citi en of India, then you are not born in
India.’’
(c) Contrapositive of “If A then B” is “If ~B then ~A”.
Hence contrapositive of “If two numbers are not equal,
then their squares are not equal” is “If squares of two
numbers are equal, then the two numbers are equal”.
(a) Statement p:
sin 120 = cos 30 =
57.
58.
59.
60.
3
Þ 2 sin 120° = 3
2
So, 1 + sin 240° - 1 - sin 240°
1- 3
1+ 3
¹ 3
2
2
Statement q:
=
So, A + B + C + D = 2p Þ
61.
A+C B+ D
+
=p
2
2
æ A+Cö
æB+Dö
Þ cos ç
÷ + cos ç 2 ÷
è 2 ø
è
ø
54.
55.
56.
æ A+Cö
æ A+Cö
- cos ç
= cos ç
÷
÷=0
è 2 ø
è 2 ø
Therefore, statement p is false and statement q is true.
(a) p ® q
then ~ q ® ~ p
\ If the square of two numbers are equal, then the
numbers are equal.
(c) Contrapositive of p ® q is given by ~ q ® ~ p
So (c) is the right option.
(a) Contrapositive of P :
T is not divisible by 2 Þ T is not odd number
T Þ F : F (V1)
62.
Contra positive Q :
T is not odd number Þ T is not a prime number
F Þ F : T (V2)
(d) Suman is brilliant and dishonest can be expressed as
PÙ ~ R
therefore given statement is equal to ( PÙ ~ R ) « Q
Negation of the above statement is ~ Q « PÙ ~ R
(d) The centre positive of the statement is “If i will come,
then it is not raining”.
(c) Given statement can be written in implication form as
I am not feeling well Þ I will go to the doctor.
Contrapositive form :
I will not go to the doctor Þ I am feeling well.
i.e. If I will not go to the doctor, then I am feeling well.
(b) let p = If it does not rain
q = I go to school
According to law of contrapositive
p Þ q º ~q Þ ~ p
i.e. ~q = I do not go to school
~p = It rains
~q Þ ~p is If I do not go to school, it rains.
(a) Let p : I become a teacher.
q : I will open a school
Negation of p ® q is ~ (p ® q) = p ^ ~q
i.e. I will become a teacher and I will not open a school.
(c) Let p : The sun is shining.
q : I shall play tennis in the afternoon.
Negation of p ® q is : ( p ® q ) = p Ù : q
63.
(a) Only statement given in option
(a) is true.
(b) The converse of tanx = 0 Þ x = 0 is
x = 0 Þ tan x = 0
\ Statement (b) is false
(c) : ( p Þ q) is equivalent to pÙ: q
\ Statement given in option (c) is false.
(d) No, p Ú q and p Ù q does not have the same truth
value.
64. (b) Given that P : there is a rational number x Î S such
that x > 0 .
~ P : Every rational number x Î S satisfies x £ 0 .
Downloaded from @Freebooksforjeeneet
M-229
Statistics
14
Statistics
7.
TOPIC Ć
1.
Arithmetic Mean, Geometric Mean,
Harmonic Mean, Median & Mode
Consider the data on x taking the values 0, 2, 4, 8, ..., 2n with
frequencies nC0, nC1, nC2, ..., nCn respectively. If the mean of
8.
this data is 728 , then n is equal to ______.
2n
[NA Sep. 06, 2020 (II)]
2.
The minimum value of 2sin x + 2cos x is : [Sep. 04, 2020 (II)]
-1+
3.
1
2
1-
Marks
5.
6.
9.
(b) 2 -1+ 2 (c) 21- 2
(a) 2
(d) 2
If for some x Î R, the frequency distribution of the marks
obtained by 20 students in a test is :
2
3
5
Frequency (x+1)2 2x – 5 x2 – 3x
4.
1
2
7
x
then the mean of the marks is :
[April 10, 2019 (I)]
(a) 3.2
(b) 3.0
(c) 2.5
(d) 2.8
The mean and the median of the following ten numbers in
increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and
y
35 respectively, then
is equal to: [April. 09, 2019 (II)]
x
(a) 9/4
(b) 7/2
(c) 8/3
(d) 7/3
The mean of a set of 30 observations is 75. If each other
observation is multiplied by a non- ero number l and then
each of them is decreased by 25, their mean remains the
same. The l is equal to
[Online April 15, 2018]
10
4
1
2
(a)
(b)
(c)
(d)
3
3
3
3
The mean age of 25 teachers in a school is 40 years. A
teacher retires at the age of 60 years and a new teacher is
appointed in his place. If now the mean age of the teachers
in this school is 39 years, then the age (in years) of the
newly appointed teacher is :
[Online April 8, 2017]
(a) 25
(b) 30
(c) 35
(d) 40
10.
11.
12.
13.
The mean of the data set comprising of 16 observations is
16. If one of the observation valued 16 is deleted and three
new observations valued 3, 4 and 5 are added to the data,
then the mean of the resultant data, is:
[2015]
(a) 15.8
(b) 14.0
(c) 16.8
(d) 16.0
Let the sum of the first three terms of an A. P, be 39 and the
sum of its last four terms be 178. If the first term of this
A.P. is 10, then the median of the A.P. is :
[Online April 10, 2015]
(a) 28
(b) 26.5
(c) 29.5
(d) 31
A factory is operating in two shifts, day and night, with 70
and 30 workers respectively. If per day mean wage of the
day shift workers is ` 54 and per day mean wage of all the
workers is ` 60, then per day mean wage of the night shift
workers (in `) is :
[Online April 10, 2015]
(a) 69
(b) 66
(c) 74
(d) 75
In a set of 2n distinct observations, each of the
observations below the median of all the observations is
increased by 5 and each of the remaining observations is
decreased by 3. Then the mean of the new set of
observations:
[Online April 9, 2014]
(a) increases by 1
(b) decreases by 1
(c) decreases by 2
(d) increases by 2
If the median and the range of four numbers
{x, y, 2x + y, x – y}, where 0 < y < x < 2y, are 10
and 28 respectively, then the mean of the numbers is :
[Online April 23, 2013]
(a) 18
(b) 10
(c) 5
(d) 14
The mean of a data set consisting of 20 observations is 40.
If one observation 53 was wrongly recorded as 33, then
the correct mean will be :
[Online April 9, 2013]
(a) 41
(b) 49
(c) 40.5
(d) 42.5
The median of 100 observations grouped in classes of
equal width is 25. If the median class interval is 20 - 30 and
the number of observations less than 20 is 45, then the
frequency of median class is
[Online May 19, 2012]
(a) 10
(b) 20
(c) 15
(d) 12
Downloaded from @Freebooksforjeeneet
EBD_8344
M-230
14.
Mathematics
The frequency distribution of daily working expenditure
of families in a locality is as follows:
Expenditure 0-50 50-100 100-150 150-200 200-250
in `. (x ):
No. of
24
33
37
25
b
families (f ):
21.
22.
If the mode of the distribution is ` 140, then the value of b
is
[Online May 7, 2012]
(a) 34
(b) 31
(c) 26
(d) 36
15.
The average marks of boys in class is 52 and that of girls is
42. The average marks of boys and girls combined is 50.
The percentage of boys in the class is
[2007]
(a) 80
16.
18.
19.
(c) 40
= 400 and
å xi
23.
(d) 20
= 80. Then the possible value of n
among the following is
[2005]
(a) 15
(b) 18
(c) 9
(d) 12
If in a frequency distribution, the mean and median are 21
and 22 respectively, then its mode is approximately [2005]
(a) 22.0
(b) 20.5
(c) 25.5
(d) 24.0
The median of a set of 9 distinct observations is 20.5. If
each of the largest 4 observations of the set is increased
by 2, then the median of the new set
[2003]
(a) remains the same as that of the original set
(b) is increased by 2
(c) is decreased by 2
(d) is two times the original median.
In a class of 100 students there are 70 boys whose average
marks in a subect are 75. If the average marks of the
complete class is 72, then what is the average of the girls?
[2002]
(a) 73
(b) 65
(c) 68
(d) 74
TOPIC n
24.
If
n
25.
i =1
i =1
(a) a – 1
(c)
n(a - 1)
(b) n a - 1
(d)
a -1
20-30
30-40
2
x
2
is 50, then x is equal to _____________.
[NA Sep. 04, 2020 (II)]
For the frequency distribution :
Variate (x ) :
x1
x2
x 1 …x 15
Frequency (f ) :
f1
f2
f 3 …f 15
15
where 0 < x1 < x2 < x3 < ... < x15 = 10 and
standard deviation cannot be :
(a) 4
26.
(b) 1
å fi > 0,
i =1
the
[Sep. 03, 2020 (I)]
(c) 6
(d) 2
Let xi (1 £ i £ 10) be ten observations of a random variable X.
10
10
If
å
( xi - p) = 3
and
å ( x - p)
i
2
= 9
where
i= 1
i =1
0 ¹ p Î R, then the standard deviation of these
observations is :
[Sep. 03, 2020 (II)]
(a)
å ( xi - a) = n and å ( xi - a)2 = na , (n, a > 1), then
the standard deviation of n observations x1, x2, ..., xn is :
[Sep. 06, 2020 (I)]
10-20
Frequency
Quartile, Measures of Dispersion,
Quartile Deviation, Mean
Deviation, Variance & Standard
Deviation, Coefficient of Variation
n
(c) x 2 - 10 x + 19 = 0
(d) x 2 - 20 x + 18 = 0
The mean and variance of 8 observations are 10 and 13.5,
respectively. If 6 of these observations are 5, 7, 10, 12, 14,
15, then the absolute difference of the remaining two
observations is :
[Sep. 04, 2020 (I)]
(a) 9
(b) 5
(c) 3
(d) 7
If a variance of the following frequency distribution :
Class
27.
20.
(b) 2 x 2 - 20 x + 19 = 0
(a) x 2 - 10 x + 18 = 0
Let x1 , x 2 , .............. xn be n observations such that
å xi2
17.
(b) 60
The mean and variance of 7 observations are 8 and 16,
respectively. If five observations are 2, 4, 10, 12, 14, then
the absolute difference of the remaining two observations
is :
[Sep. 05, 2020 (I)]
(a) 1
(b) 4
(c) 2
(d) 3
If the mean and the standard deviation of the data 3, 5, 7, a,
b are 5 and 2 respectively, then a and b are the roots of the
equation :
[Sep. 05, 2020 (II)]
28.
3
5
(b)
4
5
(c)
9
10
(d)
7
10
Let X = {x Î N : 1 £ x £ 17} and Y =+
{ax b : x Î X and
a, b Î R, a > 0}. If mean and variance of elements of Y
are 17 and 216 respectively then a + b is equal to :
[Sep. 02, 2020 (I)]
(a) 7
(b) –7
(c) –27
(d) 9
If the variance of the terms in an increasing A.P.,
b1 , b2 , b3 , ....., b11 is 90, then the common difference of
this A.P. is ___________.
[NA Sep. 02, 2020 (II)]
Downloaded from @Freebooksforjeeneet
M-231
Statistics
29.
Let the observations xi(1 £ i £ 10) satisfy the equations,
10
10
i =1
i =1
å ( xi - 5) = 10 and å ( xi - 5)2
30.
31.
32.
33.
34.
35.
37.
38.
(b) 2
(c) 4
(d)
40.
41.
(b) 380
(c) 525
(a)
10
3
(b)
100
3
(c)
10
3
(d)
100
3
2
5
(b) 2
(c)
3
2
A data consists of n observations:
(a)
42.
(d)
n
x1, x2, ...., xn. If
å ( xi + 1)2 = 9n
1
+ d each.
2
4
then |d| equals :
3
[Jan. 11, 2019 (I)]
If the variance of this outcome data is
2
n
and
i =1
å ( xi - 1) 2 = 5n
i =1
then the standard deviation of this data is:
[Jan. 09, 2019 (II)]
(a) 2
43.
(b) 5
(c) 5
(d) 7
The mean of five observations is 5 and their variance is
9.20. If three of the given five observations are 1, 3 and 8,
then a ratio of other two observations is:
[Jan. 10, 2019 (I)]
(a) 10 : 3
44.
If the standard deviation of the numbers –1, 0, 1, k is 5
where k > 0, then k is equal to:
[April 09, 2019 (I)]
5
(a) 2 6
(b) 2 10
(c) 4
(d) 6
3
3
The mean and variance of seven observations are 8 and
16, respectively. If 5 of the observations are 2, 4, 10, 12, 14,
then the product of the remaining two observations is :
[April 08, 2019 (I)]
(a) 45
(b) 49
(c) 48
(d) 40
A student scores the following marks in five tests: 45, 54,
41, 57, 43. His score is not known for the sixth test. If the
mean score is 48 in the six tests, then the standard
deviation of the marks in six tests is : [April. 08, 2019 (II)]
1
1
- d each, 10 items gave outcome
2
2
each and the remaining 10 items gave outcome
2
(d) 480
If the sum of the deviations of 50 observations from 30 is
50, then the mean of these observations is :
[Jan. 12, 2019 (I)]
(a) 30
(b) 51
(c) 50
(d) 31
The mean and the variance of five observations are 4 and
5.20, respectively. If three of the observations are 3, 4 and
4; then the absolute value of the difference of the other
two observations, is :
[Jan. 12, 2019 (II)]
(a) 7
(b) 5
(c) 1
(d) 3
The outcome of each of 30 items was observed; 10 items
gave an outcome
If both the mean and the standard deviation of 50
observations x1, x2, ….. , x50 are equal to 16, then the mean
of (x1 – 4)2, (x2 – 4)2, …, (x50 – 4)2 is : [April 10, 2019 (II)]
(a) 400
36.
= 40. If m and l are the
mean and the variance of the observations, x1 – 3, x2 – 3,
..., x10 – 3, then the ordered pair (m, l) is equal to:
[Jan. 9, 2020 (I)]
(a) (3, 3) (b) (6, 3)
(c) (6, 6)
(d) (3, 6)
The mean and the standard deviation (s.d.) of 10
observations are 20 and 2 respectively. Each of these 10
observations is multiplied by p and then reduced by q,
where p ¹ 0 and q ¹ 0. If the new mean and new s.d.
become half of their original values, then q is equal to:
[Jan. 8, 2020 (I)]
(a) –5
(b) 10
(c) –20
(d) –10
The mean and variance of 20 observations are found to
be 10 and 4, respectively. On rechecking, it was found
that an observation 9 was incorrect and the correct
observation was 11. Then the correct variance is:
[Jan. 8, 2020 (II)]
(a) 3.99
(b) 4.01
(c) 4.02
(d) 3.98
If the variance of the first n natural numbers is 10 and the
variance of the first m even natural numbers is 16, then
m + n is equal to ¾¾¾.
[NA Jan. 7, 2020 (I)]
If the mean and variance of eight numbers 3, 7, 9, 12, 13,
20, x and y be 10 and 25 respectively, then x × y is equal to
_________.
[NA Jan. 7, 2020 (II)]
If the data x1, x2, ……, x10 is such that the mean of first four
of these is 11, the mean of the remaining six is 16 and the
sum of squares of all of these is 2,000 ; then the standard
deviation of this data is :
[April 12, 2019 (I)]
(a) 2 2
39.
(b) 4 : 9
(c) 5 : 8
(d) 6 : 7
If mean and standard deviation of 5 observations
x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance
of 6 observations x1, x2, ..., x5 and – 50 is equal to:
[Jan. 10, 2019 (II)]
(a) 509.5
45.
(c) 582.5
(d) 507.5
5 students of a class have an average height 150 cm and
variance 18 cm2. A new student, whose height is 156 cm,
oined them. The variance (in cm2) of the height of these
six students is:
[Jan. 9, 2019 (I)]
(a) 16
46.
(b) 586.5
(b) 22
(c) 20
(d) 18
The mean and the standard deviation (s.d.) of five
observations are 9 and 0, respectively.
If one of the observations is changed such that the mean
of the new set of five observations becomes 10, then their
s.d. is?
[Online April 16, 2018]
(a) 0
(b) 4
(c) 2
Downloaded from @Freebooksforjeeneet
(d) 1
EBD_8344
M-232
47.
Mathematics
If the mean of the data : 7, 8, 9, 7, 8, 7, l, 8 is 8, then the
variance of this data is
[Online April 15, 2018]
9
(a)
8
(b) 2
9
48.
If
7
(c)
8
å (xi - 5) = 9 and
i =1
49.
9
51.
(b) 8.50
i=1
(c) 3
[2018]
(c) 8.00
(d) 9.00
(a) 3a2 – 34a + 91 = 0
(b) 3a2 – 23a + 44 = 0
(c) 3a2 – 26a + 55 = 0
(d) 3a2 – 32a + 84 = 0
The mean of 5 observations is 5 and their variance is 124.
If three of the observations are 1, 2 and 6; then the mean
deviation from the mean of the data is :
(b) 2.6
(c) 2.8
(d) 2.4
If the mean deviation of the numbers 1, 1 + d, ..., 1 + 100d
from their mean is 255, then a value of d is :
[Online April 9, 2016]
(a) 10.1
(b) 5.05
(c) 20.2
(d) 10
The variance of first 50 even natural numbers is [2014]
(a) 437
54.
57.
If the standard deviation of the numbers 2, 3, a and 11 is
3.5, then which of the following is true?
[2016]
(a) 2.5
53.
56.
(d) 9
[Online April 10, 2016]
52.
(a) X, M.D.
å (xi - 5)2 = 45 , then the standard
The sum of 100 observations and the sum of their squares
are 400 and 2475, respectively. Later on, three observations,
3, 4 and 5, were found to be incorrect. If the incorrect
observations are omitted, then the variance of the remaining
observations is :
[Online April 9, 2017]
(a) 8.25
50.
(b) 2
(b)
437
4
(c)
833
4
(d) 833
Let x , M and s2 be respectively the mean, mode and
variance of n observations x1, x2, ...., xn and di = – xi – a ,
i = 1, 2, ...., n, where a is any number.
Statement I: Variance of d1, d2,... dn is s2.
Statement II: Mean and mode of d1, d2, .... dn are - x - a
and – M – a, respectively.
[Online April 19, 2014]
(a) Statement I and Statement II are both false
(b) Statement I and Statement II are both true
(c) Statement I is true and Statement II is false
Let X and M.D. be the mean and the mean deviation about
X of n observations xi, i = 1, 2, ........, n. If each of the
observations is increased by 5, then the new mean and
the mean deviation about the new mean, respectively, are:
[Online April 12, 2014]
(d) 1
deviation of the 9 items x1, x2, ..., x9 is :
(a) 4
55.
58.
59.
(b) X + 5, M.D.
(c) X, M.D. + 5
(d) X + 5, M.D. + 5
All the students of a class performed poorly in
Mathematics. The teacher decided to give grace marks of
10 to each of the students. Which of the following statistical
measures will not change even after the grace marks were
given ?
[2013]
(a) mean (b) median (c) mode
(d) variance
In a set of 2n observations, half of them are equal to 'a' and
the remaining half are equal to ' –a'. If the standard deviation
of all the observations is 2 ; then the value of | a | is :
[Online April 25, 2013]
(a) 2
(b) 2
(c) 4
(d) 2 2
Mean of 5 observations is 7. If four of these observations
are 6, 7, 8, 10 and one is missing then the variance of all the
five observations is :
[Online April 22, 2013]
(a) 4
(b) 6
(c) 8
(d) 2
Let x1 , x2,...., xn be n observations, and let x be their
arithmetic mean and s2 be the variance.
Statement-1 : Variance of 2x1, 2x2, ..., 2xn is 4s2.
Statement-2 : Arithmetic mean 2x1, 2x2, ..., 2xn is 4 x .
[2012]
(a) Statement-1 is false, Statement-2 is true.
(b) Statement-1 is true, statement-2 is true; statement-2 is
a correct explanation for Statement-1.
(c) Statement-1 is true, statement-2 is true; statement-2 is
not a correct explanation for Statement-1.
(d) Statement-1 is true, statement-2 is false.
60. Statement 1: The variance of first n odd natural numbers
n2 - 1
3
Statement 2: The sum of first n odd natural number is n2
and the sum of square of first n odd natural numbers is
is
(
).
n 4n2 + 1
[Online May 26, 2012]
3
(a) Statement 1 is true, Statement 2 is false.
(b) Statement 1 is true, Statement 2 is true; Statement 2 is
not a correct explanation for Statement 1.
(c) Statement 1 is false, Statement 2 is true.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation for Statement 1.
(d) Statement I is false and Statement II is true
Downloaded from @Freebooksforjeeneet
M-233
Statistics
61.
If the mean of 4, 7, 2, 8, 6 and a is 7, then the mean deviation
from the median of these observations is
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is true. Statement-2 is
a correct explanation for Statement-1.
[Online May 12, 2012]
(a) 8
62.
67.
(b) 32, 4
(c) 28, 2
68.
(d) 28, 4
(b) 4
(c) 5
11
2
(b) 6
(c)
13
2
(d)
69.
5
2
66.
Suppose a population A has 100 observations 101, 102,
............., 200 and another population B has 100 obsevrations
151, 152, ................ 250. If VA and VB represent the variances
70.
(b)
9
4
(c)
4
9
2
n
(b)
2
(c) 2
(d)
1
n
Consider the following statements :
(C) Variance is independent of change of origin and scale.
n2 – 1
is
.
4
Which of these is / are correct ?
Statement-2 : The sum of first n natural numbers is
[2009]
2
3
(B) Median is not independent of change of scale
Statement-1 : The variance of first n even natural numbers
is n (n + 1)(2n + 1) .
6
(d)
(A) Mode can be computed from histogram
(d) 10.0
n(n + 1)
and the sum of squares of first n natural numbers
2
VA
is [2006]
VB
In a series of 2 n observations, half of them equal a and
remaining half equal –a. If the standard deviation of the
observations is 2, then |a| equals.
[2004]
(a)
[2009]
(c) 20.2
(b) a = 5, b = 2
(d) a = 3, b = 4
(a) 1
1 + 2d, .... 1 + 100d from their mean is 255, then d is equal to:
(b) 10.1
(a) a = 0, b = 7
(c) a = 1, b = 6
of the two populations, respectively then
If the mean deviation of the numbers 1, 1 + d,
(a) 20.0
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance
is 6.80. Then which one of the following gives possible
values of a and b?
[2008]
(d) 2
For two data sets, each of si e 5, the variances are given to
be 4 and 5 and the corresponding means are given to be 2
and 4, respectively. The variance of the combined data set
is
[2010]
(a)
65.
(d) 3
If the mean deviation about the median of the numbers a,
2a,.......,50a is 50, then | a | equals
[2011]
(a) 3
64.
(c) 1
A scientist is weighing each of 30 fishes. Their mean weight
worked out is 30 gm and a standarion deviation of 2 gm.
Later, it was found that the measuring scale was misaligned
and always under reported every fish weight by 2 gm. The
correct mean and standard deviation (in gm) of fishes are
respectively :
[2011RS]
(a) 32, 2
63.
(b) 5
71.
(a) (A), (B) and (C)
(b) Only (B)
(c) Only (A) and (B)
(d) Only (A)
[2004]
In an experiment with 15 observations on x, the following
results were available:
Sx 2 = 2830, Sx = 170
(a) Statement-1 is true, Statement-2 is true. Statement-2 is
not a correct explanation for Statement-1.
One observation that was 20 was found to be wrong and
was replaced by the correct value 30. The corrected
variance is
[2003]
(b) Statement-1 is true, Statement-2 is false.
(a) 8.33
(b) 78.00
(c) 188.66
Downloaded from @Freebooksforjeeneet
(d) 177.33
EBD_8344
M-234
1.
Mathematics
5.
(6.00)
Mean =
Sxi f i 0 × n C0 + 2 × n C1 + 2 2 × n C2 + ... + 2 n × n Cn
=
n
Sf i
C0 + n C1 + .... + n Cn
To find sum of numerator consider
75 l – 25 = 75 Þ l =
(1 + x)n = nC0 + n C1 x + nC2 x 2 + ... + nCn x n
...(i)
Put x = 2 Þ 3n - 1 = 2 × n C1 + 22 × nC2 + .... + 2 n × nCn
6.
To find sum of denominator, put x = 1 in (i) , we get
2n = n C0 + n C1 + ..... + n Cn
\
3n - 1
2n
=
728
2n
Þ 3n = 729 Þ n = 6
7.
1
2.
(d)
2sin x + 2cos x
³ (2sin x + cos x ) 2
2
Þ 2sin x + 2cos x ³ 2 × 2
(Q AM ³ GM)
sin x + cos x
2
8.
Since, -2 £ sin x + cos x £ 2
\ Minimum value of 2
1-
Þ 2sin x + 2cos x ³ 2
3.
sin x + cos x
2
1
2
=2
-
1
2
4.
10.
Median =
10
T5 + T6
34 + x
= 35 =
Þ x = 36 and y = 84
2
2
y 84 7
=
Hence, =
x 36 3
10 + 49 59
=
= 29.5
2
2
(c) Let average wage of Night shift worker is x
70 × 54 + 30 × x = 60 × 100
x = 74
(a) There are 2n observations x1, x2, ..., x2n
Median =
32+3+21 56
=
= 2.8
20
20
(d) Ten numbers in increasing order are
10, 22, 26, 29, 34, x, 42, 67, 70, y
n
[Q a1 = 10]
178
= 44.5
4
an = 44.5 + 1.5 + 3 = 49
9.
å xi = x + y + 300 = 42 Þ x + y = 120
(b) Sum of 16 observations = 16 × 16 = 256
Sum of resultant 18 observations
= 256 – 16 + (3 + 4+5) = 252
252
= 14
Mean of observations =
18
(c) a1 + a2 + a3 = 39
Þ a1 + (a1 + d) + (a1 + 2d) = 39
Their mean =
2 3 5 7
Marks
No. of students 16 1 0 3
Mean =
x1 + x2 + ....... + x25
= x = 40
25
Þ x1 + x2 + .......... + x25 = 1000
\ x2 + x2 + .......... + x25 – 60 + A = 39 ´ 25
Let A be the age of new teacher.
Þ 1000 – 60 + A = 975
Þ A = 975 – 940 = 35
(c) Let;
Þ d=3
Sum of last four term = 178
.
Average marks =
4
.
3
Þ 3a1 + 3d = 39
(d) Number of students are,
(x + 1)2 + (2x – 5) + (x2 – 3x) + x = 20
Þ 2x2 + 2x – 4 = 20 Þ x2 + x – 12 = 0
Þ (x + 4) (x – 3) = 0 Þ x = 3
\
(b) As mean is a linear operation, so if each observation
is multiplied by l and decreased by 25 then the mean
becomes 75 l – 25.
According to the question,
2n
So, mean =
x
å 2ni
i =1
Let these observations be divided into two parts
x1, x2, ..., xn and xn+1, ..., x2n
Each in 1st part 5 is added, so total of first part is
n
å xi + 5n.
i= 1
In second part 3 is subtracted from each
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M-235
Statistics
2n
So, total of second part is
å
14.
(d) Frequency distribution is given as
xi - 3n
Expenditure
No. of families (f )
0-50
24
2n
50-100
33
i =1
100-150
37
150-200
b
200-250
25
i = n +1
Total of 2n terms are
n
2n
i =1
i = n +1
å xi + 5n + å
xi - 3n =
å xi + 2n
2n
x + 2n
x
= å i +1
Mean = å i
n
2
2
n
i =1
i =1
So, it increase by 1.
(d) Since 0 < y < x < 2y
2n
11.
...(i)
Clearly, modal class is 100-150, as the maximum
frequency occurs in this class.
Given, Mode = 140
We have
f 0 - f –1
Mode = l +
´i
2 f 0 - f –1 - f1
where
l = 100, f0 = 37, f–1 = 33, f1 = b
i = 50
Thus, we get
...(ii)
é 37 - 33 ù
140 = 100 + ê
ú ´ 50
ë 2 ( 37 ) - 33 - b û
x
x
Þ x- y <
2
2
\ x – y < y < x < 2x + y
\ y>
Hence median =
y+x
= 10
2
Þ x + y = 20
And range = (2x + y) – (x – y) = x + 2y
But range = 28
\ x + 2y = 28
From equations (i) and (ii),
x = 12, y = 8
\ Mean =
( x - y ) + y + x + (2 x + y ) 4 x + y
=
4
4
= x+
12.
13.
15.
y
8
= 12 + = 14
4
4
20 ´ 40 - 33 + 55
= 41.1
(a) Correct mean =
20
Nearest option : (a) 41
4
é
ù
200
´ 50 = 100 +
= 100 + ê
ú
74
33
b
ë
û
41 - b
Þ 5740 = 4300 + 40b Þ b = 36
(a) Let the number of boys be x and girls be y.
Þ 52x + 42y = 50(x + y)
Þ 52x – 50x = 50y – 42y
Þ 2x = 8y Þ
x
4
x 4
=
= Þ
x+ y 5
y 1
\ Required % of boys =
(a) Median is given as
N
-F
M = l+ 2
´C
f
where
l = lower limit of the median - class
f = frequency of the median class
N = total frequency
F = cumulative frequency of the class ust before the
median class
C = length of median class
Now, given, M = 25, N = 100, F = 45,
C = 20 – 30 = 10, l = 20.
\ By using formula, we have
50 – 45
25 = 20 +
´ 10
f
25 – 20 =
50
50
Þ5=
Þ f = 10
f
f
x
´ 100
x+ y
4
´ 100 = 80 %
5
(b) We know that for positive real numbers x1, x2, ...., xn,
A.M. of kth powers of xi ³ kth the power of A.M. of xi
=
16.
Þ
17.
å x12 ³ æç å x1 ö÷
n
è
n ø
2
Þ
400 æ 80 ö
³ç ÷
è nø
n
2
Þ n ³ 16 . So only possible value for
n = 18
(d) We know that Mode = 3 Median – 2Mean
3 × 22 – 2 × 21 = 66 – 42 = 24
th
18.
æ 9 +1ö
th
(a) n = 9 then median term = ç
÷ = 5 term. That
2
è
ø
means four observation followed by it. If last four
observations are increased by 2. The median is 5th
observation which is remaining unchanged.
\ There will be no change in median.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-236
19.
20.
Mathematics
Þ x + y = 80 - 63
(b) Total student = 100
Total marks of 70 boys = 75 ´ 70 = 5250
Þ Total marks of girls = 7200 – 5250 = 1950
Number of girls = 100 – 70 = 30
1950
= 65
Average of girls =
30
(d) Standard deviation
æ
ö
å ( xi - a) ç å ( xi - a) ÷
÷
= i =1
- ç i =1
ç
÷
n
n
ç
÷
è
ø
n
n
Þ x + y = 17
Q var( x ) = 13.5
=
2
2
[Q n, a > 1]
24.
21.
na æ n ö
- ç ÷ = a -1
n è nø
(c) Let two remaining observations are x1, x2.
2 + 4 + 10 + 12 + 14 + x1 + x2
= 8 (given)
7
So, x =
Þ x1 + y1 = 14
...(i)
Þ 460 +
+
x22
25.
= (16 + 64) ´ 7
Þ x12 + x22 = 100
...(ii)
x
x=
Sf i xi 30 + 70 + 25 x
=
= 25
Sfi
4+ x
Q ( x + y)2 = x 2 + y 2 + 2 xy Þ xy = 48
...(iii)
Sfi xi2
- ( x )2
S fi
450 + 625 x + 2450
- 625
4+ x
2900 + 625 x
Þ 50 x = 200
4+ x
\x = 4
(c) If variate varries from a to b then variance
3+ 5+ 7 + a + b
= 5 Þ a + b = 10
5
32 + 52 + 72 + a 2 + b2
- (5)2 = 4
Variance =
5
(c) Mean =
æ 10
ö
( xi - p)
( xi - p) ÷
ç
ç i =1
÷
i =1
-ç
÷
10
10
ç
÷
ç
÷
çè
÷ø
10
å
26.
Þ ab = 19
Hence, a and b are the roots of the equation,
x2 – 10x + 19 = 0.
(d) Let the two remaining observations be x and y.
Qx =
5 + 7 + 10 + 12 + 14 + 15 + x + y
8
Þ 10 =
63 + x + y
8
2
Þ var ( x) < 25
Þ standard deviation < 5
It is clear that standard deviation cann't be 6.
(c) S.D. =
Þ (a + b)2 - 2ab = 62
23.
2
æ 10 - 0 ö
Þ var ( x) < ç
è 2 ÷ø
Þ x- y =2 Þ |x- y| =2
Þ a 2 + b2 = 62
2
æ b - aö
var ( x) £ ç
è 2 ÷ø
Q ( x - y)2 = ( x + y )2 - 4 xy = 196 - 192 = 4
22.
2
Þ 675 =
4 + 16 + 100 + 144 + 196 + x12 + x22
=
- 64 = 16
7
x12
...(ii)
fi
Þ 50 =
æ Sx ö
Now, s =
- ç i ÷ = 16 (given)
N è N ø
2
Þ x 2 + y 2 = 169
From (i) and (ii) we get
(x, y) = (12, 5) or (5, 12)
So, | x – y | = 7.
(4)
s2 =
2
Sxi2
25 + 49 + 100 + 144 + 196 + 225 + x 2 + y 2
- (10)2
8
xi 15 25 35
2
=
...(i)
2
å
2
2
=
27.
9 æ 3ö
9
-ç ÷ = .
10 è 10 ø
10
(b) Q x =
1 + 2 + 3 + .... + 17 17 ´ 18
=
=9
17
17 ´ 2
y = ax + b =
a(1 + 2 + 3 + ..... + 17)
+ b = 17
17
Downloaded from @Freebooksforjeeneet
M-237
Statistics
Þ
a × (17 ×18)
+ b = 17 Þ 9a + b = 17 ...(i)
17 × 2
Var (x) = sA2 =
30.
Sx 2
- ( x )2
n
New mean ( x1) = px - q
=
1 + 2 + .... + 17
- (9)2
17
Þ 10 = p(20) – q
and new standard deviations s1 = |p| s
=
17 × 18 × 35
2
- ( 9) = 105 - 81 = 24
6 ×17
Þ 1 = |p| (2) Þ |p| =
2
2
2
Var (y) = a2 Var (x) = a 2 × 24 = 216
If p =
216
= 9Þ a = 3
a =
24
2
31.
\ a + b = 3 + ( -10) = -7
æ 11 ö
ç å bi ÷
i =1
Variance =
- ç i =1 ÷
ç 11 ÷
11
ç
÷
è
ø
å bi2
2
(from equation (i))
x1 + x2 + ..... + x20
= 10
20
20
å xi
Þ
æ 10
ö
å (b1 + rd ) ç å (b1 + rd ) ÷
÷
= r =0
- ç r =0
ç
÷
11
11
ç
÷
è
ø
10
2
20
= 10
...(i)
Sxi2
- ( x )2
n
Sxi2
- 100 = 4
20
Sxi2 = 104 × 20 = 2080
Þ
2
Actual mean =
æ 10 ´ 11ö
æ 10 ´ 11 ´ 21ö
11b12 + 2b1d ç
+ d2 ç
÷ø
è 2 ÷ø
è
6
=
11
10 ´ 11 ö
æ
d
11b +
ç 1
÷
2
-ç
÷
11
ç
÷
è
ø
Variance =
2
...(ii)
200 - 9 + 11 202
=
20
20
2080 - 81 + 121 æ 202 ö
-ç
÷
20
è 20 ø
2
2120
- (10.1)2 = 106 - 102.01 = 3.99
20
(18)Var (1, 2, ....., n) = 10
=
32.
2
Þ
= (b12 + 10b1d + 35d 2 ) - (b1 + 5d ) 2 = 10d 2
12 + 22 + ..... + n 2 æ 1 + 2 + ..... + n ö
-ç
÷ = 10
n
n
è
ø
2
(n + 1)(2n + 1) æ n + 1 ö
-ç
÷ = 10
6
è 2 ø
Þ n2 – 1 = 120
Þ n = 11
Var (2, 4, 6, ....., 2m) = 16 Þ Var (1, 2, ....., m) = 4
Þ m2 – 1 = 48 Þ m = 7
Þ m + n =18
Q Variance = 90 (Given)
Þ
Þ 10d 2 = 90 Þ d = 3.
(a) Mean of the observation (xi – 5) =
i =1
Variance =
Let common difference of A.P. be d
S( xi - 5)
=1
10
\ l = {Mean (xi – 5)} + 2 = 3
Variance of the observation
m = var (xi – 5) =
1
1
Þ p=±
2
2
1
, then q = – 20
2
(a) Let x1, x2, ....., x20 be 20 observations, then
Mean =
(3)
11
29.
1
, then q = 0
2
...(i)
If p = –
\ From (i), b = 17 - 9a = 17 - 27 = -10
28.
(c) Let x and s be the mean and standard deviations of
given observations.
If each observation is multiplied with p and then q is
subtracted.
S( xi - 5) 2 S ( xi - 5)
=3
10
10
33.
(52)Mean = x =
3 + 7 + 9 + 12 + 13 + 20 + x + y
= 10
8
Þ x + y = 16
Downloaded from @Freebooksforjeeneet
...(i)
EBD_8344
M-238
Mathematics
Variance = s2 =
s2 =
34.
S ( xi )2
- ( x )2 = 25
8
Þ
2
2
9 + 49 + 81+144 +169 + 400 + x + y
- 100 = 25
8
Þ x2 + y2 = 148
From eqn. (i), (x + y)2 = (16)2
Þ x2 + y2 + 2xy = 256
Using eqn. (ii), 148 + 2xy = 256
Þ xy = 52
(b) According to the question,
37.
...(ii)
12k 2
+2
16
=5 Þk =2 6
4
(c) Let the remaining numbers are a and b.
Mean ( x ) =
å xi = 2 + 4 + 10 + 12 + 14 + a + b
N
Þ a + b = 14
Variance (s2) =
x5 + x6 + ... + x10
= 16 Þ x5 + x6 + ... +x10 = 96
6
Þ
2
and x12 + x22 + ...x10
= 2000
å xi2 - ( x )2 = 16
35.
Þb2 – 14b + 48 = 0
Þs=2
Þb = 6, 8
\ a = 8, 6.
\ (a, b) = (6, 8) or (8, 6)
x1 + x2 + ....x50
= 16
50
and 162 =
2
x12 + x22 ... x50
- 162
50
38.
\ x=
( x1 - 4) 2 + ( x2 - 4) 2 + ...( x50 - 4) 2
50
2 1
2
2
2
Now, s = [(48 - 41) + (48 - 45) + (48 - 54)
6
+ (48 – 57)2 + (48 – 43)2 + (48 – 48)2
2
x 2 + x22 + ... + x50
+ 50 ´16 - 8( x1 + x2 + ... + x50 )
= 1
50
=
36.
200 100
1
= (49 + 9 + 36 + 81 + 25) =
=
6
6
3
162 (100) + (50) - 8(16 ´ 50)
= 400
50
(a) Mean of given observation =
Hence, the product of the remaining two observations
= ab = 48
(a) Q Mean score = 48
Let unknown score be x,
41 + 45 + 54 + 57 + 43 + x
= 48
6
Þ x + 240 = 288 Þ x = 48
2
Þ 2 (16)2 50 = x12 + x22 + ....x50
Required mean =
s=
k
4
Q Standard deviation = 5
\ s2 = 5
(d) Given,
1
2
Þ s = S( xi - x )
n
2
10
3
50
39.
2
2
...(ii)
Þ196 + b2 – 28b + b2 = 100
N
(a) Given, mean and standard deviation are equal to 16.
\
22 + 42 + 102 + 122 + 142 + a 2 + b2
– (8)2 = 16
7
From (i) and (ii), (14 – b)2 + b2 = 100
2
2000 æ 44 + 96 ö
=
=4
10 çè 10 ÷ø
N
Þ a2 + b2 = 100
å xi2 - ( x )2
=8
...(i)
x1 + x2 + x3 + x4
= 11 Þ x1 + x2 + x3 + x4 = 44
4
Q standard deviation, s2 =
7
å ( xi - 30) = 50
i =1
50
å xi - 50(30) = 50
2
i =1
2
æ k ö æ k ö æ k ö æ 3k ö
ç + 1÷ + ç ÷ + ç - 1 ÷ + ç ÷
4 ø è4ø è4 ø è 4 ø
=è
=5
4
50
Þ
å xi
i =1
= 1550
Downloaded from @Freebooksforjeeneet
M-239
Statistics
Mean, x =
å xi
n
50
1550
= 31
50
(a) Let two observations be x1 and x2, then
Q
x1 + x2 + 3 + 4 + 4
=4
5
Þ x1 + x2 = 9
Q
...(i)
x12 + x22 = 65
43.
...(ii)
1
S xi2 - ( x )2
30
2
2
2
ù 1
1 éæ 1
ö
æ 1ö
æ1
ö
êç - d ÷ ´ 10 + ç ÷ ´ 10 + ç + d ÷ ´ 10ú =
è
ø
è
ø
è
ø
30 êë 2
2
2
úû 4
4
1 é
1
ù 1
30 ´ + 20d 2 ú Þ
=
3 30 êë
4
û 4
4 1 2 2 1
Þ
= + d 3 4 3
4
Þ d 2 = 2 Þ |d| = 2
(b) Variance is given by,
1
1
A - 2 B2
n
n
- (5)
å xi2
i =1
= 5(25 + 9.2)
Þ (1)2 + (3)2 + (8)2 + x42 + x52 = 171
s2 =
s2 =
i =1
5
2 = 9.2 Þ
= 125 + 46 = 171
Variance of the outcomes is,
æ1
ö
1
å x2 - å x
n i =1 i çè n i=1 i ø÷
å xi2
5
1 æ1
ö 1
ç ´ 30÷ø =
30 è 2
2
s2 =
1
1
´ 6n - 2 ´ n 2 = 6 - 1 = 5
n
n
Þs= 5
(b) Since mean of x1, x2, x3, x4 and x5 is 5
Þ x1 + x2 + x3 + x4 + x5 = 25
Þ 1 + 3 + 8 + x4 + x5 = 25
Þ x4 + x5 = 13
...(i)
Q
1
1
+ d , + d , ..., 10 times
2
2
n
...(iii)
5
1 1
æ1 ö æ1 ö
(d) Outcomes are ç - d ÷ , ç - d ÷ , 0..., 10 times, , ,
è2 ø è2 ø
2 2
n
i =1
Þ s2 =
From (i) and (ii);
x1 = 8, x2 = 1
Hence, the required value of the difference of other two
observations = |x1 – x2| = 7
42.
...(ii)
å ( xi - 1)2 = 5n
Þ A + n – 2B = 5n Þ A – 2B = 4n
From (ii) and (iii),
A = 6n, B = n
26 = 41 + x12 + x22 - 80
Mean =
i =1
n
9 + 16 + 16 + x12 + x22
- 16
5
..., 10 times,
å ( xi + 1)2 = 9n
Þ A + n + 2B = 9n Þ A + 2B = 8n
N
41.
i =1
n
å xi2 - ( x ) 2
Variance =
(5 × 20) =
2
i =1
=
40.
n
Here, A = å xi and B = å xi
Þ x42 + x52 = 97
...(ii)
Þ
Þ
...(iii)
(x4 + x5)2 – 2x4 x5 = 97
2x4 x5 = 132 – 97 = 72 Þ x4x5 = 36
(i) and (iii) Þ x4 : x5 =
4
9
or
9
4
5
44.
å xi
5
5
å xi2
i =1
5
2
- (10) 2 = 3 = 9
5
Þ
2
å xi2 = 545
i =1
Then,
...(i)
6
(d) Q x = i=1 Þ å xi = 10 ´ 5 = 50 Þ å xi = 50 – 50 = 0
i =1
i =1
5
Þ
6
5
i =1
i =1
å xi2 = å xi2 + (-50)2
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EBD_8344
M-240
Mathematics
= 545 + (–50)2 = 3045
6
å xi2
Variance =
45.
i =1
6
9
2
å (x i - 5)2 = 45
Also,
æ
ö
ç å xi ÷ 3045
- ç i =1 ÷ =
- 0 = 507.5
6
ç 6 ÷
ç
÷
è
ø
6
i=1
9
9
i =1
i =1
å x i2 - 10å xi + 9(25) = 45
Þ
...(ii)
From (i) and (ii) we get,
Sx 2
(c) Q Variance = s = i - ( x )2
N
2
9
å x i2 = 360
i =1
S xi2
- (150)2
5
Þ S xi2= 90 + 112590 = 112590
Then, variance of the height of six students
Þ 18 =
2
V' =
46.
112590 + (156) æ 750 + 156 ö
-ç
÷ø
è
6
6
\ Standard deviation = Variance = 2
49.
S xi
=9
n
Þ S xi = 9 × 5 = 45
Now, standard deviation = 0
\ all the five terms are same i.e.; 9.
Now for changed observation
50.
4(9 - 10)2 + (14 - 10)2
=2
5
7+8+9+ 7+8+7 +l +8
(d) x =
=8
8
54 + l
Þ
= 8 Þ l = 10
8
Now variance = s2
1 + 0 + 1+1 + 0 + 1+ 4 + 0 8
= =1
8
8
Hence, the variance is 1.
48.
(b) Given
9
9
i=1
i=1
å (x i - 5) = 9 Þ å xi = 54 ...(i)
i =1
(d) x =
å xi2 - æç å x i ö÷
N
2
ç N ÷
è
ø
2475 æ 388 ö
97 çè 97 ÷ø
2
Þ
2 + 3 + a + 11 a
= +4
4
4
å
Þ 3.5 =
51.
å x i2 = 2475
2425 - 1552 873
=
=9
97
97
s=
=
Þ s2 =
i =1
=
=
(7 - 8) 2 + (8 - 8) 2 + (9 - 8) 2 + (7 - 8) 2 + (8 - 8) 2
+ (7 - 8) 2 + (10 - 8) 2 + (8 - 8) 2
=
8
100
2
Variance = s =
36 + x5
= 10
5
Þ x5 = 14
47.
100
å x i = 400
(d)
xnew =
S ( xi - xnew ) 2
n
ç 9 ÷
è
ø
9
360 æ 54 ö
- ç ÷ = 40 – 36 = 4
9 è9ø
Þ x=
\ snew =
2
2
=
2
= 22821 – 22801 = 20
(c) Here mean = x = 9
å xi 2 - æ å x i ö
Since, variance =
2
x i2
- x
n
( )
2
4 + 9 + a 2 +121 æ a ö
- ç + 4÷
4
è4 ø
49 4(134 + a 2 ) - ( a 2 + 256 + 32 a)
=
4
16
Þ 3a 2 - 32a + 84 = 0
(c) n = 5
x =5
variance = 124
x1 = 1, x2 = 2, x3 = 6
x =5
x1 + x2 + x3 + x4 +x5
=5
5
Downloaded from @Freebooksforjeeneet
M-241
Statistics
Þ
Þ
Þ
Þ
Þ
x4 + x5 + 9 = 25
x4 + x5 = 16
x4 + x5 + 10 – 10 = 16
(x4 – 5) + (x5 – 5) = 16 – 10
(x4 – 5) + (x5 – 5) = 6
é x1 + x2 + x3 + .... + xn ù na
= -ê
ú - n = -x - a
n
ë
û
Since, di = – xi – a and we multiply or subtract each
observation by any number the mode remains the same.
Hence mode of –xi – a i.e. di and xi are same.
Now variance of d1, d2,...., dn
S | xi - x |
N
| x4 - 5 | + | x5 - 5 |
= |x1 – 5| + |x2 – 5| + |x3 – 5| +
5
4 + 3 + 1 + 6 14
=
=
= 2.8
5
5
1
(a) x =
[1 + (1 + d) + (1 + 2d)] ..... (1 + 100d)]
101
Mean deviation =
52.
1 101
´
[1 + (1 + 100d)] = 1 + 50d
101 2
mean deviation from mean
=
=
2|d |
(1 + 2 + 3..... + 50)
101
=
2 | d | 50 ´ 51 2550
´
|d|
=
101
2
101
2550
| d | = 225 Þ| d | = 10.1
101
(d) First 50 even natural numbers are 2, 4 , 6 ....., 100
å xi2 - ( x) 2
Variance =
N
2
2
2 + 4 2 + ... + 100 2 - æ 2 + 4 + ... + 100 ö
Þ s2 =
çè
÷ø
50
50
55.
1 n
å ( xi - x )2
n i =1
Mean of d1, d2, d3, ...., dn
=
d1 + d 2 + d3 + .... + d n
n
=
( - x1 - a) + ( - x2 - a) + ( - x3 - a ) + .... + (- xn - a )
n
1 n
1 n
2
( - xi + x ) 2 = å ( x - xi ) = s2
å
n i =1
n i =1
(b) Let xi be n observations, i = 1, 2, ...n
be X + 5 and new M.D. about new mean will be M.D.
n
æ
xi ö
çQ Mean = å
÷
nø
è
i =1
56.
(d) If initially all marks were xi then
å ( xi - x )2
i
s12 =
N
Now each is increased by 10
s12 =
57.
4(12 + 22 + 32 + .... + 50 2 )
- (51)2
50
æ 50 ´ 51 ´ 101ö
= 4ç
- (51)2
è 50 ´ 6 ÷ø
= 3434 – 2601 Þ s2 = 833
x1 + x2 + x3 + ... + xn
(b) x =
n
s2 =
1 n
å [ - xi - a + x + a]2
n i =1
about X .
If each observation is increased by 5 then new mean will
å [( xi +10) -( x +10) ]2
i
N
=
å ( xi - x )
i
N
2
= s12
Hence, variance will not change even after the grace marks
were given.
(a) Clearly mean A = 0
Now, standard deviation s =
=
54.
=
Let X be the mean and M.D be the mean deviation
=
53.
1 n
å [di - ( - x - a)]2
n i =1
=
1
[|1 – (1 + 50d) | + | (1 + d) – (1 + 50d) |.....|
101
[1 + 100d] – (1 + 50d)|]
=
=
2=
S( x - A)2
2n
(a - 0) 2 + ( a - 0) 2 + ...... + (0 - a) 2 + ......
2n
a 2 .2n
=|a|
2n
Hence, | a | = 2
(d) Let 5th observation be x.
Given mean = 7
=
58.
6 + 7 + 8 + 10 + x
5
Þ x=4
\ 7=
Downloaded from @Freebooksforjeeneet
EBD_8344
M-242
Mathematics
Now, Variance
=
62.
(6 - 7) 2 + (7 - 7)2 + (8 - 7) 2 + (10 - 7) 2 + (4 - 7)2
5
12 + 02 + 12 + 32 + 32
=
=
5
59.
63.
20
= 4=2
5
60.
61.
25th obs. + 26th obs.
2
25a + 26a
\ M =
= 25.5a
2
å xi - M
M .D (M ) =
N
1
Þ 50 = [2 ´ a ´ (0.5 + 1.5 + 2.5 + ....24.5)]
50
25
Þ 2500 = 2 a ´ (25)
2
Þ a =4
Median =
(d) A.M. of 2x1, 2x2, ..., 2xn is
æ x + x 2 + .... + x n
2x1 + 2x 2 + ... + 2x n
= 2ç 1
n
n
è
ö
÷ = 2x
ø
æ
Sum of observations ö
çQ Mean =
÷
Number of observations ø
è
So statement-2 is false.
If each observations is multiply by 2 then mean multiply
by 2 and variance multiply by 22.
variance (2xi) = 22 variance (xi) = 4s2 where i = 1, 2,......n
So statement-1 is true.
(a) Statement 2 : Sum of first n odd natural numbers is
not equal to n2.
So, statement - 2 is false.
(d) Given observations are 4, 7, 2, 8, 6, a and mean is 7.
We know
Mean =
64.
4+ 7+ 2+8+ 6+ a
6
6
Now, Mean deviation =
i =1
65.
=
1
å yi2 - (4)2 = 5 Þ å yi2 = 105
5
å xi2 + å yi2 = å ( xi2 + yi2 ) = 145
(
)
2
1
æ 1
ö
xi2 + yi2 - ç å ( xi + yi ) ÷
å
10
è 10
ø
145
11
=
-9 =
10
2
101 + d(1 + 2 + 3 + ......+100)
(b) Mean =
101
d × 100 × 101
=1+
=1 + 50 d
101 × 2
Given that mean deviation from the mean = 255
1
[| 1 - (1 + 50d ) | + | (1 + d ) - (1 + 50 d ) |
101
+ | (1 + 2d ) -(1 + 50d ) | +.... +
13
2
| (1 + 100d ) - (1 + 50 d ) |] = 255
13
13
13
13
13
13
+ 7+ 2+ 8+ 6+ 15 2
2
2
2
2
2
6
5 1 9 3 1 17
+ + + + +
2 2 2 2 2 2 = 18 = 3
=
6
6
s 2y =
Þ
6
4-
1
å xi2 - (2)2 = 4 Þ å xi2 = 40;
5
=
3rd observation + 4th observation
6 + 7 13
=
=
2
2
2
å xi -
s 2x =
Þ å xi + å yi = å ( xi + yi ) = 5(2) + 5(4) = 30
Variance of combined data
4+7 + 2+8+ 6+ a
Þ a = 15
6
Now, given observations can be written in ascending order
which is 2, 4, 6, 7, 8, 15
Since, No. of observation is even
\ Median
æ 6ö
æ6 ö
çè ÷ø th observation + çè + 1÷ø th observation
2
2
=
2
(a) s2x = 4, s2y = 5, x = 2, y = 4
Þ
Þ 7=
=
(a) We know that if each observation is increase by 2
then mean is increase by 2 but S.D. remains same.
Correct mean = observed mean + 2
= 30 + 2 = 32
Correct S. D. = observed S.D. = 2
(b) Q n = 50 (even)
66.
Þ 2d [1 + 2 + 3 + ... + 50] = 101 ´ 255
50 ´ 51
= 101´ 255
Þ 2d ´
2
101 ´ 255
= 10.1
Þ d=
50 ´ 51
(c) First n even natural numbers be 2, 4, 6, 8, ..., 2n
2(1 + 2 + 3 + ... + n)
2[n (n + 1)]
=
= (n + 1)
\ x=
n
2n
Downloaded from @Freebooksforjeeneet
M-243
Statistics
S ( x – x )2
S x2
– ( x )2
=
2n
n
4S n 2
4 n (n + 1) (2n + 1)
==
– (n + 1) 2 =
– ( n + 1) 2
n
6n
é 4n + 2 – 3n – 3 ù
2 (2n + 1) ( n + 1)
– ( n + 1)2 = (n + 1) ê
=
úû
3
3
ë
2
( n + 1)( n – 1) = n - 1
=
3
3
\ Statement-1 is false. Clearly, statement - 2 is true.
69.
And Var =
67.
(d) Mean of a, b, 8, 5, 10 is 6
a + b + 8 + 5 + 10
=6
Þ
5
Þ a+b=6
Variance of a, b, 8, 5, 10 is 6.80
Þ
Standard deviation s =
2=
=
...(i)
( a - 6) 2 + (b – 6) 2 + (8 – 6) 2 + (5 – 6) 2 + (10 – 6) 2
5
= 6.80
Þ a 2 –12a + 36 + (1 – a ) 2 + 21 = 34
[using eq. (i)]
Þ 2a2 –14a + 24 = 0 Þ a2 – 7a + 12 = 0
Þ a = 3 or 4 Þ b = 4 or 3
\ The possible values of a and b are a = 3 and b = 4
or, a = 4 and b = 3
68.
(a) s 2x =
(c) Clearly sum of observations = 0,
\ mean A = 0
å di2
(Here di = deviations are taken from the
n
mean). Since population A and population B both have 100
consecutive integers, therefore both have same standard
å ( x - A)2
2n
(a - 0)2 + ( a - 0)2 + ...(0 - a) 2 + ...
[Q s = 2]
2n
a 2 .2n
=| a |
2n
70.
Hence, | a | = 2
(c) Only first statement (A) and second statements (B)
are correct.
71.
(b) Sx = 170, Sx 2 = 2830
New, Sx' = 170 + (30 - 20) = 180
New, Sx '2 = 2830 + (900 - 400)
= 2830 + 500 = 3330
Now, Variance =
1
æ1
ö
S x ' 2 -ç S x ' ÷
n
èn
ø
2
2
=
1
æ1
ö
´ 3330 - ç ´ 180 ÷ = 222 - 144 = 78.
15
è 15
ø
deviation and hence the variance is also same. \ V A = 1
VB
Downloaded from @Freebooksforjeeneet
EBD_8344
M-244
Mathematics
15
Probability
TOPIC Ć
Random Experiment, Sample
Space, Events, Probability of an
Event, Mutually Exclusive &
Exhaustive Events, Equally Likely
(a)
7
2
4
(c)
1.
Out of 11 consecutive natural numbers if three numbers
are selected at random (without repetition), then the
probability that they are in A.P. with positive common
difference, is:
[Sep. 06, 2020 (I)]
(a)
15
101
(b)
5.
5
101
5
10
(d)
33
99
If 10 different balls are to be placed in 4 distinct boxes at
random, then the probability that two of these boxes
contain exactly 2 and 3 balls is :
[Jan. 9, 2020 (II)]
(a)
(c)
3.
965
211
945
210
(b)
965
210
(d)
945
211
6.
4.
3
3
(c)
(d)
10
20
Let S = {1, 2, ....., 20}. A subset B of S is said to be “nice”,
if the sum of the elements of B is 203. Than the probability
that a randomly chosen subset of S is “nice” is :
[Jan. 11, 2019 (II)]
7.
(d)
5
220
6
1
, then the number of children in each
12
[Online April 16, 2018]
(b) 6
(d) 5
A box ¢A¢ contanis 2 white, 3 red and 2 black balls. Another
box ¢B¢ contains 4 white, 2 red and 3 black balls. If two
balls are drawn at random, without replacement, from a
randomly selected box and one ball turns out to be white
while the other ball turns out to be red, then the probability
that both balls are drawn from box ¢B¢ is
[Online April 15, 2018]
(a)
7
16
(b)
9
32
(c)
7
8
(d)
9
16
[April 12, 2019 (I)]
1
(b)
5
(b)
2
220
Two different families A and B are blessed with equal
number of children. There are 3 tickets to be distributed
amongst the children of these families so that no child
gets more than one ticket.
If the probability that all the tickets go to the children of
family is?
(a) 4
(c) 3
If three of the six vertices of a regular hexa on are chosen
at random, then the probability that the triangle formed
with these chosen vertices is equilateral is :
1
(a)
10
20
the family B is
(c)
2.
20
If the lengths of the sides of a triangle are decided by
the three throws of a single fair die, then the probability
that the triangle is of maximum area given that it is an
isosceles triangle, is :
[Online April 11, 2015]
(a)
1
21
(b)
1
27
(c)
1
15
(d)
1
26
Downloaded from @Freebooksforjeeneet
M-245
Probability
8.
9.
A number x is chosen at random from the set {1, 2, 3, 4, ....,
100}. Define the event: A = the chosen number x satisfies
Statement -2 : If the four chosen numbers form an AP,
then the set of all possible values of common difference is
( x - 10 )( x - 50) ³ 0
( x - 30 )
(±1, ±2, ±3, ±4, ±5) .
Then P (A) is:
[Online April 12, 2014]
(a) 0.71
(b) 0.70
(c) 0.51
(d) 0.20
A set S contains 7 elements. A non-empty subset A of S
and an element x of S are chosen at random. Then the
probability that x Î A is:
[Online April 11, 2014]
1
(a)
2
(c)
10.
63
128
(a) Statement -1 is true, Statement -2 is true; Statement -2
is not a correct explanation for Statement -1
(b) Statement -1 is true, Statment -2 is false
(c) Statement -1 is false, Statment -2 is true.
(d) Statement -1 is true, Statement -2 is true ; Statement 2 is a correct explanation for Statement -1.
14.
64
(b)
127
(d)
31
128
There are two balls in an urn. Each ball can be either white
or black. If a white ball is put into the urn and there after a
ball is drawn at random from the urn, then the probability
that it is white is
[Online May 26, 2012]
(a)
1
4
(b)
2
3
15.
1
1
(d)
5
3
If six students, including two particular students A and B,
stand in a row, then the probability that A and B are
separated with one student in between them is
[Online May 19, 2012]
An urn contains nine balls of which three are red, four are
blue and two are green. Three balls are drawn at random
without replacement from the urn. The probability that the
three balls have different colours is
[2010]
(a)
2
7
(b)
1
21
(c)
2
23
(d)
1
3
Five horses are in a race. Mr. A selects two of the horses
at random and bets on them. The probability that Mr. A
selected the winning horse is
[2003]
(c)
11.
(a)
8
15
2
(c)
15
12.
(b)
4
15
(b)
4
5
(c)
3
5
(d)
1
5
TOPIC n
n
åi
{1, 2, 3, ..... , 1000}. The probability that
i =1
n
The probabilities of three events A, B and C are given by
P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5. If P (A È B) = 0.8, P
(A Ç C) = 0.3, P (A Ç B Ç C) = 0.2, P (B Ç C) = b and
P (A È B È C) = a, where 0.85 £ a £ 0.95 , then b lies in
the interval:
[Sep. 06, 2020 (II)]
(a) [0.35, 0.36]
(b) [0.25, 0.35]
(c) [0.20, 0.25]
(d) [0.36, 0.40]
17.
Let A and B be two events such that the probability that
2
is an integer
åi
[Online May 12, 2012]
(a) 0.331
(b) 0.333
(c) 0.334
(d) 0.332
Four numbers are chosen at random (without replacement)
from the set {1, 2, 3, ...20}.
[2010]
Statement -1: The probability that the chosen numbers
when arranged in some order will form an AP is
1
.
85
Odds Against & Odds in Favour of
an Event, Addition Theorem,
Boole's Inequality, Demorgan's Law
16.
i =1
13.
2
5
1
(d)
15
A number n is randomly selected from the set
is
(a)
exactly one of them occurs is
A or B occurs is
2
and the probability that
5
1
, then the probability of both of them
2
occur together is:
(a) 0.02
(c) 0.01
[Jan. 8, 2020 (II)]
(b) 0.20
(d) 0.10
Downloaded from @Freebooksforjeeneet
EBD_8344
M-246
18.
19.
Mathematics
In a class of 60 students, 40 opted for NCC, 30 opted for
NSS and 20 opted for both NCC and NSS. If one of these
students is selected at random, then the probability that the
student selected has opted neither for NCC nor for NSS is :
[Jan. 12, 2019 (II)]
(a)
1
6
(b)
1
3
(c)
2
3
(d)
5
6
For three events A, B and C,
P(Exactly one of A or B occurs)
= P(Exactly one of B or C occurs)
= P(Exactly one of C or A occurs) =
23.
1- x
3x + 1
and P(B) =
, then the set of
3
4
possible values of x lies in the interval :
[Online April 25, 2013]
that P(A) =
24.
1
and
4
(b)
7
32
7
7
(d)
16
64
From a group of 10 men and 5 women, four member
committees are to be formed each of which must contain at
least one woman. Then the probability for these committees
to have more women than men, is :
[Online April 9, 2017]
(a)
21
220
(b)
25.
2
1
(d)
23
11
If 12 identical balls are to be placed in 3 identical boxes,
then the probability that one of the boxes contains exactly
3 balls is :
(2015)
12
11
æ 1ö
æ 1ö
(a) 220 ç ÷
(b) 22 ç ÷
è 3ø
è 3ø
55 æ 2 ö
(c)
ç ÷
3 è 3ø
22.
If
A
11
and
æ 2ö
(d) 55 ç ÷
è 3ø
B
are
two
Let
X
and
Y
are
two
events
such
that
[Online May 7, 2012]
(a) Statement 1 is false, Statement 2 is true.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation of Statement 1.
(c) Statement 1 is true, Statement 2 is false.
(d) Statement 1 is true, Statement 2 is true; Statement 2 is
a correct explanation of Statement 1.
A die is thrown. Let A be the event that the number
obtained is greater than 3. Let B be the event that the
number obtained is less than 5. Then P(AÈB) is [2008]
3
5
(c) 1
26.
events
(d)
2
5
1 - 2x
3x + 1
1- x
, P( B) =
and P (C ) =
The set
2
3
4
of possible values of x are in the interval.
[2003]
P ( A) =
(a) [0 , 1]
such
(b) 0
Events A, B, C are mutually exclusive events such that
10
that
P ( A È B) = P ( A Ç B ) , then the incorrect statement
amongst the following statements is:
[Online April 9, 2014]
(a) A and B are equally likely
(b) P ( A Ç B¢ ) = 0
é 7 4ù
(d) ê - , ú
ë 9 9û
(a)
3
11
(c)
21.
é 1 5ù
(c) ê - , ú
ë 3 9û
Statement 2: P ( X ) + P (Y ) = 2P ( X Ç Y )
(c)
20.
é1 2 ù
(b) ê , ú
ë3 3û
Statement 1: P ( X Ç Y ') = P ( X 'Ç Y ) = 0
1
.
16
Then the probability that at least one of the events
occurs, is :
[2017]
3
16
(a) [0, 1]
P ( X È Y ) = P ( X Ç Y ).
P(All the three events occur simultaneously) =
(a)
If the events A and B are mutually exclusive events such
é1 2 ù
(c) ê 3 , 3 ú
û
ë
27.
é1 1 ù
(b) ê 3 , 2 ú
û
ë
é 1 13 ù
(d) ê 3 , 3 ú
û
ë
A and B are events such that P(A È B)=3/4, P(A Ç B)=1/4,
P( A ) =2/3 then P ( A Ç B) is
(a) 5/12
(b) 3/8
(c) 5/8
(d) 1/4
(c) P ( A¢ Ç B ) = 0
(d) P(A) + P(B) = 1
Downloaded from @Freebooksforjeeneet
[2002]
M-247
Probability
1.
(c) For an A.P. 2b = a + c (even), so both a and c even
numbers or odd numbers from given numbers and b number
will be fixed automatically.
4.
C2 + 5 C2
25
5
=
=
165 33
C3
(Bonus) Total number of ways placing 10 different balls in
4 distinct boxes = 410
Since, two of the 4 distinct boxes contains exactly 2 and 3
balls.
Then, there are three cases to place exactly 2 and 3 balls in
2 of the 4 boxes.
Case-1: When boxes contains balls in order 2, 3, 0, 5
Then, number of ways of placing the balls
Required probability =
2.
6
Hence, required probability =
+
10!
(2!)2 ´ 2! ´ (3!)2 ´ 2!
3.
25 ´ 17 ´ 945
10
=
C3
(2) Since total number of subsets of the set S = 220
5
220
(d) Let the number of children in each family be x.
Thus the total number of children in both the families are
2x
Now, it is given that 3 tickets are distributed amongst the
children of these two families.
Thus, the probability that all the three tickets go to the
children in family B
5.
x
=
Þ
C3
1
=
C3 12
2x
1
x ( x - 1) ( x - 2)
=
2 x (2 x - 1) (2 x - 2) 12
( x - 2) 1
=
(2 x - 1) 6
Þ x=5
Thus, the number of children in each family is 5.
(a) Probability of drawing a white ball and then a red ball
Þ
´ 4!
6.
4
from bag B is given by
15
2
D
C
7.
B
9
=
2
9
C1 ´ 3C1
2
=
7
C2
Hence, the probability of drawing a white ball and then a
bag A is given by
red ball from bag B =
A
C1 ´ 2 C1
C2
Probability of drawing a white ball and then a red ball from
17 ´ 945
F
20 ´ 21
= 210
2
Hence required probability =
4
2
(1) Total no. of triangles = 6C3
Favorable no. of triangle i.e, equilateral triangles (DAEC
and DBDF) = 2.
E
1
10
Then, find the sets which has sum 7.
(1) {7}
(2) {1, 6}
(3) {2, 5}
(4) {3, 4}
(5) {1, 2, 4}
Then, there is only 5 sets which has sum 203
= 25 × 17 × 945
Hence, the required probability
=
=
Now, the sum of all number from 1 to 20 =
11
10!
=
´ 4!
2! ´ 3! ´ 0! ´ 5!
Case-2: When boxes contains ball in order 2, 3, 1, 4.
Then, number of ways of placing the balls
10!
=
´ 4!
2! ´ 3! ´1! ´ 4!
Case-3: When boxes contains ball in order 2, 3, 2, 3
Then, number of ways of placing the balls
10!
=
´ 4!
(2!)2 ´ (3!)2 ´ 2! ´ 3!
Therefore, number of ways of placing the balls that
contains exactly 2 and 3 balls.
10!
10!
=
´ 4! +
´ 4!
2! ´ 3! ´ 0! ´ 5!
2! ´ 3! ´1! ´ 4!
2
6
7
2
9
2 2
+
7 9
=
2´7
7
=
18 + 14 16
(b) Favourable case = (6, 6, 6)
Total case = {(1, 1, 1) (2, 2, 1), (2, 2, 2), (2, 2, 3), (3, 3, 1) .....
(3, 3, 5) (4, 4, 1).... (4, 4, 6) (5, 5, 1).... (5, 5, 6)
(6, 6, 1).... (6, 6, 6)}
Downloaded from @Freebooksforjeeneet
EBD_8344
M-248
Mathematics
which satisfies condition a + b > c
Number of total case = 27
Probability =
8.
(a)
Given
Now the group of three students (student A, student B
and a student in between A and B) and the remaining
3 students can be stand in a row in 4! ways.
Hence total number of ways to stand in a row such that
A and B are separated with one student in between them
= 4 × 2 × 4!
Now total number of ways to stand 6 student stand in
a row without any restriction = 6!
Hence required probability
4 ´ 2 ´ 4! 4 ´ 2 4
=
=
=
6!
6 ´ 5 15
1
27
( x - 10)( x - 50)
³0
( x - 30)
Let x ³ 10, x ³ 50 equation will be true " x ³ 50
æ x - 50 ö
³ 0, " x Î[10, 30)
as ç
è x - 30 ÷ø
( x - 10)( x - 50)
³ 0 " x Î[10, 30)
x - 30
Total value of x between 10 to 30 is 20.
Total values of x between 50 to 100 including 50 and 100
is 51.
Total values of x = 51 + 20 = 71
n
å i2
12.
i =1
71
= 0.71
100
(b) Let S = {x1, x2, x3, x4, x5, x6, x7}
Let the chosen element be xi.
Total number of subsets of S = 27 = 128
No. of non-empty subsets of S = 128 – 1 = 127
We need to find number of those subsets that contains xi.
For n = 1, 2, 3, ......, 1000
P (A) =
9.
Value of
of
x1 x2 ------- xi ---- x7
For those subsets containing xi, each element has 2
choices.
i.e., (included or not included) in subset,
However as the subset must contain xi, xi has only one
choice. (included one)
So, total no. of subsets containing
xi = 2 × 2 × 2 × 2 × 1 × 2 × 2 = 64
64
127
10. (b) Total possible event when one ball is taken out
= 3C1
Let E : The event of 1 white ball coming out
No. of ways to 1 white ball coming out
= 2C1
\ P(E) =
11.
2
C1
3
C1
=
2
3
(b) Consider a group of three students A, B and an
other student in between A and B. Choice for a student
between A and B is 4. A and B can interchange their
places in the group in 2 ways.
3 5 7
2001
æ3 ö
, , ,........,
only first term ç = 1÷ , fourth
3 3 3
3
è3 ø
Hence, out of 1000 values of
2n + 1
,
3
total number of integral values of
2n + 1
3
= 333 + 1 = 334
No. of subsets containing xi
Total no. of non-empty subsets
=
2n + 1 3 5 7
2001
= , , ,........,
respectively. Out
3
3 3 3
3
æ9
ö
æ 2001
ö
term ç = 3 ÷ , 667th term ç
= 667 ÷ are integers.
3
3
è
ø
è
ø
2 2 2 2 1 2 2
Required prob =
(c)
n(n + 1)(2 n + 1)
2n + 1
6
=
=
n(n + 1)
3
åi
2
i =1
n
334
= 0.334
1000
(b) Four numbers are chosen from {1, 2, 3...20}
n(S) = 20C4
Statement-1:
Common difference is 1; total number of ways = 17
common difference is 2; total number of ways = 14
common difference is 3; total number of ways = 11
common difference is 4; total number of ways = 8
common difference is 5; total number of ways = 5
common difference is 6; total number of ways = 2
\
13.
Required probability =
Prob. =
17 + 14 + 11 + 8 + 5 + 2
20
C4
=
1
85
Statement -2 is false, because common difference can be 6
also.
Downloaded from @Freebooksforjeeneet
M-249
Probability
14.
P (Exactly one of B or C occurs)
n( S ) = 9 C3
(a)
1
4
P (Exactly one of C or A occurs)
= P(B) + P (C) – 2P (B Ç C) =
n( E ) = 3C1 ´ 4 C1 ´ 2 C1
3´ 4´ 2
Probability =
15.
9
=
C3
24 ´ 3!
24 ´ 6
2
´ 6! =
=
9!
9 ´8´ 7 7
(a) Let 5 horses are H1, H2, H3, H4 and H5.
Total ways of selecting pair of horses be
Adding (1), (2) and (3),we get
= 5C2 = 10[i.e. H1H2 ,H1H3 , H1H4 , H1 H5 ,
2SP(A) – 2SP (A Ç B) =
3
4
\ SP(A) – SP (A Ç B) =
3
8
H 2H3 , H 2 H 4 , H 2H5 , H3H 4 , H3H5 , H 4H5 ]
\ P (A È B È C)
= SP (A) – SP (A Ç B) + P (A Ç B Ç C)
(b) Q P ( A Ç B) = P ( A) + P ( B ) - P ( A È B )
= 1 – 0.8 = 0.2
3 1
7
+
=
8 16 16
(c) Probability of 4 member committee which contain
atleast one woman.
Þ P(3M, 1W) + P(2M, 2W) + P(1M, 3W) + P(0M, 4W)
=
Now,
Q P ( A È B È C ) = P ( A) + P ( B ) + P (C ) - P ( A Ç B )
20.
- P( B Ç C ) - P(C Ç A) + P ( A Ç B Ç C )
Þ a = 0.6 + 0.4 + 0.5 - 0.2 - b - 0.3 + 0.2
Þ
Þ b = 1.2 - a
Q a Î[0.85, 0.95] then b Î[0.25, 0.35]
2
17. (4) P(exactly one) =
5
Þ
P(A) + P(B) – 2P(A Ç B) =
P(A or B) = P(A È B) =
18.
2
5
P(A) + P(B) – P(A Ç B) =
\
P(A Ç B) =
1
2
(1) P = Set of students who opted for NCC
Q = Set of Students who opted for NSS
n(P) = 40, n(Q) = 30, n(P Ç Q) = 20
n(P È Q) = n(P) + n(Q) – n(P Ç Q)
= 40 + 30 – 20
= 50
\ Hence, required probability = 1 -
1
4
C2 5C2
15
C4
+
10
C4
C1 5 C3
15
+
C4
10
C0 5 C4
15
C4
P (1M,3W) + P ( 0M,4W)
P ( 3M,1W) + P ( 2M,2W) + P (1M,3W) + P ( 0M,4W)
105
1
1365
=
=
1155 11
1365
21.
(c) Note:- The question should state ‘3 different’ boxes
instead of ‘3 identical boxes’ and one particular box has 3
balls. Then the solution would be:
12
50
60
Required probability =
=
22.
...(1)
10
1155
1365
Probability of committees to have more women than
men.
1
6
(c) P (exactly one of A or B occurs)
= P(A) + P (B) – 2P (A Ç B) =
15
+
Þ
=
19.
C3 5 C1
600 450 100
5
+
+
+
1365 1365 1365 1365
=
1 2 5-4 1
- =
= = 0.10
2 5
10 10
10
Þ
\
1
2
Þ
...(3)
1
16
Now, P (A Ç B Ç C) =
4 2
=
Hence required probability =
10 5
16.
1
4
= P(C) + P(A) – 2P (C Ç A) =
Any horse can win the race in 4 ways
(e.g. for H1 : H1H2, H1H3, H1H4, H1H5)
...(2)
(d)
C3 ´ 29
312
55 æ 2 ö
ç ÷
3 è 3ø
11
Let A and B be two events such that
P(A È B) = P(A Ç B)
Downloaded from @Freebooksforjeeneet
EBD_8344
M-250
Mathematics
\ A Ç B = {4}
and P(A È B) = P (A) + P (B) – P(A Ç B)
option (a) : since P(A È B) = P(A Ç B) (given)
therefore A and B are equally likely
Suppose option (b) and option (c) are correct.
1
6
\ P(A È B) = P(A) + P(B) – P(A Ç B)
Þ P( A Ç B) =
\ P(A Ç B¢ ) = 0 and P(A¢ Ç B) = 0
=
Þ P(A) - P(A Ç B) = 0 and P(B) - P(A Ç B) = 0
Þ P (A) = P(A Ç B) and P(B) = P(A Ç B)
26.
Thus P(A) = P (B) = P(A Ç B) = P(A È B)
P(A È B) = P(A) + P (B) – P(A Ç B)
Þ 0 £ 3x + 1 £ 1, ³ –1 £ 3 x £ 2
3
1
2
Þ - £x£
3
3
1- x
0£
£ 1 Þ -3 £ x £ 1
4
1 - 2x
and 0 £
£ 1 Þ -1 £ 2 x £ 1
2
1
1
Þ- £x£
2
2
Also for mutually exclusive events A, B, C,
= P(A Ç B) + P(A Ç B) – P(A Ç B)
= P(A Ç B)
which is true from given condition
Hence, option (a), (b) and (c) are correct.
(c) Since events A and B are mutually exclusive
\ P(A) + P(B) = 1
3x + 1 1 - x
+
=1
3
4
Þ 12x + 4 + 3 – 3x = 12
Þ
24.
5
9
Þ
P( A È B È C) =
3x + 1 1 - x 1 - 2 x
+
+
3
4
2
1
13
Þ £x£
3
3
From (i), (ii), (iii) and (iv), we get
P ( X Ç Y ) = P ( X ) + P (Y ) - P ( X Ç Y ) (from (1)
...(iv)
1
1
é1 1 ù
£ x £ Þ xÎê , ú
3
2
ë3 2 û
P ( X ) + P (Y ) = 2 P ( X Ç Y )
Hence, Statement - 2 is true.
27.
Now, P ( X Ç Y ') = P ( X ) - P ( X Ç Y )
(a) We know that P (A È B) = P (A) + P (B) – P (A Ç B)
3
1
=1 – P( A ) + P(B) –
4
4
éëQ P ( A) = 1 - P( A)ùû
and P ( X 'Ç Y ) = P (Y ) - P ( X Ç Y )
Þ
This implies statement-1 is also true.
(c) A (number is greater than 3) = {4, 5, 6}
2
2
+ P(B) Þ P(B) = ;
3
3
Now, P( A Ç B ) = P(B) – P ( A Ç B )
5
2 1
= – = .
3 4 12
Þ P( A) =
...(iii)
1 + 3x 1 - x 1 - 2 x
+
+
£1
3
4
2
0 £ 13 - 3x £ 12 Þ 1 £ 3x £ 13
P ( X È Y ) = P ( X ) + P (Y ) - P ( X Ç Y )
25.
...(ii)
\ 0£
...(1)
We know
Þ
...(i)
P ( A È B È C ) = P ( A) + P ( B ) + P ( C )
é 1 5ù
\ x Î ê- , ú
ë 3 9û
(b) Let X and Y be two events such that
P ( X ÈY ) = P ( X ÇY )
3x + 1
1- x
and
, P( B) =
3
4
1 - 2x
2
We know that 0 £ P ( E ) £ 1
Also, we know
Þ 9x = 5 Þ x =
(b) Given that P ( A) =
P (C ) =
[Q Given P(A Ç B) = P(A È B) ]
23.
1 2 1 3 + 4 –1
+ – =
=1
2 3 6
6
Þ 1=1–
3 1
=
6 2
B (number is less than 5) = {1, 2, 3, 4} Þ P( B) =
4 2
=
6 3
Downloaded from @Freebooksforjeeneet
M-251
Relations and Functions
16
Relations and
Functions
TOPIC Ć
1.
Let A = {a, b, c} and B = {1, 2, 3, 4}. Then the number of
elements in the set C = { f : A ® B | 2 Î f ( A) and f is not
one-one} is ___________.
[NA Sep. 05, 2020 (II)]
2.
4.
6.
f : ( 0, ¥ ) ® ( 0, ¥ ) be defined by
Let a function
f ( x) = 1-
3.
(a)
(b)
(c)
(d)
Types of Relations, Inverse of a
Relation, Mappings, Mapping of
Functions, Kinds of Mapping of
Functions
1
. Then f is :
x
[Jan. 11, 2019 (II)]
(a) not inective but it is surective
(b) inective only
(c) neither inective nor surective
(d) both inective as well as surective
The number of functions f from {1, 2, 3, ..., 20} onto
{1, 2, 3, ...., 20} such that f (k) is a multiple of 3, whenever k
is a multiple of 4 is :
[Jan. 11, 2019 (II)]
5
(a) 6 × (15)!
(b) 5! × 6!
(c) (15)! × 6!
(d) 56 × 15
Let N be the set of natural numbers and two functions
f and g be defined as f, g : N ® N such that
7.
8.
ì n +1
if n is odd
ïï
f (n) = í 2
ï n
if n is even
ïî 2
5.
tion f: A ® R as f(x) =
2x
, then f is:[Jan. 09, 2019 (II)]
x -1
x
é 1 1ù
The function f : R ® ê - , ú defined as f(x) =
, is :
1 + x2
ë 2 2û
[2017]
(a) neither inective nor surective
(b) invertible
(c) inective but not surective
(d) surective but not inective
éx ù
The function f : N ® N defined by f ( x ) = x - 5 ê ú , where
ë5û
N is set of natural numbers and [x] denotes the greatest
integer less than or equal to x, is :
[Online April 9, 2017]
(a) one-one and onto.
(b) one-one but not onto.
(c) onto but not one-one.
(d) neither one-one nor onto.
Let A = {x1, x2, ........., x7} and B = {y1, y2, y3} be two sets
containing seven and three distinct elements respectively.
Then the total number of functions f : A ® B that are
onto, if there exist exactly three elements x in A such that
f(x) = y2, is equal to :
(Online April 11, 2015)
(a) 14.7C3 (b) 16.7C3
n
and g(n) = n – (– 1) . Then fog is:
[Jan. 10, 2019 (II)]
(a) onto but not one-one.
(b) one-one but not onto.
(c) both one-one and onto.
(d) neither one-one nor onto.
Let A = {xÎR : x is not a positive integer}. Define a func-
not inective
neither inective nor surective
surective but not inective
inective but not surective
9.
(c) 14.7C2
(d) 12.7C2
x -1
Let f : R ® R be defined by f(x) = x + 1 then f is:
[Online April 19, 2014]
(a)
(b)
(c)
(d)
both one-one and onto
one-one but not onto
onto but not one-one
neither one-one nor onto.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-252
10.
11.
12.
13.
14.
15.
16.
Mathematics
Let P be the relation defined on the set of all real numbers
such that
P = {(a, b) : sec2a – tan2b = 1}. Then P is:
[Online April 9, 2014]
(a) reflexive and symmetric but not transitive.
(b) reflexive and transitive but not symmetric.
(c) symmetric and transitive but not reflexive.
(d) an equivalence relation.
Let R = {(x, y) : x, y Î N and x2 – 4xy + 3y2 = 0}, where N
is the set of all natural numbers. Then the relation R is :
[Online April 23, 2013]
(a) reflexive but neither symmetric nor transitive.
(b) symmetric and transitive.
(c) reflexive and symmetric,
(d) reflexive and transitive.
Let R = {(3, 3) (5, 5), (9, 9), (12, 12), (5, 12), (3, 9), (3, 12), (3, 5)}
be a relation on the set A = {3, 5, 9, 12}. Then, R is :
[Online April 22, 2013]
(a) reflexive, symmetric but not transitive.
(b) symmetric, transitive but not reflexive.
(c) an equivalence relation.
(d) reflexive, transitive but not symmetric.
Let A = {1, 2, 3, 4} and R : A ® A be the relation defined
by R = {(l, 1), (2, 3), (3, 4), (4, 2)}. The correct statement is :
[Online April 9, 2013]
(a) R does not have an inverse.
(b) R is not a one to one function.
(c) R is an onto function.
(d) R is not a function.
If P(S) denotes the set of all subsets of a given set S, then
the number of one-to-one functions from the set
S = {1, 2, 3} to the set P(S) is
[Online May 19, 2012]
(a) 24
(b) 8
(c) 336
(d) 320
If A = {x Î z + : x < 10 and x is a multiple of 3 or 4}, where
z+ is the set of positive integers, then the total number of
symmetric relations on A is
[Online May 12, 2012]
(a) 25
(b) 215
(c) 210
(d) 220
Let R be the set of real numbers.
[2011]
Statement-1: A = {(x, y) Î R × R : y – x is an integer} is an
equivalence relation on R.
Statement-2: B = {(x, y) Î R × R : x = ay for some rational
number a} is an equivalence relation on R.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation for Statement-1.
17.
Consider the following relations:
R = {(x, y) | x, y are real numbers and x = wy for some
æ m pö
rational number w}; S = {ç , ÷ | m,n, p and q are
è n qø
18.
19.
20.
21.
integers such that n, q ¹ 0 and qm = pn}.
Then
[2010]
(a) Neither R nor S is an equivalence relation
(b) S is an equivalence relation but R is not an equivalence
relation
(c) R and S both are equivalence relations
(d) R is an equivalence relation but S is not an equivalence
relation
Let R be the real line. Consider the following subsets of
the plane R × R:
S ={(x, y): y = x + 1 and 0 < x < 2}
T ={(x, y): x – y is an integer},
Which one of the following is true?
[2008]
(a) Neither S nor T is an equivalence relation on R
(b) Both S and T are equivalence relation on R
(c) S is an equivalence relation on R but T is not
(d) T is an equivalence relation on R but S is not
Let W denote the words in the English dictionary. Define
the relation R by R = {(x, y) Î W × W| the words x and y
have at least one letter in common.} Then R is
[2006]
(a) not reflexive, symmetric and transitive
(b) relexive, symmetric and not transitive
(c) reflexive, symmetric and transitive
(d) reflexive, not symmetric and transitive
Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9),
(3, 12), (3, 6)} be a relation on the set
A = {3, 6, 9, 12}. The relation is
[2005]
(a) reflexive and transitive only
(b) reflexive only
(c) an equivalence relation
(d) reflexive and symmetric only
Let f : (– 1, 1) ® B, be a function defined by
2x
-1
f (x) = tan
1 - x2
B is the interval
æ
è
[2005]
é pö
pö
2ø
(a) ç 0, ÷
(b) ê0, ÷
ë 2ø
(c) éê- p , p ùú
2 2
(d) ç - , ÷
ë
22.
, then f is both one - one and onto when
û
æ p pö
è 2 2ø
Let R = {(1,3), (4, 2), (2, 4), (2, 3), (3,1)} be a relation on the
set A = {1, 2,3, 4}. . The relation R is
(a) reflexive
(c) not symmetric
(b) transitive
(d) a function
Downloaded from @Freebooksforjeeneet
[2004]
M-253
Relations and Functions
23.
If f : R ® S , defined by f ( x) = sin x - 3 cos x + 1, is
onto, then the interval of S is
24.
28.
[2004]
æpö
If f (x) = ((hof)og) (x), then f ç ÷ is equal to :
è3ø
[April 12, 2019 (I)]
p
11p
7p
5p
(a) tan
(b) tan
(c) tan
(d) tan
12
12
12
12
(a) [ –1, 3] (b) [–1, 1] (c) [ 0, 1]
(d) [0, 3]
A function f from the set of natural numbers to integers
defined by
[2003]
ìn -1
ïï 2 , when n is odd
is
f (n) = í
ï - n , when n is even
ïî 2
29.
(b) one-one but not onto
(c) onto but not one-one
30.
(d) one-one and onto both.
25.
1
æ 1+ x ö
loge ç
è 1 - x ÷ø
4
(b)
1
æ 1- x ö
(log8 e)log e ç
è 1 + x ÷ø
4
(c)
1
æ 1- x ö
loge ç
è 1 + x ÷ø
4
(d)
1
æ 1+ x ö
(log8 e)log e ç
è 1 - x ÷ø
4
31.
32.
æ 5ö
If g(x) = x2 + x – 1 and (gof) (x) = 4x2 – 10x + 5, then f çè ÷ø
4
27.
3
2
[Jan. 7, 2020 (I)]
(b) -
1
2
(c)
1
2
(d) -
3
2
For a suitably chosen real constant a, let a function,
a-x
. Further supa+x
pose that for any real number x ¹ – a and f (x) ¹ – a,
1
f (x) (c) f2 (x)
(d) f1 (x)
x 3
Let N denote the set of all natural numbers. Define two
binary relations on N as R1 = {(x, y) Î N × N : 2x + y = 10}
and R2 = {(x, y) Î N × N : x + 2y = 10}. Then
[Online April 16, 2018]
(a) Both R1 and R2 are transitive relations
(b) Both R1 and R2 are symmetric relations
(c) Range of R2 is {1, 2, 3, 4}
(d) Range of R1 is {2, 4, 8}
Consider the following two binary relations on the set
A = {a, b, c} : R1 = {(c, a) (b, b) , (a, c), (c, c), (b, c), (a, a)}
and R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c). Then
[Online April 15, 2018]
(a) R2 is symmetric but it is not transitive
(b) Both R1 and R2 are transitive
(c) Both R1 and R2 are not symmetric
(d) R1 is not symmetric but it is transitive
(a) f3 (x)
[Jan. 8, 2020 (I)]
(a)
(a)
1
, f (x) = 1 – x and
x 2
1
be three given functions. If a function, J(x)
1- x
satisfies (f2oJof1) (x) = f3(x) then J(x) is equal to:
[Jan. 09, 2019 (I)]
8 2 x - 8 -2 x
The inverse function of f ( x) = 2 x -2 x , x Î(-1,1), is
8 +8
is equal to:
For xÎ R – {0, 1}, let f1 (x) =
f3 (x) =
Composite Functions & Relations,
Inverse of a Function, Binary
Operations
________.
26.
Let f ( x) = x 2 , x Î R . For any A Í R , define g(A) =
{x Î R : f ( x) Î A} . If S = [0, 4], then which one of the
following statements is not true ?
[April 10, 2019 (I)]
(a) g (f (S)) ¹ S
(b) f (g (S)) = S
(c) g (f (S)) = g (S)
(d) f (g (S)) ¹ f (S)
(a) neither one -one nor onto
TOPIC n
æ 3ö
1- x2
.
For x Î ç 0, ÷ , let f(x) = x , g(x) = tan x and h(x) =
è 2ø
1 + x2
33.
(b)
Let f : A ® B be a function defined as f (x) =
x -1
, where
x-2
f : R –{– a}® R be defined by f (x) =
A = R – {2} and B = R – {1}. Then f is
[Online April 15, 2018]
æ 1ö
( fof ) (x) = x. Then f ç - ÷ is equal to:
è 2ø
(a) invertible and f –1 (y) =
2y +1
y -1
(b) invertible and f –1 (y) =
3y - 1
y -1
[Sep. 06, 2020 (II)]
1
(a)
3
(c) no invertible
1
(b) 3
(c) – 3
(d) 3
(d) invertible and f –1 (y) =
2y -1
y -1
Downloaded from @Freebooksforjeeneet
EBD_8344
M-254
34.
Mathematics
Let f (x) = 210·x + 1 and g(x) = 310·x – 1. If (fog)(x)=x, then
x is equal to :
[Online April 8, 2017]
(a)
310 - 1
(b)
310 - 2 -10
-10
(c)
35.
1- 3
(d)
210 - 3-10
38.
f ( x) = ( x -1) +1, ( x ³1) .
2
210 - 1
{
-10
f -1 ( x ) = 1 + x - 1, x ³ 1.
1
For x Î R, x ¹ 0 , let f0(x) =
and fn+1 (x) = f0(fn(x)),
1- x
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
NOT a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
æ 2ö
æ 3ö
n = 0, 1, 2, .... Then the value of f100(3) + f1 çè ÷ø + f 2 çè ÷ø is
3
2
(a)
36.
[Online April 9, 2016]
8
3
(b)
4
3
(c)
5
3
(d)
If g is the inverse of a function f and f ' ( x ) =
g ¢ ( x ) is equal to:
(a)
37.
1
1
3
39.
1
1+ x
5
, then
[2014]
(b) 1 + { g ( x )}
5
1 + { g ( x )}
5
}
Statement - 2 : f is a biection and
310 - 2 -10
equal to :
[2011RS]
-1
Statement - 1 : The set x : f ( x ) = f ( x ) = {1, 2} .
210 - 3-10
1- 2
Let f be a function defined by
(c) 1 + x5
(d) 5x4
Let A and B be non empty sets in R and f : A ® B is a
biective function.
[Online May 26, 2012]
Statement 1: f is an onto function.
Statement 2: There exists a function g : B ® A such that
fog = IB.
(a) Statement 1 is true, Statement 2 is false.
(b) Statement 1 is true, Statement 2 is true; Statement 2 is
a correct explanation for Statement 1.
(c) Statement 1 is false, Statement 2 is true.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation for Statement 1.
40.
Let f(x) = ( x + 1)2 –1, x ³ –1
Statement -1 : The set {x : f(x) = f –1(x) = {0, –1}
Statement-2 : f is a biection.
[2009]
(a) Statement-1 is true, Statement-2 is true. Statement-2 is
not a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is true. Statement-2 is
a correct explanation for Statement-1.
Let f: N®Y be a function defined as f(x) = 4x + 3 where
Y = {y Î N : y = 4x + 3 for some x Î N}.
Show that f is invertible and its inverse is
[2008]
(a) g ( y ) =
3y + 4
3
(b) g ( y ) = 4 +
(c) g ( y ) =
y+3
4
(d) g ( y ) =
Downloaded from @Freebooksforjeeneet
y+3
4
y –3
4
M-255
Relations and Functions
1.
2.
(19.00)
The desired functions will contain either one element or
two elements in its codomain of which '2' always belongs
to f (A).
\ The set B can be {2}, {1, 2}, {2, 3}, {2, 4}
Total number of functions = 1 + (23 – 2)3 = 19.
(Bonus) f : (0, ฀) ® (0, ฀)
f (x) = 1 –
Q
1
is not a function
x
f (1) = 0 and 1 Î domain but 0 Ï co-domain
Hence, f (x) is not a function.
3.
ì1,
ï1,
ï
ï 2,
ï
ï 2,
ï
í3,
ï3,
f(g(n)) = ï
ï:
ï:
ï
ïî:
(c) Domain and codomain = {1, 2, 3, ¼, 20}.
5.
n=2
n=3
n=4
n=5
n=6
:
:
:
Þ fog is onto but not one - one
(d) As A = {x Î R : x is not a positive integer}
2x
x -1
There are five multiple of 4 as 4, 8, 12, 16 and 20.
A function f : A ® R given by f(x) =
and there are 6 multiple of 3 as 3, 6, 9, 12, 15, 18.
f(x1) = f(x2) Û x1 = x2
So, f is one-one.
As f(x) ¹ 2 for any x Î A Þ f is not onto.
Hence f is inective but not surective.
Since, when ever k is multiple of 4 then f(k) is multiple of 3
then total number of arrangement
= 6c5 × 5! = 6!
Remaining 15 elements can be arranged in 15! ways.
6.
é 1 1ù
(d) We have f : R ® ê - , ú ,
ë 2 2û
Since, for every input, there is an output
Þ function f (k) in onto
\
4.
n =1
f (x) =
x
1 + x2
Total number of arrangement = 15! . 6!
ì n +1
ïï 2 , if n is odd
(a) f(n) = íï n
,
if n is even
ïî 2
ì 2, n = 1
ï1, n = 2
ï
ï 4, n = 3
í
g(n) = ï3, n = 4
ï6, n = 5
ï
î5, n = 6
Then,
Þ f ¢(x) =
(1 + x 2 ).1 - x.2x
2 2
(1 + x )
–
=
-(x + 1)(x - 1)
(1 + x 2 ) 2
+
x = –1
–
x=1
sign of f ¢ (x)
Þ f ¢ (x) changes sign in different intervals.
\ Not inective
Now y =
x
1 + x2
Þ y + yx2 = x
Þ yx2 – x + y = 0
For y ¹ 0, D = 1 – 4y2 ³ 0
Þ
ì g ( n) + 1
ïï 2 , if g ( n) is odd
í
f(g(n)) = ï g (n)
,
if g ( n) is even
ïî 2
"x Î R
é -1 1 ù
y Î ê , ú - {0}
ë 2 2û
For y = 0 Þ x = 0
é -1 1 ù
Range is ê , ú
ë 2 2û
Þ Surective but not inective
\
Downloaded from @Freebooksforjeeneet
EBD_8344
M-256
7.
8.
9.
Mathematics
(d) f (1) = 1 - 5 [1 5] = 1 üï
ý ® Many one
f ( 6 ) = 6 - 5 [6 5] = 1ïþ
f (10) = 10 – 5(2) = 0 which is not in co–domain.
Neither one–one nor onto.
(a) Number of onto function such that exactly three
1
elements in x Î A such that f(x) = is equal to
2
= 7C3, {24 – 2} = 14. 7C3
x -1
x +1
for one-one function if f (x1) = f (x2) then
x1 must be equal to x2
Let f (x1) = f (x2)
(c)
f ( x) =
x1 - 1 x2 - 1
=
x1 + 1 x2 + 1
x1 x2 + x1 - x2 - 1 = x1 x2 - x1 + x2 - 1
Þ x1 - x2 = x2 - x1
2 x1 = 2 x2
x1 = x2
x1 = x2 , x1 = – x2
here x1 has two values therefore function is many one
function.
x -1
For onto : f ( x ) =
x +1
for every value of f (x) there is a value of x in domain
set.
If f (x) is negative then x = 0
for all positive value of f (x), domain contain atleast one
element. Hence f (x) is onto function.
10. (d) P = {( a, b) : sec 2 a - tan 2 b = 1}
For reflexive :
sec 2 a - tan 2 a = 1 (true " a)
For symmetric :
sec2 b – tan2 a = 1
L.H.S
1 + tan 2 b - (sec 2 a - 1) = 1 + tan 2 b - sec 2 a + 1
= – (sec2 a – tan2b) + 2
= – 1 +2 = 1
So, Relation is symmetric
For transitive :
if sec2 a – tan2 b = 1 and sec2 b – tan2 c = 1
sec2 a – tan2 c = (1 + tan2 b) – (sec2 b – 1)
= –sec2b + tan2b + 2
=–1+2=1
So, Relation is transitive.
Hence, Relation P is an equivalence relation
(d) R = {(x, y) : x, y Î N and x2 – 4xy + 3y2 = 0}
Now, x2 – 4xy + 3y2 = 0
Þ (x – y) (x – 3y) = 0
\ x = y or x = 3y
\ R = {(1, 1), (3, 1), (2, 2), (6, 2), (3, 3),
(9, 3),......}
Since (1, 1), (2, 2), (3, 3),...... are present in the relation,
therefore R is reflexive.
Since (3, 1) is an element of R but (1, 3) is not the element of
R, therefore R is not symmetric
Here (3, 1) Î R and (1, 1) Î R Þ (3, 1) Î R
(6, 2) Î R and (2, 2) Î R Þ (6, 2) Î R
For all such (a, b) Î R and (b, c) Î R
Þ (a, c) Î R
Hence R is transitive.
12. (d) Let R = {(3, 3), (5, 5), (9, 9), (12, 12), (5, 12), (3, 9), (3, 12),
(3, 5)} be a relation on set
A = {3, 5, 9, 12}
Clearly, every element of A is related to itself.
Therefore, it is a reflexive.
Now, R is not symmetry because 3 is related to 5 but 5 is
not related to 3.
Also R is transitive relation because it satisfies the property
that if a R b and b R c then a R c.
13. (c) Domain = {1, 2, 3, 4}
Range = {1, 2, 3, 4}
\ Domain = Range
Hence the relation R is onto function.
14. (c) Let S = {1, 2, 3} Þ n(S) = 3
Now, P (S) = set of all subsets of S
total no. of subsets = 23 = 8
\ n[P(S)] = 8
Now, number of one-to-one functions from S ® P(S) is
11.
8!
= 8 × 7 × 6 = 336.
5!
15. (b) A relation on a set A is said to be symmetric iff
8P =
3
(a, b) Î A Þ (b, a) Î A, " a, b Î A
Here A = {3, 4, 6,8,9}
Number of order pairs of A ´ A = 5 ´ 5 = 25
Divide 25 order pairs of A × A in 3 parts as follows :
Part – A : (3, 3), (4, 4), (6, 6), (8, 8), (9, 9)
Par t – B : (3, 4), (3, 6), (3, 8), (3, 9), (4, 6),
(4, 8),(4, 9), (6, 8), (6, 9), (8, 9)
Part – C : (4, 3), (6, 3), (8, 3),(9, 3), (6, 4), (8, 4), (9, 4),
(8, 6), (9, 6), (9, 8)
In part – A, both components of each order pair are same.
In part – B, both components are different but not two
such order pairs are present in which first component of
one order pair is the second component of another order
pair and vice-versa.
Downloaded from @Freebooksforjeeneet
M-257
Relations and Functions
In part–C, only reverse of the order pairs of part –B are
present i.e., if (a, b) is present in part – B, then (b, a) will be
present in part –C
For example (3, 4) is present in part – B and (4, 3) present in
part –C.
Number of order pair in A, B and C are 5, 10 and 10
respectively.
In any symmetric relation on set A, if any order pair of part
–B is present then its reverse order pair of
part –C will must be also present.
Hence number of symmetric relation on set A is equal to
the number of all relations on a set D, which contains all
the order pairs of part –A and part– B.
Now n(D) = n(A) + n(B) = 5 + 10 = 15
Hence number of all relations on set D = (2)15
Þ Number of symmetric relations on set D = (2)15
16. (a) Q x – x = 0 Î I (\ R is reflexive)
Let (x, y) Î R as x – y and y – x Î I (Q R is symmetric)
Now x – y Î I and y – z Î I Þ x – z Î I
So, R is transative.
Hence R is equivalence.
Similarly as x = ay for a = 1. B is reflexive symmetric and
transative. Hence B is equivalence.
Both relations are equivalence but not the correct
explanation.
17. (b) Let x R y.
Þx = wy Þy=
x
w
Þ (y, x) Ï R
R is not symmetric
m p
Let S : S
n q
p m
Þ qm = pn Þ =
q n
Q
m m
=
\ reflexive.
n
n
m p
p m
=
Þ q = n \ symmetric
n q
Let
m p p r
S , S
n q q s
Þ
qm = pn, ps = rq
p m r
= =
Þ
q n s
Þ ms = rn transitive.
S is an equivalence relation.
18. (d) Given that
S = {(x , y) : y = x + 1 and 0 < x < 2}
Q x ¹ x + 1 for any x Î(0, 2)
Þ (x, x) Ï S
So, S is not reflexive.
Hence, S in not an equivalence relation.
Given T ={x, y): x – y is an integer}
Q x – x = 0 is an integer, " x Î R
So, T is reflexive.
Let (x, y) Î T Þ x – y is an integer then
y – x is also an integer Þ (y, x) Î R
\T is symmetric
If x – y is an integer and y – z is an integer then
(x – y) + (y– z) = x – z is also an integer.
\ T is transitive
Hence T is an equivalence relation.
19. (b) Clearly ( x, x) Î R, "x ÎW
Q All letter are common in some word. So R is reflexive.
Let ( x, y ) Î R , then ( y , x) Î R as x and y have at least one
letter in common. So, R is symmetric.
But R is not transitive for example
Let x = BOY, y = TOY and z = THREE
then ( x, y ) Î R (O, Y are common) and (y, z) Î R (T is
common) but (x, z) Ï R. (as no letter is common)
20. (a) R is reflexive and transitive only.
Here (3, 3), (6, 6), (9, 9), (12, 12) Î R
[So, reflexive]
(3, 6), (6, 12), (3, 12) Î R [So, transitive].
Q (3, 6) Î R but (6, 3) Ï R
[So, non-symmetric]
æ 2x ö
21. (d) Given f (x) = tan -1 ç
= 2tan–1x
è 1 - x 2 ÷ø
for x Î (-1, 1)
æ -p p ö
,
If x Î( -1, 1) Þ tan -1 x Î ç
è 4 4 ÷ø
æ -p p ö
,
Þ 2 tan -1 x Î ç
è 2 2 ÷ø
æ p pö
Clearly, range of f (x) = ç - , ÷
è 2 2ø
For f to be onto, codomain = range
æ p pö
\ Co-domain of function = B = ç - , ÷ .
è 2 2ø
22. (c) Q (1, 1) Ï R Þ R is not reflexive
Q (2, 3) Î R but (3, 2) Ï R
\ R is not symmetric
Q (4, 2) and (2, 4) Î R but (4, 4) Ï R
Þ R is not transitive
23. (a) Given that f ( x) is onto
\ range of f (x) = codomain = S
Downloaded from @Freebooksforjeeneet
EBD_8344
M-258
Mathematics
26. (b) (gof) (x) = g(f(x)) = f 2(x) + f(x) – 1
Now, f (x) = sin x - 3cos x + 1
2
5
5
æ æ 5 öö
æ5ö
g ç f ç ÷ ÷ = 4 ç ÷ - 10. + 5 = 4
4
4
4
è ø
è è øø
pö
æ
= 2sin ç x - ÷ + 1
è
3ø
pö
æ
we know that -1 £ sin ç x - ÷ £ 1
è
3ø
pö
æ
-1 £ 2sin ç x - ÷ + 1 £ 3
è
3ø
\ f ( x) Î[ -1, 3] = S
24. (d) We have f : N ® I
Let x and y are two even natural numbers,
-x - y
=
Þx=y
2
2
\ f (n) is one-one for even natural number.
Let x and y are two odd natural numbers and
and f ( x) = f ( y ) Þ
x -1 y - 1
=
Þx=y
2
2
\ f (n) is one-one for odd natural number.
Hence f is one-one.
f ( x) = f ( y) Þ
Let y =
n –1
Þ 2 y +1 = n
2
[Q g(f(x)) = 4x2 – 10x +5]
æ æ 5 öö
æ5ö
æ5ö
g ç f ç ÷ ÷ = f 2 ç ÷ + f ç ÷ -1
4
4
è ø
è4ø
è è øø
5
æ5ö
- = f 2ç ÷+
4
è4ø
æ5ö
æ5ö 1
f 2ç ÷+ f ç ÷+ =0
4
è ø
è4ø 4
æ
ç
è
1
æ5ö
tç ÷ =4
2
è ø
–n
Þ –2 y = n
2
This shows that n is always even number for y Î I.
..........(ii)
From (i) and (ii)
Range of f = I = codomain
\ f is onto.
Þ
and y =
y = 2x
8 + 8-2 x
2x
1+ y 8
=
1 - y 8-2 x
Þ
84 x =
1+ y
1- y
Þ
æ1+ y ö
4 x = log8 ç
÷
è1- y ø
Þ
æ 1+ y ö
1
x = log 8 ç
÷
4
è1- y ø
\
1
æ1+ x ö
f -1( x) = log8 ç
÷
4
è 1- x ø
a - ax
a(1 - x ) a - x
= f ( x) Þ
=
Þ a =1
a+ x
1+ x
1+ x
1- x
æ 1ö
Þ f ç- ÷ = 3
è 2ø
1+ x
28. (b) Q f (x) = (( hof ) og)(x)
æ æ æ p öö ö
æpö
Q f ç 3 ÷ = h ç f ç g ç 3 ÷ ÷ ÷ = h( f ( 3)) = h(31/4 )
è ø
è è è øøø
=
82 x - 8-2 x
æ a - xö
a-ç
è a + x ÷ø
f ( f ( x )) =
=x
æ a - xö
a+ç
è a + x ÷ø
\ f ( x) =
Hence f is one one and onto both.
25. (a)
2
æ5ö 1ö
f ç ÷+ ÷ =0
è4ø 2ø
27. (d)
This shows that n is always odd number for y Î I.
..........(i)
æ5ö
f ç ÷ -1
è4ø
1- 3
1
= - (1 + 3 - 2 3) =
2
1+ 3
3 - 2 = -(- 3 + 2)
pö
æ
11p
= – tan 15 = tan (180 – 15 ) = tan ç p - ÷ = tan
12
12
è
ø
29. (c) f (x) = x2 ; x Î R
g (A) = {x Î R : f (x) Î A} S = [0, 4]
g (S) = {x Î R : f (x) Î S}
= {x Î R : 0 £ x2 £ 4} = {x Î R : – 2 £ x £ 2}
\ g (S) ¹ S \ f (g (S)) ¹ f (S)
g (f (S)) = {x Î R : f (x) Î f (S)}
= {x Î R : x2 Î S2} = {x Î R : 0 £ x2 £ 16}
= {x Î R : – 4 £ x £ 4}
\ g (f (S)) ¹ g (S)
\ g (f (S)) = g (S) is incorrect.
Downloaded from @Freebooksforjeeneet
M-259
Relations and Functions
Now, for R2 , (b, a) Î R2, (a, c) Î R2 but (b, c) Ï R2.
Similarly, for R1 , (b, c) Î R1, (c, a) Î R1 but (b, a) Ï R1.
Therefore, neither R1 nor R2 is transitive.
33. (d) Suppose y = f (x)
30. (a) The given relation is
(f2o Jo f1) (x) = f3(x) =
Þ (f2oJ) (f1(x)) =
1
1- x
1
æ 1ö
1
Þ (f2o J) çè ÷ø = 1 - 1
x
x
x
Þ (f2o J)(x) =
x -1
Þ f2 (J(x)) =
x
x -1
Þ 1 – J(x) =
x
x -1
\ J(x) = 1 -
1
1- x
Þ y=
=
1
x
1
-1
x
1ù
é
êëQ f1 ( x ) = x úû
é1
ù
êë x is replaced by x úû
Þ yx – 2y = x – 1
Þ (y – 1) x = 2y – 1
Þ x = f –1 (y) =
2y -1
y -1
As the function is invertible on the given domain and its
inverse can be obtained as above.
34. (d) f (g(x)) = x
Þ f (310x – 1) = 210 (310 . x – 1) + 1 = x
Þ 210 (310x – 1) + 1 = x
Þ x (610 – 1) = 210 – 1
[Q f2(x) = 1 – x]
Þ x=
1
x
=
= f 3 ( x)
x -1 1- x
31. (c) Here,
R1 = {(x, y) Î N × N : 2x + y = 10} and
R2 = {(x, y) Î N × N : x + 2y = 10}
For R1; 2x + y = 10 and x, y Î N
So, possible values for x and y are:
x = 1, y = 8 i.e. (1, 8);
x = 2, y = 6 i.e. (2, 6);
x = 3, y = 4 i.e. (3, 4) and
x = 4, y = 2 i.e. (4, 2).
R1 = {(1, 8), (2, 6), (3, 4), (4, 2)}
Therefore, Range of R1 is {2, 4, 6, 8}
R1 is not symmetric
Also, R1 is not transitive because (3, 4), (4, 2) Î R1 but
(3, 2) Ï R1
Thus, options A, B and D are incorrect.
For R2; x + 2y = 10 and x, y Î N
So, possible values for x and y are:
x = 8, y = 1 i.e. (8, 1);
x = 6, y = 2 i.e. (6, 2);
x = 4, y = 3 i.e. (4, 3) and
x = 2, y = 4 i.e. (2, 4)
R2 = {(8, 1), (6, 2), (4, 3), (2, 4)}
Therefore, Range of R2 is {1, 2, 3, 4}
R2 is not symmetric and transitive.
32. (a) Both R1 and R2 are symmetric as
For any (x, y) Î R1, we have
(y, x) Î R1 and similarly for R2
x -1
x-2
210 - 1
610 - 1
=
1- 2-10
310 - 2-10
35. (c) f1 (x) = f0 + 1 (x) = f0 (f0 (x)) =
f2 (x) = f1 + 1 (x) = f0 (f1 (x)) =
1
1
11- x
1
=x
x -1
1x
f3 (x) = f2 + 1 (x) = f0 (f2 (x)) = f0 (x) =
f4 (x) = f3 + 1 (x) = f0 (f3 (x)) =
\ f0 = f3 = f6 = .......... =
1
1- x
x -1
x
1
1- x
x -1
x
f2 = f5 = f8 = .......... = x
f1 = f4 = f7 = .......... =
2
3 -1 2 æ 2 ö 3 - 1
1
=f100 (3) =
= f1 ç ÷ =
3
3 è 3ø
2
2
3
æ 3ö 3
f2 ç ÷ =
è 2ø 2
\
æ 2ö
f100 (3) + f1 ç ÷ + f2
è 3ø
æ 3ö
5
ç ÷=
2
3
è ø
Downloaded from @Freebooksforjeeneet
=
x -1
x
EBD_8344
M-260
Mathematics
Now f ( x ) = f -1 ( x )
36. (b) Since f (x) and g(x) are inverse of each other
\
Þ
Þ f ( x ) = x Þ ( x - 1) + 1 = x
1
f '( x )
g'( f (x)) =
2
1 ö
æ
çèQ f ¢ ( x) =
÷
1 + x5 ø
g '( f ( x)) = 1 + x5
Here x = g(y)
\
g ¢( y ) = 1 + [ g ( y )]
5
Þ g ¢ ( x ) = 1 + ( g ( x) ) 5
37. (d) Let A and B be non-empty sets in R.
Let f : A ® B is biective function.
Clearly statement - 1 is true which says that f is an onto
function.
Statement - 2 is also true statement but it is not the
correct explanation for statement-1
38. (a) Given f is a biective function
\
Þ x 2 - 3 x + 2 = 0 Þ x = 1, 2
Hence statement-1 is correct
39. (d) Given that f (x) = (x + 1)2 –1, x ³ –1
Clearly Df = [–1, ¥ ) but co-domain is not given. Therefore
f (x) is onto.
Let f (x1) = f (x2)
Þ (x1 + 1)2 – 1 = (x2 + 1)2 – 1
Þ x1 = x2
\ f (x) is one-one, hence f (x) is biection
Q (x + 1) being something +ve, " x > –1
\ f –1(x) will exist.
Let (x + 1)2 –1 = y
Þ x +1 =
Þ x =–1+
2
Þ x = 1±
Þ
2
y - 1 Þ f -1 ( y ) = 1 ± y - 1
f -1 ( x ) = 1 + x - 1 {\ x ³ 1}
Hence statement-2 is correct
y +1
Þ (x + 1)2 – 1 =
y = ( x - 1) + 1 Þ ( x - 1) = y - 1
2
(+ve square root as x +1 ³ 0 )
Þ f –1 (x) = x + 1 – 1
Then f (x) = f –1 (x)
f :[1, ¥) ® [1, ¥)
f ( x ) = ( x - 1) + 1, x ³ 1
Let
y +1
40.
x +1 –1
Þ (x + 1)2 = x + 1 Þ (x + 1)4 = (x + 1)
Þ (x + 1) [ (x + 1)3 – 1] = 0 Þ x = – 1, 0
\ The statement-1 and statement-2 both are true.
(d) Clearly f (x) = 4x + 3 is one one and onto, so it is
invertible.
Let f (x) = 4x + 3 = y
Þx=
y –3
4
\ g ( y) =
y –3
4
Downloaded from @Freebooksforjeeneet
M-261
Inverse Trigonometric Functions
17
Inverse Trigonometric
Functions
TOPIC Ć
1.
Trigometric Functions & Their
Inverses, Domain & Range of Inverse
Trigonometric Functions, Principal
Value of Inverse Trigonometric
Functions, Intervals for Inverse
Trigonometric Functions
7.
æ 9 ö
(d) sin–1 ç
÷
è 5 10 ø
æ 9ö
(c) tan–1 ç ÷
è 14 ø
A value of x satisfying the equation
[tan–1x], is :
[Online April 9, 2017]
1
2
(b) –1
(c) 0
(d)
4.
5.
8.
1
2
3p
p
p
3p
(b)
(c) (d)
4
4
4
4
The number of solutions of the equation,
sin–1 x = 2 tan –1x (in principal values) is :
[Online April 22, 2013]
(a) 1
(b) 4
(c) 2
(d) 3
A value of tan
(a)
p
4
-1
(b)
æ æ
æ 2 ööö
ç sin ç cos -1 ç
÷ is
ç 3 ÷÷ ÷÷ ÷
ç
ç
è
øøø
è è
[Online May 19, 2012]
p
2
(c)
p
3
(d)
p
6
p
(b) êé 0, ÷ö
ë 2ø
(c) [0, p]
p p
(d) æç - , ö÷
è 2 2ø
The domain of the function f ( x) =
is
9 - x2
(b) [2, 3)
(d) [2, 3]
1
2
(c) all real values of a
9.
sin -1 ( x - 3)
[2004]
The trigonometric equation sin -1 x = 2 sin -1 a has a
solution for
[2003]
(a) a £
43p ö
æ
The principal value of tan -1 ç cot
÷ is:
4 ø
è
[Online April 19, 2014]
(a) -
(a) é - p , p ö
ê 4 2÷
ë
ø
(a) [1, 2]
(c) [1, 2 ]
sin[cot–1(1 + x) ] = cos
(a) 3.
æ 9 ö
(b) cos–1 ç
÷
è 5 10 ø
2
æx ö
f ( x ) = 4- x + cos -1 ç - 1÷ + log (cos x) , is defined, is
è2 ø
[2007]
[April 8, 2019 (I)]
æ 9 ö
(a) tan–1 ç
÷
è 5 10 ø
The largest interval lying in æç -p , p ö÷ for which the
è 2 2ø
function,
æ 3ö
æ1ö
p
If a = cos–1 ç ÷ , b = tan–1 ç ÷ , where 0 < a, b < , then
2
è5ø
è3ø
a – b is equal to :
2.
6.
(b)
1
1
< a <
2
2
(d) a <
1
2
cot -1 ( cos a ) - tan -1 ( cos a ) = x ,
then sin x =
æaö
è2ø
10.
(a) tan2 ç ÷
a
(b) cot2 çæ ÷ö
(c) tan a
(d) cot ç 2 ÷
è ø
è2ø
æaö
The domain of sin-1 [log3 (x/3)] is
(a) [1, 9]
(b) [–1, 9]
(c) [–9, 1]
[2002]
(d) [–9, –1]
Downloaded from @Freebooksforjeeneet
[2002]
EBD_8344
M-262
Mathematics
TOPIC n
11.
4
5
16 ö
æ
2p - ç sin -1 + sin -1 + sin -1 ÷ is equal to :
5
13
65 ø
è
[Sep. 03, 2020 (I)]
(a)
12.
p
2
(b)
5p
4
(c)
3p
2
(d)
(b)
( cot 2, ¥ )
(c)
( -¥, cot 5) È ( cot 2, ¥ )
(d) (cot 5, cot 4)
17.
7p
4
n
æ 19
æ
öö
The value of cot ç å cot -1 ç1 + å 2p ÷ ÷ is:
ç
÷÷
ç n =1
è p =1 øø
è
[Jan. 10, 2019 (II)]
If S is the sum of the first 10 terms of the series
(a)
21
19
(b)
19
21
æ1ö
æ1ö
æ1ö
æ 1ö
tan -1 ç ÷ + tan -1 ç ÷ + tan -1 ç ÷ + tan -1 ç ÷ + . . . . ,
è 3ø
è7ø
è 13 ø
è 21 ø
then tan (S) is equal to:
[Sep. 05, 2020 (I)]
(c)
22
23
(d)
23
22
(a)
5
6
(c) -
13.
Properties of Inverse Trigonometric
Functions, Infinite Series of Inverse
Trigonometric Functions
6
5
(b)
5
11
(d)
10
11
18.
19.
If x = sin –1 (sin10) and y = cos–1 (cos10), then y – x is equal
to:
[Jan. 09, 2019 (II)]
(a) 0
(b) 10
(c) 7 p
(d) p
3ö
pæ
-1 æ 2 ö
–1 æ 3 ö
If cos ç ÷ + cos ç ÷ = ç x > ÷ ,then x is equal
4ø
è 3x ø
è 4x ø 2 è
to:
-1 æ 12 ö
-1 æ 3 ö
The value of sin ç ÷ - sin ç ÷ is equal to :
13
è ø
è 5ø
(a)
[April 12, 2019 (I)]
-1 æ 63 ö
(a) p - sin ç ÷
è 65 ø
(c)
14.
15.
p
æ 9 ö
- cos -1 ç ÷
2
è 65 ø
(b)
p
æ 56 ö
- sin -1 ç ÷
2
è 65 ø
-1 æ 33 ö
(d) p - cos ç ÷
è 65 ø
y
x < , then for all x, y, 4x2 –4xy cosa + y2 is equal to:
2
[April 10, 2019 (II)]
(a) 4 sin2a
(b) 2 sin2a
(c) 4 sin2a – 2x2y2
(d) 4 cos2 a + 2x2y2
Considering only the principal values of inverse functions,
All x satisfying the inequality (cot–1x)2 – 7 (cot–1x) + 10 >
0, lie in the interval :
[Jan. 11, 2019 (II)]
(a)
( -¥, cot 5) È ( cot 4, cot 2 )
145
12
145
10
(b)
(c)
21.
(d)
145
11
[Online April 8, 2017]
(a)
p 1
+ cos -1 x 2
4 2
(b)
p
+ cos -1 x 2
4
(c)
p 1
- cos -1 x 2
4 2
(d)
p
- cos -1 x 2
4
Let
æ 2x ö
tan –1 y = tan –1 x + tan –1 ç
è 1 - x 2 ÷ø ,
where or x <
(a)
(c)
22.
146
12
é 1 + x2 + 1 - x 2 ù
1
ú , | x |< , x ¹ 0 ,
The value of tan–1 ê
2
ê 1 + x 2 - 1 - x2 ú
ë
û
is equal to
-1
-1 y
= a , where – 1 < x < 1, –2 < y < 2,
If cos x - cos
2
pü
ì
-1
-1
the set A = í x ³ 0 : tan ( 2 x ) + tan ( 3x ) = ý
4þ
î
[Jan. 12, 2019 (I)]
(a) contains two elements
(b) contains more than two elements
(c) is a singleton
(d) is an empty set
16.
20.
[Jan. 09, 2019 (I)]
1
. Then a value of y is :
3
3x - x 3
1 + 3x 2
3x – x 3
1 – 3x 2
(b)
3x + x 3
1 + 3x 2
(d)
3x + x 3
1 - 3x 2
[2015]
æ 2x ö
2 tan -1 x + sin -1 ç
÷ , x > 1 th en
è 1+ x 2 ø
f (5) is equal to :
[Online April 10, 2015]
If f (x) =
(a) tan -1 æç 65 ö÷
è 156 ø
(c) p
p
2
(d) 4 tan–1(5)
(b)
Downloaded from @Freebooksforjeeneet
M-263
Inverse Trigonometric Functions
23.
Statement I: The equation (sin–1x)3 + (cos–1 x)3 – ap3 = 0
1
has a solution for all a ³
.
32
p
Statement II: For any x Î R , sin -1 x + cos-1 x = and
2
26.
æ
ö
1
+ tan -1 ç
÷ , then tan S is equal to :
1
(
19)(
20)
+
n
+
n
+
è
ø
[Online April 23, 2013]
20
n
(b)
(a)
2
401 + 20 n
n + 20n + 1
2
24.
25.
pö
9 p2
æ
0 £ ç sin -1 x - ÷ £
[Online April 12, 2014]
4ø
16
è
(a) Both statements I and II are true.
(b) Both statements I and II are false.
(c) Statement I is true and statement II is false.
(d) Statement I is false and statement II is true.
If x, y, z are in A.P. and tan –1x, tan–1y and tan–1z are also in
A.P., then
[2013]
(a) x = y = z
(b) 2x = 3y = 6z
(c) 6x = 3y = 2z
(d) 6x = 4y = 3z
(c)
27.
28.
æ 1 ö
,1÷
(b) ç
è 2 ø
(c) (0, 1)
æ
3ö
(d) çç 0, 2 ÷÷
è
ø
29.
20
(d)
2
n
401 + 20 n
n + 20n + 1
A value of x for which sin (cot–1(1 + x)) = cos (tan–1 x), is
:
[Online April 9, 2013]
(a) -
Let x Î (0, 1). The set of all x such that
sin–1x > cos–1x, is the interval: [Online April 25, 2013]
æ1 1 ö
(a) ç ,
÷
è2 2ø
1
æ
ö
1
ö
-1 æ
S = tan -1 ç
÷ + tan ç 2
÷ + ...
2
è n + n +1 ø
è n + 3n + 3 ø
1
2
(b) 1
(c) 0
(d)
1
2
æ xö
æ 5ö p
If sin -1 ç ÷ + cosec -1 ç ÷ = , then the values of x is
è 5ø
è 4ø 2
(a) 4
(b) 5
[2007]
(c) 1
(d) 3
y
If cos -1 x - cos -1 = a , then 4 x 2 - 4 xy cos a + y 2 is
2
equal to
[2005]
(a) 2 sin 2 a
(b) 4
(c) 4 sin 2 a
(d) – 4 sin 2 a
Downloaded from @Freebooksforjeeneet
EBD_8344
M-264
1.
Mathematics
(d) Q cos a =
Þ tan a =
4
3
and tan b =
1
3
3
4
, then sin a =
5
5
æ p 3p ö ù
-1 é
= tan ê tan çè 2 - 4 ÷ø ú
ë
û
4.
tan a - tan b
Q tan (a – b) = 1 + tan a.tan b
4 1
1
9
=3 3=
4 13 =
13
1+
9
9
\
Also, x = 1 gives angle value as
æ 9 ö
æ9ö
a – b = tan–1 ç ÷ = sin –1 ç
÷
è 13 ø
è 5 10 ø
(a)
5.
(
sin écot -1 (1 + x ) ù = cos tan -1 x
ë
û
)
1+ x
1
Let; cot l = 1 + x
tan b = x
Þ sin l = cos b
3.
1
x2 + 2x + 2
=
1
1 1 + x2
Þ
x2 + 2 x + 2 = x2 + 1
Þ
x = -1 2
3p ù
-1 é
= tan êcot ú
4û
ë
[Q cot (2np + q) = cot q]
sin q = 1 - cos 2 q = 1 -
2
3
2
=
3
1
3
6.
p
-1 é 1 ù
-1 æ 1 ö
= tan ê ú = tan ç
÷ =
è 3ø 6
ë 3û
(b) Given that
2
æx ö
f (x) = 4- x + cos -1 ç - 1÷ + log(cos x)
è2 ø
æx ö
f (x) is defined if – 1 £ ç - 1÷ £ 1 and cos x > 0
è2 ø
(c) Consider
é 43p ù
é æ
3p ö ù
tan -1 êcot
= tan -1 êcot ç10 p + ÷ ú
4 úû
ë
4 øû
ë è
2
= q Þ cos q =
3
é æ -1 2 ö ù
-1
-1
\ tan êsin çç cos
÷÷ ú = tan [sin q]
3
ø úû
ëê è
b
1+x
Þ
Þ
2
x
l
é æ
2 öù
(d) Consider tan -1 êsin ç cos-1
÷ú
ç
3 ÷ø úû
ëê è
Let cos -1
2 + 2x + x2
1
p
5p
and
4
4
5p
is outside the principal value.
4
æ 13 ö
= cos–1 çè 5 10 ÷ø
2.
p 3p
2p - 3p -p
=
=
2 4
4
4
(a) Given equation is
sin–1 x = 2 tan–1 x
Now, this equation has only one solution.
p
\ LHS = sin–1 1 =
2
p p
–1
and RHS = 2 tan 1 = 2 ´ =
4 2
=
Þ
Þ
\
p
p
x
£ 2 and - < x <
2
2
2
p
p
0 £ x £ 4 and - < x <
2
2
0£
é pö
x Î ê 0, ÷
ë 2ø
Downloaded from @Freebooksforjeeneet
M-265
Inverse Trigonometric Functions
7.
8.
(b)
f ( x) =
sin -1 ( x - 3)
9 - x2
11.
is defined
When -1 £ x - 3 £ 1 Þ 2 £ x £ 4
....(i)
and 9 - x 2 > 0 Þ -3 < x < 3
from (i) and (ii),
we get 2 £ x < 3 \ Domain = [2, 3)
....(ii)
(a) Given that sin -1 x = 2 sin -1 a
Þ-
p
p
£ 2sin -1 a £
2
2
p
p
-1
1
Þ - £ sin -1 a £ Þ
£a£
4
4
2
2
1
\ a £
2
9.
(a) Given that, cot–1 ( cos a ) – tan–1 ( cos a ) = x
æ 1
tan–1 çç
è cos a
ö
÷÷ – tan–1 ( cos a ) = x
ø
1
Þ
tan–1
Þ tan–1
1+
cos a
1
1 - cos a
2 cos a
Þ tan x =
Þ cot x =
- cos a
cos a
. cos a
4
5
16 ö
æ
2p - ç sin -1 + sin -1 + sin -1 ÷
è
5
13
65 ø
4
5
16 ö
æ
= 2p - ç tan -1 + tan -1 + tan -1 ÷
è
3
12
63 ø
é
-1 4
-1 4 ù
êëQ sin 5 = tan 3 úû
ì
ü
æ 4 5 ö
ïï -1 ç 3 + 12 ÷
ï
-1 16 ï
= 2 p - í tan ç
÷ + tan 63 ý
4
5
ï
ï
ç1- × ÷
è
ïî
ïþ
3 12 ø
p
p
£ sin -1 x £
2
2
We know that -
(c)
63
16 ö
æ
= 2p - ç tan -1 + tan -1 ÷
è
16
63ø
63
63ö
æ
= 2p - ç tan -1 + cot -1 ÷
è
16
16 ø
= 2p -
12. (a)
p 3p
.
=
2
2
1
1
æ 1ö
S = tan -1 ç ÷ + tan -1 + tan -1 + .... upto 10 terms
è 3ø
7
13
æ 2 -1 ö
æ 3- 2 ö
= tan -1 ç
+ tan -1 ç
è 1 + 2 ×1÷ø
è 1 + 3 × 2 ÷ø
=x
æ 4-3 ö
æ 11 - 10 ö
+ tan -1 ç
+ ...... + tan -1 ç
è 1 + 3 × 4 ÷ø
è 1 + 11×10 ÷ø
=x
= (tan -1 2 - tan -1 1) + (tan -1 3 - tan -1 2) +
1 - cos a
(tan -1 4 - tan -1 3) + .... + (tan -1 11 - tan -1 10)
2 cos a
æ 11 - 1 ö
æ 5ö
= tan -1 11 - tan -1 1 = tan -1 ç
= tan -1 ç ÷
è 1 + 11×1÷ø
è 6ø
2 cos a B
=
1 - cos a P
\ tan( S ) =
P = (1– cosa) and B = 2 cos a
H = P2 + B2 = 1 + cos a
1 - cos a 1 - (1 - 2 sin a / 2)
=
1 + cos a 1 + 2 cos 2 a / 2 - 1
a
or sin x = tan2
2
2
Þ sin x =
æ
æ xö ö
10. (a) f (x) = sin–1 ç log 3 ç ÷ ÷
è 3ø ø
è
We know that domain of sin–1x is x Î [–1, 1]
x
x
\ -1 £ log3 çæ ÷ö £ 1 Þ 3-1 £ £ 31
3
è 3ø
13. (b)
5
6
æ3ö
æ 12 ö
æ 3 5 12 4 ö
- sin -1 ç ÷ + sin -1 ç ÷ = - sin -1 ç ´ - ´ ÷
è5ø
è 13 ø
è 5 13 13 5 ø
(Q xy e” 0 and x2 + y2 d” 1)
{
}
é
2
2 ù
-1
-1
-1
êëQ sin x - sin y = sin x 1 - y - y 1 - x úû
æ -33 ö
æ 33 ö
= sin –1 ç
÷ = sin –1 ç 65 ÷
65
è
ø
è ø
p
æ 56 ö
-1 æ 56 ö
= cos–1 ç ÷ = - sin ç ÷
65
2
è ø
è 65 ø
Þ 1 £ x £ 9 or x Î [1, 9]
Downloaded from @Freebooksforjeeneet
EBD_8344
M-266
Mathematics
-1
-1
14. (a) Given, cos x - cos
æ 19
ö
-1
= cot ç å cot (1 + n (n + 1))÷
è n=1
ø
y
=a
2
æ xy
y 2 ö÷
Þ cos -1 ç + 1 - x 2 . 1 =a
4 ÷
ç 2
è
ø
Þ
æ 19
-1 æ ( n + 1) - n ö ö
= cot ç å tan çè 1 + (n + 1)n ÷ø ÷
è n =1
ø
é -1
ù
-1 æ 1 ö
êcot x = tan çè x ÷ø : for x > 0 ú
ë
û
1 - x2 4 - y 2
xy
+
= cos q
2
2
Þ xy + 1 - x 2 4 - y 2 = 2cos a
Þ (xy – 2 cos a)2 = (1 – x2) (4 – y2)
Þ x2y2 + 4 cos2 a – 4xy cos a = 4 – y2 – 4x2 + x2y2
Þ 4x2 – 4xy cos a + y2 = 4 sin2 a
15. (c) Consider, tan–1(2x) + tan–1(3x) =
Þ
æ 5x ö p
tan -1 ç
è 1 - 6 x 2 ÷ø = 4
Þ
5x
= 1 Þ 5x = 1 – 6x2
1 - 6 x2
p
4
æ 19
-1
-1 ö
cot
= ç å (tan ( n + 1) - tan n)÷
è n =1
ø
= cot(tan–1 20 – tan–11)
æ -1 æ 20 - 1 ö ö
÷
= cot çè tan çè
1 + 20 ´ 1ø ÷ø
21
æ -1 æ 19 ö ö
-1 æ 21ö
= cot çè tan çè ø÷ ø÷ = cot cot èç ø÷ =
21
19
19
18. (d) x = sin –1(sin 10)
Þ x = 3p – 10
Þ 6x + 5x – 1 = 0
Þ (6x – 1)(x + 1) = 0
2
Þ x=
p
p
ì
ï3p - < 10 < 3p +
2
2
í
ïîÞ 3p - x = 10
and y = cos–1(cos 10)
1
(as x ³ 0)
6
Therefore, A is a singleton set.
ì3p < 10 < 4p
í
îÞ 4 p - x = 10
Þ y = 4p – 10
\ y – x = (4p – 10) – (3p – 10) = p
16. (b)
19. (a)
3ö
æ 2ö
æ 3ö p æ
cos -1 ç ÷ + cos-1 ç ÷ = ; ç x > ÷
è 3x ø
è 4xø 2 è
4ø
Þ
æ 2ö p
æ 3ö
cos -1 ç ÷ = - cos -1 ç ÷
è 3x ø 2
è 4x ø
Þ
æ 2ö
æ 3ö é
cos -1 ç ÷ = sin -1 ç ÷ êQ sin -1 x + cos -1 x =
è 3x ø
è 4x ø ë
pù
2 úû
3
-1 æ 3 ö
Put sin ç ÷ = q Þ sin q =
è 4x ø
4x
(cot–1 x)2 – 7(cot–1 x) + 10 > 0
(cot–1 x – 5) (cot–1 – 2) > 0
cot–1 x Î (–฀, 2) È (5, ฀)
¼(1)
But cot–1 x lies in (0, p)
Now, from equation (1)
cot–1 x Î (0, 2)
2
Þ cos q = 1 - sin q = 1 -
17. (a)
æ 19
n
æ
öö
cat ç å cot -1 ç1 + å 2p ÷ ÷
ç n =1
ç p =1 ÷ ÷
è
øø
è
16x2
æ 16 x 2 - 9 ö
÷÷
Þ q = cos -1 çç
è
ø
4x
Now, it is clear from the graph
x Î (cot 2, ฀)
9
\
æ 16 x 2 - 9 ö
-1
æ 2ö
÷
cos -1 ç ÷ = cos ç
çè
4x
è 3x ø
ø÷
Þ
2
145
16 x 2 - 9
64 + 81
=
Þ x=±
Þ x2 =
3x
144
4x
9 ´ 16
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M-267
Inverse Trigonometric Functions
20. (a)
Þ
pö
æ -1
Þ ç sin x - ÷
è
4ø
3ö
æ
çèQ x > ÷ø
4
145
12
Þ x=
Let x = cos 2q;
Þ
é 1+ x2
1 - x2
tan -1 ê
ê 1+ x2
1 - x2
ë
pö
æ -1
p2
Þ ç sin x - ÷ =
(32a - 1)
è
4ø
48
Putting this value in equation (1)
ù
ú
ú
û
0£
é1 + tan q ù tan - 1 é tan æ p + q ö ù
Þ tan–1 ê
ç
÷ú
ê
ú=
è4
øû
ë
ë1 - tan q û
p 1
+ cos -1 x 2
4 2
é 2x ù
21. (c) Given that, tan–1y = tan -1 x + tan -1 ê
ú
ë1 - x 2 û
= tan -1 x + 2 tan -1 x = 3 tan–1x
3ù
é
-1 3x - x
tan–1y = tan ê
2ú
ëê 1 - 3x ûú
3x - x 3
1 - 3x 2
æ 2x ö
22. (c) f (x) = 2 tan–1 x + sin –1 çè
÷
1 + x2 ø
Þ f (x) = 2 tan–1x + p – 2 tan–1x
Þ f (x) = p
Þ f (5) = p
23. (a)
é p pù
sin -1x Î ê - , ú
ë 2 2û
p2
9
(32 a - 1) £ p 2
48
16
Þ 0 £ 32a - 1 £ 27
1
7
£a£
32
8
Statement-I is also true
24. (a) Since, x, y, z are in A.P.
\ 2y = x + z
Also, we have
2 tan–1 y = tan–1x + tan–1 (z)
æ 2y ö
-1 æ x + z ö
Þ tan–1 ç
÷ = tan çè 1 - xz ÷ø
è 1 - y2 ø
x+ z
x+z
(Q 2y = x + z)
1
- xz
1- y
Þ y2 = xz or x + z = 0 Þ x = y = z = 0
25. (b) Given sin–1 x > cos–1 x where x Î (0, 1)
p
Þ sin–1 x > - sin -1 x
2
p
p
-1
Þ sin x >
Þ 2 sin–1 x >
2
4
p
1
Þ x > sin Þ x >
4
2
p
Maximum value of sin–1 x is
2
Þ
2
1
1
1
+ tan -1
+ tan -1
+ ...... +
1+ 2
1 + 2´ 3
1 + 3´ 4
tan -1
2
9
pö
æ
0 £ ç sin -1 x - ÷ £ p 2
è
4ø
16
=
æ 1 ö
So, maximum value of x is 1. So, x Î ç
, 1÷ .
è 2 ø
26. (c) We know that,
tan -1
pö p
3p æ -1
£ ç sin x - ÷ £
Þ 4 è
4ø 4
..(1)
3
(sin -1 x)3 + (cos -1 x )3 = ap
Þ (sin–1x + cos–1x) [(sin–1x + cos–1x)2
– 3sin–1x cos–1x] = ap3
p2
- 3sin -1 x cos -1 x = 2ap2
4
p2
-1 æ p
-1 ö
Þ sin x ç - sin x÷ =
(1 - 8a )
è2
ø
12
1
1
+ tan -1
+ ...... +
1 + (n - 1)n
1 + n(n + 1)
tan -1
Statement II is true
Þ
p2
p2
(8a - 1) +
12
16
1
-1 2
q = cos x
2
é 1 + cos 2q + 1 - cos 2q ù
=ê
ú
ë 1 + cos 2q - 1 - cos 2q û
Þ y=
=
2
2
=
2
Þ tan -1
1
n + 19
= tan -1
1 + (n + 19) (n + 20)
n + 21
n -1
1
+ tan -1
1 + n(n + 1)
n +1
+ tan -1
1
1
+...... +
1 + (n + 1) (n + 2)
1 + (n + 19) ( n + 20)
-1
= tan
n + 19
n + 21
Downloaded from @Freebooksforjeeneet
EBD_8344
M-268
Mathematics
Þ tan -1
1
1
+ tan -1
+ ...... +
1 + n(n + 1)
1 + (n + 1)( n + 2)
1
n -1
-1 n + 19
- tan -1
= tan
1 + ( n + 19) ( n + 20)
n + 21
n +1
1
1
æ
ö
ö
-1 æ
Þ tan-1 ç
÷ + tan ç 2
÷ + ...... +
2
+
+
+
+
n
n
1
n
3
n
3
è
ø
è
ø
1
tan -1
1 + ( n + 19) ( n + 20)
= tan
-1
æ n + 19 n - 1 ö
ç n + 21 - n + 1 ÷
20
-1
=S
ç
÷ = tan
2
n
n
+
19
1
n
+
20n + 1
´
çç 1 +
÷÷
n + 21 n + 1 ø
è
20
\ tan -1 S =
2
n + 20n + 1
27. (a) sin (cot–1 (1 + x)) = cos (tan –1 x)
Þ cosec2 (cot–1 (1 + x)) = sec2 (tan–1 x)
Þ 1 + [cot (cot–1 (1 + x))]2 = 1 + [tan (tan–1 x)]2
1
Þ (1 + x)2 = x2 Þ x = 2
x
5ö p
æ
ö
æ
28. (d) sin -1 ç ÷ + cosec -1 ç ÷ =
è 5ø
è 4ø 2
Þ
æ xö p
æ 5ö
sin -1 ç ÷ = - cosec -1 ç ÷
è 5ø 2
è 4ø
Þ
æ xö p
æ 4ö
sin -1 ç ÷ = - sin -1 ç ÷
è 5ø 2
è 5ø
[Q sin -1 x + cos -1 x = p / 2]
Þ
æ xö
æ 4ö
sin -1 ç ÷ = cos -1 ç ÷
è 5ø
è 5ø
sin –1
x
æ4ö
= sin –1 1 – ç ÷
5
è5ø
-1
Þ sin
2
éQ cos –1x = sin –1 1 – x 2 ù
êë
úû
3
x
x 3
= sin -1 Þ
=
5
5
5 5
Þ x=3
29. (c)
cos -1 x - cos-1
y
=a
2
Þ
æ
ö
æ
xy
y2 ö
cos - 1 ç + (1 - x 2 ) ç1 - ÷ ÷ = a
4 ø÷
çè 2
è
ø
Þ
cos
Þ
xy + 4 – y 2 – 4 x 2 + x 2 y 2 = 2cos a
Þ
4 – y 2 – 4 x 2 + x 2 y 2 = 2cos a – xy
-1
æ xy + 4 - y 2 - 4 x 2 + x2 y 2 ö
ç
÷ =a
2
çè
÷ø
Squaring both sides, we get
Þ 4 - y 2 - 4 x 2 + x 2 y 2 = 4 cos 2 a + x 2 y 2 - 4 xy cos a
Þ
4 x 2 + y 2 - 4 xy cosa = 4 sin 2 a .
Downloaded from @Freebooksforjeeneet
M-269
Matrices
18
Matrices
TOPIC Ć
1.
Order of Matrices, Types of Matrices,
Addition & Subtraction of Matrices,
Scalar Multiplication of Matrices,
Multiplication of Matrices
Let a be a root of the equation x2 + x + 1 = 0
1
and the matrix A =
3
é1 1
ê
ê1 a
ê
2
ë1 a
1ù
ú
a 2 ú , then the matrix A31
ú
a4 û
is equal to:
(a) A
2.
5.
(c) A2
(a) 10
6.
(d) A3
é cos q i sin q ù æ
p ö
éa b ù
5
If A = ê
÷ and A = ê
ú, çq =
ú,
24 ø
ëi sin q cos q û è
ëc d û
7.
where i = -1, then which one of the following is not
true?
[Sep. 04, 2020 (I)]
(a) 0 £ a 2 + b2 £ 1
(b) a 2 - d 2 = 0
(c) a - c = 1
1
(d) a - b =
2
2
3.
2
2
8.
é x 1ù
4
Let A = ê
ú , x Î R and A = [aij]. If a11 = 109, then a22
1
0
ë
û
is equal to __________.
4.
2
[NA Sep. 03, 2020 (I)]
q 21 + q31
is equal to :
q32
[Jan. 12, 2019 (I)]
(c) 15
(d) 9
such that Q – P5 = I3. Then
[Jan. 7, 2020 (I)]
(b) I3
é1 0 0ù
Let P = êê3 1 0úú and Q = [qi] be two 3 × 3 matrices
êë9 3 1úû
9.
(b) 135
é1 0 0 ù
Let A = êê1 1 0 úú and B = A20. Then the sum of the
ëê1 1 1 úû
elements of the first column of B is?
[Online April 16, 2018]
(a) 211
(b) 210
(c) 231
(d) 251
é0 -1ù
A=ê
ú , then which one of the following
ë1 0 û
statements is not correct?
[Online April 10, 2015]
(a) A2 + I = A(A2 – I)
(b) A4 – I = A2 + I
(c) A3 + I = A(A3 – I)
(d) A3 – I = A(A – I)
If
é yù
6
é1 2 x ù
ê x ú be such that AB = é ù ,
B
=
If A = ê
and
ê8 ú
ú
ê
ú
ë û
ë3 - 1 2 û
ëê 1 ûú
then:
[Online April 12, 2014]
(a) y = 2x
(b) y = – 2x
(c) y = x
(d) y = – x
If p, q, r are 3 real numbers satisfying the matrix equation,
écos a - sin a ù
é0 -1ù
Let A = ê
ú , (aÎ R) such that A32 = ê1 0 ú .
sin
a
cos
a
ë
û
ë
û
Then a value of a is :
[April 8, 2019 (I)]
é3 4 1ù
[ p q r ] êê 3 2 3úú = [3 0 1] then
êë 2 0 2úû
p
(a)
32
2p + q – r equals :
(a) – 3
(c) 4
(b) 0
p
(c)
64
p
(d)
16
[Online April 22, 2013]
(b) – 1
(d) 2
Downloaded from @Freebooksforjeeneet
EBD_8344
M-270
10.
Mathematics
The matrix A2 + 4A – 5I, where I is identity matrix and
17.
é1 2 ù
A= ê
ú , equals
ë 4 -3û
11.
12.
13.
14.
15.
[Online April 9, 2013]
[2003]
(a) a = 2 ab, b = a 2 + b 2
é2 1ù
(a) 4 ê
ú
ë2 0û
é 0 -1ù
(b) 4 ê
ú
ë2 2 û
(b) a = a 2 + b 2 , b = ab
é2 1ù
(c) 32 ê
ú
ë2 0û
é1 1 ù
(d) 32 ê
ú
ë1 0 û
(d) a = a 2 + b 2 , b = a 2 - b 2 .
é 1 0 0ù
If A = ê 2 1 0 ú and B =
ê
ú
ëê -3 2 1 úû
é 1 0 0ù
ê -2 1 0ú then AB
ê
ú
ëê 7 -2 1úû
equals
[Online May 26, 2012]
(a) I
(b) A
(c) B
(d) 0
If w ¹ 1is the complex cube root of unity and matrix
éω 0 ù
H = ê
ú , then H70 is equal to
[2011RS]
ë 0 ωû
(a) 0
(b) –H
(c) H2
(d) H
The number of 3 × 3 non-singular matrices, with four entries
as 1 and all other entries as 0, is
[2010]
(a) 5
(b) 6
(c) at least 7
(d) less than 4
æ 1 2ö
æ a 0ö
Let A = ç
and B = ç
, a, b Î N. Then [2006]
è 3 4÷ø
è 0 b÷ø
(a) there cannot exist any B such that AB = BA
(b) there exist more than one but finite number of B¢s
such that AB = BA
(c) there exists exactly one B such that AB = BA
(d) there exist infinitely many B¢s such that AB = BA
If A and B are square matrices of si e n × n such that
A2 - B 2 = ( A - B )( A + B ) , then which of the following
will be always true?
[2006]
(a) A = B
(b) AB = BA
(c) either of A or B is a ero matrix
(d) either of A or B is identity matrix
16.
éa b ù
éa b ù
If A = ê
and A2 = ê
ú
ú , then
ëb a û
ëb aû
(c) a = a 2 + b 2 , b = 2ab
TOPIC n
18.
(a) 19.
3
3
1
3
(b)
1
3
(c) 3
(d)
2
3
The number of all 3 ´ 3 matrices A, with enteries from the
set {–1, 0, 1} such that the sum of the diagonal elements
of AAT is 3, is _______.
20.
[NA Jan. 8, 2020 (I)]
æ 2 2ö
æ 1 0ö
If A = ç 9 4÷ and I = ç 0 1÷ , then 10A–1 is equal to:
è
ø
è
ø
[Jan. 8, 2020 (II)]
(a) A – 4I
21.
(b) 6I – A
(c) A – 6I
(d) 4I – A
If A is a symmetric matrix and B is a skew-symmetrix matrix
é2 3 ù
such that A + B = ê 5 -1ú , then AB is equal to :
ë
û
[April 12, 2019 (I)]
(a) An = nA – (n – 1) I
n
(d) A = 2 n - 1 A + (n – 1) I
be all non- ero and satisfy
ATA = I, then a value of abc can be : [Sep. 02, 2020 (II)]
following holds for all n ³ 1, by the principle of
mathematical induction
[2005]
(c) An = nA + (n – 1) I
a, b, c Î R
æa b cö
ç
÷
A
=
If
the
matrix
a + b + c = 2.
ç b c a ÷ satisfies
çc a b÷
è
ø
3
é1 0ù
é1 0 ù
If A = ê
ú and I = ê
ú , then which one of the
1
1
ë
û
ë0 1û
(b) An = 2n - 1 A – (n – 1) I
Let
Transpose of Matrices, Symmetric
& Skew Symmetric Matrices,
Inverse of a Matrix by Elementary
Row Operations
22.
é - 4 -1ù
(a) ê -1 4 ú
ë
û
é 4 -2 ù
(b) ê -1 - 4ú
ë
û
é 4 -2 ù
(c) ê1 - 4ú
ë
û
é - 4 2ù
(d) ê 1 4ú
ë
û
æ 0 2y 1 ö
ç
÷
The total number of matrices A = ç 2 x y -1÷ , (x, y Î
x
y
2
1
è
ø
R, x ¹ y) for which ATA = 3I3 is:
(a) 2
(b) 3
(c) 6
Downloaded from @Freebooksforjeeneet
[April 09, 2019 (II)]
(d) 4
M-271
Matrices
23.
æ 0 2q r ö
Let A = ç p q -r ÷ . If AAT = I3, then |p| is :
ç
÷
ç p -q r ÷
è
ø
26.
[Jan. 11, 2019 (I)]
(a)
1
(b)
5
1
3
27.
(c)
24.
25.
1
(d)
2
1
6
For two 3 × 3 matrices A and B, let A + B = 2BT and 3A + 2B =
I3, where BT is the transpose of B and I3 is 3 × 3 identity
matrix. Then :
[Online April 9, 2017]
(a) 5A + 10B = 2I3
(b) 10A + 5B = 3I3
(c) B + 2A = I3
(d) 3A + 6B = 2I3
é 3
ê
ê 2
If P = ê 1
êë 2
1 ù
ú
2 ú
é1 1ù
APT, then
3 ú , A = êë 0 1úû and Q = PAP
ú
2 û
PT Q2015 P is ;
[Online April 9, 2016]
é 0 2015ù
(a) ê 0
0 úû
ë
0 ù
é 2015
(b) ê 1
ú
2015
ë
û
é1 2015ù
(c) ê 0
1 úû
ë
1 ù
é 2015
(d) ê 0
2015úû
ë
28.
29.
é1 2 2 ù
If A = ê 2 1 -2ú is a matrix satisfying the equation
ê
ú
êë a 2 b úû
AAT = 9I, where I is 3 × 3 identity matrix, then the ordered
pair (a, b) is equal to:
[2015]
(a) (2, 1)
(b) (–2, – 1)
(c) (2, – 1)
(d) (–2, 1)
Let A and B be any two 3 × 3 matrices. If A is symmetric
and B is skewsymmetric, then the matrix AB – BA is:
(a) skewsymmetric
[Online April 19, 2014]
(b) symmetric
(c) neither symmetric nor skewsymmetric
(d) I or – I, where I is an identity matrix.
æ a + 1ö
æ a - 1ö
ç
÷
If A = 0 , B = ç 0 ÷ be two matrices, then ABT is a
ç
÷
ç
÷
è 0 ø
è 0 ø
non- ero matrix for |a| not equal to [Online May 7, 2012]
(a) 2
(b) 0
(c) 1
(d) 3
Let A and B be two symmetric matrices of order 3. [2011]
Statement-1: A(BA) and (AB)A are symmetric matrices.
Statement-2: AB is symmetric matrix if matrix multiplication
of A with B is commutative.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation for Statement-1.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-272
1.
Mathematics
(d) Solution of x2 + x + 1 = 0 is w, w2
So, a = w and
w4 = w3.w = 1.w = w
é1 1
ê
1
A2 = ê1 w
3ê
ê1 w2
ë
1 ù é1 1
úê
w2 ú ê1 w
úê
w ú ê1 w2
ûë
Given that ( x2 + 1) 2 + x 2 = 109
x 4 + 3x 2 - 108 = 0
Þ A4 = I
Þ A30 = A28 × A3 = A3
2.
Þ ( x2 + 12)( x2 - 9) = 0
1 ù é1 0 0ù
ú
w2 ú = êê0 0 1 úú
ú
w ú êë0 1 0úû
û
\ x2 = 9
a22 = x 2 + 1 = 9 + 1 = 10.
4.
é cos q i sin q ù
(d) Q A = ê
ú
ëi sin q cos q û
é cos a - sin a ù é cos a - sin a ù
A2 = ê a
cos a úû êë sin a cos a úû
ë sin
é cos nq i sin nq ù
\ An = ê
ú , n ÎN
ëi sin nq cos nq û
écos 2a - sin 2a ù
=ê
ú
ë sin 2a cos 2a û
éa b ù
Q A5 = ê
ú
ëc d û
écos 4a - sin 4a ù
Similarly, A4 = A2 .A2 = ê sin 4a cos 4a ú
ë
û
é cos5q i sin 5q ù é a b ù
\ A5 = ê
ú=ê
ú
ëi sin 5q cos5q û ë c d û
é cos32a - sin 32a ù é0 -1ù
and so on A32 = ê sin 32a cos32a ú = ê1 0 ú
ë
û ë
û
\ a = cos 5q, b = i sin 5q = c, d = cos 5q
Then sin 32a = 1 and cos 32a = 0
\ a 2 - b 2 = cos 2 5q + sin 2 5q = 1
Þ 32a = np + (– 1)n
a 2 - c 2 = cos 2 5q + sin 2 5q = 1
Þ a=
a - d = cos 5q - cos 5q = 1
2
2
2
2
a 2 + b 2 = cos 2 5q - sin 2 5q = cos10q = cos
10 p
24
\ a 2 - b2 =
1
is wrong.
2
(10)
é x 1ù é x 1ù é x 2 + 1 xù
A2 = ê
ú
úê
ú=ê
1û
ë1 0û ë 1 0û ë x
é x2 + 1 xù é x 2 + 1 x ù
A4 = ê
úê
ú
1û ë x
1û
ë x
é( x 2 + 1)2 + x 2
=ê
2
êë x ( x + 1) + x
x ( x 2 + 1) + x ù
ú
x 2 + 1 úû
p
p
and 32a = 2np +
2
2
p
np p
np
+ (-1)n
+
and a =
where n Î z
32
64
16 64
Put n = 0, a =
p
64
p
.
64
é1 0 0ù é1 0 0ù é 1 0 0ù
ê 3 1 0ú ê 3 1 0 ú = ê 6 1 0 ú
úê
ú ê
ú
(a) P2 = ê
êë9 3 1úû êë9 3 1úû êë27 6 1 úû
Hence, required value of a is
5p
< 1 Þ 0 £ a2 + b2 £ 1
and 0 < cos
12
3.
é cos a - sin a ù
(c) A = ê
ú
ë sin a cos a û
5.
é 1 0 0ù é 1 0 0ù é 1 0 0ù
ê 6 1 0ú ê 6 1 0ú = ê12 1 0ú
úê
ú ê
ú
Þ P4 = ê
êë 27 6 1 úû êë 27 6 1 úû êë90 12 1úû
0 0ù
é 1 0 0ù é1 0 0ù é 1
ê12 1 0ú ê3 1 0ú = ê 15 1 0ú
úê
ú ê
ú
Þ P5 = ê
êë90 12 1úû êë9 3 1úû êë135 15 1 úû
Q Q – P5 = I3
Downloaded from @Freebooksforjeeneet
M-273
Matrices
\
é -1 0 ù
2
A2 = ê
ú Þ A = -I
ë 0 -1û
0 0ù
é 2
ê 15 2 0ú
ú
Q = I3 + P5 = ê
ëê135 15 2úû
é 0 1ù
A3 = ê
ú
ë -1 0 û
q21 + q31 15 + 135
= 10
q32 = 15
6.
é1 0ù
A4 = ê
ú =I
ë0 1 û
A2 + I = A3 – A
– I + I = A3 – A
é1 0 0 ù
(c) Here A = êê1 1 0 úú
êë1 1 1 úû
\
é1 0 0 ù
é1 0 0 ù
A2 = A. A = êê1 1 0 úú × êê1 1 0 úú
êë1 1 1 úû
êë1 1 1 úû
A3 ¹ A
8.
é yù
êxú
é1 2 x ù
(a) Let A = ê3 -1 2 ú and B = ê ú
ë
û
ëê1 ûú
é1 0 0ù
= êê 2 1 0úú
êë 3 2 1 úû
also
A3 = A2. A =
é yù
é1 2 x ù ê ú
x
AB = êë3 -1 2úû ê ú
ëê1 ûú
é 6ù
é y + 2x + xù
Þ ê8 ú = ê 3 y - x + 2 ú
ë û
ë
û
é1 0 0ù é1 0 0 ù
ê 2 1 0ú × ê1 1 0 ú
ê
ú ê
ú
êë 3 2 1 úû êë1 1 1 úû
é 6ù
é y + 3x ù
Þ ê8 ú = ê 3 y - x + 2 ú
ë û
ë
û
Þ y + 3x = 6 and 3y – x = 6
On solving, we get
é1 0 0ù
= êê 3 1 0úú
êë6 3 1 úû
é1 0 0ù é1 0 0 ù
and, A4 = A3. A = êê 3 1 0úú × êê1 1 0 úú
êë6 3 1 úû êë1 1 1 úû
é 1 0 0ù
= êê 4 1 0 úú
êë10 4 1 úû
On observing the pattern, we come to a conclusion that,
é
ê
1
ê
A =ê
n
ê n (n + 1)
ê
ë
2
9.
[p
q
é3
r ] ê3
ê2
ë
4
2
0
1ù
3 ú = [3
2 úû
Þ [3 p + 3q + 2r 4 p + 2q
0
1]
p + 3q + 2r ] = [3 0 1]
Þ 3p + 3q + 2r = 3
4p + 2q = 0 Þ q = – 2p
p + 3q + 2r = 1
On solving (i), (ii) and (iii), we get
p = 1, q = – 2, r = 3
\ 2p + q – r = 2(1) + (– 2) – (3) = – 3.
10. (a) A2 + 4A – 5I = A × A + 4A – 5I
0 0ù
é 1
ê
ú
\
ê 20 1 0ú
êë 210 20 1úû
Therefore, sum of first column of A20 = [1 + 20 + 210] = 231
(a) Given that
é1
= ê4
ë
é 0 -1ù
A =ê
ú
ë1 0 û
é 9 + 4 -5
= ê -8 + 16 - 0
ë
A20 =
7.
ù
0 0ú
ú
1 0ú
ú
n 1ú
û
6
12
and y =
5
5
Þ y = 2x
(a) Given
x=
é9
= ê -8
ë
2 ù é1
´
-3úû êë 4
2ù
é1
+4ê
-3úû
ë4
-4ù é 4
+
17 úû êë16
...(i)
...(ii)
...(iii)
2 ù é1
-5
-3úû êë0
8 ù é5
-12 úû êë0
-4 + 8 - 0 ù é8
=
17 - 12 - 5úû êë8
Downloaded from @Freebooksforjeeneet
0ù
1 úû
0ù
5úû
4ù
0 úû =
é2
4ê
ë2
1ù
0 úû
EBD_8344
M-274
11. (a)
Mathematics
é n 0ù én - 1 0 ù
Now nA – (n – 1) I = ê n n ú - ê 0
n - 1úû
ë
û ë
é 1 0 0ù
é 1 0 0ù
ê
ú
A = 2 1 0 , B = ê -2 1 0ú
ê
ú
ê
ú
êë -3 2 1úû
êë 7 -2 1úû
é 1 0ù
n
= ê n 1ú = A
ë
û
é1 0 0ù
AB = ê0 1 0ú = I
ê
ú
êë0 0 1úû
12. (d)
\ nA - ( n - 1) I = An
2
é ω 0 ù é ω 0 ù éω 0 ù
ú
H2 = ê
úê
ú =ê
ω2 úû
ë0 ωû ë 0 ωû êë0
é a2 + b2
=ê
êë 2 ab
éωk 0ù
We observed that H k = ê
ú
êë0 ω úû
éω
\ H 70 = ê
êë 0
70
13. (c)
0 ù éω w 0 ù éω
ú=ê
ú=ê
ω70 úû êë0 w69ωúû ë0
69
0ù
=H
ω úû
[Q w3n = 1]
blanks will be filled by 5 eros and 1 one.
14.
15.
1ù
...úú are 6 non-singular matrices.
...úû
are more than 7, non-singular 3 × 3
é1
(d) Given that A = ê
ë3
é a 2b ù
AB = ê
ú
ë3a 4b û
2ù
4 úû
éa
B = ê
ë0
0ù
b úû
é a 0 ù é1 2 ù é a 2 a ù
BA = ê
úê
ú=ê
ú
ë 0 b û ë3 4û ë3b 4b û
Hence, AB = BA only when a = b
\ There can be infinitely many B¢s for which AB = BA
(b) Given that A2 - B 2 = ( A - B )( A + B )
A2 - B 2 = A2 + AB - BA - B 2
Þ AB = BA
é1 0 ù
16. (a) Given that A = ê
ú
ë1 1 û
é 1 0 ù 3 é1 0 ù
A2 = ê
ú , A = ê3 1 ú
ë2 1 û
ë
û
é 1 0ù
Therefore we observed that An = ê
ú
ë n 1û
éa bù éa bù
ê b a ú êb a ú
ë
ûë
û
2ab ù
ú
a 2 + b 2 úû
a = a2 + b2; b = 2ab
é 1 ... ...ù
ê... 1 ...ú
ê
ú are 6 non-singular matrices because 6
êë... ... 1 úû
é... ...
Similarly, ê... 1
ê
êë 1 ...
Total = 6 + 6 = 12
So, required cases
matrices.
éa b ù
A2 = ê
ú Þ A× A =
ëb aû
17. (c)
18. (b) Given : AT A = I
éa b c ù é a b c ù é1 0 0ù
Þ ê b c a ú êb c a ú = ê 0 1 0 ú
ê
úê
ú ê
ú
êë c a b úû êë c a b úû êë0 0 1úû
éSa 2 Sab Sab ù é1 0 0ù
ê
ú
Þ ê Sab Sa 2 Sab ú = ê0 1 0ú
ê
ú
ê
2ú
ê0 0 1 ûú
ab
ab
a
ë
S
S
S
úû
ëê
So, Sa 2 = 1 and Sab = 0
Now, a3 + b3 + c 3 - 3abc
= (a + b + c )(a 2 + b 2 + c 2 - ab - bc - ca )
= (a + b + c)(1 - 0)
= (a + b + c )2 = Sa 2 + 2Sab = ±1
Þ 2 - 3abc = 1 Þ abc =
1
3
or 2 – 3abc = –1 Þ abc = 1.
19. (672) Let A = [aii]3×3
It is given that sum of diagonal elements of AAT is 3 i.e.,
tr(AAT) = 3
a211 + a212 + a213 + a221 + ..... + a233 = 3
Possible cases are
®1ü
0, 0, 0, 0, 0, 0, 1, 1, 1
0, 0, 0, 0, 0, 0, - 1, - 1, - 1 ® 1 ïï 9
ý C6 ´ 8 = 84 ´8 = 672
® 3ï
0, 0, 0, 0, 0, 0, 1, 1, - 1
® 3ïþ
0, 0, 0, 0, 0, 0, - 1, 1, - 1
20. (c) Characteristics equation of matrix ‘A’ is |A – lI| = 0
2-l
2
9
4-l
= 0 Þ l2 – 6l –10 = 0
Downloaded from @Freebooksforjeeneet
M-275
Matrices
\ A2 – 6A – 10I = 0
Þ A–1(A2) – 6A–1 – 10IA–1 = 0
Þ 10A–1 = A – 6I
éa c ù
21. (b) Let A = ê
ú and B =
ë c bû
é a
Then, A + B = ê ëc d
é0
ê -d
ë
dù
0 úû
c + d ù é2 3 ù
=
b úû êë5 -1úû
On comparing each term,
é 2 4 ù é 0 -1ù é 4 -2 ù
Now, AB = ê
úê
ú =ê
ú
ë 4 -1û ë1 0 û ë -1 -4 û
22. (d) Given, ATA = 3I
6 y2
0
\
| p|=
24. (b) AT + BT = 2B
Q [(A + B)T = (2BT)T]
Also,
3
1
and y = ±
8
2
é 0 2q r ù
ê p q -r ú
ú
23. (c) A = ê
êë p - q r úû
p
pù
é 0 2q r ù é 0
ê p q - r ú ´ ê2q q - q ú
ú ê
ú
A × AT = ê
êë p - q r úû êë r - r r úû
2q 2 - r 2
p2 + q2 + r 2
p2 - q2 - r 2
Given, AAT = I
\ 4q2 + r2 = p2 + q2 + r2 = 1
Þ p2 – 3q2 = 0 and r2 = 1 – 4q2
and 2q2 – r2 = 0 Þ r2 = 2q2
ö
÷ = 2BT
÷
ø
3BT - AT
2
3A + 2B = I3
Þ B=
Number of combinations of (x, y) = 2 × 2 = 4
é 4q 2 + r 2
ê 2 2
ê 2q - r
=ê
2
2
ëê -2 q + r
æ BT + AT
ç
ç
2
è
Þ 2A + AT = 3BT Þ A =
Þ
\
AT + BT
= A+
2
...(i)
ö æ AT + BT ö
÷ + 2ç
÷=I
÷ ç
÷ 3
2
ø è
ø
Þ 11BT – AT = 2I3
Add (i) and (ii)
35B = 7I3
0ù é 3 0 0 ù
ú
0ú = êê 0 3 0 úú
ú
3ú êë 0 0 3 úû
û
Þ 8x2 = 3 and 6y2 = 3 Þ x = ±
1
.
2
æ 3BT - AT
Þ 3 çç
2
è
é 0 2x 2x ù é 0 2 y 1 ù
ê 2 y y - y ú ê 2 x y -1ú = 31
ê
úê
ú
êë 1 -1 1 úû êë 2 x - y 1 úû
0
1 2 1
1
2
p2 = , q = and r =
2
6
3
Þ B=
a = 2, b = – 1, c – d = 5, c + d = 3
Þ a = 2, b = – 1, c = 4, d = – 1
é8x 2
ê
Þ ê 0
ê
ê 0
ë
\
-2q 2 + r 2 ù
ú
p2 - q2 - r 2 ú
ú
p 2 + q 2 + r 2 ûú
11
....(ii)
I3
I
Þ 11 3 - A = 2I3
5
5
I3
I
- 2I3 = A Þ A = 3
5
5
Q 5A = 5B = I3
Q
Þ 10A + 5B = 3I3
é 3
ê
25. (c) P = ê 2
ê 1
êë 2
1 ù
ú
2 ú T
P =
3ú
ú
2 û
é
ê
ê
ê
ê
ë
3
2
1
2
-1 ù
ú
2 ú
3ú
ú
2 û
PPT = PTP = I
Q2015 = (PAPT) (PAPT) ––––– (2015 terms)
= PA2015PT
PTQ2015P = A2015
é1 1 ù é1 1 ù é 1 2ù
A2 = ê
úê
ú=ê
ú
ë0 1û ë0 1û ë0 1û
é 1 2ù é1 1ù é1 3 ù
A3 = ê
úê
ú=ê
ú
ë 0 1 û ë0 1û ë0 1û
é 1
\ A2015 = ê
ë 0
2015 ù
1 úû
Downloaded from @Freebooksforjeeneet
EBD_8344
M-276
Mathematics
26. (b) Given that AAT = 9I
27. (b) Let A be symmetric matrix and B be skew symmetric
matrix.
\ AT = A and BT = –B
Consider
(AB – BA)T = (AB)T – (BA)T
= BTAT – ATBT = (–B) (A) – (A) (–B)
= –BA + AB = AB – BA
This shows AB – BA is symmetric matrix.
é 1 2 2 ù é 1 2 a ù é 9 0 0ù
ê 2 1 -2ú ê2 1 2ú = ê0 9 0ú
ê
úê
ú ê
ú
êë a 2 b úû êë2 -2 b úû êë0 0 9 úû
Þ
é 1+ 4 + 4
2+ 2- 4
a + 4 + 2b ù
ê
ú
4 +1+ 4
2a + 2 - 2b ú
ê 2+2-4
ê
2
2ú
ë a + 4 + 2b 2a + 2 - 2b a + 4 + b û
Þ a + 4 + 2b = 0 Þ a + 2b = – 4
2a + 2 – 2b = 0 Þ 2a – 2b = – 2
Þ a – b = –1
Subtract (ii) from (i)
a + 2b = –4
a – b = –1
– + +
3b = –3
b= –1
and a = – 2
(a, b) = (–2, –1)
é 9 0 0ù
= ê 0 9 0ú
ê
ú
êë0 0 9úû
...(i)
...(ii)
æ a –1ö
æ a + 1ö
28. (c) Let A = ç 0 ÷ , B = ç 0 ÷
ç
÷
ç
÷
è 0 ø
è 0 ø
be two matrices.
æ a 2 –1 0 0ö
æ a –1ö
ç
÷
0 0÷
ABT = ç 0 ÷ ( a + 1 0 0) = ç 0
ç
÷
çè 0
0 0÷ø
è 0 ø
Thus, ABT is non- ero matrix for | a | ¹ 1
29. (a) Given that A and B are symmetric matrix
A¢ = A
B¢ = B
Now (A(BA))¢ = (BA)¢A¢ = (A¢B¢)A¢ = (AB)A = A(BA)
(\ product of matrices are associative)
Similarly, ((AB)A)¢ = A¢(B¢A¢) = A (BA) = (AB)A
So, A(BA) and (AB)A are symmetric matrices.
Again (AB)¢ = B¢A¢ = BA
Now if BA = AB, then AB is symmetric matrix.
Downloaded from @Freebooksforjeeneet
M-277
Determinants
19
Determinants
TOPIC Ć
1.
2.
Minor & Co-factor of an Element of a
Determinant, Value of a Determinant,
Property of Determinant of Matrices,
Singular & Non-Singular Matrices,
Multiplication of two Determinants
p
cos q
and A = é - sin q
êë
5
then det (B):
(a) is one
(b)
(c) is ero
(d)
Let q =
value of
(a) f (–50) = 501
(b) f (–50) = –1
(c) f (50) = –501
(d) f (50) = 1
sin q cos q
x
1 and
If D1 = - sin q - x
x
cos q
1
(b) D1 - D 2 = x (cos 2q - cos 4q)
(c) D1 ´ D 2 = - 2( x3 + x - 1)
If
(c) - 3
8.
x - 4 2x
2x
2x
= (A + Bx)(x - A) 2 , then the
[2018]
ì
ü
0
cos x - sin x
ï
ï
0
cos x = 0 ý , then
If S = í x Î [ 0, 2p] : sin x
ï
ï
cos x sin x
0
î
þ
æp
xÎS
9.
3
x -4
ö
å tan çè 3 + x ÷ø
[April 10, 2019 (I)]
(d)
2x
2x
ordered pair (A, B) is equal to :
(a) (– 4, 3)
(b) (– 4, 5)
(c) (4, 5)
(d) (– 4, – 5)
sin 2q cos 2q
x
D 2 = - sin 2q
-x
1 ,x ¹ 0 ;
cos 2q
1
x
æ pö
then for all qÎ ç 0, ÷ :
è 2ø
(a) D1 - D 2 = -2x3
(d) –4
[Jan. 10, 2019 (II)]
(n) - 2 3
(a) 2 3
7.
(c) 1
det (A)
is:
b
x - 4 2x
x + 2 x +1
x + 3 x + 2 , then : [Jan. 9, 2020 (II)]
x+ 4 x+3
(d) D1 + D2 = -2x3
(b) 0
[April 10, 2019 (II)]
b
1ù
é2
ê
ú
2
Let A = ê b b + 1 b ú where b > 0. Then the minimum
ê1
b
2úû
ë
Let a – 2b + c = 1.
x+a
If f(x) = x + b
x+c
4.
6.
[Sep. 03, 2020 (I)]
(d) 9
(c) –3
-1
x - 3 = 0, is equal to :
x+2
(a) 6
sin q ù
4
cos qûú . If B = A + A ,
[Sep. 06, 2020 (II)]
lies in (2, 3)
lies in (1, 2)
The sum of the real roots of the equation
x -6
2 -3x
-3 2 x
x - 2 2x - 3 3x - 4
If D = 2 x - 3 3x - 4 4 x - 5 = Ax3 + Bx2 + Cx + D,
3x - 5 5x - 8 10 x - 17
then B + C is equal to :
(a) –1
(b) 1
3.
5.
is equal to
[Online April 8, 2017]
(a) 4 + 2 3
(b) -2 + 3
(c) -2 - 3
(d) -4 - 2 3
é -4 -1ù
If A = ê3 1 ú , then the determinant of the matrix
ë
û
(A2016 – 2A2015 – A2014) is : [Online April 10, 2016]
(a) –175
(b) 2014
(c) 2016
(d) –25
Downloaded from @Freebooksforjeeneet
EBD_8344
M-278
Mathematics
16.
x2 + x
10.
x–2
3x
3x – 3 = ax –12, then ‘a’ is
2
if 2 x + 3x –1
x + 2x + 3
2
11.
x +1
2 x –1 2 x –1
equal to :
[Online April 11, 2015]
(a) 24
(b) –12
(c) –24
(d) 12
The least value of the product xyz for which the
x 1 1
determinant
1
y 1
is non-negative, is :
17.
1 1
[Online April 10, 2015]
12.
(a) -2 2
(b) –1
(c) -16 2
(d) –8
1
cos q
1
1
- cos q and
If f(q) = - sin q
sin q
1
-1
A and B are respectively the maximum and the minimum
values of f(q), then (A, B) is equal to:
[Online April 12, 2014]
(a) (3, – 1)
(c)
13.
14.
15.
(2 +
2, 2 - 2
)
(b)
( 4, 2 - 2 )
(d)
(2 +
18.
)
2, -1
If B is a 3 × 3 matrix such that B 2 = 0, then
det. [(I + B)50 – 50B] is equal to: [Online April 9, 2014]
(a) 1
(b) 2
(c) 3
(d) 50
ìæ a11 a12 ö
ü
Let S = íç
: aij Î{0,1, 2}, a11 = a22 ý
÷
îè a21 a22 ø
þ
Then the number of non-singular matrices in the set S is :
[Online April 25, 2013]
(a) 27
(b) 24
(c) 10
(d) 20
Let A, other than I or – I, be a 2 × 2 real matrix such that
A2 = I, I being the unit matrix. Let Tr (A) be the sum of
diagonal elements of A.
[Online April 23, 2013]
Statement-1: Tr (A) = 0
Statement-2: det (A) = – 1
(a) Statement-1 is true; Statement-2 is false.
(b) Statement-1 is true; Statement-2 is true; Statement-2
is not a correct explanation for Statement-1.
(c) Statement-1 is true; Statement-2 is true; Statement-2
is a correct explanation for Statement-1.
(d) Statement-1 is false; Statement-2 is true.
19.
Statement - 1:
[2011RS]
Determinant of a skew-symmetric matrix of order 3 is ero.
Statement - 2 :
For any matrix A, det (A)T= det (A) and det (– A) = – det (A).
Where det (B) denotes the determinant of matrix B. Then :
(a) Both statements are true
(b) Both statements are false
(c) Statement-1 is false and statement-2 is true
(d) Statement-1 is true and statement-2 is false
Let A be a 2 × 2 matrix with non- ero entries and let A2 = I ,
where I is 2 × 2 identity matrix. Define
Tr(A) = sum of diagonal elements of A and
|A| = determinant of matrix A.
Statement - 1 : Tr(A) = 0.
Statement -2 : |A| = 1.
[2010]
(a) Statement-1 is true, Statement-2 is true ; Statement-2
is not a correct explanation for Statement -1.
(b) Statement -1 is true, Statement -2 is false.
(c) Statement -1 is false, Statement -2 is true .
(d) Statement - 1 is true, Statement 2 is true ; Statement -2
is a correct explanation for Statement -1.
Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2
identity matrix. Denote by tr(A), the sum of diagonal entries
of a. Assume that A2 = I.
[2008]
Statement-1 : If A ¹ I and A ¹ –I, then det (A) = –1
Statement-2 : If A ¹ I and A ¹ –I, then tr (A) ¹ 0.
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2
is a correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2
is not a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
5 5a a
2
Let A = 0 a 5a . If A = 25 , then a equals [2007]
0 0
5
(a) 1/5
(c) 52
20.
(b) 5
(d) 1
If 1, w, w 2 are the cube roots of unity, then
1
wn
w 2n
D = wn
w2n
1
w
(a) w 2
(c) 1
2n
1
w
is equal to
n
(b) 0
(d) w
Downloaded from @Freebooksforjeeneet
[2003]
M-279
Determinants
TOPIC n
21.
27.
Properties of Determinants,
Area of a Triangle
é1 1
ê2 b
A= ê
2
ë4 b
If the minimum and the maximum values of the function
ép pù
f : ê , ú ® R , defined by
ë4 2û
tively, then the ordered pair (m, M) is equal to :
[Sep. 05, 2020 (I)]
(b) (– 4, 0 )
(a) (0, 2 2 )
(c) (– 4, 4)
(d) (0, 4)
If a + x = b + y = c + z + 1, where a, b, c, x, y, z are non- ero
x a+ y
x+a
distinct real numbers, then y b + y
z c+ y
z+c
23.
28.
29.
line, 3x + y – 4l = 0, then a value of l is: [Jan. 8, 2020 (I)]
24.
(c) 1
(d) –3
30.
Let A = [aij] and B = [bij] be two 3 ´ 3 real matrices such that
25.
(a) 1/3
(b) 3
(c) 1/81
A value of q Î (0, p/3), for which
[Jan. 7, 2020 (II)]
(d) 1/9
26.
p
p
7p
7p
(a)
(b)
(c)
(d)
9
18
24
36
Let a and b be the roots of the equation x2 + x + 1 = 0. Then
for y �벜 0 in R,
y +1
a
b
a
y+b
1
b
1
y+a
(d) [3, 2 + 23/4]
sin q
1 ù
é 1
æ 3p 5p ö
ú
If A = ê - sin q
1
sin
qú ; then for all qÎ ç , ÷,
ê
è 4 4 ø
êë -1
1 úû
- sin q
[Jan. 12, 2019 (II)]
æ 3ù
(c) ç 0, ú
è 2û
a–b–c
2a
2b
b–c–a
2b
2c
2c
c–a–b
If
æ3 ù
(d) ç ,3ú
è2 û
2a
(a) abc
(b) – (a + b +c)
(c) 2 (a + b +c)
(d) – 2 (a + b +c)
Let dÎR, and
q Î [0, 2p]. If the minimum value of det (A) is 8, then a
value of d is:
[Jan 10, 2019 (I)]
1 + cos 2 q
sin 2 q
4 cos 6q
2
cos q 1 + sin 2 q
4 cos 6q
= 0, is :
cos 2 q
sin 2 q 1 + 4cos 6q
[April 12, 2019 (II)]
(c) [4, 6]
é- 2
ù
4+d
(sin q) -2
ê
ú
1 (sin q) + 2
d
ú,
A= ê
ê 5 (2 sin q) - d (- sin q) + 2 + 2d ú
ë
û
bij = (3)(i + j – 2) aij, where i, j = 1, 2, 3. If the determinant of B
is 81, then the determinant of A is:
(b) (2 + 23/4, 4)
= (a + b + c) (x + a + b + c)2, x ¹ 0 and a + b + c ¹ 0, then x
is equal to :
[Jan. 11, 2019 (II)]
such that the area of DPAB = 5 sq. units and it lies on the
(b) 3
(a) [2, 3)
æ 5ù
é5 ö
(a) ç 1, ú (b) ê , 4 ÷
è 2û
ë2 ø
Let two points be A(l, – 1) and B(0, 2). If a point P(x¢, y¢) be
(a) 4
[April 08, 2019 (II)]
det (A) lies in the interval :
y + b is equal to :
[Sep. 05, 2020 (II)]
(b) y (a – b)
(d) y (a – c)
(a) y (b – a)
(c) 0
1ù
cú
ú . If det(A)?[2, 16], then c lies in the
c2 û
interval :
- sin 2 q -1 - sin 2 q 1
f (q) = - cos 2 q -1 - cos 2 q 1 are m and M respec12
10
-2
22.
Let the numbers 2, b, c be in an A.P. and
(a) – 5
(c) 2
31.
(
(b) – 7
)
2 +1
(d) 2
(
2 +2
)
Let a1, a2, a3, ..., a10 be in G.P. with ai > 0 for i = 1, 2, ..., 10 and
S be the set of pairs (r, k), r, kÎN (the set of natural numbers) for which
is equal to:
log e a1r a 2k
log e a 2r a 3k
log e a 3r a k4
log e a r4 a 5k
log e a 5r a 6k
log e a 6r a 7k = 0
log e a r7 a 8k
log e a 8r a 9k
k
log e a 9r a10
Then the number of elements in S, is : [Jan. 10, 2019 (II)]
(a) y(y2 – 1)
(c) y3
[April 09, 2019 (I)]
2
(b) y(y – 3)
(d) y3 – 1
(a) 4
(b) infinitely many
(c) 2
(d) 10
Downloaded from @Freebooksforjeeneet
EBD_8344
M-280
32.
Mathematics
If
éet
ù
e - t cos t
e - t sin t
ê t
ú
A = êe -e - t cos t - e- t sin t -e - t sin t + e - t cos t ú
ê t
ú
2e- t sin t
-2e- t cos t
êë e
úû
then A is:
[Jan. 09, 2019 (II)]
(a) invertible for all tÎR.
(b) invertible only if t = p.
(c) not invertible for any tÎR.
(d) invertible only if t =
33.
34.
38.
1
39.
b
35.
36.
[2017]
é p pù
= 0 in the interval ê - , ú is :
ë 4 4û
cos x
cos x sin x
sin x
sin x
(b) 4
41.
1 + f (1) 1 + f ( 2 )
2
2
then K is equal to:
37.
If D r =
b
1
c ,l ¹ 0
1
[Online April 9, 2013]
ab
2
ab
c +a
ac
bc
ac
2
= ka2b2c2, then k is equal to
bc
a 2 + b2
2
(c) ab
(d)
r
2r - 1
3r - 2
n
2
n -1
a
1
n(n - 1) (n - 1)2
2
1
ab
1
(n - 1)(3n - 4)
2
n -1
then the value of
å Dr
r =1
42.
[2014]
(b) –1
(b) 3
-2a
[Online April 9, 2016]
(c) 2
(d) 3
1 + f (1) 1 + f ( 2 ) 1 + f ( 3) = K (1 - a ) (1 - b ) ( a - b)
,
1 + f ( 2 ) 1 + f ( 3) 1 + f ( 4 )
(a) 1
= kl a
1
c a is :
(a) 1
If a, b ¹ 0, and f ( n ) = a n + b n and
3
c2
sin x
sin x
(a) 1
b2
(a) non - negative
(b) negative
(c) positive
(d) non-positive
If a, b, c, are non ero complex numbers satisfying
a2 + b2 + c2 = 0 and
(a) 1
(b) –
(c)
(d) –1
The number of distinct real roots of the equaiton,
cos x sin x
a2
c
b2 + c 2
w7
2
c a b
2
w2
2
then k is equal to:
[Online April 12, 2014]
(a) 4labc (b) – 4labc (c) 4l2
(d) – 4l2
If a, b, c are sides of a scalene triangle, then the value of
a b
-3 . If 1 -w - 1 w = 3k, then k is equal to :
1
c2
(a + l ) ( b + l) (c + l )
( a - l )2 ( b - l)2 ( c - l )2
40.
1
2
b2
2
3ö
1ö
æ 1ö
æ
æ
æ 3ö
(a) ç 2, ÷ (b) ç 2, - ÷ (c) ç1, ÷
(d) ç1, - ÷
è 2ø
è
è
è 4ø
4ø
2ø
Let w be a complex number such that 2w + 1 = where =
1
depends only on a
depends only on n
depends both on a and n
is independent of both a and n
a2
p
.
2
Let k be an integer such that triangle with vertices
(k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the
orthocentre of this triangle is at the point :
[2017]
(a)
(b)
(c)
(d)
If
[Online April 19, 2014]
43.
44.
(c) 4
[Online May 19, 2012]
(d) 2
a+b a+c
If b + a -2b b + c = a ( a + b)( b + c)( c + a ) ¹ 0
c + a b + c -2c
then a is equal to
[Online May 12, 2012]
(a) a + b + c
(b) abc
(c) 4
(d) 1
The area of the triangle whose vertices are complex numbers
z, iz, z + iz in the Argand diagram is [Online May 12, 2012]
(a) 2|z|2
(b) 1/2|z|2
(c) 4|z|2
(d) |z|2
The area of triangle formed by the lines oining the vertex
of the parabola, x2 = 8y, to the extremities of its latus rectum
is
[Online May 12, 2012]
(a) 2
(b) 8
(c) 1
(d) 4
Let a, b, c be such that b(a + c) ¹ 0 if
[2009]
a + 1 a –1
a +1
b +1
c –1
–b b + 1 b –1
a –1
b -1
c + 1 = 0,
a
c
c –1 c + 1
+
(-1)
then the value of n is :
(a) any even integer
(c) any integer
n+2
a (-1)
n +1
b ( -1) n c
(b) any odd integer
(d) ero
Downloaded from @Freebooksforjeeneet
M-281
Determinants
45.
1 1
1
1 for x ¹ 0, y ¹ 0 , then D is
If D = 1 1 + x
1 1 1+ y
(a)
(b)
(c)
(d)
46.
divisible by x but not y
divisible by y but not x
divisible by neither x nor y
divisible by both x and y
Matrix, Rank of a Matrix
[2007]
51.
D = log an + 3
log an + 6
log an + 1
log an + 2
log an + 4
log an + 5
log an + 7
log an + 8
is equal to
(a) 1
(b) 0
2
2
pair, (| l |, m) is equal to :
(d) 2
If a + b + c = – 2 and
[2005]
52.
(1 + b 2 ) x (1 + c 2 ) x
2
2
2
f (x) = (1 + a ) x 1 + b x (1 + c ) x ,
(1 + a2 ) x (1 + b 2 ) x 1 + c 2 x
then f (x) is a polynomial of degree
(a) 1
(b) 0
(c) 3
48.
49.
log an
log an+1
log an+ 2
log an+ 3
log an+ 4
log an +5
log an+ 6
log an+ 7
log an+8
54.
c
log l p 1
then log m q 1 equals
log n r 1
é1 1 2ù
ê
ú
If the matrices A = ê1 3 4ú , B = ad A
êë1 -1 3úû
(c) 72
(d) 2
é 5 2a 1 ù
ê
ú
If B = ê 0 2 1 ú is the inverse of a 3 × 3 matrix A, then
ë a 3 -1û
é1 1ù é1 2 ù é1 3ù
é1 n – 1ù é1 78 ù
=
,
If ê 0 1ú × ê0 1 ú × ê 0 1 ú ............ ê 0
1 úû êë0 1 úû
ë
û ë
û ë
û
ë
é 1 0ù
(a) ê12 1 ú
ë
û
[2002]
55.
[April 09, 2019 (II)]
é1 -13ù
(b) ê
ú
ë0 1 û
é1 -12 ù
é 1 0ù
(c) ê0 1 ú
(d) ê13 1 ú
ë
û
ë
û
Let A and B be two invertible matrices of order 3 × 3. If
det (ABAT) = 8 and det (AB–1) = 8, then det (BA–1 BT) is
equal to :
[Jan. 11, 2019 (II)]
[2002]
(b) 2
(d) 0
(b) 16
ad B
is equal to : [Jan. 9, 2020 (I)]
C
é1 n ù
  � 矄 ᨨ ê 0 1 ú �矄
ë
û
ax + b
(a) +ve
(b) (ac-b2)(ax2+2bx+c)
(c) –ve
(d) 0
l, m, n are the pth, qth and rth term of a G. P. all positive,
(a) –1
(c) 1
æ 1ö
(d) ç 9, ÷
è 81 ø
the sum of all values of a for which det (A) + 1 = 0, is :
[April 12, 2019 (I)]
(a) 0
(b) –1
(c) 1
(d) 2
, is
bx + c is equal to
ax + b bx + c
0
b
50.
53.
(a) –2
(b) 1
(c) 2
(d) 0
If a > 0 and discriminant of ax2+2bx+c is –ve, then
b
(c) (3, 81)
(a) 8
If a1, a2 , a3 ,......, an ,.... are in G.P., then the value of the
determinant
[2004]
a
æ 1ö
(b) ç 9, ÷
è 9ø
and C = 3A, then
(d) 2
[Sep. 03, 2020 (II)]
æ 1ö
(a) ç 3, ÷
è 81 ø
[2005]
(c) 4
2
1 + a2 x
é 2 -1 1 ù
= ê -1 0 2 úú
Let A be a 3 × 3 matrix such that ad A ê
and
êë 1 -2 -1úû
B = ad(ad A). If | A |= l and | ( B-1 )T |= m, then the ordered
If a1 , a2 , a3 , ............, an , ...... are in G. P., then the
determinant
log an
47.
Adjoint of a Matrix, Inverse of a
TOPIC Đ Matrix, Some Special Cases of
(a)
1
4
(b) 1
(c)
1
16
(d) 16
Downloaded from @Freebooksforjeeneet
EBD_8344
M-282
Mathematics
écos q - sin qù
ú , then the matrix A–50 when
56. If A = ê
ë sin q cos q û
q=
p
, is equal to:
12
é
ê
ê
(a) ê
ê
ë
1
2
3
2
é 3
ê
ê 2
(c) ê 1
êë 2
57.
58.
59.
60.
61.
62.
3ù
- ú
2 ú
1 ú
ú
2 û
1 ù
ú
2 ú
3ú
ú
2 û
[Jan 09, 2019 (I)]
63.
é
ê
ê
(b) ê
ê
ë
3
2
1
2
1ù
- ú
2ú
3ú
ú
2 û
é 1
ê
ê 2
(d) ê 3
êë 2
3ù
ú
2 ú
1 ú
ú
2 û
(a) ±
64.
65.
é 72 -84 ù
(b) ê
ú
ë -63 51 û
é 51 63ù
(c) ê
ú
ë84 72 û
é 51 84 ù
(d) ê
ú
ë 63 72 û
Let A be any 3 × 3 invertible matrix. Then which one of the
following is not always true ?
[Online April 8, 2017]
(a) ad (A)= |A| . A–1
(b) ad (ad(A)) = |A|.A
(c) ad (ad(A)) = |A|2 .(ad(A))–1
(d) ad (ad(A)) = |A|.(ad(A))–1
(b) ±
1
25
(c) ±1
(d) ±5
If A is an 3 × 3 non-singular matrix such that AA' = A'A and
B = A–1A', then BB' equals:
[2014]
(b)
( B )¢
-1
(c) I + B
(d) I
Let A be a 3 × 3 matrix such that
é 1 2 3ù é 0 0 1 ù
A êê0 2 3úú = êê1 0 0úú
êë0 1 1úû êë0 1 0úû
|3A| = 108. Then A2 equals
[Online April 15, 2018]
0ù
é 4 – 32 ù
é 4
(a) ê
(b) ê
ú
ú
0
36
–
32
36
ë
û
ë
û
é 36 0 ù
é36 – 32 ù
(c) ê
(d) ê
ú
4 úû
ë – 32 4 û
ë0
Suppose A is any 3 × 3 non-singular matrix and
(A – 3I) (A – 5I) = O, where I = I3 and O = O3. If aA + bA –1 = 4I,
then a + b is equal to
[Online April 15, 2018]
(a) 8
(b) 12
(c) 13
(d) 7
é 72 -63ù
(a) ê
ú
ë -84 51 û
1
5
(a) B –1
é1 2ù
Let A be a matrix such that A . ê
ú is a scalar matrix and
ë 0 3û
é 2 -3ù
2
If A = ê
ú , then ad (3A + 12A) is equal to : [2017]
ë -4 1 û
Then :
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement–I is true, but Statement-II is false.
(d) Statement I is false, but Statement-II is true.
If A is a 3 × 3 matrix such that |5.adA| = 5, then |A| is
equal to :
[Online April 11, 2015]
Then A–1 is:
66.
67.
68.
[Online April 11, 2014]
é3 1 2 ù
ê3 0 2 ú
ú
(a) ê
ëê1 0 1 úû
é3 2 1 ù
ê3 2 0 ú
ú
(b) ê
ëê1 1 0 úû
é 0 1 3ù
ê 0 2 3ú
ú
(c) ê
êë1 1 1úû
é 1 2 3ù
ê0 1 1ú
ú
(d) ê
êë0 2 3úû
é 1 a 3ù
If P = êê 1 3 3úú is the adoint of a 3 × 3 matrix A and
êë 2 4 4úû
|A| = 4, then a is equal to :
[2013]
(a) 4
(b) 11
(c) 5
(d) 0
Let P and Q be 3 ´ 3 matrices P ¹ Q. If P3= Q3 and
P2Q = Q2P then determinant of (P2 + Q2) is equal to :
(a) – 2
(b) 1
[2012]
(c) 0
(d) – 1
æ1 0 0ö
Let A = çç 2 1 0 ÷÷ . If u1 and u2 are column matrices such
ç 3 2 1÷
è
ø
é5a - b ù
T
If A = ê
ú and A ad A = A A , then 5a + b is equal to:
ë3 2û
[2016]
(a) 4
(b) 13
(c) –1
(d) 5
Let A be a 3 × 3 matrix such that A2 – 5A + 7I = 0.
æ0ö
æ1ö
ç ÷
that Au1 = çç 0 ÷÷ and Au2 = ç 1 ÷ , then u1 + u2 is equal to :
ç0÷
ç0÷
è ø
è ø
[2012]
1
(5I - A) .
7
Statement II : the polynomial A3 – 2A2 – 3A + a can be
reduced to 5 (A – 4I).
[Online April 10, 2016]
æ -1 ö
ç ÷
(a) ç 1 ÷
ç0÷
è ø
-1
Statement–I : A =
æ -1 ö
ç ÷
1
(b) ç ÷
ç -1 ÷
è ø
æ -1 ö
ç ÷
(c) ç -1 ÷
ç0÷
è ø
Downloaded from @Freebooksforjeeneet
æ1ö
ç ÷
(d) ç -1 ÷
ç -1 ÷
è ø
M-283
Determinants
69.
If AT denotes the transpose of the matrix
é0 0 a ù
A = ê0 b c ú ,
ê
ú
ëêd e f úû
where a, b, c, d, e and f are integers such that abd ¹ 0, then
the number of such matrices for which A–1 = AT is
[Online May 19, 2012]
(a) 2(3!)
(b) 3(2!)
(c) 23
(d) 32
70.
71.
73.
75.
76.
R = { ( A, B ) A = P BP for some invertible matrix P}
Statement-1 : R is equivalence relation.
Statement-2 : For any two invertible 3 ´ 3 matrices M and
77.
-1
(c) A – I
[2005]
(d) I – A
æ 1 -1 1 ö
æ 4 2 2ö
ç
÷
Let A = 2 1 -3 . and B = ç -5 0 a ÷ . If B is the
ç
÷
ç
÷
è1 1 1 ø
è 1 -2 3 ø
[2004]
(d) –2
æ 0 0 -1ö
Let A = ç 0 -1 0 ÷ . The only correct
ç
÷
è -1 0 0 ø
statement about the matrix A is
[2004]
(a) A2 = I
(b) A = (–1) I, where I is a unit matrix
-1
(c) A does not exist
(d) A is a ero matrix
Solution of System of Linear
TOPIC Ė Equations
= N -1 M -1 .
(a) Statement-1 is true, statement-2 is true and statement2 is a correct explanation for statement-1.
(b) Statement-1 is true, statement-2 is true; statement-2 is
not a correct explanation for statement-1.
(c) Statement-1 is true, stement-2 is false.
(d) Statement-1 is false, statement-2 is true.
Let A be a 2 × 2 matrix
Statement -1 : ad (ad A) = A
Statement -2 : |ad A |= |A|
[2009]
(a) Statement-1 is true, Statement-2 is true. Statement-2
is not a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false.
(c) Statement -1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement -2 is true.
Statement-2 is a correct explanation for Statement-1.
Let A be a square matrix all of whose entries are integers.
Then which one of the following is true? [2008]
(a) If det A = ± 1, then A–1 exists but all its entries are not
necessarily integers
(b) If det A ¹ ± 1, then A–1 exists and all its entries are non
integers
(c) If det A = ± 1, then A–1 exists but all its entries are
integers
(d) If det A = ± 1, then A–1 need not exists
(b) A
inverse of matrix A, then a is
(a) 5
(b) –1
(c) 2
é0 g ù
[Online May 12, 2012]
ê d 0 ú , respectively.
ë
û
Statement 1: AB – BA is always an invertible matrix.
Statement 2: AB – BA is never an identity matrix.
(a) Statement 1 is true, Statement 2 is false.
(b) Statement 1 is false, Statement 2 is true.
(c) Statement 1 is true, Statement 2 is true; Statement 2 is
a correct explanation of Statement 1.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation of Statement 1.
Consider the following relation R on the set of real square
matrices of order 3.
[2011RS]
N, ( MN )
2
If A – A + I = 0 , then the inverse of A is
(a) A + I
éa 0 ù
Let A and B be real matrices of the form ê
ú and
ë 0 bû
–1
72.
74.
The values of l and m for which the system of linear
equations
[Sep. 06, 2020 (I)]
x+y+z=2
x + 2y + 3z = 5
x + 3y + lz = m
has infinitely many solutions are, respectively :
(a) 6 and 8
(b) 5 and 7
(c) 5 and 8
78.
(d) 4 and 9
The sum of distinct values of l for whcih the system of
equations
(l - 1) x + (3l + 1) y + 2lz = 0
(l - 1) x + (4l - 2) y + (l + 3) z = 0
2x + (3l + 1) y + 3(l – 1) z = 0,
has non- ero solutions, is ______. [NA Sep. 06, 2020 (II)]
79.
Let l Î R . The system of linear equations
2 x1 - 4 x2 + lx3 = 1
x1 - 6 x2 + x3 = 2
lx1 - 10 x2 + 4 x3 = 3
(a) exactly one negative value of l
(b) exactly one positive value of l
(c) every value of l
(d) exactly two value of l
Downloaded from @Freebooksforjeeneet
[Sep. 05, 2020 (I)]
EBD_8344
M-284
80.
Mathematics
If the system of linear equations
85.
x + y + 3z = 0
2x - y + 2z = 2
x + 3y + k2z = 0
x - 2 y + lz = -4
3x + y + 3z = 0
x + ly + z = 4
has a non- ero solution (x, y, z) for some k ÎR, then
æ yö
x + ç ÷ is equal to :
è zø
(a) – 3
81.
(b) 9
has no solution. Then the set S
(a) contains more than two elements.
(b) is an empty set.
(c) is a singleton.
(d) contains exactly two elements.
[Sep. 05, 2020 (II)]
(c) 3
(d) – 9
If the system of equations x - 2 y + 3 z = 9 , 2 x + y + z = b
86.
x - 7 y + az = 24, has infinitely many solutions, then
a – b is equal to __________.
[NA Sep. 04, 2020 (I)]
82. Suppose the vectors x1, x2 and x3 are the solutions of the
system of linear equations, Ax = b when the vector b on
the right side is equal to b1, b2 and b3 respectively. If
é1ù
é 0ù
é0ù
é1ù
é 0ù
x1 = ê1ú , x2 = ê 2ú , x3 = ê0ú , b1 = ê0ú , b2 = ê 2ú
êú
ê ú
ê ú
ê ú
ê ú
êë1 ûú
ëê1ûú
ëê1ûú
ëê0ûú
ëê 0ûú
0
é ù
b3= ê0ú , then the determinant of A is equal to :
ê ú
êë 2úû
(a) 4
and
87.
83.
(d)
3
2
If the system of equations
(d) only the trivial solution.
88.
For which of the following ordered pairs (m, d), the system
of linear equations
x + 2y + 3z = 1
3x + 4y + 5z = m
4x + 4y + 4z = d
is inconsistent?
[Jan. 8, 2020 (I)]
(a) (4, 3)
(b) (4, 6)
(c) (1, 0)
(d) (3, 4)
89.
The system of linear equations
x+ y+ z = 2
2x + 4 y - z = 6
3x + 2 y + lz = m
has infinitely many solutions, then : [Sep. 04, 2020 (II)]
84.
(a) l + 2m = 14
(b) 2l - m = 5
(c) l - 2m = -5
(d) 2l + m = 14
Let S be the set of all integer solutions, (x, y, z), of the
system of equations
x - 2 y + 5z = 0
The following system of linear equations
7x + 6y – 2z = 0
3x + 4y + 2z = 0
x – 2y – 6z = 0, has
[Jan. 9, 2020 (II)]
(a) infinitely many solutions, (x, y, z) satisfying y = 2z.
(b) no solution.
(c) infinitely many solutions, (x, y, z) satisfying x = 2z.
(b) 2
1
2
Let A = {X = (x, y, z)T : PX = 0 and x2 + y 2 + z 2 = 1},
é1 2 1ù
where P = êê -2 3 -4 úú , then the set A :
êë 1 9 -1úû
[Sep. 02, 2020 (II)]
(a) is a singleton
(b) is an empty set
(c) contains more than two elements
(d) contains exactly two elements
[Sep. 04, 2020 (II)]
(c)
Let S be the set of all l Î R for which the system of linear
equations
[Sep. 02, 2020 (I)]
lx + 2y + 2z = 5
2lx + 3y + 5z = 8
4x + ly + 6z = 10 has:
-2 x + 4 y + z = 0
-7 x + 14 y + 9 z = 0
2
2
2
such that 15 £ x + y + z £ 150. Then, the number of
elements in the set S is equal to ____________.
[NA Sep. 03, 2020 (II)]
[Jan. 8, 2020 (II)]
(a) no solution when l = 8
(b) a unique solution when l = –8
(c) no solution when l = 2
(d) infinitely many solutions when l = 2
Downloaded from @Freebooksforjeeneet
M-285
Determinants
90.
95.
If the system of linear equations
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0,
solution, then:
[Jan. 7, 2020 (I)]
1 1 1
, , are in A.P..
a b c
(b) a, b, c are in G.P.
(c) a + b + c = 0
(d) a, b, c are in A.P.
91.
If the system of linear equations,
x+y+z=6
x + 2y + 3z = 10
3x + 2y + lz = m
has more than two solutions, then m – l2 is equal to
_________.
92.
[NA Jan. 7, 2020 (II)]
If the system of linear equations
x + y+ = 5
x + 2y + 2 = 6
x + 3y + l = m, (l, m Î R), has infinitely many solutions,
then the value of l + m is :
[April 10, 2019 (I)]
(a) 12
93.
(b) 9
(c) 7
(d) 10
Let l be a real number for which the system of linear
equations:
x+y+z=6
4x + ly – lz = l –2
3x + 2y – 4z = –5
has infinitely many solutions. Then l is a root of the
quadratic equation :
[April 10, 2019 (II)]
94.
(a) l2 + 3l – 4 = 0
(b) l2 – 3l – 4 = 0
(c) l2 +l – 6 = 0
(d) l2 – l – 6 = 0
If the system of equations 2x + 3y – z =0, x + ky – 2z = 0 and
2x – y + z = 0 has a non-trivial solution (x, y, z), then
x y z
+ + + k is equal to:
y z x
(a)
3
4
(b)
1
(c) 2
(d) 0
2
96. If the system of linear equations
x – 2y + kz = 1
2x + y + z = 2
3x – y – kz = 3
has a solution (x, y, z), z ¹ 0, then (x, y) lies on the straight
line whose equation is :
[April 08, 2019 (II)]
(a) 3x – 4y – 1 = 0
(b) 4x – 3y – 4 = 0
(c) 4x – 3y – 1 = 0
(d) 3x – 4y – 4 = 0
97. An ordered pair (a, b) for which the system of linear
equations
(1 +a) x + by + = 2
ax + (1 + b)y + = 3
ax + by + 2 = 2
has a unique solution, is :
[Jan. 12, 2019 (I)]
(a) (2, 4)
(b) (–3, 1)
(c) (–4, 2)
(d) (1, – 3)
98. The set of all values of l for which the system of linear
equations
x – 2y – 2 = lx
x + 2y + = ly
–x – y = l2
has a non-trivial solution :
[Jan. 12, 2019 (II)]
(a) is a singleton
(b) contains exactly two elements
(c) is an empty set
(d) contains more than two elements
99. If the system of linear equations
2x + 2y + 3 = a
3x – y + 5 = b
x – 3y + 2 = c
where, a, b, c are non- ero real numbers, has more than one
solution, then :
[Jan. 11, 2019 (I)]
(a) b – c + a = 0
(b) b – c – a = 0
(c) a + b + c = 0
(d) b + c – a = 0
100. The number of values of q Î (0, p) for which the system of
(a) –1
where a, b, c ÎR are non- ero and distinct; has a non- ero
(a)
The greatest value of c Î R for which the system of linear
equations
x – cy – cz = 0; cx – y + cz = 0; cx + cy – z = 0
has a non-trivial solution, is :
[April 08, 2019 (I)]
1
2
[April 09, 2019 (II)]
(c) -
1
4
(d) –4
(b)
linear equations
x + 3y + 7z = 0
– x + 4y + 7z = 0
(sin 3q)x + (cos 2q)y + 2z = 0
has a non-trivial solution, is:
[Jan. 10, 2019 (II)]
(a) three
(b) two
(c) four
(d) one
Downloaded from @Freebooksforjeeneet
EBD_8344
M-286
Mathematics
101. If the system of equations
[Jan 10, 2019 (I)]
x+y+z=5
x + 2y + 3z = 9
x + 3y + az = b
has infinitely many solutions, then b – a equals:
(a) 21
(b) 8
(c) 18
102. If the system of linear equations
x – 4y + 7z = g
3y – 5z = h
– 2x + 5y – 9z = k
is consistent, then :
(a) g + 2h + k = 0
(b) g + h + 2k = 0
(c) 2g + h + k = 0
(d) g + h + k = 0
103. If the system of linear equations
x + ky + 3 = 0
3x + ky – 2 = 0
2x + 4y – 3 = 0
(d) 5
[Jan. 09, 2019 (II)]
104.
105.
106.
107.
y2
109.
110.
x
has a non- ero solution (x, y, ), then
108.
is equal to :
[2018]
(a) 10
(b) – 30
(c) 30
(d) – 10
The number of values of k for which the system of linear
equations, (k + 2) x + 10y = k, kx + (k + 3) y = k – 1 has no
solution, is
[Online April 16, 2018]
(a) Infinitely many
(b) 3
(c) 1
(d) 2
Let S be the set of all real values of k for which the system
of linear equations
x+y+z=2
2x + y – z = 3
3x + 2y + kz = 4
has a unique solution. Then S is [Online April 15, 2018]
(a) an empty set
(b) equal to R – {0}
(c) equal to {0}
(d) equal to R
If the system of linear equations
x + ay + z = 3
x + 2y + 2z = 6
x + 5y + 3z = b
has no solution, then
[Online April 15, 2018]
(a) a = 1, b ¹ 9
(b) a ¹ – 1, b = 9
(c) a = – 1, b = 9
(d) a = – 1, b ¹ 9
If S is the set of distinct values of ‘b’ for which the
following system of linear equations
[2017]
x + y+ = 1
x + ay + = 1
ax + by + = 0
has no solution, then S is :
111.
112.
113.
(a) a singleton
(b) an empty set
(c) an infinite set
(d) a finite set containing two or more elements
The number of real values of l for which the system of
linear equations
2x + 4y – l = 0
4x + ly + 2 = 0
lx + 2y + 2 = 0
has infinitely many solutions, is : [Online April 8, 2017]
(a) 0
(b) 1
(c) 2
(d) 3
The system of linear equations
x + ly – = 0
lx – y – = 0
x + y– l = 0
has a non-trivial solution for:
[2016]
(a) exactly two values of l.
(b) exactly three values of l.
(c) infinitely many values of l.
(d) exactly one value of l.
The set of all values of l for which the system of linear
equations :
[2015]
2x1 – 2x2 + x3 = lx1
2x1 – 3x2 + 2x3 = lx2
–x1 + 2x2 = lx3
has a non-trivial solution,
(a) contains two elements.
(b) contains more than two elements
(c) is an empty set.
(d) is a singleton
If a, b, c are non- ero real numbers and if the system of
equations
[Online April 9, 2014]
(a – 1)x = y + ,
(b – 1)y = + x,
(c – 1) = x + y,
has a non-trivial solution, then ab + bc + ca equals:
(a) a + b + c
(b) abc
(c) 1
(d) – 1
The number of values of k, for which the system of equations:
(k + 1) x + 8y = 4k
kx + (k + 3)y = 3k – 1
has no solution, is
[2013]
(a) infinite
(b) 1
(c) 2
(d) 3
Consider the system of equations :
x + ay = 0, y + az = 0 and z + ax = 0. Then the set of all real
values of ‘a’ for which the system has a unique solution
is:
[Online April 25, 2013]
(a) R – {1}
(b) R – { – 1}
(c) {1, – 1}
(d) {1, 0, –1}
Downloaded from @Freebooksforjeeneet
M-287
Determinants
114. Statement-1: The system of linear equations
x + (sin a) y + (cos a) z = 0
x + (cos a) y + (sin a) z = 0
x – (sin a) y – (cos a) z = 0
has a non-trivial solution for only one value of a lying in
æ pö
the interval ç 0, ÷ .
è 2ø
Statement-2: The equation in a
cos a
sin a
cos a
sin a
cos a
sin a
117. If the system of equations
[Online May 7, 2012]
x+y+z=6
x + 2y + 3z = 10
x + 2y + lz = 0
has a unique solution, then l is not equal to
(a) 1
(b) 0
(c) 2
(d) 3
118. If the trivial solution is the only solution of the system of
equations
[2011RS]
x - ky + z = 0
kx + 3 y - kz = 0
=0
cos a - sin a - cos a
æ pö
has only one solution lying in the interval ç 0, ÷ .
è 2ø
[Online April 23, 2013]
(a) Statement-1 is true, Statement-2 is true, Statement-2 is
not correct explantion for Statement-1.
(b) Statement-1 is true, Statement-2 is true, Statement-2 is
a correct explantion for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statememt-1 is false, Statement-2 is true.
115. If the system of linear equations :
x1 + 2x2 + 3x3 = 6
x1 + 3x2 + 5x3 = 9
2x1 + 5x2 + ax3 = b
is consistent and has infinite number of solutions, then :
[Online April 22, 2013]
(a) a = 8, b can be any real number
(b) b = 15, a can be any real number
(c) a Î R - {8} and b Î R - {15}
(d) a = 8, b = l5
116. Statement 1: If the system of equations x + ky + 3z = 0,
3x + ky – 2z = 0, 2x + 3y – 4z = 0 has a non-trivial solution,
31
then the value of k is
.
2
Statement 2: A system of three homogeneous equations
in three variables has a non trivial solution if the determinant
of the coefficient matrix is ero. [Online May 26, 2012]
(a) Statement 1 is false, Statement 2 is true.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation for Statement 1.
(c) Statement 1 is true, Statement 2 is true, , Statement 2 is
not a correct explanation for Statement 1.
(d) Statement 1 is true, Statement 2 is false.
3x + y - z = 0
then the set of all values of k is :
(a) R - { 2, -3}
(b) R - { 2}
(c) R - { -3}
(d)
{ 2, -3}
119. The number of values of k for which the linear equations
4x + ky + 2z = 0 , kx + 4y + z = 0 and 2x + 2y + z = 0 possess
a non- ero solution is
[2011]
(a) 2
(b) 1
(c) ero
(d) 3
120. Consider the system of linear equations;
[2010]
x1 + 2x2 + x3 = 3
2x1 + 3x2 + x3 = 3
3x1 + 5x2 + 2x3 = 1
The system has
(a) exactly 3 solutions
(b) a unique solution
(c) no solution
(d) infinite number of solutions
121. Let a, b, c be any real numbers. Suppose that there are real
numbers x, y, z not all ero such that x = cy + bz, y = az + cx,
and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to
(a) 2
(b) –1
[2008]
(c) 0
(d) 1
122. The system of equations
ax+y + = a –1
x + a y+ = a – 1
x+ y+ a = a –1
has infinite solutions, if a is
[2005]
(a) – 2
(b) either – 2 or 1
(c) not – 2
(d) 1
123. If the system of linear equations
[2003]
x + 2ay + az = 0 ; x + 3by + bz = 0 ;
x + 4cy + cz = 0 has a non - ero solution, then a, b, c.
(a)
(b)
(c)
(d)
satisfy a + 2b + 3c = 0
are in A.P
are in G..P
are in H.P.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-288
1.
Mathematics
R1 = R1 + R3 – 2R2
é cos q sin q ù
(d) Q A = ê
ú
ë - sin q cos q û
1
Þ
é cos nq sin nq ù
\ An = ê
ú , n ÎN
ë - sin nq cos nq û
4.
(d) D1 =
- sin q
-x
1
cos q
1
x
= (x – x2 – 1) – sin q (– x sin q – cos q)
+ cos q (– sin q + x cos q)
= – x3 – x + x sin2q + sin q cos q – cos q sin q + x cos2q
= – x3 – x + x = – x3
Similarly, D2 = – x3 Then, D1 + D2 = – 2x3
p
4p ù
sin + sin
5
5 ú
ú
p
4p ú
cos + cos
5
5 úû
0 1
p
Then, det( B ) = 2 sin æç ö÷ ×
è 5 ø -1 0
5.
10 - 2 5 2.35
»
» 1.175
2
2
2x - 3
3x - 4
D = 2 x - 3 3x - 4
x-2
4x - 5
3x - 5 5x - 8 10 x - 17
x-2
x -1
x -1
Þ D = 2x - 3
x -1
x -1
éC3 ® C3 - C2 ù
êC ® C - C ú
ë 2
2
1û
x -1
0
0
Þ D = x -1
3x - 5 2 x - 3 5x - 9
[ R2 ® R2 - R1 ]
Þ D = -( x - 1)[( x - 1)(5 x - 9) - ( x - 1)(2 x - 3)]
= -3x + 12 x - 15x + 6
So, B + C = –3
(a) |A| =
3.
(d) If f ( x ) = x + b
x+c
x+2
x +1
x +3 x +2
x+4
-3x
x-3 = 0
2x
x+2
x +3
b
0
=0
1
1
b b +1 b
1
b
2
= 2(2b2 + 2 – b2) – b(2b – b) + 1(b2 – b2 – 1)
= 2b2 + 4 – b2 – 1 = b2 + 3
| A|
3
=b +
b
b
Q
2
x+a
-1
2
Þ D = -( x - 1)[(5x 2 - 14 x + 9) - (2 x 2 - 5 x + 3)]
3
2
-3
2
6.
x -1
-6
\ sum of real roots =
3x - 5 2 x - 3 5x - 9
x-2
(b) Given
x
On expanding,
x (– 3x2 – 6x – 2x2 + 6x) – 6 (– 3x + 9 – 2x – 4)
– (4x – 9x) = 0
2
Þ x (– 5x ) – 6 (– 5x + 5) – 4x + 9x = 0
Þ x3 – 7x + 6 = 0
Q all the roots are real.
\ det B Î (1, 2)
(c)
x+3
sin q cos q
x
é cos q sin q ù é cos 4q sin 4q ù
=ê
ú+ê
ú
ë - sin q cos q û ë - sin 4q cos 4q û
2.
x+4
Þ f(x) = 1 Þ f(50) = 1
\B = A+ A
p
4p
é
ê cos 5 + cos 5
\B = ê
ê - sin p - sin 4p
êë
5
5
0
x+3 x+2
x+c
4
=
0
f ( x) = x + b
\
3
1
b ³ æ b 3ö 2 Þ b + 3 ³ 2 3
çè ÷ø
2
b
b
b+
| A|
³2 3
b
Minimum value of
| A|
is 2 3.
b
Downloaded from @Freebooksforjeeneet
M-289
Determinants
x -4
7.
(b) Here, 2x
2x
2x
x-4
2x = (A+Bx) (x–A)2
20
= | A|2014 -15
x -4
2x
-4
0
0
Put x = 0 Þ 0
0
-4
0 = A3 Þ A3 = (-4)3
0
Now, A2016 – 2A2015 – A2014 = A2014 (A2 – 2A – I)
Þ |A2016 – 2A2015 – A2014 | = | A2014 ||A2 – 2A –I|
2x
-4
x2 + x
10. (a)
Let 2x + 3x - 1
2
x + 2x + 3
x-4
2x
x-4
2x
2x
\
2x
2x
2x
1-
4
x
2x
0
x-4
11.
1
®0
x
tan p 3 + tan x
3.tan x
xÎ s
1
y 1 ³0
12. (c) Let f (q) =
1
cos q
1
- sin q
1
- cos q
-1
sin q
1
2
= (1 + sin q cos q) - cos q( - sin q - cos q) + 1( - sin q + 1)
= 1 + sin q cos q + sin q cos q + cos2 q - sin 2 q + 1
= 2 + 2sin q cos q + cos 2q
= 2 + sin 2q + cos 2q ...(1)
Now, maximum value of (1)
3 +1
3
3 +1 1+ 3
´
Þ
3 1+ 3
å 1-
is 2 + 12 + 12 = 2 + 2
and minimum value of (1) is
1+ 3+ 2 3
-2
xÎs
å
2 - 12 + 12 = 2 - 2 .
= -2 - 3
9.
(d)
xy – x – y – + 2 ³ 0
xy + 2 ³ x + y + > 3 (xy )1/3
xy + 2 – 3(xy )1/3 ³ 0
ut(xy ) = t3
t3 – 3t + 2 ³ 0
(t + 2) (t – 1)2 ³ 0
[t = – 2] t3 = – 8
æp
ö
å tan çè 3 + x ÷ø = xå
Î s 1 - tan p
xÎ s
= –a – 12
1 1
\ ordered pair (A, B) is (–4, 5)
(c) Since the given determinant is equal to ero.
Þ 0 (0 – cos x sin x) – cos x(0 – cos2x) – sin x
(sin2x – 0) = 0
Þ cos3x – sin3x = 0
Þ tan3 = 1 Þ tan x = 1
xÎ s
0
x 1 1
4
4
2x = (B- )(1 + ) 2
x
x
4
1x
2 2 1
å 1-
2x - 1 2x - 1
Þ –3 (6 + 6) = –a – 12 Þ – 36 + 12 = a
Þ a = 24
1 2 2
\
3x - 3 = ax – 12
-3
-3 -3
2
Þ 2 1 2 =BÞ B =5
8.
0
-2 -3
2x
Now take x ® ¥ , then
3x
x -2
Put x = – 1 , we get
2x = (Bx - 4)(x + 4) 2
Now take x common from both the sides
4
1x
x +1
2
Þ A = -4
Þ 2x
2x
5
-5 = – 25
é -4 -1ù
é -4 -1ù é -4
(d) A = ê 3 1 ú Þ A2 = ê 3 1 ú ê 3
ë
û
ë
ûë
é13 3 ù
= ê -9 -2 ú and |A| = 1.
ë
û
13. (a) det [(I + B)50 – 50B]
-1ù
1 úû
= det [50C0 I +
50
50
C1 B +
50
50
C2 B2 +
50
50
C3 B3 + ...
+ C50 B B – 50B]
{All terms having Bn, 2 £ n £ 50
will be ero because given that B2 = 0}
= det [I + 50B – 50B] = det [I] = 1
Downloaded from @Freebooksforjeeneet
EBD_8344
M-290
Mathematics
14. (d) The matrices in the form
é a11 a12 ù
êë a21 a22 úû , aij Î {0, 1, 2}, a11 = a12 are
0 /1/ 2ù é 1
0 /1/ 2 ù é 2
0 /1/ 2ù
é 0
,
,
êë 0 /1/ 2
0 úû êë0 /1/ 2
1 úû êë0 /1/ 2
2 úû
At any place, 0/1/2 means 0, 1 or 2 will be the element at
that place.
Hence there are total 27 = 3 × 3 + 3 × 3 + 3 × 3) matrices of the
above form. Out of which the matrices which are singular
are
é 0 0 / 1/ 2 ù é 0 0 ù é1 1ù é 2 2ù
,
,
,
êë 0
0 úû êë1/ 2 0 úû êë1 1úû êë 2 2úû
Hence there are total 7(= 3 + 2 + 1 + 1) singular matrices.
Therefore number of all non-singular matrices in the given
form = 27 – 7 = 20
15. (b)
é a b ù é a b ù é1 0ù
ê c d ú ê c d ú = ê0 1ú
ë
ûë
û ë
û
é a 2 + bc ab + bd ù é1 0 ù
ê
ú=ê
ú
êë ac + cd bc + d 2 úû ë 0 1 û
b(a + d) = 0, b = 0 or a = –d
… (1)
c(a + d) = 0, c = 0 or a = – d
… (2)
a2 + bc = 1, bc + d2 = 1
… (3)
‘a’ and ‘d’ are diagonal elements a + d = 0
statement-1 is correct.
Now, det( A) =ad - bc
Now, from (3) a 2 + bc = 1 and d 2 + bc = 1
So, a 2 - d 2 = 0
Adding a 2 + d 2 + 2bc = 2
Þ (a + d )2 - 2ad + 2bc = 2
éa b ù
17. (b) Let A = ê
ú where a, b, c, d ¹ 0
ëc d û
éa b ù éa b ù
A2 = ê
úê
ú=I
ëc d û ëc d û
éa 2 + bc ab + bd ù é1 0 ù
Þ A2 = ê
ú=ê
ú
êë ac + cd bc + d 2 úû ë 0 1 û
Þ a 2 + bc = 1, bc + d 2 = 1
ab + bd = ac + cd = 0
c ¹ 0 and b ¹ 0 Þ a + d = 0
| A |= ad - bc = -a 2 - bc = -1
Also if A ¹ I, then tr(A) = a + d = 0.
\ Statement-1 true and statement-2 false.
éa b ù
18. (d) Let A = ê
ú
ëc d û
Given that A2 = I
é a 2 + bc ab + bd ù é1 0 ù
ê
ú=ê
ú
êë ac + cd bc + d 2 úû ë 0 1 û
Þ a2 + bc = 1 and ab + bd =0
ac + cd = 0 and bc + d2 = 1
From these four equations,
a2 + bc = bc + d2 Þ a2 = d2
and b(a + d) = 0 = c(a + d) Þ a = – d
|A| = ad – bc = –a2 – bc = –1
Also if A ¹ I then tr(A) = a + d = 0
\ Statement 2 is false.
Þ
é 5 5a a ù
19. (a) Given that A = ê 0 a 5a ú and | A2 | = 25
ê
ú
5û
ë0 0
é5 5a a ù é5 5a a ù
\ A2 = ê0 a 5a ú ê0 a 5a ú
ê
úê
ú
5 û ë0 0
5û
ë0 0
or 0 - 2(ad - bc) = 2
é 25 25a + 5a 2
ê
a2
=ê0
êë 0
0
So, ad - bc = 1 Þ det( A) = –1
So, statement – 2 is also true.
But statement – 2 is not the correct explanation of
statement-1.
16. (d) We know that determinant of skew symmetric matrix
of odd order is ero.
So, statement-1 is true.
( )
\ | A2 | = 25 (25a 2 )
\ 25 = 25 (25a 2 ) Þ | a | =
T
We know that det A = det (A).
det (– A) = – (– 1)n det (A).
where A is a n ´ n order matrix.
So, statement-2 is false.
5a + 25a 2 + 5a ù
ú
5a 2 + 25a ú
úû
25
20. (b)
1
wn
w 2n
D = wn
w 2n
1
1
wn
w
2n
1
5
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M-291
Determinants
Expand through R1
(
)
(
)
(
3n
n
2n
2n
+ w 2n w n - w 4n
= 1 w -1 - w w - w
)
= w 3n - 1 - 0 + w 3n - w 6 n
= 1 - 1 + 1 - 1 = 0 éQ w3n = 1ù
ë
û
25. (a) C1 ® C1 + C2
2
sin 2 q
4cos 6q
2
2 1 + sin q
4 cos 6q = 0
1
sin 2 q 1 + 4 cos 6q
R1 ® R1 – R2 , R2 ® R2 – R3
21. (b) Applying C2 ® C2 - C1
0
-1
0
- sin 2 q
-1
1
1
1
-1
f (q) = - cos 2 q
-1
1
1 sin q (1 + 4cos 6q)
-2 -2
12
On expanding, we get
æ p pö
= 4 cos 2q, q Î ç , ÷
è 4 2ø
2p 4p
p 2p
or
Þ q = or
3
3
9
9
26. (c) Let a = w and b = w2 are roots of x2 + x + 1 = 0
Therefore, 6q =
Max. f (q) = M = 0
Min. f (q) = m = -4
& Let D = w
22. (b) Use properties of determinant
y b+ y
z
c+ y
x a
x+a
y +b = y b
y +b + y y 1 y +b
z +c
z
c
z 1 z +c
é R2 ® R2 - R1 , ù
ê
ú
ë R3 ® R3 - R1 û
0
z-x 0
y + w2
1
1
y+w
w
-1
=D
Applying C1 ® C1 + C2 + C3
y + 1 + w + w2
1 x+a
=0+ y y- x 0
w2
2
x 1 x +a
z+c
x
w
y +1
So, (m, M ) = ( -4, 0)
x+a
2 + 4 cos 6q = 0
æ pö
1
cos 6q = - Q qÎ ç 0, ÷ Þ 6q (0, 2p)
2
è 3ø
= 4(cos2 q - sin 2 q)
x a+ y
=0
2
D = y + 1 + w + w2
w
w2
y + w2
1
1
y+w
1 + w + w2 + y
= - y ( x - y ) = - y (b - a) = y (a - b)
0
23. (b)
D=
y
2 1
1
1 -1 1 = 5
2
x¢ y ¢ 1
D= y
y
w
w2
y + w2
1
1
y+w
(Q 1 + w + w2 = 0)
Þ – 2(1 – x¢) + (y ¢ + x¢) = ± 10
Þ – 2 + 2x¢ + y¢ + x¢ = ± 10
1
Þ 3x¢ + y¢ = 12 or 3x¢ + y¢ = – 8
\
w
D = y 1 y + w2
l = 3, – 2
1
1
w2
1
y+w
24. (d) It is given that |B| = 81
b11
\
b12
B = b21 b22
b31 b32
30 a11
31 a12
32 a13
b23 = 31 a21
32 a22
33 a23
b13
b33
Þ 81 = 33 × 32 × 31 |A|
1
Þ 34 = 36 |A| Þ A =
9
2
3
3 a31 3 a32
4
3 a33
Applying R2 ® R2 - R1 & R3 ® R3 - R1
=
D y
y + w2 - w
1 - w2
1- w
y + w - w2
=
Þ D y é y - (w - w2 )( y + (w - w2 ) - (1 - w)(1 - w2 ) ù
ë
û
=
Þ D y é y 2 - (w - w2 )2 - 1 + w2 + w - w3 ù
ë
û
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EBD_8344
M-292
Mathematics
Þ D = y é y 2 - w2 - w4 + 2w3 - 1 + w2 + w4 - w3 ù
ë
û
C1 ® C1 – C3, C2 ® C2 – C3
0
0
0
– b –c –a
c +a +b
c+a+b
(Q w4 = w)
D = (a + b + c)
Þ D = y (y2) = y3
27. (c) Consider, | A | =
1
1
1
2
b
c
4 b2
c2
2b
c –a –b
= (a + b + c) (a + b + c)2
Hence, x = –2(a + b + c)
C2 ® C2 - C1 , C3 ® C3 - C1
|A|=
1
-2
4+d
sin q - 2
1
sin q + 2
d
5
2sin q - d
- sin q + 2 + 2d
30. (a) det(A) =
1
0
0
2
b-2
c-2
4 (b - 2)(b + 2) (c - 2)(c + 2)
= (b – 2) (c – 2) (c – b)
\ 2, b, c are in A.P.
\ (b – 2) = (c – b) = d and c – 2 = 2d
Þ | A | = d.2d.d = 2d3
Applying R3 ® R3 - 2 R2 + R1 we get
det (A) =
-2
4+d
sin q - 2
1
sin q + 2
d
1
0
0
d(4 + d) – (sin2q – 4)
Q| A |Î [2,16] Þ 1 £ d 3 £ 8 Þ 1 £ d £ 2
=
4 £ 2d + 2 £ 6 Þ 4 £ c £ 6
Þ det (A) = d2 + 4d + 4 – sin2q = (d + 2)2 – sin2q
sin q
1
- sin q
1
sin q
-1
- sin q
1
0
0
2
- sin q
1
sin q
-1
- sin q
1
28. (d) |A| =
=
Minimum value of det (A) is attained when sin2q = 1
1
\
Þ d = –5 or 1
31. (b) Let common ratio of G.P. be R
Þ a2 = a1R, a3 = a1R2, ... a10 = a1R9
R1 ® R1 + R3
C1 ® C1 – C2, C2 ® C2 – C3
2(sin2q + 1)
=
æ 3p 5p ö
æ 1ö
Since, q Î çè , ÷ø Þ sin2q Î çè 0, ÷ø
4 4
2
\
det(A) Î [2, 3)
D=
æ3 ù
[2, 3) Ì çè , 3ú
2 û
2a
2b
b–c –a
2b
2c
2c
c –a –b
a +b+c
a +b+c
2b
b–c–a
2b
2c
2c
c –a –b
= (a + b + c)
æ ar ak ö
ln ç 2r 3k ÷
è a3 a4 ø
ln a3r a4k
æ ar ak ö
ln ç 4r 5k ÷
è a5 a6 ø
æ ar ak ö
ln ç 5r 6k ÷
è a6 a7 ø
ln a6r a7k
æ ar ak ö
ln ç r8 k9 ÷
è a9 a10 ø
k
ln a9r a10
a7r a8k
a8r a9k
2a
R1 ® R1 + R2 + R3
D=
æ ar ak ö
ln ç 1r 2k ÷
è a2 a3 ø
ln
a –b–c
29. (d) D =
(d + 2)2 –1 = 8 Þ (d + 2)2 = 9 Þ d + 2 = ± 3
a +b+c
1
Rr +k
1
ln r + k
R
D=
1
ln r + k
R
ln
1
R r +k
1
ln r +k
R
1
ln r +k
R
ln
ln a3r a4k
ln a6r a7k = 0
k
ln a9r a10
1
1
1
"r , K Î N
2b
b–c–a
2b
Hence, number of elements in S is infinitely many.
2c
2c
c – a –b
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M-293
Determinants
æ b - 2 öæ 8 ö = –1
ç
÷ç ÷
è a + 2 øè 3 ø
Þ 3a + 8b = 1
Solving (1) and (2), we get
32. (a) det(A) = |A|
=
\
et
e -t cos t
e -t sin t
et
- e -t cos t - e- t sin t
- e- t sin t + e- t cos t
et
2e- t sin t
-2e -t cos t
1 cos t
=
æ 1ö
orthocentre is ç 2, ÷
è 2ø
34. (b) Given 2w + 1 = z;
et × e -t × e -t 1 - cost - sint - sin t + cos t
-2cos t
1 2sin t
3i - 1
2
Þ w is complex cube root of unity
Applying R1 ® R1 + R2 + R3
and =
0 2cos t + sin t
=
2sin t - cos t
e -t 0 - cos t - 3sin t - sin t + 3cos t
1
-2cos t
2sin t
1
2
a = 2, b =
sin t
... (2)
R1 ® R1 - R2
R2 ® R2 + R3
3
3i Þ w =
0
0
2
1 -w - 1 w2
=
0 -5sin t
=
5cos t
e -t 0 - cos t - 3sin t - sin t + 3cos t R1 ® R1 + 2 R2
1
-2cos t
2sin t
w2
w
1
= 3 (–1 – w – w) = –3 (1 + 2w) = – 3
Þ k=–
cos x sin x
= e–t[(–5 sin t)(–sin t + 3 cos t) – 5 cos t (–cos t – 3 sin t)
= 5e–t ¹ 0, V t Î R
\ A is invertible.
33. (a) Let A (k, –3k), B(5, k) and C(–k + 2),
we have
k
1
5
2
-k
cos x sin x = 0
sin x
sin x
cos x
cos x - sin x sin x - cos x
1 = 28
1
sin x
Þ 5k + 13k – 46 = 0
or 5k2 + 13k + 66 = 0
Now, 5k2 + 13k – 46 = 0
0
cos x - sin x sin x - cos x = 0
sin x
cos x
0
C2 ® C2 + C3
2
cos x - sin x sin x - cos x
-23
-13 ± 1089
;k = 2
\ k=
Þ k=
5
10
since k is an integer, \ k = 2
Also 5k2 + 13k + 66 = 0
0
0
sin x
sin x
0
sin x - cos x = 0
cos x
Expanding using second row
2 sin x (sin x – cos x)2 = 0
sin x = 0 or sin x = cos x
p
x = 0 or x =
4
36. (a) Consider
-13 ± -1151
10
So no real solution exist
A(2, –6), B(5, 2) and C(–2, 2)
For orthocentre H (a, b)
BH ^ AC
Þ k=
æ b - 2 öæ 8 ö
\ ç
÷ç ÷ = –1
è a - 5 øè -4 ø
Þ a – 2b = 1
Also CH ^ AB
sin x
R1 ® R1 – R2
R2 ® R2 – R3
-3k 1
k
2
35. (c)
sin x
...(1)
=
3
1 + f (1)
1 + f (1)
1 + f (2)
1 + f (2)
1 + f (3)
1 + f (2)
1 + f (3)
1 + f (4)
1+1 +1
1+ a + b
1 + a 2 + b2
1+ a + b
1 + a 2 + b2
1 + a 3 + b3
1 + a 2 + b2
1 + a3 + b3
1 + a 4 + b4
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EBD_8344
M-294
Mathematics
=
1
1
1
a
1
b
1
a
2
2
1
1
1
1
a
b
1
a2
b2
=
b
1
´ 1
1
a
1
b
1
2
2
a
b
a2
1
[Q |A| = |A |]
( a - l )2
r =1
n -1
å
a2
= [(1 – a) (1 – b) (a – b)]2
(a - l )
r = 1 + 2 + 3 + ... + (n – 1) =
å
r =1
n -1
r =1
n -1
å
r =1
å
Dr =
(n - 1)(3n - 4)
2
S (2 r - 1)
S (3r - 2)
n
2
n(n - 1)
2
n -1
a
(n - 1)2
( n - 1)(3n - 4)
2
D r consists of (n – 1) determinants in L.H.S. and
in R.H.S every constituent of first row consists of
(n – 1) elements and hence it can be splitted into sum
of (n – 1) determinants.
n(n - 1)
(n - 1)(3n - 4)
(n - 1)2
2
2
n -1
n
n -1
a
\ å Dr =
2
r =1
n(n - 1)
(n - 1)(3n - 4)
(n - 1)2
2
2
=0
(Q R1 and R3 are identical)
n -1
Hence, value of
å
r =1
38. (c) Let D =
Dr is independent of both 'a' and 'n'.
a2
b2
c2
(a + l)2
(b + l )2
(c + l ) 2
(a - l)2
(b - l )2
(c - l ) 2
Apply R2 ® R2 – R3
(b - l )
4c l
2
(c - l ) 2
Taking out 4 common from R2
a2
b2
al
c2
bl
2
2
cl
a + l - 2al b + l - 2bl c + l 2 - 2cl
2
2
Apply R3 ® [R3 – (R1 – 2R2)]
a2
Sr
c2
4bl
2
2
(3r - 2) = 1 + 4 + 7 + .. + (3n – 3 – 2)
=
(c - l ) 2
(Q ( x + y )2 - ( x - y )2 = 4 xy )
=4
= (n – 1)2
n -1
\
n (n - 1)
2
b2
4al
=
(2 r - 1) = 1 + 3 + 5 + ... + [2 (n – 1) – 2]
r =1
(b - l )2
2
n -1
å
c2
2
2
2
2
2
2
D = (a + l ) - (a - l ) (b + l ) - (b - l ) (c + l ) - (c - l )
So, K = 1
37. (d)
b2
b2
c2
= 4 al bl
cl
l
2
l
2
l2
Taking out l common from R2 and l2 from R3.
a2
b2
c2
a2
b2
c2
= 4l ( l ) a
1
b
1
c = kl a
1
1
b
1
c
1
2
Þ k = 4l2
39. (b)
a
b
c
b
c
a
c
a +b +c
a =
b
b
c
1
= (a + b + c) b
c
0
= (a + b + c) b - c
c-a
1
c
a
a+b+c
c
a
a+b+c
a
b
1
a
b
0
c-a
a -b
1
a
b
= (a + b + c) [ab + bc + ca – a2 – b2 – c2]
= – (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2]
Since a, b, c are sides of a scalene triangle, therefore at
least two of the a, b, c will be unequal.
\ (a – b)2 + (b – c)2 + (c – a)2 > 0
Also a + b + c > 0
\ – (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2] < 0
Downloaded from @Freebooksforjeeneet
M-295
Determinants
b2 + c 2
ab
ab
c2 + a 2
40. (c) Let D =
ac
0
2 ( a + b) a - c
D = 2 ( a + b)
0
b-c
ac
bc
a-c
2
a + b2
bc
Multiply C1 by a, C2 by b and C3 by c and hence divide by
abc.
(
a b2 + c 2
=
1
abc
)
ab2
ac 2
(
a 2b
b c2 + a 2
a2c
b 2c
)
bc2
(
c a 2 + b2
)
Take out a, b, c common from R1, R2 and R3 respectively.
\
D=
b2 + c 2
b2
c2
a2
c2 + a 2
c2
a2
b2
a 2 + b2
abc
abc
Apply C1® C1 – C2 – C3
b2
c2
c2 + a 2
c2
b2
a2 + b2
0
D = -2c 2
-2b2
b2
c2
c2 + a 2
c2
b2
a2 + b2
0
= -2 c 2
b2
1
x
y
1
1
-y
x
\ Area of triangle =
2
x- y x+ y 1
1
= [x (x – x – y) – y (– y – x + y) + 1 (– yx – y2 – x2 + xy)]
2
1
1
[– xy + xy – y2 – x2] = (x2 + y2)
2
2
(Q Area can not be negative)
(
1 2
2
z
Q z = x + iy , z = x 2 + y 2
2
43. (b) Given parabola is x2 = 8y
Þ 4a = 8 Þ a = 2
To find: Area of DABC
A = (– 2a, a) = (– 4, 2)
B = (2a, a) = (4, 2)
C = (0, 0)
=
0
b2
c2
2
= -2 c
a2
0 = – 2 [– b2 (c2a2) + c2 (– a2b2)]
b2
0
a2
= 2a2b2c2 + 2a2b2c2 = 4a2b2c2
But D = ka2b2c2 \ k = 4
-2c
On expanding, we get
D = – 2 (a + b) {– 2c [2(a + b)] – (a – c) (b – c)}
+ (a – c) [2(a + b) (b – c)]
D = 8c (a + b) (a + b) + 4 (a + b) (a – c) (b – c)
= 4 (a + b) [2ac + 2bc + ab – bc – ac + c2]
= 4 (a + b) [ac + bc + ab + c2]
= 4(a + b) [c(a + c) + b (a + c)]
= 4 (a + b) (b + c) (c + a)
= a (a + b) (b + c) (c + a)
Hence, a = 4
42. (b) Vertices of triangle in complex form is
z, iz, z + iz
In cartesian form vertices are
(x, y), (– y, x) and (x – y, x + y)
=
Apply C2 – C1 and C3 – C1
-2a
b-c
A
B
(– 2a, a)
a+b a+c
)
(2a, a)
41. (c) Let D = b + a -2b b + c
c + a b + c -2c
C (0, 0)
Applying C1 + C3 and C2 + C3
-a + c
D=
2a + b + c a + c
2b + a + c
-b + c
b+c
a-c
b-c
-2c
Now, applying R1 + R3 and R2 + R3
-4 2 1
1
1
4 2 1 = [– 4 (2) – 2(4) + 1(0)]
\ Area =
2
2
0 0 1
=
-16
= -8 » 8 sq. unit (Q area cannot be negative )
2
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EBD_8344
M-296
Mathematics
44. (b)
46. (b) Let r be the common ratio of an G.P., then
a +1
a -1
a a +1 a -1
-b b + 1 b - 1 +
c
c -1 c +1
a
b +1
b -1
c -1
c +1 = 0
(-1) n+ 2 a ( -1)n +1 b (-1)n c
a + 1 a -1
a +1 a -1
Þ -b b + 1 b - 1 + b + 1 b - 1
c c -1 c + 1
c -1 c + 1
( -1)
n +2
( -1)
n+1
b =0
Þ
( -1)n + 2 a
-b b + 1 b - 1 c
c -1 c +1
( -1)
n+ 2
(-1)
a -1 a + 1
( -b) b - 1 b + 1 = 0
n+2
c
c + 1 c -1
Þ
a + 1 a -1
a
Þ
a + 1 a -1
-b b + 1 b - 1 + (-1)n+ 2 -b b + 1 b - 1 = 0
c c -1 c +1
c c -1 c + 1
a
a + 1 a -1
é1 + (-1)n+ 2 ù -b b + 1 b - 1 = 0
ë
û
c c -1 c + 1
C2 – C1, C3 – C1
a
Þ
Þ
Þ
Þ
Þ
Þ
log an +5
log an+ 6
log an+ 7
log an+8
log a1r n -1
log a1r n
log a1r n +1
= log a1r n+ 2
log a1r n +3
log a1r n + 4
log a1r n+ 5
log a1r n+ 6
log a1r n+ 7
log a1 + ( n - 1) log r
log a1 + n log r
log a1 + (n + 1) log r
= log a1 + ( n + 2) log r log a1 + (n + 3) log r log a1 + (n + 4) log r
log a1 + ( n + 5) log r log a1 + (n + 6)log r log a2 + (n + 7) log r
Applying C3 ® C3 + C1, we get
log a1 + ( n - 1) log r log a1 + n log r
2 [log a1 + n log r ]
= log a1 + ( n + 2)log r log a1 + (n + 3) log r 2 [log a1 + ( n + 3) log r ]
log a1 + ( n + 5) log r log a1 + (n + 6) log r 2 [log a1 + (n + 6) log r ]
=0
C2 Û C3
a
log an+ 2
( -1) n c
C1 Û C3
a + 1 a -1
log an+1
log an+ 4
a
(Taking transpose of second determinant)
a
log an
log an+ 3
-1
1
47. (d) Applying, C1 ® C1 + C2 + C3 , we get
1 + (a 2 + b 2 + c 2 + 2) x (1 + b 2 ) x (1 + c 2 ) x
2
2
2
f (x) = 1 + (a + b + c + 2) x
1 + b2 x
(1 + c 2 ) x
1 + (a 2 + b 2 + c 2 + 2) x (1 + b 2 ) x
1 + c2 x
[Q a2 + b2 + c2 = –2]
1 (1 + b2 ) x (1 + c 2 ) x
= 1
1 + b2 x
(1 + c 2 ) x
1 (1 + b2 ) x
1 + c2 x
é1 + (-1)n+ 2 ù -b 2b + 1 2b - 1 = 0 R + R
1
3
ë
û
c
-1
1
Applying, R2 ® R2 - R1 , R3 ® R3 - R1
0
0
a+c
é1 + ( -1) n+ 2 ù -b 2b + 1 2b - 1 = 0
ë
û
-1
c
1
\ f (x) = 0
0
n+2
[1+ (– 1) ](a + c) (2b + 1+ 2b – 1) = 0
4b (a + c) [1 + (–1)n + 2] = 0
1 + (–1)n + 2 = 0 as b (a + c) ¹ 0
n should be an odd integer.
1
1
1
1
45. (d) Given that, D = 1 1 + x
1 1 1+ y
Applying R2 ® R2 – R1 and R3 ® R3 – R1
1 1 1
\ D = 0 x 0 = xy
0 0 y
Hence, D is divisible by both x and y
1 (1 + b2 ) x (1 + c 2 ) x
1- x
0
0
1- x
f (x) = ( x - 1)2
Hence degree = 2.
48. (d) Let r be the common ratio of an G.P., then
log an
log an+1
log an+ 2
log an+ 3
log an+ 4
log an +5
log an+ 6
log an+ 7
log an+8
log a1r n-1
log a1r n
log a1r n+1
= log a1r n + 2
log a1r n+ 3
log a1r n + 4
log a1r n +5
log a1r n+ 6
log a1r n + 7
log a1 + ( n - 1) log r log a1 + n log r
log a1 + (n + 1) log r
= log a1 + ( n + 2) log r log a1 + (n + 3) log r log a1 + (n + 4) log r
log a1 + ( n + 5) log r log a1 + (n + 6)log r log a2 + (n + 7) log r
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M-297
Determinants
Applying C3 ® C3 + C1, we get
5
Now, | B | =
log a1 + n log r
2 [log a1 + n log r ]
= log a1 + ( n + 2)log r log a1 + (n + 3) log r 2 [log a1 + ( n + 3) log r ]
log a1 + ( n + 5) log r log a1 + (n + 6) log r 2 [log a1 + (n + 6) log r ]
log a1 + ( n - 1) log r
49. (c) Given that
b
ax + b
b
c
bx + c
ax + b bx + c
Þ
Þ
0
Applying R3 ® R3 – (xR1 + R2);
bx + c
= b c
0 0 -(ax + 2bx + c)
= (ax2 + 2bx + c)(b2 – ac) = (+)(–) = –ve.
[Given that discriminant of ax2 + 2bx + c is –ve
\ 4b2 – 4ac < 0 Þ b2 – ac < 0]
50. (d) l = ARp –1 Þ log l = log A + (p – 1) log R
m = ARq –1 Þ log m = log A + (q – 1) log R
n = ARr –1 Þ log n = log A + (r – 1) log R
[Q | ad A | = | A |
= 2a2 – 2a – 25
1
2a 2 - 2a - 25
2a 2 - 2a - 24
2a 2 - 2a - 25
+1 = 0
=0
(n - 1)n
= 78 Þ n2 – n – 15 = 0
2
Þ n = 13
Þ
p 1
q 1 =0
r
é1 n ù é1 13ù
Now, the matrix ê
ú=ê
ú
ë 0 1 û ë0 1 û
é1 13ù é1 -13ù
Then, the required inverse of ê
ú =ê
ú
ë 0 1 û ë0 1 û
55. (c) Let |A| = a, |B| = b
1
Þ |AT| = a |A–1| =
| ad A | = | A |2 = 9
n -1
1
-1
é1 1 + 2 + 3 + ... + ( n - 1) ù é1 78 ù
Þê
ú = ê0 1 ú
1
ë0
û ë
û
log l p 1 log A + ( p - 1) log R p 1
Now, log m q 1 = log A + (q - 1) log R q 1
log n r 1 log A + (r - 1) log R r 1
Operating
51. (a)
2
3
é1 1ù é1 2 ù é1 3ù é1 4 ù é1 n - 1ù é1 78 ù
=
ê 0 1ú ê0 1 ú ê0 1 ú ê0 1 ú ... ê 0
1 úû êë 0 1 úû
ë
ûë
ûë
ûë
û ë
2
0
C1 – (log R)C2 + (log R – log A) C3 = 0
0
0
a
Þ a = 4, – 3 Þ Sum of values = 1
54. (b)
ax + b
a b
1
Given, det. (A) + 1 = 0
=0
a
2a
Q
1
1
, |BT| = b, |B–1| =
a
b
|ABAT| = 8 Þ |A| |B| |AT| = 8¼(1)
Þ a.b.a = 8 Þ a2b = 8
]
Þ | A | = ±3 = l Þ | l | = 3
Q
Þ | B | = | ad A |2 = 81
|AB–1| = 8 Þ |A| |B–1| = 8 Þ a .
1
=8
b
¼(2)
From (1) & (2)
-1 T
m = | (B ) | = | B
52. (a)
1
A =1
1
3
-1
1
1
=
| = |B| =
| B | 81
-1
a = 4, b =
2
4 = ((9 + 4) - 1(3 - 4) + 2(-1 - 3))
Then, |BA–1BT| = |B| |A–1| |BT| = b .
1 -1 3
= 13 + 1 – 8 = 6
2
|adjB| = |adj(adjA)| = |A|(n – 1) = |A|4 = (36)2
3
|C| = |3A| = 3 × 6
Hence,
53.
| adjB | 36 ´ 36
= 3
=8
|C |
3 ´6
(c) Q B = A = Þ | B | =
–1
1
| A|
1
2
56.
1
1
b2
.b=
=
a
16
a
écos q - sin q ù
(c) A = ê sin q cos q ú Þ | A| = 1
ë
û
T
é cos q sin q ù
é+ cos q - sin q ù
ad(A) = ê
ú = ê - sin q cos q ú
sin
cos
+
q
+
q
ë
û
ë
û
é cos q sin q ù
Þ A–1 = ê - sin q cos q ú = B
ë
û
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EBD_8344
M-298
Mathematics
é cos q sin q ù é cos q sin q ù
B2 = ê - sin q cos q ú ê- sin q cos q ú
ë
ûë
û
é cos 2q sin 2q ù
= ê- sin 2q cos 2q ú
ë
û
é36 – 32 ù
Þ A2 = ê
4 úû
ë0
é cos3q sin 3q ù
Þ B3 = ê - sin3q cos3q ú
ë
û
For k = – 6
é cos(50q) sin(50q) ù
Þ A–50 = B50 = ê- sin(50q) cos(50q) ú
ë
û
é 3
ê
2
( A-50 ) p ê
q= = ê 1
12
êë 2
1 ù
ú
2 ú
3ú
ú
2 û
é
3ù
pö
p
æ 50 p ö
æ
÷ = cos ç 4 p + ÷ = cos =
êQ cos ç
ú
12
6
6
2
è
ø
è
ø
ë
û
57.
é6 – 4ù
A= ê
ú ... From (1)
ë0 2 û
é1 2ù
(d) Since A . ê
ú is a scalar matrix and |3A| = 108
ë 0 3û
ék 0 ù
suppose the scalar matrix is ê
ú
ë0 k û
é1 2ù é k 0 ù
\A. ê
ú =ê
ú
ë 0 3û ë 0 k û
é k 0 ù é1 2 ù
ÞA= ê
ú ê
ú
ë 0 k û ë0 3 û
–1
1 é k 0 ù é 3 – 2ù
ê
úê
ú
3 ë0 k û ë0 1 û
2ù
é
1 – ú
ék 0 ù ê
3
Þ A=ê
ú
úê
k
0
1
ú
ë
û ê0
3 úû
ëê
58. (a) We have
(A – 3I) (A – 5I) = O
Þ A2 – 8A + 15I = O
Multiplying both sides by A – 1, we get;
A – 1 A . A – 8A – 1 A + 15A – 1 I = A – 1 O
Þ A – 8I + 15A – 1 = O
A + 15A – 1 = 8I
A 15 A -1
+
= 4I
2
2
1 15 16
+ =
=8
2 2
2
é 2 -3ù
59. (c) We have A = ê
ú
ë -4 1 û
\
a+b=
é 48 -27 ù
ê -36 39 ú
ë
û
é 48 -27 ù é 24 -36 ù
\ 3A2 + 12A = ê -36 39 ú + ê -48 12 ú
ë
û ë
û
é 72 -63ù
=ê
ú
ë -84 51 û
é 51 63ù
ad (3A2 + 12A) = ê
ú
ë84 72 û
60. (b)
61. (d) Given that A(ad A) = A AT
2 ù
é
êk – 3 k ú
Þ A=ê
ú
k ú ... (1)
ê0
êë
3 úû
–1
Pre-multiply by A
both side, we get
Þ A A (ad A) = A A A
–1
–1
T
T
ad A = A
Q |3 A| = 108
Þ 108 =
é36 – 32 ù
Þ A2 = ê
4 úû
ë0
é 16 -9 ù
Þ A2 = ê -12 13 ú Þ 3A2 =
ë
û
é 24 -36 ù
Also 12A = ê 48 12 ú
ëû
[ \ AB = C Þ ABB–1 = CB–1 Þ A = CB–1]
Þ A=
é– 6 4 ù
ÞA=ê
ú .... From (1)
ë 0 – 2û
3k
– 2k
0
k
Þ 3k2 = 108 Þ k2 = 36
For k = 6
é 2 b ù é 5a 3 ù
Þê
ú=ê
ú
ë -3 5a û ë- b 2 û
Þk =±6
2
Þ a = and b = 3
5
Þ 5a + b = 5
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M-299
Determinants
62. (a) A2 – 5A = – 7I
AAA–1 – 5AA–1 = –7I A–1
AI – 5I = –7A–1
A – 5I = –7A–1
1
A–1 = (5I – A)
7
A3 – 2A2 – 3A + I = A (5A – 7I) – 2A2 – 3A + I
= 5A2 – 7A – 2A2 – 3A + I = 3A2 – 10A + I
= 3 (5A – 7I) – 10A + I = 5A – 20I = 5(A – 4I)
63. (a) |5. ad A | = 5 Þ 53. |A|3–1 = 5
Þ 125 |A|2 = 5 Þ |A| = ±
1
5
64. (d) BB' = B ( A-1 A ') ' = B ( A ') '( A-1 ) '
= BA (A–1)' = ( A -1 A ')( A( A -1) ')
= A–1A .A'.(A–1)'
{as AA' = A'A}
–1
2
= I(A A)' = I × I = I = I
é0 0 1ù
é1 2 3ù
65. (a) Given A ê0 2 3ú = êê1 0 0úú
ê
ú
êë0 1 0úû
ëê0 1 1ûú
Applying C1 « C3
é3
A ê3
ê
ëê1
Again
é1
2 1ù
ê
ú
2 0 = ê0
ú
êë0
1 0ûú
Applying C2
0 0ù
0 1ú
ú
1 0úû
« C3
é3 1 2ù
é1
ê3 0 2ú
ê0
ú = ê
A ê
êë1 0 1úû
êë0
pre-multiplying both
0 0ù
1 0ú
ú
0 1úû
sides by A–1
é3 1 2ù
A–1 A êê3 0 2úú = A–1
ëê1 0 1úû
é 1 0 0ù
ê 0 1 0ú
ê
ú
êë0 0 1úû
é3 1 2ù
ê3 0 2ú
ú = A–1 I = A–1
I ê
ëê1 0 1úû
(Q A–1A = I and I = Identity matrix)
é3 1 2ù
ê3 0 2ú
ú
Hence, A = ê
ëê1 0 1úû
66. (b) | P | = 1(12 – 12) – a(4 – 6) + 3(4 – 6) = 2a – 6
Now, ad A = P Þ | ad A | = | P |
Þ | A |2 = | P |
Þ | P | = 16
Þ 2a – 6 = 16
Þ a = 11
–1
67. (c) Given that P3 = Q3
and P2Q = Q2P
Subtracting (1) and (2), we get
P3 – P2Q = Q3 – Q2P
Þ P2 (P–Q) + Q2 (P – Q) = 0
Þ (P2 + Q2) (P–Q) = 0
2
2
Q P ¹ Q, \ P + Q = 0
Hence |P2 + Q2| = 0
...(1)
...(2)
æ1ö
æ0ö
ç ÷
ç ÷
68. (d) Let Au1 = ç 0 ÷ and Au2 = ç 1 ÷
ç0÷
ç0÷
è ø
è ø
æ 1ö æ0ö
ç ÷ ç ÷
Then, Au1 + Au2 = ç 0 ÷ + ç 1 ÷
ç0÷ ç0÷
è ø è ø
Þ
æ 1ö
ç ÷
A ( u1 + u2 ) = ç 1 ÷
ç0÷
è ø
...(1)
æ1 0 0ö
ç
÷
A
=
Given that
ç2 1 0÷
ç 3 2 1÷
è
ø
Þ |A| = 1(1) – 0 (2) + 0 (4 – 3) = 1
C11 = 1
C21 = 0 C31 = 0
C12 = –2 C22 = 1 C32 = 0
C13 = 1
C23 = –2 C33 = 1
é 1 0 0ù
\ adjA = êê -2 0 0 úú
êë 1 -2 1 úû
We know,
A-1 =
1
adjA
A
Þ A -1 = adj ( A )
(Q
A = 1)
Now, from equation (1), we have
u1 + u2 = A
æ 1ö
÷
ç 1÷
ç0÷
è ø
-1 ç
é 1 0 0ù æ 1 ö é 1 ù
ç ÷
= êê -2 1 0 úú ç 1 ÷ = êê -1úú
êë 1 -2 1úû èç 0 ø÷ êë -1úû
é0 0
ê
69. (c) A = ê 0 b
êë d e
aù
ú
c ú , |A| = – abd ¹ 0
f úû
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EBD_8344
M-300
Mathematics
c11 = + (bf – ce), c12 = – (– cd) = cd, c13 = + (– bd) = –bd
c21 = –(–ea) = ae, c22 = + (–ad) = –ad, c23 = –(0) = 0
c31 = + (–ab) = – ab, c32 = – (0) = 0, c33 = 0
é (bf - ce) ae
ê
cd
-ad
Ad A = ê
êë -bd
0
A– 1 =
1
1
(ad A) =
| A|
abd
-ab ù
ú
0 ú
0 úû
ébf - ce ae
ê
-ad
ê cd
êë -bd
0
-ab ù
ú
0 ú
0 úû
Now A–1 = AT
ébf - ce ae
1 ê
cd
-ad
-abd ê
êë -bd
0
ébf - ce ae
ê
- ad
Þ ê cd
0
ëê -bd
é0 0 d ù
-ab ù
ê
ú
ú
0 ú = ê0 b e ú
êë a c f úû
0 úû
é 0
- ab ù
ê
ú
0 ú = ê 0
ê
0 ûú
ê -a 2bd
ë
0
-ab 2d
- abcd
- abd 2 ù
ú
-abde ú
ú
- abdf ú
û
\ bf – ce = ae = cd = 0
...(i)
2
2
2
abd = ab, ab d = ad, a bd = bd
...(ii)
abde = abcd = abdf = 0
...(iii)
From (ii),
(abd 2 ). (ab2d). (a2bd) = ab. ad. bd
Þ (abd)4 – (abd)2 = 0
Þ (abd)2 [(abd)2 – 1] = 0
...(iv)
Q abd ¹ 0 , \ abd = ±1
From (iii) and (iv),
e=c=f=0
...(v)
From (i) and (v),
bf = ae = cd = 0
...(vi)
From (iv), (v) and (vi), it is clear that a, b, d can be any
non- ero integer such that abd = ± 1
But it is only possible, if a = b = d = ± 1
Hence, there are 2 choices for each a, b and d.
there fore, there are 2×2×2 choices for a, b and d. Hence
number of required matrices = 2×2×2=(2)3
éa 0ù
70. (a) Let A and B be real matrices such that A = ê
ú
ë 0 bû
é0 g ù
and B = ê
ú
ëd 0û
é 0 gb ù
and BA = ê
ú
ë da 0 û
Statement - 1 :
é 0
g ( a - b)ù
AB - BA = ê
ú
d
b
a
0 û
(
)
ë
AB - BA = ( a - b) gd ¹ 0
2
\ AB – BA is always an invertible matrix.
Hence, statement - 1 is true.
But AB – BA can be identity matrix if g = – d or d = – g
So, statement - 2 is false.
71. (b) For reflexive
é0 0 d ù
ê
ú
T
A = ê0 b e ú
êë a c f úû
Þ
é 0 ag ù
Now, AB = ê
ú
ëbd 0 û
A = P -1 AP is true,
For P = I, which is an invertible matrix.
( A, A) Î R
\ R is reflexive.
For symmetry
As ( A, B) ÎR for matrix P
A = P-1 BP
Þ PAP-1 = B
Þ B = PAP -1
Þ
( )
B = P –1
-1
A (P–1)
\ (B, A) Î R for matrix P -1
\ R is symmetric.
For transitivity
A = P-1 BP
and B = P–1CP
Þ
Þ
Þ
(
)
A = P –1 P -1CP P
( ) CP
A = (P ) C (P )
A = P -1
2 -1
2
2
2
2
\ (A, C) ÎR for matrix P
R
is
transitive.
\
So R is equivalence.
So, statement-1 is true.
We know that if A and B are two invertible matrices of
order n, then
(AB)–1 = B–1 A–1
So, statement-2 is true.
72. (a) We know that if A is square matrix of order n then
adj (adj A) = | A | n–2 A.
= | A |0 A = A
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M-301
Determinants
Also | adj A | = | A | n–1 = | A | 2–1 = | A |
\ Both the statements are true but statement-2 is not a
correct explanation for statement-1 .
73. (c) Given that all entries of square matrix A are integers,
therefore all cofactors should also be integers.
If det A = ± 1 then A–1 exists. Also all entries of A–1 are
integers.
74. (d) Given that A2 – A + I = 0
Pre-multiply by A–1 both side, we get
A-1 A2 - A-1 A + A-1.I = A-1.0
-1
Þ A - I + A-1 = 0 or A = I - A .
é 4 2 2ù
75. (a) Given that 10 B = ê -5 0 a ú
ê
ú
êë 1 -2 3 úû
é 4 2 2ù
1
Þ B = ê -5 0 a ú
ú
10 ê
ëê 1 -2 3 úû
Given that B = A-1 Þ AB = I
Þ
é 1 -1 1 ù é 4 2 2 ù é 1 0 0 ù
1 ê
úê
ú ê
ú
2 1 -3ú ê -5 0 a ú = ê0 1 0 ú
10 ê
êë 1 1 1 úû êë 1 -2 3 úû êë0 0 1 úû
é10 0 5 - 2 ù é1 0 0ù
1 ê
0 10 -5 + a úú = êê0 1 0úú
Þ
10 ê
êë 0 0 5 + a úû êë0 0 1 úû
Þ
5- a
=0Þa=5
10
é 0 0 -1ù
ê
ú
76. (a) Given that A = ê 0 -1 0 ú
êë -1 0 0 úû
clearly A ¹ 0. Also |A| = -1 ¹ 0
é -1 0 0 ù
ê 0 -1 0 ú ¹ A
-1
I
=
(
1)
\ A exists, further
ê
ú
ëê 0 0 -1ûú
é 0 0 -1ù é 0 0 -1ù
Also A = ê 0 -1 0 ú ê 0 -1 0 ú
ê
úê
ú
êë -1 0 0 úû êë -1 0 0 úû
2
é1 0 0ù
= ê0 1 0ú = I
ê
ú
êë0 0 1úû
1 1 1
77. (c)
D= 1 2 3 =0Þl =5
1 3 l
2 1 1
Dx = 5 2 3 = 0 Þ m = 8
m 3 5
78. (3.00)
For non- ero solution, D = 0
l - 1 3l + 1
Þ l - 1 4l - 2
2
2l
l+3
=0
3l + 1 3(l - 1)
Þ 6l 3 - 36l 2 + 54l = 0
Þ 6l[l 2 - 6l + 9] = 0
Þ l = 0, l = 3
[Distinct values]
Then, the sum of distinct values of l = 0 + 3 = 3.
2 -4 l
79. (a) Q 1 -6 1 = 0 Þ 3l 2 - 7l - 12 = 0
l -10 4
2
Þ l = 3 or 3
1 -4 l
D1 = 2 -6 1 = 2(3 - l )
3 -10 4
2
\ When l = - , D1 ¹ 0.
3
2
Hence, equations will be inconsistent when l = - .
3
80. (a) Since, system of linear equations has non- ero
solution
\D = 0
1 1 3
Þ 1 3 k2 = 0
3 1
3
Þ 1(9 - k 2 ) - 1(3 - 3k 2 ) + 3(1 - 9) = 0
Þ 9 - k 2 - 3 + 3k 2 - 24 = 0
Þ 2k 2 = 18 Þ k 2 = 9, k = ±3
So, equations are
x + y + 3z = 0
x + 3y + 9z = 0
3x + y + 3z = 0
Now, from equation (i) – (ii),
y
-2 y - 6 z = 0 Þ y = -3z Þ = -3
z
Now, from equation (i) – (iii),
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...(i)
...(ii)
...(iii)
...(iv)
EBD_8344
M-302
Mathematics
-2 x = 0 Þ x = 0
Put z = 0 in equation (i), we get x = 2y
Q15 £ x2 + y 2 + z 2 £ 150
y
= 0 - 3 = -3
z
So, x +
Þ 15 £ 4 y 2 + y 2 £ 150
81. (5.00)
For infinitely many solutions,
[Q x = 2 y, z = 0]
D = D1 = D 2 = D 3 = 0
Þ 3 £ y 2 £ 30
1 -2 3
D= 2
Þ y = ±2, ± 3, ± 4, ± 5
Þ 8 solutions.
1 =0
1
1 -7 a
2 -1 2
Þ (a + 7) - 2(1 - 2a) + 3(-15) = 0
85. (d)
D = 1 -2 l = -(l - 1)(2l + 1)
Þ a =8
1
1 -2
D3 = 2
9
D1 = -4 -2 l = -2(l 2 + 6l - 4)
1 -7 24
Þ (24 + 7b) - 2(b - 48) + 9(-15) = 0
Þ b =3
82.
\ a - b = 5.
(b) Given that Ax = b has solutions x1, x2, x3 and b is
equal to b1, b2 and b3
\ x1 + y1 + z1 = 1
Þ 2 y1 + z1 = 2 Þ z1 = 2
Determinant of coefficient matrix
1
2 4 -1 = 0
[Q Equation has many solutions]
l
But given that x2 + y 2 + z 2 = 1
7
\k = ±
9
Þ -15 + 6 + 2l = 0 Þ l =
2
1 1
\ DZ = 2 4
86. (d) Q | P | = 1( -3 + 36) - 2(2 + 4) + 1( -18 - 3) = 0
Given that PX = 0
\ System of equations
Let z = k ÎR and solve above equations, we get
11k
2k
x= , y=
,z=k
7
7
0 0 1
3 2
1
and D1 ¹ 0
2
1ü
ì
Hence, S = í1, - ý
2þ
î
\ D = 0 Þ l = 1, -
and x + 9 y - z = 0 has infinitely many solution.
| A| = 0 2 1 = 2
1 1
4 l 1
For no solution D = 0 and at least one of D1, D2 and D3 is
non- ero.
x + 2 y + z = 0 ; 2x - 3 y + 4z = 0
1 1 1
83. (d)
1
-1 2
2
b =0
1
l
2
6 =0 Þ m = 5
3 2 2m
\ 2l + m = 14.
84. (8)
The given system of equations
x - 2 y + 5z = 0
...(i)
-2 x + 4 y + z = 0
...(ii)
-7 x + 14 y + 9 z = 0
...(iii)
From equation, 2 × (i) + (ii) Þ z = 0
174
\ Two solutions only.
87. (c) The given system of linear equations
7x + 6y – 2z = 0
3x + 4y + 2z = 0
x – 2y – 6z = 0
Now, determinant of coefficient matrix
7
6
-2
D=3
4
2
1 -2 -6
= 7(– 20) – 6(–20) – 2(–10)
= – 140 + 120 + 20 = 0
So, there are infinite non-trivial solutions.
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...(i)
...(ii)
...(iii)
M-303
Determinants
From eqn. (i) + 3 × (iii); we get
10x – 20z = 0 Þ x = 2z
Hence, there are infinitely many solutions (x, y, z) satisfying
x = 2z.
88. (a) From the given linear equation, we get
1 2 3
D = 3 4 5 ( R3 ® R3 - 2 R2 + 3R3 )
4 4 4
1 2 3
= 3 4 5 =0
0 0 0
Now, let P3 = 4x + 4y + 4z – d = 0. If the system has solutions
it will have infinite solution.
So, P3 º aP1 + bP2
Hence, 3a + b = 4 and 4a + 2b = 4
Þ a = 2 and b = – 2
So, for infinite solution 2m – 2 = d
Þ For 2m ¹ d + 2 system is inconsistent
l
89. (c)
2 2
D = 2l 3 5
l 6
4
D = l2 + 6l – 16
D = (l + 8) (2 – l)
For no solutions, D = 0
Þ l = – 8, 2
when l = 2
5
2 2
D1 = 8
3 5
Þ
1 1 1
, , in A.P.
a b c
91. (13) x + y + z = 6
...(i)
x + 2y + 3z = 10
...(ii)
3x + 2y + lz = µ
...(iii)
From (i) and (ii),
If z = 0 Þ x + y = 6 and x + 2y = 10
Þ y = 4, x = 2
(2, 4, 0)
If y = 0 Þ x + z = 6 and x + 3z = 10
Þ z = 2 and x = 4
(4, 0, 2)
So, 3x + 2y + lz = m, must pass through (2, 4, 0) and (4, 0, 2)
So, 6 + 8 = m Þ m = 14
and 12 + 2l = m
12 + 2l = 14 Þ l = 1
So, m – l2 = 14 –1 = 13
92. (d) Given system of linear equations: x + y + z = 5;
Þ
x + 2y + 2z = 6 and x + 3y + lz = m have infinite solution.
\ D = 0, Dx = Dy = Dz = 0
1 1 1
Now, D =
Dy =
There exist no solutions for l = 2
90. (a) For non- ero solution
1 6 2
1 m 3
=0Þ
1 1
b =0
2 4c
c
1
5
1
0
1
1
0 m-5 2
Here,
1
D = 4 l -l = 0
-4
3 2
C1 ® C1 – C2 & C2 ® C2 – C3
1 2a a
1 3b b = 0
c
Þ (3bc – 4bc) – (2ac – 4ac) + (2ab – 3ab) = 0
Þ – bc + 2ac – ab = 0
Þ ab + bc + 2ac
=0
Þ 1 (2 – m + 5) = 0 Þ m = 7
\ l + m = 10
93. (d) Q system of equations has infinitely many solutions.
\ D = D1 = D2 = D3 = 0
2 2a a
2 3b
=0
1 5 1
= 5[18 – 10] – 2 [48 – 50] + 2 (16 – 30]
= 40 + 4 – 28 ¹ 0
1 4c
1 2 2
1 3 l
Þ 1 (2l – 6) – 1 (l – 2) + 1 (3 – 2) = 0
Þ 2l – 6 – l + 2 + 1 = 0 Þ l = 3
10 2 6
Þ
2 1 1
= +
b a c
0
0
1
Þ 4 - l 2l -l = 0
1
6
-4
Þl=3
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EBD_8344
M-304
Mathematics
6
Now, for l = 3,
1
1
D1 = l - 2 l -l = 0
-5
1
6
2
= 1 (– k + 1) + 2 (– 2k – 3) + k (– 2 – 3)
= – k + 1 – 4k – 6 – 5k = – 10k – 5 = – 5(2k + 1)
-4
D1 =
1
3
-5
1 1
-4
D2 =
1 1
6
For l = 3, D3 = 4 l l - 2 = 0
3 2 -5
2
3
-1
Þ D =0Þ 1
k
-2 = 0 Þ k =
2 -1
\ equations are
1
9
2
2x + 3y – z = 0
2x – y + z = 0
2x + 9y – 4z = 0
...(i)
...(ii)
...(iii)
By (i) – (ii), 2y = z
\ z = – 4x and 2x + y = 0
-1 1
9 1
x y z
+ + +k =
+ -4+ =
y z x
2 2
2 2
95. (b) If the system of equations has non-trivial solutions,
then the determinant of coefficient matrix is ero
c
-1
Þ
Þ
Þ
Þ
c
3 3 -k
-2 k
1
3 -1 -k
-2 1
2 1
= 0, D3 =
2
3 -1 3
=0
10 - 3l
2l
and y = –
10
5
\ (x, y) must lie on line 4x – 3y – 4 = 0
97. (a) Q The system of linear equations has a unique solution.
\ D¹0
Let z = l ¹ 0 then x =
1+ a
D= a
a
b
1
1+ b 1 ¹ 0
b
2
1+ a + b +1
Þ
b
1
a +1+ b +1 b + 1 1 ¹ 0
a+b+2
b
2
b
1
[C1 ® C1 + C2 + C3]
(a + b + 2) 1 b + 1 1 ¹ 0
1
b
2
1 b 1
Þ
(a + b + 2) 0 1 0 ¹ 0
0 0 1
R2 ® R2 - R1
R3 ® R3 - R1
Þ (a + b + 2) 1(1) ¹ 0
Þ a+b+2¹0
1
Hence, the greatest value of c is for which the system
2
of linear equations has non-trivial solution.
96. (b) Given system of linear equations,
x – 2y + kz = 1, 2x + y + z = 2, 3x – y – kz = 3
D=
1
1
2
\ System of equation has infinite many solutions.
Þ
=0
1
Þ c = – 1 or
2
2 1
k
2 2 1
1
(1 – c2) + c (– c – c2) – c (c2 + c) = 0
(1 + c) (1 – c) – 2c2 (1 + c) = 0
(1 + c) (1 – c – 2c2) = 0
(1 + c)2 (1 – 2c) = 0
1
= – 5 (2k + 1)
Þ –5 (2k + 1) = 0 Þ k = -
1 -c -c
c
1
Qz ¹ 0 Þ D = 0
\ for l = 3, system of equations has infinitely many
solutions.
94. (b) Given system of equations has a non-trivial solution.
c -1
2 1
3 -1 -k
4 l - 2 -l = 0
For l = 3, D2 =
\
-2 k
1
Q
Ordered pair (2, 4) satisfies this condition
\
a = 2 and b = 4.
98. (a) Consider the given system of linear equations
x(1 –l) – 2y – 2z = 0
x + (2 – l)y + z = 0
–x – y – lz= 0
Now, for a non-trivial solution, the determinant of coefficient matrix is ero.
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M-305
Determinants
1- l
-2
-2
1 5
1
1
2- l
1
1 9
3
-1
-1
-l
D2 =
=0
Þ (1 – l)3 = 0
l=1
99. (b) Q System of equations has more than one solution
\ D = D1 = D2 = D3 = 0 for infinite solution
2
a
D1 =
3
b -1 5
c -3 2
= a(13) + 2(5c – 2b) + 3(–3b + c)
= 13a – 13b + 13c = 0
b–c–a=0
100. (b) Since, the system of linear equations has, non-trivial
solution then determinant of coefficient matrix = 0
sin 3q cos 2q 2
i.e.,
= 9a - 3b - 5 (a – 3) + 1(b - 9)
= 9a - 3b - 5a + 15 + b - 9 = 4a - 2b + 6
1 1 5
D3 =
1 2 9
1 3 b
= 2b – 27 – b + 9 + 5 = b – 13
Since, the system of equations has infinite many solutions.
Hence,
D1 = D 2 = D 3 = D = 0
Þ a = 5, b = 13 Þ b – a = 8
102. (c) Consider the system of linear equations
i.e, a – b + c = 0
or
1 b a
1
3
7
-1
4
7
=0
sin3q(21– 28) – cos2q(7 + 7) + 2 (4 + 3) = 0
sin3q + 2cos2q – 2 = 0
x – 4y + 7z = g
...(i)
3y – 5z = h
...(ii)
–2x + 5y – 9z = k
...(iii)
Multiply equation (i) by 2 and add equation (i), equation
(ii) and equation (iii)
Þ 0 = 2g + h + k. \ 2g + h + k = 0
then system of equation is consistent.
103. (a) For non ero solution of the system of linear
equations;
1 k
3
4sin3q + 4sin 2q – 3sinq = 0
3 k
-2 = 0
sinq (4sin2q + 4sinq – 3) = 0
2 4
-3
3sinq – 4sin3q + 2 – 4sin 2q – 2 = 0
sinq (4sin2q + 6sinq – 2sinq – 3) = 0
sinq [2sinq (2sinq – 1) + 3 (2sinq – 1)] = 0
sinq (2sinq – 1) (2sinq + 3) = 0
sin q = 0, sin q =
1
2
3ö
æ
çèQ sin q ¹ - ÷ø
2
p 5p
,
6 6
Hence, for two values of q, system of equations has nontrivial solution
q=
1 1
101. (b) D = 1 2
1
3 = 2a – 9 – a + 3 + 1 = a – 5
1 3 a
5 1
1
D1 = 9 2 3 = 5(2a – 9) – 1(9a – 3b) + (27 – 2b)
b 3 a
= 10a - 45 - 9a + 3b + 27 - 2b
= a + b -18
Þ k = 11
Now equations become
x + 11y + 3 = 0
3x + 11y – 2 = 0
2x + 4y – 3 = 0
Adding equations (1) & (3) we get
3x + 15y = 0
Þ x = –5y
Now put x = –5y in equation (1), we get
–5y + 11y + 3 = 0
Þ = –2y
\
x
( -5 y)( -2 y)
= 10
y2
104. (c) Here, the equations are;
(k + 2) x + 10y = k
& kx + (k + 3)y = k – 1.
These equations can be written in the form of Ax = B as
y
2
=
...(1)
...(2)
...(3)
é k + 2 10 ù é x ù é k ù
=
ê k
k + 3úû êë y úû êë k – 1úû
ë
For the system to have no solution
|A| = 0
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EBD_8344
M-306
Mathematics
é k + 2 10 ù
= 0 Þ (k + 2) (k + 3) – k × 10 = 0
Þ ê
k + 3úû
ë k
Þ k2 – 5k + 6 = (k – 2) (k – 3) = 0
\ k = 2, 3
For k = 2, equations become:
4x + 10 y = 2
& 2x + 5y = 1
& hence infinite number of solutions.
For k = 3, equations becomes;
5x + 10y = 3
3x + 6y = 2
& hence no solution.
\ required number of values of k is 1
105. (b) The system of linear equations is:
x+y+z=2
2x + y – z = 3
3x + 2y + kz = 4
As, system has unique solution.
1 1 1
So, 2 1 –1 ¹ 0
3 2 k
Þ k + 2 – (2k + 3) + 1 ¹ 0
Þk¹0
Hence, k Î R – {0} º S
106. (d) As the system of equations has no solution then D
should be ero and at least one of D1, D2 and D3 should
not be ero.
1 a 1
\
D= 1 2 2 =0
2
\
-l
4
4 l
2 =0
l
2
2
Þ l + 4l – 40 = 0
l has only 1 real root.
109. (b) For non-trivial solution,
3
l
1
-1
l -1 -1 = 0
1
-l
1
Þ -l(l + 1)(l - 1) = 0 Þ l = 0, +1, –1
2x1 - 2x 2 + x 3 = lx1ü
ï
2x1 - 3x 2 + 2x 3 = lx 2 ý
- x1 + 2x 2 = lx3 ïþ
110. (a)
Þ (2 – l)x1 – 2x2 + x3 = 0
2x1 – (3 + l) x2 + 2x3 = 0
– x1 + 2x2 – lx3 = 0
For non-trivial solution,
D=0
2-l
-2
2
-(3 + l )
-1
2
i.e.
1
2 =0
-l
Þ (2 – l) [l(3 + l) – 4] + 2[–2l + 2] + 1[4 – (3 + l)] = 0
Þ l3 + l2 – 5l + 3 = 0
Þ l = 1, 1, 3
Hence l has 2 values.
111. (b) Given system of equations can be written as
(a - 1) x - y - z = 0
1 5 3
Þ –a–1=0 Þ a=–1
- x + (b - 1) y - z = 0
- x - y + (c - 1) z = 0
For non-trivial solution, we have
1 3 1
D2 = 1 6 2 ¹ 0
1 b 3
Þ b¹9
1 1 1
107. (a) D = 1 a 1 = 0
a b 1
a -1
-1
-1
-1
b -1
-1
-1
-1
c -1
=0
R2 ® R2 – R3
a - 1 -1
Þ 1 [a – b] – 1 [1 – a] + 1 [b – a ] = 0
Þ (a – 1)2 = 0
Þa=1
For a = 1, First two equations are identical
i.e., x + y + z = 1
To have no solution with x + by + z = 0
b=1
So b = {1} Þ It is singleton set.
108. (b) Since the given system of linear equations has
infinitely many solutions.
2
-1
-c
0
b
-1
-1 c - 1
=0
C2 ® C2 – C3
a -1
0
-1
0
b+c
-c
-1
-c
c -1
=0
Apply R3 ® R3 – R1
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M-307
Determinants
a -1
0
-a
0
-1
b + c -c
-c
=0
c
Þ (a - 1)[bc + c 2 - c 2 ] - 1[a(b + c)] = 0
Þ (a - 1)[bc ] - ab - ac = 0
Þ abc - bc - ab - ac = 0
Þ ab + bc + ca = abc
112. (b) Since, system of equations have no solution
k +1
8
4k
\
=
¹
(Q System has no solution)
k
k + 3 3k - 1
Þ k2 + 4k + 3 = 8k Þ k2– 4k + 3 = 0
Þ k = 1, 3
If k = 1 then
8
4.1
¹
which is false
1+ 3
2
8 4.3
and if k = 3 then ¹
which is true, therefore k = 3
6 9 -1
Hence for only one value of k. System has no solution.
113. (b) Given system of equations is homogeneous which is
x + ay = 0
y + az = 0
z + ax = 0
It can be written in matrix form as
æ1 a 0ö
A =ç0 1 a÷
ça 0 1÷
è
ø
Now, | A | = [1 – a(– a2)] = 1 + a3 ¹ 0
So, system has only trivial solution.
Now, | A | = 0 only when a = – 1
So, system of equations has infinitely many solutions
which is not possible because it is given that system has a
unique solution.
Hence set of all real values of ‘a’ is R – {– 1}.
114. (c)
p æ pö
in ç 0, ÷
4 è 2ø
\ D1 = 2(sin a) × 0 = 0,
a=
1 sin a cos a
D1 = 1 cos a sin a
1 - sin a cos a
0 sin a - cos a cos a - sin a
= 0 cos a + sin a sin a - cos a
1
- sin a
cos a
= (sin a – cos a)2 – (cos2 a – sin2 a)
= sin2 a + cos2 a – 2 sin a . cos a – cos2 a + sin2 a
= 2 sin2 a – 2 sin a . cos a
= 2 sin a (sin a – cos a)
Now, sin a – cos a = 0 for only
æ pö
since value of sin a is finite for a Î ç 0, ÷
è 2ø
Hence non-trivivial solution for only one value of a in
æ pö
ç 0, ÷
è 2ø
cos a sin a
cos a
sin a cos a
sin a = 0
cos a - sin a - cos a
Þ
0
sin a
cos a
0
cos a
sin a = 0
2cos a - sin a - cos a
Þ 2 cos a (sin2 a – cos2 a) = 0
\ cos a = 0 or sin2 a – cos2 a = 0
æ pö
But cos a = 0 not possible for any value of a Î ç 0, ÷
è 2ø
\ sin2 a – cos2 a = 0 Þ sin a = – cos a, which is also not
æ pö
possible for any value of a Î ç 0, ÷
è 2ø
Hence, there is no solution.
115. (d) Given system of equations can be written in matrix
form as AX = B where
æ1 2 3ö
æ6ö
A = ç 1 3 5 ÷ and B = ç 9 ÷
ç2 5 a÷
çb÷
è
ø
è ø
Since, system is consistent and has infinitely many
solutions
\ (ad. A) B = 0
æ 3a - 25 15 - 2a 1 ö æ 6 ö æ 0 ö
a - 6 -2 ÷ ç 9 ÷ = ç 0 ÷
Þ ç 10 - a
ç -1
1 ÷ø çè b ÷ø çè 0 ÷ø
-1
è
Þ – 6 – 9 + b = 0 Þ b = 15
and 6(10 – a) + 9(a – 6) – 2(b) = 0
Þ 60 – 6a + 9a – 54 – 30 = 0
Þ 3a = 24 Þ a = 8
Hence, a = 8, b = 15.
116. (a) Given system of equations is
x + ky + 3z = 0
3x + ky – 2z = 0
2x + 3y – 4z = 0
Since, system has non-trivial solution
\
1 k
3
3 k
-2 = 0
2 3 -4
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EBD_8344
M-308
Mathematics
Þ 1 (– 4k + 6) – k(– 12 + 4) + 3 (9 – 2k) = 0
Þ 4k + 33 – 6k = 0 Þ k =
33
2
Hence, statement - 1 is false.
Statement-2 is the property.
It is a true statement.
117. (d) Given system of equations is
x+y+z=6
x + 2y + 3z = 10
x + 2y + lz = 0
It has unique solution.
\
1 1 1
1 2 3 ¹0
1 2 l
Þ 1(2l – 6) – 1 (l – 3) + 1(2 – 2) ¹ 0
Þ 2l – 6 – l + 3 ¹ 0 Þ l – 3 ¹ 0 Þ l ¹ 3
118. (a)
x - ky + z = 0
kx + 3 y - kz = 0
3x + y - z = 0
The given that system of equations have trivial solution,
1 -k 1
\
k 3 -k ¹ 0
3 1 -1
Þ 1(-3 + k ) + k (-k + 3k ) + 1(k - 9) ¹ 0
Þ k - 3 + 2k 2 + k - 9 ¹ 0
Þ k 2 + k - 6 ¹ 0 Þ k = -3, k ¹ 2
So, the equation will have only trivial solution,
when k Î R – {2, – 3}
119. (a) Given that system of equations have non- ero solution
D=0
4 k 2
Þ
k 4 1 =0
2 2 1
Þ 4(4 - 2) - k (k - 2) + 2(2k - 8) = 0
Þ
8 - k 2 + 2k + 4k - 16 = 0
k 2 - 6k + 8 = 0
Þ (k - 4)(k - 2) = 0 Þ k = 4, 2
120. (c)
1 2 1
D= 2 3 1 =0
3 5 2
3 2 1
Dx = 3 3 1 ¹ 0
1 5 2
Þ Given system, does not have any solution.
Þ No solution
121. (d) The given equations are
–x + cy + bz = 0
cx –y + az = 0
bx + ay – z = 0
Given that x, y, z are not all ero
\ The above system have non- ero solution
–1 c b
Þ D = 0 Þ c –1 a = 0
b a –1
Þ –1(1– a2) – c(– c – ab) + b(ac + b) = 0
Þ –1 + a2 + b2 + c2 + 2abc = 0
Þ a2 + b2 + c2 + 2abc = 1
122. (a) ax + y + = a – 1;
x + a y + = a – 1;
x+ y+ a = a –1
a 1
1
D= 1 a
1
1
1
a
2
= a( a - 1) - 1(a - 1) + 1(1 - a)
= a (a - 1)(a + 1) - 1(a - 1) - 1(a - 1)
= (a - 1)[a 2 + a - 1 - 1]
2
= (a - 1)[a + a - 2]
= (a – 1) [a 2 + 2a - a - 2]
= (a - 1)[a (a + 2) - 1(a + 2)]
= ( a - 1) ( a + 2 )
Q Equations has infinite solutions
\ D=0
Þ (a - 1) = 0, a + 2 = 0
Þ a = – 2, 1;
But a ¹ 1 .
\a=–2
123. (d) For homogeneous system of equations to have non
ero solution, D = 0
2
1 2a a
1 3b b = 0 C ®
-
-
1 4c c
Applying C2 ® C2 – 2C3
1
Þ 1
0
b
a
b = 0 R3 ® R3 - R1, R2 ® R2 - R1
1 2c c
1
0
Þ 0
b
a
b-a =0
0 2c c - a
Þ bc – ab = 2bc – 2ac
2 1 1
Þ = +
b a c
\ a, b, c are in Harmonic Progression.
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M-309
Continuity and Differentiability
20
Continuity and
Differentiability
5.
TOPIC Ć Continuity
1.
2.
ì 2 cos x – 1
p
,x ¹
ïï
cot x – 1
4
í
f(x) = ï
p
k,
x=
ïî
4
is continuous, then k is equal to:
éxù
Let f (x) = x. ê ú , for –10 < x < 10, where [t] denotes the
ë 2û
greatest integer function. Then the number of points of
discontinuity of f is equal to ______.
[NA Sep. 05, 2020 (I)]
If a function f (x) defined by
ìae x + be - x , - 1 £ x < 1
ïï
f ( x) = í cx 2
, 1 £ x £ 3 be continuous for some
ï 2
ïî ax + 2cx , 3 < x £ 4
6.
(c)
3.
4.
e 2 - 3e + 13
e
e 2 + 3e + 13
(b)
(d)
f(x) and xlim
f(x) exist but are not equal.
(c) Both xlim
®4®4+
e
(d) xlim
f(x) exists but xlim
f(x) does not exist.
®4®4+
e 2 - 3e - 13
e
7.
(c)
If the function
{
a | p - x | +1, x £ 5
f(x) = b | x - p | +3, x > 5
e 2 - 3e + 13
é4ù
x ú = A.
Let [t] denote the greatest integer £ t and xlim
®0 ê
ëxû
Then the function, f(x) = [x2] sin(px) is discontinuous, when
x is equal to :
[Jan. 9, 2020 (II)]
(a)
1
(b) 1
(c) 1
(d)
2
2
éxù
If f ( x) = [ x] - ê ú , x Î R, where [x] denotes the greatest
ë4û
integer function, then:
[April 09, 2019 (II)]
(a) f is continuous at x = 4.
f(x) exists but xlim
f(x) does not exist.
(b) xlim
®4+
®4-
[Sep. 02, 2020 (I)]
(a)
A +1
(b)
A+5
A + 21
(d)
A
[April 09, 2019 (I)]
(a) 2
a, b, c Î R and f '(0) + f '(2) = e, then the value of a is :
1
æ p pö
If the function f defined on ç , ÷ by
è6 3ø
is continuous at x = 5, then the value of a – b is:
[April 09, 2019 (II)]
2
-2
2
(b)
(c)
p+5
p+5
p -5
Let f : [– 1, 3] ® R be defined as
(a)
8.
(d)
2
5-p
æ 1 1ö
If the function f defined on ç - , ÷ by
è 3 3ø
ì x + [ x ] , - 1£ x < 1
ï
f(x) = í x + x , 1 £ x < 2
ï x + [ x ] , 2 £ x £ 3,
î
ì1
æ 1 + 3x ö
, when x ¹ 0
ï log
f(x) = í x e çè 1 - 2 x ø÷
is continuous, then k
, when x = 0
ïk
î
is equal to __________.
[NA Jan. 7, 2020 (II)]
where [t] denotes the greatest integer less than or equal
to t. Then, f is discontinuous at : [April 08, 2019 (II)]
(a) only one point
(b) only two points
(c) only three points
(d) four or more points
Downloaded from @Freebooksforjeeneet
EBD_8344
M-310
9.
Mathematics
Let f : R ® R be a function defined as
14.
if
x £1
ì 5,
ï a + bx, if 1 < x < 3
ï
f ( x) = í
ï b + 5 x, if 3 £ x < 5
ïî 30,
x³5
if
Then, f is :
[Jan 09, 2019 (I)]
(a) continuous if a = 5 and b = 5
(b) continuous if a = – 5 and b = 10
(c) continous if a = 0 and b = 5
(d) not continuous for any values of a and b
10. If the function f defined as
f (x) =
15.
k -1
1
x e2 x - 1
11.
(b) (3, 2)
æ1 ö
(c) ç , 2 ÷
è3 ø
is continuous at x = p, then k equals:
[Online April 19, 2014]
(d) (2, 1)
1
ì
ï
2- x
Let f (x) = í ( x - 1) , x > 1, x ¹ 2
ïî
k,
x=2
(a) 0
16.
12.
e –1
(a)
(b) e
(c)
The value of k for which the function
13.
(b)
2
5
(c)
3
5
(d) -
17.
2
5
, 0 £ x <1
, 1£ x < 2
,
2£x<¥
is continuous in the interval [0, ¥) , then an ordered pair
(a, b) is :
[Online April 10, 2016]
(a) ( - 2,1 - 3)
(b) ( 2, -1 + 3)
(c) ( 2,1 - 3)
(d) ( - 2,1 + 3)
continuous
and
(d)
1
4
æ9ö 2
fç ÷=
,
è2ø 9
then
[Online April 9, 2014]
8
(c) 0
(d)
9
æ - x, -1 £ x < 0
ç
1 - x, 0 £ x < 1
Statement 2 : f (x) = ç
ç 1 + x, 1 £ x < 2
ç
è 2 + x, 2 £ x £ 3
Let a, b Î R, (a ¹ 0) . if the function f defined as
ì 2x 2
ï
ï a
ï
ía
f(x) = ï 2
ï 2b - 4b
ïî x 3
(c) 2
2
9
(b)
9
2
Consider the function :
f (x) = [ x] + | 1 – x |, -1 £ x £ 3 where [x] is the greatest
integer function.
Statement 1 : f is not continuous at x = 0, 1, 2 and 3.
(a)
[Online April 9, 2017]
17
20
is
1
2
æ 1 - cos 3x ö
lim f ç
÷ is equal to:
è
ø
x2
(d) 1
tan 4x
ì
ï æç 4 ö÷ tan 5x , 0 < x < p
p
ï
2
f (x) = í è 5 ø2
p is continuous at x = , is :
2
, x =
ï k+ 5
2
ïî
(a)
If f(x)
(b)
x ®0
The value of k for which f is continuous at x = 2 is
[Online April 15, 2018]
e –2
ì
ï
(e x - 1)
ï
, x¹0
ïï æ x ö
æ xö
sin
log ç1 + ÷
If f(x) = í çè k ÷ø
è 4ø
ï
, x =0
12
ï
ï
ïî
is a continuous function then the value of k is:
(a) 4
(b) 1
(c) 3
(d) 2
If the function
ì 2 + cos x - 1
,x ¹ p
ï
f ( x ) = í ( p - x)2
ï
,x = p
îk
x ¹ 0, is continuous at x = 0,
then the ordered pair (k, f (0)) is equal to?
[Online April 16, 2018]
(a) (3, 1)
Let k be a non– ero real number.
[Online April 11, 2015]
18.
[Online April 25, 2013]
(a) Statement 1 is true ; Statement 2 is false,
(b) Statement 1 is true; Statement 2 is true; Statement 2 is
not correct explanation for Statement 1.
(c) Statement 1 is true; Statement 2 is true; Statement It is
a correct explanation for Statement 1.
(d) Statement 1 is false; Statement 2 is true.
Let f be a composite function of x defined by
1
1
, u ( x) =
.
x -1
u2 + u - 2
Then the number of points x where f is discontinuous is :
[Online April 23, 2013]
(a) 4
(b) 3
(c) 2
(d) 1
f (u ) =
Downloaded from @Freebooksforjeeneet
M-311
Continuity and Differentiability
19.
20.
21.
Let f (x) = – 1 + | x – 2 |, and g (x) = 1 – | x |; then the set of
all points where fog is discontinuous is :
[Online April 22, 2013]
(a) {0, 2}
(b) {0, 1, 2}
(c) {0}
(d) an empty set
If f : R ® R is a function defined by f (x) = [x]
æ 2x - 1 ö
cos ç
÷ p , where [x] denotes the greatest integer
è 2 ø
function, then f is .
[2012]
(a) continuous for every real x.
(b) discontinuous only at x = 0
(c) discontinuous only at non- ero integral values of x.
(d) continuous only at x = 0.
Let f : [1, 3] ® R be a function satisfying
Statement - 1 : f (x) is continuous on R.
Statement - 2 : f1 ( x) and f 2 ( x) are continuous on R.
24.
ì
ï sin( p + 1) x + sin x , x < 0
x
ï
ï
f ( x) = í q
, x = 0 is continuous for all x in R,
ï
2
ï x+ x - x
,x > 0
ïî
x3 / 2
x
£ f ( x) £ 6 - x , for all x ¹ 2 and f (2) = 1,
[ x]
are
where R is the set of all real numbers and [x] denotes the
largest integer less than or equal to x.
Statement 1: lim f ( x ) exists. [Online May 19, 2012]
x® 2 -
22.
Statement 2: f is continuous at x = 2.
(a) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation for Statement 1.
(b) Statement 1 is false, Statement 2 is true.
(c) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation for Statement 1.
(d) Statement 1 is true, Statement 2 is false.
Statement 1: A function f : R ® R is continuous at x0 if and
25.
Statement 2: A function f : R ® R is discontinuous at x0 if
x ® x0
23.
x ® x0
26.
[Online May 12, 2012]
(a) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation of Statement 1.
(b) Statement 1 is false, Statement 2 is true.
(c) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation of Statement 1.
(d) Statement 1 is true, Statement 2 is false.
Define f (x) as the product of two real functions
[2011RS]
ì 1
ïsin , if x ¹ 0
f1 ( x ) = x, x Î R, and f 2 ( x ) = í x
ï0,
if x = 0
î
as follows :
ïì f1 ( x ) . f 2 ( x) , if x = 0
f ( x) = í
if x = 0
ïî 0
3
1
(b) p = - , q =
2
2
1
3
(c) p = , q =
2
2
1
3
(d) p = , q = 2
2
The function f : R /{0} ® R given by
[2007]
1
2
x e2 x - 1
can be made continuous at x = 0 by defining f (0) as
(a) 0
(b) 1
(c) 2
(d) – 1
x® x0
and only if, lim f ( x ) exists and lim f ( x ) ¹ f ( x0 ) .
5
1
(a) p = , q =
2
2
f ( x) =
only if lim f ( x ) exists and lim f ( x ) = f ( x0 ).
x ® x0
(a) Statement -1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
NOT a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
The values of p and q for which the function
[2011]
Let f ( x) =
p
1 - tan x
é pù
, x ¹ , x Î ê 0, ú .
4x - p
4
ë 2û
é pù
æ pö
If f (x) is continuous in ê 0, ú , then f çè ÷ø is
2
4
ë
û
(a) –1
(c) 27.
1
2
(b)
[2004]
1
2
(d) 1
f is defined in [-5, 5] as
f (x) = x if x is rational
= – x if x is irrational. Then
(a) f (x) is continuous at every x, except x = 0
(b) f (x) is discontinuous at every x, except x = 0
(c) f (x) is continuous everywhere
(d) f (x) is discontinuous everywhere
Downloaded from @Freebooksforjeeneet
[2002]
EBD_8344
M-312
Mathematics
34.
TOPIC n Differentiability
28.
29.
is the identity function. If for some a, b Î R, g¢ (a) = 5 and
Let f : R ® R be a function defined by f (x) = max{x, x2}. Let
S denote the set of all points in R, where f is not
differentiable. Then:
[Sep. 06, 2020 (II)]
(a) {0, 1}
(b) {0}
(c) f (an empty set)
(d) {1}
2
ì
If the function f ( x) í k1 ( x - p) - 1, x £ p is twice difx>p
î k2 cos x,
ferentiable, then the ordered pair (k1, k2) is equal to:
g (a) = b, then f ¢ (b) is equal to:
(a)
35.
that:
(b)
(a) f (5) + f '(5) £ 26
33.
f '(5) + f ''(5) £ 20
36.
in the interval:
ì sin(a + 2) x + sin x
; x<0
ï
x
ïï
; x =0
b
If f ( x) = í
ï
2 1/3
1/3
ï ( x + 3x ) - x
; x>0
ïî
x 4/3
37.
[Jan. 9, 2020 (I)]
(d) –2
[Jan. 7, 2020 (I)]
(a) (– ฀ , 20]
(b) [–3, 11]
(c) (– ฀, 11]
(d) [–6, 20]
Let S be the set of points where the function,
f(x) = |2 – |x – 3||, xÎR, is not differentiable.
Then å f(f(x)) is equal to ¾¾¾.
xÎS
38.
[NA Jan. 7, 2020 (I)]
ì sin (p + 1) x + sin x
,x < 0
ï
x
ï
q
,x=0
If f (x) = í
ï
2
x+x - x
ï
, x >0
î
x3/2
is continuous at x = 0, then the ordered pair (p, q) is equal to:
[April 10, 2019 (I)]
39.
is continuous at x = 0, then a + 2b is equal to:
(c) 0
Let the function, f: [–7, 0] ® R be continuous on [ –7, 0]
all xÎ(–7, 0), then for all such functions f, f ¢(–1) + f(0) lies
(d) f (5) £ 10
ìp
-1
ïï 4 + tan x, | x |£ 1
The function f ( x) = í
ï 1 (| x | -1) , | x |> 1
ïî 2
is :
[Sep. 04, 2020 (II)]
(a) continuous on R – {1} and differentiable on
R – {–1, 1}.
(b) both continuous and differentiable on R – {1}.
(c) continuous on R – {–1} and differentiable on
R – {–1, 1}.
(d) both continuous and differentiable on R – {–1}.
(b) –1
f (1) - f (c )
= f ¢(c)
1- c
and differentiable on (–7, 0). If f(–7) = –3 and f ¢ (x) d” 2, for
(b) f (5) + f '(5) ³ 28
f ( x)
= 1, then f '(3) is equal to ___________.
x
[NA Sep. 04, 2020 (I)]
(a) 1
[Jan. 8, 2020 (II)]
(d) | f (c) – f (1)| < | f ¢(c)|
f ( x + y ) = f ( x) + f ( y ) + xy 2 + x 2 y , for all real x and y. If
32.
2
5
(c) | f (c) + f (1)| < (1 + c) |f ¢(c)|
Suppose a differentiable function f (x) satisfies the identity
x ®0
(d)
every f in S, there exists a c Î (0,1), depending on f, such
æ1
ö
(c) ç , -1÷
(d) (1, 1)
2
è
ø
Let f be a twice differentiable function on (1, 6). If f (2) = 8,
lim
(c) 5
Let S be the set of all functions f : [0,1] ® R, which are
(b) (1, 0)
f '(2) = 5, f '(x) ³ 1 and f ''( x ) ³ 4, for all x Î (1, 6), then :
[Sep. 04, 2020 (I)]
31.
(b) 1
(a) | f (c) – f (l)| < (l – c)|f ¢(c)|
æ1 ö
(a) ç ,1÷
è2 ø
(c)
1
5
[Jan. 9, 2020 (II)]
continuous on [0, 1] and differentiable on (0,1). Then for
[Sep. 05, 2020 (I)]
30.
Let f and g be differentiable functions on R such that fog
æ 3 1ö
(a) ç - , - ÷
è 2 2ø
æ 1 3ö
(b) ç - , ÷
è 2 2ø
æ 3 1ö
(c) ç - , ÷
è 2 2ø
æ5 1ö
(d) ç , ÷
è2 2ø
Let f(x) = loge (sinx), (0 < x < p) and g(x) = sin–1 (e–x), (x > 0).
If a is a positive real number such that a = (fog)¢ (a) and
b = (fog) ( a), then:
[April 10, 2019 (II)]
(a) aa2 + ba + a = 0
(b) aa2 – ba – a =1
(c) aa2 – ba – a = 0
(d) aa2 + ba – a = – 2a2
Downloaded from @Freebooksforjeeneet
M-313
Continuity and Differentiability
40.
Let f : R ® R be differentiable at c Î R and f (c) = 0. If
g (x) = f (x) , then at x = c, g is :
(a) is an empty set
(b) equals {– 2, – 1, 0, 1, 2}
[April 10, 2019 (I)]
(a) not differentiable if f '(c) = 0
(c) equals {– 2, – 1, 1, 2}
(b) differentiable if f "(c) ¹ 0
(d) equals {– 2, 2}
(c) differentiable if f ' (c) = 0
48.
(d) not differentiable
41.
Let f(x) = 15 – |x – 10|; x Î R. Then the set of all values of x,
at which the function, g(x) = f (f(x)) is not
differentiable, is:
42.
(b) {10, 15}
(c) {5, 10, 15, 20}
(d) {10}
f (f (f (x))) + (f (x))2 at x = 1 is :
43.
44.
49.
If f (1) = 1, f¢(1) = 3, then the derivative of
[April 08, 2019 (II)]
(a) 33
(b) 12
(c) 15
(d) 9
Let f be a differentiable function such that f(1) = 2 and
f ¢ (x) = f (x) for all x Î R. If h (x) = f ( f (x)), then h¢ (1) is equal
to :
[Jan. 12, 2019 (II)]
(a) 2e2
(b) 4e
(c) 2e
(d) 4e2
-2 £ x < 0
ìï -1,
and
Let f ( x ) = í 2
ïî x - 1, 0 £ x £ 2
50.
51.
(1 + 2e )
2 4+ e
2
( 2e –1)
(b)
(1 + 2e )
(c)
46.
47.
4+e
2
2 4+ e
(a)
52.
2
ìïmax {| x |, x 2 }
| x|£ 2
Let f (x) = í
2 <| x|£ 4
ïî 8 - 2 | x |,
Let S be the set of points in the interval (– 4, 4) at which f
is not differentiable. Then S:
[Jan 10, 2019 (I)]
p+2
2
-p - 2
2
If the function.
[Online April 9, 2016]
(b)
p-2
2
(d) –1 – cos–1(2)
ïì k x + 1, 0 £ x £ 3
g(x) = í
is differentiable, then the
ïî m x + 2, 3 < x £ 5
value of k + m is :
10
(a)
3
4 + e2
Let K be the set of all real values of x where the function
f (x) = sin | x | – | x | + 2 (x – p) cos | x | is not differentiable.
Then the set K is equal to :
[Jan. 11, 2019 (II)]
(a) f (an empty set)
(b) {p}
(c) {0}
(d) {0, p}
a
is equal to :
b
(c)
e
(d)
Let S = { t Î R : f (x) = | x - p | (e|x| - 1)sin | x | is not
differentiable at t}. Then the set S is equal to :
[2018]
(a) {0}
(b) {p}
(c) {0, p}
(d) f (an empty set)
Let S = {(l, m) Î R × R : f (t) = (|l|e|t| – m). sin (2|t|), t Î R, is
a differentiable function}. Then S is a subest of?
[Online April 15, 2018]
(a) R × [0, ¥)
(b) (–¥, 0) × R
(c) [0, ¥) × R
(d) R × (–¥, 0)
If the function
x = 1, then
dy
If x log e ( log e x ) - x 2 + y 2 = 4 ( y > 0 ) , then
at x = e is
dx
equal to :
[Jan. 11, 2019 (I)]
(a)
}
1 - x 2 . If K be the set of all points at which f
x <1
ìï- x,
f(x) = í
is differentiable at
-1
ïîa + cos (x + b), 1 £ x £ 2
g(x) = |f(x)| + f(|x|). Then, in the interval (–2, 2), g is :
[Jan. 11, 2019 (I)]
(a) differentiable at all points
(b) not continuous
(c) not differentiable at two points
(d) not differentiable at one point
45.
{ - | x |, -
is not differentiable, then K has exactly:
[Jan. 10, 2019 (II)]
(a) five elements
(b) one element
(c) three elements
(d) two elements
[April 09, 2019 (I)]
(a) {5, 10, 15}
Let f : (– 1, 1) ® R be a function defined by f (x) = max
(c) 2
53.
[2015]
(b) 4
(d)
16
5
Let f: R ® R be a function such that f ( x ) £ x 2 , for all
x Î R . Then, at x = 0, f is:
[Online April 19, 2014]
(a) continuous but not differentiable.
(b) continuous as well as differentiable.
(c) neither continuous nor differentiable.
(d) differentiable but not continuous.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-314
54.
Mathematics
Let f, g: R ® R be two functions defined by
60.
ì
æ1ö
ï x sin ç ÷ , x ¹ 0
f (x) = í
èxø
, and g(x) = x f(x)
ï0,
,x = 0
î
55.
56.
57.
58.
59.
f (x) = min {x + 1, x + 1} ,Then which of the following is true?
(a) f (x) is differentiable everywhere
(b) f (x) is not differentiable at x = 0
Statement I: f is a continuous function at x = 0.
Statement II: g is a differentiable function at x = 0.
[Online April 12, 2014]
(a) Both statement I and II are false.
(b) Both statement I and II are true.
(c) Statement I is true, statement II is false.
(d) Statement I is false, statement II is true.
Consider the function, f (x) = | x – 2 | + | x – 5 |, x Î R.
Statement-1 : f ¢(4) = 0
Statement-2 : f is continuous in [2,5], differentiable in (2,5)
and f (2) = f (5).
[2012]
(a) Statement-1 is false, Statement-2 is true.
(b) Statement-1 is true, statement-2 is true; statement-2 is
a correct explanation for Statement-1.
(c) Statement-1 is true, statement-2 is true; statement-2 is
not a correct explanation for Statement-1.
(d) Statement-1 is true, statement-2 is false.
If f(x) = a |sinx| + be |x| + c|x|3, where a, b, c Î R, is
differentiable at x = 0, then
[Online May 26, 2012]
(a) a = 0, b and c are any real numbers
(b) c = 0, a = 0, b is any real number
(c) b = 0, c = 0, a is any real number
(d) a = 0, b = 0, c is any real number
If x + | y | = 2y, then y as a function of x, at x = 0 is
[Online May 7, 2012]
(a) differentiable but not continuous
(b) continuous but not differentiable
(c) continuous as well as differentiable
(d) neither continuous nor differentiable
If function f (x) is differentiable at x = a,
x2 f (a) - a2 f ( x)
then xlim
is :
®a
x-a
Let f : R ® R be a function defined by
[2011RS]
(a) -a 2 f ' ( a )
(b) a f (a ) - a 2 f ' ( a )
(c) 2af ( a) - a 2 f ' ( a)
2
(d) 2a f (a) +a f '(a)
1
ì
if x ¹ 1
ï( x –1) sin
Let f ( x) = í
x –1
ïî
0
if x = 1
Then which one of the following is true?
(a) f is neither differentiable at x = 0 nor at x =1
(b) f is differentiable at x = 0 and at x =1
(c) f is differentiable at x = 0 but not at x = 1
(d) f is differentiable at x = 1 but not at x = 0
(c) f (x) ³ 1 for all x Î R
(d) f (x) is not differentiable at x = 1
61.
The set of points where f ( x ) =
x
is differentiable is
1+ | x |
[2006]
(a) (-¥,0) È (0, ¥)
(b) (-¥,-1) È (-1, ¥)
(c) (-¥, ¥)
(d) (0, ¥)
62.
If f is a real valued differentiable function satisfying
63.
| f (x) – f (y) | £ ( x - y )2 , x, y Î R and f (0) = 0, then f (1)
equals
[2005]
(a) – 1
(b) 0
(c) 2
(d) 1
Suppose f (x) is differentiable at x = 1 and
lim
1
h®0 h
f (1 + h) = 5 , then f '(1) equals
(a) 3
64.
(b) 4
(c) 5
ì æ 1 1ö
ï -ç + ÷
If f ( x ) = í xe è x x ø , x ¹ 0
ï0
,x = 0
î
(a)
(b)
(c)
(d)
[2005]
(d) 6
then f(x) is
discontinuous every where
[2003]
continuous as well as differentiable for all x
continuous for all x but not differentiable at x = 0
neither differentiable nor continuous at x = 0
TOPIC Đ
65.
[2008]
[2007]
Chain Rule of Differentiation,
Differentiation of Explicit & Implicit
Functions, Parametric & Composite
Functions, Logarithmic & Exponential
Functions, Inverse Functions,
Differentiation by Trigonometric
Substitution
æ 1 + x2 - 1 ö
÷ with respect to
The derivative of tan -1 ç
ç
÷
x
è
ø
æ 2x 1- x2
tan -1 ç
ç 1 - 2x2
è
(a)
2 3
5
(b)
ö
÷ at x = 1 is :
÷
2
ø
3
12
(c)
2 3
3
Downloaded from @Freebooksforjeeneet
[Sep. 05, 2020 (II)]
(d)
3
10
M-315
Continuity and Differentiability
66.
If (a + 2b cos x)(a - 2b cos y) = a 2 - b 2 , where a > b > 0,
dx æ p p ö
then
at ç , ÷ is :
dy è 4 4 ø
a -b
a - 2b
(a)
(b)
a+b
a + 2b
[Sep. 04, 2020 (I)]
a+b
(c)
a -b
73.
æ -1 æ 3 cos x + sin x ö ö
æ pö
dy
÷÷ ÷÷ , x Î ç 0, ÷ then
If 2y = çç cot çç
2
dx
è
ø
è cos x - 3 sin x ø ø
è
is equal to :
[April 08, 2019 (I)]
ì 3p p 3p p ü
(b) í- , - , , ý
4 4 4þ
î 4
ì p p p pü
(c) í- , - , , ý
î 2 4 4 2þ
ì 3p p p 3p ü
(d) í- , - , , ý
2 2 4 þ
î 4
3
4
4
dy
ü
at x = 0 is
í cos kx - sin kx ý, then
dx
5
î5
þ
[NA Sep. 02, 2020 (II)]
3
8
(c)
3
2
(d) -
75.
4
3
(d) -
(c) –4
1
4
2
= k – x 1 - y where k is a constant and
5
(b) 2
(c)
2
5
(d)
(a) – 4
77.
(a)
2
equal to :
78.
(b) 30
(c) – 2
p
, is:
4
1
3 2
If x =
æ 1 1 ö
(b) ç - , 2 ÷
è e e ø
(b)
2cosec
–1
t
1
(c)
6 2
and y =
æ1 1 ö
(c) ç , 2 ÷
èe e ø
æ 1 1ö
(d) ç - , - 2 ÷
è e e ø
x
-1 æ sin x - cos x ö
The derivative of tan ç
÷ , with respect to ,
è sin x + cos x ø
2
æ
æ p öö
where ç x Î ç 0, 2 ÷ ÷ is :
è
øø
è
[April 12, 2019 (II)]
equal to.
(a)
y
x
(b) –
cos x
79.
d2y
dx 2
at
[Jan. 09, 2019 (II)]
[April 12, 2019 (I)]
æ1 1 ö
(a) ç , - 2 ÷
èe e ø
(d) 8
If x = 3 tan t and y = 3 sec t, then the value of
t=
5
2
æ dy d y ö
If ey + xy = e, the ordered pair çç dx , 2 ÷÷ at x = 0 is
dx ø
è
(b) loge 2x
x log e 2 x + log e 2
(d) x loge 2x
x
Let f : R ® R be a function such that
f (x) = x3 + x2f¢(1) + xf ²(2) + f ¢²(3), xÎR. Then f (2) equals:
[Jan 10, 2019 (I)]
[Jan. 7, 2020 (II)]
5
(a) 4
2
(c)
76.
dy
1
1
æ 1ö
at x = , is equal to:
y ç ÷ = - . Then
dx
2
è 2ø
4
dy
is
dx
[Jan. 12, 2019 (I)]
For x > 1, if ( 2 x )2 y = 4e 2 x- 2 y , then (1 + log e 2 x )
equal to :
x log e 2 x - log e 2
(a)
x
[Jan. 7, 2020 (I)]
Let y = y(x) be a function of x satisfying
y 1 - x2
74.
3
4
1
æ tan a + cot a ö
æ 3p ö
If y (a) = 2 ç
÷ + 2 , a Î ç 4 , p ÷ , then
2
è
ø
è 1 + tan a ø sin a
(b)
(a)
[Jan. 9, 2020 (II)]
(b) -
(a) 4
72.
(d) 2
2
-1 ì 3
5p
dy
at a =
is:
6
da
71.
1
2
pü
ì p
(a) í- , 0, ý
4
4þ
î
(a)
70.
(c)
d2y
at q = p is :
dx 2
å k cos
k =1
69.
2
3
If x = 2sinq – sin2q and y = 2cosq – cos2q, q Î [0, 2p], then
If y =
___________.
68.
2a + b
(d)
2a - b
(b)
p
p
p
p
– x (b) x –
(c)
–x
(d) 2x –
6
6
3
3
Let S be the set of all points in (–p,p) at which the function
f(x) = min {sinx, cosx} is not differentiable. Then S is a
subset of which of the following?
[Jan. 12, 2019 (I)]
6
67.
(a) 1
y
x
x
2
If f (x) = 2 sin x x
tan x
x
3
2 2
(d)
1
6
dy
is
dx
[Online April 16, 2018]
x
x
(c) –
(d)
y
y
2sec
–1
t
(| t | ³ 1), then
1
2 x , then lim
x®0
1
f ¢ (x)
x
[Online April 15, 2018]
(a)
(b)
(c)
(d)
Exists and is equal to – 2
Does not exist
Exist and is equal to 0
Exists and is equal to 2
Downloaded from @Freebooksforjeeneet
EBD_8344
M-316
80.
Mathematics
æ 2 ´ 3x ö
æ 1ö
If f (x) = sin -1 çç
÷ , then f ¢ ç - ÷ equals.
x ÷
1
9
+
è 2ø
è
ø
[Online April 15, 2018]
(b) - 3 log e 3
3 log e 3
(a)
(c) - 3 loge 3
81.
If x2 + y2 +
82.
(b) – 32
x.g ( x ) , then g(x) equals :
(c)
83.
84.
3
(b)
1 + 9x3
3x x
then g¢(7) equals:
(a) 85.
(b)
1
2
(c)
1
3
1
13
dy
at x = 1 is equal to :
dx
1
(b)
2
4
3
91.
(d)
(c)
1
2
(d)
2
94.
[Online April 22, 2013]
.x ¹ 0, -2 . Then
1
(1 - x )
d
[ f -1 ( x )] (wherever
dx
[Online April 9, 2013]
(1 - x )2
-3
(d)
2
x2
3
(b)
2
x2 + y2
(1 - x )2
( 3 - 2 x)
2
12
é æ 2 x + 3ö ù
sin êlog ç
÷ú
ë è 3 - 2x ø û
2
é æ 2 x + 3ö ù
cos êlog ç
÷ú
ë è 3 - 2x ø û
If x m . y n = ( x + y ) m+ n , then
y
x
(b) x + y
xy
y +L to ¥
dy
is
dx
(c) xy
[2006]
(d)
x
y
dy
is
[2004]
dx
1
x
1+ x
1- x
(a)
(b)
(c)
(d)
x
x
x
1+ x
Let f (x) be a polynomial function of second degree.
If x = e y + e
, x > 0, then
If f(1) = f(-1) and a, b, c are in A. P , then f '(a), f ¢ (b), f '(c)
are in
[2003]
(a) Arithmetic -Geometric Progression
(b) A.P
(c) G..P
(d) H.P.
3
4
2
y2
(d)
( 3 - 2 x)
Let f : (–1, 1) ® R be a differentiable function with f (0) = –
1 and f ¢ (0) = 1. Let g(x) = [f (2f (x) + 2)]2. Then g¢(0) =
[2010]
(a) –4
(b) 0
(c) –2
(d) 4
Let y be an implicit function of x defined by
x2x – 2xx cot y – 1= 0. Then y¢(1) equals
[2009]
(a) 1
(b) log 2
(c) –log 2 (d) –1
(a)
[2013]
-1
æ pö
cos-1 t
For a > 0, t Î ç 0, ÷ , let x = asin t and y = a
,
è 2ø
dy
Then, 1 + æç ö÷ equals :
è dx ø
90.
92.
x2 y 2
+
= 1 and y3 = 16x intersect at right
a
4
angles, then a value of a is :
[Online April 23, 2013]
(b)
(1 - x )
(d)
93.
(c) 1
2
12
(d) -
x2 + y2
(c)
æ 2 x + 3ö
dy
If f ¢(x) = sin (log x) and y = f ç
equals
, then
è 3 - 2 x ÷ø
dx
[Online May 12, 2012]
(c)
3
1
, and g(x) is its inverse function,
2
[Online April 12, 2014]
1
13
x2 - x
x2
-1
3x
If the curves
(a) 2
87.
1
3
If y = sec(tan–1x), then
(a)
86.
9
1 + 9x3
1 - 9x
1 - 9x
For x Î R, f(x) = |log2 – sinx| and g(x) = f(f(x)), then :
[2016]
(a) g'(0) = – cos(log2)
(b) g is differentiable at x = 0 and g'(0) = – sin(log2)
(c) g is not differentiable at x = 0
(d) g'(0) = cos(log2)
If f (x) = x2 – x + 5, x >
y2
y2
é
æ 2x + 3ö ù
(a) sin ê log ç
è 3 - 2 x ÷ø úû
ë
12
(b)
( 3 - 2 x)2
[2017]
(d)
3
(c)
89.
(b)
x + 2x
it is defined) is equal to :
d2 y
at the point
dx 2
[Online April 15, 2018]
(c) – 2
(d) 4
x2
Let f (x) =
(a)
æ 6x x ö
æ 1ö
If for x Îç 0, ÷ , the derivative of tan -1 ç
÷ is
è 4ø
è 1 - 9x 3 ø
(a)
88.
3 loge 3
(d)
sin y = 4, then the value of
(– 2, 0) is
(a) – 34
(a)
95.
If f ( x + y ) = f ( x ). f ( y )"x. y and f (5) = 2,
f '(0) = 3, then f ¢ (5) is
(a) 0
(b) 1
[2002]
(c) 6
Downloaded from @Freebooksforjeeneet
(d) 2
M-317
Continuity and Differentiability
TOPIC Ė
96.
Differentiation of Infinite Series,
Successive Differentiation, nth
Derivative of Some Standard
Functions, Leibnitz’s Theorem,
Rolle’s Theorem, Lagrange’s Mean
Value Theorem
For all twice differentiable functios f : R®R, with
f(0) = f(1) = f’(0) = 0
[Sep. 06, 2020 (II)]
f "(0) = 0
æ p pö
2
2
If y + log e (cos x) = y, x Î ç - , ÷ , then :
è 2 2ø
[Sep. 03, 2020 (I)]
(a) y ''(0) = 0
(b) | y '(0) | + | y ''(0) |= 1
(c) | y ''(0) |= 2
(d) | y '(0) | + | y ''(0) |= 3
If c is a point at which Rolle’s theorem holds for the
æ x2 + a ö
function, f ( x ) = log e ç
÷ in the interval [3, 4], where
è 7x ø
a Î R, then f ²(c) is equal to:
(a) -
1
12
1
12
(b)
[Jan. 8, 2020 (I)]
(c) -
1
24
(d)
3
7
1
99.
dy æ y ö 3
Let x + y = a , (a, k > 0) and
+ ç ÷ = 0, then k is:
dx è x ø
k
k
k
[Jan. 7, 2020 (I)]
3
(a)
2
4
(b)
3
2
(c)
3
1
(d)
3
100. The value of c in the Lagrange’s mean value theorem for
the function f(x) = x3 – 4x2 + 8x + 11, when x Î [0,1] is:
[Jan. 7, 2020 (II)]
(a)
4- 5
3
(b)
4- 7
3
(c)
2
3
(d)
7-2
3
1
101. If 2 x = y 5 + y
-
1
5
d2y
dy
+ lx + ky = 0,
dx
dx 2
[Online April 9, 2017]
(c) 26
(d) – 26
= and (x2 – 1)
then l + k is equal to :
(a) – 23
(b) – 24
d2 y
dx 2
15
é x + x 2 - 1ù
ëê
ûú
=
+x
dy
dx
15
+ éê x - x 2 - 1ùú
ë
û
is equal to
,
then
[Online April 8, 2017]
1
, then 2b + c equals :
2
[Online April 10, 2015]
(a) –3
(b) –1
(c) 2
(d) 1
105. If f and g are differentiable functions in [0, 1] satisfying
+ cx, x Î [–1, 1], at the point x =
(d) f "(x) = 0, at every point x Î (0,1)
98.
y
(a) 12 y
(b) 224 y2 (c) 225 y2
(d) 225 y
104. If Rolle’s theorem holds for the function f (x) 2x3 + bx2
(b) f "(x) = 0, for some x Î (0,1)
97.
103. If
(x 2 - 1)
(a) f "( x ) ¹ 0 at every point x Î (0,1)
(c)
102. Let f be a polynomial function such that f (3x) = f ¢ (x) , f ²
(x), for all x Î R. Then :
[Online April 9, 2017]
(a) f (b) + f ¢ (b) = 28
(b) f ² (b) – f ¢ (b) = 0
(c) f ² (b) – f ¢ (b) = 4
(d) f (b) – f ¢ (b) + f ² (b) = 10
f (0) = 2 = g(1), g(0) = 0 and f (1) = 6, then for some c Î]0,1[
[2014]
(a) f ¢(c) = g¢(c)
(b) f ¢(c) = 2g ¢(c)
(c) 2f ¢(c) = g ¢(c)
(d) 2f ¢(c) = 3g¢(c)
106. Let f(x) = x|x|, g(x) = sin x and h(x) = (gof) (x). Then
[Online April 11, 2014]
(a) h(x) is not differentiable at x = 0.
(b) h(x) is differentiable at x = 0, but h¢(x) is not continuous
at x = 0
(c) h¢(x) is continuous at x = 0 but it is not differentiable at
x= 0
(d) h¢(x) is differentiable at x = 0
107. Let for i = 1, 2, 3, pi(x) be a polynomial of degree 2 in x, p¢i(x)
and p¢¢i(x) be the first and second order derivatives of pi(x)
respectively. Let,
é p ( x ) p ¢ ( x ) p ¢¢ ( x ) ù
1
1
ê 1
ú
ê
¢
A ( x ) = p2 ( x ) p2 ( x ) p2¢¢ ( x ) ú
ê
ú
ê p3 ( x ) p3¢ ( x ) p3¢¢ ( x ) ú
ë
û
and B(x) = [A(x)]T A(x). Then determinant of B(x):
[Online April 11, 2014]
(a) is a polynomial of degree 6 in x.
(b) is a polynomial of degree 3 in x.
(c) is a polynomial of degree 2 in x.
(d) does not depend on x.
108. If the Rolle’s theorem h olds for the function
f(x) = 2x3 + ax2 + bx in the interval [– 1, 1] for the point
1
, then the value of 2a + b is: [Online April 9, 2014]
2
(a) 1
(b) – 1
(c) 2
(d) – 2
c=
Downloaded from @Freebooksforjeeneet
EBD_8344
M-318
Mathematics
109. If f (x) = sin (sin x) and f "(x) + tan x f¢ (x) + g(x) = 0, then
g(x) is :
[Online April 23, 2013]
(a) cos2 x cos (sin x)
(b) sin2 x cos (cos x)
(c) sin2 x sin (cos x)
(d) cos2 x sin (sin x)
110. Consider a quadratic equation ax2 + bx + c = 0, where
x3
x2
+ b + cx .
3
2
[Online May 19, 2012]
Statement 1: The quadratic equation has at least one root
in the interval (0, 1).
Statement 2: The Rolle’s theorem is applicable to function
g(x) on the interval [0, 1].
(a) Statement 1 is false, Statement 2 is true.
(b) Statement 1 is true, Statement 2 is false.
(c) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation for Statement 1.
(d) Statement 1 is true, Statement 2 is true, , Statement 2 is
a correct explanation for Statement 1.
2a + 3b + 6c = 0 and let g ( x ) = a
2
114. Let f be differentiable for all x. If f (1) = – 2 and
f '( x) ³ 2 for x Î [1, 6], then
(a) f (6) ³ 8(b) f (6) < 8
dy 2
equals :
æ d2y ö
(a) - ç 2 ÷
è dx ø
-1
[2011]
æ dy ö
çè ÷ø
dx
-3
æ d 2 y ö æ dy ö -3
ç 2 ÷ çè dx ÷ø
(c)
è dx ø
æ d 2 y ö æ dy ö -2
(b) ç 2 ÷ çè dx ÷ø
è dx ø
æd2yö
(d) ç 2 ÷
è dx ø
-1
(b) loge3
(c) 2 log3e
(d) f (6) = 5
a1 ¹ 0, n ³ 2, has a positive root x = a , then the equation
n -2
nan x n -1 + (n – 1) an -1 x
+ ......... + a1 = 0 has a positive
root, which is
[2005]
(a) greater than a
(b) smaller than a
(c) greater than or equal to a
(d) equal to a
116. If 2a + 3b + 6c = 0, then at least one root of the equation
ax 2 + bx + c = 0 lies in the interval
(a) (1, 3) (b) (1, 2)
(c) (2, 3)
[2004]
(d) (0, 1)
117. If f ( x) = x n , then the value of
f (1) -
[2003]
f ' (1) f ' ' (1) f ' ' ' (1)
( -1) n f n (1)
+
+ ..........
is
n!
1!
2!
3!
(b) 2 n
(d) 0.
(c) 2 n - 1
118. Let f (a) = g (a) = k and their nth derivatives
(a) 1
f n (a) , g n (a) exist and are not equal for some n. Further
if
112. Let f (x) = x | x | and g (x) = sin x.
Statement-1 : gof is differentiable at x = 0 and its derivative
is continuous at that point.
Statement-2 : gof is twice differentiable at x = 0. [2009]
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is false.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is true; Statement-2 is
a correct explanation for Statement-1.
113. A value of c for which conclusion of Mean Value Theorem
holds for the function f (x) = loge x on the interval [1, 3] is
[2007]
(a) log3e
(c) f (6) < 5
115. If the equation an x n + an -1 x n -1 + ............. + a1 x = 0
d x
111.
[2005]
(d)
1
log e
2 3
lim
x ®a
f ( a ) g ( x) - f ( a ) - g ( a ) f ( x ) + f ( a )
=4
g ( x) - f ( x)
then the value of k is
(a) 0
(b) 4
[2003]
(d) 1
(c) 2
119. If y = (x + 1 + x 2 )n, then (1 + x2)
d2y
dx2
+x
dy
is [2002]
dx
(a) n2y
(b) – n2y
(c) –y
(d) 2x2y
120. If 2a + 3b + 6c = 0, (a, b, c Î R) then the quadratic equation
ax2 + bx + c = 0 has
[2002]
(a) at least one root in [0, 1]
(b) at least one root in [2, 3]
(c) at least one root in [4, 5]
(d) None of these
Downloaded from @Freebooksforjeeneet
M-319
Continuity and Differentiability
1.
(8) We know [x] discontinuous for x Î Z
éxù
x
f ( x ) = x ê ú may be discontinuous where
is an
2
2
ë û
integer.
So, points of discontinuity are,
4.
æ 3ln(1 + 3x) 2ln(1 - 2 x) ö
= lim ç
÷
3x
-2 x
x ®0 è
ø
=3+ 2=5
Q f(x) will be continuous
but at x = 0
lim f ( x ) = 0 = f (0) = lim f ( x)
2.
x ®0
æ ln(1 + 3x) ln(1 - 2 x) ö
= lim ç
÷
x
x
x ®0 è
ø
x = ±2, ± 4, ± 6, ± 8 and 0
x ® 0+
æ 1 æ 1 + 3x ö ö
lim f ( x) = lim ç ln ç
÷÷
x®0 è x è 1 - 2 x ø ø
(5)
x ® 0-
\
So, f (x) will be discontinuous at x = ±2, ±4, ±6 and ±8.
(d) Since, function f (x) is continuous at x = 1, 3
5.
\ f (1) = f (1+ )
k = f (0) = lim f ( x ) = 5
x®0
(b) Since, f(x) is continuous, then
lim f ( x)
Þ ae + be -1 = c
x®
...(i)
p
4
f (3) = f (3+ )
lim
2 cos x - 1
=k
cot x - 1
Þ 9c = 9a + 6c Þ c = 3a
From (i) and (ii),
...(ii)
p
x®
4
b = ae(3 - e)
...(iii)
Now by L- hospital’s rule
é ae x - be - x
ê
f '( x ) = ê 2cx
ê 2ax + 2c
ë
æ 1 ö
2ç
÷
è 2 ø =k Þ k = 1
=
Þ
k
lim
2
p
2
( 2 )2
x ® c osec x
-1 < x < 1
2 sin x
1< x < 3
3< x< 4
4
f '(0) = a - b, f '(2) = 4c
6.
Given, f '(0) + f '(2)= e
a - b + 4c= e
From eqs. (i), (ii), (iii) and (iv),
...(iv)
é4ù
f ( 4 ) = [ 4] - ê ú = 4 - 1 = 3
ë4û
Þ 13a - 3ae + ae 2 = e
Þa=
(a)
Q LHL = f (4) = RHL
\ f (x) is continuous at x = 4
e
e - 3e + 13
2
é 4 ì 4 üù
é4ù
lim x ê ú = A Þ lim x ê - í ý ú = A
x ®0 ë x î x þ û
x ®0 ë x û
7.
4
Þ lim 4 - x ìí üý = A Þ 4 – 0 = A
x ®0
îxþ
As, f (x) = [x2]sin(px) will be discontinuous at non-integers
And, when x = A + 1 Þ x = 5 ,
which is not an integer.
Hence, f (x) is discontinuous when x is equal to
æ
éxùö
(a) L.H.L. lim- ç [ x ] - ê 4 ú ÷ = 3 - 0 = 3
ë ûø
x®4 è
éxù
R.H.L. lim+ [ x ] - ê ú = 4 - 1 = 3
ë4û
x ®4
a - 3ae + ae 2 + 12a = e
3.
æpö
= fç ÷
è4ø
A +1
(d) R.H.L. lim+ b ( x - p ) + 3 = ( 5 - p ) b + 3
x ®5
f (5) = L.H.L. lim- a ( p - x ) + 1 = a ( 5 - p ) + 1
x ®5
Q function is continuous at x = 5
\ LHL = RHL
(5 – p) b + 3 = (5 – p) a + 1
Þ 2 = (a – b) (5 – p) Þ a - b =
2
5- p
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EBD_8344
M-320
8.
Mathematics
For the limit to exist, power of x in the numerator should be
greater than or equal to the power of x in the denominator.
Therefore, coefficient of x in numerator is equal to ero
Þ 3–k=0
Þ k=3
So the limit reduces to
(c) Given function is,
ì| x | +[ x], -1 £ x < 1
ï
x + | x |, 1 £ x < 2
f (x) = í
ï x + [ x ],
2£ x£3
î
ì - x - 1,
ï x,
ïï
= í 2 x,
ï x + 2,
ï
ïî6,
-1 £ x < 0
0 £ x <1
æ 4 8x
ö
( x2 ) ç +
+ ... ÷
2! 3!
è
ø
lim
2
x ®0
æ
ö
4 x 8x
( x 2 ) çç 2 +
+
+ ... ÷÷
2!
3!
è
ø
1£ x < 2
2£ x<3
x=3
Þ f (–1) = 0, f (–1+) = 0;
f (0–) = –1, f (0) = 0, f (0+) = 0;
f (1–) = 1, f (1) = 2, f (1+) = 2;
f (2–) = 4, f (2) = 4, f (2+) = 4;
f (3–) = 5, f (3) = 6
f (x) is discontinuous at x = {0, 1, 3}
Hence, f (x) is discontinuous at only three points.
9.
(d) Let f(x) is continuous at x = 1, then
f(1–) = f(1) = f(1+)
Þ 5=a+b
...(1)
Let f(x) is continuous at x = 3, then
f(3–) = f(3) = f(3+)
Þ a + 3b = b + 15 ...(2)
Let f(x) is continuous at x = 5, then
f(5–) = f(5) = f(5+)
Þ b + 25 = 30
Þ b = 30 – 25 = 5
From (1), a = 0
But a = 0, b = 5 do not satisfy equation (2)
Hence, f(x) is not continuous for any values of a and b
10. (a) If the function is continuous at x = 0, then
11.
4 8x
+
+ ...
2!
3!
=1
= lim
x ®0
4 x 8 x2
2+
+
+ ...
2!
3!
Hence, f (0) = 1
(c) Since f (x) is continuous at x = 2.
\ lim f ( x ) = f (2)
x®2
Þ lim ( x - 1)
x®2
where l = lim ( x - 1 - 1) ´
x®2
æ e 2 x - 1 - kx + x ö
= lim çç
÷÷
2x
x®0
è ( x ) ( e - 1) ø
éæ
ù
ö
(2 x) 2 (2 x )3
ê ç1 + 2 x +
+
+ ... ÷÷ - 1 - kx + x ú
ç
2!
3!
ê
ú
ø
= lim ê è
ú
2
3
x ®0
æ
ö
æ
ö
ê ( x) ç 1 + 2 x + (2 x ) + (2 x ) + ... - 1÷ ú
ç
÷
÷ ÷ú
çç
ê
2!
3!
ø øû
èè
ë
é
ù
4 x2 8x 3
ê (3 - k ) x +
+
+ ... ú
2!
3!
ê
ú
= xlim
®0 ê æ
ö ú
4 x3 8 x3
2
+
+ ... ÷ ú
ê ç 2x +
2!
3!
êë è
ø úû
1
x-2
= lim
2 - x x®2 2 - x
æ x-2ö
= lim ç
÷
x ®2 è x - 2 ø
Þ k = e–1
12. (c)
Þ
lim
x ®p 2
f ( x ) = f (p 2 )
k + 2 5 =1 Þ k = 1 –
13. (c)
0
2
5
Þ
2b 2 - 4b
2x2
a
x®0
æ 1 k -1 ö
Now, lim f ( x) = lim ç - 2 x ÷
x ®0
x ®0 x
e -1 ø
è
(1¥ form)
=k
\ el = k
lim f ( x) will exist and f (0) = lim f ( x)
x®0
1
2-x
x3
a
1
2
Continuity at x = 1
2
=a Þa= ± 2
a
Continuity at x =
2 a=
2
2b - 4b
2
a=
2 2
Put a = 2
2 = b2 – 2b Þ b2 – 2b – 2 = 0
b=
2 ± 4 + 4.2 1 ± 3
=
2
So, (a, b) = ( 2,1 - 3)
Downloaded from @Freebooksforjeeneet
k=
3
5
M-321
Continuity and Differentiability
14. (c) Since f(x) is a continuous function therefore limit of
f(x) at x ® 0 = value of f(x) at 0.
=
(e x - 1)2
x ®0
æxö æ xö
sin ç ÷ log ç 1 + ÷
èk ø è 4ø
\ lim f ( x) = lim
x ®0
2æe
-1 ö
÷
x ÷ø
x
2
x ç
ç
æxö
è
´ç ÷
lim
=
x ®0
é æ x öù
æ xö è4ø
sin
log ç1 + ÷
x ê çè R ÷ø ú
4ø
è
ê
ú.
x
x
Rê
æ ö
ú
ç ÷
êë R úû
è4ø
2æe
2
-1 ö
÷ 4k
x ÷ø
x
x ç
ç
è
æ xö
x
sin log ç 1 + ÷
4ø
è
k.
x
x
k
4
on applying limit we get
4k = 12 Þ k = 3
= xlim
®0
15. (d) Since f (x) =
2 + cos x - 1
( p - x) 2
is
lim
q® 0
= lim
q®0
2 - cos q - 1
q2
(2 - cos q) - 1
q2
´
1
2 - cos q + 1
1 - cos q 1
= lim
. (Q cos 0 = 1)
q®0
2
q2
=
=
2
1
2sin 2 q / 2
2 lim sin q / 2
lim
=
2 q® 0
2 q® 0 q 2
q2
.4
4
1
4
sin x ö
æ
= 1÷
çèQ lim
ø
x® 0 x
lim1 ù
2é
x® 0
ê
ú
3x ú
9ê
sin 2
ê
2 ú
ê lim
x® 0 æ 3 x ö 2 ú
ê
çè ÷ø ú
êë
úû
2
2 é1ù 2
=
.
9 êë1úû 9
17. (a) Let f (x) = [x] + | 1 – x |, – 1 £ x £ 3
where [x] = greatest integer function.
f is not continuous at x = 0, 1, 2, 3
But in statement-2 f (x) is continuous at x = 3.
Hence, statement-1 is true and 2 is false.
1
, which is discontinous at x = 1
x -1
1
1
=
f (u) = 2
,
(
+
2)
(u - 1)
u
u +u -2
which is discontinous at u = – 2, 1
1
1
when u = – 2, then
= -2 Þ x =
2
x -1
1
=1 Þ x = 2
when u = 1, then
x -1
Hence given composite function is discontinous at three
1
points, x = 1, and 2.
2
19. (d) fog = f (g(x)) = f (1 – | x |)
m( x) =
= –1 + 1- | x | -2
= –1 + - | x | -1 = -1 + | x | +1
Let fog = y
\ y = -1 + | x | +1
Þ y=
Þ y=
æ 9ö
2
16. (b) Given that f çè ÷ø =
2
9
sin x ü
ì
= 1ý
í lim
î x®0 x
þ
=
18. (b)
Continuous at x = p
\ L.H.L = R.H.L = f (p)
Let (p – x) = q, q ® 0 when x ® p
\
=
4
æ 1 ö
lim
9 ´ 2 x®0 ç sin 2 3 x ÷
ç
÷
2 ÷
ç
2
ç æ 3x ö ÷
çè çè 2 ÷ø ÷ø
{
{
-1 + x + 1, x ³ 0
-1 - x + 1, x < 0
x, x ³ 0
- x, x < 0
LHL at (x = 0) = lim (- x) = 0
x®0
æ
ö
x
æ 1 - cos 3x ö
lim f ç
= lim ç
÷
÷
2
0
x
®
1
cos
3
x
ø
x®0 è
è
ø
x
2
æ 9 2 4ö
æ
ö
.x .
x2
1
ç
lim ç
9÷
÷
lim ç 4
= x ®0 ç
2 3x ÷ =
3
x÷
x
®
0
2
sin
2
2
ç
÷
ç sin
÷
è
2 ø
è
2ø
RHL at (x = 0) = lim ( x ) = 0
x ®0
When x = 0, then y = 0
Hence, LHL at (x = 0) = RHL at (x = 0)
= value of y at (x = 0)
Hence y is continuous at x = 0.
Clearly at all other point y continuous. Therefore, the set
of all points where fog is discontinuous is an empty set.
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EBD_8344
M-322
Mathematics
2x - 1 ö
20. (a) Let f ( x ) = [ x ] cos æç
÷
è 2 ø
We know that [x] is discontinuous at all integral points
and cos x is continuous at xÎ R.
So, check at x = n, n Î I
æ 2x - 1 ö
L.H.L = lim- [ x ] cos ç
÷p
è 2 ø
x ®n
æ 2n –1ö
π=0
= ( n – 1) cos ç
è 2 ÷ø
(Q [x] is the greatest integer function)
æ 2x - 1 ö
R.H.L = lim+ [ x ] cos ç
÷p
è 2 ø
x ®n
æ 2n - 1 ö
= n cos ç
÷p = 0
è 2 ø
Now, value of the function at x = n is
f (n) = 0
Since, L.H.L = R.H.L. = f (n)
æ 2x - 1 ö
\ f (x) = [x] cos ç
÷ is continuous for every real x.
è 2 ø
21. (d) Consider
x
£ f ( x) £ 6 - x
[ x]
x
2
=
=2
[
x
]
1
x®2
x ® 2-
\ lim
x® 2-
6- x = 2
x ®0
( at x = 0)
= lim
h®0
sin{( p + 1)( - h)} - sinh
=p+1+1=p+2
-h
R.H .L = lim+ f ( x)
( at x = 0)
= lim
x®0
x + x2 – x
h ®0
3/ 2
Þ
f (0) = lim
x®0
Hence by Sandwich theorem lim f ( x ) does not exists.
x ® 2+
Therefore f is not continuous at x = 2. Thus statement-1
is true but statement-2 is not true
22. (d) Statement - 1 is true.
It is the definition of continuity.
Statement - 2 is false.
ïì x sin (1/ x ) , x ¹ 0
23. (c) Given that f ( x ) = í
ïî 0 , x = 0
At x = 0
ì
æ 1ö ü
LHL = lim– í - h sin çè - ÷ø ý
h þ
h®0 î
= 0 × a finite quantity between – 1
and 1= 0
RHL = lim h sin
h®0+
1
=0
h
Also, f (0) = 0
Thus LHL = RHL = f ( 0)
\ f ( x) is continuous on R.
but f 2 ( x) is not continuous at x = 0
=
1
1
=
1+1 2
1
2
x e2 x - 1
( e2 x - 1) - 2 x
2x
4e 2 x
x ® 0 2( xe
[By Sandwich theorem]
x
= 1 , lim+ 6 - x = 2
Now lim+
[
]
x
x ®2
x®2
2
é0
ù
; ê form ú
0
ë
û
x(e - 1)
\ Applying, L'Hospital rule
Differentiate two times, we get
= lim
f (0) = lim
f (x) = 2
x + x2 + x
´
x
x+ x + x
f (0) = 2
Given that f(x) is continuous at x = 0
1
\p+2=q=
2
3
1
Þ p = - ,q =
2
2
1
2
25. (b) Given, f (x) = - 2 x
is continuous at x = 0
x e -1
x®0
Þ limÞ lim
L.H .L = lim- f ( x)
24. (b)
= lim
2x
2 + e 2 x .1) + e 2 x .2
é0
ù
êë 0 form úû
4e 2 x
x ® 0 4 xe 2 x
+ 2e
2x
+ 2e
2x
4e 2 x
4.e0
=
=1
x ® 0 4( xe 2 x + e 2 x )
4(0 + e0 )
= lim
26. (c) Given that f ( x) =
1 - tan x
is continuous in
4x - p
æ pö
\ f ç ÷ = lim f ( x ) = lim f ( x)
è 4ø
pp+
x®
x®
4
æp
ö
lim f ( x ) = lim f ç + h÷
è
ø
4
®
h
0
p
x®
é pù
ê0, 2 ú
ë
û
4
4
æp
ö
1 + tan h
1 - tan ç + h÷
1è4
ø
1 - tan h
= lim
, h > 0 = lim
h ®0 æ p
ö
4h
h
®
0
4 ç + h÷ - p
è4
ø
-2
tan h -2
1 é
tan q ù
= lim
=
=.
= 1ú
Q lim
4
2 êë q® 0 q
h ®0 1 - tan h 4 h
û
27. (b) Let a is a rational number other than 0, in [–5, 5],
then f (a) = a and lim f ( x) = - a
x®a
\ x ® a– and x ® a+ is tends to irrational number
\ f (x) is discontinuous at any rational number
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M-323
Continuity and Differentiability
and also f ''( x ) ³ 4
If a is irrational number, then
f (a) = – a and lim f ( x) = a
x®a
\ f (x) is not continuous at any irrational number. For x = 0,
Þ
f '(5) - f '(2)
³ 4 Þ f '(5) ³ 12 + f '(2)
5-2
Þ f '(5) ³ 17
lim f ( x) = f (0) = 0
x®0
Hence, f (5) + f '(5) ³ 28
\ f (x) is continuous at x = 0
31. (10.00)
28. (a)
2
2
y=x
y=x
y=x
f ( x + y ) = f ( x) + f ( y ) + xy 2 + x 2 y
Differentiate w.r.t. x :
f '( x + y ) = f '( x ) + 0 + y 2 + 2 xy
(0, 0)
(1, 0)
Put y = – x
f '(0) = f '( x) + x2 - 2 x2
{
f ( x) = max . x, x 2
Q lim
x®0
}
f ( x)
= 1 Þ f (0) = 0
x
\ f '(0) = 1
ì x2 ,
x<0
ï
Þ f ( x ) = í x, 0 £ x < 1
ï 2
x ³1
îx ,
29. (a) f (x) is differentiable then, f (x) is also continuous.
\ lim f ( x) = lim f ( x) = f ( p )
x®p -
Þ -1 = - K 2 Þ K 2 = 1
...(ii)
From equations (i) and (ii),
f '( x) = ( x2 + 1) Þ f '(3) = 10.
\ f ( x) is not differentiable at x = 0, 1
x ®p +
32. (a)
ì - x -1
x < -1
ï 2 ,
ï
ïp
f ( x) = í + tan -1 x, - 1 £ x £ 1
ï4
ï1
x >1
ï 2 ( x - 1),
î
y
ì 2 K ( x - p) : x £ p
\ f '( x) = í 1
x>p
î - K 2 sin x
Then, lim f ( x ) = lim f ( x) = 0
x ®p +
...(i)
x ®p –
; x£p
ì 2 K1
f ''( x) = í
K
cos
x
;
x>p
î 2
1
- ( x + 1)
2
x'
p
+ tan - 1 x
4
(–1, 0) (0, 0)
x -1
2
x
(1, 0)
Then, lim f ( x ) = lim f ( x )
x ®p +
x ®p -
Þ 2 K1 = K 2 Þ K1 =
1
2
æ1 ö
So, (K1, K2) = ç , 1÷
è2 ø
30. (b) Let f be twice differentiable function
Q f ' ( x) ³ 1
Þ
f (5) - f (2)
³1
3
Þ f (5) ³ 3 + f (2)
Þ f (5) ³ 3 + 8 Þ f (5) ³ 11
y'
It is clear from above graph that,
f (x) is discontinuous at x = 1.
i.e. continuous on R – {1}
f (x) is non-differentiable at x = – 1, 1
i.e. differentiable on R – {–1, 1}.
33. (c) LHL = lim
x ®0
sin(a + 2) x + sin x
x
æ sin( a + 2) x ö
sin x
= lim ç
=a+3
÷ ( a + 2) + lim
x ®0 è ( a + 2) x ø
x ®0 x
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EBD_8344
M-324
Mathematics
f (0) = b
1
æ
ö
ç (1 + 3h) 3 - 1 ÷
÷ =1
RHL = lim ç
h
h ®0 ç
÷÷
ç
è
ø
Q Function f(x) is continuous
\
lim f ( x ) = lim f ( x) = f (0)
x ®0 -
x ®0 +
\ a+3=1 Þ a=–2
and b = 1
Hence, a + 2b = 0
34. (a) It is given that functions f and g are differentiable
and fog is identity function.
\ (fog)(x) = x Þ f (g(x)) = x
Differentiating both sides, we get
f ¢(g(x)) × g¢(x) = 1
Now, put x = a, then
f ¢(g(a)) × g¢(a) = 1
f ¢(b) × 5 = 1
f (0 - h ) = Lim sin( p + 1)(-h) + sin(-h)
f (0–) = hLim
®0
h ®0
-h
é - sin( p + 1)h sin h ù
+
= Lim ê
h úû
-h
h® 0 ë
= Lim
h® 0
= (p + 1) + 1 = p + 2
And f (0+) = Lim
f (0 + h) =
h ®0
...(2)
h +h- h
2
h3/2
1
Lim
=
h® 0
(h) 2 éë h + 1 - 1ùû
æ 1ö
hç h2 ÷
ç ÷
è ø
h +1 -1
h + 1 + 1 Lim h + 1 - 1
´
=
h
h® 0
h + 1 + 1 h®0 h( h + 1 + 1)
= Lim
= Lim
1
f ¢(b) =
5
sin( p + 1) h
sin h
´ ( p + 1) + Lim
h ®0 h
h ( p + 1)
h® 0
1
1
1
=
=
h +1 +1 1+1 2
...(3)
Now, from equation (1),
35. (Bonus) For a constant function f(x), option (1), (3) and
(4) doesn’t hold and by LMVT theorem, option (2) is
incorrect.
36. (a) From, LMVT for x Î [–7, –1]
f (-1) + 3
f ( -1) - f ( -7)
£2 Þ
£ 2 Þ f(–1) £ 9
( -1 + 7)
6
1
2
æ 3 1ö
1
-3
and p =
\ (p, q) = ç - , ÷
2
2
è 2 2ø
39. (b) f (x) = ln (sin x), g (x) = sin–1 (e–x)
Þq=
Þ f (g(x)) = ln (sin (sin–1 e–x)) = – x
Þ f (g(x)) = – a
From, LMVT for x Î [–7, 0]
But given that (fog) (a) = b
f (0) - f ( -7)
£2
(0 + 7)
\ – a = b and f ‘ (g (a)) = a, i.e., a = – 1
f (0) + 3
£ 2 Þ f(0) £ 11
7
\ f (0) + f(–1) £ 20
37. (c) Q f(x) is non differentiable at x = 1, 3, 5
[Q |x – 3| is not differentiable at x = 3]
S f(f(x)) = f(f(1) + f(f(3)) + f(f(5))
=1+1+1=3
ì
ï
ï sin(p + 1)x + sin x
ï
x
ï
38. (c) f (x) = í
q
ï
2
ï x +x - x
ï
3
ï
î
x2
f (0–) = f (0) = f (0+) Þ p + 2 = q =
x<0
x = 0 is continuous at x = 0
x>0
Therefore, f (0–) = f (0) = f (0+) ...(1)
\ aa2 – ba – a = – a2 + a2 – (– 1)
Þ aa2 – ba – a = 1.
g ( x ) - g (c )
x -c
| f ( x ) | - | f (c ) |
Þ g '(c) = lim
x-c
x ®c
Since, f (c) = 0
| f ( x) |
Then, g '(c) = lim
x ®c x - c
f (x)
; if f (x) > 0
Þ g '(c) = lim
x®c x - c
- f ( x)
; if f (x) < 0
and g '(c) = lim
x ®c x - c
Þ g '(c) = f '(c) = – f ‘ (c)
Þ 2f '(c) = 0 Þ f '(c) = 0
Hence, g (x) is differentiable if f ¢(c) = 0
40. (c) g '(c) = lim
x ®c
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M-325
Continuity and Differentiability
41. (a) Since, f(x) = 15 – |(10 – x)|
\ g(x) = f(f(x)) = 15 – |10 – [15 – |10 – x|]|
= 15 – ||10 – x| – 5|
\ Then, the points where function g(x) is Nondifferentiable are
10 – x = 0 and |10 – x| = 5
Þ x = 10 and x – 10 = ± 5
Þ x = 10 and x = 15, 5
42. (a) Let g (x) = f ( f ( f (x))) + (f(x))2
Differentiating both sides w.r.t. x, we get
g' (x) = f '( f ( f (x))) f '( f (x)) f '(x) + 2 f (x) f '(x)
g' (1) = f '( f ( f (1))) f '( f (1)) f '(1) + 2 f (1) f '(1)
= f '( f (1)) f '(1) f '(1) + 2 f (1) f '(1)
= 3 × 3 × 3 + 2 × 1 × 3 = 27 + 6 = 33
43. (b) Since, f ¢(x) = f(x)
ì x2 ,
-2 £ x < 0
ï
0 £ x <1
= í 0,
ï 2
î 2( x - 1), 1 £ x £ 2
g¢(0–) = 0, g¢(0+) = 0, g¢(1–) = 0, g¢(1+) = 4
Þ g(x) is non-differentiable at x = 1
Þ g(x) is not differentiable at one point.
45. (b) Consider the equation,
x loge (loge x) – x2 + y2 = 4
Differentiate both sides w.r.t. x,
loge (log e x ) + x ×
log e (log e x) +
f ¢( x)
f ¢( x)
Þ f ( x ) = dx Þ f ( x ) dx = òdx
When x = e, y =
4 + e 2 . Put these values in (1),
0 + 1 - 2e + 2 4 + e 2
dy
=0
dx
2e - 1
dy
.
=
dx
2 4 + e2
46. (a) f (x) = sin |x| – |x| + 2(x – p) cos |x|
There are two cases,
Case (1), x > 0
f (x) = sin x – x + 2(x – p) cos x
2 c
=e
e
Then, from eqn (1)
Þ
f ¢(x) = cos x – 1 + 2(1 – 0) cos x – 2 sin (x – p)
f ¢(x) = 3 cos x – 2(x – p) sin x – 1
2 x
f(x) = e
e
Þ
dy
1
- 2x + 2 y
x × log e x
dx = 0
dy
1
- 2x + 2 y
dx = 0 ...(1)
log e x
f ¢( x )
Then,
=1
f ( x)
Þ ln |f(x)| = x + c
f(x) = ±ex + c
...(1)
Since, the given condition
f(1) = 2
From eqn (1) f(x) = ex + c = ecex
Then,
f(1) = ec × e1
c
Þ 2=e ×e
Then, function f(x) is differentiable for all x > 0
Case (2) x < 0
2 x
f ¢(x) = e
e
f(x) = – sin x + x + 2(x – p) cos x
Nowh(x) = f(f(x))
Þ h¢(x) = f ¢(f(x)) × f ¢(x)
h¢(1) = f ¢(2) × f ¢(1) =
44.
2
-2 £ x < 0
ïì 1 + x - 1,
Þ g(x) = í 2
2
ïî( x - 1) + | x - 1|, 0 £ x £ 2
2 2 2
e × × e = 4e
e
e
-2 £ x < 0
ìï -1,
(d) f(x) = í 2
ïî x - 1, 0 £ x £ 2
-2 £ | x | < 0
ìï -1,
Then, f(|x|) = í 2
ïî| x | -1, 0 £ | x | £ 2
2
Þ f(|x|) = x – 1, –2 £ x £ 2
f ¢(x) = – cos x + 1 – 2(x – p) sin x + 2 cos x
f ¢(x) = cos x + 1 – 2(x – p) sin x
Then, function f(x) is differentiable for all x < 0
Now check for x = 0
f ¢(0+) R.H.D. = 3 – 1 = 2
f ¢(0–) L.H.D. = 1 + 1 = 2
L.H.D. = R.H.D.
Then, function f(x) is differentiable for x = 0. So it is
differentiable everywhere
Hence, k = f
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EBD_8344
M-326
Mathematics
| x| £ 2
ïìmax{| x |, x 2 }
47. (b) Given f(x) = í
x
8
2
|
|
2
< |x | £ 4
ïî
y = x2
y = 8 - 2|x|
y = x2
y = 8 - 2|x|
y = |x| y = |x|
50.
Q f(x) is not differentiable at –2, –1, 0, 1 and 2.
\ S = {–2, –1, 0, 1, 2}
48. (c) Consider the function
{
2
f(x) = max - | x |, - 1 - x
}
Now, the graph of the function
x
1
2
x
1
2
From the graph, it is clear that f(x) is not differentiable at x
= 0, -
1 1
,
2 2
ìï(| l | et – m) sin 2t, t > 0
=í
-t
ïî (| l | e – m) (– sin 2t ), t < 0
ìï(| l | et ) sin 2t + (| l | et – m) (2cos 2t ), t > 0
f ¢ (t) = í
-t
-t
ïî| l | e sin 2t + (| l | e – m) (– 2cos 2t ), t < 0
As, f (t) is differentiable
\ LHD = RHD at t = 0
Þ |l| . sin 2 (0) + (|l|e0 – m)2 cos (0)
= |l|e–0 . sin 2 (0) – 2 cos (0) (|l|e– 0 – m)
Þ 0 + (|l| – m)2 = 0 – 2 (|l| – m)
Þ 4 (|l| – m) = 0
Þ |l| = m
So, S º (l, m) = {l Î R & m Î [0, ¥)}
Therefore set S is subset of R × [0, ¥)
ìï- x
x <1
51. (a) f (x) = í
-1
ïîa + cos ( x + b) 1 £ x £ 2
f (x) is continuous
Þ
1 ü
ì 1
, 0,
ý
Then, K = í2þ
î 2
Hence, K has exactly three elements.
49. (d)
| - h - p | (e|-h| - 1)sin | - h | - 0
=0
-h
h®0
\ RHD = LHD
Therefore, function is differentiable.
at x = 0.
Since, the function f(x) is differentiable at all the points
including p and 0.
i.e., f(x) is every where differentiable .
Therefore, there is no element in the set S.
Þ S = f (an empty set)
(a) S = {(l, m) Î R × R : f (t) = (|l|e|t| – m) sin (2|t|), t Î R
f (t) = (|l|e| t | – m) sin (2|t|)
L.H.D. = lim
f (x) =| x - p | (e|x| - 1)sin | x |
Check differentiability of f(x) at x = p and x = 0
at x = p :
| p + h- p | (e|x + h| - 1) sin | p + h | - 0
h
h ®0
R.H.D. = lim
| p - h - p | (e|p-h| - 1)sin | p - h | - 0
=0
-h
h ®0
Q RHD = LHD
Therefore, function is differentiable at x = p
at x = 0:
L.H.D = lim
| h - p | (e|h| - 1) sin | h | - 0
=0
R.H.D. = lim
h
h ®0
–1
lim f (x) = lim a + cos (x + b) = f (x)
x ®1+
x ®1-
Þ –1 = a + cos–1 (1 + b)
cos–1 (1 + b) = – 1 – a
f (x) is differentiate
Þ LHD = RHD
Þ –1=
.....(a)
-1
1 - (1 + b) 2
Þ 1 – (1 + b)2 = 1 Þ b = – 1
From (a) Þ cos–1(0) = – 1 – a
\
–1–a=
a = –1 –
\
p
2
-p - 2
p
Þ a=
2
2
.....(c)
a
p+2
=
b
2
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.....(b)
M-327
Continuity and Differentiability
52. (c) Since g (x) is differentiable, it will be continuous at x = 3
\ lim g(x) = lim g(x)
+
x ®3
x®3
2k = 3m + 2
Also g(x) is differentiable at x = 0
...(1)
For g(x)
\ lim– g¢ (x) = lim+ g¢(x)
x ®3
ì
æ 1ö ü
LHL = lim í - h 2 sin ç ÷ ý
è hø þ
h® 0 - î
x ®3
k
2 3 +1
1
ì 2
ï x sin , x ¹ 0
x
g(x) = í
ïî0,
x=0
=m
= 02 × a finite quantity between –1 and 1 = 0
k= 4m
Solving (1) and (2), we get
æ 1ö
RHL = lim h2 sin ç ÷ = 0
+
è hø
h® 0
...(2)
Also g(0) = 0
\ g(x) is continuous at x = 0
2
8
m= , k=
5
5
k+m =2
55. (c)
53. (b) Let | f (x) | £ x2, "x Î R
Now, at x = 0, | f (0) | £ 0
Þ f (0) = 0
f ( h ) - f (0)
f (h)
= lim
\ f ¢ (0) = lim
h-0
h® 0
h® 0 h
Now,
f (h)
£|h|
h
ìx - 2 ,
=í
î2 - x ,
...(1)
2
(Q | f ( x ) | £ x )
f (h)
£| h |
Þ - | h |£
h
f (h)
®0
...(2)
h
(using sandwich Theorem)
\ from (1) and (2), we get f ¢(0) = 0,
i.e. – f (x) is differentiable, at x = 0
Since, differentiability Þ Continutity
\ | f (x) | £ x2, for all x Î R is continuous as well as
differentiable at x = 0.
Þ lim
h® 0
ì
æ 1ö
x sin ç ÷ , x ¹ 0
54. (b) f (x) = ïí
è xø
ï0,
x=0
î
and g(x) = x f (x)
For f (x)
ì
æ 1ö ü
LHL = lim- í - h sin ç - ÷ ý
è hø þ
h® 0 î
= 0 × a finite quantity between –1 and 1 = 0
RHL = lim+ h sin
h® 0
1
=0
h
Also, f (0) = 0
Thus LHL = RHL = f(0)
\ f (x) is continuous at x = 0
ìx - 2 ,
f ( x) = x - 2 = í
î2 - x ,
x-2³0
x-2£0
x³2
x£2
Similarly,
ìx - 5 , x ³ 5
f ( x) = x - 5 = í
î5 - x , x £ 5
\
f ( x) = x - 2 + x - 5
= { x - 2 + 5 - x = 3, 2 £ x £ 5}
Thus f (x) = 3 , 2 £ x £ 5
f ¢ (x) = 0 , 2 < x < 5
f ¢ (4) = 0
Q Statement-1 is true
Y
2
5
X
Since f (x) = 3, 2 £ x £ 5 is constant function.
So, it continuous in 2, 5 and differentiable in (2, 5)
Q f (2) = 0 + |2 – 5| = 3
and f (5) = |5 – 2| + 0 = 3 statement-2 is also true.
56. (d) |sin x| and e|x| are not differentiable at x = 0 and |x|3
is differentiable at x = 0.
\ for f (x) to be differentiable at x = 0, we must have
a = 0, b = 0 and c is any real number.
57. (b) Given x + | y | = 2y
Þ x + y = 2y or x – y = 2y
Þ x = y or x = 3y
Downloaded from @Freebooksforjeeneet
EBD_8344
M-328
Mathematics
This represent a straight line which passes through origin.
Hence, x + | y | = 2y is continuous at x = 0.
Now, we check differentiability at x = 0
x + | y | = 2y Þ x + y = 2y, y ³ 0
x – y = 2y, y < 0
ìï x,
Thus, f ( x ) = í x
,
îï 3
Now, L.H.D. = lim
y < 0üï
ý
y ³ 0ï
þ
h ®0
R.H.D = lim
h®0+
Y
y=–x+1
(0, 1)
X
(–1, 0)
Y’
h
2 xf (a) - a2 f ¢ ( x)
= xlim
= 2af (a) - a2 f '(a)
®a
1
ì
æ 1 ö
, if x ¹ 1
ï(x – 1) sin ç
59. (c) Given that, f (x) = í
è x – 1÷ø
ïî 0
, if x = 1
At x = 1
f (1 + h) – f (1)
h
1
h sin – 0
1
h
= lim
= lim sin = a finite number
h
h
h ®0
h ®0
Let this finite number be l
f (1 - h) - f (1)
-h
æ 1 ö
-h sin ç ÷
æ 1 ö
è -h ø
= lim
= lim sin ç
÷
h ®0
è -h ø
-h
h ®0
L.H.D. = lim
h®0
æ 1ö
= - lim sin çè ÷ø = – (a finite number) = – l
h
h®0
Thus R.H.D ¹ L.H.D
\ f is not differentiable at x = 1
y=x+1
X’
f ( x + h) - f ( x )
x 2 f ( a ) - a 2 f ( x)
(c) xlim
®a
x-a
Applying L-Hospital rule
h®0
\ f is differentiable at x = 0
60. (a) f (x) = min {x + 1, | x | + 1}
Þ f (x) = x + 1 " x Î R
x+h-x
= -1
-h
Since, L.H.D ¹ R.H.D. at x = 0
\ given function is not differentiable at x = 0
R.H.D. = lim
1
x –1
æ 1 öù
–
cos ç
÷ú
2
(x –1) (x –1)
è x – 1 ø úû x = 0
= –sin 1 + cos 1
-h
x+h x
3
3 = lim 1 = 1
= lim
h
h®0+ 3 3
h® 0+
58.
f '(0) = sin
f ( x + h) - f ( x )
h®0-
= lim-
At x = 0
Since f (x) = x + 1 is polynomial function
Hence, f (x) is differentiable everywhere for all x Î R.
61. (c)
ì x
ïï1 - x , x < 0
f ( x) = í
ï x , x³0
îï1 + x
f(x) =
x
is not define at x ¹ 1 but here x < 0 and f (x)
1- x
x
is not define at x = –1 but here x > 0. So, f (x) is
1+ x
continuous for x Î R.
=
ì x
, x<0
ï
2
ï (1 - x)
and f '( x ) = í
ï x , x³0
ïî (1 + x) 2
\
f '( x) exist at everywhere.
62. (b) Given that | f (x) – f (y) | £ (x – y)2, x, y Î R ...(i) and
f (0) = 0
f ( x + h ) - f ( x)
f '(x) = lim
h
h®0
f ( x + h ) - f ( x)
(h)2
| f '( x) | = lim
£ lim
h
h®0
h®0 h
Þ | f '( x) | £ 0 Þ f '( x) = 0
Þ f (x) = constant
As f (0) = 0
Þ f (1) = 0.
f (1 + h) - f (1)
;
h
Given that function is differentiable so it is continuous
also
63. (c)
f '(1) = lim
h®0
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M-329
Continuity and Differentiability
and lim
h®0
f (1 + h)
= 5 and hence f (1) = 0
h
Hence, f '(1) = lim
h®0
f (1 + h)
=5
h
( - 2b sin x)(a - 2b cos y) + (a + 2b cos x)
64. (c) Given that f (0) = 0; f ( x ) = xe
æ 1 1ö
-ç + ÷
è x xø
h
R.H.L. = lim (0 + h)e -2 / h = lim
h ®0 e 2 / h
h® 0
L.H.L. = lim (0 - h)e
æ 1 1ö
-ç - ÷
è h hø
=0
( 2b sin y )
Þ
pö
è 4 4 ø÷
therefore, f (x) is continuous at x = 0.
Now, R.H.D = lim
(0 + h)e
æ 1 1ö
-ç + ÷
è h hø
h® 0
L.H.D. = lim
-0
h
(0 - h)e
æ 1 1ö
-ç - ÷
è h hø
-0
-h
¹
therefore, L.H.D. R.H.D.
f (x) is not differentiable at x = 0.
h® 0
=0
=1
Put x = tan q Þ q = tan -1 x
qö
æ sec q - 1 ö
-1 æ
\ u = tan -1 ç
÷ = tan ç tan ÷
2ø
è tan q ø
è
q 1
= = tan -1 x
2 2
du 1
1
\
= ´
dx 2 (1 + x 2 )
æ 2 x 1 - x2
Let v = tan -1 ç
ç 1 - 2x2
è
ö
÷
÷
ø
Put x = sin f Þ f = sin -1 x
-1 æ
2 sin f cos f ö
-1
v = tan ç
÷ = tan (tan 2f)
è cos 2f ø
= 2f = 2sin -1 x
dv
1
=2
dx
1 - x2
du du / dx
1 - x2
=
=
dv dv / dx 4(1 + x 2 )
3
æ du ö
\ç ÷
=
è dv øçæ x = 1 ÷ö 10
è
2ø
=
a-b
dx a + b
Þ
=
a+b
dy a - b
6
4
ü
-1 ì 3
67. (91) y = å k cos í 5 cos kx - 5 sin kx ý
î
þ
k =1
Let cos a =
æ 1 + x2 - 1 ö
÷
65. (d) Let u = tan -1 ç
ç
÷
x
è
ø
dy
=0
dx
dy ( 2b sin x )(a - 2b cos y )
=
dx (a + 2b cos x )( 2b sin y )
é dy ù
\ê ú
ë dx û æç p ,
=0
h® 0
(a + 2b cos x)(a - 2b cos y) = a 2 - b 2
Differentiating both sides,
66. (c)
3
4
and sin a =
5
5
6
\ y = å k cos -1{cos a cos kx - sin a sin kx}
k =1
6
= å k cos -1 (cos(kx + a))
k =1
6
6
k =1
k =1
= å k (kx + a) = å (k 2 x + ak )
\
6
dy
6(7)(13)
= å k2 =
= 91.
dx k =1
6
68. (Bonus) It is given that
x = 2sinq – sin2q
y = 2cosq – cos2q
Differentiating (i) w.r.t. q, we get
...(i)
...(ii)
dx
= 2cos q - 2cos2q
dq
Differentiating (ii) w.r.t. q; we get
dy
= -2sin q + 2sin 2q
dq
From (ii) 揤 (i), we get
dy sin 2q - sin q
=
dx cos q - cos2q
\
3q
q
2sin .cos
2
2 = cot 3q
=
q
3q
2
2sin .sin
2
2
Again, differentiating eqn. (iii), we get
d2y
dx
2
=
3q d q
-3
cosec 2 .
2
2 dx
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...(iii)
EBD_8344
M-330
Mathematics
Putting y = 1 and x = 0 in (ii),
-3
3q
cosec 2
d y
2
= 2
dx 2 2(cos q - cos 2q)
2
d2y
2
dx (q = p)
69. (a)
=-
dy
1
dy
=+0+1=0Þ
dx
dx
e
On differentiating (ii) w. r. to x,
e
3
3
=
4(-1 - 1) 8
ey
2sin a cos a
+
2
cos a sin a = 2 cos a + 1
2
sin a cos a sin 2 a
sec a
y (a ) =
= 2cot a + cosec 2 a = 2cot a + 1 + cot 2 a
dy
æ dy ö
=4
= cosec 2a Þ ç
÷
da
è d a øa= 5p
1
-1
-1
Þ xy =
70. (b) Given, x = , y =
2
4
8
y.
2 1 - x2
+ y¢ 1 - x
xy
+ y¢ 1 - x 2 = - 1 - y 2 +
Þ
æ
xy
y¢ ç 1 - x 2 ç
1- y2
è
Þ
Þ
1- x
2
æ
ç 3
1
+
y¢ç
2
15
ç
8.
ç
è
4
xy. y ¢
1- y
ey
æ 45 + 1 ö
(1 + 45)
y¢ çç
÷÷ = 4 3
è 2 15 ø
dy
dy
+ x + y = 0 ...(ii)
dx
dx
2
ö
÷ = xy - 1 - y 2
÷
1 - x2
ø
ö
÷
-1
15
÷=
4
3
÷
÷ 8.
ø
2
y¢ = -
1
dy
= - in (iii),
dx
e
d2y
1 2
d2y 1
+
=
=
0
Þ
dx 2 e 2
dx 2 e e
é p
æ p p öù
êQ 4 - x Î ç - 4 , 4 ÷ ú
è
øû
ë
p
4
Now, differentiate w.r.t. y,
5
2
71. (b) Given, ey + xy = e
...(i)
Putting x = 0 in (i), Þ
ey = e Þ y = 1
On differentiating (i) w. r. to x
\
e
Let y = Þ f (y) = 2 y -
ì
x.(-2 y ) üï
ï
y¢ ý
= - í1. 1 - y 2 +
ïî
2 1 - y 2 ïþ
-
Putting y = 1, x = 0 and
...(iii)
p
æp
ö
So, f (x) = - ç - x ÷ = x 4
è4
ø
2
Þ
dy y dy
d 2 y dy dy
.e . + x 2 + +
=0
dx
dx
dx dx
dx
æ
æp
öö
tan
- x ÷÷
= – tan–1 çè çè 4
øø
6
1.( -2 x )
dx 2
+
æ dy d 2 y ö æ 1 1 ö
ç
Hence, ç dx , 2 ÷÷ º ç - e , 2 ÷
dx ø è
e ø
è
æ tan x - 1 ö
72. (d) f (x) = tan–1 çè tan x + 1 ÷ø
é
æ 3p öù
êQ a Î ç 4 , p ÷ú
è
øû
ë
= |1 + cot a| = – 1 – cot a
d2y
df ( y )
= 2.
dy
é
æ 3
öù
1
cos x + sin x ÷ ú
ê
ç
1
2
2
ê cot ç
÷ú
73. (none) 2y = ê
3
ç1
÷ú
x
x
cos
sin
ç
÷ú
êë
è2
øû
2
é
æ
æp
ö öù
ê
ç cos ç - x ÷ ÷ú
è6
ø ÷ú
ê cot -1 ç
Þ 2y = ê
ç sin æ p - x ö ÷ú
ç
÷÷
ç
ê
è6
ø øúû
è
ë
2
2
2
é -1 æ æ p
ö öù
Þ 2y = ê cot ç cot ç - x ÷ ÷ ú Q
ø øû
è è6
ë
p
æ p pö
- x Îç- , ÷
6
è 3 6ø
ìæ 7p
p
ö
æ -p ö
,0 ÷
ïç
- x ÷ , if - x Î ç
ïè 6
ø
è 3 ø
6
í
2
Þ 2y = ïæ p
p
ö
æ pö
x
, if - x Î ç 0, ÷
ç
÷
ï
6
ø
è 0ø
îè 6
2
Þ
ì
7p
ïx - 6
ï
dy í
= ï
p
dx
xïî
6
æp pö
if x Î ç , ÷
è6 2ø
æ pö
if x Î ç 0, ÷
è 6ø
Downloaded from @Freebooksforjeeneet
M-331
Continuity and Differentiability
74. (b)
f "'(x) = 6 Þ f "'(3) = 6
f(x) = min{sinx, cosx}
y = sin x
y
y=1
-
p
2
3p
4
–p
-
p
2
p
4
p
x
Q
f(x) = x3 + f '(1)x2 + f " (2) x + f "'(3)
\
f '(1) = a Þ 3 + 2a + b = a Þ a + b = – 3
...(1)
also f "(2) = b Þ 12 + 2a = b Þ 2a – b = – 12
...(2)
and f "(3) = c Þ c = 6
Add (1) and (2)
3a = – 15 Þ a = – 5 Þ b = 2
y = –1
y = cos x
Q
f(x) is not differentiable at x = -
\
ì 3p p ü
S = í- , ý
î 4 4þ
3p p
,
4 4
Þ f(x) = x3 – 5x2 + 2x + 6
Þ f(2) = 8 – 20 + 4 + 6 = –2
77. (b) Q x = 3 tant Þ
and y = 3 sect Þ
ì 3p -p 3p p ü
, , ý
Þ S Í í- ,
î 4 4 4 4þ
75. (a) Consider the equation,
dy dy / dt
dy tan t
= sin t
=
\
=
dx dx / dt
dx sec t
\
d2y d
dt
(sin t) ×
2 =
dt
dx
dx
Taking log on both sides
= cos t ×
…(1)
Differentiating both sides w.r.t. x,
2y
2
=
...(2)
1 æ ln 2 + x ö
dy
(1 + ln 2 x) = 1 - ç
x è 1 + ln 2 x ÷ø
dx
Þ
(1 + ln 2 x)2
æ x + ln 2 ö
dy
= 1 + ln(2 x) - çè
÷
x ø
dx
x ln(2 x) - ln 2
=
x
3
2
76. (c) Let f(x) = x + ax + bx + c
f '(x) = 3x2 + 2ax + b Þ f '(1) = 3 + 2a + b
f " (x) = 6x + 2a Þ f "(2) = 12 + 2a
dy
=
dt
\
3
1
6 2
dx
=
dt
78. (b) Here,
From (1) and (2),
1
3 sec 2 t
pö 1 æ 1 ö
d2y æ
ç at t = ÷ø = × çè
÷
4
3
dx 2 è
2ø
\
dy
dy
1
2 + 2 ln(2 x ) = 0 + 2 - 2
dx
dx
2x
2 y 2x - 2 y
dy
(1 + ln(2 x ) = 2 =
x
x
dx
dy
= 3 sect × tant
dt
Q
(2x)2y = 4e2x–2y
2y ln(2x) = ln4 + (2x – 2y)
dx
= 3 sec2t
dt
1
2 2
sec –1t
1
2 2
cosec –1t
2sec –1t log 2.
2cosec –1t log 2.
1
x x2 -1
dy
–1t
–1t
2sec
dy dt – 2cosec
=
=
–1t
–1t
dx dx
2cosec
2sec
dt
dy
dy dt
=
=–
dx dx
dt
2sec –1t
2
cosec –1t
=
–y
x
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-1
x x2 - 1
EBD_8344
M-332
Mathematics
cos x
x
1
2
2x
æ dy ö
Þç ÷
=4
è dx ø( -2,0)
x
1
Again differentiating equation (1) w. r. t to x, we get
79. (a) f (x) = 2 sin x x
tan x
= cos x (x2 – 2x2) – x (2 sin x – 2x tan x)
+1(2x sin x – x2 tan x)
= – x2 cos x – 2x sin x + 2x2 tan x + 2x sin x – x2 tan x
= x2 tan x – x2 cos x = x2 (tan x – cos x)
Þ f ¢(x) = 2x (tan x – cos x) + x2 (sec2 x + sin x)
f ¢ ( x)
\ lim
=
x ®0
x
2x (tan x – cos x) + x2 (sec2 x + sin x)
x® 0
x
lim
= 2 (0 – 1) + 0 = – 2
x®0
-1
Þ (2 y + cos y)
dx 2
æ 2 ´ 3x ö
çç
x ÷
÷
è1+ 9 ø
Þ
æ 2 tan t ö
= sin–1 (sin 2t) = 2t
ç
2 ÷
t
+
1
tan
è
ø
Þ
= 2 tan–1 (3x)
2
æ -1 ö
1+ ç3 2 ÷
ç
÷
è
ø
2
1
´3 2
=
. loge 3
dy
dy
+ cos y
=0
dx
dx
dy
Þ 2 x + (2 y + cos y )
=0
dx
dy
– 2x
=
dx 2 y + cos y
– 2´ – 2
æ dy ö
=
At ( – 2, 0), ç ÷
è dx ø( -2,0) 2 ´ 0 + cos 0
4
æ dy ö
Þç ÷
=
è dx ø( -2,0) 0 + 1
2
2
d2y
dx
2
d2y
dx 2
=
– 2 – 2 ´ 16
1
= – 34
æ 2.(3x 3/ 2 ) ö
–1
3/2
÷
= tan–1 çç
3/ 2 2 ÷ = 2 tan (3x )
è 1 - (3x ) ø
æ 3ö
As 3x3/2 Î ç 0, ÷
è 8ø
1
´ 3 ´ log e 3
= 3 ´ loge 3
2
81. (a) Given, x2 + y2 + sin y = 4
After differentiating the above equation w. r. t. x we get
Þ
æ dy ö
= - 2 – (2 – sin y ) ç ÷
dx
è dx ø
2
– 2 – (2 – 0) ´ 42
2 ´ 0 +1
1
3ù
é
3/ 2 1
< Þ 0 < 3x 3/ 2 < ú
êQ 0 < x < 4 Þ 0 < x
8
8û
ë
=
2x + 2 y
d2y
æ 6x x ö
æ 1ö
82. (b) Let F(x) = tan–1 çç
3 ÷÷ where x Î ç 0, ÷ .
è 1 - 9x ø
è 4ø
2
´ 3x . log e 3
So, f ¢ (x) =
1 + (3x ) 2
æ 1ö
\ f ¢ç - ÷ =
è 2ø
2
d2y
æ dy ö
Þ 2 + (2 - sin y ) ç ÷ + (2 y + cos y) 2 = 0
dx
è dx ø
d2y
Suppose 3x = tan t
Þ f (x) = sin–1
2
So, at (– 2, 0),
f ¢ (x)
=–2
x
80. (a) Since f (x) = sin
2
d2 y
d2y
æ dy ö
æ dy ö
2 + 2 ç ÷ + 2 y 2 – sin y ç ÷ + cos y 2 = 0
dx
dx
è dx ø
è dx ø
æ dy ö
– 2 – (2 – sin y ) ç ÷
d2y
è dx ø
Þ 2 =
2 y + cos y
dx
= xlim
(tan x – cos x) + x (sec2 x + sin x)
®0
So, lim
... (2)
... (1)
1
9
dF(x)
3
x
=2×
× 3 × × x1/2 =
3
dx
2
1 + 9x
1 + 9x 3
On comparing
So
\ g(x) =
9
1 + 9x 3
83. (d) g (x) = f (f (x))
In the neighbourhood of x = 0,
f(x) = | log2 – sin x| = (log 2 – sin x)
\ g (x) = |log 2 – sin| log 2 – sin x ||
= (log 2 – sin(log 2 – sin x))
\ g (x) is differentiable
and g'(x) = – cos(log 2 – sin x) (– cos x)
Þ g'(0) = cos (log 2)
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M-333
Continuity and Differentiability
84. (c) f (x) = y = x2 – x + 5
1 1
x2 – x + - + 5 = y
4 4
87. (d) Let x = a sin
-1
1ö
19
æ
=y
çè x - ÷ø +
2
4
x-
2
= y-
g¢(x) =
g¢(7) =
1
2
1
19
+ x2
4
1
19
4
1
=
2
28 - 19
2
=
1
3
2 dy - log a dt
. =
.
y dx
1 - t 2 dx
Þ
2 dy - log a 2 1 - t 2
. =
´
y dx
x log a
1- t2
Þ
dy
y
=dx
x
Þ
Þ
Þ
Q
Þ
Þ
2
dy - 4 x
=
dx
ay
...(i)
dy
dy 16
= 16 Þ
=
dx
dx 3 y 2
Since curves intersects at right angles
- 4 x 16
´
= -1 Þ 3ay3 = 64x
ay 3 y 2
64 x
4
=
Þ a=
3 ´ 16 x 3
x2 - x
x2 + 2 x
(x + 2x) y = x2– x
x(x + 2)y = x(x – 1)
x[(x + 2)y – (x – 1)] = 0
x ¹ 0, \ (x + 2)y – (x – 1) = 0
xy + 2y – x + 1 = 0
x(y – 1) = – (2y + 1)
2
2x + 1
2 y +1
Þ f –1(x) =
1- x
1- y
2(1 - x) - (2 x + 1)( -1)
d
( f -1 ( x )) =
dx
(1 - x) 2
2 - 2x + 2 x + 1
3
=
=
(1 - x) 2
(1 - x ) 2
\
2 x 2 y dy
x2 y 2
+
. =0
+
=1 Þ
a
4 dx
a
4
y3 = 16x Þ 3 y 2 .
\
Þ
88. (b) Let y =
Þ y = 1 + x2
1
dy
· 2x
Þ
=
dx
2 1 + x2
At x = 1,
1
dy
=
.
dx
2
Þ
...(1)
x2 + y2
æ dy ö
æ -y ö
Hence, 1 + ç ÷ = 1 + ç
÷ =
è dx ø
è x ø
x2
æ
ö
85. (a) Let y = sec(tan–1 x) = sec ç sec –1 1 + x 2 ÷
è
ø
86. (b)
2 1 - t 2 dt
=
x log a
dx
2
19
2 x4
1
2 7-
Þ
-1
1
19
+ y2
4
g(x) =
2
log a dt
.
=
x
1 - t 2 dx
Now, let y = acos t
Þ 2 log y = cos–1 t. log a
1
19
x = ± y2
4
x=
Þ
19
4
1
19
= ± y2
4
As x >
t
t
Þ x2 = a sin
Þ 2 log x = sin–1 t . log a
2
1ö
æ
çè x - ÷ø
2
-1
...(ii)
x=
æ 2 x + 3ö
89. (c) Let f ¢(x) = sin [log x] and y = f ç
è 3 - 2 x ÷ø
Now,
dy
æ 2 x + 3 ö d æ 2 x + 3ö
= f 'ç
.
è 3 - 2 x ÷ø dx çè 3 - 2 x ÷ø
dx
é æ 2 x + 3ö ù éë( 6 - 4 x ) - ( -4 x - 6) ùû
= sin êlog ç
÷ú
ë è 3 - 2x ø û
(3 - 2 x)2
=
12
( 3 - 2 x)
2
é æ 2 x + 3ö ù
.sin ê log ç
÷ú
ë è 3 - 2x ø û
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(from (1)
EBD_8344
M-334
Mathematics
90. (a) Given that g(x) = [f (2 f (x)) + 2]2
\
æ d
ö
g '( x ) = 2 ( f (2 f ( x) + 2) ) ç ( f (2 f ( x) + 2) ) ÷
è dx
ø
= 2 f (2 f ( x) + 2) f '(2 f ( x)) + 2).(2 f '( x))
Þ
94. (b)
91. (d) x2x – 2xx cot y – 1 = 0
Þ 2 cot y = xx – x – x
Let u = xx
1
Þ 2 cot y = u –
u
Differentiating both sides with respect to x, we get
Now f '(a ); f '(b ) and f '(c)
are 2 a( a ); 2a (b); 2a (c )
i.e. 2a2, 2ab, 2ac.
Þ If a, b, c are in A.P. then f '(a); f '(b) and f '(c ) are
also in A.P.
95. (c) Given that f (x + y) = f (x) ´ f (y)
Differentiate with respect to x, treating y as constant
1 du
= 1 + log x
u dx
f ¢ (x + y) = f ¢ (x) f (y)
Putting x = 0 and y = x, we get f '(x)= f '(0) f (x) ;
du
= x x (1 + log x)
dx
\ We get
Þ
Put n = 1 in eqn. x2x – 2xx cot y – 1 = 0, gives
1 – 2 cot y – 1 = 0
Þ cot y = 0
\ Putting x = 1 and cot y = 0 in eqn. (i), we get
y ' (1) =
(1 + 1) (1 + 0)
-2(1 + 0)
= -1
92. (a) x . y n = ( x + y ) m+ n
taking log both sides
Þ m ln x + n ln y = (m + n) ln (x + y)
Differentiating both sides, we get
m
\
Þ
Þ f ¢ (5) = 3 f (5) = 3 × 2 = 6.
96. (b) Let f : R ® R, with f (0) = f (1) = 0 and f '(0) = 0
Q f ( x) is differentiable and continuous and
f (0) = f (1) = 0.
dy
= (1 + x -2 x ).x x (1 + log x)
dx
)
m n dy m + n æ dy ö
+
=
ç1 + ÷
x y dx x + y è dx ø
æ m m + n ö æ m + n n ö dy
çè x - x + y ø÷ = çè x + y - y ÷ø dx
Þ
my - nx æ my - nx ö dy
=
x( x + y ) èç y ( x + y ) ø÷ dx
Þ
dy y
=
dx x
f ( x) = ax 2 + bx + c
\ f ( x) = ax 2 + c or f '( x ) = 2ax
Now u = xx Taking log both sides
Þ log u = x log x
(
dy 1
1- x
= -1 =
dx x
x
1 dy
=
+1
x dx
Þ a + b + c = a - b + c or b = 0
dy æ
1 ö du
– 2cosec y
= ç1 +
÷
dx è u 2 ø dx
x x + x - x (1 + log x)
dy
=
Þ
dx
-2(1 + cot 2 y )
Þ x = e y+ x .
f (1) = f (-1)
2
– 2 cosec2 y
y +L¥
log x = y + x differentiating both side Þ
\
g '(0) = 2 f (2 f (0) + 2. f '(2 f (0) + 2)
2
.2 f '(0) = 4 f (0)( f '(0)) 2 = 4(–1)(1) = – 4
Þ
93. (c) Given that x = e y + e
Taking log both sides.
…(i)
Then by Rolle's theorem, f '(c) = 0, c Î (0, 1)
Now again
Q f '(c ) = 0, f '(0) = 0
Then, again by Rolle's theorem,
f ''( x) = 0 for some x Î (0, 1)
97. (c)
y 2 + 2log e (cos x) = y
Þ 2 yy '- 2 tan x = y '
From (i), y(0) = 0 or 1
\ y '(0) = 0
Again differentiating (ii) we get,
2( y ')2 + 2 yy ''- 2sec 2 x = y ''
Put x = 0, y(0) = 0, 1 and y'(0) = 0,
we get, |y'' (0)| = 2.
98. (b) Since, Rolle’s theorem is applicable
\ f(a) = f(b)
f(3) = f(4) Þ a = 12
f ¢( x ) =
x 2 - 12
x ( x 2 + 12)
As f ¢(c) = 0 (by Rolle’s theorem)
1
x = ± 12 , \ c = 12 , \ f ¢¢(c) = 12
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...(i)
...(ii)
M-335
Continuity and Differentiability
99. (c)
Þ
Þ
Þ
k .x k -1 + k . y k -1
æxö
dy
= -ç ÷
dx
è yø
dy æ x ö
+ç ÷
dx è y ø
k -1 = -
dy
=0
dx
k -1
1
3
f (1) - f (0) 16 - 11
=
1- 0
1
= 3c2 – 8c + 8
Þ 3c2 – 8c + 3 = 0
c=
27 a = 18a 2
3
, b = 0, c = 0, d = 0
2
f (x) =
3 3
x ,
2
9
f ¢( x ) = x 2 , f ' ( x ) = 9x
2
Þ f ¢ (2) = 18
and f ² (2) = 18
Þ f ² (b) – f ¢ (b) = 0
{
} + {x -
2
103. (d) y = x + x - 1
æ 1 -4 5 1 -6 5 ö dy
- y
ç5 y
÷ dx = 2
5
è
ø
Þ
y¢ y1 5 - y -1 5 = 10 y
Þ
y1/5 + y -1/5 = 2x
(
1/5
)
-y
-1/5
2
= 4x - 4
Þ
y
Þ
y¢ çæ 2 x 2 - 1 ÷ö = 10 y
è
ø
Þ
2x
y ¢¢ çæ 2 x 2 - 1 ÷ö + y ¢2
= 10 y ¢
è
ø
2 x2 - 1
Þ
y ¢¢ ( x 2 - 1) + xy ¢ = 5 x 2 - 1 ( y ¢)
Þ
y ¢¢( x 2 - 1) + xy ¢ - 25 y = 0
l = 1, k = -25
Let f ( x) = ax3 + bx 2 + cx + d
Þ
f (3x) = 27ax3 + 9bx 2 + 3cx + d
Þ
f ¢ ( x ) = 3ax 2 + 2bx + c
15
}
x2 -1
15
Differentiate w.r.t. 'x'
(
)
14 é
dy
= 15 x + x 2 - 1 ê1 +
dx
êë
ù
ú
x - 1 úû
x
2
(
+15 x - x 2 - 1
dy
=
dx
Þ
Þ
102. (b)
Þ
8± 2 7 4± 7
=
6
3
4- 7
Î (0,1)
3
101. (b) y1/5 + y–1/5 = 2x
\
f ( 3x ) = f ¢ ( x ) f ¢¢ ( x )
Þ
f ¢(c) =
c=
Þ
=0
1 2
k =1- =
3 3
100. (b) Since, f (x) is a polynomial function.
\ It is continuous and differentiable in [0, 1]
Here, f (0) = 11, f(1) = 1 – 4 + 8 + 11 = 16
f ¢ (x) = 3x2 – 8x + 8
Þ
f ¢¢ ( x ) = 6ax + 2b
Þ a=
k -1
Þ
\
Þ
15
x2 - 1
.y
)
14 æ
ç1 è
ö
÷
x -1ø
x
2
....(i)
dy
= 15 y
dx
Again differentiating both sides w.r.t. x
x2 - 1 .
Þ
dy
d2y
dy
+ x 2 - 1 2 = 15
2
dx
dx
dx
x -1
x
Þ
.
x
dy
d2y
+ ( x2 - 1) 2
dx
dx
= 15 x 2 - 1 .
15
. y = 225 y
x2 - 1
104. (b) Conduction for Rolls theorem
f (1) = f (– 1)
æ 1ö
and f ¢ ç ÷ = 0
è 2ø
1
2
2b + c = – 1
105. (b) Since, f and g both are continuous function on [0, 1]
c = –2 and b =
and differentiable on (0, 1) then $ c Î (0,1) such that
f ¢(c ) =
f (1) - f (0) 6 - 2
=
=4
1
1
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EBD_8344
M-336
Mathematics
f (a) = f (1) = 2 (1)3 + a (1)2 + b (1) = 2 + a + b
f (b) = f(–1) = 2 (–1)3 + a (–1)2 + b (–1)
= –2 + a – b
f(a) = f(b)
2+a+b=–2+a–b
2b = – 4
b=–2
1
(given that c = )
2
f ¢ (x) = 6x2 + 2ax + b
g (1) - g (0) 2 - 0
=
=2
and g ¢(c ) =
1
1
Thus, we get
f ¢ ( c ) = 2 g ¢ ( c)
106. (c) Let f (x) = x|x| = x|x|, g(x) = sin x
and h (x) = gof (x) = g[f (x)]
2
ïì sin x ,
í
\ h(x) =
2
ïî - sin x ,
x³0
x<0
2
ïì 2 x cos x , x ³ 0
í
Now, h¢ (x) =
2
ïî -2 x cos x , x < 0
at x =
Since, L.H.L and R.H.L at x = 0 of h¢ (x) is equal to 0
therefore h¢ (x) is continuous at x = 0
Now, suppose h¢ (x) is differentiable
2
2
2
ïì2(cos x + 2 x ( - sin x ), x ³ 0
\ h¢¢(x) = í
2
2
2
ïî 2( - cos x + 2 x sin x ), x < 0
Since, L.H.L and R.H.L at x = 0 of h¢¢ (x) are different
therefore h¢¢ (x) is not continuous.
Þ h¢¢(x) is not differentiable
Þ our assumption is wrong
Hence h¢(x) is not differentiable at x = 0.
107. (a) Let p1(x) = a1x2 + b1x + c1
p2(x) = a2x2 + b2x + c2
and p3(x) = a3x2 + b3x + c3
where a1, a2, a3, b1, b2, b3, c1, c2, c3 are real numbers.
\
é a1 x 2 + b1x + c1
ê
A(x) = ê a2 x 2 + b2 x + c2
ê
ê a3 x 2 + b3 x + c3
ë
2a1x + b1
2a2 x + b2
2a3 x + b3
é a1x 2 + b1x + c1 a2 x 2 + b2 x + c2
ê
B(x) = ê 2a1x + b1
2a2 x + b2
ê 2a
2
a2
ë 1
é a1x 2 + b1x + c1
ê
´ ê a2 x 2 + b2 x + c2
ê
ê a3 x 2 + b3 x + c3
ë
2a1x + b1
2a2 x + b2
2a3 x + b3
2a1 ù
ú
2 a2 ú
ú
2a3 ú
û
a3 x 2 + b3 x + c3 ù
ú
2a3 x + b2
ú
ú
2a3
û
2a1 ù
ú
2a2 ú
ú
2a3 ú
û
It is clear from the above multiplication, the degree of
determinant of B(x) can not be less than 4.
108. (b) f (x) = 2x3 + ax2 + bx
let, a = – 1, b = 1
Given that f (x) satisfy Rolle’s theorem in interval [–1, 1]
f (x) must satisfy two conditions.
(1) f (a) = f (b)
(2) f ¢ (c) = 0
(c should be between a and b)
1
, f ¢ (x) = 0
2
2
æ 1ö
æ 1ö
0 = 6 ç ÷ + 2a ç ÷ + b
è 2ø
è 2ø
3
+a+b = 0
2
3
+a-2 = 0
2
a=2–
3
1
=
2
2
1
–2=1–2=–1
2
109. (d) f (x) = sin (sin x)
Þ f ¢(x) = cos (sin x) . cos x
Þ f ²(x) = – sin (sin x) . cos2 x + cos (sin x). (– sin x)
= – cos2 x . sin (sin x) – sin x . cos (sin x)
Now f ²(x) + tan x . f ¢(x) + g (x) = 0
Þ g(x) = cos2 x . sin (sin x) + sin x . cos (sin x)
– tan x . cos x . cos (sin x)
Þ g(x) = cos2 x . sin (sin x).
2a + b = 2 ×
ax3
x2
+ b. + cx
3
2
g¢(x) = ax2 + bx + c
Given: ax2 + bx + c = 0 and 2a + 3b + 6c = 0
Statement-2:
110. (d) Let g(x) =
(i)
g(0) = 0 and g(1) =
=
a b
2a + 3b + 6c
+ +c =
3 2
6
0
=0
6
Þ g(0) = g(1)
(ii) g is continuous on [0, 1] and differentiable on (0, 1)
\ By Rolle’s theorem $ k Î(0,1) such that g¢(k) = 0
This holds the statement 2. Also, from statement-2,we can
say ax2 + bx + c = 0 has at least one root in (0, 1).
Thus statement-1 and 2 both are true and statement-2 is a
correct explanation for statement-1.
Downloaded from @Freebooksforjeeneet
M-337
Continuity and Differentiability
111. (c)
d 2x
dy
2
=
d æ dx ö d æ dx ö dx = d æ 1 ö dx
ç
÷
ç ÷= ç ÷
dx è dy / dx ø dy
dy è dy ø dx è dy ø dy
é
ù
æ 1ö
dç ÷
è xø
d2 y 1 ê
1
1ú
êQ
. 2.
==– ú
2
dx
x2 ú
æ dy ö dx dy ê
êë
úû
ç ÷
dx
è dx ø
=–
d y
3
2
æ dy ö dx
ç ÷
è dx ø
112. (b) Given that f (x) = x | x | and g (x) = sin x
So that
go f (x) = g ( f (x)) = g (x | x | ) = sin x | x |
2
ì
2
ïìsin (– x ), if x < 0 ïí – sin x , if x < 0
= í
=
2
2
îï sin x , if x ³ 0
ïîsin ( x ), if x ³ 0
f (6) ³ 10 – 2 Þ
f (6) ³ 8.
....(i)
Given a1 ¹ 0 Þ f (0) = 0
Again f (x) has root a, Þ f (a ) = 0
\ f (0) = f (a)
\ By Rolle’s theorem f ¢(x) = 0 has root between (0, a)
Hence f ¢ ( x ) has a positive root smaller than a.
Þ f (1) =
a b
+ +c
3 2
2 a + 3b + 6c
= 0 (given)
6
\ f (0) = f (1)
\ f (x) satisfies all conditions of Rolle’s theorem therefore
f ¢(x) = 0 has a root in (0, 1)
i.e. ax 2 + bx + c = 0 has at lease one root in (0, 1)
117. (d) Given that f ( x ) = x n Þ f (1) = 1
f ' ( x ) = nx n -1 Þ f ' (1) = n
f '' ( x ) = n ( n - 1) x n - 2 Þ f '' (1) = n ( n - 1)
........................................................
c Î [a, b]
Given f (x) = logex
nan x n -1 + (n – 1) an -1 x n - 2 + ....+ a1 = 0 = f ¢(x)
f (0) = 0 and f (1) =
Here
L(go f )'' (0) = – 2 and R (go f )'' (0) = 2
Q L(go f)'' (0) ¹ R (go f )'' (0)
Þ go f (x) is not twice differentiable at x = 0.
\ Statement - 1 is true but statement -2 is false.
113. (c) Using Lagrange's Mean Value Theorem
Let f (x) be a function defined on [a, b]
f (b ) - f ( a )
b-a
The other given equation,
ax3 bx 2
+
+ cx
3
2
Being polynomial, it is continuous and differentiable, also,
2
2
2
ïì – 2 cos x + 4 x sin x , x < 0
Now (go f )'' (x) = í
2
2
2
ïî 2 cos x - 4 x sin x , x ³ 0
\
Þ
f (6) ³ 10 + f (1)
f ( x) =
Here we observe
L (go f )¢ (0) = 0 = R (go f )¢ (0)
Þ go f is differentiable at x = 0
and (go f )¢ is continuous at x = 0
\
Þ
116. (d) Let us define a function
2
ïì – 2 x cos x , if x < 0
\ (go f )¢ (x) = í
2
ïî 2 x cos x , if x ³ 0
then, f '(c) =
f (6) - f (1)
= f '(c) ³ 2
5
115. (b) Let f (x) = an x n + an -1 x n -1 + ...... + a1 x = 0
2
1
114. (a) As f (1) = – 2 & f '( x) ³ 2 " x Î [1, 6]
Applying Lagrange’s mean value theorem
1
\ f ' (x) =
x
equation (i) become
1 f (3) - f (1)
=
c
3-1
f n ( x ) = n ! Þ f n (1) = n !
f (1) =1-
Þ
1 log e 3 - log e 1 log e 3
=
=
c
2
2
Þ
c=
2
loge 3 Þ c = 2 log3e
( -1) n f n (1)
f '(1) f "(1) f "'(1)
+
+ ... +
1!
2!
3!
n!
n n ( n - 1) n ( n - 1)( n - 2)
n n!
+
+ ¼+ ( -1)
1!
2!
3!
n!
= nC0 - nC1 + n C2 - n C3 + ¼+ ( -1)
= (1 – 1)n = 0
Downloaded from @Freebooksforjeeneet
n n
Cn
EBD_8344
M-338
Mathematics
lim
118. (b)
x® a
f (a) g ¢ ( x ) - g (a ) f ¢ ( x )
=4
g ¢( x ) - f ¢ ( x )
Þ 1 + x 2 y1 = ny (Q y1 =
(By Applying L’ Hospital rule)
lim
x® a
k g ¢ ( x) - k f ¢ ( x)
=4
g ¢ ( x) - f ¢ ( x)
\ k = 4.
we get (1 + x 2 ) y12 = n2 y 2
Differentiating it w.r. to x,
(1 + x 2 )2 y1 y2 + y12 .2 x = n 2 .2 yy 1
119. (a) Given that y = ( x + 1 + x )
2 n
...(i)
Þ (1 + x2)y2 + xy1 = n2y
Differentiating both sides w.r. to x
dy
æ 1
ö
= n( x + 1 + x 2 )n -1 ç 1 + (1 + x 2 ) -1/2 . 2 x ÷
è 2
ø
dx
120. (a) Let f (x) =
dy
( 1 + x + x)
= n( x + 1 + x 2 ) n -1
dx
1 + x2
or
f (1) =
1[. So by Rolle’s theorem, f ¢ (x) = 0.
1 + x2
dy
= ny
dx
a b
2a + 3b + 6c
+ +c =
=0
3 2
6
Also f (x) is continuous and differentiable in [0, 1] and [0,
n( 1 + x 2 + x ) n
1 + x2
ax3 bx 2
+
+ cx
3
2
Þ f (0) = 0 and
2
=
dy
) Squaring both sides,
dx
[from (i)]
i.e ax 2 + bx + c = 0 has at least one root in
[0, 1].
Downloaded from @Freebooksforjeeneet
21
Applications
of Derivatives
5.
TOPIC Ć Rate of Change of Quantities
1.
2.
3.
The position of a moving car at time t is given by
f (t) = at2 + bt + c, t > 0, where a, b and c are real numbers greater
than 1. Then the average speed of the car over the time interval
[t1, t2] is attained at the point :
[Sep. 06, 2020 (I)]
(a) (t2 – t1)/2
(b) a(t2 – t1) + b
(c) (t1 + t2)/2
(d) 2a(t1 + t2) + b
If the surface area of a cube is increasing at a rate of 3.6
cm2/sec, retaining its shape; then the rate of change of its
volume (in cm3/sec.), when the length of a side of the cube
is 10 cm, is :
[Sep. 03, 2020 (II)]
(a) 18
(b) 10
(c) 20
(d) 9
If a function f (x) defined by
[Sep. 02, 2020 (I)]
ìae x + be - x , - 1 £ x < 1
ïï
f ( x) = í cx 2
, 1 £ x £ 3 be continuous for some
ï 2
ïî ax + 2cx , 3 < x £ 4
a, b, c Î R and f '(0) + f '(2) = e, then the value of a is :
4.
1
(a)
e 2 - 3e + 13
(c)
e + 3e + 13
e
2
(a)
e 2 - 3e - 13
(d)
e - 3e + 13
6.
(b)
25
3
25
(d) 25
3
A spherical iron ball of radius 10 cm is coated with a layer of
ice of uniform thickness that melts at a rate of 50 cm3/min.
When the thickness of the ice is 5 cm, then the rate at
which the thickness (in cm/min) of the ice decreases, is :
[April 10, 2019 (II)]
(a)
1
18p
(b)
1
36p
5
1
(d)
6p
9p
A water tank has the shape of an inverted right circular
cone, whose semi-vertical angle is tan –1 . Water is poured
into it at a constant rate of 5 cubic meter per minute. Then
the rate (in m/min.), at which the level of water is rising at
the instant when the depth of water in the tank is 10m; is:
[April 09, 2019 (II)]
(a) 1/15 p
(b) 1/10 p
(c) 2/p
(d) 1/5 p
If the volume of a spherical ball is increasing at the rate of
4p cc/sec, then the rate of increase of its radius (in cm/sec),
when the volume is 288 p cc,
[Online April 19, 2014]
(c)
7.
e
2
A spherical iron ball of 10 cm radius is coated with a layer
of ice of uniform thickness that melts at a rate of 50 cm3/
min. When the thickness of ice is 5 cm, then the rate (in
cm/min.) at which of the thickness of ice decreases, is:
[Jan. 9, 2020 (I)]
5
1
(a)
(b)
6p
54p
1
1
(c)
(d)
36p
18p
25 3
(c)
e
(b)
A 2 m ladder leans against a vertical wall. If the top of the
ladder begins to slide down the wall at the rate 25 cm/sec.,
then the rate (in cm/sec.) at which the bottom of the ladder
slides away from the wall on the hori ontal ground when
the top of the ladder is 1 m above the ground is:
[April 12, 2019 (I)]
8.
(a)
1
6
(b)
1
9
(c)
1
36
(d)
1
24
Downloaded from @Freebooksforjeeneet
EBD_8344
M-340
9.
Two ships A and B are sailing straight away from a fixed
point O along routes such that ÐAOB is always 120 . At a
certain instance, OA = 8 km, OB = 6 km and the ship A is
sailing at the rate of 20 km/hr while the ship B sailing at the
rate of 30 km/hr. Then the distance between A and B is
changing at the rate (in km/hr): [Online April 11, 2014]
(a)
(c)
10.
11.
12.
13.
14.
15.
Mathematics
260
37
80
37
16.
17.
260
(b)
37
(d)
If a circular iron sheet of radius 30 cm is heated such that
its area increases at the uniform rate of 6p cm2/hr, then the
rate (in mm/hr) at which the radius of the circular sheet
increases is
[Online May 7, 2012]
(a) 1.0
(b) 0.1
(c) 1.1
(d) 2.0
Two points A and B move from rest along a straight line
with constant acceleration f and f ' respectively. If A takes
m sec. more than B and describes ‘n’units more than B in
acquiring the same speed then
[2005]
80
37
(a) ( f - f ')m2 = ff ' n
A spherical balloon is being inflated at the rate of 35cc/
min. The rate of increase in the surface area (in cm 2/min.)
of the balloon when its diameter is 14 cm, is :
[Online April 25, 2013]
(b) ( f + f ')m2 = ff ' n
(a) 10
18.
1
ff ' m2
2
A li ard, at an initial distance of 21 cm behind an insect,
19.
moves from rest with an acceleration of 2 cm / s 2 and
pursues the insect which is crawling uniformly along a
straight line at a speed of 20 cm/s. Then the li ard will
catch the insect after
[2005]
(a) 20 s
(b) 1 s
(c) 21 s
(d) 24 s
A spherical iron ball 10 cm in radius is coated with a layer
(b)
(c)
10
(c) 100
(d) 10 10
If the surface area of a sphere of radius r is increasing
uniformly at the rate 8 cm2/s, then the rate of change of its
volume is :
[Online April 9, 2013]
(a) constant
(b) proportional to r
(c) proportional to r2
(d) proportional to r
A spherical balloon is filled with 4500p cubic meters of
helium gas. If a leak in the balloon causes the gas to escape
at the rate of 72p cubic meters per minute, then the rate (in
meters per minute) at which the radius of the balloon
decreases 49 minutes after the leakage began is:
7
9
(a)
(b)
[2012]
9
7
2
9
(c)
(d)
9
2
If a metallic circular plate of radius 50 cm is heated so that
its radius increases at the rate of 1 mm per hour, then the
rate at which, the area of the plate increases (in cm 2/hour)
is
[Online May 26, 2012]
(a) 5 p
(b) 10 p
(c) 100 p
(d) 50 p
The weight W of a certain stock of fish is given by W = nw,
where n is the si e of stock and w is the average weight of
a fish. If n and w change with time t as n = 2t2 + 3 and
w = t2 – t + 2, then the rate of change of W with respect to
t at t = 1 is
[Online May 19, 2012]
(a) 1
(b) 8
(c) 13
(d) 5
Consider a rectangle whose length is increasing at the
uniform rate of 2 m/sec, breadth is decreasing at the uniform
rate of 3 m/sec and the area is decreasing at the uniform
rate of 5 m2/sec. If after some time the breadth of the
rectangle is 2 m then the length of the rectangle is
[Online May 12, 2012]
(a) 2 m
(b) 4 m
(c) 1 m
(d) 3 m
1
( f + f ')m = ff ' n 2
2
(d) ( f '- f )n =
of ice of uniform thickness that melts at a rate of 50 cm 3 /min.
When the thickness of ice is 5 cm,then the rate at which
the thickness of ice decreases is
[2005]
20.
(a)
1
cm/min.
36 p
(b)
1
cm/min.
18 p
(c)
1
cm/min.
54 p
(d)
5
cm/min
6p
A point on the parabola y 2 = 18x at which the ordinate
increases at twice the rate of the abscissa is
[2004]
æ9 9ö
(b) (2, - 4)
æ -9 9ö
(d) (2, 4)
(a) ç 8 , 2 ÷
è
ø
(c) ç 8 , 2 ÷
è
ø
TOPIC n Increasing & Decreasing Functions
21.
The function, f ( x) = (3x - 7) x 2 / 3 , x Î R, is increasing
for all x lying in :
[Sep. 03, 2020 (I)]
14
(a) (-¥, 0) È æç , ¥ ö÷
è 15
ø
æ3
ö
(b) (-¥, 0) È ç , ¥ ÷
è7
ø
14 ö
æ
(c) ç -¥, ÷
15 ø
è
14 ö
æ
(d) ç -¥, - ÷ È (0, ¥)
15 ø
è
Downloaded from @Freebooksforjeeneet
M-341
Applications of Derivatives
22.
Let f be any function continuous on [a, b] and twice
differentiable on (a, b). If for all x Î (a, b), f 2 (x) > 0 and
f ²(x) < 0, then for any c Î (a, b),
f (c ) - f ( a )
is greater
f (b) - f ( c )
than:
23.
[Jan. 9, 2020 (I)]
(a)
b+ a
b-a
(b) 1
(c)
b-c
c-a
(d)
c-a
b-c
29.
é p pù
Let f(x) = x cos–1 (–sin |x|), x Î ê - , ú , then which of the
ë 2 2û
following is true?
[Jan. 8, 2020 (I)]
æ p ö
æ pö
(a) f ’ is increasing in çè - ,0÷ø and decreasing in çè 0, ÷ø
2
2
24.
a +x
2
-
d-x
b + (d - x)
2
2
, x Î R where a, b
and d are non- ero real constants. Then :
[Jan. 11, 2019 (II)]
(a) f is an increasing function of x
(b) f is a decreasing function of x
(c) f ¢ is not a continuous function of x
(d) f is neither increasing nor decreasing function of x
The function f defined by
f(x) = x3 – 3x2 + 5x + 7, is :
[Online April 9, 2017]
(a) increasing in R.
(b) decreasing in R.
(c) decreasing in (0, ¥) and increasing in (– ¥ , 0).
(d) increasing in (0, ¥) and decreasing in (– ¥, 0).
Let f(x) = sin4x + cos4x. Then f is an increasing function in
the interval :
ù 5p 3p ù
(a) ú , ú
û 8 4û
ù p 5p ù
(b) ú , ú
û2 8 û
æ p ö
æ pö
(d) f ’ is decreasing in çè - ,0÷ø and increasing in çè 0, ÷ø
2
2
ùp pù
(c) ú , ú
û 4 2û
ù pù
(d) ú 0, ú
û 4û
Let f(x) = ex – x and g(x) = x2 – x, x Î R. Then the set of all
x Î R, where the function h(x) = (fog) (x) is increasing, is :
[April 10, 2019 (I)]
é -1 ù
(d) ê ,0ú È [1, ¥ )
ë2 û
If the function f : R – {1, –1} ® A defin ed by
x2
1 - x2
( x - 1) 2
(a) –7
(c) 7
32.
, is surective, then A is equal to:
[April 09, 2019 (I)]
(a) R – {–1}
(b) [0, 벜)
(c) R – [–1, 0)
(d) R – (–1, 0)
Let f: [0 : 2] ® R be a twice differentiable function such
that f''(x) > 0, for all xÎ(0, 2). If f(x) = f(x) + f(2 – x), then f is :
[April 08, 2019 (I)]
(a) increasing on (0, 1) and decreasing on (1, 2).
(b) decreasing on (0, 2)
(c) decreasing on (0, 1) and increasing on (1, 2).
(d) increasing on (0, 2)
If the function f given by
f (x) = x3 – 3(a – 2)x2 + 3ax + 7, for some aÎR is increasing
in (0, 1] and decreasing in [1, 5), then a root of the equation,
f ( x ) - 14
31.
é 1ù
(b) ê 0, ú È [1, ¥ )
ë 2û
[0,¥ )
f(x) =
27.
x
2
p
2
(c) f ’ is not differentiable at x = 0
(c)
26.
30.
Let f ( x ) =
(b) f’ (0) = –
é -1ù é 1 ö
(a) ê -1, ú È ê , ¥ ÷
2 û ë2 ø
ë
25.
28.
= 0( x ¹ 1) is
[Jan. 12, 2019 (II)]
(b) 5
(d) 6
33.
34.
Let f and g be two differentiable functions on R such that
f ¢(x) > 0 and g¢(x) < 0 for all x Î R . Then for all x:
[Online April 12, 2014]
(a) f (g (x)) > f (g (x – 1)) (b) f (g (x)) > f (g (x + 1))
(c) g(f (x)) > g (f (x – 1)) (d) g(f (x)) < g (f (x + 1))
The real number k for which the equation, 2x3 + 3x + k = 0
has two distinct real roots in [0, 1]
[2013]
(a) lies between 1 and 2
(b) lies between 2 and 3
(c) lies between .1 and 0
(d) does not exist.
Statement-1: The function x2 (ex + e–x) is increasing for
all x > 0.
Statement-2: The functions x2ex and x2e–x are increasing
for all x > 0 and the sum of two increasing functions in any
interval (a, b) is an increasing function in (a, b).
[Online April 22, 2013]
(a) Statement-1 is false; Statement-2 is true.
(b) Statement-1is true; Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1.
(c) Statement-1 is true; Statement-2 is false.
(d) Statement-1is true; Statement-2 is true; Statement-2 is
a correct explanation for statement-1.
Statement-1: The equation x log x = 2 – x is satisfied by at
least one value of x lying between 1 and 2.
Statement-2: The function f (x) = x log x is an increasing
function in [l, 2] and g (x) = 2 – x is a decreasing function in
[1, 2] and the graphs represented by these functions
intersect at a point in [1, 2]
[Online April 9, 2013]
Downloaded from @Freebooksforjeeneet
EBD_8344
M-342
35.
36.
37.
38.
39.
Mathematics
(a) Statement-1 is true; Statement-2 is true; Statement-2
is a correct explanation for Statement-1.
(b) Statement-1 is true; Statement-2 is true; Statement-2
is not correct explanation for Statement-1.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is false.
If f(x) = xex(1 – x), x Î R , then f(x) is
[Online May 12, 2012]
(a) decreasing on [–1/2, 1]
(b) decreasing on R
(c) increasing on [–1/2, 1]
(d) increasing on R
For real x, let f (x) = x3 + 5x + 1, then
[2009]
(a) f is onto R but not one-one
(b) f is one-one and onto R
(c) f is neither one-one nor onto R
(d) f is one-one but not onto R
How many real solutions does the equation
x7 + 14x5 + 16x3 + 30x – 560 = 0 have?
[2008]
(a) 7
(b) 1
(c) 3
(d) 5
The function f (x) = tan –1(sin x + cos x) is an increasing
function in
[2007]
p
(a) æç 0, ö÷
è 2ø
æ p pö
(b) ç - , ÷
è 2 2ø
(c) æç p , p ö÷
è 4 2ø
p p
(d) æç - , ö÷
è 2 4ø
41.
curve x4 e y + 2 y + 1 = 3 at the point (1, 0)?
42.
x - 3x + 3x + 3
(b) [2, ¥ )
2 x3 - 3 x 2 - 12 x + 6
1ù
æ
(c) ç - ¥, ú
3û
è
3x 2 - 2 x + 1
(d) (– ¥ , – 4)
x3 + 6 x 2 + 6
3
2
43.
a
is equal to _____. [NA Sep. 05, 2020 (II)]
b
If the tangent to the curve, y = ex at a point (c, ec) and the
normal to the parabola, y2 = 4x at the point (1, 2) intersect
at the same point on the x-axis, then the value of c is
____________.
[NA Sep. 03, 2020 (II)]
44.
If y =
6
e -1
(a)
e
(c)
æ 1 ö
ç
÷
eè 1-e ø
(b)
æ 1 ö
ç
÷
eè e-1 ø
e
(d)
e -1
-1 ì 3
4
dy
ü
at x = 0 is
í cos kx - sin kx ý, then
dx
5
î5
þ
___________.
45.
[NA Sep. 02, 2020 (II)]
Let the normal at a point P on the curve y2 – 3x2 + y + 10 = 0
æ 3ö
intersect the y-axis at ç 0, ÷ . If m is the slope of the
è 2ø
tangent at P to the curve, then |m| is equal to ______.
[NA Jan. 8, 2020 (I)]
46. The length of the perpendicular from the origin, on the
normal to the curve, x 2 + 2xy – 3y2 = 0 at the point (2, 2)
is:
[Jan. 8, 2020 (II)]
(a)
2
(c) 2
47.
48.
If the tangent to the curve, y = f (x) = xlogex, (x > 0) at a
point (c, f(c)) is parallel to the line segement oining the
points (1, 0) and (e, e), then c is equal to:
[Sep. 06, 2020 (II)]
å k cos
k =1
TOPIC Đ Tangents & Normals
40.
[Sep. 05, 2020 (II)]
(a) (2, 2)
(b) (2, 6)
(c) (– 2, 6)
(d) (– 2, 4)
If the lines x + y = a and x – y = b touch the curve
y = x2 – 3x + 2 at the points where the curve intersects the
x-axis, then
A function is matched below against an interval where it is
supposed to be increasing. Which of the following pairs is
incorrectly matched?
[2005]
Interval
Function
(a) (– ¥ , ¥ )
Which of the following points lies on the tangent to the
49.
(b) 4 2
(d) 2 2
(
)
x
, x Î R, x ¹ ± 3 ,
x2 - 3
at a point (a, b) (0, 0) on it is parallel to the line
2x + 6y – 11 = 0, then :
[April 10, 2019 (II)]
(a) |6a + 2b| = 19
(b) |6a + 2b| = 9
(c) |2a + 6b| = 19
(d) |2a + 6b| = 11
If the tangent to the curve, y = x3 + ax – b at the point
(1, –5) is perpendicular to the line, – x + y + 4 = 0, then
which one of the following points lies on the curve?
[April 09, 2019 (I)]
(a) (–2, 1)
(b) (–2, 2)
(c) (2, –1)
(d) (2, –2)
Let S be the set of all values of x for which the tangent to
the curve y = f(x) = x3 – x2 – 2x at (x, y) is parallel to the line
segment oining the points (1, f(1)) and (– 1, f(– 1)), then S
is equal to:
[April 09, 2019 (I)]
If the tangent to the curve y =
ì1 ü
(a) í ,1ý
î3 þ
1
(b) ìí– , –1üý
î 3 þ
ì1 ü
(c) í , –1ý
î3 þ
ì 1 ü
(d) í– ,1ý
î 3 þ
Downloaded from @Freebooksforjeeneet
M-343
Applications of Derivatives
50.
51.
The tangent and the normal lines at the point ( 3 , 1) to
the circle x2 + y2 = 4 and the x-axis form a triangle. The area
of this triangle (in square units) is : [April 08, 2019 (II)]
(a)
4
3
(b)
(c)
2
3
(d)
1
3
(c)
1
3
56.
The maximum area (in sq. units) of a rectangle having its
base on the x-axis and its other two vertices on the parabola,
y = 12 – x2 such that the rectangle lies inside the parabola,
is:
[Jan. 12, 2019 (I)]
(a) 36
53.
(d) 18 3
The tangent to the curve y = x2 – 5x + 5, parallel to the line
2y = 4x + 1, also passes through the point :
[Jan. 12, 2019 (II)]
æ7 1ö
(a) ç , ÷
è2 4ø
æ1 ö
(b) ç , -7 ÷
è8 ø
æ 1 ö
(c) ç - , 7 ÷
è 8 ø
æ1 7ö
(d) ç , ÷
è4 2ø
curve y = x , ( x > 0), is:
(a)
(c)
54.
5
2
3
2
57.
(d)
59.
The tangent to the curve, y = xe
60.
x2
(d)
1
2
7
3
8
15
8
(d)
17
(b)
If the curves y2 = 6x,9x 2 + by 2 = 16 intersect each other
at right angles, then the value of b is :
[2018]
7
2
(b) 4
9
(d) 6
2
Let P be a point on the parabola,x2 = 4y. If the distance of
P from the centre of the circle, x2 + y2 + 6x + 8 = 0 is
minimum, then the equation of the tangent to the parabola
at P, is
[Online April 16, 2018]
(a) x + 4y – 2 = 0
(b) x + 2y = 0
(c) x + y + 1 = 0
(d) x – y + 3 = 0
If the tangents drawn to the hyperbola 4y2 = x2 + 1 intersect
the co-ordinate axes at the distinct points A and B, then
the locus of the mid point of AB is[Online April 15, 2018]
(a) x2 – 4y2 + 16 x2 y2 = 0
(b) 4x2 – y2 + 16 x2 y2 = 0
(c) 4x2 – y2 – 16 x2 y2 = 0
(d) x2 – 4y2 – 16 x2 y2 = 0
If b is one of the angles between the normals to the ellipse,
x 2 + 3y 2 = 9 at the points (3cosq, 3 sin q) and
passing through the
(– 3sin q,
point (1, e) also passes through the point:
[Jan. 10, 2019 (II)]
55.
1
3
(c)
3
2
5
4
7
3
(b)
If q denotes the acute angle between the curves,
y = 10 – x2 and y = 2 + x2 at a point of their intersection,
then |tan q| is equal to:
[Jan. 09, 2019 (I)]
(a)
[Jan. 10, 2019 (I)]
(b)
1
6
4
9
7
(c)
17
58.
æ3 ö
The shortest distance between the point ç , 0 ÷ and the
è2 ø
5
6
(a)
(b) 20 2
(c) 32
52.
(a)
(a) (2, 3e)
æ4
ö
(b) ç , 2e ÷
è3
ø
æ5
ö
(c) ç , 2e ÷
3
è
ø
(d) (3, 6e)
(a)
A helicopter is flyin g along the curve given by
æ1 ö
y – x3/2 = 7, (x ³ 0). A soldier positioned at the point ç , 7 ÷
è2 ø
wants to shoot down the helicopter when it is nearest to
him. Then this nearest distance is:
[Jan. 10, 2019 (II)]
1
(b)
2
3
3
4
3
A normal to the hyperbola, 4x2 – 9y2 = 36 meets the coordinate axes x and y at A and B, respectively. If the
parallelogram OABP (O being the origin) is formed, then
the locus of P is
[Online April 15, 2018]
(a) 4x2 – 9y2 = 121
(b) 4x2 + 9y2 = 121
(c) 9x2 – 4y2 = 169
(d) 9x2 + 4y2 = 169
(c)
61.
2
2 cot b
æ pö
3 cos q); Î ç 0, ÷ ; then
is equal to
sin 2q
è 2ø
[Online April 15, 2018]
(d)
Downloaded from @Freebooksforjeeneet
EBD_8344
M-344
62.
Mathematics
The normal to the curve y(x – 2)(x – 3) = x + 6 at the point
where the curve intersects the y-axis passes through the
point:
[2017]
æ 1 1ö
(a) çè , ÷ø
2 3
63.
64.
1
. If one of its directices is x = – 4, then the
2
æ 3ö
equation of the normal to it at ç1, ÷ is :
[2017]
è 2ø
(a) x + 2y = 4
(b) 2y – x = 2
(c) 4x – 2y = 1
(d) 4x + 2y = 7
A tangent to the curve, y = f (x) at P(x, y) meets x-axis at A
and y-axis at B. If AP : BP = 1 : 3 and f (a) = 1 , then the curve
also passes through the point : [Online April 9, 2017]
æ 1ö
æ 1 ö
(c) ç 2, ÷
(d) ç 3, ÷
è 8ø
è 28 ø
The tangent at the point (2, –2) to the curve,
x2y2 – 2x = 4 (1–y) does not pass through the point :
[Online April 8, 2017]
æ 1ö
(a) ç 4, ÷
è 3ø
(c) (– 4, – 9)
66.
The normal to the curve, x2 + 2xy – 3y2 = 0, at (1, 1) [2015]
(a) meets the curve again in the third quadrant.
(b) meets the curve again in the fourth quadrant.
(c) does not meet the curve again.
(d) meets the curve again in the second quadrant.
The equation of a normal to the curve,
70.
æp
ö
sin y = x sin ç + y ÷ at x = 0, is :
è3
ø
[Online April 11, 2015]
72.
(a) 2x –
3y = 0
(b) 2x +
3y = 0
(c) 2y –
3x = 0
(d) 2y +
3x = 0
If the tangent to the conic, y – 6 = x2 at (2, 10) touches the
circle, x2 + y2 + 8x – 2y = k (for some fixed k) at a point
(a, b) ; then (a, b) is :
[Online April 10, 2015]
æ 7 6ö
(a) çè - , ÷ø
17 17
æ 4 1ö
(b) çè - , ÷ø
17 17
æ 6 10 ö
(c) çè - , ÷ø
17 17
æ 8 2ö
(d) çè - , ÷ø
17 17
The distance, from the origin, of the normal to the curve,
p
, is :
4
[Online April 10, 2015]
(b) (8, 5)
x = 2 cost + 2t sint, y = 2 sint – 2t cost at t =
(d) (– 2, – 7)
(a) 2
Consider
(b) 4
(c)
æ 1 + sin x
f ( x ) = tan -1 ç
ç 1 - sin x
è
ö
æ pö
÷÷ , x Î ç 0, ÷ .
è 2ø
ø
A normal to y = f(x) at x =
67.
69.
71.
æ1 ö
(b) ç , 4 ÷
è2 ø
æ1
ö
(a) ç , 24 ÷
è3
ø
65.
If the tangent at a point P, with parameter t, on the curve
x = 4t2 + 3, y = 8t3 – 1, t Î R, meets the curve again at a
point Q, then the coordinates of Q are :
[Online April 9, 2016]
(a) (16t2 + 3, –64t3 – 1) (b) (4t2 + 3, –8t3 – 2)
(c) (t2 + 3, t3 – 1)
(d) (t2 + 3, – t3 – 1)
æ 1 1ö
(b) ç - , - ÷
è 2 2ø
æ 1 1ö
æ 1 1ö
(c) ç , ÷
(d) ç , - ÷
è 2 2ø
è 2 3ø
The eccentricity of an ellipse whose centre is at the
origin is
68.
[2016]
73.
p
also passes through the point:
6
æp ö
(a) ç , 0 ÷
è6 ø
æp ö
(b) ç , 0 ÷
è4 ø
(c) (0, 0)
æ 2p ö
(d) ç 0, ÷
è 3 ø
3
. If P is
4
2
a point on C, such that the tangent at P has slope , then
3
a point through which the normal at P passes, is :
[Online April 10, 2016]
(a) (1, 7)
(b) (3, –4)
(c) (4, –3)
(d) (2, 3)
Let C be a curve given by y(x) = 1 +
(d) 2 2
2
For the curve y = 3 sinq cosq, x = eq sin q, 0 £ q £ p, the
tangent is parallel to x-axis when q is:
[Online April 11, 2014]
(a)
3p
4
(b)
p
2
74.
p
p
(d)
6
4
If an equation of a tangent to th e curve,
y – cos(x + f), – 1 -1 £ x £ 1 + p, is x + 2y = k then k is equal
to :
[Online April 25, 2013]
(a) l
(b) 2
75.
p
p
(d)
4
2
The equation of the normal to the parabola,
x2 = 8y at x = 4 is
[Online May 19, 2012]
(a) x + 2y = 0
(b) x + y = 2
(c) x – 2y = 0
(d) x + y = 6
(c)
4x - 3, x >
(c)
Downloaded from @Freebooksforjeeneet
M-345
Applications of Derivatives
76.
77.
78.
79.
80.
4
The equation of the tangent to the curve y = x + 2 , that
x
is parallel to the x-axis, is
[2010]
(a) y = 1
(b) y = 2
(c) y = 3
(d) y = 0
Angle between the tangents to the curve y = x 2 - 5 x + 6
at the points (2, 0) and (3, 0) is
[2006]
p
(a) p
(b)
2
p
p
(c)
(d)
6
4
The normal to the curve
[2005]
x = a (cos q + q sin q ), y = a (sin q – q cos q ) at any
point q is such that
(a) it passes through the origin
83.
[Sep. 06, 2020 (II)]
æ 1 1ö
(a) ç - , ÷ - {0}
è 2 2ø
84.
æ 1 1ö
æ 3 3ö
(c) ç - , ÷
(d) ç - , ÷ - {0}
è 2 2ø
è 2 2ø
If x = 1 is a critical point of the function
[Sep. 05, 2020 (II)]
(a) x = 1 and x = -
2
are local minima of f.
3
p
(b) it makes an angle
+ q with the x- axis
2
(b) x = 1 and x = -
2
are local maxima of f.
3
æ p
ö
(c) it passes through ç a , - a÷
è 2
ø
(d) It is at a constant distance from the origin
(c) x = 1 is a local maxima and x = -
The normal to the curve x = a(1 + cos q), y = a sinq at ‘q’
always passes through the fixed point
[2004]
(a) (a, a)
(b) (0, a)
(c) (0, 0)
(d) (a, 0)
85.
A function y = f ( x ) has a second order derivative
(a) ( x + 1)2
(b) ( x - 1)3
(c) ( x + 1)3
(d) ( x - 1)2
1 + sin 2 x
sin 2 x
1 + cos 2 x
sin 2 x
sin 2 x
2
cos x
2
sin x
1 + sin 2 x
Then the ordered pair (m, M) is equal to :
(a) (– 3, 3)
(b) (– 3, – 1)
(c) (– 4, – 1)
(d) (1, 3)
Let AD and BC be two vertical poles at A and B respectively
on a hori ontal ground. If AD = 8 m, BC = 11 m and AB = 10
m; then the distance (in meters) of a point M on AB from
the point A such that MD2 + MC2 is minimum is ______.
[NA Sep. 06, 2020 (I)]
2
is a local maxima of f.
3
The area (in sq. units) of the largest rectangle ABCD whose
vertices A and B lie on the x-axis and vertices C and D lie
on the parabola, y = x2 – 1 below the x-axis, is :
[Sep. 04, 2020 (II)]
(a)
(c)
86.
Let m and M be respectively the minimum and maximum
values of
[Sep. 06, 2020 (I)]
cos 2 x
2
is a local minima of f.
3
(d) x = 1 is a local minima and x = -
TOPIC Ė Approximations, Maxima & Minima
82.
æ 3 3ö
(b) ç - , ÷
è 2 2ø
f ( x) = (3x 2 + ax - 2 - a)e x , then :
f "( x) = 6( x - 1). If its graph passes through the point
(2,1) and at that point the tangent to the graph is y = 3x –
5, then the function is
[2004]
81.
The set of all real values of l for which the function
æ p pö
f ( x) = (1 - cos2 x ).(l + sin x) , x Î ç - , ÷ , has exactly
2 2ø
one maxima and exactly minima, is: è
87.
88.
2
3 3
4
3
(b)
(d)
1
3 3
4
3 3
Suppose f(x) is a polynomial of degree four, having critical
points at –1, 0, 1. If T = {x Î R | f ( x ) = f (0)}, then the
sum of squares of all the elements of T is :
[Sep. 03, 2020 (II)]
(a) 4
(b) 6
(c) 2
(d) 8
Let f(x) be a polynomial of degree 3 such that f(–1) = 10,
f(1)= –6, f(x) has a critical point at x = –1 and f ¢(x) has a
critical point at x = 1. Then f(x) has a local minima at
x = ________.
[NA Jan. 8, 2020 (II)]
Let f(x) be a polynomial of degree 5 such that x = ±1 are its
critical points. If = 4, then which one of the following is
not true ?
[Jan. 7, 2020 (II)]
(a) f is an odd function.
(b) f(l) – 4f(–l) = 4.
(c) x = 1 is a point of maxima and x = –1 is a point of
minima of f.
(d) x = 1 is a point of minima and x = –1 is a point of
maxima of f.
Downloaded from @Freebooksforjeeneet
EBD_8344
M-346
89.
Mathematics
If m is the minimum value of k for which the function
96.
f ( x ) = x kx - x 2 is increasing in the interval [0,3] and M
is the maximum value of f in [0,3] when k = m, then the
ordered pair (m, M) is equal to :
[April 12, 2019 (I)]
(a)
(c)
90.
91.
92.
(b)
(4,3 2)
(d)
(3,3 3)
2
3
(b)
3
(d) 3
6
(c) 2 3
(b)
19
79
2
(d) 31
34
Let P(4, –4) and Q(9, 6) be two points on the parabola,
y2 = 4x and let this X be any point arc POQ of this
parabola, where O is vertex of the parabola, such that
the area of DPXQ is maximum. Then this minimum area
(in sq. units) is:
[Jan. 12, 2019 (I)]
(a)
75
2
(b)
125
4
625
125
(d)
4
2
The maximum value of the function f(x) = 3x3 – 18x2 + 27 x – 40
(c)
95.
{
is :
[Jan. 11, 2019 (II)]
}
2
on the set S = x Î R : x + 30 £ 11x is :
m+n
1
(d)
6mn
4
The maximum volume (in cu.m) of the right circular cone
having slant height 3 m is:
[Jan. 09, 2019 (I)]
(a) 6 p
(b) 3 3 p
(c)
97.
(c)
98.
4
p
3
(b) – 222
(d) 222
(d) 2 3 p
1
1
Let f (x) = x 2 +
and g(x) = x - , x Î R - {-1,0,1} .
2
x
x
f (x)
If h(x) =
, then the local minimum value of h(x) is :
g(x)
[2018]
(a) – 3
(b) -2 2
(c) 2 2
(d) 3
Let M and m be respectively the absolute maximum and
the absolute minimum values of the function,
f (x) = 2x3 – 9x2 + 12x + 5 in the interval [0, 3]. Then
M – m is equal to
[Online April 16, 2018]
(a) 1
(b) 5
(c) 4
(d) 9
100. If a right circularcone having maximum volume, is inscribed
in a sphere of radius 3 cm, then the curved surface area
(in cm2) of this cone is
[Online April 15, 2018]
(b) 6 2p
(c) 6 3p
(d) 8 2p
101. Twenty metres of wire is available for fencing off a flowerbed in the form of a circular sector. Then the maximum area
(in sq. m) of the flower-bed, is :
[2017]
(a) 30
(b) 12.5
(c) 10
(d) 25
102. A wire of length 2 units is cut into two parts which are bent
respectively to form a square of side = x units and a circle
of radius = r units. If the sum of the areas of the square and
the circle so formed is minimum, then:
[2016]
(a) x = 2r
(b) 2x = r
(c) 2x = (p+ 4)r
(d) (4– p) x= pr
103. The minimum distance of a point on the curve y = x2 – 4
from the origin is :
[Online April 9, 2016]
(a)
15
2
(b)
19
2
(c)
15
2
(d)
19
2
[Jan. 11, 2019 (I)]
(a) – 122
(c) 122
1
(b)
2
(a) 8 3p
(c)
94.
(1 + x2m )(1 + y2n )
99.
pö
æ
The maximum value of 3cos q + 5sin ç q - ÷ for any real
6ø
è
value of q is:
[Jan. 12, 2019 (I)]
(a)
xm y n
The maximum value of the expression
(a) 1
(5,3 6)
Let a1, a2, a3, …. be an A. P. with a6 = 2. Then the common
difference of this A.P., which maximises the product a1 a 4
a 5, is :
[April 10, 2019 (II)]
3
8
(b)
(a)
2
5
6
2
(c)
(d)
5
3
If S1 and S2 are respectively the sets of local minimum and
local maximum points of the function,
f(x) = 9x4 + 12x3 – 36x2 + 25, xÎR, then :
[April 08, 2019 (I)]
(a) S1 = {–2}; S2 = {0, 1} (b) S1 = {–2, 0}; S2 = {1}
(c) S1 = {–2, 1}; S2 = {0} (d) S1 = {–1}; S2 = {0, 2}
The height of a right circular cylinder of maximum volume
inscribed in a sphere of radius 3 is : [April 08, 2019 (II)]
(a)
93.
(4,3 3)
Let x, y be positive real numbers and m, n positive integers.
Downloaded from @Freebooksforjeeneet
M-347
Applications of Derivatives
104. Let k and K be the minimum and the maximum values of
the function f(x) =
(1 + x )0.6
1 + x 0.6
in [0, 1] respectively, then
the ordered pair (k, K) is equal to :
[Online April 11, 2015]
(a) (2–0.4, 1)
(b) (2–0.4, 20.6)
(c) (2–0.6, 1)
(d) (1, 20.6)
105. From the top of a 64 metres high tower, a stone is thrown
upwards vertically with the velocity of 48 m/s. The
greatest height (in metres) attained by the stone,
assuming the value of the gravitational acceleration g =
32 m s2, is:
[Online April 11, 2015]
(a) 128
(b) 88
(c) 112
(d) 100
106. If x = –1 and x = 2 are extreme points of
f ( x ) = a log x + b x 2 + x then
(a) a = 2, b = -
1
2
[2014]
(b) a = 2, b =
1
2
1
1
(d) a = -6, b = 2
2
107. The minimum area of a triangle formed by any tangent to
(c) a = -6, b =
x 2 y2
+
= 1 and the co-ordinate axes is:
16 81
[Online April 12, 2014]
(a) 12
(b) 18
(c) 26
(d) 36
108. The volume of the largest possible right circular cylinder
the ellipse
that can be inscribed in a sphere of radius = 3 is:
[Online April 11, 2014]
(a)
4
3p
3
(c) 4p
(b)
8
3p
3
(d) 2p
bö
æ
109. The cost of running a bus from A to B, is ` ç av + ÷ ,
vø
è
where v km/h is the average speed of the bus. When the
bus travels at 30 km/h, the cost comes out to be ` 75 while
at 40 km/h, it is ` 65. Then the most economical speed (in
km/ h) of the bus is :
[Online April 23, 2013]
(a) 45
(b) 50
(c) 60
(d) 40
110. The maximum area of a right angled triangle with
hypotenuse h is :
[Online April 22, 2013]
(a)
(c)
h2
2 2
h2
2
(b)
h2
2
(d)
h2
4
111. Let a, b Î R be such that the function f given by
f (x) = ln | x | + bx2 + ax, x ¹ 0 has extreme values at x = –1
and x = 2
Statement-1 : f has local maximum at x = –1 and at x = 2.
1
-1
Statement-2 : a = and b =
[2012]
2
4
(a) Statement-1 is false, Statement-2 is true.
(b) Statement-1 is true, statement-2 is true; statement-2 is
a correct explanation for Statement-1.
(c) Statement-1 is true, statement-2 is true; statement-2 is
not a correct explanation for Statement-1.
(d) Statement-1 is true, statement-2 is false.
112. A line is drawn through the point (1,2) to meet the
coordinate axes at P and Q such that it forms a triangle
OPQ, where O is the origin. If the area of the triangle OPQ
is least, then the slope of the line PQ is :
[2012]
1
(a) (b) – 4
4
1
(c) – 2
(d) 2
113. Let f: ( -¥, ¥ ) ® ( -¥, ¥ ) be defined by
f(x) = x3 + 1.
[Online May 26, 2012]
Statement 1: The function f has a local extremum at x = 0
Statement 2: The function f is continuous and
differentiable on ( -¥, ¥ ) and f ¢(0) = 0
(a) Statement 1 is true, Statement 2 is false.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
a correct explanation for Statement 1.
(c) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation for Statement 1.
(d) Statement 1 is false, Statement 2 is true.
114. Let f be a function defined by [2011RS]
ì tan x
, x¹0
ï
f (x) = í x
ï1,
x=0
î
Statement - 1 : x = 0 is point of minima of f
Statement - 2 : f ¢ ( 0) = 0.
(a) Statement-1 is true, statement-2 is true; statement-2 is
a correct explanation for statement-1.
(b) Statement-1 is true, statement-2 is true; statement-2 is
NOT a correct explanation for statement-1.
(c) Statement-1 is true, statement-2 is false.
(d) Statement-1 is false, statement-2 is true.
x
æ 5p ö
115. For x Î ç 0, ÷ , define f ( x) = ò t sin t dt . Then f has
è 2 ø
0
[2011]
(a) local minimum at p and 2p
(b) local minimum at p and local maximum at 2p
(c) local maximum at p and local minimum at 2p
(d) local maximum at p and 2p
Downloaded from @Freebooksforjeeneet
EBD_8344
M-348
Mathematics
116. Let f : R ® R be a continuous function defined by
(a) The cubic has minima at
1
f ( x) = x
e + 2e - x
[2010]
1
Statement -1 : f (c) = , for some c Î R.
3
Statement -2 : 0 < f (x) £
1
, for all x Î R
2 2
(a) Statement -1 is true, Statement -2 is true ; Statement 2 is not a correct explanation for Statement -1.
(b) Statement -1 is true, Statement -2 is false.
(c) Statement -1 is false, Statement -2 is true .
(d) Statement - 1 is true, Statement 2 is true ; Statement -2
is a correct explanation for Statement -1.
117. Let f : R ® R be defined by
f ( x) =
{ k2-x+23,x, ifif xx £>--11
If f has a local minimum at x = – 1 , then a possible value of
k is
[2010]
1
(a) 0
(b) 2
(c) –1
(d) 1
118. Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the
only real root of P' (x) = 0. If P(–1) < P(1), then in the interval
[ –1, 1] :
[2009]
(a) P(–1) is not minimum but P(1) is the maximum of P
(b) P(–1) is the minimum but P(1) is not the maximum of P
(c) Neither P(–1) is the minimum nor P(1) is the maximum
of P
(d) P(–1) is the minimum and P(1) is the maximum of P
119. Suppose the cubic x3 – px + q has three distinct real roots
where p > 0 and q > 0. Then which one of the following
holds?
[2008]
(b) The cubic has minima at –
p
p
and maxima at –
3
3
p
and maxima at
3
(c) The cubic has minima at both
p
p
and –
3
3
(d) The cubic has maxima at both
p
p
and –
3
3
p
3
x 2
+ has a local minimum at
2 x
(a) x = 2
(b) x = -2
[2006]
(c) x = 0
(d) x = 1
121. The real number x when added to its inverse gives the
minimum value of the sum at x equal to
(a) –2
(b) 2
[2003]
(c) 1
(d) –1
120. The function f ( x ) =
122. If the function f ( x) = 2 x3 - 9ax 2 + 12a 2 x + 1 , where
a > 0 , attains its maximum and minimum at p and q
respectively such that p 2 = q , then a equals
[2003]
1
(b) 3
2
(c) 1
(d) 2
123. The maximum distance from origin of a point on the curve
(a)
æ at ö
æ at ö
x = a sin t–b sin çè ÷ø , y = a cos t – b cos çè ÷ø , both
b
b
a, b > 0 is
(a) a – b
(b) a + b
(c) a 2 + b2
(d)
a 2 - b2
Downloaded from @Freebooksforjeeneet
[2002]
M-349
Applications of Derivatives
1.
2.
From (i) and (ii), 50 = 4p(10 + x)
f (t2 ) - f (t1 )
t2 - t1
t1 + t2
2at + b = a(t1 + t2 ) + b Þ t =
2
(d) Let the side of cube be a.
(c) Average speed = f '(t ) =
S = 6a 2 Þ
dS
da
da
= 12a ×
Þ 3.6 = 12 a ×
dt
dt
dt
5.
Þ
dx
1
cm / min
=
dt 18p
[Q thickness of ice x = 5]
(b) According to the question,
...(i)
When y = 1 Þ x = 3
Diff. equation (i) w. r. t. t,
dV
da
æ 3 ö
= 3a 2 ×
= 3(10)2 × ç
=9
è 100 ÷ø
dt
dt
(d) Since, function f (x) is continuous at x = 1, 3
V = a3 Þ
2x
+
\ f (1) = f (1 )
Þ ae + be -1 = c
50 = 4p(10 + 5)2
dy
= – 25 at y = 1
dt
By Pythagoras theorem, x2 + y2 = 4
da
da
Þ 12(10)
= 3.6 Þ
= 0.03
dt
dt
3.
dx
dt
Þ
dx
dy
+ 2y
=0
dt
dt
...(i)
f (3) = f (3+ )
Þ 9c = 9a + 6c Þ c = 3a
From (i) and (ii),
...(ii)
b = ae(3 - e)
...(iii)
é ae x - be - x
ê
f '( x ) = ê 2cx
ê 2ax + 2c
ë
y
x
dx
dy
dx
=0Þ 3
Þ x +y
+ (– 25) = 0
dt
dt
dt
-1 < x < 1
1< x < 3
3< x< 4
f '(0) = a - b, f '(2) = 4c
6.
Given, f '(0) + f '(2)= e
a - b + 4c= e
From eqs. (i), (ii), (iii) and (iv),
dVice
= 50
dt
4
4
Vice = p(10 + r )3 - p(10)3
3
3
dV 4
dr
dr
Þ
= p3(10 + r )2
= 4p(10 + r )2
dt 3
dt
dt
Þ 13a - 3ae + ae 2 = e
4.
dx 25
Þ dt =
cm/s
3
(a) Given that ice melts at a rate of 50 cm3/min.
\
...(iv)
a - 3ae + ae 2 + 12a = e
Þa=
2
e
e 2 - 3e + 13
(d) Let the thickness of ice layer be = x cm
Total volume V =
10
4
p(10 + x)3
3
dV
dx
= 4p(10 + x )2
dt
dt
Since, it is given that
dV
= 50 cm3 / min
dt
5
r
...(i)
Substitute r = 5,
...(ii)
50 = 4p(225)
dr Þ dr = 50
1
cm/min
=
dt 4p(225) 18p
dt
Downloaded from @Freebooksforjeeneet
EBD_8344
M-350
7.
Mathematics
9.
(d)
r
A
(a)
tan q = 1/2
q
120
10 m
O
3
5 m /min
B
Let OA = x km, OB = y km, AB = R
(AB)2 = (OA)2 + (OB)2 – 2 (OA) (OB) cos 120
æ 1ö
R2 = x2 + y2 – 2 xy ç - ÷ = x 2 + y 2 + xy
è 2ø
Given that water is poured into the tank at a constant
rate of 5 m3/minute.
R at x = 6 km, and y = 8 km
dv
= 5m3 / min
dt
Volume of the tank is,
\
R=
62 + 82 + 6 ´ 8 = 2 37
Differentiating equation (i) with respect to t
1
V = pr 2 h
...(i)
3
where r is radius and h is height at any time.
By the diagram,
tan q =
2R
=
r 1
=
h 2
dh 2dr
Þ h = 2r Þ
=
dt
dt
Differentiate eq. (i) w.r.t. ‘t’, we get
...(ii)
10.
(c)
dV
= 5 in the above equation.
dt
Volume of sphere V =
4 3
pr
3
4
288 ´ 3
p(r 3 ) Þ
= r3
3
4
Þ 216 = r3
Þr= 6
Hence,
dr 1
=
.
dt 36
(a) Volume of sphere V =
4 3
pr
3
35
dr
dr
or
=
dt
dt 4pr 2
35 = 4pr 2 .
...(i)
Surface area of sphere = S = 4pr2
dS
dr
dr
= 4p´ 2 r ´
= 8pr .
dt
dt
dt
...(i)
dv
4
2 dr
= .3pr .
dt
3
dt
dr
4p = 4pr2.
dt
1
dr
=
dt
r2
Since, V = 288p, therefore from (i), we have
288p =
1
[2 ´ 8 ´ 20 + 2 ´ 6 ´ 30 + (8 ´ 30 + 6 ´ 20)]
2R
dV 4
dr
= . p . 3r 2 .
dt 3
dt
75p dh
dh 1
5=
Þ
= m/min.
3 dt
dt 5p
8.
dx
dy æ dy
dx ö
dR
= 2x + 2 y + ç x + y ÷
dt
dt è dt
dt ø
dt
1
260
dR
[1040] =
=
2 ´ 2 37
37
dt
dV 1 æ
dr
dh ö
= p2r h + pr 2 ÷
dt 3 çè
dt
dt ø
Putting h = 10, r = 5 and
...(i)
dS 70
=
dt
r
(By using (i))
Now, diameter = 14 cm, r = 7
\
11.
dS
= 10
dt
(d) V =
dV
dr
4 3
pr Þ
= 4pr 2 .
dt
dt
3
S = 4pr2 Þ
Þ 8 = 8pr
dS
dr
= 8pr .
dt
dt
1
dr
dr
Þ
=
dt
dt pr
Downloaded from @Freebooksforjeeneet
...(i)
M-351
Applications of Derivatives
Putting the value of
dr
in (i), we get
dt
dV
1
= 4pr 2 ´
= 4r
dt
pr
Þ
12.
dV
is proportional to r.
dt
(c) Volume of spherical balloon = V = 4 pr 3
3
Differentiate both the side, w.r.t 't' we get,
dV
æ dr ö
...(i)
= 4pr 2 ç ÷
è dt ø
dt
\ After 49 min,
Volume = (4500 – 49 ´ 72)p = (4500 – 3528)p = 972 p m3
Þ V = 972 p m3
4 3
pr
3
Þ r3 = 3 ´ 243 = 3 ´ 35 = 36 = (32)3
Þ r=9
\
972 p =
Given
Putting
\
13.
dV
= 72 p
dt
dV
= 72p and r = 9, we get
dt
dA
dl
db
= -5 ,
=2,
= -3
dt
dt
dt
We know, A = l × b
Given :
dA
db
dl
= l. + b.
= -3l + 2b
dt
dt
dt
Þ – 5 = – 3l + 2b.
When b = 2, we have
9
– 5 = – 3l + 4 Þ l = = 3m
3
16. (b) Let A = pr2.
Þ
dA
dr
= 2pr .
dt
dt
6p = 2p ( 30 ) .
Also, given
1
dr
= 1mm = cm
dt
10
\ A = pr2
dA
dr
1
= 2 pr
= 2p.50. = 10 p
dt
dt
10
Hence, area of plate increases in 10p cm2/hour.
14. (c) Let W = nw
dW
dw
dn
=n
+ w.
Þ
...(i)
dt
dt
dt
2
Given : w = t – t + 2 and n = 2t2 + 3
Þ
dw
dn
= 2t - 1 and
= 4t
dt
dt
\ Equation (i)
Þ
dw
= (2t2 + 3) (2t – 1) + (t2 – t + 2) (4t)
dt
Thus,
dW
= (2 + 3) (2 – 1) + (2) (4)
dt t =1
= 5 (1) + 8 = 13
dr
dt
3 dr
dr 1
=
Þ
=
= 0.1
30 dt
dt 10
Thus, the rate at which the radius of the circular sheet
increases is 0.1
17. (d)
Þ
u=0
f
t+m
s+n
u=0
f¢
t
s
A
æ dr ö
72 p = 4 p´ 9 ´ 9 ç ÷
è dt ø
dr æ 2 ö
=ç ÷
Þ
dt è 9 ø
(b) Let A = pr2 be area of metalic circular plate of
r = 50 cm.
Þ
15. (d) Let A be the area, b be the breadth and l be the length
of the rectangle.
B
v
v
As per question if point B moves s distance in t time then
point A moves (s + n) distance in time (t + m) after which
both have same velocity v.
Then using equation v = u + at we get
v = f (t + m) = f 't Þ t =
f m
f '- f
....(i)
Using equation v 2 = u 2 + 2 , as we get
v2 = 2 f ( s + n) = 2 f ' s
Þs=
f n
f '- f
....(ii)
1 2
Also for point B using the eqn s = ut + at , we get
2
1
2
s = f 't
2
Substituting values of t and s from equations (i) and (ii)
in the above relation, we get
f n
f 2m 2
1
= f'
f '- f 2 ( f '- f ) 2
Þ ( f '- f ) n =
1
ff ' m 2
2
Downloaded from @Freebooksforjeeneet
EBD_8344
M-352
Mathematics
18. (c) Let the li ard catches the insect after time t then
distance covered by li ard = 21cm + distance covered by
insect
19.
Þ
1 2
ft = 4 ´ t + 21
2
Þ
1
´ 2 ´ t 2 = 20 ´ t + 21
2
Þ
dr
= 50
dt
dy
dy 9
20. (a) Given y 2 = 18 x Þ 2 y
= 18 Þ
=
dx
dx y
dy 2dx
dy
=2
=
Þ
dt
dt
dx
æ
æp
öö
æp
ö
= x ç p - ç - sin -1 (sin | x |) ÷ ÷ = x ç + | x | ÷
è2
øø
è2
ø
è
x³0
x<0
ìp
+ 2x , x ³ 0
ïï 2
f ¢( x) = í
ï p - 2x , x < 0
ïî 2
æ -p ö
ç 2 ,0 ÷ .
è
ø
9
8
æ9 9ö
\ Required point is ç , ÷
è8 2ø
f ( x) = (3x - 7) × x
24. (b) Given functions are, f (x) = ex – x and g (x) = x2 – x
f (g (x)) = e( x
2/3
2
- x)
- (x 2 - x)
Given f (g (x)) is increasing function.
\(f (g (x)))’ = e( x
2
f '( x ) = 3 x 2 / 3 + (3 x - 7) × x -1/ 3
3
=
(Q f (x) is increasing)
æ pö
Hence, f ¢(x) is increasing in ç 0, ÷ and decreasing in
è 2ø
9
9
= 2Þ y =
y
2
Putting in y 2 = 18 x Þ x =
21. (a)
f ( c) - f ( a ) c - a
>
f (b ) - f ( c ) b - c
ì æp
ö
ïxç 2 + x ÷ ,
ï è
ø
f ( x) = í
ïxæ p - x ö ,
÷
ïî çè 2
ø
dr
50
1
cm/min
Þ dt = 4 p (15) 2 =
18p
Þ
f (c) - f (a) f (b) - f (c )
>
c-a
b-c
23. (d) f ¢(x) = x (p – cos–1 (sin|x|))
dv
d æ 4 3ö
= 50 cm3/min Þ
ç pr ÷ø = 50
dt
dt è 3
ATQ
f ( b) - f ( c )
= f ¢(b), bÎ (c, b)
b-c
Qf ²(x) < 0 Þ f ¢(x) is decreasing
f ¢(a) > f ¢(b) Þ
Þ t 2 - 20t - 21 = 0 Þ t = 21sec
(b) Given that
Total radius r = 10 + 5 = 15 cm
2
Þ 4pr
Again by LMVT for x Î [c, b]
= (2 x - 1)e( x
15 x - 14
2
2
- x)
-x)
´ (2 x - 1) - 2 x + 1
+ 1 - 2 x = (2 x - 1)[e ( x
2
- x)
- 1] ³ 0
For (f (g (x)))’ ³ 0,
3 x1/ 3
+
–
0
+
14
15
(2 x - 1) &[e( x
2
- x)
negative
+ve
For increasing function
æ 14 ö
f '(x) > 0 then x Î (-¥, 0) È ç , ¥÷
è 15 ø
22. (d) Since, function f (x) is twice differentiable and
continuous in x Î [a, b]. Then, by LMVT for x Î [a, c]
- 1] are either both positive or
0
–ve
1
2
é 1ù
x Î ê0, ú È [1, ¥)
ë 2û
f ( c ) - f (a )
= f ¢(a ), a Î (a, c )
c-a
Downloaded from @Freebooksforjeeneet
+
1
M-353
Applications of Derivatives
25. (c)
f ( x) =
Þ f ( - x) =
f ¢(- x) =
Þ 1 – 2a + 4 + a = 0
Þ a=5
Then, f(x) = x3 – 9x2 + 15x + 7
Now,
x2
1- x2
x2
1 - x2
= f ( x)
f ( x) - 14
=0
( x - 1)2
2x
(1 - x 2 )2
f(x) increases in x Î (10, ¥)
Also f(0) = 0 and
lim f ( x ) = -1 and f(x) is even function
x ®± ¥
Set A = R – [–1, 0)
And the graph of function f(x) is
Þ
x3 - 9 x 2 + 15 x + 7 - 14
=0
( x - 1)2
Þ
( x - 1)2 ( x - 7)
=0Þx=7
( x - 1) 2
x
=
0
–1
x
28. (a) f (x) =
2
a +x
2
a +x
2
b + ( x – d )2
f '(x) – f '(2 – x) > 0
Þ
f '(x) > f '(2 – x)
But f ''(x) > 0 Þf '(x) is an increasing function
Then, f '(x) > f '(2 – x) > 0
Þ x>2–x
Þ x>1
Hence, f (x) is increasing on (1, 2) and decreasing on (0, 1).
27. (c) f(x) = x3 – 3 (a – 2) x2 + 3ax + 7, f(0) = 7
2 a 2 + x2
b2 + ( x – d )2 –
+
=
(
a2
=
(b
2
(a
2
+x
)
2 3/2
+
)
b2
(b
2
+ ( x – d )2
)
3/2
Þ
f ¢(x) > 0, ฀ x Î R
Þ
f (x) is increasing function.
>0
Hence, f(x) is increasing function.
f ( x ) = x3 - 3 x 2 + 5 x + 7
For increasing
f ¢ ( x ) = 3x2 - 6 x + 5 > 0
Þ xÎ R
For decreasing
f ¢ ( x ) = 3x2 - 6 x + 5 < 0
x= 1
Þ
f ¢(x) = 3x2 – 6(a – 2) x + 3a
f ¢(1) = 0
30. (c) f (x) = sin4 x + cos4 x
f '(x) = 4sin3 x cos x + 4cos3 x (– sin x)
= 4sin x cos x (sin2 x – cos2 x)
= – 2sin 2x cos 2x = – sin 4x
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( x – d ) 2( x – d )
2 b 2 + ( x – d )2
+ ( x – d )2
a 2 + x 2 – x 2 b2 + ( x – d )2 – ( x – d )2
+
3/2
(a 2 + x 2 )3/2
b 2 + ( x – d )2
29. (a)
f (x) = 0 at x = 1
x(2 x)
(a 2 + x 2 )
–1
Þ
b + (d – x)2
2
2
f ¢(x) =
26. (c) f (x) = f (x) + f (2 – x)
Now, differentiate w.r.t. x,
f '(x) = f '(x) – f '(2 – x)
For f (x) to be increasing f '(x) > 0
(d – x)
–
(x – d )
+
a 2 + x2 –
1
2
)
EBD_8344
M-354
Mathematics
Thus statement -1 and 2 both are true and statement-2 is
a correct explanation of statement 1.
f (x) is increasing when f '(x) > 0
Þ – sin 4x > 0 Þ sin 4x < 0
Þ
f ( x ) = xe x(1- x) , x Î R
35. (c)
x Î çæ p , p ù
è 4 2 úû
f ' ( x ) = e x(1- x) . é1 + x – 2 x 2 ù
ë
û
x(1- x ) é 2
. 2 x - x –1ù
= -e
ë
û
y = sin 4x
p
4
= -2e
p
2
x(1- x )
éæ
ù
1ö
. êç x + ÷ ( x - 1)ú
è
ø
2
ë
û
f ' ( x ) = -2e x(1- x ) . A
1ö
æ
where A = çè x + ÷ø ( x - 1)
2
31. (b) Since f ¢(x) > 0 and g¢(x) < 0, therefore
f(x) is increasing function and g(x) is decreasing function.
Þ f (x + 1) > f (x) and g (x + 1) < g (x)
Þ g [f (x + 1)] < g [ f (x)] and f [g (x + 1)] < f [g (x)]
Hence option (b) is correct.
32. (d) f (x) = 2x3 + 3x + k
f'(x) = 6x2 + 3 > 0 " x Î R (Q x2 > 0)
Þ f (x) is strictly increasing function
Þ f (x) = 0 has only one real root, so two roots are not
possible.
33. (c) Let y = x2 . e–x
For increasing function,
dy
> 0 Þ x [(2 – x) e–x] > 0
dx
Q x > 0, \ (2 – x) e–x > 0
Þ (2 – x)
1
1
36. (b) Given that f (x) = x3 + 5x + 1
\ f ' (x) = 3x2 + 5 > 0, " x Î R
Þ f (x) is strictly increasing on R
Þ f (x) is one one
\ Being a polynomial f (x) is continuous and increasing.
f ( x ) = -¥
on R with xlim
®¥
Þ f '(x) = 7x6 + 70x4 + 48x2 + 30 > 0, " x Î R
Also lim f ( x) = ¥ and lim f ( x ) = – ¥
x ®¥
f(x) = x log x, xÎ [1, 2]
1
g(x) = 2–x, x Î [1, 2]
1
2
x® – ¥
From (i) and (ii) clear that the curve
y = f (x) crosses x-axis only once.
\ f (x) = 0 has exactly one real root.
38. (d) Given that f (x) = tan–1 (sin x + cos x)
Differentiate w.r. to x
Y
log 4
...(i)
Þ f is an increasing function on R
34. (a) f (x) = x log x, f (1) = 0, f (2) = 4
g(x) = 2 – x, g(1) = 1, g(2) = 0
log 10 > log 4 Þ 1 > log 4
O
é 1 ù
\ f(x) is increasing on ê - ,1ú
ë 2 û
R
\ Range of f = ( - ¥, ¥) =
Hence f is onto also. So, f is one one and onto R.
37. (b) Let f (x) = x7 + 14x5 + 16x3 + 30x –560
< 0 , but it is not possible
ex
Hence the statement-2 is false.
X¢
é 1 ù
Hence, f ¢(x) is +ve in ê - ,1ú
ë 2 û
x ®¥
e
For 0 < x < 2, (2 – x) < 0
\
é 1 ù
be opposite to the sign of A which is –ve in ê - ,1ú
ë 2 û
and lim f ( x ) = ¥
>0
x
Now, exponential function is always +ve and f ¢(x) will
X
f '(x) =
1
1 + (sin x + cos x) 2
.(cos x - sin x )
Y¢
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...(ii)
M-355
Applications of Derivatives
æ 1
ö
1
2. ç
cos x sin x ÷
è 2
ø
2
=
(4 x3 + x 4 × y ')e y +
1 + (sin x + cos x) 2
p
p
æ
ö
2 ç cos .cos x - sin .sin x÷
è
ø
4
4
=
1 + (sin x + cos x) 2
\
f '(x) =
y - 0 = -2( x - 1) Þ 2 x + y = 2
p
p p
< x+ <
2
4 2
\
3p
p
<x<
4
4
Hence, f (x) is increasing when
Þ -
dy
æ dy ö
æ dy ö
= 2 x - 3; ç ÷
= -1 and ç
=1
÷
è dx ø ( x = 1)
dx
è dx ø ( x = 2 )
Equation of tangent at A(1, 0),
y = -1( x - 1) Þ x + y = 1
Equation of tangent at B(2, 0),
æ p pö
n Îç - , ÷
è 2 4ø
y = 1( x - 2) Þ x - y = 2
39. (c) From option (c), f (x) = 3 x 2 - 2 x + 1 is increasing
So a = 1 and b = 2
Þ
when f '( x ) = 6x – 2 ³ 0
40.
- 4 x 3e y
æ dy ö
Þç ÷ =
è dx ø æ 1
ö
+ e y x4 ÷
ç
è y +1
ø
Only point (–2, 6) lies on the tangent.
42. (0.50)
The given curve y = (x – 1)(x – 2), intersects the x-axis at
A(1, 0) and B(2, 0).
pö
æ
\ f ' (x) > 0 Þ cos ç x + ÷ > 0
è
4ø
Þ
=0
\ Equation of tangent;
2
Given that f (x) is increasing
Þ-
1+ y
æ dy ö
Þç ÷
= -2
è dx ø(1,0)
pö
æ
2 cos ç x + ÷
è
4ø
1 + (sin x + cos x )
y'
x Î[1/ 3, ¥)
a 1
= = 0.5.
b 2
43. (4)
1ù
æ
\ f (x) is incorrectly matched with ç -¥, ú
3û
è
(b) The given tangent to the curve is,
For (1, 2) of y 2 = 4 x Þ t = 1, a = 1
Equation of normal to the parabola
y = x loge x
Þ x + y = 3 intersect x-axis at (3, 0)
(x > 0)
Þ
dy
= 1 + log e x
dx
Þ
dy ù
= 1 + loge c
dx úû x = c
dy
= ex
dx
Equation of tangent to the curve
y = ex Þ
(slope)
Q The tangent is parallel to line oining (1, 0), (e, e)
\1 + log e c =
e-0
e -1
1
e
Þ log e c =
- 1 Þ log e c =
e -1
e -1
1
e e -1
Þ tx + y = 2at + at 3
Þc=
4
y
41. (c) The given curve is, x × e + 2 y + 1 = 3
Differentiating w.r.t. x, we get
Þ y - e c = ec ( x - c)
Q Tangent to the curve and normal to the parabola
intersect at same point.
\ 0 - ec = ec (3 - c) Þ c = 4.
44. (91)
y=
6
ì3
4
ü
å k cos -1 íî 5 cos kx - 5 sin kxýþ
k =1
Let cos a =
3
4
and sin a =
5
5
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EBD_8344
M-356
Mathematics
6
\ y = å k cos -1{cos a cos kx - sin a sin kx}
k =1
3 (a2 + 3) = (a2 – 3)2 Þ a2 = 9
6
= å k cos -1 (cos(kx + a))
And, b =
k =1
6
6
k =1
k =1
= å k (kx + a) = å (k 2 x + ak )
\
6
6(7)(13)
dy
= å k2 =
= 91.
dx k =1
6
45. (4.0)P º (x1, y1)
2yy¢ – 6x + y¢ = 0
Þ
æ 6 x1 ö
y¢ = ç
÷
è 1 + 2 y1 ø
[By point slope form, y – y1 = m(x – x1)]
9 – 6y1 = 1 + 2y1
y1 = 1
x1 = ± 2
æ ±12 ö
Slope of tangent (m) = ç
÷ =±4
è 3 ø
\ |m| = 4
46. (d) Given equation of curve is
x2 + 2xy – 3y2 = 0
Þ 2x + 2y + 2xy¢ – 6yy¢ = 0
Þ x + y + xy¢ –3yy¢ = 0
Þ y¢(x – 3y) = – (x + y)
dy x + y
=
dx 3 y - x
- dx x - 3 y
=
Slope of normal =
dy
x - 3y
2-6
=–1
2+2
Equation of normal to curve = y – 2 = – 1 (x – 2)
Þ x+y=4
\ Perpendicular distance from origin
Normal at point (2, 2) =
=
47
0+ 0- 4
2
a
a
Þ =6
b
b
1
2
These values of a and b satisfies |6a + 2b| = 19
48. (d) y = x3 + ax – b
Since, the point (1, –5) lies on the curve.
Þ1+a– b= –5
Þ a – b = –6
Þ a = ±3, b = ±
...(i)
Since, required line is perpendicular to y = x – 4, then
slope of tangent at the point P (1, –5) = –1
3+a=–1
a=–4
b=2
the equation of the curve is y = x3 – 4x – 2
(2, –2) lies on the curve
49. (d) y = f(x) = x3 – x2 – 2x
dy
- 3x 2 - 2 x - 2
dx
f(1) = 1 – 1 – 2 = – 2,
f(–1) = –1 –1 + 2 = 0
Since the tangent to the curve is parallel to the line
segment oining the points (1, –2) (–1, 0)
Since their slopes are equal
Þ 3x 2 - 2 x - 2 =
-2 - 0
2
Þ x = 1,
-1
3
ì -1 ü
Hence, the required set S = í ,1ý
î3 þ
50. (c) Equation of tangent to circle at point ( 3,1) is
3x + y = 4
P( 3,1)
=2 2
2
(a) Given curve is, y =
Þ
Þ a2 - 3 =
æ dy ö
ç ÷
è dx øat x =1 = 3 + a
\
Þ
x
a2 - 3
dy
= 3x 2 + a
dx
æ3
ö
ç 2 - y1 ÷
æ 1 + 2 y1 ö
ç
÷ = -ç
÷
x
è 6 x1 ø
1 ÷
ç
è
ø
Þ
Þ
\
a2 - 3
2
1
dy
= 2
=- =dx (a,b) (a - 3) 2
6
3
x
2
x -3
dy ( x 2 - 3) - x (2 x ) - x 2 - 3
=
=
dx
( x 2 - 3)2
( x 2 - 3) 2
O
æ 4 ö
Aç
,0÷
è 5 ø
M
3x + y = 4
2
2
x +y =4
Downloaded from @Freebooksforjeeneet
M-357
Applications of Derivatives
æ 4 ö
,0 ÷
coordinates of the point A = ç
è 3 ø
Area =
y2 = x
53. (a)
1 4
2
1
´1 =
´ OA ´ PM = ´
sq. units
2
3
3
2
tö
æ1
P ç t2, ÷
2ø
è4
51. (c) Given, the equation of parabola is,
x2 = 12 – y
æ3 ö
Q ç ,0 ÷
è2 ø
y
(0, 12)
(–t, 12–t2)
(t, 12–t2)
x
1
Here the curve is parabola with a = .
4
æ t2 t ö
Let P(at2, 2at) or P ç , ÷ be a point on the curve.
è 4 2ø
Now, y2 = x
1
dy
dy
Þ2 y dx = 1 = dx =
2 x
Area of the rectangle = (2t) (12 – t2)
A = 24t – 2t3
1
æ dy ö
=
Þçè dx ÷ø
at p t
dA
= 24 – 6t2
dt
\
tö
1 2ö
æ
æ
çè y - ÷ø = -t çè x - t ÷ø
4
2
dA
Put
= 0 Þ 24 – 6t2 = 0
dt
Þ t = ±2
–
+
–2
52.
1 1 3
Þy = -tx + t + t
2 4
–
c =-
...(i)
æ3 ö
For minimum PQ, (i) passes through Q çè , 0÷ø
2
+2
At t = 2, area is maximum = 24(2) – 2(2)3
= 48 – 16 = 32 sq. units
(b) Q Tangent to the given curve is parallel to line 2y = 4x + 1
\ Slope of tangent (m) = 2
Then, the equation of tangent will be of the form
y = 2x + c
...(i)
2
Q Line (i) and curve y = x – 5x + 5 has only one point
of intersection.
\ 2x + c = x2 – 5x + 5
x2 – 7x + (5 – c) = 0
\ D = 49 – 4(5 – c) = 0
Þ
equation of normal at P to y2 = x is,
29
4
-3 t t 3
t + + = 0 Þ –4t + t3 = 0
2
2 4
Þ t(t2 – 4) = 0 Þ t = –2, 0, 2
Q t ³ 0 Þ t = 0, 2
If t = 0, P(0, 0) Þ AP =
If t = 2, P(1, 1) Þ AP =
3
2
5
2
æ3 ö
5
Shortest distance çè , 0÷ø and y = x is
2
2
54. (b) The equation of curve y = xe x
Hence, the equation of tangent: y = 2x –
29
4
Þ
dy
2
2
= e x .1 + x.e x .2 x
dx
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2
EBD_8344
M-358
Mathematics
2
Since (1, e) lies on the curve y = xe x , then equation of
tangent at (1, e) is
(
2
x
2
y – e = e (1 + 2 x )
)
x =1
( x - 1)
y – e = 3e(x – 1)
3ex – y = 2e
So, equation of tangent to the curve passes through the
Adding eqn (i) and (ii), we get
2y = 12 Þ y = 6
Then, from eqn (i)
Differentiate equation (i) with respect to x
55. (c) f(x) = y = x3/2 + 7
Þ
y = 2 + x2 ...(ii)
x = ±2
æ4
ö
point çè , 2e÷ø
3
Þ
56. (b) Since, the equation of curves are
y = 10 – x2 ...(i)
dy
3
x >0
Þ
dx
2
f(x) is increasing function " x > 0
dy
= –2x Þ
dx
æ dy ö
èç dx ÷ø (2, 6) = 4 and
At (–2, 6), tan q =
\ |tan q| =
(
)
(4) - ( -4)
8
8
=
Þ tan q =
1 + (4)( -4) -15
15
8
15
57. (c) Let curve intersect each other at point P(x1, y1)
Y
mTP = mat P = –1
y2 = 6x
Þ
æ
ö
1
3/ 2
ç x1 ÷ 3 2
´
x
ç
1 ÷ 2 1 = –1
x èç 1 2 ø÷
Þ
x12
2
- =x -1
1
3
2
Þ
-3x12 = 2x1 – 1 Þ 3 x12 + 2 x1 - 1 = 0
Þ
3x12 + 3x1 - x1 - 1 = 0
Þ
3x1(x1 + 1) – 1(x1 + 1) = 0
Þ
x1 =
1
3
P(x1, y1)
X
9x2 + by 2 = 16
(Q x1 > 0)
1 ö
æ1
Þ P çè 3 ,7 +
÷
3 3ø
TP =
æ dy ö
çè ÷ø
dx ( -2, 6) = –4
æ ( -4) - (4) ö 8
At (2, 6) tan q = çè 1 + ( -4) ´ (4) ÷ø = 15
1
,7 T
2
3/ 2
Let P x1 , x1 + 7
æ dy ö
çè ÷ø
dx ( -2, 6) = 4
Differentiate equation (ii) with respect to x
dy
= 2x Þ
dx
(0, 7)
æ dy ö
çè ÷ø
dx (2, 6) = –4 and
1
1 1 7
+
=
27 36 6 3
Since, point of intersection is on both the curves, then
y12 = 6x1
...(i)
and 9x12 + by12 = 16
...(ii)
Now, find the slope of tangent to both the curves at the
point of intersection P(x1, y1)
For slope of curves:
Curve (i):
æ dy ö
çè ÷ø
dx (x
1,y1 )
= m1 =
3
y1
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M-359
Applications of Derivatives
Equation of tangent at (x1, y1) is
y = mx + c
Curve (ii):
9x
æ dy ö
= m2 = - 1
and çè dx ÷ø
by1
(x1,y1 )
Þ y=
Since, both the curves intersect each other at right angle
then,
m1m 2 = -1 Þ
27x1
by12
= 1 Þ b = 27
x1
1 9
=
6 2
58. (c) Let P(2t, t2) be any point on the parabola.
Centre of the given circle C = (– g, – f ) = (–3, 0)
For PC to be minimum, it must be the normal to the
parabola at P.
y2 - y1 t 2 - 0
=
x2 - x1 2t + 3
Also, slope of tangent to parabola at P =
\ Slope of normal =
dy x
= =t
dx 2
-1
t
t 2 - 0 -1
=
2t + 3 t
Þ t 3 + 2t + 3 = 0
Þ (t + 1) (t 2 – t + 3) = 0
\ Real roots of above equation is
t=–1
Coordinate of P = (2t, t 2) = (– 2, 1)
Slope of tangent to parabola at P = t = – 1
Therefore, equation of tangent is:
(y – 1) = (– 1) (x + 2)
Þ x+y+1=0
59. (d) Equation of hyperbola is :
4y2 = x2 + 1
Þ – x2 + 4y2 = 1
x2
12
+
y2
æ1ö
ç ÷
è2ø
2
=1
\ a = 1, b = 1
2
4 y12 – x12
4 y1
Therefore, y =
dy
= 2 x1
dx
x
dy 2 x1
Þ
=
= 1
dx 8 y1 4 y1
=
1
4 y1
1
x1
x+
4 y1
4 y1
Þ 4y1y = x1 x + 1
æ
1 ö
B ç 0,
÷
è 4 y1 ø
Let midpoint of AB is (h, k)
\ h=
–1
2 x1
Þ x1 =
–1
1
& y1 =
2h
8k
2
2
æ 1 ö æ – 1ö
Thus, 4 ç ÷ = ç ÷ + 1
è 8k ø è 2h ø
Þ
1
16 k 2
Þ1=
=
1
4h2
+1
16k 2
+ 16k 2
4h 2
Þ h2 = 4k2 + 16 h2 k.
So, required equation is
x2 – 4y2 – 16 x2 y2 = 0
60. (b) Since, x2 + 3y2 = 9
Þ 2x + 6y
Now, tangent to the curve at point (x1, y1) is given by
4 ´ 2 y1
ÞC =
x1 x1
+c
4 y1
æ –1 ö
, 0 ÷ and y axis at
which intersects x axis at A ç
è x1 ø
\
Þ –
As tangent passes through (x1, y1)
\ y1 =
y12
\ from equation (i), b = 27 ´
Slope of line PC =
x1
×x+c
4 y1
Þ
dy
=0
dx
dy – x
=
dx 3 y
Slope of normal is –
dx
3y
=
dy
x
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EBD_8344
M-360
Mathematics
æ
dx ö
÷
è dy ø(3cos q,
Þ ç–
æ dx ö
& ç–
÷
è dy ø( -3sin q,
=
3sin q)
=
3 3 sin q
= 3 tan q = m1
3 cos q
4(x0)2 – 9(y0)2 = 36
- 9x
4y
and y0 =
13
13
From equation (ii), we get
9x2 – 4y2 = 169
Hence, locus of point P is : 9x2 – 4y2 = 169
As, b is the anagle between the normals to the given ellipse
then
=
So, tan b
=
x+6
62. (c) We have y = (x - 2)(x - 3)
At y-axis, x = 0 Þ y = 1
On differentiating, we get
m1 – m2
1 + m1m2
3 tan q + 3 cot q
1 – 3 tan q cot q
...(ii)
From equation (i): x0 =
3 cos q)
3 3 cos q
= – 3 cot q = m2
– 3 sin q
tan b =
æ - 13x0 13 y0 ö
,
...(i)
So, P (x, y) º ç
4 ÷ø
è 9
Q (x0, y0) lies on hyperbola, therefore
=
3 tan q + 3 cot q
1–3
3
tan q + cot q
2
dy (x 2 - 5x + 6) (1) - (x + 6) (2x - 5)
=
dx
(x 2 - 5x + 6)2
dy
= 1 at point (0, 1)
dx
Þ
1
3 sin q cos q
=
+
cot b
2 cos q sin q
\ Slope of normal = – 1
Now equation of normal is y – 1 = –1 (x – 0)
Þ y–1=–x
x + y= 1
Þ
1
3
1
=
cot b
2 sin q cos q
æ1 1ö
\ ç , ÷ satisfy it.
è2 2ø
Þ
1
3
=
cot b sin 2 q
Þ
2 cot b
2
=
sin 2 q
3
61. (c) Given, 4x2 – 9y2 = 36
After differentiating w.r.t. x, we get
4.2.x – 9.2.y.
dy
=0
dx
dy 4 x
=
dx 9 y
-9 y
So, slope of normal =
4x
Now, equation of normal at point (x0, y0) is given by
Þ Slope of tangent =
y – y0 =
-9 y0
(x – x0)
4 x0
63. (c) Eccentricity of ellipse =
Now, –
1
a
=–4Þa=4× =2Þa=2
2
e
æ 1ö
We have b2 = a2 (1 – e2) = a2 ç 1 - ÷
è 4ø
3
=3
4
\ Equation of ellipse is
=4×
x 2 y2
+
=1
4
3
Now differentiating, we get
x 2y
+ ´ y¢ = 0 Þ y¢ = – 3x
2 3
4y
As normal intersects X axis at A, Then
Þ
æ 13x0 ö
æ 13 y0 ö
Aº ç
, 0 ÷ and B º ç 0,
4 ÷ø
9
è
ø
è
As OABP is a parallelogram
3 2
1
y¢ (1,3/ 2) = - ´ = 4 3
2
æ 13 y0 ö
º Midpoint of AP
\ midpoint of OB º ç 0,
8 ÷ø
è
1
2
Slope of normal = 2
Downloaded from @Freebooksforjeeneet
M-361
Applications of Derivatives
65. (d) x2y2 – 2x = 4 – 4y
Differentiate w.r.t. 'x'
æ 3ö
\ Equation of normal at ç1, ÷ is
è 2ø
2xy2 + 2y . x2 .
3
y – = 2 (x – 1) Þ 2y – 3 = 4x – 4
2
\ 4x – 2y = 1
64. (c)
Y
B
,
(x
y)
X
Let y = f (x) be a curve
slope of tangent = f ¢ (x)
Equation of tangent (Y – y) = f ¢ (x) (X – x)
Put Y = 0
æ
y ö
X = çç x ÷
f ¢ ( x ) ÷ø
è
Put X = 0
Y = y - x f ¢ (x)
Þ
æ
ö
y
, 0 ÷÷
A = çç x ¢
f
x
(
)
è
ø
and B = (0, y – x f ¢ (x))
\ AP : PB = 1 : 3
3æ
y ö
çx ÷
4 çè
f ¢ ( x ) ÷ø
Þ
x =
Þ
dy -3y
-3y
=
Þ
x=
dx
x
¢
f (x)
dy -3 dx
C
=
Þ y= 3
y
x
x
Q
\
f (a) = 1 Þ C = 1
1
y= 3
x
dy
(2y . x2 + 4) = 2 – 2x . y2
dx
Þ
2 - 2 ´ 2 ´ 4 -14 7
dy
=
=
=
dx 2, -2 2 ( -2 ) ´ 4 + 4 -12 6
\
Equation of tangent is
7
or 7 x - 6 y = 26
6
\ (–2, –7) does not passes through the required tangent.
A
Þ
Þ
( y + 2) = ( x - 2)
P
Þ
æ 1ö
is required curve and ç 2, ÷ passing
è 8ø
66. (d)
æ 1 + sin x ö
f ( x) = tan –1 ç
è 1– sin x ÷ø
æ
ç
= tan –1 ç
ç
ç
è
2ö
x
xö
æ
xö
æ
çè sin + cos ÷ø ÷
1 + tan
÷
2
2 ÷
–1 ç
2
÷
2 ÷ = tan ç
x
x
x
æ
ö
ç 1 – tan ÷
çè sin – cos ÷ø ÷
è
2ø
2 ø
x
æ
æ p xö ö
= tan –1 ç tan ç + ÷ ÷
è 4 2ø ø
è
Þ y=
p x
dy 1
+
Þ
=
dx 2
4 2
Slope of normal =
–1
= –2
dy
æ ö
çè ÷ø
dx
æp p pö
Equation of normal at çè , + ÷ø
6 4 12
pö
æp pö
æ
y – ç + ÷ = –2 ç x – ÷
è 4 12 ø
è
6ø
y–
4p
2p
= –2 x +
12
6
y–
p
p
= –2 x +
3
3
y = –2 x +
2p
3
æ 2p ö
This equation is satisfied only by the point ç 0, ÷
è 3ø
67. (a)
1
through y = 3
x
dy
dy
– 2 = –4 .
dx
dx
Þ
Þ
dy
1
2
=
´4=
3
dx 2 4 x - 3
4x – 3 = 9
x=3
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EBD_8344
M-362
Mathematics
71. (d) x2 – y + 6 = 0
So, y = 4
Equation of normal at P (3, 4) is
2x –
3
y – 4 = – (x – 3)
2
i.e. 2y – 8 = – 3x + 9
Þ 3x + 2y – 17 = 0
This line is satisfied by the point (1, 7)
68. (d) P (4t2 + 3,8t3 – 1)
dy
=4
dx ( x, y )= (2,10)
equation of tangent
y – 10 = 4(x – z)
4x – y + z = 0
tangent passes through (a, b)
4a – b + z = 0 Þ b = 4a + z
and 2x + 2yy' + 8 – 2y' = 0
dy / dt dy
=
= 3t (slope of tangent at P)
dt / dt dx
Let Q = (4l2 + 3,8l3 – 1)
slope of PQ = 3t
8t 3 - 8l3
2
2
2 x + 8 2a + 8
=
=4
2 - 2 y 2 - 2b
from (i) and (ii)
= 3t
a=
-t
2
\ Q [t2 + 3, – t3 – 1].
69. (b) Given curve is
x2 + 2xy – 3y2 = 0
Differentiatew.r.t. x
...(i)
so,
dy
dy
+ 2y - 6y
=0
dx
dx
Equation of normal at (1, 1) is
y=2–x
Solving eqs. (i) and (ii), we get
x=1,3
Point of intersection (1, 1), (3, –1)
Normal cuts the curve again in 4th quadrant.
æp
ö
70. (b) Given curve is sin y = x sin ç + y÷
è3
ø
Diff with respect to x, we get
æp
ö
æp
ö dy
dy
cos y
= sin çè + y÷ø + x cos çè + y÷ø
3
3
dx
dx
æp
ö
sin ç + y÷
è3
ø
æp
ö
cos y - x cos ç + y ÷
è3
ø
dy
3
at (0, 0) =
dx
2
dx
= -2 sin t + 2 [ t cos t + sin t ]
dt
dy
= 2t sin t
dx
...(ii)
dy 2t sin t
=
dx 2t cos t
dy
= tan t
dx
æ dy ö
=1
çè ÷ø
dx t =p / 4
so the slope of the normal is – 1
At t = p /4x =
2+
p
2 2
and
y = 2 - p/ 2 2
the equation of normal is
dy
Þ
=
dx
3 y=0
–8
2
,b =
17
17
dy
= 2 cos t - 2 [ - t sin t + cos t ]
dt
æ dy ö
=1
ç ÷
è dx ø(1, 1)
Þ Equation of normal is y – 0 = -
...(ii)
æ -8 2 ö
ç , ÷
è 17 17 ø
72. (a) Given that
x = 2 cos t + 2t sin t
t = l (or) l =
Þ 2x +
...(i)
y¢ =
4t - 4l
Þ t3 – 3l2t + 2l3 = 0
(t – l) . (t2 + tl – 2l2) = 0
(t – l)2 . (t + 2l) = 0
2x + 2x
dy
dy
=0Þ
= 2x
dx
dx
éy ë
2
3
(
y- 2+
(x – 0)
)
( (
2 - p / 2 2 ù = -1 éê x û
ë
p
2 2
))
2 + p / 2 2 ùú
û
= -x + 2 + p / 2 2
x + y = 2 2 , so the distance from the origin is 2
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M-363
Applications of Derivatives
73. (c) Given, y = 3 sin q.cos q
Þ x=
dy
= 3[sin q(- sin q) + cos q(cos q)]
dq
Thus x + 2y = k
dy
2
2
= 3[cos q - sin q] = 3 cos 2q
dq
and x = eq sin q
Now
...(ii)
3(cos 2 q - sin 2 q)
dy
3cos 2q
= q
= q
dx
e (sin q + cos q)
e (sin q + cos q)
3( cos q + sin q )(cos q - sin q)
dy
=
eq ( sin q + cos q )
dx
3(cos q - sin q)
dy
=
dx
eq
dy 2 x x dy ù
=
= , ú
=1
4 dx û x = 4
dx 8
1
= -1
dy
dx
Euqation of normal at x = 4 is
y – 2 = – 1 (x – 4)
Þy = – x + 4 + 2 = – x + 6
Þx+ y = 6
76. (c) Since the tangent is parallel to x-axis,
Slope of normal = -
8
dy
= 0 Þ 1- 3 = 0 Þ x = 2 Þ y = 3
dx
x
Equation of the tangent is y – 3 = 0 (x – 2)
Þ y=3
dy
= 0
dx
3(cos q - sin q)
dy
= 2 x - 5 \ m1 = (2 x - 5)(2, 0) = -1 ,
dx
m2 = (2 x - 5)(3, 0) = 1 Þ m1m2 = -1
77. (b)
eq
or cos q – sin q = 0 Þ cos q = sin q
tan p
p
Þ q=
4
4
i.e. the tangents are perpendicular to each other.
78. (d) Given x = a ( cos q + q sin q)
74. (d) Let y = cos (x + y)
dy
æ dy ö
= - sin ( x + y ) ç1 + ÷
Þ
dx
è dx ø
Now, given equation of tangent is
x + 2y = k
So,
...(i)
\
Given tangent is parallel to x-axis then
Þ Slope =
p
=k
2
75. (d) x2 = 8y
When, x = 4, then y = 2
dx
= eq (sin q + cos q)
dq
Dividing (i) by (i)
Þ tan q = 1 Þ tan q =
Þ
...(i)
dx
= eq cos q + sin q eq
dq
0=
p
and y = 0
2
-1
2
Þ
dx
= a ( - sin q + sin q + q cos q )
dq
Þ
dx
= aq cos q
dq
...(i)
.....(i)
y = a ( sin q - q cos q)
dy
= a [ cos q - cos q + q sin q ]
dq
dy -1
put this value in (i), we get
=
dx 2
dy
= aq sin q
dq
From equations (i) and (ii) we get
Þ
.....(ii)
-1
æ 1ö
= - sin ( x + y ) ç1 - ÷
2
è 2ø
dy
= tan q Þ Slope of normal = – cot q
dx
Þ sin (x + y) = 1
Equation of normal at ' q ' is
y – a (sin q – q cos q) = – cot q (x – a (cos q + q sin q))
Þ y sin q – a sin 2 q + a q cos q sin q
Þ x+ y =
p
p
Þ y= -x
2
2
p
Now, - x = cos (x + y)
2
= – x cos q + a cos 2 q + a q sin q cos q
Þ x cos q + y sin q = a
Clearly this is an equation of straight line which is at a
constant distance ‘a’ from origin.
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EBD_8344
M-364
Mathematics
79. (d) Since, x = a (1 + cos q)
æ 3p ö
M = f ç ÷ = -2 + 1 = -1
è 4ø
dx
= -a sin q and y = a sin q
dq
dy
Þ
= a cos q
dq
Þ
So, (m, M ) = ( -3, - 1)
82. (5)
dy
= - cot q.
dx
\ The slope of the normal at q = tan q
\ The equation of the normal at q is
C
\
D
8
y - a sin q = tan q( x - a - a cos q)
Þ y cos q – a sin q cos q = x sinq – a sinq – a sinq cosq
Þ x sin q - y cos q = a sin q
Þ y = ( x - a ) tan q
f ¢¢ ( x) = 6( x - 1). Inegrating, we get
f ¢ ( x) = 3 x 2 - 6 x + c
Slope at (2, 1) = f ¢(2) = c = 3
[Q slope of tangent at (2,1) is 3]
\ f ¢ ( x ) = 3 x 2 - 6 x + 3 = 3( x - 1) 2
Inegrating again, we get f ( x ) = ( x - 1)3 + D
The curve passes through (2, 1)
Let f ( x) = 2
2
sin x
cos 2 q
= 0 - sin 2 x
1
M
10–x
B
\ (MD)2 + (MC ) 2 = 64 + x 2 + 121 + (10 - x) 2 = f ( x)
(say)
f '( x) = 2 x - 2(10 - x) = 0
Þ 4 x = 20 Þ x = 5
f ''( x ) = 2 - 2( -1) > 0
\ f (x) is minimum at x = 5 m.
83. (d)
f ( x) = (1 - cos 2 x)(l + sin x) = sin 2 x(l + sin x)
Þ f ( x) = l sin 2 x + sin 3 x ...(i)
2l
Þ x = a (let)
3
So, f (x) will change its sign at x = 0, a because there is
Þ sin x = 0 and sin x = -
sin 2 x
sin 2 x
1
sin 2 x 1 + sin 2 x
R1 ® R1 - 2 R3 ; R2 ® R2 - 2 R3
0
x
Þ f '( x ) = sin x cos x [2l + 3sin x] = 0
Þ 1 = (2 - 1)3 + D Þ D = 0
\ f (x) = ( x – 1)3
81. (b) C1 ® C1 + C2
2 1 + sin 2 x
A
Let AM = x m
which always passes through (a, 0)
80. (b)
11
æ -p p ö
exactly one maxima and one minima in ç
,
è 2 2 ÷ø
–
-p
2
- (2 + sin 2 x)
+
a
0
- (2 + sin 2 x) = -2 - 2 sin 2 x
sin x
–
f '( x) = -2cos 2 x = 0
Þ cos 2 x = 0 Þ x =
-
p 3p
,
4 4
æ pö
So, f '' ç ÷ = 4 > 0
è 4ø
Þ -1 £ -
(minima)
æ pö
m = f ç ÷ = -2 - 1 = -3
è 4ø
æ 3p ö
f '' ç ÷ = -4 < 0
è 4ø
+
a
p
2
Now, sin x = -
f ''( x) = 4sin 2 x
+
p
2
OR
1 + sin 2 x
2
–
–
0
+
p
2
2l
3
2l
3
3
£ 1 Þ - £ l £ - {0}
3
2
2
Q If l = 0 Þ f ( x) = sin 3 x (from (i))
Which is monotonic, then no maxima/minima
(maxima)
æ 3
So, l Î ç - ,
è 2
3ö
÷ - {0}
2ø
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M-365
Applications of Derivatives
84. (d) The given function
Q f ( x ) = f (0)
f ( x ) = (3x 2 + ax - 2 - a)e x
f '( x) = (6 x + a)e + (3x + ax - 2 - a)e
2
x
Þk
x
( x4 - 2 x2 )
+C = C
4
Þ x2 ( x2 - 2) = 0
f '( x) = [3x 2 + (a + 6) x - 2]e x
Q x = 1 is critical point :
Þ x = 0, 2, - 2
\ f '(1) = 0
Þ T = {0, 2, - 2}
Þ (3 + a + 6 - 2) × e = 0
87. (3) Let f(x) = ax3 + bx2 + cx + d
(Q e > 0)
Þ a = -7
\ f '( x) = (3x2 - x - 2)e x
= (3x + 2)( x - 1)e x
+
–2/3
\x = -
1
and x = 1 is point of local minima.
85. (d) Area of rectangle ABCD
A = 2 x × ( x - 1) = 2 x - 2 x
b=
-3
9
,c=4
4
Þ
f(x) = a(x3 – 3x2 – 9x) + d
f ¢(x) =
æ d 2 Aö
-12
= (12 x) Þ ç 2 ÷
=
<0
è dx ø x = -1
dx
3
y = x2–1
x
C
(x, x2–1)
y'
\ Maximum area =
2
3 3
-
2
3
=
4
3 3
86. (a) Q The critical points are –1, 0, 1
\ f '( x) = k × x( x + 1)( x - 1) = k ( x 3 - x)
æ x4 x2 ö
Þ f ( x) = k ç - ÷ + C
2ø
è 4
3
æ
ax5 + bx 4 + cx 3 ö
lim ç 2 +
÷=4
÷
x ®0 çè
x3
ø
3
D
(–x, x2–1)
–
88. (d) f(x) = ax5 + bx4 + cx3
y
B
x = 3, – 1
–1
Local minima exist at x = 3
2
A
3 2
(x – 2x – 3) = 0
4
졔
1
dA
=0Þ x= ±
dx
3
d2A
Þ f (0) = C
...(ii)
Þ
For maximum area
x'
a+b+c+d=–6
3
dA
= 6x2 - 2
dx
\
...(i)
1
35
a= ,d =
4
4
2
is point of local maxima.
3
2
– a + b – c + d = 10
Solving equations (i) and (ii), we get
+
–
f(–1) = 10 and f(1) = – 6
Þ 2+c=4 Þ c=2
f ¢(x) = 5ax4 + 4bx3 + 6x2
= x2(5ax2 + 4bx + 6)
Since, x = ± 1 are the critical points,
\ f ¢(1) = 0 Þ 5a + 4b + 6 = 0
f ¢(–1) = 0 Þ 5a – 4b + 6 = 0
From eqns. (i) and (ii),
b = 0 and a = -
...(i)
...(ii)
6
5
-6 5
x + 2 x3
5
f ¢(x) = – 6x4 + 6x2 = 6x2 (–x2 + 1)
= – 6x2 (x + 1) (x – 1)
–
졔
f ( x) =
\
–
–1
1
f(x) has minima at x = – 1 and maxima at x = 1
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EBD_8344
M-366
Mathematics
89. (b) Given function f (x) = x kx - x 2 = kx 3 - x 4
Differentiating w. r. t. x,
(3kx 2 - 4 x3 )
f ' (x) =
2 kx3 - x 4
r
3
³ 0 for x Î [0, 3]
h
2
[Q f (x) is increasing in [0, 3]]
Þ3k – 4x ³ 0 Þ 3k ³ 4x
i.e., 3k ³ 4x for x Î [0, 3]
\k ³ 4 i.e., m = 4
Putting k = 4 in the function, f (x) = x 4 x - x 2
For max. value, f ‘ (x) = 0
2
3
i.e. 12 x - 4 x = 0 Þ x = 3
2 4 x3 - x 4
y = 3 3 i.e., M = 3 3
90. (b) a6 = a + 5d = 2
Here, a is first term of A.P and d is common difference
Let A = a1 a4 a5 = a (a + 3d) (a + 4d)
= a (2 – 2d) (2 – d)
A = (2 – 5d) (4 – 6d + 2d2)
dA
=0
By
dd
(2 – 5d) (– 6 + 4d) + (4 – 6d + 2d2) (– 5) = 0
8 2
– 15d2 + 34d – 16 = 0 Þ d = ,
5 3
8 d A
,
<0.
5 dd 2
8
Hence d =
5
91. (c) f (x) = 9x4 + 12x3 – 36x2 + 25
f '(x) = 36[x3 + x2 – 2x] = 36x (x – 1) (x + 2)
–
–
+
–2
0
æ
h2 ö
p 3
V = ph çç 9 - 4 ÷÷ Þ V = 9ph - h
4
è
ø
Differentiating w.r.t. h,
dV
3
= 9p - ph2
dh
4
For maxima/minima,
dV
= 0 Þ h = 12
dh
and
d 2V
3
= - ph
2
dh
æ d 2V
ç 2
ç dh
è
2
For d =
h2
=9
4
Now, volume of cylinder, V = pr2h
Substitute the value of r2 from equation (i),
\ r2 +
+
1
Here at –2 & 1, f '(x) changes from negative value to
positive value.
Þ –2 & 1 are local minimum points. At 0, f '(x) changes
from positive value to negative value.
Þ 0 is the local maximum point.
Hence, S1 = {–2, 1} and S2 = {0}
92. (c) Let radius of base and height of cylinder be r and h
respectively.
2
ö
<0
÷
÷
øh = 12
Volume is maximum when h = 2 3
93. (a) Let, the functions is,
f(q) = 3cos q + 5sin q × cos
= 3cos q + 5 ´
p
p
- 5sin cos q
6
6
3
1
sin q - 5 ´ cos q
2
2
5ö
3
æ
sin q
= çè 3 - ÷ø cos q + 5 ´
2
2
=
1
5 3
cos q +
sin q
2
2
max f(q) =
1 25
76
+ ´3 =
= 19
4 4
4
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...(i)
M-367
Applications of Derivatives
97. (d)
94. (b)
X(t2, 2t)
Q(9, 6)
h2 + r2 = l2 = 9 ...(i)
Volume of cone
P(4, –4)
1 2
pr h ...(ii)
3
From (i) and (ii),
V=
Parametric equations of the parabola y2 = 4x are,
x = t2 and y = 2t.
1
2
ÞV = p(9 - h )h
3
t2
1
4
Area DPXQ = 2
9
dv 1 (
1
3
= p 9 - 3h 2 )
ÞV = p(9h - h ) Þ
dh 3
3
For maxima/minima,
2t 1
-4 1
6
1
dV
1
2
= 0 Þ p(9 - 3h ) = 0
dh
3
= –5t2 + 5t + 30
= –5(t2 – t – 6)
Þh = ± 3 Þ h = 3
éæ 1 ö 2 25 ù
= -5 êçè t - 2 ÷ø - 4 ú
ëê
ûú
For maximum area t =
Now;
1
2
æ 25 ö 125
\ maximum area = 5 çè ÷ø =
4
4
95. (c) Consider the function,
f(x) = 3x(x – 3)2 – 40
NowS = {x Î ฀R : x2 + 30 £ 11
1x}
So x2 – 11x + 30 £ 0
Þ x ฀Î [5, 6]
\ f(x) will have maximum value for x = 6
The maximum value of function is,
f(6) = 3 × 6 × 3 × 3 – 40 = 122.
96. (c) A =
xm y n
1
=
(1+ x )(1+ y 2n ) ( x – m + x m )( y – n + y n )
2m
1
xm + y – m
³ ( x m . x – m ) 2 Þ xm + x–m ³ 2
2
In the same way, y–n + yn ³ 2
Then, (xm + x–m) (y–n + yn) ³ 4
Þ
1
1
£
–m
–n
m
n
( x + x )( y + y ) 4
(Q h > 0)
d 2V 1
= p ( -6 h )
dh2 3
æ d 2V ö
Here, èç dh 2 ø÷
at h =
<0
3
Then, h =
3 is point of maxima
Hence, the required maximum volume is,
V=
1
p(9 - 3) 3 = 2 3p
3
1
2
1ö
2
æ
x
= çx - ÷ +
98. (c) Here, h(x) =
1
1
è
xø
xxx
x
x2 +
When x \
x-
1
<0
x
1
+
x
2
x-
1
x
£ -2 2
Hence, -2 2 will be local maximum value of h(x).
When x -
1
>0
x
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EBD_8344
M-368
\
Mathematics
x-
1
2
+
³2 2
x x- 1
x
Þ b2 = 2hr – h2 = 2 ×
Hence, 2 2 will be local minimum value of h(x).
99. (a) Here, f (x) = 2x3 – 9x2 + 12x + 5
Þ
f ¢(x) = 6x2 – 18x + 12 = 0
For maxima or minima put f ¢(x) = 0
Þ x2 – 3x + 2 = 0
Þ x = 1 or x = 2
Now, f ²(x) = 12x – 18
Þ
f ²(1) = 12(1) – 18 = – 6 < 0
Hence, f (x) has maxima at x = 1
\ maximum value = M = f (1) = 2 – 9 + 12 + 5 = 10.
And, f ²(2) = 12(2) – 18 = 6 > 0.
Hence, f (x) has minima at x = 2.
\ minimum value = m = f (2)
= 2(8) – 9(4) + 12(2) + 5 = 9
\ M – m = 10 – 9 =1
100. (a) Sphere of radius r = 3 cm
Let b, h be base radius and height of cone respectively.
So, volume of cone =
1 2
pb h
2
=
4r
16r 2
8r 2 16r 2
r–
=
–
3
9
3
9
(24 – 16) r 2 8r 2
=
9
9
Þb=
2 2
r = 2 2 cm
3
Therefore curved surface area = pbl
= pb h 2 + r 2 = p2 2 42 + 8 = 8 3p cm2
101. (d) We have
Total length = r + r + rq = 20
Þ 2r + r q = 20
Þq=
20 - 2r
r
q
´ pr 2
2p
A = Area =
=
...(i)
1 2
1 æ 20 - 2r ö
r q = r2 ç
÷
2
2 è r ø
r
q
A = 10r – r2
For A to be maximum
A
r
h–r
B
b
C
1
1
ph é2hr – h 2 ù = é 2h 2 r – h 3 ù
ë
û 3ë
û
3
dv 1 é
= 4hr – 3h 2 ùû = 0 Þ h ( 4r – 3h ) = 0
dh 3 ë
= –2 < 0
dr 2
\ For r = 5 A is maximum
From (i)
20 - 2(5) 10
=
=2
θ =
5
5
2
´ p(5) 2 = 25 sq. m
A=
2p
102. (a) 4x + 2pr = 2
Þ 2x + pr = 1
2
2
S = x + pr
2
æ 1 - pr ö
2
S=ç
÷ + pr
è 2 ø
d 2v
1
= [ 4r – 6 h ]
2
3
dh
At h =
qr
d 2A
In right angled D ABC by Pythagoras theorem
...(i)
(h – r)2 + b2 = r2
2
2
2
2
2
2
Þ b = r – (h – r) = r – (h – 2hr + r ) = 2hr – h2
\ Volume (v) =
dA
= 0 Þ 10 – 2r = 0
dr
Þr=5
4r d 2 v 1 é
4r
ù 1
,
=
4r –
´ 6 ú = [ 4 r – 8r ] < 0
3 dh 2 3 êë
3
û 3
Þ maximum volume ocurs at h =
As from (i),
(h – r)2 + b2 = r2
4r 4
= ´ 3 = 4 cm
3 3
dS
æ 1 - pr öæ -p ö
= 2ç
֍
÷ + 2pr
dr
è 2 øè 2 ø
Þ
-p p2 r
+
+ 2pr = 0
2
2
Þx=
2
p+ 4
Þr=
1
p+ 4
Þ x = 2r
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r
M-369
Applications of Derivatives
106. (a) Let f (x) = a log | x | + bx2 + x
Differentiate both side,
a 2 + (a 2 - 4)2
D2 = a2 + a4 + 16 - 8a2 = a4 - 7a2 + 16
103. (a) D =
a
+ 2bx + 1
x
Since x = –1 and x = 2 are extreme points therefore
f ¢ ( x) =
dD 2
= 4a3 - 14a = 0
da
2a(2a2 - 7) = 0
a2 =
f ¢ ( x ) = 0 at these points.
Put x = –1 and x = 2 in f ¢( x ) , we get
7
2
O
107. (d)
15
2
104. (a) Let f(x) =
3
3
5
(1 + x )2
3
3ù
é
3 ê 1+ x 5
(1 + x) 5 ú
= ê
ú
2
2
5ê
ú
5
5
x
ëê (1 + x)
ûú
2
+ x -1- x
2
x) 5
=
x 5 -1
2
2
x 5 (1 + x) 5
æ 16 ö
B ç , 0÷
è h ø
-2
Let (h, k) be the point on ellipse through which tangent
is passing.
Equation of tangent at (h, k) =
at y = 0, x =
16
h
at x = 0, y =
81
k
Area of AOB =
xh yk
+
= 1
16 81
1 æ 16 ö æ 81ö 648
´ç ÷ ´ç ÷ =
2 è h ø è k ø hk
<0
Also, f (0) = 1 Þ f(x) Î [2 , 1]
f (a) = 2–0.4
105. (d) Let ‘u’ be the velocity
\ u = 4 8 m/s, Given, g = 32
At maximum height v = 0
Now, we know v2 = u2 – 2gh
Þ 0 = (48)2 – 2 (32)h Þ h = 36
Maximum height = 36 + 64 = 100 mt
–0.4
O
3
- (1 + x) 5 (x 5 )
5
2
éæ
3ö
3 -2 ù
5
3ê
5 ÷ (1 + x )
5x 5 ú
ç
1
+
x
(1
+
x)
= 5 êç
ú
÷ø
ëêè
ûú
2
x 5 (1 +
æ 81ö
A ç 0, ÷
è kø
and x Î [0, 1]
3
1+ x 5
5
Þ f¢(x) =
1
2
(h, k)
3
(1 + x) 5
3
2
3
5
5
(1 + x ) (1 + x)
=
...(ii)
\ a=2
49 49
15
49
+ 16 = - + 16 =
4
2
4
4
2
x5
a
+ 4b + 1 = 0 Þ a +8b = –2
2
6b = -3 Þ b = -
2
D=
...(i)
On solving (i) and (ii), we get
(a , a – 4)
D2 =
– a –2b + 1 = 0 Þ a +2b = 1
A2 =
(648)2
...(i)
h2k 2
(h, k) must satisfy equation of ellipse
h2 k 2
+
= 1
16 81
16
(81 - k 2 )
81
Putting value of h2 in equation (i)
h2 =
Downloaded from @Freebooksforjeeneet
EBD_8344
M-370
2
A =
Mathematics
81(648) 2
16 ´ k 2 (81 - k 2 )
=
Now, for largest possible right circular cylinder the
volume must be maximum
a
81k 2 - k 4
dV
= 0
dh
Now, Differentiating eq. (2) w.r.t. h
\ For maximum volume,
differentiating w.r. to k
-1 ö
æ
(162k - 4 k 3 )
2AA¢ = a ç
è 81k 2 - k 4 ÷ø
3
dV
= 3p - ph 2
dh
4
2AA¢ = –2A (81k – 4k3) Þ A¢= – 81k – 4k3
Put A¢ = 0
Þ 162k – 4k3 = 0, k (162 – 4k2) = 0
Þ
k = 0, k = ±
3 2
3
or 3p - ph = 0 Þ 3p = p h 2
4
4
Þ h2 = 4 Þ h = 2
Now, volume (V) of the cylinder
9
2
A¢¢ = – (81 – 12k2)
For both value of k, A¢¢ = 405 > 0
Area will be minimum for k = ±
h2 =
æ
h2 ö
= p ç 3 - ÷ h = p (6 - 2) = 4p
4ø
è
b
109. (c) Let cost C = av +
v
According to given question,
9
2
16
(81 - k 2 ) = 8
81
b
= 75
… (i)
30
b
40a +
= 65
… (ii)
40
On solving (i) and (ii), we get
30 a +
h = ±2 2
Area of triangle AOB =
648 ´ 2
= 36 sq unit
2 2 ´9
108. (c) Given, radius of sphere = 3
Now, In DOAB, by Pythagoras theorem
(OA)2 = (OB)2 + (AB)2
a=
Now, C = av +
Þ
h/2
3
b
v
dC
b
=a- 2
dv
v
dC
b
=0Þ a- 2 =0
dv
v
O
h
1
and b = 1800
2
b
= 3600 Þ v = 60 kmph
a
110. (d) Let base = b
Þv=
B
C
A
r
2
æ hö
2
( 3)2 = çè ÷ø + r
2
h
h2 - b 2
2
3=
h
h2
2
+ r2 Þ r = 3 4
4
...(i)
Now, volume of cylinder = pr2h
æ
h2 ö
V = pç3 - ÷ h
4ø
è
V = 3 ph -
ph3
4
b
Altitude (or perpendicular) =
(using eq. (i))
...(ii)
Area, A =
Þ
h2 - b2
1
1
2
2
× base × altitude = ´ b ´ h - b
2
2
dA 1 é 2
- 2b ù
=
h - b2 + b .
ú
db 2 ê
2 h2 - b2 û
ë
Downloaded from @Freebooksforjeeneet
M-371
Applications of Derivatives
Þ y – 2 = m (–1)
Þ y = 2 – m and OQ = y = 2 – m
1 é h2 - 2b2 ù
= ê
ú
2 êë h2 - b2 ûú
Put
Area of DPOQ =
dA
=0, Þ
db
b=
h
2
1
´ base ´ height )
2
4 öù
1é
4
ù 1é æ
= ê 2 - m - + 2 ú = ê4 - ç m + ÷ú
m øû
2ë
m
û 2ë è
m 2
= 2- 2 m
(Q Area of D =
h2 h2
1 h
´
´ h2 =
2
2
4
2
Maximum area =
111. (b) Given that, f ( x ) = ln x + bx 2 + ax
1
+ 2bx + a
x
At x = –1, f '(x) = –1– 2b + a =0
\ f Ἐ(x) =
Þ
Q
a – 2b = 1
a + 4b = -
Þ
...(i)
f '(x) =
At x = 2,
=
1
2
Now, f ' ( m ) =
1 x 1 2- x + x
- + =
x 2 2
2x
–
0
Now, f ἘἘ ( m ) =
– 2 = m (x – 1) Þ x – 1 =
Þ
x=
-4
m3
1
<0
2
1
f ἘἘ ( m ) m = - 2 = > 0
2
Area will be least at m = –2
Hence, slope of PQ is –2.
f ἘἘ ( m ) m = 2 = -
2 –
¥
So maxima at x = –1, 2
112. (c) Equation of a line passing through (x1,y1) having
slope m is given by y – y1 = m (x –x1)
Since the line PQ is passing through (1,2) therefore its
equation is (y – 2) = m (x – 1)
where m is the slope of the line PQ.
Now, point P (x,0) will also satisfy the equation of PQ
\ y –2 = m (x –1) Þ 0 – 2 = m (x – 1)
Þ
-1
+
2 m2
Put f ¢ (m) = 0
Þ m2 = 4 Þ m = ± 2
+
–1
m 2
2 m
2
Let Area = f (m) = 2 -
2
+
P
O
...(ii)
- ( x + 1)( x - 2 )
- x 2 + x + 2 -( x 2 - x - 2)
=
=
2x
2x
2x
–¥
(1,2)
1
+ 4b + a = 0
2
1
1
On solving (i) and (ii) we get a = , b = 2
4
Thus, f ' ( x ) =
1æ
2ö
1
(OP)(OQ ) = ç1 - ÷ ( 2 - m )
2 è mø
2
-2
m
-2
+1
m
113. (d) Let f : (-¥, ¥) ® (-¥, ¥) be defined by f(x) = x3 + 1.
Clearly, f(x) is symmetric along y = 1 and it has neither
maxima nor minima.
\ Statement-1 is false.
Hence, option (d) is correct.
114. (b)
ì tan x
, x¹0
ï
f ( x) = í x
ï 1, x = 0
î
For x > 0
Also, OP = ( x - 0 ) 2 + ( 0 - 0 )2 = x = -2 + 1
m
Similarly, point Q (0,y) will satisfy equation of PQ
\ y –2 = m (x– 1)
tan x > x
tan x
>1
x
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EBD_8344
M-372
Mathematics
Y
ì k - 2 x, if x £ -1
f ( x) = í
î 2 x + 3, if x > -1
117. (c)
y = tan x
y=x
X´
O
2x + 3
k–2x
X
1
–1
Y´
For x < 0 Þ tan x < x
tan x
Þ
>1
x
f ( 0) = 1 at x = 0
Þ x = 0 is the point of minima
So, Statement 1 is true. Statement 2 is also true.
f '( x) = x sin x
115. (c)
f '( x) = 0
Þ x = 0 or sin x = 0
Þ x = 2 p, p
f ''( x ) = x cos x +
=
1
2 x
1
sin x
2 x
Þ 9a2 – 32 b < 0 Þ b >
(2 x cos x + sin x)
Hence, local maxima at x = p
At x = 2p , f ''( x) > 0
Hence local minima at x = 2p
f '( x ) =
e x + 2e - x
=
ex
e2 x + 2
(e 2 x + 2) 2
f '( x) = 0 Þ e
Þ e2 x = 2
Qf"
1
(e 2 x + 2) e x - 2e 2 x .e x
2x
+ 2 = 2e
p
dy
= 0 Þ 3x2 – p = 0 Þ x = ±
3
dx
( 2 ) = +ve
\ Maximum values of f (x) =
1
2 2
1
Since, 0 < < 1
3 2 2
2
1
=
4
2 2
d2y
dx
2
= 6x
"x Î R
1
Þ for some c Î R , f (c ) =
3
"x > 0
\ P (x) is an increasing function on (0,1)
\ P (0) < P (a)
Similarly we can prove P (x) is decreasing on (– 1, 0)
\ P (– 1) > P (0)
So we can conclude that
Max P (x) = P (1) and Min P (x) = P (0)
Þ P(–1) is not minimum but P (1) is the maximum of P.
119. (a) Let y = x3 – px + q Þ dy = 3 x 2 – p
dx
For maxima and minima
2x
Þ ex = 2
Þ 0 < f ( x) £
9a 2
>0
32
Hence a, b > 0
Þ P' (x) = 4 x 3 + 3ax2 + 2bx > 0
At x = p, f ''( x ) < 0
116. (d) Given f ( x) =
Clear that f (x) is minimum at (–1, 1)
\ f (–1) = 1
1=k+2Þ k=–1
118. (a) Given that P (x) = x4 + ax3 + bx2 + cx + d
Þ P' (x) = 4 x3 + 3ax2 + 2bx + c
But given P' (0) = 0 Þ c = 0
\ P(x) = x4 + ax3 + bx2 + d
Again given that P ( – 1) < P (1)
Þ 1–a+b+d<1+a+b+d
Þ a>0
Now P ' (x) = 4x 3 + 3ax 2 +2bx = x (4x 2 + 3ax + 2b)
As P' (x) = 0, there is only one solution x = 0, therefore
4x2 + 3ax + 2b = 0 should not have any real roots i.e. D < 0
d2y
dx 2 x = p
= + ve and
3
\ y has minimum at x =
x= –
d2y
dx 2 x = – p
= – ve
3
p
and maximum at
3
p
3
Downloaded from @Freebooksforjeeneet
M-373
Applications of Derivatives
1 2
x 2
+ Þ f '( x) = - 2 = 0
2 x
2 x
Þ x2 = 4 Þ x = 2, – 2;
120. (a) Given f (x) =
Now, f ''( x) =
4
x3
f ''( x )] x =2 = + ve Þ f ( x ) has local min at x = 2.
1
1
dy
Þ
= 1x
dx
x2
For maxima. or minima.,
121. (c) ATQ, y = x +
1-
1
x2
= 0 Þ x = ±1
Þ x = a or x = 2a.
f ''( x ) = 12 x - 18a
f "(a) = – 6a < 0 \ f(x) is max. at x = a,
f "(2a) = 6a > 0
\ f(x) is min. at x = 2a
\ p = a and q = 2a
ATQ, p 2 = q
\ a 2 = 2a Þ a = 2or a = 0
but a > 0, therefore, a = 2.
123. (b) We know that distance of origin from
(x, y) =
x2 + y 2
æ at ö
a 2 + b2 - 2ab cos ç t - ÷ ;
è
bø
æ d2 yö
d2 y 2
= 3 Þç 2 ÷ = 2> 0
2
dx
x
è dx ø x =1
=
\ y is minimum at x = 1
£ a 2 + b 2 + 2ab
122. (d)
f ( x) = 2 x3 - 9ax 2 + 12a 2 x + 1
2
2
f '( x ) = 6 x - 18ax + 12a ;
For maxima or minima.
éì
ù
æ at ö ü
= -1ú = a + b
ê í cos ç t - ÷ ý
è
b ø þ min
êë î
úû
\ Maximum distance from origin = a + b
6 x 2 - 18ax + 12a2 = 0 Þ x 2 - 3ax + 2a2 = 0
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EBD_8344
M-374
Mathematics
22
Integrals
Standard Integrals, Integration by
5.
TOPIC Ć Substitution, Integration by Parts
Let f ( x) = ò
x
(1 + x)2
dx ( x ³ 0). Then f (3) – f (1) is equal
to :
1.
2.
3.
æ ( x -1) 2
ö
t cos(t 2 )dt ÷
ç ò0
lim ç
[Sep. 06, 2020 (I)]
÷
x ®1 ç ( x - 1)sin( x - 1) ÷
ç
÷
è
ø
1
(a) is equal to
(b) is equal to 1
2
1
(c) is equal to –
(d) does not exist
2
-x
(e x + e - x )
x
6.
-x
dx = g ( x)e(e + e ) + c ,
If ò (e + 2e - e - 1) e
where c is a constant of integeration, then g (0) is equal to:
[Sep. 05, 2020 (I)]
(a) e
(b) e 2
(c) 1
(d) 2
2x
If
x
cos q
ò 5 + 7 sin q - 2 cos 2 qd q = A loge | B(q) | +C ,
C is a constant of integration, then
2sin q + 1
sin q + 3
(b)
2sin q + 1
5(sin q + 3)
(c)
5(sin q + 3)
2 sin q + 1
(d)
5(2 sin q + 1)
sin q + 3
7.
3
p 1
+ +
12 2 4
(b)
3
p 1
+ 6 2 4
(c) -
3
p 1
+ +
6 2 4
(d)
3
p 1
+ 12 2 4
æ
x ö
-1
If sin -1 çç
÷÷ dx = A( x) tan ( x ) + B ( x ) + C , wher e
1
+
x
è
ø
C is a constant of integration, then the ordered pair (A(x),
B(x)) can be :
[Sep. 03, 2020 (II)]
ò
(a) ( x + 1, - x )
(b) ( x + 1,
x)
(c) ( x - 1, - x )
(d) ( x - 1,
x)
x
æ
ö
The integral ò ç
÷ dx is equal to
è x sin x + cos x ø
(where C is a constant of integration) : [Sep. 04, 2020 (I)]
x sec x
+C
(a) tan x x sin x + cos x
x tan x
+C
(b) sec x +
x sin x + cos x
x tan x
+C
(c) sec x x sin x + cos x
x sec x
(d) tan x +
+C
x sin x + cos x
The integral
ò ( x + 4)8/7 ( x - 3)6/7
is equal to:
(where C is a constant of integration)
1/7
æ x-3 ö
(a) ç
÷
è x+4ø
+C
1æ x-3 ö
2 çè x + 4 ÷ø
+C
dq
8.
If
ò cos2 q(tan 2q + sec 2q)
[Jan. 9, 2020 (I)]
æ x -3ö
(b) - ç
÷
è x+4ø
3/7
(c)
(d) -
-1/7
+C
1 æ x-3 ö
13 çè x + 4 ÷ø
-13/7
+C
= ltanq + 2loge|f(q)| + C where
C is a constant of integration, then the ordered pair
(l, f(q)) is equal to:
[Jan. 9, 2020 (II)]
(a) (1, 1 – tanq)
(b) (–1, 1 – tanq)
(c) (–1, 1 + tanq)
(d) (1, 1 + tanq)
2
4.
(a) -
dx
where
B (q )
can be :
A
[Sep. 05, 2020 (II)]
(a)
[Sep. 04, 2020 (I)]
9.
If
cos x dx
ò sin3 x(1 + sin 6 x)2/3 = f ( x)(1 + sin
6
1/l
x)
+ c where c is
æ pö
a constant of integration, then lf çè ÷ø is equal to:
3
[Jan. 8, 2020 (II)]
(a) -
9
8
(b) 2
(c)
9
8
Downloaded from @Freebooksforjeeneet
(d) –2
M-375
Integrals
10.
11.
2 x3 - 1
ò x4 + x dx is equal to :
(Here C is a constant of integration) [April 12, 2019 (I)]
The integral
(a)
x3 + 1
1
log e
+C
2
x2
(c)
log e
x3 + 1
+C
x
(b)
1
( x 3 + 1)2
+C
log e
2
x3
(d)
log e
x3 + 1
x2
+C
Let a Î (0, p/2) be fixed. If the integral
tan x + tan a
ò tan x - tan a dx = A(x) cos2a+B(x) sin2a+C, where C is
a constant of integration, then the functions A(x) and
B(x) are respectively :
[April 12, 2019 (II)]
(a)
x + a and loge sin( x + a)
(b)
x - a and loge sin( x - a)
(c)
x - a and loge cos( x - a)
(d)
x + a and loge sin( x - a)
15.
13.
14.
If
16.
1
17.
ò ( x2 - 2 x + 10) 2
x -1 ö
f ( x)
æ
ö
= A ç tan -1 çæ
÷+
÷ + C where C is a
è 3 ø x 2 - 2 x + 10 ø
è
constant of integration, then :
[April 10, 2019 (I)]
1
(a) A =
and f(x) = 3 (x – 1)
54
1
(b) A =
and f(x) = 3 (x – 1)
81
1
(c) A =
and f(x) = 9 (x – 1)
27
1
(d) A =
and f(x) = 9 (x – 1)2
54
If f(x) is a non- ero polynomial of degree four, having
local extreme points at x = –1, 0, 1; then the set
S = {x Î R : f(x) = f(0)}contains exactly:
[April 09, 2019 (I)]
(a) four irrational numbers.
(b) four rational numbers.
(c) two irrational and two rational numbers.
(d) two irrational and one rational number.
The integral ò sec 2/3 x cosec4/3 xdx is equal to:
[April 09, 2019 (I)]
3
(a) –3 tan–1/3 x + C
(b) – tan–4/3 x + C
4
(c) –3 cot–1/3 x + C
(d) 3 tan–1/3 x + C
(Here C is a constant of integration)
5x
ò 2x dx is equal to :
sin
2
(where c is a constant of integration.) [April 08, 2019 (I)]
(a) 2x + sinx + 2 sin2x + c (b) x + 2 sinx + 2 sin2x + c
(c) x + 2 sinx + sin2x + c (d) 2x + sinx + sin2x + c
sin
dx
12.
x
If ò esec (sec x tan x f ( x ) + (sec x tan x + sec 2 x)) dx
= esec x f(x) + C, then a possible choice of f(x) is:
[April 09, 2019 (II)]
1
1
(a) sec x + tan x +
(b) sec x - tan x 2
2
1
1
(c) sec x + x tan x (d) x sec x + tan x +
2
2
18.
dx
6 3
= xf ( x ) (1 + x ) + C , where C is a
If ò 3
x (1 + x 6 )2/3
constant of integration, then the function f(x) is equal to :
[April 08, 2019 (II)]
3
1
1
1
(a) 2
(b) – 3
(c) – 2
(d) – 3
x
6x
2x
2x
Theintegral ò cos ( log e x )dx is equal to :
(where C is a constant of integration) [Jan. 12, 2019 (I)]
x
ésin ( log e x ) - cos ( log e x ) ùû + C
2ë
(a)
(b) x éëcos ( log e x ) + sin ( log e x ) ùû + C
x
écos ( loge x ) + sin ( log e x )ûù + C
2ë
(c)
(d) x éëcos ( log e x ) - sin ( log e x ) ùû + C
19.
The integral
3 x13 + 2 x11
ò (2 x 4 + 3x 2 + 1) 4 dx
is equal to:
(where C is a constant of integration) [Jan. 12, 2019 (II)]
x4
(a)
6(2 x 4 + 3 x 2 + 1)3
If
x12
4
6(2 x + 3 x 2 + 1)3
+C
x12
+C
+ C (d)
(2 x 4 + 3 x 2 + 1) 3
(2 x 4 + 3 x 2 + 1) 3
x4
(c)
20.
+ C (b)
ò
1 - x2
x4
m
dx = A(x) æç 1 - x 2 ö÷ + C , for a suitable
è
ø
chosen integer m and a function A (x), where C is a constant
of integration, then (A(x))m equals :
[Jan. 11, 2019 (I)]
(a)
-1
27x
9
(b)
-1
3x
3
(c)
1
27x
Downloaded from @Freebooksforjeeneet
6
(d)
1
9x 4
EBD_8344
M-376
21.
If
Mathematics
x +1
ò
2x -1
of integration, then f(x) is equal to:
(a)
1
( x + 1)
3
(b)
2
( x - 4)
3
(c)
22.
dx = f ( x ) 2 x - 1 + C , where C is a constant
25.
sin 2 x cos 2 x
ò (sin5 x + cos3 x sin 2 x + sin 3x cos2 x + cos5 x)2 dx
[Jan. 11, 2019 (II)]
to:
2
( x + 2)
3
(a)
1
(d) ( x + 4 )
3
1
Then
ò
sin n + 1 q
26.
dq is equal to:
(a)
n æ
1
ç1 ÷
n 2 - 1 è sin n - 1 q ø
n æ
1
ö
(b) 2
ç1 n -1 ÷
n +1è
sin
qø
(c)
1
æ
ö
ç1 +
÷
n2 - 1 è
sin n - 1 q ø
n
n +1
n
+C
n +1
23.
n æ
1
ö n
+C
(d) 2
ç1 ÷
n - 1 è sin n + 1 q ø
(where C is a constant of integration)
For x2 ¹ np + 1, nÎN (the set of natural numbers), the
integral
[Jan. 09, 2019(I)]
òx
2 sin ( x
(a) log e
(b)
(
- 1) + sin 2 ( x
) dx
is equal to:
- 1)
28.
2 sin ( x 2 - 1) - sin 2 x 2 - 1
2
(
2
29.
)
1
sec 2 x 2 - 1 + c
2
æ x2 - 1 ö
(d) log e sec çç 2 ÷÷ + c
è
ø
(where c is a constant of integration)
5 x8 + 7 x 6
(x
2
+ 1 + 2x
7
)
2
(c)
1
2
1
2
30.
dx, ( x ³ 0 ) ,
and f(0) = 0, then the value of f(1) is:
(a) –
(b) –
(d)
1
1 + cot 3 x
+C
-1
1
+C
+C
(d)
1 + cot 3 x
3(1 + tan3 x)
(where C is a constant of integration)
If
K
æ x – 4ö
If f ç
÷ = 2x + 1, (x Î R = {1, – 2}), then ò f (x) dx is
è x + 2ø
equal to (where C is a constant of integration)
[Online April 15, 2018]
(a) 12 loge |1 – x| – 3x + c
(b) – 12 loge |1 – x| – 3x + c
(c) – 12 loge |1 – x| + 3x + c
(d) 12 loge |1 – x| + 3x + c
æ x +3ö
dx = A 7 - 6 x - x 2 + B sin -1 ç
÷+C
è 4 ø
7 - 6x - x
(where C is a constant of integration), then the ordered
pair (A, B) is equal to
[Online April 15, 2018]
(a) (– 2, – 1)
(b) (2, – 1)
(c) (– 2, 1)
(d) (2, 1)
ò
2x + 5
2
4
æ 3x - 4 ö
If f ç
÷ = x + 2, x ¹ - , and
3
è 3x + 4 ø
(A, B) is equal to :
[Online April 9, 2017]
(where C is a constant of integration)
1
log e | sec ( x 2 - 1) | + c
2
If f ( x ) = ò
(b)
ò f (x) dx = A log |1 – x | + Bx + C, then the ordered pair
2
1
2 æ x -1ö
log
sec
ç
e
(c) 2
ç 2 ÷÷ + c
è
ø
24.
+C
æ K tan x + 1 ö
tan -1 ç
÷ + C,
A
A
è
ø
(C is a constant of integration), then the ordered pair
(K, A) is euqal to
[Online April 16, 2018]
(a) (2, 3) (b) (2, 1)
(c) (– 2, 1) (d) (– 2, 3)
27.
+C
-1
3(1 + tan 3 x)
tan x
+C
n +1
n
[2018]
ò 1 + tan x + tan 2 x dx = x -
[Jan 10, 2019(I)]
n +1
ö n
is equal
(c)
p
Let n ³ 2 be a natural number and 0 < q <
2
(sin n q + sin q) n cos q
The integral
1
4
1
4
[Jan. 09, 2019 (II)]
æ8 2ö
(a) ç , ÷
è3 3ø
æ 8 2ö
(b) ç - , ÷
è 3 3ø
æ 8 2ö
(c) ç - , - ÷
è 3 3ø
æ8 2ö
(d) ç , - ÷
è3 3ø
The integral
ò
1 + 2 cot x(cosec x + cot x)dx
pö
æ
[Online April 8, 2017]
ç 0 < x < ÷ is equal to :
2ø
è
(where C is a constant of integration)
(a) 2 log sin
x
+C
2
(b) 4 log sin
x
+C
2
(c) 2 log cos
x
+C
2
(d) 4 log cos
x
+C
2
Downloaded from @Freebooksforjeeneet
M-377
Integrals
31.
If
ò cos3 x
= (tan x) A + C(tan x) B + k , where k
37.
2sin 2x
is a constant of integration, then A + B + C equals :
[Online April 9, 2016]
16
5
(a)
32.
dx
ò
If
27
10
(b)
7
10
(c)
log(t + 1 + t 2 )
dt =
21
5
(d)
1
( g (t )) 2 + C , where C is a
2
1+ t2
constant, then g(b) is equal to :
[Online April 11, 2015]
(a)
1
5
(c) 2 log(2 + 5)
38.
(c)
34.
( x + 1) e
1
x
+c
(b) - xe
( x - 1) e
x+
1
x
+c
(d) xe
The integral ò
2
3
x+
1
x
1
x
x + cos3 x
39.
+c
+c
40.
)
2
dx is equal to:
[Online April 12, 2014]
(a)
1
(1 + cot x)
3
3
(c)
35.
36.
sin x
(1 + cos x)
3
(b) -
+c
(d) -
+c
(
1
3 1 + tan 3 x
)
(
cos x
+c
41.
3
3 1 + sin 3 x
)
+c
æ 1 - x2 ö
x
cos
ç
÷
The integral ò
ç 1 + x 2 ÷ dx (x > 0) is equal to:
è
ø
[Online April 11, 2014]
(a) – x + (1 + x2) tan–1 x + c
(b) x – (1 + x2) cot–1 x + c
(c) – x + (1 + x2) cot–1x + c
(d) x – (1 + x2) tan– 1 x + c
sin8 x - cos8 x
(1 - 2sin2 x cos2 x)
1
sin 2x + c
2
1
(c) - sin x + c
2
(c)
1 3
x y ( x3 ) - ò x 2 y ( x3 )dx + C
3
1é 3
x y ( x 3 ) - ò x3 y ( x3 ) dx ùû + C
3ë
If the integral
1
1
1
1
(b)
(c)
(d) 16
16
8
8
5tan
x
If the
ò tan x - 2 dx = x + a ln sin x - 2cos x + k, then a is
equal to :
[2012]
(a) – 1
(b) – 2
(c) 1
(d) 2
æ x 2 + sin 2 x ö
2
If f ( x ) = ò ç
÷ sec x dx and f (0) = 0, then f (1)
è 1 + x2 ø
equals
[Online May 19, 2012]
p
(b) tan1 + 1
(a) tan1 4
p
p
(c)
(d) 1 4
4
x2 - x
The integral of 3
w.r.t. x is
x - x2 + x - 1
[Online May 12, 2012]
(
)
1
log x 2 + 1 + C
2
(
)
(b)
1
log x 2 - 1 + C
2
2
(d) log x - 1 + C
42.
Let f(x) be an indefinite integral of cos3x.
Statement 1:f(x) is a periodic function of period p.
Statement 2: cos3x is a periodic function.
[Online May 7, 2012]
(a) Statement 1 is true, Statement 2 is false.
(b) Both the Statements are true, but Statement 2 is not
the correct explanation of Statement 1.
(c) Both the Statements are true, and Statement 2 is correct
explanation of Statement 1.
(d) Statement 1 is false, Statement 2 is true.
43.
The value of
dx is equal to:
1
(b) - sin 2x + c
2
(d) - sin 2 x + c
1 3
x y ( x3 ) - 3ò x 3 y ( x 3 )dx + C
3
2
(c) log x + 1 + C
[Online April 9, 2014]
(a)
(b)
(a)
-1
ò
1é 3
x y ( x 3 ) - ò x 2 y ( x3 )dx ùû + C
3ë
(a) -
[2014]
2
sin x cos x
(sin
x+
(a)
where k is an arbitrary constant, then A is equal to :
[Online April 25, 2013]
(d) log(2 + 5)
x+
f ( x 3 )dx is equal to [2013]
cos8 x + 1
x+
The integral ò æç1 + x - 1 ö÷ e x dx is equal to
xø
è
(a)
5
ò cot 2 x - tan 2 x dx = A cos8 x + k ,
1
33.
ò f ( x)dx = y(x), then ò x
(d)
1
log(2 + 5)
(b)
2
log(2 + 5)
If
2ò
sin xdx
is
pö
æ
sin ç x – ÷
è
4ø
Downloaded from @Freebooksforjeeneet
[2008]
EBD_8344
M-378
Mathematics
pö
æ
(a) x + log | cos ç x – ÷ | + c
è
4ø
pö
æ
(b) x – log | sin ç x – ÷ | + c
è
4ø
TOPIC n
pö
æ
(c) x + log | sin ç x – ÷ | +c
è
4ø
Expressions of ex
pö
æ
(d) x – log | cos ç x - ÷ | + c
è
4ø
44.
dx
ò cos x +
3 sin x
equals
[2007]
49.
(a) e (4e +1)
(c) e (4e–1)
A value of a such that
ò
a
(b) log tan æç x - p ö÷ + C
è 2 12 ø
50.
dx
ò cos x - sin x
If
[2004]
æ xö
log cot ç ÷ + C
è 2ø
2
2
(d)
46.
is equal to
52.
æ x 3p ö
log tan ç - ÷ + C
è2 8 ø
æ x pö
log tan ç - ÷ + C
è 2 8ø
2
1
sin x
ò sin( x - a) dx = Ax + B log sin( x - a), +C, then value
of (A, B) is
47.
51.
1
1
(c)
(b) (cos a, sin a)
(c) (- sin a, cos a)
(d) (sin a, cos a)
f(x) and g(x) are two differentiable functions on [0, 2] such
that f ¢¢( x ) - g ¢¢ ( x) = 0, f ¢(1) = 2 g ¢(1) = 4 f(2) = 3g(2) = 9
then f (x)–g(x) at x = 3/2 is
[2002]
(b) 2
(c) 10
1
2
(c) -
1
2
(d) 2
dx = g ( x)e - x + c , where c is a constant of
2
(d) 5
(c) -
(b) 1
5
2
(d) -
1
2
1 - 4 x3
f ( x) + C, where C is a
e
48
constant of integration, then f (x) is equal to:
[Jan. 10, 2019 (II)]
(a) – 2x3 – 1
(b) – 4x3 – 1
(c) – 2x3 + 1
(d) 4x3 + 1
If
òx
5 - 4 x3
e
The integral
dx =
dx
ò
is equal to :
(1 + x ) x - x 2
(where C is a constant of integration)
[Online April 10, 2016]
(a) -2
1+ x
+C
1- x
(b) -
1- x
+C
1+ x
(c) -2
1- x
+C
1+ x
(d) 2
1+ x
+C
1- x
[2004]
(a) (- cos a, sin a)
(a) 0
5 - x2
òx e
(a) –1
æ x 3p ö
log tan ç + ÷ + C
è2 8 ø
2
(b)
(b)
integration, then g(–1) is equal to : [April 10, 2019 (II)]
1
(a)
If
[Sep. 06, 2020 (II)]
(b) 4e2 – 1
(d) e (2e – 1)
dx
æ9ö
= log e ç ÷ is : [April 12, 2019 (II)]
( x + a)( x + a + 1)
è8ø
(a) –2
1
æx pö
(d)
log tan ç - ÷ + C
2
è 2 12 ø
45.
x (2 + log e x) dx equals:
The integral
a+1
1
x p
log tan çæ + ÷ö + C
2
è 2 12 ø
2 x x
.
ò1 e
48.
æx pö
(a) log tan ç + ÷ + C
è 2 12 ø
(c)
Integration of the Forms: òex(f(x) +
f'(x))dx, òekx(df(x) + f'(x))dx,
Integration by Partial Fractions,
Integration of Some Special
Irrational Algebraic Functions,
Integration of Different
dx
53.
The integral
ò x 2 (x 4 + 1)3/4
equals :
[2015]
1
(a)
1
- (x 4 + 1) 4
+c
æ x 4 + 1ö 4
(b) - ç 4 ÷ + c
è x ø
1
æ x 4 + 1ö 4
(c) ç 4 ÷ + c
è x ø
1
(d) (x 4 + 1) 4 + c
Downloaded from @Freebooksforjeeneet
M-379
Integrals
54.
ò
The integral
dx
3
( x + 1) 4
(x
5
- 2) 4
Evaluation of Definite Integral
is equal to :
TOPIC Đ by Substitution, Properties of
Definite Integrals
[Online April 10, 2015]
1
x +1ö 4
4
(a) - çæ
÷
3 è x - 2ø
1
x +1ö 4
(b) 4 æç
è x - 2 ÷ø
+C
1
60.
+C
1
ò
x
5m -1
+ 2x
4m-1
( x 2m + xm + 1)
3
56.
5m
(
)
2m ( x 5m + x 4m )
2
( x 2m + xm + 1)
The integral
ò
2
(b)
x
(
)
( x5m - x4m )
2
2m ( x 2m + x m + 1)
(d)
(a) 1
63.
(b) 0
ò
equals :
(c)
xe
x
1+ x
2
+C
(d)
[Sep. 04, 2020 (II)]
7
1
1
9
(b) (c) (d)
18
9
18
2
Let {x} and [x] denote the fractional part of x and the
greatest integer £ x respectively of a real number x. If
n
and 10(n2 – n), (n Î N, n > 1) are
three consecutive terms of a G.P., then n is equal to
_______________.
[NA Sep. 04, 2020 (II)]
p
65.
ò | p- | x || dx is equal to :
(a)
66.
2p2
(b) 2p2
If the value of the integral
[Sep. 03, 2020 (I)]
(c) p2
ò
(1 - x 2 )3/ 2
k is equal to :
x
dx is
k
, then
6
(b) 2 3 + p
(d) 3 2 - p
(c) 3 2 + p
+C
p2
2
[Sep. 03, 2020 (II)]
(a) 2 3 - p
+C
(d)
x2
1/ 2
0
[2005]
(log x)2 + 1
3
2
-p
2
ïì (log x - 1) ïü
ò íï1 + (log x)2 ýï dx is equal to
î
þ
x2 + 1
(d)
tan 3 x × sin 2 3 x (2 sec 2 x × sin 2 3 x + 3 tan x × sin 6 x ) dx
ò0 {x} dx, ò0 [ x] dx
[Online April 9, 2013]
(b) ln | x | + p (x) + c
(d) x + p (x) + c
(b)
1
2
(a)
ò x + x7 = p( x) then, ò x + x7 dx is equal to:
+C
(c)
is equal to :
x6
x
[Sep. 04, 2020 (I)]
p /6
-1
x - x + 1 cot -1 x
e
dx = A( x )ecot x + C , then A(x) is
If ò 2
x +1
equal to :
[Online April 22, 2013]
(a) – x
(b) x
(c) 1 - x (d) 1 + x
log x
[Sep. 05, 2020 (I)]
The integral
n
(log x)2 + 1
5051
5050
p /3
2
(a)
sin x
-p / 2 1 + e
dx is:
(d)
0
2
(c) - x log 1 - 2 - x2 + c (d) x log 1 - 2 + x + c
59.
5050
5051
3
2 - x 2 + 2 - x2
(a) ln | x | – p (x) + c
(c) x – p (x) + c
[Sep. 06, 2020 (I)]
Then ò ( g ( x) - f ( x)) dx is equal to :
2
64.
If
dx such
62.
(a) log 1 + 2 + x 2 + c (b) - log 1 + 2 - x 2 + c
58.
)
p
3p
p
(b) p
(c)
(d)
2
2
4
Let f ( x) =| x - 2 | and g ( x) = f ( f ( x)), x Î [0, 4].
4m
2m x 2m + x m + 1
xdx
dx
1
ò
(c)
50 101
The value of
[Online April 23, 2013]
57.
5050
5049
1
ò0 (1 - x
(a)
2m x 2m + x m + 1
(c)
(b)
p/2
[Online April 19, 2014]
x
5049
5050
)
61.
dx = f ( x ) + c ,
then f(x) is:
(a)
50 100
dx and I2 =
that I2 = aI1 then a equals to :
(a)
x - 2 ö4
4 x - 2 ö4
(c) 4 çæ
(d) - æç
÷ +C
÷ +C
3 è x +1 ø
è x +1 ø
55. If m is a non- ero number and
1
ò0 (1 - x
If I1 =
2
67.
The integral
ò | x - 1| - x dx
is equal to ________.
0
[NA Sep. 02, 2020 (I)]
Downloaded from @Freebooksforjeeneet
EBD_8344
M-380
68.
Mathematics
Let [t] denote the greatest integer less than or equal to t.
Then the value of
ò
2
75.
| 2 x - [3x] | dx is ___________.
1
69.
[NA Sep. 02, 2020 (II)]
If for all real triplets (a, b, c), f(x) = a + bx + cx2; then
If
1
ò
f ( x ) dx is equal to:
Let f : R ® R be a continuously differentiable function
1
such that f(2) = 6 and f’(2) = .
48
ì
æ 1 öü
(a) 2 í3 f (1) + 2 f ç ÷ý
è 2 øþ
î
1ì
æ 1 öü
(b) 2 í f (1) + 3 f ç 2 ÷ ý
è øþ
î
1ì
æ 1 öü
(c) 3 í f (0) + f ç 2 ÷ ý
è øþ
î
1ì
æ 1 öü
(d) 6 í f (0) + f (1) + 4 f ç 2 ÷ý
è øþ
î
The value of
ò
0
x sin8 x
dx is equal to:
8
sin x + cos8 x
(b) 2p2
(a) 2p
2
If I =
ò
1
(a)
72.
dx
2 x - 9 x + 12 x + 4
3
2
1
1
< I2 <
8
4
[Jan. 9, 2020 (I)]
(d) 4p
(c) p2
, then:
76.
1
1
1
1
< I2 <
< I2 <
(c)
(d)
16
9
6
2
If f (a + b + 1 – x) = f(x), for all x, where a and b are fixed
a
is equal to:
cot x
1
2
(b) 1
(a)
òa+1 f ( x)dx
(c)
òa-1 f ( x + 1)dx
b -1
òa-1 f ( x)dx
(d)
òa+1 f ( x + 1)dx
òe
- a| x|
dx = 5 , is :
The value of
ò éësin 2x (1 + cos3x )ùû dx , where [t] denotes
0
78.
The integral
p/3
òp/6 sec
2/3
(a) 3 - 3
(c) 37/6 - 35/6
(b) 3 - 31/3
(d) 35/3 - 31/3
2/3
4/3
p /2
79.
The value of
ò
0
(a)
p-2
8
sin 3 x
dx is:
sin x + cos x
(b) p - 1
4
80.
q2
ò cos
2
3q dq is equal to:
p
(d)
9
p -1
2
[April 09, 2019 (II)]
p 1
p
- loge 2
- loge 2
(b)
2 2
4
p
p 1
- loge 2
- loge 2
(c)
(d)
2
4 2
If f : R ® R is a differentiable function and f (2) = 6, then
(a)
81.
f (x)
lim
x ®2
82.
ò
6
2t dt
( x - 2) is:
[April 09, 2019 (II)]
(a) 24 f ¢ (2) (b) 2 f ¢ (2) (c) 0
(d) 12 f ¢ (2)
2 - x cos x
and g ( x) = loge x, (x > 0) then the
If f ( x) =
2 + x cos x
p
4
value of the integral
ò
g ( f ( x))dx is :
p
4
q1
p 1
+
(c)
3 6
(d)
-1
2
4
The value of the integral ò x cot (1 - x + x )dx is:
[Jan. 7, 2020 (II)]
2p
(b)
3
p-2
4
0
(a) loge2
5
2cot q + 4 = 0 , then
sin q
(c)
[April 9, 2019 (I)]
1
[Jan. 7, 2020 (II)]
2
x cos ec 4/3 x dx is equal to :
[April 10, 2019 (II)]
5/6
-1
3
4
(b) loge æç ÷ö (c) loge
(d) loge æç ÷ö
2
è 2ø
è 3ø
If q1 and q2 be respectively the smallest and the largest
values of q in (0,2p) – {p} which satisfy the equation,
(d) –1
the greatest integer function, is:
[April 10, 2019 (I)]
(a) p
(b) –p
(c) –2p
(d) 2p
b +1
2
1
2
2p
77.
b -1
(b)
The value of a for which 4a
p
(a)
3
(c)
[Jan. 7, 2020 (I)]
b +1
[April 12, 2019 (I)]
(d) 36
(c) 12
ò02 cot x + cosec x dx = m (p + n) , then m.n is equal to :
(a) -
1
1
< I2 <
9
8
(b)
If
b
74.
(b) 24
[April 12, 2019 (I)]
[Jan. 8, 2020 (II)]
1
positive real numbers, then a + b ò x(f (x) + f (x + l))dx
73.
x ®2
x
2p
71.
4t3dt = (x – 2) g (x), then lim g(x) is equal to :
(a) 18
[Jan. 9, 2020 (I)]
0
70.
f ( x)
ò6
(a) loge3
(b) logee
(c) loge2
Downloaded from @Freebooksforjeeneet
[April 8, 2019 (I)]
(d) loge1
M-381
Integrals
x
83.
( 0, 2 )
( 2, - 2 )
(a)
Let f(x) = ò g ( t ) dt , where g is a non- ero even function. If
0
(c)
x
f(x + 5) = g(x), then
ò f ( t ) dt equals : [April 08, 2019 (II)]
0
5
5
g ( t ) dt
(d) 5
ò
ò f ( x ) g ( x ) dx
0
[Jan. 12, 2019 (I)]
a
a
(a) 4 ò f ( x ) dx
(b)
0
a
0
eì
x
2x
ïæ x ö
æ e ö üï
The integral ò íç ÷ - ç ÷ ý loge x dx is equal to :
è x ø þï
ïè e ø
1î
[Jan. 12, 2019 (II)]
(a)
1
1
-e- 2
2
e
1 1 1
(b) - + 2 e 2e2
(c)
3 1 1
- 2 e 2e 2
(d)
3
1
-e- 2
2
2e
sin 2 x
ò é x ù 1 dx
-2
êë p úû + 2
(where [x] denotes the greatest integer less than or equal
to x) is :
[Jan. 11, 2019 (I)]
(a) 0
(b) sin 4
(c) 4
(d) 4 –sin 4
The value of
p /4
òp/6 sin 2 x
dx
( tan5 x + cot5 x )
(c)
88.
1 æp
-1 æ 1 ö ö
÷÷
(b) 10 ç 4 - tan ç
è 9 3 øø
è
ò cos x
3
dx is:
[Jan 9, 2019 (I)]
(b)
2
ò f ( x ) dx is equal to :
[Jan. 09, 2019 (II)]
0
(a) 1
(b) 2
(c)
1
2
tan q
1
dq = 1 , ( k > 0 ) then the value of k
2k
sec
q
2
0
is:
[Jan. 09, 2019 (II)]
1
(a) 4
(b)
(c) 1
(d) 2
2
If
ò
p
2
94.
The value of
(a)
95.
p
2
If f (x) =
(a)
(b)
(c)
(d)
ò
sin 2 x
p 1+ 2
2
(b) 4p
x
dx is :
(c)
[2018]
p
4
(d)
p
8
x
ò0 t (sin x - sin t ) dt then
[Online April 16, 2018]
f ²¢(x) + f ¢(x) = cos x – 2x sin x
f ²¢(x) + f ²(x) – f ¢(x) = cos x
f ²¢(x) – f ²(x) = cos x – 2x sin x
f ²¢ (x) + f ²(x) = sin x
b
Let I = ò ( x 4 - 2 x 2 )dx. If I is minimum then the ordered
a
pair (a, b) is:
(d) 0
p /3
93.
equals :
1æp
-1 æ 1 ö ö
÷÷
(d) 5 ç 4 - tan ç
è 3 3 øø
è
p
40
dx
, where [t] denotes the
[
]
+
[sin
x
x] + 4
- p /2
ò
1
[Jan. 11, 2019 (II)]
1
æ 1 ö
tan -1 ç
(a)
÷
20
è9 3ø
[Jan. 10, 2019 (II)]
6
(d)
25
4
2
4
(c)
(d) 3
3
3
92. Let f be a differentiable function from R to R such that |f (x)
– f (y)| £ 2|x – y|3/2, for all x, y, Î R. If f (0) = 1 then
(a) 0
The value of the integral
The integral
4
(c)
5
0
2
87.
The value of
p
91.
(d) –3ò f ( x ) dx
0
86.
18
(b)
25
a
(c) 2 ò f ( x ) dx
85.
ò f ( x ) dx
0
)
greatest integer less than or equal to t, is:
[Jan. 10, 2019 (II)]
1
1
(a)
(7p + 5)
(7p - 5)
(b)
12
12
3
3
(4p - 3)
(4p - 3)
(c)
(d)
20
10
a
is equal to:
2, 2
p /2
90.
Let f and g be continuous functions on [0, a] such that
f(x) = f(a – x) and g(x) + g(a – x) = 4, then
)
x
24
(a)
25
g ( t ) dt
2, 0
f (t) dt = x 2 + ò t 2 f (t) dt, then f ¢ (1/2) is:
x +5
5
84.
g ( t ) dt
5
x +5
ò
ò
(b)
x +5
(c) 2
ò
0
x +5
g ( t ) dt
ò
(a)
If
(d)
((-
1
x
89.
(b)
[Jan 10, 2019 (I)]
Downloaded from @Freebooksforjeeneet
EBD_8344
M-382
96.
Mathematics
The value of integral
ò
3p
4
p
4
104. For x Î R, x ¹ 0, if y(x) is a differentiable function such
x
dx is
1 + sin x
[Online April 15, 2018]
p
(a)
( 2 + 1)
2
(c) 2p ( 2 - 1)
97.
1 -x
e
0
If I1 = ò
(d) p 2
1 - x3
e
0
cos2 x dx and I3 = ò
[Online April 15, 2018]
(b) I3 > I1 > I2
(d) I3 > I2 > I1
ò–2p sin
4
2
æ
æ 2 + sin x ö ö
x çç1 + log ç
÷ ÷÷ dx
è 2 – sin x ø ø
è
is
[Online April 15, 2018]
3
p
16
(a)
(b) 0
(c)
99.
dx
is equal to :
[2017]
p
4
(a) –1
(b) –2
ò tan
(c) 2
(d) 4
n
æ 1 ö
æ 1 ö
æ1 ö
(a) çè - , 0÷ø (b) çè - ,1÷ø (c) çè , 0÷ø
5
5
5
2
101. If
dx
ò
(b) 2
102. The integral
p
4
p
12
(tan x + cot x)3
dx equals :
15
64
(b)
2x12 + 5x 9
103. The integral ò
(x
x
(a)
(c)
+ x3 + 1
5
(
(
)
5
)
2 x5 + x3 +1
-x
(c)
5
)
x5 + x3 +1
2
2
+C
+C
3
13
32
15
256
(d)
(b)
(a)
1
3
(b) 6
(c) 7
1
1
0
0
-1
[Online April 9, 2016]
p
+ log 2
2
p
- log 4
(c)
2
107. The integral
(
)
2 x5 + x3 +1
10
(d)
4
(b) log2
(d) log4
[2015]
log x
2
ò log x 2 + log(36 - 12 x + x 2 ) dx is equal to :
2
(a) 1
(b) 6
(c) 2
(d) 4
108. Let f : R ® R be a function such that
f(2 – x) = f(2 + x) and f(4 – x) = f(4 + x), for all xÎR and
ò f (x)
dx = 5. Then the value of
x
(
)
2 x5 + x3 +1
2
50
ò10 f (x)
dx is :
0
[Online April 11, 2015]
(a) 125
(b) 80
(c) 100
(d) 200
109. Let f : (–1, 1) ® R be a continuous function. If
ò
f (t ) dt =
0
1
2
æ 3ö
3
x, then f ç
ç 2 ÷÷ is equal to :
2
è
ø
[Online April 11, 2015]
(b)
110. For x > 0, let f (x) =
2
(d) 3
(1 - x + x 2 )dx is equal to :
3
2
x
-x
C -x
e
x3
4 ë
û
denotes the greatest integer less than or equal to x, is :
[Online April 10, 2016]
[2016]
10
(d)
ò é x 2 - 28x + 196ù + [x 2 ] , where [x]
(a)
dx is equal to :
C -x
e
x
[x 2 ]dx
sin x
[Online April 8, 2017]
15
128
1
1
(c)
2
[Online April 9, 2017]
(c) 3
(d) 4
8cos 2x
ò
æ1 ö
(d) çè , -1÷ø
5
k
then k is equal to :
k +5
=
3
1 2
( x - 2 x + 4) 2
(a) 1
(a)
10
(a)
x dx,(n > 1) . I4 + I6 = a tan5x + bx5 + C,
where C is constant of integration, then the ordered pair
(a, b) is equal to :
[2017]
100. Let In =
C
(b)
[Online April 10, 2016]
2
-1
-1
106. If 2 ò tan xdx = ò cot (1 - x + x )dx , then
3
4
(d)
1
e x
x2
105. The value of the integral
1
ò 1 + cos x
The integral
3
p
8
1
Cx 3e x
ò0 tan
3p
4
1
(a)
cos2 x dx; I2 = ò
The value of the integral
1
(b) p ( 2 - 1)
p
98.
x
(where C is a constant)
1 - x2
e
0
dx; then
(a) I2 > I3 > I1
(c) I2 > I1 > I3
x
that x ò y(t)dt = (x + 1) ò ty(t)dt , then y(x) equals :
1
+C
log t
ò 1+ t
(c)
3
2
3
1
dt. Then f (x) + f æç ö÷ is equal to:
è xø
[Online April 10, 2015]
(a)
1
( log x ) 2
4
(b) log x
(c)
1
( log x ) 2
2
(d)
+C
(d)
1
log x 2
4
Downloaded from @Freebooksforjeeneet
M-383
Integrals
p
111. The integral
ò
1 + 4sin 2
0
x
x
- 4sin dx equals:
2
2
[2014]
(c) p - 4
1
x
value of the integral
e
b
Statement-2 :
e
dt, x > 0 then the
t
t
ò t + a dt, where a > 0, is:
[Online April 19, 2014]
(b) e -a éë F ( x + a ) - F ( a ) ùû
sin 2 x
ò
(c) ea éë F ( x + a ) - F (1 + a ) ùû
cos 2 x
-1
ò ( f ( x ) + x ) dx
-p
(b) 0
p/2
120. The value of
-p / 2 1 + 2
x
dx is :
(a) p
(b)
p
2
(c) 4p
(b) 0
(d) -
(c) –1
p
4
ò
tan 2 x dx is equal to :
p
2
[Online April 22, 2013]
(b) log 2
(c) 2 log 2
(d) log 2
e
122. If x = ò
1
0
1
2 ln
ò
(a) log 2 2
y
[Online April 11, 2014]
(c) – 9 e
(d) 10
(b) 10e
0
(1 + 2x )
1 + 4x
2
dt
1+ t2
(a) y
x
ò t dt , x Î R, which are parallel to the line y = 2x, are
d2y
dx 2
is equal to :
(b)
x
1+ y
p
p
p
p
ln2
ln2 (d)
ln2
ln2 (b)
(a)
(c)
8
16
32
4
117. The intercepts on x-axis made by tangents to the curve,
, then
[Online April 9, 2013]
(c)
dx , equals:
[Online April 9, 2014]
2
1+ y 2
(d) y 2
x
ò
123. If g (x) = cos 4t dt , then g (x + p) equals
0
g ( x)
(a)
g ( p)
(c) g (x) – g (p)
(b) g (x) + g (p)
(d) g (x) . g (p)
0
equal to :
(d)
7p / 3
121. The integral
[Online April 12, 2014]
115. If for n ³ 1, Pn = ò ( log x )n dx , then P10 – 90P8 is equal to:
y=
sin 2 x
p
4
7p / 4
is equal to:
0
116. The integral
ò
(d) -
(c) 1
[Online April 23, 2013]
p
(a) – 9
cos -1 ( t ) dt equals :
0
= p2 –
p
p
p
(a) p
(b)
(c)
(d)
3
6
2
114. If [ ] denotes the greatest integer function, then the integral
p
2
ò
sin ( t )dt +
[Online April 25, 2013]
æ pö
t2, for all t ³ – p, then f ç - ÷ is equal to:
è 3ø
[Online April 12, 2014]
(a)
[2013]
a
p
, the value of
2
p
(a)
4
t
ò [ cos x ] dx
ò
0
éëF ( x + a ) - F (1 + a ) ùû
113. If for a continuous function f(x),
b
f ( x )dx = ò f (a + b - x )dx.
(a) Statement-1 is true; Statement-2 is true; Statement-2
is a correct explanation for Statement-1.
(b) Statement-1 is true; Statement-2 is true; Statement-2
is not a correct explanation for Statement-1.
(c) Statement-1 is true; Statement-2 is false.
(d) Statement-1 is false; Statement-2 is true.
119. For 0 £ x £
(a) e éë F ( x ) - F (1 + a ) ùû
a
(d) e
is equal to p/6
a
1
-a
dx
p / 6 1 + tan x
x t
112. Let function F be defined as F(x) = ò
p /3
ò
p
(b) 4 3 - 4 3
2p
- 4-4 3
(d)
3
(a) 4 3 - 4
(a) ± 1
(b) ± 2
(c) ± 3
(d) ± 4
118. Statement-1 : The value of the integral
[2013]
Downloaded from @Freebooksforjeeneet
[2012]
EBD_8344
M-384
Mathematics
124. If [x] is the greatest integer £ x, then the value of the integral
æ 2
æ 2 - xöö
dx is [Online May 26, 2012]
ò ç éë x ùû + log ç
è 2 + x ÷ø ÷ø
è
-0.9
(a) 0.486 (b) 0.243
(c) 1.8
(d) 0
0.9
125. The value of the integral
ò
éë x - 2 [ x ]ûù dx ,
2
2
(d) I > and J < 2
and J > 2
3
3
133. The solution for x of the equation
x
[Online May 12, 2012]
(b) 2[G (p 4) - G(p 16)]
(c) p[G (1 2) - G (1 4)]
(d) G (1
x
127. If ò t f ( t ) dt = sin x - x cos x e
2) - G(1 2)
(b)
1
F(e) equals
(a) 1
(b) 2
3
2
3
4
(d)
5
4
1
(c) [a ] f ([a ]) - { f (1) + f (2) + .............. f ( a)}
(d) af ([ a]) - { f (1) + f (2) + .............. f ( a )}
-
p
2
ò
[( x + p )3 + cos 2 ( x + 3p )]dx is equal to
[2010]
0
(a) 21
(b) 41
(c) 42
(d)
(a)
p4
32
ò [cot x] dx , where [ . ] denotes the greatest integer
0
function, is equal to :
(a) 1
(b) –1
1
132. Let I = ò
0
sin x
x
p
(c) 2
1
dx and J = ò
0
the following is true?
cos x
x
p
(d)
2
(b)
p4 p
p
+
(c)
32 2
2
ò xf (sin x)dx
p
-1
4
is equal to
[2006]
0
p
p
(a) p ò f (cos x )dx
(b) p ò f (sin x )dx
0
0
(c) p
2
p /2
ò
(d) p
f (sin x)dx
6
3
1
2
ò
f (cos x )dx
0
138. The value of integral,
I =ò
(a)
p /2
0
[2009]
dx. Then which one of
(d)
p
137.
41
p
131.
[2006]
3p
2
1
equals
[2006]
(b) [a ] f (a) - { f (1) + f (2) + .............. f ([a ])}
136.
dx is
[2011]
1+ x2
p
p
log 2 (b)
log 2 (c) log 2
(a)
(d) p log 2
8
2
130. Let p(x) be a function defin ed on R such that
p¢(x) = p¢(1 – x), for all x Î [0, 1], p (0) = 1 and p (1) = 41.
ò p( x) dx
(d) 0,
greatest integer not exceeding x is
(a) af (a) - { f (1) + f (2) + .............. f ([ a])}
[2011 RS]
(c)
[2007]
(c) 1/2
135. The value of ò [ x ] f '( x )dx , a > 1 where [x] denotes the
0
Then
(d) None of these
log t
1
dt , Then
134. Let F(x) = f (x) + f æç ö÷ ,where f ( x) = ò
1+ t
è xø
l
8log(1 + x)
ò
129. The value of
(b) 2 2
a
x2
, for all x Î R - { 0} ,
2
x éë x 2 ùû dx is :.
(a) 0
3
2
(a)
æ pö
[Online May 7, 2012]
then the value of f ç ÷ is
è 6ø
(a) 1/2
(b) 1
(c) 0
(d) – 1/2
128. Let [.] denote the greatest integer function then the value
0
[2007]
(c) 2
(a) G (p 4) - G (p 16)
ò
p
is
2
=
x
2 tan(p x 2 )
dx is equal to
ò .e
14 x
of
2
2 t t -1
, x Î (0, p 2), then
12
1.5
dt
ò
where [.] denotes the greatest integer function is
[Online May 19, 2012]
(a) 0.9
(b) 1.8
(c) – 0.9
(d) 0
d
e
126. If
G( x ) =
dx
x
2
and J < 2
3
(b) I <
(c) I <
0
tan x
2
and J > 2
3
(a) I >
0.9
(b)
3
2
x
9- x + x
(c) 2
[2008]
Downloaded from @Freebooksforjeeneet
dx is
(d) 1
[2006]
M-385
Integrals
p
(a) a p
cos 2 x
ò
139. The value of
- p 1+ a
147. Let f(x) be a function satisfying f '(x) = f(x) with
f(0)=1 and g(x) be a function that satisfies
2
1
1
p
a
(c)
x
140. If I1 = ò 2 dx , I 2 =
ò f ( x ) g ( x ) dx, is
(d) 2p
3
0
2
2
x
ò 2 dx and
(a) e +
e2 5
+
2 2
(b) e -
e2 5
2 2
(c) e +
e2 3
2 2
(d) e -
e2 3
- .
2 2
1
3
x
I 4 = ò 2 dx then
[2005]
b
148. If f (a + b - x) = f ( x) then ò xf ( x )dx is equal to [2003]
1
a
(a) I 2 > I1 (b) I1 > I 2 (c) I3 = I 4 (d) I3 > I 4
141. Let f : R ® R be a differentiable function having f (2) = 6,
æ 1 ö
f '(2) = ç ÷ . Then lim
x®2
è 48 ø
(a) 24
142. If f ( x) =
(b) 36
e
f ( x)
ò
6
b
3
4t
dt equals
x-2
(c) 12
(a)
a+b
a+b b
f (a + b + x )dx (b)
ò f (b - x)dx
ò
2
2 a
a
(c)
a+b b
ò f ( x)dx
2 a
[2005]
(d) 18
f ( a)
x
1+ e
ò
, I1 =
x
ò
2
ò sec tdt
f (-a )
x ®0
(a) 0
g{x (1 - x )}dx,
I2
is
I1
(b) –3
p
p/2
0
0
ò xf (sin x)dx = A ò
(a) 2p
(b) p
ò
(c) –1
p
4
1 + sin 2 x
(b) 1
(c) 2
[2004]
(d) 0
151.
dx is
(b) F (t ) = 1 - te -t (1 + t )
(c) F (t ) = e t - (1 + t )
(d) F (t ) = tet .
ò
p 2 x (1 + sin x )
1 + cos 2 x
[2004]
p2
4
(a)
(d) 0
dx is
(b) p 2
[2002]
(c)
ero
(d)
p
2
2
ò |1- x
(b)
[2003]
(a) F (t ) = te -t
-p
2
|dx is
[2004]
152.
14
3
ò [x
2
]dx is
[2002]
0
-2
1
3
(d) 1
0
3
(a)
(c) 2
t
(d) 2
f (sin x )dx, then A is
(sin x + cos x )2
0
145. The value of
(b) 3
[2003]
F (t ) = ò f (t - y ) g ( y ) dy , then
[2004]
(c)
p/2
144. The value of I =
(a) 3
is
x sin x
150. If f ( y ) = e y , g ( y ) = y; y > 0 and
then the value of
143. If
0
149. The value of lim
f (-a )
(a) 1
b-a b
ò f ( x )dx .
2 a
(d)
x2
xg{x (1 - x )}dx
f ( a)
and I 2 =
[2003]
0
x
ò 2 dx , I3 =
0
2
[2005]
f(x) + g(x) = x 2 . Then the value of the integral
p
2
(b)
1
x
dx , a > 0, is
(c)
7
3
(d)
(a) 2 –
28
3
p/4
[2003]
153. I n =
0
ò
tan n x dx then lim n[ I n + I n + 2 ] equals
n®¥
0
(a)
1
1
+
n +1 n + 2
(b)
1
n +1
(c)
1
n+2
(d)
1
1
.
n +1 n + 2
2
(d) - 2 - 3 + 5
2 –1
(c)
1
146. The value of the integral I = ò x(1 - x ) n dx is
(b) 2 +
2
(a) 1
2
154.
10 p
ò0
(b) 1
(c) ¥
(d) ero
| sin x | dx is
(a) 20
(b) 8
[2002]
[2002]
(c) 10
Downloaded from @Freebooksforjeeneet
(d) 18
EBD_8344
M-386
Mathematics
TOPIC Ė
1
Reduction Formulae for Definite
Integration, Gamma & Beta
Function, Walli's Formula,
Summation of Series by
Integration
æ ( n + 1) (n + 2)...3n ö n
159. lim ç
÷ is equal to:
n ®¥ è
n 2n
ø
(a)
155. Let a function f : [0, 5] ® R be continuous, f (1) = 3 and
F be defined as:
x
x
F(x) =
òt
2
g (t )dt , where g(t) =
ò f (u )du
Then for the function F, the point x = 1 is :
[Jan. 9, 2020 (II)]
(a) a point of local minima.
(b) not a critical point.
(c) a point of local maxima.
(d) a point of inflection.
æ (n + 1)1/3 (n + 2)1/3
(2n)1/3 ö
+
+
+
lim
.....
ç
÷ is equal to :
156. n ®¥ ç
4/3
n 4/3
n 4/3 ÷ø
è n
[April 10, 2019 (I)]
3 4/3 3
4 4/3
(2)
(2) (a)
(b)
4
4
3
3 4/3 4
4 3/4
(2)
(2) (c)
(d)
2
3
3
n
n
1 ö
æ n
+
+ ..... + ÷ is equal
157. lim ç 2 2 + 2
2
2
2
n ®¥ è n + 1
5n ø
n +2
n +3
to :
[Jan. 12, 2019 (II)]
p
4
158. If lim
n ®¥ ( n
(b) tan–1(3) (c)
p
2
+ 1)a -1 [ (na + 2) + .....(na + n) ]
=
1
60
for some positive real number a, then a is equal to :
[Online April 9, 2017]
(a) 7
(b) 8
(c)
15
2
18
e
(d)
17
2
(d)
4
dx
é1
161. nLim
ê
®¥
ë n2
27
e2
is a polynomial of degree
sin 6 x
[Online May 26, 2012]
(b) 5 in tan x
(d) 3 in cot x
(a) 5 in cot x
(c) 3 in tan x
ù
1
2
4
1
sec 2 2 + 2 sec 2 2 ............ + sec 2 1ú
n
n
n
n
û
equals
(a)
[2005]
1
sec 1
2
1
cosec 1
2
1
tan 1
(d)
2
(b)
(c) tan 1
162.
Lim
n®¥
n
163. lim
164.
[2004]
(b) e – 1
1 + 24 + 34 + ...n 4
n®¥
(a)
r
1
å n e n is
r =1
(a) e + 1
(d) tan–1(2)
1a + 2a + ......... + na
(b) 3 log 3 – 2
e2
160. f ( x ) = ò
1
1
(a)
(c)
9
[2016]
n
1
5
lim
n ®¥
5
(b)
(c) 1 – e
- lim
1 + 23 + 33 + ...n3
(c) Zero
1 p + 2 p + 3 p + ..... + n p
n p +1
[2003]
n5
n®¥
1
30
(d) e
is
(d)
1
4
[2002]
(a)
1
p +1
(b)
1
1- p
(c)
1
1
p p -1
(d)
1
p+2
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M-387
Integrals
1.
(Bonus)
1
sin( x - 1) 4
lim 2
x ®1 ( x - 1) sin( x - 1)
4.
Q
Let x – 1 = h when x ® 1 then h ® 0
lim
sin h4
h4
h® 0
´
h
´ h2 = 1 ´ 1 ´ 0 = 0
sin h
x2
ò ( x sin x + cos x) 2 dx
(a)
d
( x sin x + cos x ) = x cos x
dx
=ò
x cos x
æ x ö
ç
÷ dx
( x sin x + cos x ) 2 è cos x ø
2.
(d)
ò (e
+ 2e x - e - x - 1) × e( e
2x
I = ò ( e 2 x + e x - 1) × e( e
= ò e x (e x + 1 - e - x ) × e (e
= ò (e - e
x
-x
+ 1)e
+ e- x )
x
x
x
+ e- x )
dx + e( e
dx + e
=
dx
dx + ò ( e x - e - x ) e( e
+ e- x )
(e x + e- x + x )
x
x
+ e- x )
= e( e
x
+e
-x
x
+ x)
= (e x + 1). e(e
+ e- x )
= et + e (e
x
-ò
+ e- x )
(ex + e- x )
+ e- x )
+ e(e
x
+e
+ e- x )
-x
)
+C
ò 5 + 7 sin q - 2 cos
2
q
dq =
5 + 7t - 2 + 2t 2
(d)
x
ò (1 + x)2 dx ( x > 0)
2 tan 2 q × sec 2 q
sec 4 q
sin 2q
+C
2
f ( x) = q -
1 2t + 1
1 2sin q + 1
= ln
+ C = ln
+C
5
5
sin q + 3
t +3
tan q
1 + tan 2 q
+ C = tan -1 x -
Now f (3) - f (1) = tan -1 ( 3) -
1
2sin q + 1
and A =
5
2(sin q + 3)
B(q) 5(2sin q + 1)
=
A
(sin q + 3)
d q = ò 2sin 2 qd q
1
2 tan q
Þ f ( x) = q - ´
+C
2 1 + tan 2 q
ò
Þ
- x sec x
+ tan x + C
x sin x + cos x
= q-
dt
1
t+
dt
1
1
2 +C
Þ
= ln
2 æ 7 ö2 æ 5 ö2 5 t + 3
çt + ÷ -ç ÷
è 4ø è4ø
\ B (q ) =
=
I =ò
(d) Let sin q = t Þ cos q d q = dt
cos q
x é
-1
ù
+ sec2 x dx
cos x êë x sin x + cos x úû ò
Put x = tan 2 q Þ 2 x dx = 2 tan q sec 2 q d q
+C
So, g (x) = 1 + ex and g (0) = 2
3.
x sin x + cos x é
-1
ù
dx
ê
2
cos x
ë x sin x + cos x úû
=
+C
5.
x
x é
-1
ù
cos x êë x sin x + cos x úû
dx
Let e x + e - x + x = t Þ (e x + e - x + 1)dx = dt
= ò et dt + e (e
I
II
(No any option is correct)
=
6.
x
+C
1+ x
3
1
- tan -1 (1) +
1+ 3
2
3
p 1
+ 12 2 4
æ
x ö
-1
(a) I = sin -1 ç
x × 1 dx
ç 1 + x ÷÷ dx = tan
è
ø
ò
ò
I
Downloaded from @Freebooksforjeeneet
II
EBD_8344
M-388
Mathematics
= x tan -1 x - ò
= x tan
1
1
×
.x dx + C
1+ x 2 x
9.
(Put x = t 2 Þ dx = 2t dt )
= x tan -1 x - ò
t
2
I =ò
dt + C
1+ t2
cos x dx
ò sin 3 x(1 + sin6 x)2 3
= f (x) (1 + sin6x)1/l + c
If sin x = t
then, cos x dx = dt
1 t × 2t dt
x- ò
+C
2 1+ t 2
-1
(d) Let I =
dt
2
t3 1 + t 6 3
= x tan -1 x - t + tan -1 t + C
Put 1 +
= x tan -1 x - x + tan -1 x + C
=ò
(
)
1
= r3 Þ
t6
...(i)
dt
2
7æ
1 ö3
t ç1 + 6 ÷
è t ø
dt
t7
=
-1 2
r dr
2
= ( x + 1) tan -1 x - x + C
-
Þ A( x) = x + 1 Þ B( x) = - x
7.
(a) I = ò
dx
1
-6
7
1
dx
( x + 4)2
f(x) = –
x-3 7
=t ,
x+4
Differentiate on both sides, we get
( x + 4)
dx = 7t 6dt
2
10.
Hence, I = ò t -6t 6 dt = t + C = æç
÷ +C
è x+4ø
(c) I = ò
=ò
(c) Given integral, I =
Þ I=
dq
= log
cos 2 q(tan 2q + sec 2q)
11.
sec2 q
1 + tan 2 q
2
+
2 tan q
dq
ò
(2 x 3 - 1) dx
4
x +x
=ò
(2 x - x -2 )dx
x 2 + x -1
sec 2 q(1 - tan 2 q)
(1 + tan q)2
du
òu
= log | u | + c = log| x2 + x–1 | + c
x3 + 1
+c
x
(b) Given integral
sin( x + a )
tan x + tan a
ò tan x - tan adx = ò sin( x - a) dx
2
1 - tan q 1 - tan q
=ò
[from eqn. (i)]
Put x2 + x–1 = u Þ (2x – x–2)dx = du
1
x - 3 ö7
8.
1
cosec2x and l = 3
2
æpö
\ lf ç ÷ = -2
è3ø
Let
7
1
1
1 æ sin 6 x + 1 ö 3
=- ç
(1 + sin 6 x ) 3 + c
÷ +c =2 çè sin 6 x ÷ø
2sin 2 x
( x + 4)8/7 ( x - 3)6/7
æ x-3ö
= òç
÷
è x+4ø
1 r 2 dr
1
=- r +c
2 ò r2
2
Let x – a = t Þ dx = dt
dq
=
sec 2 q(1 - tan q)
dq
1 + tan q
Let tanq = t Þ sec2q dq = dt, then
=ò
2 ö
æ 1- t ö
æ
I = òç
dt = ò ç -1 +
dt
÷
1 + t ÷ø
è 1+ t ø
è
= – t + 2 log (1 + t) + C
= – tanq + 2 log (1 + tanq) + C
Hence, by comparison l = – 1 and f (x) = 1 + tanq
ò
sin(t + 2a)
dt = ò [cos 2a + sin 2a. cot t ]dt
sin t
= cos 2a . t + sin 2a . log | sin t | + c
= (x – a) . cos 2a + sin 2a . log | sin (x – a) | + c
dx
12.
(a) Let I =
Let (x – 1)2 = 9 tan2 q
Þ tan q =
dx
ò ( x2 - 2 x + 10)2 = ò (( x - 1)2 + 9) 2
x -1
3
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...(i)
M-389
Integrals
After differentiating equation ...(i), we get
2 (x – 1) dx = 18 tan q sec2q dq
\ I=
15.
òe
sec x
18tan q sec2 qd q
ò 2 ´ 3tan q ´ 81sec4 q
I=
1
1 1
cos 2 qd q =
´ (1 + cos 2q)d q
ò
27
27 2 ò
I=
1 ì sin 2q ü
íq +
ý+c
54 î
2 þ
(
\ f (x) = sec x + tan x + C
16.
= ò (3 - 2(1 - cos 2 x ) + 2cos x) dx
= ò (1 + 2 cos x + 2cos 2 x ) dx
4
2
x6 ) 3
=ò
dx
2
x 7 (1 + x -6 ) 3
1
1
6 3
1
= - (1 + x -6 ) 3 + C = - 1 (1 + x ) + C
2
2
2
x
sec 2 xdx
4
tan 3 x
Put tan x =
Þ sec2 x dx = d
=-1
dx
ò
2
1
æ 1 ö t dt
Now, I = ò ç - ÷ 2 = - t + C
2
2
è
ø t
(a) I = sec 3 x.cosec 3 dx
ò
Þ I =òz
(d) Let,
dx æ 1 ö 2
Put 1 + x–6 = t3 Þ – 6–7 dx = 3t2 dt Þ 7 = ç - ÷ t dt
è 2ø
x
\ S = {0, - 2, 2}
4
3
= x + 2 sin x + sin 2x + c
x3 (1 +
x2 æ x2 ö
Þ 2 çç 2 - 1÷÷ = 0 Þ x = 0,0, 2, - 2
è
ø
-
sin 3 x + sin 2 x
dx
sin x
[Qsin 2x = 2 sin x cos x and sin 3x = 3 sin x – 4 sin3x]
17.
æx
x ö
Q f(x) = f(0) Þ K çç 4 - 2 ÷÷ = 0
è
ø
ò
x
5x
2cos .sin
2
2 dx
ò
x
x
2cos .sin
2
2
= ò (3 - 4sin 2 x + 2cos x)dx
æ x4 x 2 ö
Þ f(x) = K çç 4 - 2 ÷÷ + C (using integration)
è
ø
Þ f(0) = C
2
æ 5x ö
sin ç ÷
è 2 ø dx
(c) ò
=
æxö
sin ç ÷
è 2ø
=
1
and f (x) = 3 (x – 1)
54
(d) Since, function f(x) have local extreem points at
x = –1, 0, 1. Then
f(x) = K (x + 1) x (x – 1)
= K (x3 – x)
I=ò
)
f ( x ) = ò ( sec x tan x ) + sec 2 x dx
we get: A =
2
( ( g ' ( x ) f ( x ) ) + f '( x ) ) dx = e g ( x ) ´ f ( x ) + C
Our comparing above equation by equation (i),
é -1 æ x - 1 ö
ù
f ( x)
Compare it with A ê tan ç
÷+ 2
ú+c,
b
è
ø x - 2 x + 10 û
ë
14.
2
g x
Q òe ( )
1 é -1 æ x - 1 ö
3( x - 1) ù
I = 54 ê tan ç 3 ÷ + 2
ú+c
è
ø x - 2 x + 10 û
ë
4
(sec x tan x f ( x ) + (sec x + tan x + sec x )) dx
= esec x f ( x ) + C ...(i)
é
æ x -1 ö ù
2ç
ê
÷ ú
1 ê -1 æ x - 1 ö 1
3 ø ú
+ ´ è
+c
tan ç
÷
2ú
I = 54 ê
è 3 ø 2
æ x -1 ö ú
ê
1+ ç
÷
êë
è 3 ø úû
13.
(a) Given,
-1
z3
× dz =
+ C Þ I = -3(tan x ) 3 + C
æ -1 ö
ç ÷
è 3 ø
1
2 x3
x(1 +
1
6 3
x )
Hence, f (x) = -
+C
1
2x3
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EBD_8344
M-390
18.
Mathematics
(c) Let the integral, I = ò cos(ln x) dx
Þ I = cos(ln x ).x - ò
3
1æ 1
ö2
= - ç 2 - 1÷ + C
ø
3è x
- sin(ln x )
.x dx
x
3
1 1
= - × 3 × (1 - x 2 ) 2 + C
3 x
= x cos(ln x) + ò sin(ln x)dx
(
)
3
-1
1 - x2 + C
3
3x
Compare both sides,
cos(ln x )
.x dx
= x cos (ln x) + sin(ln x).x – ò
x
=
= x cos(ln x) + sin(ln x) . x – I
Þ A(x) = -
x
[cos (ln x) + sin (ln x )] + C
2
Þ (A(x))3 =
Þ I=
19.
3x13 + 2 x11
dx =
(b) I = ò
(2 x4 + 3x2 + 1) 4
ò
3x13 + 2 x11
3
1ö
æ
x16 ç 2 + 2 + 4 ÷
è
x
x ø
4
dx
21.
Let 2 +
I=
I=
ò
Let
I=
+C =ò
x+4
3
Hence, f (x) =
1
22.
n
n
(a) Let, I = ò (sin q - sin q) cos qd q
n+1
sin q
Let sin q = u
Þ cos q dq = du
1- x
dx
x4
2
1
n
n
\ I = ò (u - u) du
n +1
u
1
-1
x2
dx
x3
1
1 ön
æ
1
ç1 ÷
= u - n (1 - u1-n ) n du
= è u n -1 ø
du ò
ò un
1
- 1 = u2
x2
Þ -
(t 2 + 3)
t 3 3t
dt = + + C
2
6 2
= f (x). 2 x –1 + C
1
x12
+C
6 (2 x4 + 3x 2 + 1)3
2
ò
1
dt
-4+1
2 = -1 t
+C
2 -4 + 1
t4
( 1- x )
2 x -1 = t
3
-
m
x +1
dx
2x – 1
ò
2
= (2 x –1) + 3 (2 x –1) 2 + C
6
2
æ x + 4ö
= 2 x –1 çè
÷ +C
3 ø
-1
1
1
´
+C
3
2 (-3) æ
3
1ö
çè 2 + 2 + 4 ø÷
x
x
(a) A( x)
=
ò
-1
27x 9
\ 2x – 1 = t2 Þ dx = tdt
3
1
æ 3 2ö
+ 4 = t, -2 ç 3 + 5 ÷ dx = dt
2
èx
x ø
x
x
Then, I =
(c) Let I =
Put
3
2
+
x3 x5
dx
ò
4
I= æ
3
1ö
çè 2 + 2 + 4 ÷ø
x
x
20.
1
and m = 3
3x3
Þ 2I = x(cos (ln x) + sin (ln x)) + C
2 2u du
=
x3
dx
Let 1 – u1 – n = v
Þ –(1 – n)u–n du = dv
dx
= –u du
x3
Þ u–n du =
m
3
u
A( x) æç 1 - x 2 ö÷ + C = ò (-u 2 )du = - + C
è
ø
3
dv
n -1
Downloaded from @Freebooksforjeeneet
M-391
Integrals
\ I=ò
1
vn
n
= 2 v
n -1
=
23.
25.
1
+1
vn
1
×
dv
×
= n -1 1
+1
n -1
n
n +1
n
=ò
n +1
ö n
æ
+C =
ç1 ÷
n2 - 1 è u n -1 ø
n
n+1
ö n
n æ
1
ç1 - n-1 ø÷
è
n -1
sin q
2
1
=ò
+C
sin2 xcos2 x
(sin5x+ cos3xsin2x+ sin3xcos 2 x + cos5x)2
sin 2 x cos2 x
[(sin 2 x + cos 2 x)(sin 3 x + cos3 x)]2
=ò
sin 2 x cos 2 x
3
+C
2sin( x 2 - 1) - 2sin( x 2 - 1)cos( x 2 - 1)
dx
2sin( x 2 - 1) + 2sin(x 2 - 1)cos( x 2 - 1)
\ I=
26.
ÞI=
1 - cos( x - 1)
dx
1 + cos( x 2 - 1)
2
æ x - 1ö
Þ I = ò x tan ç 2 ÷ dx ,
è
ø
= x-ò
2
æ x 2 - 1ö
1
2 æ x - 1ö
or I = ln sec ç 2 ÷ + c = 2 ln sec ç 2 ÷ + c
è
ø
è
ø
=ò
Þ f(x) = ò
- dt
= ò - t -2 dt = t -1 + c
t2
Þ f(x) =
1
+ c, f (0) = 0 Þ c = 0
2 + x -7 + x -5
\ f(1) =
1
4
dt
1
1
t2 + t + +1 4
4
æ 1ö æ 3ö
÷
ç t + ÷ + çç
è 2 ø è 2 ÷ø
2
2
æ 2 tan x + 1 ö
tan -1 ç
÷+C
3
3
è
ø
\ A = 3 and K = 2
5x + 7 x
dx
(2 + x -7 + x -5 )2
Let 2 + x–7 + x–5 = t
Þ (–7x–8 – 5x–6)dx = dt
sec2 xdx
1 + tan x + tan 2 x
dt
ÞI= x-
-8
27.
dx
tan x
æ 1ö
çt + ÷
2
-1
2 ÷+C
tan ç
ÞI= x3
ç 3 ÷
ç
÷
è 2 ø
5 x8 + 7 x 6
dx, x ³ 0
( x 2 + 1 + 2 x7 )2
5 x8 + 7 x 6
dx
= ò 14 -5
x ( x + x -7 + 2)2
-6
(1 + tan 3x) 2
tan x + 1 + tan 2 x
(1 + tan 2 x)
dx
ò 1 + tan x + tan 2 x
tan x + 1 + tan 2 x
\I= x-ò
2x
x2 - 1
dx = dt
=t Þ
2
2
(a) f(x) = ò
ò
tan 2 x × sec 2 x
Put tan x = t Þ sec2 x . dx = dt
\ I = ò | tan(t ) | dt = ln | sec t | + C
24.
dx = ò
dx
ò 1 + tan x + tan 2 x dx
ÞI= x-ò
2
Now let
2
dx
-1
1 dt
1
+C
=- + C =
ò
2
3 t
3t
3(1 + tan3 x)
(a) Let I =
(\ sin 2 q = 2sin q cos q)
Þ I=ò x
3
(sin x + cos x)
Now, put (1 + tan3x) = t
Þ 3 tan2x sec2x dx = dt
(c, d ) Consider the given integral
I=ò x
(a) Let I
2
(b) Suppose,
x–4
= y Þ x – 4 = y (x + 2)
x+2
Þ x (1 – y) = 2y + 4 Þ x =
2y + 4
1– y
æ 2y + 4 ö
So, f (y) = 2 ç
÷ +1
è 1– y ø
æ 2x + 4 ö
3x + 9
Now, f (x) = 2 ç
÷ +1 =
x
1– x
1
–
è
ø
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EBD_8344
M-392
3 ( x + 3) 3 (x –1 + 4)
12
=
= – 3+
1– x
1– x
1– x
ò
\ ò f (x) dx = – 12 loge |1 – x| – 3x + c
(a) Q 7 – 6x – x2 = 16 – (x + 3)2
Þ I =
ò
and
d
(7 – 6x – x2) = – 2x – 6
dx
Þ I =
ò cosec x + cot x . dx
So,
ò
-ò
2x + 5
7 - 6x - x
dx = ò
2
1
16 - ( x + 3)2
2x + 6
7 - 6x - x2
Þ I = log 1 - cos x + C
dx
Þ I = log 2sin 2
4
3x - 4 ö
(b) f çæ
÷ = x + 2, x ¹ 3
è 3x + 4 ø
Consider
3x - 4
=t
3x + 4
31.
Þ 3x - 4 = 3tx + 4t
Þ x=
4t + 4
+2
3 - 3t
Þ I = 2log sin
x
+ C1
2
dx
ò cos3 x
(a)
dx
=
ò 2 cos4 x
4sin x cos x
Let tan x = t2 Þ sec2 x = 1 + t4
sec2 x dx = 2t dt
sec 4 x dx
ò2
tan x
=
ò
2 tan x
(1 + t 4 )2t dt
t5
= ò (1 + t 4 ) dt = t + + k
2t
5
ò
2x - 10
ò f ( x ) dx = ò 3x - 3 dx
=
1
tan x + tan 5 / 2 x + k ét = tan x ù
ë
û
5
dx
2x
dx - 10ò
= ò
3x - 3
3x - 3
A=
1
1
5
,B= ,C=
5
2
2
2 x -1
2 dx 10 dx
dx + ò
= ò
3 x -1
3 x - 1 3 ò x -1
A+B+C=
16
5
(d) Let I =
ò
2x 8
- ln ( x - 1) + C
3 3
32.
log(t + 1 + t 2 )
1 + t2
8
2
Here, A = - , B =
3
3
put u = log(t + 1 + t 2 )
æ 8 2ö
\ (A, B) = ç - , ÷
è 3 3ø
du =
(a) Let, I =
Þ I=
ò
ò
1 + 2 cot x cosec x + 2 cot 2 x . dx
sin 2 x + 2cos x + 2cos2 x
sin 2 x
tan x
sec 2 x (sec2 x dx )
2x - 10
3x - 3
=
30.
x
+ log 2 + C
2
=
Þ f (x) =
\
x
+C
2
Þ I = log sin 2
=
10 - 2t
3 - 3t
Þ f (t) =
1 + cos x
. dx
sin x
Þ I = log cosec x - cot x + log sin x + C
dx
æ x+3ö
= -2 7 - 6 x - x 2 - sin -1 ç
÷+C
è 4 ø
Therefore, A = – 2, & B = – 1.
29.
1 + 2cos x + cos2 x
. dx
sin x
Þ I=
=
28.
Mathematics
dt
é t + 1+ t2 ù
ú =
.ê
t + 1 + t 2 êë 1 + t 2 úû
1
\ I = ò u du =
. dx
Since, I =
u2
+c
2
1
[g(t)]2 + c
2
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1
1+ t2
dt
M-393
Integrals
\ g(t) = log (t + 1 + t 2 )
35.
Put t = 2
g (b) = log (2 + 5)
33.
= òe
= x.e
x+ 1 x
= x.e
= xe
34.
\I=2
1ö
æ
(d) Let I = ò ç1 + x - ÷ e
è
xø
x+ 1 x
x+
x+
1
x
1
x
x+ 1
1ö
æ
-ò x ç 1 - ÷ e
è
x2 ø
x+ 1
x dx
1 ö x+ 1
æ
+ ò ç x - ÷ e x dx
è
xø
é x2
ù
1 x2 + 1
1
1
dx + ò
dx ú + c
I = 2 ê tan 1 x - ò
2
2 x2 + 1
2 1+ x
êë 2
úû
é x2
ù
1
1
I = 2 ê tan -1 x - ò 1.dx + tan -1 x ú + c
2
2
ëê 2
ûú
+C
sin 2 x cos 2 x
ò (sin 3 x + cos3 x)2 dx
é x2
ù
x 1
I = 2 ê tan -1 x - + tan -1 x ú + c
2
2
2
ëê
ûú
2
-1
-1
tan
tan
x
x
+
x
x
+
c
I=
ö
÷ dx
è cos3 x (1 + tan 3 x) ø÷
sin x. cos x
or I = - x + ( x 2 + 1) tan -1 x + c
2
æ sin x.sec 2 x ö
= òç
÷ dx
è (1 + tan3 x ) ø
36.
(b) Let I =
Put 1 + tan 3 x = t
dt = 3tan 2 x sec2 x dx or dx =
\I=
I=
ò
sin 2 x.sec4 x
t
2
dt
=
3tan 2 x sec 2 x
2
3 tan x sec x
1 dt 1 -2
= ò t dt
3 ò t2
3
1 é t -2+1 ù
-1 é1ù
+c
I = 3 ê -2 + 1ú + c =
3 êë t úû
êë
úû
1
3(1 + tan 3 x)
ò
(sin 4 x + cos 4 x)(sin 4 x - cos 4 x)
=
dt
1 sin 2 x.sec4 x
´
= 3ò
2
2
t
sin x sec 4 x
or I = -
ò
=
2
+c
sin8 x - cos8 x
ò 1 - 2sin 2 x cos 2 x dx
(sin 4 x )2 - (cos 4 x)2
dt
´
1 sin 2 x.sec 4 x
dt
´
3ò
sin 2 x
t2
´ sec2 x
cos 2 x
\I=
x dx
é x2
1 x2 + 1 - 1 ù
I = 2 ê tan -1 x - ò 2
dx ú + c
2 x +1
ëê 2
ûú
1 ö x+ 1
æ
1 ö x+ 1
æ
- ò ç x - ÷ e x dx + ò ç x - ÷ e x dx
è
xø
è
xø
2
ò çç
-1
æ 1 - x2 ö
ç
÷ dx
è 1 + x2 ø
é x2
x2 ù
1
-1
´
dx ú + c
I = 2 ê tan x - ò
1 + x 2 2 úû
êë 2
æ sin x.cos x ö
ò çè sin3 x + cos3 x ÷ø dx
æ
I=
ò IIx .tanI
-1
é
ù
æ d
ö
I = 2 ê tan -1 x ò xdx - ò ç (tan -1 x÷ ò xdx)dx ú
è dx
ø
ë
û
2
I=
ò x cos
Applying Integration by parts
x dx
1 ö x+ 1
æ
dx + ò ç x - ÷ e x dx
è
xø
(b) Let I =
(a) Let I =
ò
1 - 2sin 2 x cos 2 x
1 - 2sin 2 x cos2 x
dx
[(sin 2 x + cos 2 x )2 - 2sin 2 x cos 2 x ]
[(sin 2 x + cos 2 x ][sin 2 x - cos 2 x]
1 - 2sin 2 x cos 2 x
= - ò cos 2 x dx =
37.
dx
(c) Let
Let I =
- sin 2 x
1
+ c = - sin 2 x + c
2
2
ò f ( x)dx = y ( x)
òx
5
dx
f ( x3 )dx
put x3 = t
Þ 3x2dx = dt
1
3 · x 2 · x 3 · f ( x3 ) · dx
3ò
1
1
= ò tf (t )dt = éët ò f (t )dt - ò f (t )dt ùû
3
3
I=
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EBD_8344
M-394
Mathematics
1é
t y (t ) - ò y (t )dt ùû
3ë
1 3
3
2
3
= éë x y ( x ) - 3ò x y ( x )dx ùû + c
3
1 3
3
2
3
= x y ( x ) - ò x y ( x )dx + c
3
=
38.
(a) Let I = ò
=
cos 2 x sin 2 x
sin 2 x cos 2 x
cos 2 2 x - sin 2 2 x 2 cos 4 x
=
sin 2 x cos 2 x
sin 4 x
2 cos 2 4 x
2cos 2 4 x . sin 4 x
dx = ò
dx
2cos 4 x
2 cos 4 x
sin 4 x
1
1 cos8 x
1
+ k = - .cos 8 x + k
= ò sin 8 x dx = 2
2 8
16
1
Now, - .cos 8 x + k = A cos 8 x + k
16
1
Þ A=16
sin x
5
5 tan x
x dx
cos
dx = ò
(d) ò
sin x
tan x - 2
-2
cos x
cos x
æ 5sin x
ö
dx
= òç
´
è cos x sin x - 2cos x ÷ø
5sin x dx
=ò
sin x - 2 cos x
æ 4sin x + sin x + 2cos x - 2cos x ö
= òç
÷ø dx
è
sin x - 2cos x
\ I=ò
39.
=ò
ò
ò
1+ x
sin 2 x
cos 2 x dx
2
2
x sec 2 x + tan 2 x
1 + x2
(
dx
)
x 2 1 + tan 2 x + tan 2 x
=
ò
=
ò
=
ò 1 + x2
=
ò
1+ x
dx
2
x 2 + tan 2 x(1 + x2 )
1 + x2
x2
dx
dx + ò tan 2 x dx
x2 +1 -1
1 + x2
(
)
dx + ò sec 2 x - 1 dx
dx
2
+ ò sec 2 xdx - ò dx
1+ x
= – tan–1 x + tan x + c
Given : f (0) = 0
Þ f (0) = – tan –10 + tan0 + c Þ c = 0
\ f (x) = – tan–1x + tan x
Now, f (1) = – tan–1(1) + tan 1 = tan 1 –
41.
p
4
x2 - x
(a) Let I = ò
dx
x3 - x 2 + x - 1
= ò
x ( x - 1)
x 2 ( x - 1) + ( x - 1)
dx = ò
x dx
x2 + 1
=
1 2 x dx
ò
2 ( x 2 + 1)
Let x2 + 1 = t Þ 2x dx = dt
dx
sin x - 2 cos x
( sin x - 2 cos x ) + 2 ( cos x + 2sin x ) dx
=ò
( sin x - 2 cos x )
cos x + 2sin x
= ò dx + 2ò
dx = I1 + I2
sin x - 2 cos x
cos x + 2sin x
where, I1 = ò dx and I 2 = 2ò
dx
sin x - 2 cos x
Let sin x – 2cos x = t
Þ (cos x + 2sin x) dx = dt
dt
\ I 2 = 2ò = 2 ln t + C = 2 ln (sin x–2cos x) + C
t
Hence, I1 + I 2 = dx + 2ln ( sin x - 2cos x ) + c
ò
x 2 sec 2 x +
= ò 1 dx - ò
( sin x - 2 cos x) + ( 4 sin x + 2 cos x )
sin x - 2cos x
æ cos x + 2sin x ö
dx + 2ò ç
dx
=ò
è sin x - 2cos x ÷ø
sin x - 2cos x
æ x 2 + sin 2 x ö
2
(a) Let f (x) = ò ç
÷ sec x dx
è 1 + x2 ø
=
cos 8x + 1
dx
cot 2 x - tan 2 x
Now, Dr = cot 2x – tan 2x =
=
40.
\I=
(
)
1
log x 2 + 1 + c
2
where ‘c’ is the constant of integration.
(d) Statement - 2: cos3x is a periodic function.
It is a true statement.
Statement - 1
=
42.
1 dt 1
ò = log t + c
2 t
2
æ cos 3x 3cos x ö
+
Given f ( x) = ò cos3 x dx = ò ç
÷ dx
è 4
4 ø
=
1 sin 3 x 3
1
3
+ sin x = sin 3x + sin x
12
4
4 3
4
Now, period of
1
2p
sin 3x =
12
3
= x + 2ln |(sin x – 2 cos x)| + k Þ a = 2
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M-395
Integrals
Period of
3
sin x = 2p
4
æ p x pö
log tan ç + + ÷ + C
è 4 2 8ø
1
=
2
L.C.M. ( 2p, 2p ) 2p
= 2p
Hence period of f(x) = HCF of 1,3 =
( )
1
43.
Thus, f(x) is a periodic function of period 2p.
Hence, Statement - 1 is false.
sin xdx
(c) Let I = 2 ò
pö
æ
sin ç x – ÷
è
4ø
Let x –
é
æ p xö ù
êQ ò sec x dx = log tan çè + ÷ø ú
4 2 û
ë
46.
p
= t Þ dx = dt
4
= x–
47.
p
pö
æ
+ log sin ç x – ÷ + c1
è
4
4ø
(c) I = ò
Þ I =ò
sin( x - a + a)
dx
sin( x - a)
ò sin( x - a) dx =ò
=ò
sin( x - a ) cos a + cos( x - a)sin a
dx
sin( x - a )
= (cos a) x + (sin a) logsin( x - a) + C
Comparing with Ax + B logsin(x – a) + c
\ A = cosa, B = sin a
(d) Q f ''(x) – g¢¢ (x) = 0
Integrating, f ¢ (x) – g¢ (x) = c;
Þ f ¢ (1)– g¢ (1) = c Þ 4 – 2 = c Þ c = 2.
pö
æ
pö
æ
= x + log sin èç x – ÷ø + c çè where c = c1 – ÷ø
4
4
44.
sin x
(b)
= ò {cos a + sin a cot( x - a )}dx
æ pö
sin ç t + ÷
è 4 ø = 2 æ sin t + cos t ö dt
Þ I = 2ò
dt
ò ç sin t ÷ø
2 è
sin t
Þ I = ò (1 + cot t )dt = t + log |sin t| + c1
æ x 3p ö
log tan ç + ÷ + C
è2 8 ø
2
1
=
\ f ¢ (x) – g¢ (x) = 2;
Integrating, f (x) – g (x) = 2x + c1
dx
Þ f (2) – g(2) = 4 + c1 Þ 9 – 3 = 4 + c1;
Þ c1 = 2 \ f (x) – g(x) = 2x + 2
At x = 3/2, f (x) – g(x) = 3 + 2 = 5.
cos x + 3 sin x
dx
é1
ù
3
2 ê cos x +
sin x ú
2
2
ë
û
48.
2
(c) I = ò e x x x (2 + loge x ) dx
1
2
=
dx
1
1
dx
= .ò
pö
p
2
2ò é p
ù
æ
sin ç x + ÷
êësin 6 cos x + cos 6 sin x úû
è
6ø
Þ I=
1
2
= ò e x [ x x + x x (1 + loge x)] dx
1
Q ò e x ( f ( x) + f '( x)) dx = e x f ( x) + c
1
pö
æ
. cosec ç x + ÷ dx
è
2ò
6ø
2
We know that
\ I = éëe x x x ùû
1
ò cosec x dx = log | (tan x / 2) | + C
= e2 ´ 4 - e ´ 1 = 4e2 - e = e(4e - 1)
\ I=
45.
I = ò e x x x [1 + (1 + loge x)] dx
(a)
=ò
1
. log tan
2
dx
æx pö
çè + ÷ø + C
2 12
ò cos x - sin x =ò
dx
1
æ 1
ö
2ç
cos x sin x÷
è 2
ø
2
1
pö
dx
æ
=
ò sec çè x + 4 ÷ø dx
pö
æ
2
2 cos ç x + ÷
è
4ø
a+1
49.
ò
(a)
a
dx
( x + a )( x + a + 1)
a+1
=
1
é 1
ù
ê x + a - x + a + 1 ú dx [Using partial fraction]
ë
û
a
ò
a+1
æ ( x + a) ö
÷
= log ç
è ( x + a + 1) ø a
= log
æ 2a + 1 2a + 1 ö
.
= log ç
÷
è 2a + 2 2a ø
9
(Given)
8
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EBD_8344
M-396
Mathematics
(2a + 1)2 9
= Þ
So,
a(a + 1) 2
2
\
2
8a + 8a + 2 = 9a + 9a
Þ a2 + a – 2 = 0 Þ a = 1, – 2
50.
=
2
(c) Let, 1 = ò x2 .e - x dx
t 2 .et dt
-1 t 2
e (t - 2t + 2)c
=
( -2)
2
\ g ( x) =
51.
(b) I =
54.
-1 4
-5
( x + 2 x 2 + 2) Þ g (-1) =
2
2
(b)
ò
5 -4x
ò x e dx
3
Put 1 +
x =tÞ
Þ 1=
2dt
2
( x - 2)
1
2 x
55.
x 2 (x 4 + 1)3/ 4
(b) ò
=ò
dx = dt
2
x +1
=t
x-2
put
,
( x - 2 )2
=
dt
dx
=-
-1 dt
1 -3
dt
= - ò t lt
=
3 ò t 3/ 4
3
4
3
x 5m -1 + 2 x 4 m -1
(x
2m
m
+ x + 1)
3
x - m -1 + 2 x -2 m -1
(1 + x - m + x -2 m )3
dx = ò
x5 m -1 + 2 x 4 m -1
x
6m
(1 + x - m + x -2 m )3
dx
Put t = 1 + x–m + x–2m
dt
= - mx - m -1 - 2mx -2 m -1
dx
dt
= ( x - m -1 + 2 x -2 m -1 ) dx
Þ
-m
\
\ ò
=
=
2-t
1- x
+c =– 2
+c
=– 2
t
1+ x
dx
( x - 2 )5/ 4
é -3 +1 ù
1/ 4
1 ê t 4 ú -4 é x + 1 ù
+c
ú =
= ê
ê
ú
3 ê -3 ú
3 ë x - 2û
+1
ëê 4
ûú
t 2t - t 2
1
-1
Again put t = Þ dt = 2 dz
z
2
-1
dz
- dz
z2
Þ I=2 ò
= 2ò
1 2 1
2z - 1
z z z2
2
= – 2 2 z -1 + c = – 2 - 1 + c
t
(b) I = ò
3/ 4
dx
I=
53.
( x + 1)
( x - 2)
1 q
1
q
q
ò 48 qe d q = 48 [q e - e ] + C
ò
dx
-3
1 -4 x3
e
(-4 x 3 - 1) + C
48
Then, by comparison
f(x) = –4x3 – 1
dx
(c) I = ò
(1 + x ). x 1 - x
+C
3/ 4
æ x +1 ö
ç
÷
è x-2ø
dq
Þ x2 dx = 12
52.
ò
1/ 4
dx
Put –4x3 = q
Þ –12x2 dx = dq
I=
-1
´ 4(1 + y)1/ 4 = -(1 + x -4 )1/ 4 + C
4
æ x 4 + 1ö
= –ç 4 ÷
è x ø
Put – x2 = t Þ – 2x dx = dt
1= ò
-1
x 3dy
-1
dy
= ò
I= 4 ò 3
3/ 4
4 (1 + y)3/ 4
x (1 + y)
\
dx
=
ò x 3 (1 + x -4 )3/ 4
–4
Let x = y
Þ –4x–3 dx = dy Þ dx =
-1 3
x dy
4
56.
x5 m -1 + 2 x 4 m -1
(x
2m
m
+ x + 1)
3
dx =
1 -3
1
+C
ò t dt =
-m
2mt 2
1
2m(1 + x - m + x -2 m ) 2
x 4m
2m( x 2 m + x m + 1) 2
f ( x) =
(b) I =
ò
+C
+C
x 4m
2 m( x 2 m + x m + 1) 2
x dx
2 - x 2 + 2 - x2
Put t = 2 - x 2 ,
dt
1
=
. (-2 x)
dx 2 2 - x 2
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dx
M-397
Integrals
Þ – t dt = x dx
(-t ) dt
\ I=ò
= - log
57.
=-ò
t2 + t
Þ
1
dt = - log | t + 1|
t +1
Þa=
2 - x2 + 1 + c
x 2 - x + 1 cot -1 x
dx
.e
x2 + 1
Put x = cot t Þ – cosec2 t dt = dx
Now, 1 + cot2 t = cosec2 t
e (cot t - cot t + 1)
I=ò
\
2
t
61.
(1 + cot t )
0
ò e (cot t - cosec
t
2
-1
(a)
=
x6
ò
x+ x
7
(d)
0
6
x(1 + x )
-1
x
p/2
+C
=
ò (1 + (log x)2 )2
dx = ò
dx =
ò
(1 + x ) - 1
x (1 + x 6 )
dx
62.
1 + ( log x ) - 2 log x
2
2 ù2
é1 + ( log x )
ë
û
é 1
2t ù
t
= òe ê
ú dt
2
(1 + t 2 ) 2 ûú
ëê1 + t
60.
+c =
x
1
= f (x) × e x + cù
û
63.
I 2 = ò (1 - x )
0
1
=
dx - ò x x (1 - x50 )100 dx
4244
3
0 14
II
1
50 101
1 (1 - x )
é x
ù
(1 - x50 )101 ú - ò
I 2 = I1 + ê
dx
0
5050
ë 5050
û0
I 2 = I1 + 0 -
I2
5050
1 + esin x
dx =
p
2
f ( x) < 2
f ( x) ³ 2
x<2
x<2
x³2
x³2
3
3
0
2
0
p/3 é1
1
d (tan 4 x )
d (sin 4 3 x) ù
× sin 4 3 x + tan 4 x ×
údx
dx
2
dx
ë
û
òp / 6 ê 2
1 p/3
d (tan 4 x × sin 4 3x )dx
2 òp / 6
1
p /3
×1
é tan 4 x sin 4 3x ù
9× 0 9
-1
=ê
=
=
ú
2
2
2
18
ë
ûp / 6
49
I
1 + esin x
2
(c)
0
50 100
1
æ 1
ö
+
çè
÷ dx
1 + esin x 1 + e - sin x ø
= ò x dx + ò (4 - x ) dx - ò | x - 2 | dx = 1
1
0
dx
3
(c) I 2 = (1 - x50 )101dx = (1 - x50 )(1 - x50 )100 dx
ò
ò
1
0
1
1 + esin x
ì 2 - x, x < 2
(a) f ( x ) = | x - 2 | = í
î x - 2, x ³ 2
+c
1 + (log x) 2
ò
0
ò
e
dx +
sin x
\ ò [ g ( x) - f ( x)] dx
é Which is of the form e x ( f ( x ) + f ' ( x ) ) dx
ò
ë
1+ t2
p/2
1
ì 2 - (2 - x ), 2 - x < 2,
ï(2 - x ) - 2, 2 - x ³ 2,
ï
=í
ï 2 - ( x - 2), x - 2 < 2,
ïî( x - 2) - 2, x - 2 ³ 2,
ì -x 0 < x £ 0
ï x
0< x<2
ï
=í
ï4 - x 2 £ x < 4
ïî x - 4
x³4
dx
é et
2t et ù
\ I= òê
ú dt
2
(1 + t 2 )2 úû
êë1 + t
=
dx
sin x
-p / 2 1 + e
ì2 - f ( x),
g ( x) = f ( f ( x)) = í
î f ( x) - 2,
é
ù
1
2 log x
dx
= òê
2
2 2ú
ëê (1 + (log x) ) (1 + (log x) ) ûú
t
ò
0
6
1
1
dx – ò
dx = ln | x | – p(x) + c
x
x + x7
ò
(log x - 1) 2
59.
x6
dx = ò
ò
=
1
ò
-p / 2 1 + e
t ) dt = et cot t + C
= ecot x ( x) + C º A(x) . ecot
Þ A(x) = x
58.
ò
p/2
= - ò et (cosec 2 t - cot t ) dt
=
(c) I =
=
(- cosec 2t ) dt
2
5050
5051
p/2
ò
(b) Let I =
5051
5050
I 2 = I1 Þ I 2 =
I1
5050
5051
64.
(21)
n
1
0
0
ò {x} dx =nò x × dx =
n
2
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EBD_8344
M-398
Mathematics
n
n
Þ ò [ x ] dx = ò ( x - {x})dx =
0
0
1
n2 n
2 2
= [ x - x 2 ]02 + [ x 2 - x]11 + [ x ]12
2
According to the questions,
=
n n2 - n
,
, 10(n2 - n) are in GP
2
2
68.
(1)
2
ò1 | 2 x - [3x] | dx
2
æ n2 - n ö
n
2
\ç
÷ = 2 ´ 10(n - n)
2
è
ø
2
= ò | 3x - [3 x] - x | dx
Þ n = 21n Þ n = 21.
1
2
65.
1 1
1 1
3
æ 1 1ö
- + (1 - 1) - ç - ÷ + 2 - 1 = + + 1 =
è 4 2ø
2 4
4 4
2
2
p
(c) I = ò | p- | x || dx
-p
2
= ò | {3x} - x | dx = ò ( x - {3x})dx
[Q | p- | x || is even]
1
1
2
2
= ò xdx - ò {3x} dx
p
= 2ò | p- | x || dx
1
0
1
2
p
1/ 3
é x2 ù
= ê ú - 3 ò 3 x dx
ë 2 û1
0
= 2ò (p - x) dx
0
p
é
æ
p2 ö
x2 ù
= 2 ê px - ú = 2 ç p 2 - ÷ = p 2 .
2 û0
2ø
è
ë
1/ 3
=
é x2 ù
(4 - 1)
3 1
-9ê ú = - =1
2
2
2
2
ë û0
1
66.
k 2
x2
dx
=ò
(a)
6 0 (1 - x 2 )3/ 2
1
69.
0
Let x = sin q; dx = cos q d q,
1
2
then
p
6
x2
ò (1 - x2 )3/ 2 dx = ò
0
0
2
(d) ò (a + bx + cx )dx = ax +
sin 2 q cos q
cos3 q
bx 2 cx 3
+
2
3
1
=a+
0
b c
+
2 3
b c
æ1ö
Now, f (1) = a + b + c, f (0) = a and f ç ÷ = a + +
2 4
è2ø
dq
1æ
æ 1 öö
Now, 6 ç f (1) + f (0) + 4 f ç 2 ÷ ÷
è øø
è
p
b c öö
1æ
æ
= ç a + b + c + a + 4ç a + + ÷÷
6è
2 4 øø
è
k 6 sin 2 q
\ =ò
× cos q d q
6 0 cos3 q
Þ
Þ
67.
p
6
k
= tan 2 q d q = ò (sec 2 q - 1) d q
6 ò0
0
k
æ 1 pö 2 3 - x
= (tan q - q)p0 / 6 = ç
=
è 3 6 ÷ø
6
6
Þ k =2 3-p
(1.50)
2
ò0
b c
1
= (6a + 3b + 2c ) = a + +
6
2 3
p
6
1
Hence,
1ì
æ 1 öü
ò f ( x) = 6 íî f (0) + f (1) + 4 f çè 2 ÷ø ýþ
0
2p
70.
(c)
x sin8 x
ò sin8 x + cos8 x dx
0
1
2
| x - 1| - x dx = ò 1 - x - x dx + ò | x - 1 - x dx
0
1
1
0
1/ 2
= ò (1 - 2 x )dx + ò
1
2
(2 x - 1)dx + ò dx
1
pé
x sin8 x
(2p - x )sin8 x ù
= òê 8
+
dx
8
8
8 ú
0 ëê sin x + cos x sin x + cos x ûú
a
a
é 2a
ù
êQ ò f ( x)dx = ò f ( x)dx + ò f (2a - x )dx ú
êë 0
úû
0
0
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M-399
Integrals
p
b
2 p sin8 x
=ò
2 I = 2 ò f ( x)dx
dx
8
8
0 sin x + cos x
= 2p
ù
sin8 x
cos8 x
+
ê
ò ê sin8 x + cos8 x sin 8 x + cos8 x úú dx
û
0 ë
p /2
ò
1dx = 2p ´
0
(b) f ( x ) =
f ¢( x ) =
=
b -1
p /2 é
= 2p
71.
a
\
p
= p2
2
73.
1
3
2 x - 9 x + 12 x + 4
ö
-1 æ
(6 x 2 - 18 x + 12)
ç
÷
2 çè (2 x3 - 9 x 2 - 12 x + 4)3/2 ÷ø
-6( x - 1)( x - 2)
Put e -a = t
Þ 4t2 + 4t – 3 = 0
Þ
74.
1
1
< I2 <
9
8
e-a =
1
Þ a = loge2
2
(a) 2cot 2 q -
5
+4=0
sin q
5
+4=0
sin 2 q sin q
Þ 2cos2q – 5 sinq + 4 sin2q = 0, sinq ¹ 0
Þ 2sin2q – 5 sinq + 2 = 0
Þ (2 sinq – 1) (sinq – 2) = 0
1
x[ f ( x ) + f ( x + 1)]dx
(a + b) ò
...(i)
a
x®a+b–x
b
\ sin q =
1
(a + b - x)[ f (a + b - x) + f (a + b + 1 - x)]dx
( a + b) ò
a
1
2
Þ q=
5p /6
b
1
I=
(a + b - x )[ f ( x + 1) + f ( x)]dx
( a + b) ò
\
...(ii)
ò
5p /6
cos 2 3q d q =
p /6
[Q put x ® x + 1 in f(a + b +
1– x) = f(x)]
=
Add (i) and (ii)
ò
p /6
a
p 5p
,
6 6
1 + cos 6q
dq
2
5p /6
1 é sin 6q ù
1 é 5p p 1
ù
q+
= ê - + (0 - 0) ú
2 êë
6 úû p /6
2ë 6 6 6
û
1 4p p
= . =
2 6 3
b
2 I = ò [ f ( x + 1) + f ( x)]dx
f ( x)
a
b
b
a
a
75.
2 I = ò f ( x + 1)dx + ò f ( x )dx
(a) Given,
ò
4t 3dt = ( x - 2) g ( x)
6
Differentiating both sides,
4(f (x))3. f ‘ (x) = g’ (x)(x – 2) + g(x)
b
= ò f (a + b + 1 - x )dx + ò f ( x )dx
a
Þ (2t + 3) (2t – 1) = 0
2 cos2 q
b
b
ö æ e -2a - 1 ö üï
÷-ç
÷ =5
÷ ç a ÷ý
ø è
ø þï
Þ 4(2 - e -a - e -2a ) = 5
1
1
<I<
3
8
I=
[Q Put x ® x + 1]
2
ìï 0
ü
ax
-ax ï
(a) 4a í ò e dx + ò e dx ý = 5
îï -1
þï
0
ìïæ 1 - e -a
Þ 4a íçç a
îïè
2(2 x3 - 9 x 2 + 12 x + 4)3/2
(c) I =
f ( x + 1)dx
2ü
ì ax 0
e -ax ï
ïe
4
a
+
í
ý =5
Þ
ïî a -1 -a 0 ïþ
2
1
1
f (1) = and f (2) =
3
8
It is increasing function
72.
ò
a -1
a
Putting x = 2,
4(6)3.1
g ( x) = 18
= g (2) Þ xlim
®2
48
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EBD_8344
M-400
Mathematics
p /2
76.
ò
(d)
0
p /2
=
ò
0
=
p /2
cot x dx
cot x + cosec x
cot x dx
=
1 + cos x
[ x ]0p /2 -
1
æ
ö
ò çè1 - 1 + cos x ÷ødx
0
1
2cos2
0
x
2
dx
p 1
= 2-2
Use the property
p /2
p/2
ò
0
I=
x
sec 2 dx
2
ò
0
p
æp ö
= - [1] = ç - 1÷ = mp + mn
2
è2 ø
ò [sin 2 x(1 + cos 3x)]dx
0
0
Q ò f ( x) = ò f (a - x )dx
...(2)
80.
ò [- sin 2x(1 + cos 3x)]dx
...(2)
0
From (1) + (2), we get;
æ 1
ö
ç 1 - sin(2 x ) ÷dx
2
è
ø
0
ò
p /2
1é
1
ù
x + cos 2 x ú
ê
2ë
4
û0
Þ I=
p -1
4
1
2p
\I=
0
ÞI =
...(1)
0
a
a
f ( x ) dx = ò f (a - x )dx
cos3 x dx
sin x + cos x
Þ 2I =
2p
a
a
ò0
...(1)
Adding equation (1) and (2), we get
\ m = , n = – 2, Hence, mn = – 1
(b) I =
sin 3 xdx
sin x + cos x
p /2
p /2
p æ
xö
= - ç tan ÷
2 è
2 ø0
77.
ò
0
p /2
p /2
ò
79.
(b) Let I =
(
1
)
1
ö
-1
-1 æ
2
4
(0) ò x cot 1 - x + x dx = ò x tan ç
4
2÷
è 1+ x - x ø
0
0
(
(
)
)
æ x2 - x2 - 1 ö
1
÷ dx
= ò x tan -1 ç
ç 1 + x2 x 2 - 1 ÷
ç
÷
0
è
ø
2p
2I =
ò (-1)dx Þ 2I = -( x)02p Þ I = – p
=
0
p
3
78.
(c) Let,
p
3
=
4
I = ò sec 3 x.cosec 3 x dx
p
6
1dx
ò
p
6
2
=
4
p
3
ò
p
6
cos 2 x.tan 3 x
sec2 xdx
4
tan 3 x
Let tan x = u
I=
3 -4
u 3 du
ò
ò
p
6
0
-1
Put x = t Þ 2xdx = dt in the first integral
and x2 – 1 = k Þ 2xdx = dk in the second integral.
2
1.dx
2
4
cos 3 x.sin 3 x
=
1
1
0
0
1
1
1tan -1 tdt - ò 1tan -1 kdk
2ò
2
1 1
ö
t
1æ
dt ÷
= ç t tan -1 t - ò
2ç
1 + t 2 ÷ø
0 0
è
0
ö
1
k
1æ
÷
dk
- ç k tan -1 k - ò
2
÷
0
2ç
1
+
k
è
ø
-1
3
=
-1
é 1
ù
1
-1
1 ú
ê -6
6
= -3 ê3 - -1 ú = -3(3 - 3 6 )
ê
ú
36 û
ë
1
=
0
é -1 ù
3 êu 3 ú
ê
ú1
ë
û
1
3
1
p
3
1
1
1
1tan -1 t 2 dt - ò 1tan -1 k dk
2ò
2
7
5
0 öö
1 ö 1æ p æ1
1æp æ1
= ç - ç ln 1 + t 2 ÷ - çç - - ç ln 1 + k 2
÷÷
2 çè 4 è 2
-1 ø ÷ø
0 ø 2è 4 è 2
(
)
(
æp 1
ö æ -p 1
ö p 1
= ç - ln 2 ÷ - ç
- 10 - ln 2 ÷ = - ln 2
è8 4
ø è 8 4
ø 4 2
= 3(3 6 - 3 6 ) = (36 - 36 )
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)
M-401
Integrals
81.
f (0) = 0, g (x) is even Þ f (x) is odd
(d) Using L’ Hospital rule and Leibnit theorem, we get
x +5
f ( x)
ò
2tdt
6
lim
( x - 2)
x ®2
Putting x = 2, 2f (2) f ‘ (2) = 12f ‘ (2)
82.
x +5
éëQ f ( 2 ) = 6 ùû
...(i)
(c) f(x) = f(a – x)
I=
b
=
a
a
ò0
a
ò0
f ( x ) g ( x ) dx
f (a - x ) × g (a - x ) dx
éQ b f ( x ) dx = b f (a + b - x) dx ù
òa
êë òa
úû
p/4
æ 2 + x cos x ö
I = ò log ç 2 - x cos x ÷ dx
è
ø
-p/4
ò0
Þ I=
ò0 4 f ( x) dx - ò0
Þ I=
ò0 4 f ( x) dx - I
...(ii)
ò
-p /4
æ 2 - x cos x 2 + x cos x ö
log ç
.
÷ dx
è 2 + x cos x 2 - x cos x ø
ò
-p /2
85.
Q g is a non- ero even function.
\ g (– x) = g (x),
Given, f (x + 5) = g (x)
From (i) f ‘ (x) = g (x)
ìæ
eï
xö
(d) I = ò1 íçè e ÷ø
ïî
2x
æ xö
Let ç ÷ = t
è eø
...(ii)
...(iii)
æ xö
Þ x ln çè ÷ø = ln t
e
x
æ e ö üï
- ç ÷ ý loge x dx
è xø ï
þ
x
x
Þ x (ln x – 1) = ln t
On differentiating both sides w.r. tx we get
0
lnx.dx =
ò f (t )dt ,
x +5
Put t = l – 5 Þ I =
ò
Q f (x + 5) = g (x)
Þ f (– x + 5) = g (– x) = g (x)
x +5
f (l - 5) d l
dt
t
1
When x = e then t = 1 and when x = 1 then t = .
e
f (l - 5) d l
5
5
a
ò0 4 f ( x) dx
...(i)
0
ò
a
a
(a) f (x) = ò g ( g )dt ,
I=
f ( x ) × g ( x) dx
0
x
Let, I =
a
Þ I = 2ò f ( x) dx
log(1)dx = 0
Þ I = 0 = log 1
83.
f ( x)[4 - g ( x)] dx
a
Þ 2I =
p /2
2I =
a
Þ I=
Adding (i) and (ii)
2I =
g (t )dt (from (iv))
Let, the integral,
Then, equation (i) becomes,
p/4
ò
x+ 5
g(x) + g(a – x) = 4
f ( x )dx = ò f (a + b - x )dx
a
g (l)d l =
5
84.
æ 2 - x cos x ö
Let I = ò log ç 2 + x cos x ÷ dx
è
ø
-p/4
ò
5
ò
ÞI=
p/4
Use the property
- f (5 - l )d l
5
æ 2 - x cos x ö
(d) g ( f (x)) = log ç
÷,x>0
è 2 + x cos x ø
b
ò
\I=
2 f ( x ) f '( x ) - 0
= lim
1
x®2
...(iv)
1 æ 2 1ö dt
1æ
1ö
I = ò1 çè t - t ø÷ × t = ò1 çè t - t 2 ÷ø dt
e
e
1
æ t 2 1ö
æ1 ö æ 1
ö 3
1
+
= çè 2 t ø÷ 1 = èç + 1ø÷ - èç 2 + eø÷ = - e - 2
2
2
2e
2e
e
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EBD_8344
M-402
86.
Mathematics
sin 2 x
éxù 1
êë p úû + 2
(a) Let f (x) =
So, f(–x) =
Þ f(–x) =
sin 2 (- x )
é -x ù 1
êë p úû + 2
For a = 0, b =
For a = - 2 , b = 0
I=
sin 2 x
sin 2 x
=
= - f ( x)
1 éxù
éxù 1
-1 - ê ú +
- -ê ú
2 ëp û
ëpû 2
I=
f ( x ) dx = 0
-2
87.
ò
(b) I =
=
ò
p
4
p
6
p
4
p
6
I=
(
dx
sin 2 x tan 5 x + cot 5 x
16 2
.
15
(
)
2
tan x .sec x
2
sin x æ
2
tan 5 x + 1ö
è
ø
cos x
(
\ I is minimum when (a, b) = - 2, 2
89.
)
ò
1
f (t )dt = x 2 + ò t 2 f (t )dt
x
Þ f(x) = 2x – x f(x)
)
Þ f ¢(x) =
5 tan 4 x . sec2 x dx = dt.
æ 1 ö
p
then t ® ç ÷
è 3ø
6
æ 1ö
2 ç1 - ÷
è 4ø
5
f ¢(1/2) = æ 1 ö
çè1 + ÷ø
4
1 1
dt
5
\ I = 10 òæ 1 ö t 2 +1
ç ÷
è 3ø
p
2
90
(c) I =
1 æp
–1 æ 1 ö ö
= ç – tan çè
÷
10 è 4
9 3 ø ÷ø
2
=
3 16 24
´
=
2 25 25
dx
ò [ x] + [sin x] + 4
-p
2
-1
0
1
4
2
(d) I = ò ( x - 2 x ) dx
a
2
dI
= x4 – 2x2 = 0 (for minimum)
dx
Þ x = 0, ± 2
3 ùb
éx
2x
Also, I = ê 5 - 3 ú
ë
ûa
5
p
2
dx
dx
dx
dx
+
+
+
= -òp -2 - 1 + 4 ò -1 - 1 + 4 ò 0 + 0 + 4 ò 1 + 0 + 4
0
1
-1
b
Þ
2(1 - x2 )
(1 + x 2 )2
Then,
p
When x ®
then t ® 1
4
88.
2x
1 + x2
Þ f(x) =
Let tan4 x = t.
and x ®
)
2
p
(
(a)
0
1 4 tan 4 x .sec2 x
dx.
2
= 2 òp
tan 5 x +1
6
2
-16 2
.
15
x
5
2,b=– 2
For a = – 2 , b =
2
ò
-8 2
.
15
For a =
Þ f(x) is odd function
Hence,
-8 2
15
I=
Q [–x] = –1 – [x]
2
1
1æ p ö
pö 1
æ
= çè -1 + ÷ø + (0 + 1) + (1 - 0) + çè - 1ø÷
2
2
4
5 2
=
3p 9
3
=
(4p - 3)
5 20 20
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M-403
Integrals
p
91.
3
(b) I = ò | cos x | dx
=
2æ
1 ö
1÷
k çè
2ø
=
2 -1
k
0
p/2
=2
ò
cos3 x dx
0
2
=
4
p/2
ò
(3cos x + cos3 x) dx
\ k=2
0
[Q cos 3q = 4cos3 q – 3cos q]
94.
p/2
1é
sin3 x ù
= ê3sin x +
2ë
3 úû0
(a) Q f : R ® R
and |f(x) – f(y)| £ 2 × |x – y|3/2
b
a
a
sin 2 x
ò
-p /2 1 + 2
p /2
ò
Þ 2I = 2 ´
0
1
dx
...(ii)
-p /2
95.
(a) f (x) =
ò
0
1
2k
p /3
ò
0
1
2k
2
= k
ò
1
ò
1
2
sin 2 x dx
x
ò0 t (sin x - sin t ) × dt
x
x
0
0
= sin x ò t × dt - ò t sin t × dt
=
tan q
dq
2 k sec q
Þ f (x) =
sin q
dq
cos q
f ¢(x) =
x2
x
sin x + [t cos t ] 0 + sin x
2
x2
sin x + x cos x + sin x
2
x2
cos x + 2 cos x
2
f ²(x) = x cos x –
x2
sin x – 2 sin x
2
x2
cos x – 2 cos x
2
\ f ¢²(x) + f ¢(x) = cos x – 2x sin x
f ¢²(x) = cos x – 2x sin x –
-2t dt
t
3p
96.
1
ò
p
p
ÞI=
4
4
1
0
Let cos q = t2
\ sinq dq = –2t dt
Hence, integral becomes,
1
2
x /2
0
0
(d) Let, I =
I=
-x
( x ) dx = ò 1 dx = [ x ] = 1
p /3
=
...(i)
dx
sin 2 x dx Þ 2I = 2 ×
Þ |f ¢(x)| = 0
\ f(x) is a constant function.
Given f(0) = 1 Þ f(x) = 1
Hence, the integral
òf
x
Adding (i) and (ii), we get;
2I =
2
sin 2 x
-p /2 1 + 2
f ( x) - f ( y)
£ lim 2 x - y
Þ xlim
x® y
®y
x- y
1
93.
ò
b
p /2
I=
f ( x) - f ( y )
£ 2 x- y
x- y
Þ
p /2
(c) Let, I =
Using, ò f (x)dx = ò f (a + b - x)dx, we get :
1æ
1ö 4
= çè 3 - ÷ø =
2
3
3
92.
1
(Given)
2
= 1–
(a) Let I =
ò p4
4
dt
also let K =
x
dx
1 + sin x
x
1 + sin x
Multiplying numerator and denominator by (1 – sin x), we
get;
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EBD_8344
M-404
Mathematics
x (1 - sin x) x (1 - sin x)
=
1 - (sin x )2
(cos x)2
= x (1 – sin x) sec2 x
= x sec2 x – x sin x sec2 x = x sec2 x – x tan x sec x
p
æ
æ 2 + sin (– x) ö ö
Þ I = ò 2p sin 4 (- x )) ç1 + log ç
÷ ÷ × dx
–
è 2 – sin (– x) ø ø
è
2
K=
Now, I =
ò
3p
4
p
4
x sec2 xdx - ò
3p
4
p
4
b
b
= éQ ò f ( x ). dx = ò f (a + b – x ). dx ù
a
ëê a
ûú
x sec x tan xdx
3p
ù4
3p
ù4
4
4
=
dx
dx
é
é
= ê x tan x - ò tan xdx ú - ê x sec x - ò sec xdx ú
dx
dx
ë
ûp ë
ûp
= [ x tan x - ln | sec x | ]
=
3p
4
p
4
p
2
p
–
2
æ
æ 2 – sin x ö ö
(sin 4 x) ç 1 + log ç
÷ ÷ . dx
è 2 + sin x ø ø
è
p
2
p
–
2
æ
æ 2 + sin
sin 4 x çç1 – log ç
è 2 – sin
è
ò
ò
After adding equation (1) and (2) we get,
- [ x sec x - ln | sec x + tan x | ]
3p
4
p
4
+c
ò
2I = 2
ì é 3p
3p
3p
Þ I = í ê tan
- ln
4
4
4
ë
î
é 3p
3p
3p
3p ù ü
- ê sec
- ln sec
+ tan
ý
4
4
4 úû þ
ë4
ìé p
p
p
- í ê tan - ln
4
4
4
îë
p
2
p
–
2
p
2
0
ò
sin 4 x . dx
I =2
p
2
0
3 1
2´ ´ ´p
3p
2
2
=
sin x . dx =
2 ´2
8
ò
ò
I1 =
ò
I2 =
99.
3p
4
cos x dx;
I=
2
cos x dx and
ò
2
3
\ e- x < e- x and e- x < e- x
Þ e <e
- x2
3
98.
I=
ò
...(i)
<e
dx
ò 1 - cos x
...(ii)
p
4
b
b
a
a
Adding (i) and (ii)
3p
4
2
- x3
2I =
ò
p
4
3p
4
2
sin 2 x
dx ; I =
ò cosec
2
x dx
p
4
2
Þ e - x > e - x > e- x
Þ I3 > I2 > I1
(c) Let
p
2
p
–
2
dx
ò 1 + cos x
Using ò f (x)dx = ò f (a + b - x) dx
For x Î (0, 1)
Þ x > x2 or – x < – x2
and x2 > x3 or – x2 < – x3
-x
(c) I =
p
4
1 - x3
e dx
0
I3 =
3p
4
2
1 - x2
e
0
4
[By Gamma function]
p
= ( 2 + 1)
2
(d) Given:
1 -x
e
0
sin 4 x . dx
2I = 4
ép
p
p
p ùü
- ê sec - ln sec + tan ú ý
4
4
4
4 ûþ
ë
97.
x öö
÷ ÷ . dx .....(2)
x ø ÷ø
æ
æ 2 + sin x ö ö
sin x çç1 + log ç
÷ ÷÷ dx ..... (1)
è 2 – sin x ø ø
è
4
pù
é 3p
I = – (cot x)3pp/ 4/ 4 = – ê cot - cot ú = 2
4
4û
ë
100. (c) In =
ò tan
n
x dx, n > 1
Let I = I4 + I6
= ò (tan 4 x + tan 6 x)dx = ò tan 4 x sec2 x dx
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M-405
Integrals
Let tan x = t
Þ sec2 x dx = dt
\ I = ò t 4 dt =
t5
+C
5
1 5
tan x + C Þ On comparing, we have
5
1
a= ,b=0
5
=
101. (a) Let I = ò
dx
2
((x - 1a) 2 + 3)3 / 2
1
Let; x - 1 = 3 tan q
2
Þ dx = 3 sec . dq
Þ I=ò
0
x
3 sec 2 q dq
p6
æ
çè
(
3 tan q
2 ö 3/2
1 p6
1 p 6 sec2 q
dq = ò cos q dq
= ò0
3
3 0
sec3 q
1 1
1
1
p6
= [sin q]0 = ´ =
3 2
6
3
=
102. (a)
cos 2 x
ò
p æ 1 ö
ç
÷
12 è sin 2 x ø
=
1
4
3
=
p
4
2
ò cos 2 x ´ sin 2 x . sin ( 2 x) dx
p
12
p 4
ò
sin 4 x . (1 - cos 4 x ) dx
p 12
p
ép
ù
4
ê
ú
1
1 4
= ê ò sin 4 x - ò sin 8 x ú
4êp
2p
ú
êë12
úû
12
p 4
1 é cos4 x cos8x ù
+
4
16 úûp
= ê4ë
103. (d)
12
1 é15 ù 15
= ê ú=
4 ë 32 û 128
ò ( x5 + x3 + 1)3 dx
Dividing by x15 in numerator and denominator
ò
x
3
+
5
x6
ò
ty (t) dt +
1
ò
x
ò y(t )dt + x[ y( x) - y(1)]
1
x
= ò ty (t )dt + x[ xy ( x) - y (1)] + xy( x) - y(1)
x
x
1
1
Diff. again w.r. to x
y (x) – y (a) = xy (x) – y (a) + 2x y (x) + x2y1 (x)
(1 – 3x) y (x) = x2y1 (x)
y1 ( x) 1 - 3x
=
y( x)
x2
1dy
1 - 3x
1
=
Þ ln y = –
– 3 ln x
y dx
x2
x
1
ln (y x3) = –
x
yx3 = – e–1/x
-
y=
e
1
x
x3
or y =
ce
- 1x
x3
é x2 ù
ë û
ò é x 2 - 28 x + 196 ù + é x2 ù dx .....(a)
4 ë
û ë û
10
105. (d) I =
dx
æ
1
1ö
çè1 + 2 + 5 ÷ø
x
x
b
3
ty (t) dt
1
Differentiate w.r. to x.
2 x12 + 5 x9
2
y (t) dt = x
1
x
2
ò y(t )dt = ò ty(t )dt + x y( x) -y (1)
k =1
p
4
x
1
1
k
=
Þ k + 5 = 6k
6 k +5
Þ
ò
104. (d) x
) + ( 3 ) ÷ø
2
1
1
Let 1 + 2 + 5 = t
x
x
5 ö
æ –2 5 ö
æ 2
Þ ç
dx = dt Þ ç 3 + 6 ÷ dx = - dt
–
è x 3 x 6 ÷ø
èx
x ø
This gives,
2
5
dx
+
– dt
1
x3 x6
= ò 3 = +C
ò
3
t
2
t2
1
1ö
æ
+
+
1
çè
÷
x 2 x5 ø
1
x10
=
+
C
=
+C
2
2( x5 + x3 + 1) 2
æ 1 1ö
2ç1+ 2 + 5 ÷
è x x ø
Use
òf
a
b
(a + b – x) dx =
ò f (x) dx
a
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EBD_8344
M-406
Mathematics
Þ f (x) = f (x + 4)
Hence period is 4
é( x - 14) 2 ù
ë
û
I= ò
dx
2ù é
2ù
é
(
14)
x
+
x
4 ë û ë
û
10
.....(b)
50
14
ò
Consider
(a) + (b)
f ( x)dx = 10
10
ò
f ( x )dx = 10 [5 + 5] = 100
10
109. (d) Let f : (–1, 1) ® R be a continuous function
10 é ( x - 14) 2 ù + é x 2 ù
ë
û ë û dx
2I = ò é 2 ù é
2ù
4 ë x û + ë ( x - 14) û
sin x
Let
ò
f (t)dt =
0
3
x
2
10
ò dx Þ 2I = 6
2I =
Þ I=3
f(sin x).
4
1
1 p
106. (b) 2ò tan -1 x dx = ò æç - tan -1 (1 - x + x 2 ) ö÷ dx
0
0è 2
ø
1
1p
0
0
2 ò tan -1 x dx = ò
1
ò0 tan
-1
2
Let, I1 =
1
ò0 tan
-1
Þ f (sin x). cosx =
dx - ò tan -1 (1 - x + x 2 ) dx
(1 - x + x 2 ) dx =
1
put x =
0
1
p
- 2 ò tan -1 xdx
0
2
æ 3ö
fç ÷ =
è 2 ø
1/x
log x 2
2 log x
4
2
+ log(36 - 12x + x )
I=ò
2 log x
4
2
+ log(6 - x) 2
2 log(6 - x)
dt = -
1
2
f(x) = ò
+ log x 2
dx .
1
...(i)
2
ln t
ò 1+t dt
1
d
ln
(
+ 1)
d
x
dx
...(ii)
x
é ( ln ) 2 ù
( ln x ) 2
ln x
æ 1ö
ú =
f ( x) + f ç ÷ = ò
d =ê
è xø
2
êë 2 úû
1
1
p
Adding (i) and (ii)
4
Let t =
1
x
log(6 - x) 2
I=ò
dx
2
log x
3
1
110. (c) f æç ö÷ =
èxø
p
ép 1
ù
- 2 ê - log 2ú = log 2
2
4
2
ë
û
2
p
3
æ 3ö 1
f ç ÷. = 3
è 2 ø 2
2
xdx
p 1 x
p 1
dx = - log 2
= - ò0
4
4 2
1 + x2
By equation (a)
4
3
2
pö
p
æ
3
f çè sin ÷ø .cos =
3
3
2
.....(a)
1
1 1
-1
x dx
= éë (tan x ) x ùû - ò0
0
1 + x2
107. (a) I = ò
d
3
(sin x) =
dx
2
2I = ò dx = [ x ]2 = 2
4
2
I=1
108. (c) Let f : R ® R be a function such that f (2 – x) = f (e +
x)
Put x = 2 + x we get
f (–x) = f (4 + x) = f (4 – x)
111. (b) Let I =
ò
1 + 4 sin 2
0
p /3
x
x
- 4sin dx =
2
2
p
p
x
ò 2 sin 2 - 1 dx
0
xö
x ö
æ
æ
= ò ç 1 - 2 sin 2 ÷ dx + ò ç 2 sin 2 - 1÷ dx
è
ø
è
ø
0
p/3
x 1
x p
é
êQ sin 2 = 2 Þ 2 = 6
ë
p x 5p
5p
ù
Þx= , =
Þx=
> pú
3 2 6
3
û
Downloaded from @Freebooksforjeeneet
M-407
Integrals
p /3
xù
é
= ê x + 4cos ú
2 û0
ë
æ
p
3
3 pö
+4
- 4 +ç0 - p + 4
+ ÷
ç
3
2
2 3 ÷ø
è
p
= 4 3-43
=
112. (d) F(x) =
x
ò1
et
dt , x > 0
t
et
ò1 t + a dt
Put t + a = z Þ t = z – a; dt = dz
for t = 1, z = 1 + a
for t = x, z = x + a
x+a
ò1+ a
\I=
-a
= e ò
x+a
1+ a
e
z -a
dz
z
ez
x + a et
dz º e - a ò
dt
1+ a
z
t
= e
é 1+ a et
ù
x + a et
dt + ò
dt ú
ê-ò
1
1
t
t úû
ëê
= e [ - F (1 + a) + F ( x + a)]
= e [ F ( x + a) - F (1 + a)]
113. (a) Let
t
Þ
ò-p
Þ
t
ò-p
t
ò-p
( f ( x ) + x)dx = p2 – t2
f ( x )dx + ò
t
-p
xdx = p2 – t2
æ t 2 p2 ö
f ( x )dx + ç - ÷ = p2 – t2
2ø
è2
3 2 2
(p - t )
ò-p
2
differentiating with respect to t
Þ
t
f ( x )dx =
d é t
3 d 2 2
f ( x) dx ùú =
(p - t )
dt êë ò-p
û
2 dt
dt
d
- f ( -p) (-p ) = – 3t
dt
dt
f (t) = – 3t
f (t ).
æ pö
æ pö
f ç - ÷ = -3 ç - ÷ = p
è 3ø
è 3ø
...(1)
p
ò
[cos( p - x )]dx =
ò
[ - cos x ]dx ...(2)
0
0
On adding (1) and (2), we get
p
p
0
0
ò [cos x]dx + ò [ - cos x]dx
p
2I =
ò
0
2I =
[cos x ] + [ - cos x]dx
p
ò0
(Q [ x ] + [ - x] = -1 if x Ï Z )
- 1dx
p
2I = - x 0 = -p
-p
2
e
ò (log x )
115. (c) Pn =
n
dx
1
put log x = t then x = et and dx = et dt
Also, when x = 1, then t = log 1 = 0
and when x = e, then t = loge e = 1
1
\ Pn =
òt
n t
. e dt
0
1
(By the definition of F(x))
-a
[cos x]dx
p
I=
\ P10 =
-a
ò
0
ÞI=
é
x + a et ù
et
-a 1
dt + ò
dt ú
I = e êò1 a
1
+
t
t úû
ëê
-a
114. (d) Let I =
2I =
x
Let I =
p
p
x
é
ù
+ ê -4 cos - x ú
2
ë
ûp / 3
Now, P10
1
10 t
ò t e dt and P8 =
òt
0
0
8 t
e dt
1
1
0
0
10 t
8 t
– 90P8 = ò tI IIe dt - 90ò t e dt
1
1 9 t
1 8 t
10 t
P10 – 90 P8 = éët e ùû - 10ò t e dt - 90ò t e dt
0
0
0
P10 – 90 P8
1
1
é 1
d 9 t ù
9
8 t
t
ê
ú
10
(
)
90
e
t
e
dt
t
e
dt
=
ò
ò dt ò ú ò t e dt
êë 0
0
0
û
1 8 t ù
1 8 t
é
P10 – 90 P8 = e - 10 ê e - 9ò0 t e dt ú - 90ò0 t e dt
ë
û
1
8 t
8 t
P10 – 90 P8 = e - 10e + 90ò t e dt - 90ò t e dt
0
\ P10 - 90 P8 = -9e
1/ 2
ln(1 + 2 x)
1/ 2
ln(1 + 2 x)
dx or
2
ò 1 + (2 x)2 dx
1
+
4
x
0
0
Put 2x = tanq
116. (c) Let I =
\
ò
2dx
sec 2 qd q
= sec 2 q or dx =
dq
2
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EBD_8344
M-408
Mathematics
also when x = 0 Þ q = 0
x
p
1
and when x = Þ q = 45°or
2
4
p
4
ò
\I=
0
ln(1 + tan q)
2
1 + tan q
´
sec 2 qd q
2
1
ln(1 + tan q)
´ sec 2 qd q
2 ò 1 + tan 2 q
0
(Q 1 + tan2 q = sec2 q)
p
14
I = ò ln(1 + tan q)d q
2
0
±2
\ Equation of tangent is
y – 2 = 2(x – 2) or y + 2 = 2(x + 2)
Þ x-intercept = ± 1.
p /3
p /3
p
é
ù
tan - tan q ú
ê
1
4
I = ò ln ê1 +
ú dq
p
2
ê 1 + tan ´ tan q ú
0
4
ë
û
p
4
1
é 1 - tan q ù
dq
I = ò ln ê1 +
2
ë 1 + tan q úû
0
p
4
é1 + tan q + 1 - tan q ù
1
ln ê
ú dq
2ò
1 + tan q
ë
û
0
p
14
é 2 ù
I = ò ln ê
ú dq
2
ë1 + tan q û
0
p
14
I = ò [ln 2 - ln(1 + tan q)]d q
2
p
4
1
1
ln 2.d q - ò ln(1 + tan q)d q
2ò
2
0
0
p 4
1
-I
I = ln 2q
(from eq. (i))
2
0
1
æp ö
I + I = ln 2 çè - 0÷ø
2
4
1 p
2I = ´ ´ ln 2
2 4
p
p
2I = ln 2 or I = ln 2
8
16
dx
1 + tan x
p /3
dx
ò
=
ò
=
æp
ö
tan ç - x÷
è2
ø
p /6 1 +
(Using the property of definite integral)
I=
ò
p /6
é
1
æp
öù
ln ê1 + tan ç - q÷ ú d q
è
øû
2ò
4
ë
0
0
p
4
t dt = ± 2
0
118. (d) Let, I =
p
4
I=
ò
Points y =
p
4
I=
dy
=2
dx
Þ x=±2
But from y = 2x, \
...(i)
0
ò t dt , x ÎR
dy
therefore
= x
dx
Þ |x|=2
p
4
I=
117. (a) Since, y =
tan x dx
1 + tan x
p /6
…(i)
Also, given
p /3
I=
tan x dx
ò
…(ii)
1 + tan x
p /6
By adding (i) and (ii), we get
p /3
2I=
ò
dx
p /6
1 ép pù p
= ,
2 êë 3 6 úû 12
Statement-1 is false
Þ I=
b
Q
ò
b
f ( x )dx = ò f (a + b - x ) dx
a
a
It is fundamental property.
Statement -2 is true.
119. (a) Consider
sin 2 x
ò
-1
sin ( t ) dt +
0
cos2 x
ò
cos -1 ( t ) dt
0
Let I = f (x) after integrating and putting the limits.
f ¢(x) = sin -1 sin 2 x (2 sin x cos x) - 0
+ cos-1 cos2 x (-2 cos x sin x) - 0
\ f ¢(x) = 0 Þ f (x) = C (constant)
p
Now, we find f (x) at x =
4
1/2
\ I=
ò
0
sin -1 t dt +
1/2
ò cos
-1
t dt
0
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M-409
Integrals
1/2
ò (sin
=
-1
t + cos -1 t ) dt =
1/2
0
0
\ f ( x) =
p /2
ò
ò
Þ I=
sin 2 x
– p/2 1 + 2
p/2
x
2
sin x
-x
-p/2 1 + 2
p /2
ò
Þ I=
-p /2
y
p
p
dt = = C
2
4
1+ t2
dy
1
.
Þ 1=
2 dx
1+ y
y( x)
é
dI(x)
êQ If I( x) = ò f (t ) dt , then
= f {y( x)} .
dx
ê
f
(
)
x
ë
0
p
4
dx
1+ 2
x
{
...(i)
p p
dx , by replacing x by æç - - x ö÷
è2 2
ø
2 x .sin 2 x
dt
ò
122. (a) x =
p
4
\ Required integration =
120. (d) I =
ò
dx
...(ii)
ò
2I =
-p /2
1
Þ I=
4
=
1
4
Þ I=
1
sin x dx =
2
p /2
ò
2
(1 - cos 2 x ) dx
d2y
Þ
dx
2
=
1
2 1+ y
123. (b, c) g ( x + p ) =
-p /2
p/2
x +p
ò
cos 4t dt
x
p+ x
0
x
éæ p sin p ö æ p sin (-p) ö ù
÷ú
êçè 2 + 2 ÷ø - çè - 2 +
øû
2
ë
1 ép pù p
+
=
4 êë 2 2 úû 4
p+ x
ò
ò
0
p
cos 4t dt = ò cos 4t dt = g (x) + g (p) = g (x) – g (p)
x
0
(QFrom graph of cos 4t, g (p) = 0)
0.9
tan 2 x dx
124. (d)
7 p /3
tan x dx = - log cos x 7 p /4
ì 2
æ 2 - xö ü
í[ x ] + log ç
ý dx
è 2 + x ÷ø þ
î
-0.9
ò
0.9
=
ò
0.9
[ x 2 ] dx +
-0.9
7p
7p ù
é
= - ê log cos
- log cos
3
4 úû
ë
7p
7p
- log cos
4
3
é
p öù
æ
7p ù
é
ê cos çè 2p - 4 ÷ø ú
ê cos 4 ú
= log ê
ú
ú = log ê
ê cos æ 2p + p ö ú
ê cos 7p ú
ç
÷
3 û
ë
3 ø ûú
è
ëê
=0+
æ
pö
ç
÷
4 = log ç
÷
p÷
ç
ç
÷
3ø
è
æ 2 ö
= log ç
÷ = log 2.
è 2ø
1
2
1
2
ö
÷
÷
÷
÷
ø
ò
-0.9
0.9
ò
-0.9
= log cos
æ
ç cos
= log ç
çç cos
è
ò
p
cos 4t dt = g ( x ) + ò cos 4t dt
(it is clear from graph of cos 4t)
7 p /3
7 p /4
dy
y
=
. 1+ y2 = y
2
dx
1+ y
= ò cos 4t dt +
7 p /4
ò
. 2y .
0
7 p /3
=
2
sin 2 x ù
é
êë x + 2 úû
-p / 2
121. (d) Let I =
{
dy
= 1 - y2
dx
Adding equations (i) and (ii), we get
p /2
}
æ 2 - xö
dx
log ç
è 2 + x ÷ø
æ 2-x ö
log ç
÷ dx
è 2+ x ø
Put x = – x Þ f (x) = log
2-x
2+ x
(2 - x)
2+ x
= – log
= – f (x)
2-x
2+ x
So, it is an odd function, hence
Required integral = 0.
and f (–x) = log
a
125. (d) Since
ò [ x] = 0 where 0 £ a £ 1
0
0.9
\
}
ù
d
d
y ( x ) - f {f( x )} .
f( x ) ú
dx
dx
û
ò [ x - 2[ x]]dx = 0
0
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EBD_8344
M-410
Mathematics
a
d
e tan x
æ pö
, x Î ç 0, ÷
G ( x) =
è 2ø
dx
x
126. (a) Let
1
1
=8
Let px2 = t Þ 2px dx = dt
=8
p
4
\I= ò
p
0
4
1
p
1
p
, t = and x = , t =
2
4
4
16
p /4
p /4
é 1 - tan q ù
é 2 ù
ò log êë1 + 1 + tan q úû d q = 8 ò log êë1 + tan q úû d q
0
0
p /4
ò
=8
p
æ pö
æ pö
e tan t
dt = G (t ) 4p = G çè 4 ÷ø - G çè 16 ÷ø
t
[log 2 - log(1 + tan q)]d q
0
p14
p14
0
0
= 8log 2
16
16
0
é
æp
öù
log ê1 + tan ç - q÷ ú d q
è4
øû
ë
ò
2 2p x
2
2
2
e tan px .dx
Now, I = ò e tan px × dx = ò
2
1 px
1 x
When x =
a
f ( x) dx = ò f (a - x ) dx
0
p /4
2
4
ò
Applying
ò 1d q - 8 ò
I = 8 × (log 2)[ x]0p /4 - 8
x2
127. (d) Let ò t f ( t ) dt = sin x - x cos x 2
e
By using Leibnit rule, we get
x
p
Now, put x = , we get
6
2
2
1.5
1
129. (d) I = 8 log(1 + x ) dx
ò 1+ x2
0
Put x = tan q,
\ dx = sec 2 qd q
p /4
\ I = 8 log(1 + tan q ) .sec 2 q dq
ò 1 + tan 2 q
0
p /4
ò
0
log(1 + tan q )dq
130. (a) p '( x ) = p '(1 - x )
...(i)
...(i)
0
1
1.5
é 2ù
1.5
= ò x.0 dx + ò xdx + ò 2 xdx = 0 + ê x ú + é x 2 ù
ë
û
2
0
1
2
ë 2 û1
1
1
= ( 2 - 1) + ( 2.25 - 2) = + 0.25
2
2
1 1 3
= + =
2 4 4
I =8
[From equation (i)]
1
2
2
p
I = 8 × × log 2 - I
4
Þ 2I = 2p log 2
\ I = p log 2
Let I = ò p ( x)dx
2
2
2
2
ò x éë x ùû dx = ò x[ x ]dx + ò x éë x ùû dx + ò x éë x ùû dx
0
0
1
1
log(1 + tan q) d q
Þ p ( x ) + p (1 - x) = 42
1
1
p
æ pö
Þ f ç ÷ = sin - 1 = - 1 = –
è 6ø
6
2
2
1
ò
Þ p( x) = - p(1 - x ) + c
at x = 0
p(0) = – p(1) + c Þ 42 = c
Now, p( x ) = - p(1 - x) + 42
p æ pö p
p p
. f ç ÷ = .sin 6 è 6ø 6
6 6
1.5
p /4
0
ù d é
d éx
x2 ù
êsin x - x cos x - ú
ê ò t f ( t ) dt ú =
dx ë e
2 úû
û dx ëê
Þ x f(x) – e f(e). 0 = x sin x – x
128. (c)
log(1 + tan q) d q
Þ I = ò p(1 - x)dx
...(ii)
0
Adding eqn. (i) and (ii),
1
2 I = ò (42) dx Þ I = 21
0
131. (c) Let I =
=
p
ò0 [cot x] dx
p
1
....(i)
p
ò0 [ cot (p - x) ] dx = ò0 [- cot x] dx
....(ii)
Adding eqn s (i) & (ii),
We get
2I =
p
ò0 ([ cot x] + [ - cot x]) dx
p
= ò ( -1)dx
0
[ Q[ x] + [- x] = -1, if x Ïz and [x] + [–x] = 0, if x Î z]
= [ - x ]0p = -p
Þ I=–
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p
2
M-411
Integrals
132. (b) We know that
Þ
sin x
1
Þ
ò
x
1
x
0
Þ
1
ò
Þ
ò
dx <
2
2
ÞI<
3
3
cos x
x
ò
=ò
p
=
2
2
t t -1 2
p
=
2
Þ sec–1 x – sec–1 2 =
p
2
Þ sec–1x –
p
p
p p
Þ sec–1x = +
=
2
4
2 4
Þ sec–1x =
pö
3p
æ
3p
= sec ç p - ÷
Þ x = sec
è
4
4ø
4
Þ x = - sec
p
Þx=–
4
2
æ 1ö
134. (c) Given that F (x) = f (x) + f ç ÷ ,where
è xø
f (x) =
e log t
t
dt
1+ t
1/ e log t
t
...(1)
dt + ò
dt
1 1+ t
1+ t
1/ e log t
Let I = ò
dt
1 1+ t
dz
1
1
\ Put = z Þ - dt = dz Þ dt = –
2
t
t
z2
when t = 1 Þ z = 1 and when t = 1/e
Þ z=e
\ I=
e
ò1
log t
e (log t )(t + 1)
+ log t
dt = ò
dt
1
t (1 + t )
t (1 + t )
e t.log t
e log t
ò1
Þ F(e) =
t
dt
1
dt = dx
t
[when t = 1, x = 0 and when t = e, x = log e = 1]
\
Let log t = x
1
é x2 ù
F(e) = ê ú
ëê 2 ûú 0
\ F(e) = 1 x dx
ò
0
1
2
135. (b) Let a = k + h where k is an integer such that and 0 £ h < 1
Þ F(e) =
Þ [a] = k
a
2
3
1
1
2
\ F(e) = f (e) + f æç 1 ö÷
è eø
Þ F(e) =
e
\ ò [ x] f '( x)dx = ò1 f '( x)dx + ò 2 f '( x)dx +
x log
ò1
b
ò1 1 + t dt + ò1 t (1 + t ) dt
1
é
ù
dx
= sec -1 x ú
êQ ò
2
x x -1
ëê
ûú
b
òa f (t )dt = òa f ( x)dx ]
Now from eqn. (1)
F(e) =
dt
2
[Q log1 = 0]
log z
dz
z ( z + 1)
1
dx < 2 Þ J < 2
\ ésec t ù
ë
û
e
[By property
1
-1
=ò
log z æ dz ö
ç- ÷
( z + 1) è z ø
for xÎ(0, 1)
x
x
1
æ dz ö
çè - 2 ÷ø
z
e log t
dt
\ I= ò
1 t (t + 1)
dx < ò x –1/ 2 dx = éë 2 x ùû = 2
0
x
0
x
z +1
=ò -
1
1
<
cos x
0
133. (d)
dx < ò
x
0
1
1
é 2 x3 / 2 ù
xdx = ê
ú
êë 3 úû 0
1
cos x
Also
e (log1 - log z ).z
1
x on x Î (0, 1)
0
sin x
ò
=ò
e
sin x
0
Þ
<
x
sin x
< 1 , for x Î (0, 1)
x
e log
ò1
æ 1ö
log ç ÷
è z ø æ dz ö
1 çè z 2 ÷ø
1+
z
k +h
k
...
ò
( k - 1) f '( x)dx +
k -1
ò
kf '( x )dx
k
= {f (2) – f (1)} + 2{f (3) – f (2)} + 3{f (4) – f (3)}
+ ........ + (k – 1) {f (k) – f (k – 1)} + k{f (k + h) – f (k)}
= – f (1) – f (2) – f (3) ......... – f (k) + kf (k + h)
= [a ] f (a) - { f (1) + f (2) + f (3) + ¼ f ([a])}
-
136. (c) I =
p
2
ò
[( x + p)3 + cos 2 ( x + 3p)]dx
3p
2
Put x + p = t
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EBD_8344
M-412
Mathematics
p
2
ò [t
I=
-
3
p
2
p
2
p
2
0
0
é
ù
æp
ö
= 2 ´ 2 ò cos2 x dx = 4 ò sin 2 x dx êQ f çè - x ÷ø = f ( x) ú
2
ë
û
+ cos2 t ]dt = 2 ò cos 2 tdt
0
[Q t3 is odd and cos2 t is even function]
p
2
p
2
p
= ò (1 + cos 2t )dt = + 0
2
p
2
2
Þ I = 2 ò sin x dx = 2
0
0
p
p
0
0
137. (d) I = ò xf (sin x )dx = ò (p - x) f (sin x )dx
p
p
= p ò f (sin x ) dx - I Þ 2 I = p ò f (sin x )dx
0
0
p
p
I = ò f (sin x )dx = p
20
p /2
ò
f (sin x )dx
9- x + x
3
6
9-x+ x
3
dx
1
b
f ( x)dx = ò f (a + b - x )dx ]
ò
a
6
3
2I = ò dx = [ x]63 = 3 Þ I =
2
3
p
=
=
cos 2 x
-p 1 + a
ò
x
dx
....(1)
-p
p
2
< 2 , 0 < x <1
1
2
b
b
é
ù
ê Using ò f ( x ) dx = ò f ( a + b - x ) dx ú
êë
úû
a
a
xö
p
æ 1+ a
2
xç
÷ dx = ò cos x dx
è 1+ a x ø
-p
= 2 ò cos 2 x dx [Q f (p – x) = f(x)]
3
x
x
ò 2 dx > ò 2 dx Þ I1 > I 2
0
x
3
ò 4t dt
4t 3
dt = lim 0
141. (d) lim ò
x-2
x®2
x-2
x® 0
0
Applying L Hospital rule
f ( x)
= 4 ´ 63 ´
1
= 18
48
142. (d) f ( x) =
dx
...(2)
1+ ax
Adding equations (1) and (2) we get
ò cos
0
x2
f ( x)
ex
1+ e
x
Þ f ( - x) =
e- x
1+ e
-x
=
1
x
e +1
\ f ( x ) + f (- x ) = 1 " x Î R
f (a )
Now I1 =
ò
xg{x(1 - x )}dx
f ( - a)
-p
2I =
x3
[4 f ( x)3 f '( x)]
= 4( f (2))3 f '(2)
1
x ®2
a x cos 2 x
p
3
lim
cos 2 ( - x )
dx
ò
-x
-p 1 + a
p
1
2
and 2 > 2 , x > 1
Þ I4 > I3
… (2)
b
3
0
0
a
ò
1
2
I 3 = ò 2 x dx, I 4 = ò 2 x dx
Þ
Adding equation (1) and (2)
p
0
1
1
… (1)
dx
[Q
139. (b) Let I =
0
x3
9- x
I =ò
p
2
0
x
138. (b) I = ò
p
2
140. (b) I1 = ò 2 x dx, I 2 = ò 2 x dx,
Q2
6
)
x dx
p
æ pö
Þ I + I = 2ç ÷ = p Þ I =
è 2ø
2
= p ò f (cos x ) dx
0
2
0
0
p /2
ò (1 - cos
2
Þ I = 2 ò dx - 2 ò cos x dx
0
[Q sin(p - x) = sin x]
p
2
f (a )
=
ò
(1 - x) g{x (1 - x )}dx
f ( - a)
b
b
é
ù
ê using ò f ( x ) dx a = ò f ( a + b - x ) dx ú
êë
úû
a
a
0
Downloaded from @Freebooksforjeeneet
M-413
Integrals
f ( a)
ò
Þ
-1
f (a)
ò
g{x(1 - x)}dx -
f ( -a )
f ( - a)
= I 2 - I1 Þ 2 I1 = I 2
143. (b) Let I = ò xf (sin x )dx
...(i)
0
=
We know that
ò
a
p
0
0
f ( x)dx = ò f (a - x )dx = ò (p - x) f (sin x)dx ...(ii)
0
p
p
2
2
0
p
2
(sin x + cos x )
I=
0
0
é x n +1 x n + 2 ù
=ê
ú
êë n + 1 n + 2 úû 0
=
1
1
n +1 n + 2
f ¢( x )
=1
f ( x)
f (0) = 1 Þ f ( x ) = e x
dx
p
2
0
p
3
-2
ò|x
2
\ Integral is
ò (x
-2
0
0
]
[ ]
é e2 1 ù
2
= e - ê - ú - 2[e - e + 1] = e - e - 3
2
2
úû
2 2
ëê
b
b
a
a
b
b
Q ò f ( x)dx = ò f (a + b - x )dx
ì x 2 - 1 if
ïï
2
2
Now | x - 1|= í1 - x if
ï 2
ïî x - 1 if
x £ -1
a
b
= (a + b )ò f (a + b - x)dx - ò xf (a + b - x )dx
x ³1
3
- 1)dx + ò (1 - x )dx +ò ( x - 1)dx
2
-1
a
b
-1 £ x £ 1
1
2
1
We know that
- 1 | dx
-2
-1
1
148. (c) I = ò xf ( x)dx = ò ( a + b - x ) f ( a + b - x) dx
or I = [ - cos x + sin x ]02 = 2
| dx =
0
[ ] [
ò
2
0
1
1 1
1
= x 2e x 0 - 2 xe x - e x 0 - e 2 x 0
2
(sin x + cos x )2
dx = (sin x + cosx)dx
(sin x + cos x )
ò |1- x
1
= ò x 2 e x dx - ò e 2 x dx
2
3
1
\ ò f ( x ) g ( x) dx = ò e x ( x 2 - e x ) dx
pù
é
êQ sin x + cos x > 0 if 0 < x < 2 ú
ë
û
145. (d)
0
log f ( x ) = x + c Þ f ( x) = e x + c
2
(sin x + cos x )
ò
1
2
1 + sin 2 x
(sin x + cos x)
p
2
1
\ g(x) = x2 – f (x) = x2 – ex
0
0
0
Integrating both side we get
0
Let logx = t Þ et = x
1
Þ dx = dt Þ dx = xdt Þ et dt .
x
ò
0
147. (d) Given that f ¢ ( x ) = f ( x) Þ
\ I = p ò f (sin x )dx Þ A = p
p /2
1
1
[Q sin(p - x) = sin x]
ò
1
146. (d) I = ò x(1 - x )n dx = ò (1 - x )(1 - 1 + x)n dx
= ò (1 - x) x n dx = ò ( x n - x n +1 )dx
\ 2 I = pò f (sin x)dx = p.2 ò f (sin x)dx
144. (c) I =
æ 27 ö æ 1 ö
+ ç - 3 ÷ - ç - 1÷
è 3
ø è3 ø
2 2 4
2 28
+ + +6+ =
3 3 3
3
3
Adding (i) and (ii)
p
2
3
2ö
æ 1 ö æ 8
ö æ
= ç - + 1÷ - ç - + 2 ÷ + ç 2 - ÷
3ø
è 3 ø è 3 ø è
p
a
1
é x3
ù
é
é x3
ù
x3 ù
= ê 3 - xú + ê x - 3 ú + ê 3 - xú
êë
úû -2 êë
úû -1 êë
úû1
xg ( x(1 - x)dx
2
1
a
b
b
a
a
a
= (a + b)ò f ( x ) dx - ò xf ( x )dx
[Q Given that f(a + b – x) = f(x)]
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EBD_8344
M-414
Mathematics
b
when x = 0, t = 1 and when x = p, t = –1
a
\ I = -2 p
2I = (a + b)ò f ( x)dx
-1
b
Þ I=
( a + b)
f ( x )dx
2 ò
1
1
1
ò 1 + t 2 dt = 2p ò 1 + t 2 dt
-1
1
1
-1
-1
-1
= 2p éë tan t ùû = 2p éë tan 1 - tan ( -1) ùû
-1
a
2
d x
2
ò sec tdt
dx 0
sec 2 x 2 .2 x
= lim
149. (d) lim
x ®0 d
x ® 0 sin x + x cos x
( x sin x)
dx
(by L’ Hospital rule)
2 sec2 x 2
2 ´1
lim
=
=1
x ® 0 æ sin x
ö 1+1
cos
x
+
çè
÷ø
x
é p æ -p ö ù
p
= 2p ê 4 - ç 4 ÷ ú = 2p. 2 = p2
è
øû
ë
152. (d) We know that [x] is greatest integer function less than
equal to x
2
ò éë x
\
1
2ù
2
dx = ò é x 2 ù dx +
û
ë û
0
ò
0
é x 2 ù dx +
ë û
1
3
ò
t
2
150. (c) F (t ) = ò f (t - y ) g ( y )dy
0
t
1
t
t
= et é - ye - y - e - y ù = - et é ye - y + e - y ù
ë
û0
ë
û0
é t + 1 - et ù
= - e ét e -t + e - t - 0 - 1ù = - et ê
ú
t
ë
û
êë e
úû
t
= et - (1 + t )
p
ò-p
p
= ò-p
2 x (1 + sin x)
2
1 + cos x
1 + cos2 x
p
+ 2ò
p
x sin x dx
a
x) sin (p - x )
ò
ò0
tan n x(1 + tan 2 x )dx
1 + cos2 (p - x )
dx
sin x
put cos x = t Þ - sin x dx = dt
p/4
2
é
x n +1 ù
n
êQ ò x dx =
ú
n + 1û
ë
1- 0
1
=
n +1 n + 1
1
n [In+ In+2]
Þ nlim
®¥
n +1
n
n = lim
1
=1
= lim n.
= lim
1ö
æ
n
®¥
n
n
+
1
+
1
n ®¥
n®¥
n ç1 + ÷
è nø
\ In + In+2 =
10 p
dx
1 + cos 2 x
p sin x dx
x sin x dx
Þ I = 4pò
- 4ò
0 1 + cos 2 x
1 + cos 2 x
p sin x
Þ 2I = 4p ò
dx
0 1 + cos 2 x
I=4
ò
é tan n +1 x ù
tan x sec x dx = ê
ú
êë n + 1 úû 0
n
0
p (p - x )
3
p/4
=
= 2 ò f ( x)dx , if f (x) is even
I = 4 ò0
+ [ 3x]
x
f (x)dx = 0 , if f (x) is odd.
p (p -
3
= 5- 3 - 2
0
-a
3
2
dx
;
é x 2 ù dx
ë û
= 2 -1 + 2 3 - 2 2 + 6 - 3 3
=
x sin x
ò
ò 3dx
0
a
ò
= [ x ]1 + [ 2 x ]
3
2
2
2
2dx +
2
2
p/4
dx
-p 1 + cos 2
1 + cos2 x
We know that
Q
1
153. (b) I n + I n + 2 =
2 x dx
= 0+4 ò0
3
ò 1dx + ò
0
0
151. (b)
2
= ò 0dx +
t
= ò et - y ydy = et ò e - y ydy
0
éx2 ù +
ë û
ò
154. (a) I =
p
| sin x | dx = 10ò | sin x | dx
0
0
[Q sin(10p - x) = sin x]
p
= 10ò sin x dx
0
Q sin x > 0, for 0 < x < p.
as sin (p – x) = sinx
p /2
I = 20
ò
p /2
sin x dx = 20 [ - cos x ]0
0
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= 20
M-415
Integrals
1
. n a +1 + a1n a + a 2 n a -1 + .......
1
(a
+ 1)
158. (a) lim
=
n ®¥
1ö
60
æ
1+
÷
a -1
2ç
n
( n + 1) . n çè a +
÷
2 ø
x
155. (a) F ( x) = ò t 2 g (t )dt
1
Differentiate by using Leibnit ’s rule, we get
x
F ¢(x) = x2g(x) = x 2 ò f (u )du
...(i)
1
At x = 1,
1
F ¢(1) = 1ò f (u )du = 0
a
a
a
æ 1ö
æ 2ö
æ nö
1
çè ÷ø + çè ÷ø + ....... + çè ÷ø
n
n
n
Þ lim
=
60
n ®¥
é
n ( n + 1) ù
( n + 1)a -1 ê n 2 a +
ú
2 û
ë
1
Now, differentiate eqn (i)
x
=
F ¢¢( x ) = x 2 f ( x ) - 2 x ò f (u)du
1
At x =1,
1
156. (a)
lim
1
n(n) 3
1
n
å
n ®¥
= lim
(n + r) 3
1
Þ
Þ
Þ
Þ
(a + 1) (2a + 1) = 120
2a2 + 3a – 119 = 0
2a2 + 17a – 14a – 119 = 0
(a – 7) (2a + 17) = 0
=
1
60
17
2
æ (n + 1)(n + 2)...3n ö n
159. (d) y = lim ç
÷
n ®¥ è
n 2n
ø
1
ln y = lim
n ®¥ n
1
4ù
4
é
ê 3 (1 + x) 3 ú = 3 (2) 3 - 3
= ê4
ú
4
4
ë
û0
2n
ln y = lim
æ 1 öæ 2 ö æ 2n ö
ln ç1 + ÷ç1 + ÷ ....ç1 + ÷
n ø
è n øè n ø è
1é æ 1ö
æ 2ö
æ 2n ö ù
êln çè1 + n ÷ø + ln çè1 + n ÷ø + .... + ln çè1 + n ÷ø ú
û
n®¥ n ë
2
dx
n
å n 2 + r 2 = ò 1 + x2
157. (d) Let L = nlim
®฀
r =1
0
1
é r
ù
êëQ n ® x, r ® dx úû
= tan–12
é 1 æ 1ö ù
êa + 2 çè1 + n ÷ø ú
ë
û
1
a +1 = 1
1ö
60
æ
ça + ÷
2ø
è
Þ a = 7, -
1
dù
é r
êëQ n ® x and n ® x úû
0
2
r =1
1
3
= ò (1 + x) dx
-1
= éë tan x ùû
0
a
dx
1
r =1
n.n 3
1
æ rö
Þ
1
(n + 1) 3 + (n + 2) 3 + ... + (n + n) 3
n ®¥
a -1
n
å çè n ÷ø
1
1
1
a
= +1 =
=
1 ö 60 a + 1 60
æ
ça + ÷
2
2ø
è
òx
= 0
= f (1) – 2 × 0 = f(1)
F ²(1) = 3
Then, for F ¢(1) = 0, F ²(1) = 3 > 0
Hence, x = 1 is a point of local minima.
1
æ 1ö
çè1 + ÷ø
n
1 a
1
F ¢¢(1) = 1. f (1) - 2 ´1.ò f (u)du
1
lim 1
n ®¥
n
rö
1 2n æ
å ln ç 1 + n ÷ø = ò02 ln(1 + x)dx
n ®¥ n r =1 è
= lim
Let 1 + x = t Þ dx = dt
when x = 0, t = 1
x = 2, t = 3
æ 33 ö
3
3
æ 27 ö
ln y = ò ln t d t = [t ln t – t ]1 = ln ç 2 ÷ = ln ç 2 ÷
1
èe ø
èe ø
27
Þ y= 2
e
Downloaded from @Freebooksforjeeneet
EBD_8344
M-416
Mathematics
160. (a) Let f (x) =
f(x) =
ò cosec
6
2
1 r
2 r
lim
.
sec
=
n®¥ n n
n®¥ n
n2
n2
Þ Given limit is equal to value of integral
dx
ò sin6 x
r
sec2
2
lim
x dx
r2
1
From reduction formula, we have
In =
ò
cosec x dx = - cosec x cot x + n - 2 I n -2
n -1
n -1
n
cosec 4 x cot x 4 é –cosec2 x cot x 2 ù
+ ê
+ I2 ú
\ f(x) = 5
5 ëê
3
3 ûú
= =
8
cosec4 x cot x 4
- cosec2 x.cot x + [- cot x ]
15
5
15
-(1 + cot 2 x )2 .cot x 4
- (1 + cot 2 x)cot x
5
15
-1 é
4
1 + cot 4 x + 2cot 2 x ù cot x - écot x + cot3 x ù
û
û
5 ë
15 ë
-
x dx
0
1
1
1
1
2 x sec x 2 dx = ò sec 2 tdt
ò
2
2
or
0
[put x 2 = t ]
0
1
1
1
= (tan t )0 = tan1 .
2
2
n
r
1
162. (b) n®¥ å e n [Using definite integrals as limit of sum]
n
r =1
Lim
= ò e x dx = e - 1
0
163. (a) lim
14 + 24 + 34 + .......n4
n5
n®¥
8
cot x
15
-1
[cot x + cot 5 x + 2cot 3 x]
=
5
-4
4
8
cot x - cot 3 x - cot x
15
15
15
5
=
2 2
1
8
- (- cot x) (Q cosec2 x = 1 + cot 2 x)
15
=
ò x sec
n-2
-15
cot x 10 3
cot x - cot x
15
5
15
- cot 5 x 2 3
- cot x - cot x
=
5
3
It is a polynomial of degree 5 in cot x.
2
é1
ù
2 1
2 4
ê 2 sec 2 + 2 sec 2
ú
n
n
n
n
ú is equal to
161. (d) lim ê
3
1 2 ú
n ®¥ ê
2 9
sec
....
sec
1
+
+
+
ê
ú
n
n2
n2
ë
û
-
lim
13 + 23 + 33 + ....... + n3
n5
n®¥
4
3
1 n æ rö
1
1 æ rö
S ç ÷ - lim . lim ç ÷
n®¥ n r =1 è n ø
n®¥ n n® ¥ n è n ø
= lim
1
4
= ò x dx - lim
n ®¥
0
1
1
´ x3 dx =
n ò
0
164. (a) We have lim
n® ¥
1
é x5 ù
1
ê ú -0=
5
ëê 5 ûú 0
1p + 2 p + .... + n p
n p+1
1
é x p +1 ù
lim å p = ò x dx = ê
ú
n ®¥ r =1 n × n
ëê p + 1 ûú
n
rp
1
p
0
;
=
0
Downloaded from @Freebooksforjeeneet
1
p +1
23
Applications of
Integrals
4.
TOPIC Ć
1.
The
Curve & X-axis Between two
Ordinates, Area of the Region
Bounded by a Curve & Y-axis
Between two Abscissa
area
(in
sq.
units)
of
8
1
23
2
(b)
the
region
5.
4
2 +1
3
(a)
6.
(b) 2
(c)
4
3
(d)
1
3
Let g(x) = cos x 2 ,f (x) = x , and a, b (a < b) be the
roots of the quadratic equation 18x 2 - 9 px + p 2 = 0 . Then
the area (in sq. units) bounded by the curve y = (gof)(x)
1
{( x, y) : 0 £ y £ x + 1, 0 £ y £ x + 1, £ x £ 2} is :
2
[Sep. 03, 2020 (I)]
(a)
and the lines x = a, x = b and y = 0 , is :
2
(a)
3.
2
3
4
1
8
2 -1
2(d)
3
2
3
The area (in sq. units) of the region
(c)
2.
53
59
26
(b) 8
(c)
(d)
6
6
3
The area of the region
A = {(x, y): 0 £ y £ x |x| + 1 and – 1 £ x £ 1} in sq. units is:
[Jan. 09, 2019 (II)]
(a)
A= {( x, y) : ( x - 1)[ x] £ y £ 2 x , 0 £ x £ 2}, where [t]
denotes the greatest integer function, is :
[Sep. 05, 2020 (II)]
(a)
The area (in sq. units) of the region
A = {(x, y) ÎR × R|0 d” x d”3, 0 d” y d” 4, y d” x2 + 3x} is :
[April 8, 2019 (I)]
23
16
(b)
79
24
(c)
79
16
(d)
23
6
3
1
3
+
(b)
3 4
ò
x f ( x) dx, then :
[Online April 25, 2013]
-2
2
(a)
1
( 3 - 2)
2
1
1
( 2 - 1)
( 3 - 1)
(d)
2
2
Let f : [ – 2, 3] ® [0, ¥ ) be a continuous function such
that f (1– x) = f (x) for all x Î [-2, 3] .
If R1 is the numerical value of the area of the region
bounded by y = f (x), x = –2, x = 3 and the axis of x and
R2 =
1ö
æ
and g(x) = ç x - ÷ , x Î R. Then the area (in sq. units) of
2ø
è
the region bounded by the curves, y = f(x) and y = g(x)
between the lines, 2x = 1 and 2x =
(b)
(c)
7.
1
ì
, 0£x<
ï x
2
ï
1
ï 1
,
x=
Given: f(x) = í
2
ï 2
1
ï
ï1 - x , 2 < x £ 1
î
1
( 3 + 1)
2
[2018]
3, is :
[Jan. 9, 2020 (II)]
3 1
1
3
(c)
4 3
2 4
(d)
1
3
+
2 4
8.
(a) 3R1 = 2R2
(b) 2R1 = 3R2
(c) R1 = R2
(d) R1 = 2R2
Let f (x) be a non – negative continuous function such that
the area bounded by the curve y = f (x), x - axis and the
ordinates x =
p
p
and x = b >
is
4
4
p
æ
ö
ç b sin b + cos b + 2b ÷ . Then f
4
è
ø
Downloaded from @Freebooksforjeeneet
æ pö
ç ÷ is
è 2ø
[2005]
EBD_8344
M-418
9.
10.
Mathematics
ö
æp
(a) ç + 2 - 1÷
ø
è4
ö
æp
(b) ç - 2 + 1÷
4
ø
è
ö
æ p
(c) ç1 - - 2 ÷
4
ø
è
æ p
ö
(d) ç1 - + 2 ÷
4
è
ø
17.
18.
x + y £ 1, x ³ 0, y ³ 0} is a
:
ò xf ¢( x)dx is
[2002]
19.
0
(b) 1
(c) 5/4
(a) 2 6
Different Cases of Area Bounded
11.
12.
13.
(c) 24
(d) 4 3
The region represented by x - y £ 2 and x + y £ 2 is
bounded by a :
[April 10, 2019 (I)]
(a)
(a)
(d) rhombus of area 8 2 sq. units
The area (in sq. units) of the region bounded by the curves
y = 2x and y = |x + 1|, in the first quadrant is :
[April 10, 2019 (II)]
1
7
1
5
(b)
(c)
(d)
3
6
6
6
The area (in sq. units) of the region enclosed by the curves
y = x2 – 1 and y = 1 – x2 is equal to: [Sep. 06, 2020 (II)]
4
3
(b)
8
3
(c)
7
2
(d)
21.
16
3
(a) loge 2 +
Consider a region R = {( x, y)ÎR 2 : x 2 £ y £ 2 x}. If a line
y = a divides the area of region R into two equal parts,
then which of the following is true? [Sep. 02, 2020 (II)]
(b) 3a 2 - 8a 3/ 2 + 8 = 0
(c) 3a - 8a + 8 = 0
(d) a - 6a - 16 = 0
The area of the region, enclosed by the circle x2 + y2 = 2
which is not common to the region bounded by the
parabola y2 = x and the straight line y = x, is:
[Jan. 7, 2020 (I)]
(a) (24p – 1)
(b) (6p – 1)
(c) (12p – 1)
(d) (12p – 1)/6
The area (in sq. units) of the region
{(x, y) ÎR2|4x2 £ y £ 8x + 12} is:
[Jan. 7, 2020 (II)]
3
128
125
124
127
(b)
(c)
(d)
3
3
3
3
For a > 0, let the curves C1: y2 = ax and C2: x2= ay intersect
at origin O and a point P. Let the line x = b (0 < b < a)
intersect the chord OP and the x-axis at points Q and R,
respectively. If the line x = b bisects the area bounded by
1
, then
2
[Jan. 8, 2020 (I)]
the curves, C1 and C2, and the area of DOQR =
‘a’ satisfies the equation:
3
2
(b)
3
2
3
1
1
(d) 2 - log 2
2
e
The area (in sq. units) of the region
A = {(x, y) : x2 £ y £ x + 2} is:
[April 9, 2019 (I)]
(c)
22.
3/ 2
(a)
16.
(b) 48
1
, then l is equal to :
9
[April 12, 2019 (II)]
(a) square of side length 2 2 units
(b) rhombus of side length 2 units
(c) square of area 16 sq. units
2
15.
20.
The area (in sq. units) of the region
A = {(x, y) : |x| + |y| £ 1, 2y2 ³ |x|} is : [Sep. 06, 2020 (I)]
(a) a3 - 6a 2 + 16 = 0
14.
10
8
2
(b) 6
(c)
(d) 3
3
3
If the area (in sq. units) bounded by the parabola y2 = 4lx
and the line y = lx, l > 0, is
(d) –3/4
TOPIC n Between the Curves
2 + b, then a – b is equal to
[April 12, 2019 (I)]
(a)
2
(a) 3/2
32
34
29
31
(b)
(c)
(d)
3
3
3
3
If the area (in sq. units) of the region {(x, y) : y2 £ 4x,
(a)
The area enclosed between the curve y = log e ( x + e) and
the coordinate axes is
[2005]
(a) 1
(b) 2
(c) 3
(d) 4
If y = f(x) makes +ve intercept of 2 and 0 unit on x and y
axes and encloses an area of 3/4 square unit with the axes
then
(a) x6 – 6x3 + 4 = 0
(b) x6 – 12x3 + 4 = 0
6
3
(c) x + 6x – 4 = 0
(d) x6 – 12x3 – 4 = 0
The area (in sq. units) of the region
{(x, y) Î R2: x2 £ y £ |3 – 2x|, is:
[Jan. 8, 2020 (II)]
10
9
31
(b)
(c)
3
2
6
The area (in sq. units) of the region
(a)
23.
A = {( x, y) :
y2
£ x £ y + 4} is:
2
(d)
[April 09, 2019 (II)]
53
(b) 30
(c) 16
(d) 18
3
2
Let S(a) = {(x, y) : y £ x, 0 £ x £ a} and A(a) is area of the
region S(a). If for a l, 0 < l < 4, A(l) : A(a) = 2 : 5, then l
equals :
[April 08, 2019 (II)]
(a)
24.
13
6
1
æ 4 ö3
(a) 2 ç ÷
è 25 ø
1
2 3
(b) 2 æç ö÷
è5ø
1
æ 2 ö3
(c) 4 ç ÷
è5ø
1
4 3
(d) 4 æç ö÷
è 25 ø
Downloaded from @Freebooksforjeeneet
M-419
Applications of Integrals
25.
The area (in sq. units) of the region bounded by the
parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and x = 3,
is :
[Jan. 12, 2019 (I)]
(a)
15
21
17
15
(b)
(c)
(d)
4
2
4
2
The area (in sq. units) of the region bounded by the curve
1
2p
1 4p
+
+
(d)
2 3 3
3 3
The area (in sq. units) of the region {(x, y) : y2 ³ 2x and
x2 + y2 £ 4x, x ³ 0, y ³ 0} is :
[2016]
34.
x 2 = 4 y and the straight line x = 4 y – 2 is :
(a) p -
[Jan. 11, 2019 (I)]
27.
9
7
5
3
(b)
(c)
(d)
8
8
4
4
The area (in sq. units) in the first quadrant bounded by the
parabola, y = x2 + 1, the tangent to it at the point (2, 5) and
the coordinate axes is :
[Jan. 11, 2019 (II)]
28.
14
37
187
(c)
(d)
3
24
24
If the area enclosed between the curves y = kx2 and
x = ky2, (k > 0), is 1 square unit. Then k is:
[Jan. 10, 2019 (I)]
(a)
(a)
8
3
1
2
3
(b)
(c) 3
(d)
3
3
2
The area (in sq. units) bounded by the parabola y = x2 –1, the
tangent at the point (2, 3) to it and the y-axis is:
[Jan. 9, 2019 (I)]
29.
8
32
56
14
(b)
(c)
(d)
3
3
3
3
30. If the area of the region bounded by the curves, y = x2,
(a)
1
and the lines y = 0 and x = t (t > 1) is 1 sq. unit, then
x
t is equal to
[Online April 16, 2018]
35.
4
3
(b) e 2/3
(c)
3
2
The area (in sq. units) of the region
31.
36.
37.
38.
{(x, y) : x ³ 0, x + y £ 3, x2 £ 4y and y £ 1 +
33.
4 2
3
(b)
p 2 2
2
3
19
17
7
13
(b)
(c)
(d)
6
6
2
6
The area (in sq. units) of the region described by
{(x, y) : y2 £ 2x and y ³ 4x – 1} is
[2015]
15
9
7
5
(b)
(c)
(d)
64
32
32
64
The area (in square units) of the region bounded by the
curves y + 2x2 = 0 and y + 3x2 = 1, is equal to :
[Online April 10, 2015]
3
1
4
(b)
(c)
5
3
3
The area of the region described by
{( x, y ) : x
2
(d)
3
4
}
+ y 2 £ 1 and y 2 £ 1 - x is:
[2014]
p 2
p 2
p 4
p 4
+
+
(b)
(c)
(d)
2 3
2 3
2 3
2 3
The area of the region above the x-axis bounded by the
(a)
39.
curve y = tan x, 0 £ x £
8
3
x=
p
is:
4
(a)
1æ
1ö
ç log 2 - ÷
2è
2ø
p
and the tangent to the curve at
2
[Online April 19, 2014]
(b)
1æ
1ö
ç log 2 + ÷
2è
2ø
1
1
(1 - log 2 )
(1 + log 2 )
(d)
2
2
Let A = {(x, y): y2 £ 4x, y – 2x ³ – 4}. The area (in square
units) of the region A is:
[Online April 9, 2014]
(a) 8
(b) 9
(c) 10
(d) 11
The area (in square units) bounded by the curves
y = x , 2y – x + 3 = 0, x-axis, and lying in the first quadrant
is :
[2013]
(c)
40.
x } is :
[2017]
5
59
3
7
(b)
(c)
(d)
2
12
2
3
The area (in sq. units) of the smaller portion enclosed
between the curves, x2 + y2 = 4 and y2 = 3x, is :
[Online April 8, 2017]
(a)
2p
3
4
8
(d) p 3
3
The area (in sq. units) of the region described by
A = {(x, y)| y ³ x2 – 5x + 4, x + y ³ 1, y £ 0} is:
[Online April 9, 2016]
A=
(d) e 3/2
(d)
3
+
(a)
[Online April 15, 2018]
13
10
5
(b)
(c)
3
3
3
32. The area (in sq. units) of the region
1
(a)
{x Î R : x ³ 0, y ³ 0, y ³ x – 2 and y £ x } , is
(a)
(b)
(a)
y=
(a)
p
3
(c) p -
(b)
(a)
2 3
+
(c)
(a)
26.
1
41.
(a) 9
(b) 36
(c) 18
Downloaded from @Freebooksforjeeneet
(d)
27
4
EBD_8344
M-420
42.
Mathematics
p
,
2
[Online April 23, 2013]
The area under the curve y = | cos x – sin x |, 0 £ x £
and above x-axis is :
(b) 2 2 - 2
(a) 2 2
43.
44.
45.
52.
(c) 2 2 + 2
(d) 0
The area of the region (in sq. units), in the first quadrant
bounded by the parabola y = 9x2 and the lines x = 0, y = l
and y = 4, is :
[Online April 22, 2013]
(a) 7/9
(b) 14/3
(c) 7/3
(d) 14/9
The area bounded by the curve y = ln (x) and the lines
y = 0, y = ln (c) and x = 0 is equal to :[Online April 9, 2013]
(a) 3
(b) 3 ln (c) – 2
(c) 3 ln (c) + 2
(d) 2
y
The area between the parabolas x 2 =
and
4
x2 = 9y and the straight line y = 2 is :
[2012]
10 2
20 2
(c)
(d) 10 2
3
3
The area bounded by the parabola y2 = 4x and the line
2x – 3y + 4 = 0, in square unit, is [Online May 26, 2012]
2
1
1
(b)
(c) 1
(d)
5
3
2
The area of the region bounded by the cur ve
y = x3, and the lines, y = 8, and x = 0, is
[Online May 19, 2012]
(a) 8
(b) 12
(c) 10
(d) 16
53.
48.
49.
50.
51.
54.
55.
56.
57.
y 2 = 4 x and x2 = 4y is:
60.
16
sq units
3
32
sq units
3
(b)
(c)
8
sq. units
3
(d) 0 sq. units
(b) 4 2 - 1
(c) 4 2 + 1
(d) 4 2 - 2
The area of the region bounded by the parabola
(y – 2)2 = x –1, the tangent of the parabola at the point
(2, 3) and the x-axis is:
[2009]
(a) 6
(b) 9
(c) 12
(d) 3
The area of the plane region bounded by the curves
x + 2y2 = 0 and x + 3y2 = 1is equal to
[2008]
1
4
5
(b)
(c) 2
(d)
3
3
3
3
The area enclosed between the curves y2 = x and y = | x | is
[2007]
(a) 1/6
(b) 1/3
(c) 2/3
(d) 1
(a)
If a straight line y – x = 2 divides the region x2 + y 2 £ 4
into two parts, then the ratio of the area of the smaller part
to the area of the greater part is [Online May 12, 2012]
(a) 3p – 8 : p + 8
(b) p – 3 : 3p + 3
(c) 3p – 4 : p + 4
(d) p – 2 : 3p + 2
The area enclosed by the curves y = x 2 , y = x 3 ,
x = 0 and x = p, where p > 1, is 1/6. The p equals
[Online May 12, 2012]
(a) 8/3
(b) 16/3
(c) 2
(d) 4/3
The parabola y2 = x divides the circle x2 + y2 = 2 into two
parts whose areas are in the ratio [Online May 7, 2012]
(a) 9p + 2 : 3p – 2
(b) 9p – 2 : 3p + 2
(c) 7p – 2 : 2p – 3
(d) 7p + 2 : 3p + 2
The area bounded by the curves
[2011 RS]
(a)
1
5
square units
(d)
square unit
2
2
The area bounded by the curves y = cos x and
3p
y = sin x between the ordinates x = 0 and x =
is
2
[2010]
(a) 4 2 + 2
(a)
47.
[2011]
(c)
(a) 20 2 (b)
46.
The area of the region enclosed by the curves
1
y = x, x = e, y = and the positive x-axis is
x
3
(a) 1 square unit
(b)
square units
2
2
The parabolas y 2 = 4 x and x = 4 y divide the square
region bounded by the lines x = 4, y = 4 and the coordinate
axes. If S1 , S2 , S3 are respectively the areas of these parts
58.
numbered from top to bottom; then S1 : S2 : S3 is [2005]
(a) 1 : 2 : 1 (b) 1 : 2 : 3 (c) 2 : 1 : 2 (d) 1 : 1 : 1
The area of the region bounded by the curves
[2004]
59.
y =| x - 2 |, x = 1, x = 3 and the x-axis is
(a) 4
(b) 2
(c) 3
(d) 1
The area of the region bounded by the curves
y = x - 1 and y = 3 - x is
[2003]
(a) 6 sq. units
(b) 2 sq. units
(c) 3 sq. units
(d) 4 sq. units.
The area bounded by the curves y = lnx, y = ln |x|,y = | ln x
| and y = | ln |x| | is
[2002]
(a) 4sq. units
(b) 6 sq. units
(c) 10 sq. units
(d) none of these
Downloaded from @Freebooksforjeeneet
M-421
Applications of Integrals
1.
(a) [x] = 0 when x Î [0, 1) and [x] = 1 when x Î[1, 2)
æ1 1ö
Sç ÷
è 2 2ø
0 £ x <1
ì 0
y=í
îx -1 1 £ x < 2
æ1 ö
P = ç 0÷
è2 ø
y =2 x
y=x–1
1
æ1 ö
P = ç 0÷
è2 ø
2
Q( 3 / 2 0)
x=
3
2
2
ò
1
\ A = 2 x dx - (1)(1)
2
Required area = Area of trape ium PQRS
0
=
2.
3/ 2 2
4x
3
3/2
-
0
1 8 2 1
=
2
3
2
ò
1/2
2
1ö
æ
ç x - ÷ dx
2ø
è
3/2
3
1 æ 3 - 1 öæ 1
3 ö 1ææ
1ö ö
çç x - ÷ ÷
= çç
+
1
֍
÷
֍ 2
2 è 2 øè
2 ø÷ 3 ç è
2ø ÷
è
ø1/2
(b)
(2, 3)
2
(1,
)
1
4.
1
2
3 1
4 3
(c) Since, the relation y £ x2 + 3x represents the region
below the parabola in the 1st quadrant
=
1
2
1
2
1
2
1
(0, ) (1, )
Required area = ò ( x 2 + 1)dx + ò ( x + 1)dx
1
2
é 4 13 ù 5 79
=ê - ú+ = .
ë 3 24 û 2 24
æ 3 ö æ 3
1
3ö
(b) Coordinates of P æç ,0 ö÷ , Q ç
, 0 ÷, R ç
,1 ÷
ç
÷
ç
2 ÷ø
è2 ø è 2 ø è 2
1 1
and S æç , ö÷
è2 2ø
(0, 0) (1, 0)
(3, 0)
2
é x3
ù
é x2
ù
= ê + x ú + ê + xú
3
2
ë
û1 ë
û1
3.
(–3, 0)
Q y=4
Þ x2 + 3x = 4 Þ x = 1, – 4
\ the required area = area of shaded region
1
é x3 3 x2 ù
3
= ò ( x + 3 x) dx + ò 4.dx = ê +
ú + [4 x ]1
0
1
2 û0
ë3
1
=
2
3
59
1 3
+ +8 =
3 2
6
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EBD_8344
M-422
5.
Mathematics
(b) Given A = {(x, y) : 0 £ y £ x|x| + 1 and –1 £ x £ 1}
Differentiating w. r . t b , we get
f ( b ) = b cos b + sin b –
p
sin b +
4
2
p
æ pö
æ pö p
f ç ÷ =b× 0 + ç1- ÷ sin + 2 = 1- + 2
è 2ø
è 4ø 2
4
y
(a)
\ Area of shaded region
0
1
y=loge(x+e)
(1–e,
0)
9.
(0,1)
-1
2
0
0
1
æ x3
ö
æ x3
ö
= ç - + x÷ + ç + x÷
è 3
ø -1 è 3
ø0
x+e=0
0
Required area A =
æ1 ö æ1 ö
= 0 - çè - 1÷ø + çè + 1÷ø - (0 + 0)
3
3
6.
e
1
2
(d) Given that
-2
2
2
2
0
0
0
ò xf ¢( x)dx = xò f ¢( x)dx - ò f ( x)dx
3
f ( x) dx =
3
3
= 2 f (2) 4
4
3
3
= 0 - (Q f (2) = 0) = - .
4
4
2
= [ x f ( x)]0 -
ò (1 - x) f (1 - x) dx
-2
b
b
é
ù
ê Using ò f ( x ) dx = ò f (a + b - x) dx ú
êë
úû
a
a
11.
(d)
A
3
Þ R2 =
3
ò f ( x)dx = 4 ; Now,,
0
cos xdx = 3 - 1
2
p /6
ò
(d) We have
òx
e
e - e - 0 +1 = 1
Hence the required area is 1 square unit.
p /3
R2 =
log e ( x + e)dx
\ A = ò loge tdt = [ t log e t - t ]1
10.
3
ò
1- e
put x + e = t Þ dx = dt also when x = 1 – e, t = 1 and when
x = 0, t = e
2 4 6
= + = = 2 square units
3 3 3
(d) Here, 18x2 – 9px + p2 = 0
Þ (3x – p) (6x – p) = 0
\ Req. area =
ò
0
ydx =
1-e
p
p
Þ a= ,b=
6
3
Also, gof(x) = cosx
7.
x
0
= ò (- x + 1)dx + ò ( x + 1)dx
2
ò (1 - x) f ( x) dx
B
-2
1
2
P
1, 1
2 2
O
(Q f (x) = f (1 – x) on [– 2, 3])
3
\ R2 + R2 =
ò
3
x f ( x ) dx +
-2
ò (1 - x )
f ( x ) dx
-2
3
=
8.
Þ 2R2 = R1C(d) From given condition
b
ò
p/4
f ( x )dx = b sin b +
p
cos b + 2b
4
ò
-2
f ( x ) dx = R1
é 1
ù
1
Required area = 4 ê ò 2 2 y 2 dy + area( DPAB) ú
0
2
êë
úû
é2 3 1 1 1 1ù
é 2 1 1ù
= 4 ê éë y ùû 2 + ´ ´ ú = 4 ê ´ + ú
0
3
2
2
2
ë 3 8 8û
ëê
ûú
= 4´
5 5
= .
24 6
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M-423
Applications of Integrals
12.
(b) Required area
y
1
= 2p - ò x - xdx
y = x2 – 1
0
1
x
æ 2 x3/2 x 2 ö
= 2p - ç
- ÷
ç 3
2 ÷ø
è
0
y = 1 – x2
1
(
)
æ2 1ö
æ 1 ö 12p - 1
= 2p - ç - ÷ = 2p - ç ÷ =
3
2
6
è
ø
è6ø
Area = 2ò (1 - x 2 ) - ( x 2 - 1) dx
0
1
= 4ò (1 - x 2 ) dx
15.
(b)
0
æ
x3 ö
= 4ç x - ÷
3ø
è
13.
1
0
2 8
æ 1ö
= 4 ç1 - ÷ = 4 × = sq. units
è 3ø
3 3
(b) Let y = x2 and y = 2x
2
y=x
4
x'
O
Given curves are
4x2 = y
y = 8x + 12
From eqns. (i) and (ii),
4x2 = 8x + 12
Þ x2 – x – 3 = 0
Þ x2 – 2x – 3 = 0
Þ x2 – 3x + x – 3 = 0
Þ (x + 1) (x – 3) = 0
Þ x = – 1, 3
Required area bounded by curves is given by
y = 2x
(2, 4)
y=a
x
y'
According to question
aæ
yö
\ ò ç y - ÷ dy =
0 è
2ø
4æ
yö
y - ÷ dy
2ø
òa çè
a
3
A=
4
é
ù
é
ù
a
4
é y2 ù
é y2 ù
ê y 3/ 2 ú
ê y 3/ 2 ú
Þê
ú -ê ú = ê 3 ú -ê ú
ë 4 û0 ê
ë 4 ûa
ê 3 ú
ú
ë 2 û0
ë 2 ûa
2
)dx
-1
A=
2
1
a2 2
Þ a 3/ 2 = (8 - a3/ 2 ) - (16 - a 2 )
3
4
3
4
8 x2
4 x3
+ 12 x 2
3
3
-1
4ö
æ
= (4(9) + 36 - 36) - ç 4 - 12 + ÷
3ø
è
4 3/ 2 a 2 4
a =
3
2
3
Þ 8a 3/ 2 - 3a 2 = 8
Þ
14.
ò (8 x + 12 - 4 x
4
4 132 - 4 128
= 44 - =
=
3
3
3
3
(b) Given eqns. are, x2 = ay and y2 = ax
= 36 + 8 -
\ 3a 2 - 8a 3/ 2 + 8 = 0
(d) Total area – enclosed area between line and parabola
16.
x2 = ay
Q(
b, b
)
B(a
, a)
y
2
- 2
0
1
O A
R
2
- 2
x㦨b
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y2 㦨 ax
...(i)
...(ii)
EBD_8344
M-424
Mathematics
After solving, we get x = a, y = a
Now, coordinates of B is (a, a) and A is (0, 0)
Now, coordinates of Q is (b, b)
Þ (y + 2)2 = 8 Þ y + 2 = ±2 2
Hence, required area
1 2 1
Þ b =1
b =
2
2
Area bounded by curves and x = 1 is
=
3- 2 2
\
1æ
a
2
2
1 æ
1/2 x ö
1/2 x
a
x
dx
=
ç
÷
ç ax òç
ò
÷
ç
2 è
a ø
a
0è
0
17.
0
é 2 3/2 ù
= ê2 ´ x ú
ë 3
û0
Þ 4a a - 2 = a 3
Þ a6 + 4a3 + 4 = 16a3
Þ a6 – 12a3 + 4 = 0
(a) Point of intersection of y = x2 and y = – 2x + 3 is
obtained by x2 + 2x – 3 = 0
4
´ (3 - 2 2) 3 - 2 2 + 6 - 4 2
3
=
4
(3 - 2 2)( 2 - 1) + 6 - 4 2
3
[Q( 2 - 1)2 = 3 - 2 2]
=
4
(3 2 - 3 - 4 + 2 2) + 6 - 4 2
3
=-
X
1
1
+ (8 + 4 - 8 2)
2
=
Y
–3
1
2 x dx + ´ (2 2 - 2) ´ (2 2 - 2)
2
3- 2 2
ö
÷ dx
÷
ø
2
1 a2
a=
3
3a 6
Þ
ò
10 8
+
2 =a 2+b
3 3
10 8
+ =6
3 3
(c) Given parabola y2 = 4lx and the line y = lx
\ a = 8/3 and b = – 10/3 Þ a - b =
19.
Þ x = – 3, 1
1
So, required area =
ò (line - parabola)dz
4
l
-3
3 ù1
é
2
2 x
(3
2
x
x
)
dx
x
x
=
3
ê
ú
ò
3 ûú
ëê
-3
-3
1
=
Putting y = l in y2 = 4lx, we get x = 0,
æ 12 - 32 ö æ 13 + 33 ö
28 32
= (3)4 - 2 ç
÷-ç
÷
ç 2 ÷ ç 3 ÷ = 12 + 8 - 3 = 3
è
ø è
ø
18.
\ required area =
(b) Consider y2 = 4x and x + y = 1
y
2 l .x 3/2 lx 2
=
3/ 2
2
P (3 - 2 2 2 2 - 2)
(1 0)
ò (2
lx - lx)dx
0
y2 = 4x
O
4
l
x
=
0
32 8
3l l
8 1
= Þ l = 24
3l 9
(a) Let, C1 : | y – x | £ 2
C2 : | y + x | £ 2
=
x+y㦨1
20.
Substituting x = 1 – y in the equation of parabola,
y2 = 4(1 – y) Þ y2 + 4y – 4 = 0
4/ l
By the diagram, region is square
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4
l
M-425
Applications of Integrals
23.
(d)
(x 㦨 y 졔 4)
(8 4)
Required area
dy
(2 –2)
2
2
y 㦨 2x
ì
ü
y2
£ x £ y + 4ý
Given region, A = í( x, y ) :
î
þ
2
2
Now, length of side =
21.
22 + 22 = 2 2
4
(d)
Hence, area =
(1 2)
1
0
-2
æ
ò çç y + 4 -
-2 è
y2 ö
÷ dy
2 ÷ø
4
é y2
64 ö æ
8ö
y3 ù
æ
= ê + 4 y - ú = ç 8 + 16 - ÷ - ç 2 - 8 + ÷
2
6
6
6
ø è
ø
úû -2 è
ëê
(0 1)
x
Area = ò (( x + 1) - 2 ) dx
ò
4
x dy =
32 ö æ
4 ö 40 14 54
æ
= ç 24 - ÷ - ç -6 + ÷ =
+ =
= 18
3 ø è
3ø 3 3
3
è
(Q Area = ò ydx )
24.
y
(d)
(l l )
x ù1
é x2
2
2 ö æ -1 ö 3 1
æ1
= = ê 2 + x - ln 2 ú = ç + 1 ln 2 ÷ø çè ln 2 ÷ø 2 ln 2
êë
úû 0 è 2
x㦨l
O
22.
x
y㦨x졔2
(b)
2
y 㦨x
2
y㦨x
l
l
0
0
Area of the region = 2 ´ ò ydx = 2 ò
3
2
–1
Required area is equal to the area under the curves
y ³ x2 and y d” x + 2
2
= 2´ p2
3
3
2
4
A(l) = 2 ´ (l ´ l ) = l 2
3
3
2
\ required area
ò (( x + 2) - x
2
3
)dx
-1
2
æ x2
x3 ö
8ö æ 1
1ö 9
æ
= ç + 2x - ÷ = ç 2 + 4 - ÷ - ç + - 2 + ÷ =
ç 2
÷
3
3
2
3ø 2
è
ø
è
è
ø -1
2
Given, A(l ) = 2 Þ l = 2
8
5
A(4) 5
2
1
16
4
l = æç ö÷ 3 = 4. æç ö÷ 3
è 5ø
è 25 ø
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x dx
EBD_8344
M-426
25.
Mathematics
(d)
27.
(b)
B (2, 5)
y=x+1
(0, 2)
C
(0, 1)
O
x=3
(0, 0)
D
3
A ( , 0)
The equation of parabola x2 = y – 1
The equation of tangent at (2, 5) to parabola is
æ dy ö
y – 5 = çè dx ÷ø
(x – 2)
(2,5)
3
2
Area of the bounded region ò [( x + 2) - ( x + 1)] dx
y – 5 = 4(x – 2)
0
4x – y = 3
3
é x3 x 2
ù
= ê - + xú
3
2
ë
û0
= 9-
26.
Then, the required area
2
2
= ò {( x +1) – (4 x – 3)} dx – Area of DAOD
9
15
+3 =
2
2
0
2
(b)
1 3
2
= ò ( x – 4 x + 4) dx – 2 ´ 4 ´ 3
0
x2 = 4y
æ 1ö
ç 0, ÷
è 2ø
2
é ( x – 2)3 ù
9 37
ú – =
=ê
3
ë
û 0 8 24
Q(2, 1)
P(–1, 1/4)
(–2, 0)
28.
(b) x2 =
1
y
k
æ1 1ö
Aç , ÷
èk kø
O
y2 =
Let points of intersection of the curve and the line be
P and Q
1
x
k
æ x + 2ö
x2 = 4 çè
÷
4 ø
æ1 1ö
Two curves will intersect in the Ist quadrant at A ç , ÷
èR Rø
x2 – x – 2 = 0
x = 2, – 1
Q area of shaded region = 1.
1ö
æ
Point are (2, 1) and çè -1, ÷ø
4
\
2
éæ x + 2 ö æ x 2 ö ù
é x2 1
x3 ù
÷ø - ç ÷ ú dx = ê + x - ú
4
12 û -1
è 4 ø ûú
ë8 2
-1 ë
2
Area =
ò êêèç
2ö æ 1 1 1 ö 9
æ1
= çè + 1 - ÷ø - çè - + ø÷ =
2
3
8 2 12 8
1
kæ
ò çè
0
ö
x
- kx 2 ÷ dx = 1
k
ø
1
3 ök
1
æ
æ x3 ö k
ç 1 x2 ÷
Þ ç k × 3 ÷ - çè k × 3 ÷ø = 1
0
çè
÷
2 ø0
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M-427
Applications of Integrals
2
Þ
1
×
3 k
-
3
k2
k
=1
3k 3
=-
Þ 3k2 = 1
Þ k=±
\ k=
30.
1
3
1
(Q k > 0)
3
(a)
1
is (1, 1)
x
Area bounded by the curves is the region ABCDA is given
as:
(b) The intersection point of y = x2 and y =
Area =
Y
(0 3)
1 2
ò0 x
dx + ò
t
1
1
dx
x
1
é x3 ù
1
t
= ê ú + [ ln ( x )]1 = + ln (t )
3
ë 3 û0
(2 3)
xἘ
X
(0 –1)
y = x2
(0 –5)
yἘ
(1 , 1 )
B
Q Curve is given as :
y = x2 – 1
y = x1
C
y=0
D
A
dy
= 2x
Þ
dx
x=t
æ dy ö
=4
Þ çè dx ÷ø
(2,3)
\ equation of tangent at (2, 3)
Q area = 1
(y – 3) = 4 (x – 2)
Þ
Þ y = 4x – 5
31.
but x = 0
2
1
+ ln (t ) = 1 Þ ln (t) =
3
3
(b) The intersection point of y = x – 2 and y =
The required area
Þ y=–5
Here the curve cuts Y–axis
\ required area
1
= 4
3
=
4
ò0
xdx –
-5
x is (4, 2).
1
16
10
´2´2=
–2=
2
3
3
y
3
ò ( y + 5)dy - ò
2
Þ t = e3
y + 1 dy
(4, 2 )
-1
y=
x
3
=
1 é9
25
ù
+ 15 + 25 ú
û
4 êë 2
2
=
x
–
2
3
ù -2
1 é y2
éë ( y + 1)3/2 ùû
= ê + 5yú
-1
4ë 2
û -5 3
y
29.
32 16 8
- = sq-units.
4 3 3
=
2
1
- 2 =1
Þ
2
3k
3k
2 [ 3/2 ]
4 -0
3
O
(2, 0 )
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x
EBD_8344
M-428
Mathematics
æ 2
3 2p ö
+ ÷ ´2
=ç
2
3ø
è 3
2p ö
1 4p
æ 1
+ ÷ ´2=
+
= çè
ø
3
2 3
3 3
x+y=3
32.
(a)
y=1+ x
(1, 2)
4y = x 2
(0, 1)
A (2, 2)
P
(2, 1)
(0, 0)
(1, 0)
34.
(2, 0)
(d)
O
B
Area of shaded region
1
2
2
0
1
0
(2, –2)
x2
(1
x
)dx
(3
x)dx
dx
+
+
ò
ò
ò
=
4
Points of intersection of the two curves are (0, 0), (2, 2) and
(2, –2)
1
é 3ù
2 2 é 3 ù2
x
5
1 ê x2 ú
2 éx ù
= [ x ]0 + ê ú + [3x ]1 - ê ú - ê ú = sq.units
3
ë
û
ë
û
2
12
2
1
0
ê ú
ë 2 û0
Area = Area (OPAB) – area under parabola (0 to 2)
2
=
0
35.
y = x2 – 5x + 4
(a)
(d)
1, 3
2
=
4
y2 = 38
2
+
y
(3,0)
A1
x
33.
p ´ (2) 2
8
- ò 2 x dx = p 4
3
(4,0)
(3,–2)
1
2
x + y =1
Required area = A1 + A2
1, - 3
=
From the equations we get;
x2 + 3x – 4 = 0
Þ (x + 4) (x – 1) = 0 Þ x = –4, x = 1
1
×2×2+
2
= 2+
36.
4
ò3 ( x
2
- 5 x + 4) dx
7 19
=
sq. units
6
6
(b) Required area
when x = 1, y = 3
B æ 1 ,1ö
çè
÷
2 ø
1æ1
2
ö
2
Area = ò ç ò 3 . x dx + ò 4 - x . dx ÷ ´ 2
ç
÷
0è0
1
ø
æ æ 32
x
= ç 3 çç
çç
32
è è
O
2ö
ö æx
xö
4 - x 2 + 2 sin -1 ÷ ÷ ´ 2
÷ +ç
÷ è2
2 ø1 ÷÷
ø0
ø
1
æ æ 2 ö ìï p æ 3 p ö üïö
+ ÷ ý÷ ´ 2
= ç 3 ç ÷ + í2 . - ç
è è 3 ø ïî 2 è 2 3 ø ïþø
1
=
ò
-1/ 2
y +1
dy 4
1
ò
-1/ 2
C æ 1 - 1ö
çè , ÷ø
8 2
y2
dy
2
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M-429
Applications of Integrals
1
39.
1
ù
1 é y2
1 é y3 ù
+ yú
- ê ú
= ê
4 ëê 2
ûú -1/ 2 2 ëê 3 ûú -1/ 2
1 é 3 3 ù 9 15 9 27 9
+ =
=
=
4 êë 2 8 úû 48 32 48 96 32
(c) Solving
y + 2x2 = 0
y + 3x2 = 1
...(1)
pö
æ 2 pö æ
y – 1 = çè sec ÷ø çè x - ÷ø
4
4
=
37.
(a) The given curve is y = tanx
p
when x = , y = 1
4
Equation of tangent at P is
Y
y = 2x + 1 –
æp ö
P ç ,1÷
è4 ø
O
X¢
y = tan x
(1, – 2)
(– 1, 2)
X
M
L
Y¢
p
2
Area of shaded region
= area of OPMO – ar (DPLM)
Point of intersection (1, – 2) and (– 1, – 2)
1
((
or y = 2x + 1 –
))
) (
2
2
Area = 2 ò 1 - 3x - -2x dx
0
1
æ
x3 ö
4
2ò 1 - x 2 dx = 2 ç x - ÷ =
ç
÷
3
è
ø0 3
0
= 15 – 6 = 9 sq units
(c) Given curves are x2 + y2 = 1 and y2 = 1 –x.
Intersecting points are x = 0, 1
1
38.
=
(
)
40.
p
4 tan x
0
ò
...(2)
1
dx - (OM - OL)PM
2
p
1 ì p p - 2ü
= [ log sec x ]04 - í ý ´1
2î4
4 þ
1é
1ù
= êlog 2 - ú sq unit
2ë
2û
(b) Area of shaded portion
Y
y2 = 4x
1
4
3
2
1
1
–1
(4, 4)
X¢
1 2 3 4
–1
–2
–1
Y¢
Area of shaded portion is the required area.
So, Required Area = Area of semi-circle
+ Area bounded by parabola
1
=
1
pr 2
p
+ 2ò 1 - xdx = + 2 ò 1 - x dx
2
2
0
0
(Q radius of circle = 1)
1
é (1 - x ) 3 2 ù
p 4
p 4
p
ú = - (-1) = + sq. unit
= + 2ê
2 3
2
3
3
ê
ú
2
2 û0
ë
=
4æ
ò2 çè
y + 4ö
÷ dy 2 ø
ù
1 é y2
= ê + 4 yú
2 ëê 2
ûú
=
4
-2
4
ò-2
y2
dy
4
1 é y3 ù
- ê ú
4 ëê 3 ûú
4
-2
1
1 64 8
[{8 + 16} - {2 - 8}] - 4 ìí 3 + 3 üý = 9
2
î
þ
Downloaded from @Freebooksforjeeneet
X
p
2
EBD_8344
M-430
41.
Mathematics
(a) Given curves are
y= x
and 2y – x + 3 = 0
On solving both we get y = –1, 3
44.
…(1)
…(2)
(d) To find the point of intersection of curves y = ln (x)
and y = ln (3), put ln (x) = ln (3)
Þ ln (x) – ln (3) = 0
Þ ln (x) – ln (3) = ln (1)
x
=1 , Þ x = 3
3
Þ
Y
(9, 3)
Y
y = ln (x)
3
X
–3
2
y = ln (3)
X
O
3
Required area =
ò { (2 y + 3) - y
2
1
3
} dy
0
3
é
y3 ù
= ê y 2 + 3 y - ú = 9.
ë
3 û0
42.
3
3
Required area = ò ln (3) dx - ò ln ( x) dx
(b) y = | cos x – sin x |
0
1
= [ x ln (3) ]0 - [ x ln ( x ) - x ]1 = 2
3
Y
f(x) = cos x
g(x) = sin x
45.
(c)
Y
1 2
y = 4x2 y = 9 x
y=2
X
X
p/4
O
3
p/2
2
æ
yö
Required area = 2 ò çç 9 y ÷ dy
4 ÷ø
0è
p /4
ò
Required area = 2
0
p/4
0
= 2 [ sin x + cos x ]
2
4
(d) Required area =
ò
y =1
=
1
3
4
ò
y =1
y
dy
9
1 2
y1/2 dy = ´ ( y 3/2 )
3 3
1
2
2
= [(41/ 2 ) 3 - (11/ 2 ) 3 ] = [8 - 1]
9
9
=
2
14
´ 7 = sq. units.
9
9
5
20 2
= 2. 2 2 =
3
3
46.
4
2
é 3 1 3ù
é5 3 ù
= 2 ê2 y 2 - y 2 ú = 2 ´ ê y 2 ú
3 ú
ê
ê3 ú
ë
û0
ë
û0
é 2
ù
= 2ê
- 1ú = (2 2 - 2) sq. units
ë 2 û
43.
2
3
3ù
é2
æ
1 2
yö
dy = 2 ê ´ 3. y 2 - ´ . y 2 ú
= 2ò ç 3 y ÷
ç
2 3
ê3
ú
2 ÷ø
0è
ë
û0
2
(cos x - sin x ) dx
(b) Intersecting points are x = 1, 4
4é
æ 2x + 4ö ù
\ Required area = ò ê 2 x - èç 3 ø÷ ú dx
û
1ë
2x
= 3
=
3
4
2
2
4
1
4
2 x2
4
- x
3´2
3 1
1
4 æ 32 32 ö 1
4ù
é4
ç 4 - 1 ÷ - (16 - 1) - ê ( 4 ) - ú
3è
3
3
3
ë
û
ø
Downloaded from @Freebooksforjeeneet
M-431
Applications of Integrals
=
Now, area of II = Area of circle – area of I.
= 4p – (p – 2) = 3p + 2
4
28 - 27 1
28
(7 ) - 5 - 4 = - 9 =
=
3
3
3
3
8
47.
(b) Required Area =
ò
1/3
y
Hence, required ratio =
dy
y =0
49.
Y
area of I
p-2
=
area of II 3p + 2
(d) Given curves are y = x2 and y = x3
Also, x = 0 and x = p, p > 1
Now, intersecting point is (1, 1)
y = x3
y=8
y = x2
y = x3
(p,0)
(0.0)(1,0) x = p
X
X¢
Y¢
1
8
1 +1
y 3
= 1
+1
3
=
3 æ 43 ö
y ÷
ø0
4 çè
0
1 x3 x4
=
6 3
4
0
3é 4
3
(8) 3 - 0ùú = éë 24 ùû = 3 ´16 = 12 sq. unit.
ê
4ë
û 4
4
(d) Let I be the smaller portion and II be the greater portion
of the given figure then,
2
Þ
x+
Y
Þ
p
(
0
x 4 x3
+
4
3
p
1
1 1 1 1 1 3 p4 – 4 p3
– + + – =
6 3 4 4 3
12
p3 ( 3 p - 4)
12
= 0 Þ p3 (3p – 4) = 0
Þ p = 0 or
I
II
1
1
4
3
Since, it is given that p > 1
\ p can not be ero.
(0, 2)
X¢
)
1 æ 1 1 ö æ p4 p3 1 1ö
= ç - ÷ +ç
- + ÷
Þ
6 è 3 4ø è 4
3 4 3ø
y=
48.
=
(
)
Required Area = ò x 2 - x3 dx + ò x3 - x 2 dx
8
X
Hence, p =
(– 2, 0)
50.
4
3
2
(b)
2
(1, 1)
2
x +y =2
y =x
C
Y¢
0
2
é
ù
Area of I = ò ê 4 - x - ( x + 2 ) ú dx
û
-2 ë
O
B
A
0
0
é x2
ù
éx
4 -1 æ x ö ù
2
= ê 2 4 - x + 2 sin çè 2 ÷ø ú - ê 2 + 2 x ú
ë
û -2 ëê
ûú -2
p
é 4 ù
-1
= éë 2sin ( -1) ùû - ê - + 4ú = 2 ´ - 2 = p - 2
2
ë 2 û
D (1, –1)
Area of circle = p
( 2)
2
= 2p
Downloaded from @Freebooksforjeeneet
EBD_8344
M-432
Mathematics
Area of OCADO = 2{Area of OCAO}
= 2 {area of OCB + area of BCA}
1
2
0
1
y
x=e
= 2 ò y p dx + 2 ò yc dx
where y p =
y=x
x and yc = 2 - x 2
1
2
0
1
\ Required Area = 2 ò x dx + 2 ò
(1, 1)
A
2 - x 2 dx
1
B e, e
O
x
C
é x 2 - x2
ù
é2
ù
-1 x ú
ê
2
.1
0
+
2
+
sin
= ê3
ú
2
ê
2ú
ë
û
ë
û1
2
53.
(d)
4
ìp p 1ü 4
ì p 1 ü 3p + 2
= + 2í - - ý = + 2í - ý =
3
6
î 2 4 2þ 3
î4 2þ
3p + 2 9p - 2
=
Bigger area = 2p 6
6
9p - 2
i.e., 9p – 2 : 3p + 2
\ Required Ratio =
3p + 2
51.
cos x
sin x
p
0
p
p
4
2
5p
3p
4
2
2p
Area above x-axis = Area below x-axis
(b)
\ Required area
p
p /2
é p /4
ù
= 2 ê ò (cos x - sin x) dx + ò sin xdx - ò cos xdx ú
p /4
p /4
ëê 0
ûú
p /4
= 2 é( sin x + cos x )0
ë
p
p /2
+ ( - cos x )p /4 - ( sin x ) p /4 ù
û
éæ 1
1 ö
1 ö æ
1 öù
æ
= 2 êç
+
÷ - (0 + 1) + ç1 +
÷ - ç1 ÷ú
2ø
2ø è
2 øû
è
ëè 2
1
1 ù
é
= 2 ê 2 -1+ 1+
-1 +
2
2 úû
ë
4æ
x2 ö
Required area = ò ç 2 x - ÷ dx
4ø
0è
4
é æ x3/2 ö x3 ù
= ê2 ç
÷- ú
3 / 2 ø 12 ú
ëê è
û0
52.
= 2 [ 2 + 2 - 1] = 4 2 - 2
54.
(b)
D(0, 3)
32
16
4
64
=
- 16 = sq. units
= ´83
3
3
12
(b) Area of required region AOCBO
(2, 3)
P
A(0, 2)
1
e
1
é x2 ù
1
e
= ò xdx + ò dx = ê ú + [ log x ]1
ë 2 û0
x
0
=
1
1
3
+ 1 = sq. units
2
2
B(–4, 0) 0
C(5, 0)
For slope of tangents at (2, 3)
(y – 2)2 = x – 1
Downloaded from @Freebooksforjeeneet
M-433
Applications of Integrals
2( y - 2)
1
2
ì 2
ü
Þ 2í
´3 3 ´ ×2 2 ý
3
2
3
3
î
þ
dy
dy
1
= 1Þ
=
dx
dx 2( y - 2)
{ } { }
1
1
æ dy ö
=
=
m=ç ÷
è dx ø(2, 3) 2(3 - 2) 2
4
6-4
4
=2
= sq. units
3
3
3
(a) It is clear from the figure, area lies between y2 = x and
y= x
Intersection point y = x and y2 = x is (1, 1)
Þ 2 2-
56.
Equation of tangent
1
y - 3 = ( x - 2)
2
y=x
Y
Þ x - 2y + 4 = 0
...(i)
2
The given parabola is (y – 2) = x – 1 ...(ii)
vertex (1, 2) and it meets x-axis at (5, 0)
Then required area = Ar DBOA + Ar (OCPD) – Ar (DAPD)
=
1
×4×2+
2
y2 = x
y2
y1
X'
X
(1, 0)
(0, 0) O
1
3
ò0 xdy - 2 ´ 2 ´ 1
Y'
3
\ Required area =
é ( y - 2)3
ù
3
= 3 + ò ( y - 2) 2 + 1 dy = 3 + ê
+ yú
0
êë 3
úû 0
55.
(1, 1)
A
y = -x
1
ò0 ( y2 - y1 )dx
1
8ù
é1
= 3 + ê + 3 + ú = 3 + 6 = 9 sq. units
3
3û
ë
x
2
2
(d) Given x + 2 y = 0 Þ y = –
2
1
and x + 3 y 2 = 1 Þ y 2 = – ( x – 1)
3
On solving these two equations we get the points of
intersection as (–2, 1), (–2, –1)
é x3 / 2 x 2 ù
- ú
= ò ( x - x) dx = ê
2 úû
0
ëê 3 / 2
0
1
=
57.
2 é 3 / 2 ù1 1 é 2 ù1 = 2 - 1 = 1
- x
x
û0 2 ë û0
3 2 6
3ë
(d) On solving, we get intersection points of x 2 = 4 y
and y2 = 4x are (0, 0) and (4, 4).
y
2
x =4y
y=4 S
1
S2
Y
A (–2, 1)
(0, 0) 0
D
(1, 0)
C
X´
y2=4x
(4, 4)
S3
x=4
x
X
B
By symmetry, we observe
(–2, –1)
4
Y´
S1 = S3 = ò ydx
The required area is ACBDA, given by
0
4 2
1
0
1
ïì 1
ïü
A = 2í ò
- x dx ý
1 - x dx ò
2 -2
ïî -2 3
ïþ
=ò
0
(
)
(
4
é x3 ù
16
sq. units
ê ú =
3
êë 12 úû0
4
1
0
ìï 1 é 2
1 é2
ù
ù üï
Þ 2í
(1 - x )3/ 2 ú ( - x )3/ 2 ú ý
ê
ê
û -2
û -2 þï
2 ë3
îï 3 ë 3
ìé 1 2
ù é -1 2
Þ 2 íê´ 0 - 33/2 ú - ê
´ 0 - 23/2
û ë 2 3
îë 3 3
x
dx =
4
)ùúûýþü
é 3
ù
ê 2 x 2 x3 ú
x
- ú
Also S2 = ò ç 2 x - ÷ dx = ê
4ø
12 ú
ê 3
0è
úû 0
ëê 2
4æ
2ö
4
16 16
sq. units
´8 =
3
3
3
\ S1 : S2 : S3 = 1:1:1
=
Downloaded from @Freebooksforjeeneet
EBD_8344
M-434
58.
Mathematics
(d)
60.
Y
(a) Separate graph of each curve
y=ln|x|
y=lnx
(y = 2 – x)
(y = x – 2)
(1,0)
0
Required Area
(1,0)
3
3
é x2
ù
A = 2ò ( x - 2)dx = 2 ê - 2 x ú = 1
ëê 2
ûú 2
2
59.
(–1,0)
y=|l n x|
X
3
x
(d) Intersection point of y = x – 1 and y = 3 – x is (2, 1) and
eqns. y = –x + 1 and y = 3 + x is (–1, 2)
Y
y=x–1
(–1,2)
x
(–1,0)
0 (1,0)
–1
2
0
A=
y = –ln x
y = –ln (–x)
ò { (3 - x) - ( - x + 1)} dx + ò { (3 - x) - ( x - 1)} dx
0
=
O
y = ln (–x)
x
1
y = ln x
Required area = 4ò (-lnx )dx
2
0
–1
1
ò {(3 + x) - ( - x + 1)} dx +
-1
1
0
1
=4
0
= - 4 [ x ln x - x ]
sq. units
1
2
1
ò (2 + 2 x)dx + ò 2dx + ò (4 - 2 x)dx
-1
0
0
x2 ù +
û -1
(1,0)
1
2
= é2x +
[ 2 x] + éë 4 x - x2 ùû1
ë
= 0 - (-2 + 1) + (2 - 0) + (8 - 4) - (4 - 1)
= 1 + 2 + 4 - 3 = 4 sq. units
1
0
x
[Note: Graph of y = | f(x) | can be obtained from the graph of
the curve y = f(x) by drawing the mirror image of the portion
of the graph below x-axis, with respect to x-axis.
Hence the bounded area is as shown by combined all figure.
y
(2,1)
X
y=3–x
x
y=|l n |x||
(0,3)
y=– x+1
X'
y=3+x
(1,0)
Downloaded from @Freebooksforjeeneet
M-435
Differential Equations
24
Differential
Equations
TOPIC Ć
1.
Ordinary Differential Equations,
Order & Degree of Differential
Equations, Formation of
Differential Equations
(a) x(y¢)2 = x + 2yy¢
3.
Statement 2: The degree of a differential equation, when it
is a polynomial equation in derivatives, is the highest
positive integral power of the highest order derivative
involved in the differential equation, otherwise degree is
not defined.
[Online May 12, 2012]
(a) Statement 1 is true, Statement 2 is true, Statement 2 is
not a correct explanation of Statement 1.
(b) Statement 1 is false, Statement 2 is true.
[Jan. 8, 2020 (II)]
(b) x(y¢)2 = 2yy¢ – x
(c) xy² = y¢
(d) x(y¢)2 = x – 2yy¢
The differential equation representing the family of ellipse
having foci either on the x-axis or on the y-axis centre at
the origin and passing through the point (0, 3) is:
[Online April 16, 2018]
(a) xyy' + y2 – 9 = 0
(b) x + yy" = 0
(c) xyy" + x (y')2 – yy' = 0 (d) xyy' – y2 + 9 = 0
If the differential equation representing the family of all
circles touching x-axis at the origin is
(x
(a)
2
-y
2
1
x
2
)
(c) Statement 1 is true, Statement 2 is false.
6.
(b) 2x2
(d) Statement 1 is true, Statement 2 is true; Statement 2 is
a correct explanation of Statement 1.
The differential equation which represents the family of
curves y = c1ec2 x , where c1, and c2 are arbitrary constants,
is
[2009]
(a) y" = y'y
(b) yy" = y'
(c) yy" = (y')2
(d) y' = y2
dy
= g ( x ) y , then g(x) equals:
dx
[Online April 9, 2014]
1 2
x
2
Statement-1: The slope of the tangent at any point P on a
parabola, whose axis is the axis of x and vertex is at the
origin, is inversely proportional to the ordinate of the point P.
Statement-2: The system of parabolas y2 = 4ax satisfies
a differential equation of degree 1 and order 1.
[Online April 9, 2013]
(a) Statement-1 is true; Statement-2 is true; Statement-2
is a correct explanation for statement-1.
(b) Statement-1 is true; Statement-2 is true; Statement-2
is not a correct explanation for statement-1.
(c) 2x
4.
d2 y
dy
+ y 2 = x and
+ y = sin x are equal.
dx
dx 2
The differential equation of the family of curves,
x2 = 4b (y + b), b Î R, is:
2.
5.
(c) Statement-1 is true; Statement-2 is false.
(d) Statement-1 is false; Statement-2 is true.
Statement 1: The degrees of the differential equations
7.
The differential equation of the family of circles with fixed
radius 5 units and centre on the line y = 2 is
[2009]
(a) (x – 2)y'2 = 25 –(y – 2)2
(d)
(b) (y – 2)y'2 = 25 –(y – 2)2
(c) (y – 2)2y'2 = 25 –(y – 2)2
8.
(d) (x – 2)2 y'2 = 25 –(y – 2)2
The differential equation of all circles passing through the
origin and having their centres on the x-axis is [2007]
(a)
y 2 = x 2 + 2 xy
(c)
x 2 = y 2 + xy
dy
dx
dy
dx
(b)
y 2 = x 2 - 2 xy
dy
dx
(d)
x 2 = y 2 + 3 xy
dy
dx
Downloaded from @Freebooksforjeeneet
EBD_8344
M-436
Mathematics
9.
The differential equation whose solution is Ax 2 + By 2 = 1
10.
where A and B are arbitrary constants is of
[2006]
(a) second order and second degree
(b) first order and second degree
(c) first order and first degree
(d) second order and first degree
The differential equation representing the family of curves
y 2 = 2c ( x + c ) , where c > 0, is a parameter, is of order
11.
15.
and degree as follows :
[2005]
(a) order 1, degree 2
(b) order 1, degree 1
(c) order 1, degree 3
(d) order 2, degree 2
The differential equation for the family of circle
2
12.
13.
(b)
16.
(c) ( x 2 - y 2 ) y ¢ = 2 xy (d) 2( x 2 - y 2 ) y ¢ = xy
The degree and order of the differential equation of the
family of all parabolas whose axis is x - axis, are respectively.
[2003]
(a) 2, 3
(b) 2, 1
(c) 1, 2
(d) 3, 2.
The order and degree of the differential equation
dy ö
æ
çè1 + 3 ÷ø
dx
(a) (1,
2/3
= 4
d y
dx3
are
2
)
3
TOPIC n
1 + y 2 - 1 + x2 =
æ 1 + x 2 - 1ö
1
log e ç
÷ +C
2
çè 1 + x 2 + 1÷ø
æ2
ö
If y = ç x - 1÷ cosec x is the solution of the differential
èp
ø
p
dy
2
+ p( x ) y = cosec x , 0 < x < , then the
2
dx
p
[Sep. 06, 2020 (II)]
(a) cot x
(b) cosec x
(c) sec x
(d) tan x
If y = y (x) is the solution of the differential equation
5 + e x . dy x
+ e = 0 satisfying y (0) = 1, then a value of
2 + y dx
y(loge 13) is :
17.
3
(c) (3, 3)
14.
2( x 2 + y 2 ) y ¢ = xy
(d)
function p(x) is equal to:
2
( x 2 + y 2 ) y ¢ = 2 xy
æ 1 + x 2 - 1ö
1
1 + y 2 + 1 + x 2 = log e ç
÷ +C
2
çè 1 + x 2 + 1÷ø
equation,
x + y - 2ay = 0, where a is an arbitrary constant is
[2004]
(a)
(c)
[Sep. 05, 2020 (I)]
(a) 1
(b) – 1
(c) 0
(d) 2
The solution of the differential equation
dy
y + 3x
+ 3 = 0 is :
dx log e ( y + 3 x)
[2002]
[Sep. 04, 2020 (II)]
(b) (3, 1)
(where C is a constant of integration.)
(d) (1, 2)
1
(a) x - (log e ( y + 3 x)) 2 = C
2
(b) x - log e ( y + 3x ) = C
General & Particular Solution of
Differential Equation, Solution of
Differential Equation by the Method
of Separation of Variables, Solution
of Homogeneous Differential
Equations
The general solution of the differential equation
dy
= 0 is : [Sep. 06, 2020 (I)]
dx
(where C is a constant of integration)
1+ x 2 + y 2 + x 2 y 2 + xy
(a)
æ 1 + x 2 + 1ö
1
1 + y 2 + 1 + x 2 = log e ç
÷ +C
2
çè 1 + x 2 - 1÷ø
(b)
1 + y 2 - 1 + x2 =
æ 1 + x 2 + 1ö
1
log e ç
÷ +C
2
çè 1 + x 2 - 1÷ø
1
(c) y + 3x - (loge x )2 = C
2
(d) x - 2 log e ( y + 3x ) = C
18.
Let f : (0, ¥) ® (0, ¥) be a differentiable function such
t 2 f 2 ( x ) - x 2 f 2 (t )
= 0.
t ®x
t-x
that f (1) = e and lim
If f (x) = 1, then x is equal to :
(a)
1
e
(b) 2e
(c)
1
2e
(d) e
Downloaded from @Freebooksforjeeneet
[Sep. 04, 2020 (II)]
M-437
Differential Equations
19.
The solution curve of the differential equation,
dy
= y 2 , which passes through the point
(1 + e - x )(1 + y 2 )
dx
(0, 1), is :
[Sep. 03, 2020 (I)]
æ
æ 1 + e- x
(a) y + 1 = y ç log e ç
ç 2
ç
è
è
2
æ
æ 1+ e
(b) y 2 + 1 = y ç log e çç
ç
è 2
è
x
æ 1 + ex
(c) y = 1 + y log e çç
è 2
ö
÷÷
ø
2
ö ö
÷÷ + 2 ÷÷
ø ø
24.
ö ö
÷÷ + 2 ÷÷
ø ø
25.
(c)
p -1
4
(d)
p+ 2
4
dy
xy
= 2
; y (1) = 1; then a value of x satisfying
dx x + y 2
If
(a)
1
3e
2
(c)
2e
[Jan. 9, 2020 (II)]
e
2
(b)
(d)
3e
–1
–1
Let f (x) = (sin(tan x) + sin(cot x))2 – 1, |x| > 1. If
( 3 ) = p6 , then y (- 3) is
[Jan. 8, 2020 (I)]
If x3dy + xy dx = x 2 dy + 2 y dx; y (2) = e and x > 1, then
y(4) is equal to :
[Sep. 03, 2020 (II)]
(a)
2p
3
(b)
-
3
+ e
(a)
2
(c)
5p
6
(d)
p
3
3
e
(b)
2
26.
1
e
+ e
(d)
2
2
Let y = y(x) be the solution of the differential equation,
æ 3ö
(a) ç 2, ÷
(b) (1, –1)
è 2ø
(c) (1, 1)
(d) (2, 1)
If a curve y = f (x), passing through the point (1, 2), is the
solution of the differential equation,
1
1 + loge 2
(b)
(c) 1 + log e 2
(d)
(a)
1
1 - log e 2
-1
1 + loge 2
p
p
<x< ,
2
2
and f(0) = 0, then f(1) is equal to:
[Jan. 9, 2020 (I)]
If f 2 (x) = tan–1 (sec x + tan x), –
p
6
Let y = y(x) be a solution of the differential equation,
1 - x2
dy
+ 1 - y 2 = 0,| x |< 1.
dx
æ -1 ö
3
æ 1ö
, then y ç
If y ç ÷ =
is equal to:
è 2 ÷ø
è 2ø
2
[Jan. 8, 2020 (I)]
27.
(a)
3
2
(b) –
1
2
(c)
1
2
(d) –
3
2
If y = y(x) is the solution of the differential equation,
ey = ex such that y(0) = 0, then y(l) is equal to:
æ1ö
2 x 2 dy = (2 xy + y 2 ) dx, then f ç ÷ is equal to :
è 2ø
[Sep. 02, 2020 (II)]
23.
1
4
equal to:
2 + sin x dy
×
= - cos x, y > 0, y (0) = 1. If y (p) = a and
y + 1 dx
dy
at x = p is b, then the ordered pair (a, b) is equal to :
dx
[Sep. 02, 2020 (I)]
22.
(b)
dy 1 d
=
(sin–1 (f (x))) and y
dx 2 dx
(c)
21.
p +1
4
y(x) = e is:
æ 1 + e- x ö
(d) y 2 = 1 + y loge çç
÷÷
è 2 ø
20.
(a)
[Jan. 7, 2020 (I)]
28.
(a) l + loge 2
(b) 2 + loge 2
(c) 2e
(d) loge 2
The general solution of the differential equation (y2 – x3)
dx – xydy = 0 ( x ¹ 0) is :
(a) y2 – 2x2 + cx3 = 0
[April 12, 2019 (II)]
(b) y2 + 2x3 + cx2 = 0
(c) y2 + 2x2 + cx3 = 0
(d) y2 – 2x3 + cx2 = 0
(where c is a constant of integration)
Downloaded from @Freebooksforjeeneet
EBD_8344
M-438
29.
Mathematics
If cos x
æpö
p
dy
- y sin x = 6 x,(0 < x < ) and y ç ÷ = 0, then
dx
2
è3ø
æpö
y ç ÷ is equal to:
è6ø
30.
(a)
p2
2 3
(c)
p2
2 3
[April. 09, 2019 (II)]
p2
2
(b)
-
(d)
p2
4 3
equation,
dy
æ1ö
æ 3ö
= f (x) with y(0) = 1, then y ç ÷ + y ç ÷ is
4
dx
è ø
è 4ø
(a) x loge | y | = 2(x – 1)
equal to:
[Jan. 09, 2019 (II)]
(a) 3
(b) 4
(c) 2
(d) 5
35. The curve satisfying the differential equation,
(x2 – y2) dx + 2xydy = 0 and passing through the point
(1, 1) is
[Online April 15, 2018]
(a) a circle of radius two (b) a circle of radius one
(c) a hyprbola
(d) an ellipse
(b) x loge | y | = – 2(x – 1)
36. If (2 + sin x)
Given that the slope of the tangent to a curve y = y(x) at
2y
. If the curve passes through the
x2
centre of the circle x2 + y2 – 2x – 2y = 0, then its equation is:
[April. 08, 2019 (II)]
any point (x, y) is
dy
æ pö
+ (y + 1) cos x = 0 and y(0) = 1, then y ç ÷
dx
è 2ø
is equal to :
[2017]
2
(c) x loge | y | = – 2(x – 1)
(d) x loge | y | = x – 1
31.
(a) a circle with centre on the x-axis.
(b) an ellipse with maor axis along the y-axis.
(c) a circle with centre on the y-axis.
(d) a hyperbola with transverse axis along the x-axis.
34. Let f : [0, 1] ® R be such that f (xy) = f (x).f (y), for all
x, y Î [0, 1], and f (0) ¹ 0. If y = y(x) satisfies the differential
dy
2
= (x - y) ,
The solution of the differential equation,
dx
when y (1) = 1, is :
[Jan. 11, 2019 (II)]
2- x
= x- y
2- y
(a)
log e
(b)
1- x + y
- log e
= 2 ( x - 1)
1+ x - y
(c)
- log e
(a)
4
3
(b)
1
3
2
1
(d) 3
3
37. If a curve y = f(x) passes through the point (1, –1) and
satisfies the differential equation, y(1 + xy) dx = x dy, then
(c)
-
æ 1ö
f ç - ÷ is equal to :
è 2ø
1+ x - y
=x+ y-2
1- x + y
(a)
2
5
(c)
-
[2016]
(b)
2
5
4
5
(d) -
4
5
38. If f(x) is a differentiable function in the interval ( (0, ¥) such
2- y
= 2 ( y - 1)
(d) loge
2-x
32.
If
t 2 f (x) - x 2f (t)
= 1 , for each x > 0,
t-x
t ®x
that f(a) = 1 and lim
dy
3
1
æpö 4
æ- p pö
, x Îç
, ÷ , and y ç ÷ = ,
+
y=
2
dx cos 2 x
3
3
cos x
è
ø
è4ø 3
æ pö
then y ç - ÷ equals:
è 4ø
(a)
1
+ e6
3
[10 Jan 2019 I]
(b)
1
3
4
1 3
+e
(d)
3
3
33. The curve amongst the family of curves represented by
the differential equation, (x2 – y2) dx + 2xydy = 0 which
passes through (1, 1), is:
[Jan. 10, 2019 (II)]
(c)
-
æ3ö
then f ç ÷ is equal to :
è2ø
(a)
23
18
[Online April 9, 2016]
(b)
13
6
25
31
(d)
9
18
39. The solution of the differential equation ydx – (x + 2y2)dy
= 0 is x = f(y). If f(–1) = 1, then f(a) is equal to :
[Online April 11, 2015]
(a) 4
(b) 3
(c) 1
(d) 2
(c)
Downloaded from @Freebooksforjeeneet
M-439
Differential Equations
Statement-1: The substitution z = y2 transforms the above
equation into a first order homogenous differential
equation.
Statement-2: The solution of this differential equation is
40. If y(x) is the solution of the differential equation
( x + 2 ) dy = x 2 + 4x - 9, x ¹ -2 and y(0) = 0, then y (–4)
dx
is equal to :
[Online April 10, 2015]
(a) 0
(b) 2
(c) 1
(d) –1
41. Let the population of rabbits surviving at time t be
governed by the differential equation
dp ( t )
If p(0) = 100, then p(t) equals:
(a)
42.
600 - 500 e t 2
(b)
dt
=
y2 e
(a)
(b)
(c)
(d)
1
p ( t ) – 200.
2
[2014]
400 - 300 e -t 2
46.
æ xö
y
+ F ç ÷ , for some function F, is given by
x
è yø
y ln |cx| = x, where c is an arbitrary constant, then F(2) is
equal to:
[Online April 11, 2014]
(b)
43.
(d)
(c) 3500
44.
3
2
(b)
3
2
5
5
(c)
(d) –
2
2
Consider the differential equation :
dy
y3
=
dx 2 xy 2 - x 2
(
)
Let y (x) be a solution of
(a)
( 2 + sin x ) dy = cos x
. If y (0) = 2,
(1 + y ) dx
5
2
[Online May 7, 2012]
(b) 2
7
(d) 3
2
The curve that passes through the point (2, 3), and has the
property that the segment of any tangent to it lying between
the coordinate axes is bisected by the point of contact is
given by :
[2011RS]
6
(a) 2 y - 3x = 0
(b) y =
x
48.
2
(c)
49.
x2 + y 2 = 13
2
æxö æ yö
(d) ç ÷ + ç ÷ = 2
è 2ø è 3ø
Let I be the purchase value of an equipment and V (t) be
the value after it has been used for t years. The value V(t)
depreciates at a rate given by differential equation
dV (t )
= - k (T - t ), where k > 0 is a constant and T is the
dt
total life in years of the equipment. Then the scrap value
V(T) of the equipment is
[2011]
(a)
[Online April 22, 2013]
(d) ln 18
(c)
7
If a curve passes through the point æç 2, ö÷ and has slope
è 2ø
(a) –
1
ln 18
2
æ pö
then y ç ÷ equals
è 2ø
(d) 4500
æ
1 ö
ç 1 - 2 ÷ at any point (x, y) on it, then the ordinate of the
x ø
è
point on the curve whose abscissa is – 2 is :
[Online April 23, 2013]
45.
47.
-
dP
number of workers x is given by
= 100 – 12 x . If the
dx
firm employs 25 more workers, then the new level of
production of items is
[2013]
(a) 2500
(b) 3000
The population p (t) at time t of a certain mouse species
(c)
1
4
At present, a firm is manufacturing 2000 items. It is estimated
that the rate of change of production P w.r.t. additional
(c) – 4
Both statements are false.
Statement-1 is true and statement-2 is false.
Statement-1 is false and statement-2 is true.
Both statements are true.
satisfies the differential equation
y¢ =
(a) 4
=C.
x
dp ( t )
= 0.5 p(t) – 450. If
dt
p (0) = 850, then the time at which the population becomes
ero is :
[2012]
(a) 2ln 18
(b) ln 9
(c) 400 - 300 et 2
(d) 300 - 200 e -t 2
If the general solution of the differential equation
1
4
- y2
I-
kT 2
2
(c) e– kT
(b)
I-
k (T - t )2
2
2
(d) T -
Downloaded from @Freebooksforjeeneet
1
k
EBD_8344
M-440
50.
51.
Mathematics
dy
= y + 3 > 0 and y (0) = 2, then y (ln 2) is equal to :
dx
[2011]
(a) 5
(b) 13
(c) – 2
(d) 7
The solution of the differential equation dy = x + y
dy
x
satisfying the condition y(1) =1 is
[2008]
(a) y = ln x + x
(b) y = x ln x + x2
56.
If
xy '- y = x 2 ( x cos x + sin x), x > 0. If y (p) = p, then
æpö
æpö
y '' ç ÷ + y ç ÷ is equal to :
2
è ø
è2ø
xe(x – 1)
52.
53.
(c) y =
(d) y = x ln x + x
The normal to a curve at P(x, y) meets the x-axis at G. If the
distance of G from the origin is twice the abscissa of P,
then the curve is a
[2007]
(a) circle
(b) hyperbola
(c) ellipse
(d) parabola.
dy
= y (log y – log x + 1), then the solution of the
dx
equation is
[2005]
57.
2+
p
2
(b) 1 +
p p2
+
2 4
(c)
2+
p p2
+
2 4
(d) 1 +
p
2
If for x ³ 0, y = y(x) is the solution of the differential equation,
(x + 1)dy = ((x + 1)2 + y – 3)dx, y(2) = 0,
then y(3) is equal to _____.
58.
[NA Jan. 09, 2020 (I)]
Let y = y(x) be the solution curve of the differential
(
2
equation, y - x
) dydx = 1, satisfying y(0) = 1. This curve
æ xö
(a) y log ç ÷ = cx
è yø
æ yö
(b) x log ç ÷ = cy
è xø
intersects the x-axis at a point whose abscissa is:
æ yö
(c) log ç ÷ = cx
è xø
æ xö
(d) log ç ÷ = cy
è yø
(a) 2 – e
(b) – e
(c) 2
(d) 2 + e
The solution of the equation
d y
dx
2
= e -2 x
[Jan. 7, 2020 (II)]
[2002]
59.
æ
1ö
2
Consider the differential equation, y dx + ç x - ÷ dy = 0 .
yø
è
(a)
e -2 x
4
(b)
e -2 x
+ cx + d
4
If value of y is 1 when x = 1, then the value of x for which
y = 2, is :
[April 12, 2019 (I)]
(c)
1 -2 x
e
+ cx 2 + d
4
(d)
1 -4 x
e
+ cx + d
4
(a)
TOPIC Đ d 2 y
60.
5 1
+
2
e
(b)
1 1
+
2
e
(d)
dy
= (tan x - y )sec2 x , x Î
dx
dx 2
Let y = y(x) be the solution of the differential equation
dy
æ pö
cos x + 2 y sin x = sin 2 x , x Î ç 0, ÷ .
dx
è 2ø
If y (p / 3) = 0, then y (p / 4) is equal to :
[Sep. 05, 2020 (II)]
(a)
2- 2
(b)
(c)
2 -2
(d)
2+ 2
1
2
3 1
2
e
3
- e
2
If y = y(x) is the solution of the differential equation
(c)
55.
[Sep. 04, 2020 (I)]
(a)
If x
2
54.
Let y = y (x) be the solution of the differential equation,
æ p pö
ç - 2 , 2 ÷ , such that y (0) = 0,
è
ø
æ pö
then y ç - ÷ is equal to :
è 4ø
[April 10, 2019 (I)]
(a) e – 2
(b)
1
e
(d)
(c)
2+
1
-e
2
1
-2
e
-1
Downloaded from @Freebooksforjeeneet
M-441
Differential Equations
61.
Let y = y(x) be the solution of the differential equation,
66.
dy æ 2 x + 1 ö
1 -2
-2 x
+ç
÷ y = e , x > 0, where y (1) = e , then
dx è x ø
2
:
[Jan. 11, 2019 (I)]
æ p pö
dy
+ y tan x = 2 x + x 2 tan x , x Î ç - , ÷ , such that y (0)
dx
è 2 2ø
= 1. Then :
[April 10, 2019 (II)]
(a)
æpö
æ pö
y 'ç ÷ + y ' ç - ÷ = - 2
è4ø
è 4ø
(c)
æpö
æ pö
yç ÷ - yç - ÷ = 2
4
è ø
è 4ø
(d)
62.
2
æpö
æ pö p
+2
yç ÷ + yç- ÷ =
è4ø
è 4ø 2
(b)
(b)
y ( log e 2 ) =
log e 2
4
67.
Let f be a differentiable function such that f ¢(x) = 7 -
3 f ( x)
,
4 x
The solution of the differential equation x
æ1ö
(x > 0) and f (1) ¹ 4. Then lim x f ç ÷ :
+
x ®0
è xø
[Jan. 10, 2019 (II)]
(x ¹ 0) with y(1) = 1, is:
(a) exists and equals
dy
+ 2y = x2
dx
[April 09, 2019 (I)]
y(1) =
(a)
(b) y = x + 1
5 5 x2
dy
+ 2x(x2 + 1)y = 1 such that y(0) = 0. If
dx
68.
a
p
, then the value of ‘a’ is : [April 08, 2019 (I)]
32
1
4
(b)
1
2
æ1ö
dy
+ 2 y = x 2 satisfying y(1) = 1, then y ç ÷ is equal to:
dx
è2ø
[Jan. 09, 2019 (I)]
1
7
(a)
(b)
64
4
49
13
(c)
(d)
16
16
69. Let y - y(x) be the solution of the differential equation
x
sin x
æ pö
dy
+ y cos x = 4x, x Î (0, p) . If y ç ÷ = 0 , then
è2ø
dx
æ pö
y ç ÷ is equal to :
è6ø
dy
+ y = x log x, (x > 1). If 2y(2) = log 4 –1, then y(e) is
e
e
dx
equal to :
[Jan. 12, 2019 (I)]
(a)
(d)
x
(a)
-
e
2
(b)
-
e
2
e
e2
(d)
4
4
If a curve passes through the point (1, – 2) and has slope
x2 - 2 y
of the tangent at any point (x, y) on it as
, then the
x
curve also passes through the point :[Jan. 12, 2019 (II)]
(c) (–1, 2)
(b)
(d)
( )
( - 2,1)
-8 2
p
9 3
[2018]
(b)
8
- p2
9
4 2
4
p
- p2
(d)
9
3
9
Let y = y (x) be the solution of the differential equation
(c)
2
(c)
(a) (3, 0)
(b) exists and equals 4.
(c) does not exist.
(d) exists and equals 0.
If y = y(x) is the solution of the differential equation,
1
16
Let y = y(x) be the solution of the differential equation,
(c) 1
4
.
7
3
3 2
1
3
x2
+ 2
(d) y = x + 2
4
4x
4 4x
Let y = y(x) be the solution of the differential equation,
(x2 + 1)2
65.
y ( log e 2 ) = loge 4
æ1 ö
y ( x ) is decreasing in ç ,1÷
è2 ø
(d) y (x) is decreasing in (0, 1)
(c) y =
64.
(a)
(c)
æpö
æ pö
y¢ ç ÷ - y ¢ ç - ÷ = p - 2
è4ø
è 4ø
4 3
1
(a) y = x + 2
5
5x
63.
If y(x) is the solution of the differential equation
70.
dy
+ 2 y = f (x), where f (x) =
dx
ì1, x Î [0, 1]
í
î0, otherwise
æ 3ö
If y (0) = 0, then y ç ÷ is
è 2ø
(a)
3, 0
(c)
e2 – 1
2e
1
2e
3
[Online April 15, 2018]
(b)
(d)
e2 – 1
e3
e2 + 1
Downloaded from @Freebooksforjeeneet
2e 4
EBD_8344
M-442
71.
The curve satisfying the differential equation, ydx–(x +
3y2) dy = 0 and passing through the point (1, 1), also
passes through the point :
[Online April 8, 2017]
(a)
72.
Mathematics
æ1 1ö
ç ,- ÷
è4 2ø
(b)
æ1 1ö
(c) ç , - ÷
è 3 3ø
The solution
of
(a)
æ 1 1ö
ç- , ÷
è 3 3ø
æ1 1ö
(d) ç , ÷
è4 2ø
the differential
x
sec x + tan x
(b) y = 1 +
(b) x2 – 1
x -1
x
x2 - 1
(d)
x2 - 1
x
The general solution of the differential equation
dy 2
+ y = x2 is
[Online May 19, 2012]
dx x
(c)
78.
equaiton
(a)
dy y
tan x
p
+ sec x =
, where 0 £ x < , and y(0) = 1, is
dx 2
2y
2
given by :
[Online April 10, 2016]
(a) y2 = 1 +
1
2
x
sec x + tan x
x
x
(d) y2 = 1 –
sec x + tan x
sec x + tan x
Let y(x) be the solution of the differential equation
dy
(x log x) + y = 2x log x, (x ³ 1). Then y (e) is equal to:
dx
[2015]
(a) 2
(b) 2e
(c) e
(d) 0
74.
75.
dy
+ y tan x = sin 2x and y(0) = 1, then y(p) is equal to:
dx
[Online April 19, 2014]
(a) 1
(b) – 1
(c) – 5
(d) 5
The general solution of the differential equation,
(a)
(b)
80.
81.
(1 + x 2 )
(a)
(c)
77.
dy
+ 2 xy = 4 x 2 is
dx
(1 + x 2 ) y = x3
(1 + x 2 ) y = 3x3
[Online April 25, 2013]
(b)
(d)
3(1 + x 2 ) y = 2 x3
3(1 + x 2 ) y = 4 x3
The integrating factor of the differential equation
(
)
dy
x2 - 1
+ 2 xy = x is
dx
2 e
y e
(b)
3-
[Online May 26, 2012]
1
y
1 e
+
y e
1
1 ey
1 ey
(c) 1 + (d) 1 - +
y e
y e
Solution of the differential equation
p
cos x dy = y (sin x - y ) dx, 0 < x < is
2
(a) y sec x = tan x + c
(b) y tan x = sec x + c
(c) tan x = (sec x + c) y
(d) sec x = (tan x + c) y
Solution
of
the
differential
[2010]
equation
= 0 is
ydx + ( x + x y )dy
(a)
[2004]
log y = Cx
(b)
-
y cot x = tan x + c
(d) y cot x = x + c
y tan x = cot x + c
The equation of the curve passing through the origin and
satisfying the differential equation
x3
5
[2011RS]
2
1
+ log y = C
xy
1
1
=C
(d) + log y = C
xy
xy
The solution of the differential equation
(c)
(c)
76.
4-
1
y
1
æ dy
ö
sin 2x ç - tan x ÷ - y = 0 , is :
dx
è
ø
[Online April 12, 2014]
y tan x = x + c
x2
4
æ
1ö
y 2 dx + ç x - ÷ dy = 0. If y (1) = 1, then x is given by :
è
yø
If
(a)
y = cx 3 -
(b)
y = cx 2 +
(c) y = 1 –
73.
x2
4
x3
(d) y = cx -2 +
5
Consider the differential equation
(c)
79.
y = cx -3 -
82.
(1 + y 2 ) + ( x - e tan
(a)
xe 2 tan
-1
y
-1
= e tan
(b) ( x - 2) = ke 2 tan
(c)
(d)
2 xe tan
xe tan
-1
-1
y
y
y
)
dy
= 0 , is
dx
-1
-1
= e 2 tan
y
+k
y
-1
y
+k
= tan -1 y + k
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[2003]
M-443
Differential Equations
1.
\ Eqn is (x)2 + (y – a)2 = a2
where (0, a) is the centre of circle.
(a) Since, x2 = 4b(y + b)
x2 = 4by + 4b2
2x = 4by’
Þ x 2 + y 2 + a 2 - 2ay = a2
x
Þ b=
2 y'
So, differential equation is
æ xö
2x
.y + ç ÷
x =
y'
è y' ø
2
2x + 2 y
2
2
x2 y2
+
=1
a 2 b2
Since, it passes through (0, 3)
(c)
\
x+ y
Þ
0
9
+
=1
a 2 b2
dy
dy
-a
=0
dx
dx
dy
dx
dy
dx
=a
Put value of 'a' in eqn (1), we get
Þ b2 = 9
é dy
ù
ê y dx + x ú
2
2
x + y - 2y ê
ú =0
ê dy ú
ë dx û
\ eq. of ellipse becomes:
x2 y2
+
=1
9
a2
dy
dy
- 2 y2
- 2 xy = 0
dx
dx
differential w.r.t.x, we get;
Þ ( x2 + y 2 )
2 x 2 y dy
+
=0
9 dx
a2
Þ ( x2 + y 2 - 2 y 2 )
Þ
y
x
æ dy ö – 9
ç ÷= 2
è dx ø a
2
y d y
+
x dx 2
x
dy
= 2xy
dx
dy
= 2xy º g ( x) y
dx
Hence, g(x) = 2x
(b) Statement -1 : y2 = ± 4ax
2
2
Þ (x - y )
Again differentiating w.r.t.x, we get;
3.
dy
dy
- 2a
=0
dx
dx
Þ x+ y
x(y¢)2 = 2yy¢ + x
2.
2
...(1)
Þ x + y - 2ay = 0
Differentiate both side w.r.t 'x', we get
4.
dy
–y
dy
dx
=0
dx
x2
Þ
Þ xyy" + x(y')2 – yy' = 0
(c) Since family of all circles touching x-axis at the
origin
dy
= 4a
dx
Thus both statements are true but statement-2 is not a
correct explanation for statement-1.
(d) Statement - 1
dy
Given differential equations are
+ y 2 = x and
dx
Statement -2 : y2 = 4ax Þ 2 y
5.
(0, a)
1
dy
dy 1
= ± 2a . Þ
µ
dx
y
dx y
d2y
(0, 0)
+ y = sin x
dx 2
Their degrees are 1.
Both have equal degree.
Also, Statement - 2 is the correct explanation for Statement
- 1.
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EBD_8344
M-444
6.
(c) We have y = c1e c2 x
Differentiate it w.r. to x
Þ
y ¢ = c1c2 e c2 x = c2 y
Þ
y¢
= c2 Differentiate it w.r. to x
y
Þ
7.
Mathematics
10.
= 0 Þ y ¢¢ y = ( y ¢)2
y2
(c) Let the centre of the circle be ( h, 2)
\ Equation of circle is
11.
dy
dx
(d) Ax + By = 1
Differentiate w.r. to x
dy
Ax + By
=0
dx
Again differentiate w.r. to x
12.
2
ìï
d2y
dy
æ dy ö üï
x í - By 2 - B ç ÷ ý + By
=0
è
ø
dx
dx
dx
îï
þï
Dividing both sides by –B, we get
d2y
2
dy
æ dy ö
+ xç ÷ - y
=0
2
è dx ø
dx
dx
Therefore order 2 and degree 1.
xy
y 2 = 4a( x - h),
(c)
Differentiating 2 yy1 = 4a Þ yy1 = 2a
Again differentiating, we get
Þ y12 + yy2 = 0
Degree = 1, order = 2.
13.
14.
2
æ 4d3yö
æ
d yö
1
3
+
=ç
(c) ç
÷
÷
d xø
è
è d x3 ø
3
2
æ d3yö
æ
d yö
Þ ç1 + 3
= 16 ç
÷
÷
è
d xø
è d x3 ø
Order = 3, degree 3
3
1 + x 2 × 1 + y 2 = - xy
(a)
...(i)
....(ii)
ò
dy
dx
1 + x2
y
dx = - ò
dy
x
1 + y2
Let x = tan q Þ dx = sec 2 qd q
2
d2y
æ dy ö
+ Bç ÷ = 0
2
è dx ø
dx
From (ii) and (iii)
A + By
x + yy ¢
dy
dy
- 2a
=0 Þa=
y¢
dx
dx
Þ ( x 2 - y 2 ) y ' = 2 xy
dy
.y = 0
Þ x2 + y2 – 2x2 – 2x
dx
9.
......(1)
Þ ( x 2 + y 2 ) y '- 2 xy - 2 y 2 y ' = 0
2
2
........ (iii)
æ x + yy ¢ ö
Putting in (1) we get, x 2 + y 2 - 2 ç
y=0
è y ¢ ÷ø
( y – 2) ç ÷ + ( y – 2) = 25
è dx ø
Þ (y – 2)2 (y')2 = 25 – (y –2)2
(a) General equation of circles passing through origin and
having their centres on the x-axis is
x2 + y2 + 2gx = 0
...(i)
On differentiating w.r.t x, we get
dy
dy ö
æ
2x + 2y .
+ 2g = 0 Þ g = – ç x + y ÷
è
dx
dx ø
Putting in (i)
ì æ
dy ö ü
x2 + y2 + 2 í - ç x + y ÷ ý . x = 0
è
dx ø þ
î
2
x 2 + y 2 - 2ay = 0
Differentiate w.r. to x,
(c)
2x + 2 y
2
Þ y2 = x2 + 2xy
........ (ii)
( y - 2 xy ') 2 = 4 yy '3
Hence equation (iii) is of order 1 and degree 3.
…(1)
( x – h) 2 + ( y – 2) 2 = 25
Differentiating with respect to x, we get
dy
2( x – h) + 2( y – 2) = 0
dx
dy
Þ x – h = –( y – 2)
dx
Substituting in equation (1) we get
8.
........ (i)
Þ y 2 = 2 yy ' ( x + yy ')
On simplifying, we get
y ¢¢ y - ( y ¢ ) 2
2 æ dy ö
2
(c) y = 2c( x + c )
Differentite it w.r. to x
2 yy ' = 2c.1 or yy ' = c
[On putting value of c from (ii) in (i)]
....(iii)
Þò
Þò
sec3 qd q
2y
= -ò
dy
tan q
2 1 + y2
sin 2 q + cos 2 q
sin q × cos 2 q
dq = - 1+ y2
Þ ò (tan q × sec q + cosecq ) d q = - 1 + y 2
Þ sec q + log e | cosec q - cot q | = - 1 + y 2 + C
\ 1 + x 2 + loge
1 + x2 - 1
= - 1 + y2 + C
x
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M-445
Differential Equations
Þ 1+ y2 + 1 + x2 =
15.
1 æ 1 + x 2 + 1ö
ln ç
÷ +C
2 çè 1 + x 2 - 1÷ø
2
(a) Q y = æç x - 1ö÷ cosec x
èp
ø
1
Þ x - (ln( y + 3x )) 2 = C
2
18.
Þ lim
é2 æ 2
ù
ö
= cosec x ê - ç x - 1÷ cot x ú
è
ø
p
ëp
û
dy 2
- cosec x = y cot x
dx p
It is given that,
dy 2
- cosec x = - yp( x )
dx p
By comparision of (i) and (ii), we get
p(x) = cot x
Þ
16.
(b)
Þ f ( x) = xf '( x)
ò
...(i)
f '( x)
1
dx = ò dx
f ( x)
x
log e f ( x) = log e x + loge C
Q f (1) = e
Þ f ( x ) = Cx ,
...(ii)
Þ C = e; so f (x) = ex
When f ( x ) = 1 = ex Þ x =
5 + e x dy
×
= -e x
2 + y dx
19.
1
e
æ y 2 + 1ö
e x dx
dy
=
ò çè y 2 ÷ø
ò ex + 1
(c)
ex
dy
ò 2 + y = - ò 5 + e x dx
Þ y-
Þ log e | 2 + y | × log e | 5 + e x |= log e C
Þ| (2 + y)(5 + e ) |= C
x
1
= log e | e x + 1| +c
y
Q Passes through (0, 1).
Q y (0) = 1
\ c = - log e 2
C = 18.
æ e x + 1ö
Þ y 2 - 1 = y log e ç
÷
è 2 ø
\ (2 + y) × (5 + e ) = 18
x
When x = loge13 then (2 + y) × 18 = 18
æ e x + 1ö
Þ y 2 = 1 + y log e ç
÷
è 2 ø
Þ 2 + y = ±1
\ y = -1, - 3
20.
\ y (ln13) = -1
17.
lim
2tf 2 ( x) - 2 x 2 f (t ) × f '(t )
=0
t®x
1
Using L'Hospital's rule
dy 2
æ2
ö
= cosec x - ç x - 1÷ cosec x × cot x
èp
ø
dx p
Þ
t 2 f 2 ( x ) - x 2 f 2 (t )
=0
t-x
t® x
(a)
x3dy + xy dyx = 2 y dx + x 2 dy
(b)
(a) Let y + 3x = t
Þ ( x3 - x 2 )dy = (2 - x) y dx
dy
dt
+3=
dx
dx
Putting these value in given differential equation
Þ
Þ
Þò
dt
t
=
dx loge t
log e t
dt = ò dx
Þò
t
Þ
2
dy
2-x
dx
= 2
y
x ( x - 1)
(loge t )
= x-C
2
Let
dy
2- x
dx
=ò 2
y
x ( x - 1)
2- x
x 2 ( x - 1)
=
...(i)
A B
C
+
+
x x2 x - 1
Þ 2 - x = A( x - 1) + B( x - 1) + Cx 2
Compare the coefficients of x, x2 and constant term.
C = 1, B = –2 and A = –1
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EBD_8344
M-446
Mathematics
\ò
dy
1 ü
ì -1 2
= òí - 2 +
ý dx
y
x - 1þ
îx x
Þ ln y = - ln x +
22.
2
+ ln | x - 1 | + C
x
\ y (2) = e
Þ v+ x
[Q log e = 1]
Þ 1 = - ln 2 + 1 + 0 + C
Þ C = ln 2
Þ
Þ ln y = - ln x +
2
+ ln | x - 1| + ln 2
x
Þ
1
Þ ln y (4) = - ln 4 + + ln 3 + ln 2
2
23.
3 1/ 2
e
2
(c) The given differential equation is
Þ y (4) =
dy
- 2x
= loge x - 1
y
= tan
( - cos x)dx
2 + sin x
1
1
Þ y=
2
1 + log e 2
(a) f ¢(x) = tan–1 (sec x + tan x)
= tan
2 + sin x dy
= - cos x, y > 0
y + 1 dx
dy
cos x
dx
Þ
=y +1
2 + sin x
Integrate both sides,
ò y +1 = ò
-2
-2 x
= loge x + c Þ
= log e x + c
v
y
Hence, put x =
æ 3ö 1
æ3
ö
Þ ln y (4) = ln ç ÷ + = ln ç e1/ 2 ÷
è 2ø 2
è2
ø
[Q log m + log n = log(mn)]
dv
v2
dv
dx
Þò2 2 =ò
=v+
x
dx
2
v
Put x = 1, y = 2, we get c = –1
At x = 4,
21.
dy 2 xy + y 2
=
dx
2 x2
It is homogeneous differential equation.
\ Put y = vx
(a)
-1 æ 1 + sin x ö
ç
÷ = tan
è cos x ø
æ
æp
öö
ç 1 - cos ç 2 + x ÷ ÷
è
ø÷
ç sin æ p + x ö ÷
ç
÷ ÷
ç
è2
ø ø
è
-1 ç
æ
ç
ö
æp xö
2sin 2 ç + ÷
÷
è4 2ø
÷
ç 2sin æ p + x ö cos æ p + x ö ÷
ç
÷
ç
÷÷
ç
è4 2ø
è 4 2øø
è
-1 ç
ln | y + 1|= - ln | 2 + sin x | + ln C
æ
æ p x öö p x
= tan -1 ç tan ç + ÷ ÷ = +
è 4 2 øø 4 2
è
Þ ln | y + 1| + ln | 2 + sin x |= ln C
Integrate both sides, we get
Þ ln | ( y + 1)(2 + sin x) |= ln C
ò ( f ¢( x) ) dx = ò çè 4 + 2 ÷ø dx
æp
Q y (0) = 1 Þ ln 4 = ln C Þ C = 4
\ ( y + 1)(2 + sin x ) = 4
\y=
Q
2 - sin x
2 - sin p
Þ y (p ) =
=1
2 + sin x
2 + sin p
C = 0 Þ f ( x) =
Þ a =1
Now,
dy
dx
So,
dy (2 + sin x )( - cos x) - (2 - sin x) × cos x
=
dx
(2 + sin x )2
= 1 Þ b = 1.
x =p
p
x2
x+
+C
4
4
f(0) = 0
f ( x) =
4
-1
2 + sin x
Þy=
xö
24.
f (1) =
x2
p
x+
4
4
p +1
4
(d) The given differential equation,
dy
xy
=
dx x 2 + y 2
Ordered pair (a, b) = (1, 1).
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M-447
Differential Equations
Put y = vx Þ
Then, v + x
Þ
1 + v2
v3
27.
dy
dv
=v+ x
dx
dx
ey
\
1
dv = - dx
x
I.F. = e ò
Þ
-1 æ 1 ö
+ ln v = - ln x + c
2 çè v 2 ÷ø
28.
-
2 y2
é
êQ v =
ë
= - ln y + c
yù
x úû
Þ
x = ± 3e
29.
So, x = 3e
25.
(
= e- x
)
( ydx - xdy ) y
x2
æ yö
= xdx Þ - yd ç ÷ = xdx
èxø
2
y æ yö
1 y
- .d ç ÷ = dx Þ - æç ö÷ = x + c
1
x èxø
2è x ø
Þ 2x3 + cx2 + y2 = 0
[Here, c = 2c1]
(c) cos xdy – (sin x) y dx = 6x dx
Þ
dy 1 d
=
sin -1 f ( x )
(b)
dx 2 dx
2y = sin –1 f(x) + C = sin–1 (sin(2tan –1x)) + C
ò d ( y cos x ) = ò 6 x dx Þ
æ æ 2p ö ö
æpö
Þ 2 ç ÷ = sin -1 ç sin ç ÷ ÷ + C
6
è ø
è è 3 øø
Putting x =
p p
= +C \ C=0
3 3
(10 ) ´ æç
æ æ -2p ö ö
for x = - 3 , 2y = sin–1 ç sin ç
÷÷ + 0
è è 6 øø
So, from (1) y cos x = 3x 2 -
1- y2
+
dx
1 - x2
Now, put x =
= 0 Þ sin –1y + sin –1x = c
1
p
3
At x = , y =
Þ c=
2
2
2
Þ sin–1y = cos–1x
æ
1 öö
æ 1 ö
-1 æ
Hence, y ç ÷ = sin ç cos ç ÷÷
2ø
2 øø
è
è
è
æ
æ 1 öö 1
= sin ç p - cos -1 ç
÷÷ =
2 øø
2
è
è
p
and y = 0 in eq. (1), we get
3
-p2
1 ö 3p 2
+C ÞC
÷=
9
3
è2ø
-p
-p
Þ y=
3
6
(c) The given differential eqn. is
dy
y cos x = 3 x 2 + C ...(1)
æpö
Given, y ç ÷ = 0
è3ø
Þ 2y =
26.
-1.dx
Put x = 0, y = 0, then we get c = 1
ey – x = x + 1
y = x + loge(x + 1)
Put x = 1 \ y = 1 + loge2
(b) Given differential equation can be written as,
y2dx – xydy = x3dx
Þ
1
When x = 1, y = 1, then - = c
2
Þ x2 = y2(1 + 2 ln y)
At y = e, x2 = e2(3)
Þ
é
y dy
y
xù
êëQ e dx - e = e úû
dt
- t = ex
dx
t (e - x ) = ò e x .e- x dx Þ ey – x = x + c
-1
æ 1 1ö
ò çè v3 + v ÷ødv = ò x dx
Þ
dy dt
=
dx dx
dv
vx 2
v
= 2 2 2 =
dx x + v x
1+ v2
Þ
x2
(a) Let ey = t
y
30.
p
in the above equation,
6
3 3p2 p2
=
Þ
2
36
3
(1) Given
p2
3
-p2
3 y -3p2
=
Þ y=
2
12
2 3
dy 2 y
=
dx x 2
Integrating both sides,
ò
dy
dx
= 2ò 2
y
x
2
Þ ln | y | = - + C
...(i)
x
Equation (i) passes through the centre of the circle x2 + y2
– 2x – 2y = 0, i.e., (1, 1)
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EBD_8344
M-448
Mathematics
\ C=2
31.
\
-1
1
ln |1 - 3 y | = tan x – 1 – ln 3
3
3
2
Now, ln | y | = - + 2
x
x ln | y | = – 2 (1 – x) Þ x ln | y | = 2 (x – 1)
(b) The given differential equation
Put, x = –
dy
= (x – y)2
dx
Þ
¼(1)
Let x – y = t Þ 1 –
Þ
dy dt
=
dx dx
dy
dt
=1–
dx
dx
dt ö
æ
çè 1 – ÷ø = (t)2
dx
Þ 1 – t2 =
33.
dt
dt
Þ ò dx = ò
1– t 2
dx
æ y2 ö
= - ò dx
x ø÷
y2
= –x + C
...(1)
x
Since, the above curve passes through the point (1, 1)
1– x + y
1– y + x
dy
= sec2 x(1 – 3y)
dx
dy
1
- ln |1 - 3 y | = tan x + C ...(i)
3
Q
æpö 4
yç ÷ =
è4ø 3
Þ
p
1
- ln |1 - 4| = tan + C
3
4
Þ
1
1
- ln 3 = C + 1 Þ C = –1 – ln 3
3
3
(Given)
12
= –1 + C Þ C = 2
1
Now, the curve (1) becomes
y2 = –x2 + 2x
Þ y2 = –(x – 1)2 + 1
(x – 1)2 + y2 = 1
The above equation represents a circle with centre (1, 0)
and centre lies on x-axis.
(a) f(xy) = f(x).f(y) ...(1)
Put x = y = 0 in (1) to get f(0) = 1
Put x = y = 1 in (1) to get f(1) = 0 or f(1) = 1
f(1) = 0 is reected else y = 1 in (1) gives f(x) = 0
imply f(0) = 0.
Hence, f(0) = 1 and f(1) = 1
By first principle derivative formula,
Then,
1
dy
3
y=
+
(a) Given,
dx cos 2 x
cos 2 x
Þ
(a) (x2 – y2)dx + 2xy dy = 0
y2dx – 2xydy = x2dx
2xydy – y2dx = –x2dx
d(xy2) = –x2 dx
ò d çè
1 1– 1 – 1
ln
+ c Þ c = –1
2 1 –1+1
ò (1 - 3 y) = ò sec 2 x dx
1
1
1
- y = 6 Þ - y = e6 Þ y = ± e6
3
3
3
æ y2 ö
d ç ÷ = –dx
è xø
Q The given condition y(1) = 1
Þ
ln
xd ( y 2 ) - y 2 d ( x )
= –dx
x2
1 x – y –1
Þ –x = ln
+c
2 x – y +1
Hence, 2(x – 1) = – ln
1
æ pö
1
- ln |1 - 3 y | = tan ç - ÷ - 1 - ln 3
3
3
è 4ø
Þ
1
t –1
ln
Þ –x =
+c
2 ´1 t +1
–1 =
p
4
1
= -1 - 1 - ln 3
3
Þ ln |1 – 3y| = 6 + ln 3
Now, from equation (1)
32.
in eq. (i), we get
34.
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M-449
Differential Equations
h® 0
æ
ç
lim f ( x) ç
= h®0
ç
çè
ö
æ hö
f ç1 + ÷ - f (1)
÷
è xø
÷
h
÷
÷ø
Þ f ¢(x) =
f ( x)
f ¢ (1)
x
Þ
(2 + sin x) (y + 1) = C
At x = 0, y = 1 we have
(2 + sin 0) (1 + 1) = C
Þ C=4
4
Þ y+ 1 =
2 + sin x
4
-1
y=
2 + sin x
f ( x + h) - f ( x)
h
f ¢(x) = lim
f ¢( x ) k
= Þ ln f(x) = k ln x + c
f ( x) x
f(1) = 1 Þ ln1 = k ln 1 + c Þ c = 0
Þ ln f(x) = k ln x Þ f(x) = xk but f(0) = 1
Þ k=0
\ f(x) = 1
37.
xdy - ydx
y2
35.
1
x x2
- = + C as y(1) = –1 Þ C =
2
y 2
æ 1ö
æ 3ö 1
3
y ç ÷ + y ç ÷ = +1 + + 1 = 3
è 4ø
è 4ø 4
4
(b) (x2 – y2) dx + 2xydy = 0
dy y 2 - x 2
=
dx
2 xy
Let y = vx
Hence, y =
38.
Þ
v+x
Þ
x
æ -1 ö 4
Þfç ÷=
è 2 ø 5
x +1
2
t 2 f ( x ) - x 2 f (t )
=1
t-x
t® x
Applying L.H. rule
(d) Let L = lim
2t f ( x) - x 2 f ¢ (t )
=1
1
t® x
2x f (x) – x2 f ¢(x) = 1
solving above differential equation, we get
2 2 1
f (x) = x +
3
3x
dv v 2 x 2 - x 2
dv v 2 - 1
=
Þ v+x
=
2
dx
dx
2v
2vx
dv - v 2 - 1
=
dx
2v
Put x =
2vdv
dx
=x
v2 + 1
After integrating, we get
ln | v2 + 1| = – ln | x | + lnc
Þ
y2
c
+1=
x
x2
As curve passes through the point (1, 1), so 1 + 1 = c
Þc=2
x2 + y2 – 2x = 0, which is a circle of radius one.
36.
-2x
L = lim
dy
dv
=v+ x
dx
dx
Þ
= xdx
æxö
Þ ò -d ç ÷ = ò xdx
è yø
dy
= f(x) = 1 Þ y = x + c, y(0) = 1 Þ c = 1
dx
Þ y=x+1
\
4
1
4
æ pö
- 1 = -1 =
Now y ç ÷ =
p
3
3
2
è ø 2 + sin
2
(b) y (1 + xy)dx = xdy
(b) We have (2 + sinx)
dy
+ (y + 1) cos x = 0
dx
d
(2 + sin x)(y + 1) = 0
Þ
dx
On integrating, we get
3
2
2
39.
3 2 27 + 4 31
æ3ö 2 æ3ö 1 2
=
f ç ÷ = ´ç ÷ + ´ = + =
2 9
18
18
è2ø 3 è2ø 3 3
(b) Given differential equation is
ydx – (x + 2y2) dy = 0
Þ ydx – xdy – 2y2dy = 0
ydx - xdy
Þ
= 2dy
y2
æ xö
Þ d ç ÷ = 2dy
è yø
Integrate both the side
x
= 2y + c
Þ
y
using f(–1) = 1, we get
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EBD_8344
M-450
Mathematics
c=1
x
Þ
= 2y + 1
y
put y = 1, we get f(a) = 3
40.
(a)
dy
( x + 2 ) = x 2 + 4x - 9 x ¹ -2
dx
t
p(t )
- 200 = e 2 k
2
Using given condition p(t) = 400 – 300 et/2
Þ
42.
dy y
æ yö
= + fç ÷
è xø
dx x
dy
dv
= x +v
dx
dx
x 2 + 4x - 9
dy =
dx
x+2
or
x 2 + 4x - 9
ò dy = ò x + 2 dx
from (1) & (2), x
13 ö
æ
y = òç x + 2 ÷dx
x+2ø
è
1
dx
y = ò (x + 2) dx - 13 ò
x+2
or,
x2
+ 2x – 13 log | x + 2 | + c
2
Given that y = (0) = 0
0 = – 13 log 2 + c
ò
...(2)
dv
æ 1ö
+ v = v + fç ÷
è vø
dx
dx
dv
=
x
æ 1ö
fç ÷
è vø
Integrating both sides, we get
y=
dx
=
x
ò
dv
dv
Þ ln x + c = ò
æ 1ö
æ 1ö
fç ÷
fç ÷
è vø
è vø
(where c being constant of integration)
x
But, given y =
is the general solution
ln | cx |
x2
y=
+ 2x – 13 log |x + 2| + 13 log 2
2
y(–4) = 8 – 8 – 13 log 2 + 13 log 2 = 0
so that
(c) Given differential equation is
dp (t ) 1
= p (t ) - 200
dt
2
By separating the variable, we get
é1
ù
dp(t) = ê p(t ) - 200ú dt
2
ë
û
dp(t )
Þ
= dt
1
p (t ) - 200
2
Integrate on both the sides,
d ( p (t ))
= ò dt
ò1
p (t ) - 200
2
1
dp(t )
= ds
Let p(t ) - 200 = s Þ
2
2
d p (t )
= dt
So, ò 1
æ
ö ò
(
)
200
p
t
ç
÷
è2
ø
2ds
= ò dt Þ 2 log s = t + c
Þ ò
s
æ p (t )
ö
- 200÷ = t + c
Þ 2log ç
è 2
ø
...(1)
æ yö
Let çè ÷ø = v so that y = xv
x
dy x 2 + 4x - 9
=
dx
x+2
41.
(d) Given
x
1
=
= ln|cx| =
v
y
ò
dv
æ 1ö
fç ÷
è vø
Differentiating w.r.t v both sides, we get
æ xö
æ 1ö
y2
-1
fç ÷ =
Þ
f
=
ç
÷
è vø
è yø
v2
x2
2
when
43.
2
x
æ yö
æ 1ö
æ -1ö
= 2 i.e. f(2) = - ç ÷ = - ç ÷ = ç ÷
è xø
è 2ø
è 4ø
y
(c) Given, Rate of change is
dP
= 100 - 12 x
dx
Þ dP = (100 – 12 x ) dx
By integrating
ò dP = ò (100 - 12
x ) dx
P = 100x – 8x3/2 + C
Given when x = 0 then P = 2000
Þ C = 2000
Now when x = 25 then
P = 100 × 25 – 8 × (25)3/2 + 2000 = 4500 – 1000
Þ P = 3500
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M-451
Differential Equations
44.
Þ
ò
Putting in (i)
1
dy
=1- 2
dx
x
1 ö
æ
ç1 - 2 ÷ dx
è x ø
(a) Slope =
dy = ò
1
+ C , which is the equation of the curve
x
æ 7ö
since curve passes through the point ç 2, ÷
è 2ø
7
1
\ = 2+ +C Þ C=1
2
2
1
\ y = x + +1
x
1
-3
+1 =
when x = – 2, then y = -2 +
2
-2
(d) Given differential equation is
47.
dy
cos x dx
=ò
1+ y
2 + sin x
Þ log (1 + y) = log (2 + sin x) + log C
Þ 1 + y = C (2 + sin x)
Given y(0) = 2
3
Þ 1 + 2 = C[2 + sin 0] Þ C =
2
æ pö
Now, y ç ÷ can be found as
è 2ø
9
pö
3æ
1 + y = ç 2 + sin ÷ Þ 1 + y =
2
2è
2ø
7
Þ y=
2
æ pö 7
Hence, y ç ÷ =
è 2ø 2
(b) Equation of tangent at P
dy
Y–y=
( X - x)
dx
ò
x2
2
46.
0 = 900 - 50e 2
\ t1 = 2ln 18
(c) Given differential equation is
(2 + sin x) dy
. = cos x
(1 + y ) dx
which can be rewritten as
dy
cos x
dx
=
1 + y 2 + sin x
Integrate both the sides, we get
2
Now, y 2 e - y / x = C satisfies the given diff. equation
\ It is the solution of given diff. equation.
Thus, statement-2 is also true.
(a) Given differential equation is
dp (t )
= 0.5 p(t ) - 450
dt
dp(t ) 1
= p(t ) - 450
Þ
dt
2
dp (t ) p (t ) - 900
=
Þ
dt
2
dp(t )
= - ëé900 - p ( t ) ûù
Þ 2
dt
dp(t )
= -dt
Þ 2
900 - p(t )
Integrate both the side, we get
Þ
p(t ) = 900 - 50e 2
t1
x é xù
æxö
= ê1 - ú » F ç ÷
èzø
z2 z ë z û
Hence, statement-1 is true.
dx x
= dz z
900 - p (t ) = 50e 2
let p (t1) = 0
dz
2z
z
=
=
dx 2( xz - x 2 ) xz - x 2
Now,
Þ
t
dy
y3
=
dx 2( xy 2 - x 2 )
By substituting z = y2, we get diff. eqn. as
2
é æ 900 - p ( t ) ö ù
2 êln ç
÷ø ú = t
50
ë è
û
t
Þ y = x+
45.
Þ
48.
Y
B (0, y – xdy/dx )
P
dp(t )
- 2ò
= dt
900 - p(t ) ò
Let 900 – p (t) = u
Þ – dp (t) = du
du
= dt Þ 2 ln u = t + c ...(i)
u ò
Þ 2ln [900 – p(t)] = t + c
Given t = 0, p (0) = 850
2 ln (50) = c
2ò
X¢
( x, y)
O
A(x – y)/(dy/dx), 0)
Y¢
y
dy / dx
xdy
Y-intercept = y –
dx
Since P is mid-point of A and B
X-intercept = x -
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X
EBD_8344
M-452
Mathematics
y
xdy
= 2 x and y = 2y
dy
dx
- xdy
-y
=y
= x and
Þ
dy
dx
dx
dx
dy
Þ
=x
y
lny = – lnx + lnc
c
y=
x
Since the above line passes through the point (2, 3).
\ c=6
6
Hence y = is the required equation.
x
dV (t )
= -k (T - t )
(a)
dt
x-
49.
52.
dx
( X - x)
dy
Coordinate of G at X axis is (X, 0) (let)
\ 0 – y = – dx ( X - x )
dy
dy
Þ y
= X -x
dx
dy
Þ X=x+y
dx
dy ö
æ
\ Co-ordinate of G çè x + y , 0÷ø
dx
Given distance of G from origin = twice of the abscissa of
P.
Q distance cannot be –ve, therefore abscissa x should be
+ve
dy
dy
\ x+ y
= 2 x Þ y dx = x
dx
Þ ydy = xdx
Y–y=–
Þ ò dV (t ) = -k ò (T - t )dt
V (t ) =
k (T - t )2
+c
2
at t = 0, V (t) = I
I =
kT 2
+c
2
kT 2
2
k
Þ V (t ) = I + (t 2 - 2tT )
2
k 2
k
V (T ) = I + (T - 2T 2 ) = I - T 2
2
2
On Integrating, we have
Þc=I -
50.
(d)
dy
= y+3 Þ
dx
53.
dy
ò y + 3 = ò dx
xdy
= y (log y – log x + 1)
dx
Put y = vx
dy
xdv
xdv
=v+
Þ v+
= v (log v + 1)
dx
dx
dx
dv
dx
xdv
= v log v Þ ò
=ò
v log v
x
dx
Put log v = z
1
dz
dx
dv = dz Þ ò
=
v
z ò x
ln z = ln x + ln c
æ yö
x = cx or log v = cx or log ç ÷ = cx.
è xø
Þ ln y + 3 = x + ln5
Put x = ln 2 , then ln |y + 3| = ln 2 + ln5
Þ ln | y + 3 | = ln 10
\ y + 3 = ±10 Þ y = 7, – 13
dy x + y
y
=
= 1+
dx
x
x
It is homogeneous differential eqn.
dy
dv
=v+x
Putting y = vx and
dx
dx
we get
dv
dx
v+x
= 1+ v Þ ò
= dv
x ò
dx
Þ v = ln x + c Þ y = xlnx + cx
(c)
dy y æ
æ yö ö
=
log ç ÷ + 1÷
è xø ø
dx x çè
Given y (0) = 2, \ l n 5 = c
(d)
y 2 x2
=
+ c1
2
2
Þ x2 – y2 = –2c1
\ the curve is a hyperbola
Þ ln y +3 = x+ c
51.
As y(1) = 1
\ c = 1 So solution is y = x lnx + x
(b) Equation of normal at P(x, y) is
54.
(b)
-2 x
= e-2 x ; on integration dy = e
+c;
dx
-2
dx
d2y
2
Again integrate we get y =
55.
(c)
e -2 x
+ cx + d
4
dy
+ 2 y tan x = 2 sin x
dx
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M-453
Differential Equations
ò
I.F. = e
2 tan xdx
Q
\
= sec 2 x
The solution of the differential equation is
3
é
ù
-3
Then, y = (1 + x) ê x +
ë 1 + x úû
ò
y × I.F. = I.F. ´ 2sin x dx + C
ò
Þ y × sec 2 x = 2sin x × sec 2 x dx + C
Þ y sec2 x = 2sec x + C
When x =
3
é
ù
-3 = 3
Now, at x = 3, y = (1 + 3) ê3 +
ë 1 + 3 úû
...(i)
p
, y = 0; then C = – 4
3
58.
\ From (i), y sec 2 x = 2sec x - 4
Þy=
56.
2sec x - 4
sec 2 x
I.F. =
1
- ò dx
e x
=
æpö
Þ yç ÷ = 2 -2
è4ø
= y2e y – 2(y.e y – e y) + C
Þ x.ey = y2ey – 2yey + 2ey + C
Þ x = y2 – 2y + 2 + C. e–y
As y(0) = 1, satisfying the given differential eqn,
\ put x = 0, y = 1 in eqn. (i)
0 = 1- 2 + 2 +
Q y (p) = p Þ C = 1
2
p
æ pö p
+
y = x 2 sin x + x Þ y ç ÷ =
è 2ø
4 2
59.
p
æ pö
y '' = 2 sin x - x 2 sin x Þ y '' ç ÷ = 2 è 2ø
4
2
p2 p2 p
p
æ pö
æ pö
\ y '' ç ÷ + y ç ÷ = 2 +
+ = 2+
è 2ø
è 2ø
4
4 2
2
C=–e
y = 0, x = 0 – 0 + 2 + (– e) (e–0)
x=2–e
(b) Consider the differential equation,
1
dx æ 1 ö
Þ dy + çç 2 ÷÷ x = 3
y
y
è ø
1
dy
æ y -3ö
= (1 + x ) + ç
÷
dx
è 1+ x ø
\ x.e
1
3
dy
y = (1 + x) dx (1 + x )
(1 + x)
Put -
I.F. = e
\
-ò
1
dx
(1+ x )
=
1
(1 + x )
d æ y ö
3
ç
÷ = 1dx è 1 + x ø
(1 + x )2
é
ù
3
y = (1 + x) ê x +
+ Cú
(1 + x )
ë
û
1
dy
I.F. = ò y 2
e
=e y
(c) (x + 1)dy = ((x + 1)2 + (y – 3))dx = 0
Þ
C
e
æ
1ö
y2dx + ç x - ÷ dy = 0
yø
è
y ' = 2 x sin x + x 2 cos x + 1
57.
dx
+ Px = Q, where P = 1, Q = y2
dy
x.e y = ò ( y 2 )e y . dy = y 2 .e y - ò 2 y.e y .dy
1
x
y
= x sin x + C
x
dx
+ x = y2
dy
Now, I.F. = e ò1.dy = e y
æ yö
\ ò d ç ÷ = ò ( x cos x + sin x)dx
è xø
Þ
(a) The given differential equation is
Comparing with
dy y
- = x( x cos x + sin x)
dx x
(a)
At x = 2, y = 0
0 = 3(2 + 1 + C) Þ C = – 3
Þ
-
1
y
= òe
1
y
1
y3
dy + c
1
1
= u Þ 2 dy = du
y
y
-
x.e
1
y
= - ò ueu du + c = -ueu + eu + c
1
Þ
-
1
- æ
1 ö
x.e y = e y ç + 1 ÷ + c
èy ø
At y = 1, x = 1
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...(i)
EBD_8344
M-454
Mathematics
æpö
æ pö
Þ y¢ ç ÷ – y¢ ç - ÷ = p - 2
è4ø
è 4ø
1
1 = 2 + ce Þ c = -
1
æ 1ö 1
Þ x = ç1 + ÷ - e y
e
yø e
è
3 1
On putting y = 2, we get x = 2
e
60.
62.
dy
+ y sec2 x = sec 2 ´ tan x
dx
Given equation is linear differential equation.
2
ò dx
differential equation, then I .F = e x = x 2
Now, the solution of the linear differential equation
(a)
sec
IF = e ò
2
xdx
dy 2
+ y = x y(1) = 1 (given)
dx x
Since, the above differential equation is the linear
(c)
y ´ x 2 = ò x3dx
= e tan x
Þ y.etanx = ò e tan x sec 2 ´ tan xdx
Þ yx 2 =
Put tan x = u = sec2x dx = du
Q y (1) = 1
yetan x = ò euu du Þ yetan x = ueu – eu + c
3
1
\1´ 1 = + C Þ C =
4
4
\ Solution becomes.
Þ yetan x = (tan x – 1) etan x + c
Þ y = (tan x – 1) + c. e– tan x
\ y (0) = 0 (given) Þ 0 = – 1 + c Þ c = 1
Hence, solution of differential equation,
æ pö
yç- ÷ = – 1 – 1 + e = – 2 + e
è 4ø
61.
(d) Given differential equation is,
dy
+ y tan x = 2 x + x 2 tan x
dx
Here, P = tan x, Q = 2x + x2 tan x
tanxdx
= eln|sec x| =| sec x |
I.F. = e ò
\ y (sec x) = ò (2 x + x 2 tan x )sec xdx
=
òx
2
tan x sec xdx + ò 2 x sec xdx = x2sec x + c
Given y (0) = 1 Þ c = 1
\ y = x2 + cos x
...(i)
Now put x =
p
-p
and x =
in equation (i),
4
4
2
2
1
1
æpö p
æ pö p
yç ÷ =
+
+
and y ç - ÷ =
2
2
è 4 ø 16
è 4 ø 16
æpö
æ pö
Þ yç ÷ - yç- ÷ = 0
è4ø
è 4ø
dy
= 2 x - sin x
dx
p 1
æpö p 1
æ pö
\ y¢ ç ÷ = and y¢ ç - ÷ = - +
2
2
2
è4ø 2
è 4ø
x4
+C
4
y=
63.
3
x2
+
4 4 x2
(d) (1 + x2)2
Þ
dy
+ 2x (1 + x2) y = 1
dx
dy æ 2 x
+
dx çè 1 + x 2
1
ö
÷y =
(1 + x 2 )2
ø
Since, the above differential equation is a linear differential
equation
\
2x
I.F. = ò 1+ x2 dx = elog(1+ x 2 ) = 1 + x2
e
Then, the solution of the differential equation
Þ y (1 + x2) =
dx
ò 1 + x2 + c
Þ y (1 + x2) = tan–1 x + c
If x = 0 then y = 0 (given)
Þ 0=0+c
Þ c=0
Then, equation (1) becomes,
...(1)
Þ y (1 + x2) = tan–1 x
Now put x = 1 in above equation, then
2y =
p
4
Þ
æ p ö p
2ç
÷=
è 32 a ø 4
pù
é
êë a y (1) = 32 úû
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M-455
Differential Equations
Þ
a=
Then, equation of curve
1
4
y=
1
16
(c) Consider the differential equation,
Þ a=
64.
Since, above curve satisfies the point.
Hence, the curve passes through
dy y
+ = logex
dx x
66.
IF = e ò x dx = x
\
yx =
ò x ln x dx
2x
-2 x
y( x) × x e2 x = ò x e × e dx + c
2
Þ y(e) =
\
1 -2 2
1
e × e ×1 = + c Þ c = 0
2
2
\
y(x) =
x 2 e -2 x
×
x
2
x -2 x
y(x) = × e
2
Differentiate both sides with respect to x,
x2 - 2 y
x
dy x 2 - 2 y
=
dx
x
y¢(x) =
e -2 x
æ1 ö
(1 - 2 x ) < 0 " x Î ç , 1÷
è2 ø
2
æ1 ö
Hence, y(x) is decreasing in çè , 1÷ø
2
dy 2
+ y =x
dx x
2
x2
+c
2
1 -2
Given, y(1) = e
2
e e e
- =
2 4 4
(b) Q Slope of the tangent =
Q
y ( x) × e 2 x × x =
x
x
ln x 2
4
Þ y=
65.
= ò x dx + c
x2
x2
ln x 2
4
Þ xy =
(c) Given differential equation is,
æ 1ö
ò ç 2+ ÷ dx
è
ø
x
x
× ln x - + c
2
4
Given, 2y(2) = loge 4 – 1.
\ 2y = 2 ln 2 – 1 + c
Þ ln4 – 1 = ln4 – 1 + c
i.e. c = 0
Þ xy =
67.
dx
I.F. = e ò x = e2ln x = x 2
(b) Let y = f(x)
Solution of equation
dy æ 3 ö
+ç ÷ y =7
dx è 4 x ø
y × x2 = ò x × x 2dx
I.F. =
x4
+C
4
\ curve passes through point (1, –2)
x2y =
(1)2 (–2) =
Þ C=
14
+C
4
-9
4
)
3, 0 .
x
IF = e
= e2x + ln x = xe2x
Complete solution is given by
x2
1 x2
- ò × dx
2
x 2
2
(
dy æ
1ö
+ ç 2 + ÷ y = e–2x, x > 0
è
dx
xø
1
Q
Þ xy = ln x ×
9
x2
4 4 x2
e
ò
3
dx
4x
3
= e4
ln x
æ 3ö
ç ÷
= xè 4 ø
Solution of differential equation
3
3
y × x 4 = ò 7 × x 4 dx + C
7
7
x4
3 7×
+ C = 4x4 + C
y × x4 = æ 7ö
çè ÷ø
4
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EBD_8344
M-456
Mathematics
3
y = 4 x + Cx - 4
Then, ysin x = 2x 2 3
æ 1ö 4
Þ f ç ÷ = + Cx 4
è xø x
7ö
æ
æ 1ö
lim+ x × f ç ÷ = lim+ ç 4 + Cx 4 ÷ = 4
è xø x®0 è
x ®0
ø
Þ
68.
(c) Since, x
Þ
p2
is the solution
2
2
p2 ö
8p2
æpö æ p
\ y ç 6 ÷ = çç 2. 36 - 2 ÷÷ 2 = - 9
è ø è
ø
70.
(a) When x Î [0, 1], then
Q y (0) = 0 Þ y (x) =
dy
+ 2 y = x2
dx
dy 2
+ y =x
dx x
Here, y (1) =
1 1 –2 x
– e
2 2
1 1 –2 e 2 – 1
– e =
2 2
2e 2
2
I.F. = e ò x dx = e2 ln x = e lnx 2 = x2.
Solution of differential equation is:
When x Ï [0, 1], then
y × x2 = ò x × x 2dx
Q y (1) =
x4
+ C ...(1)
4
y(1) = 1
71.
3
C=
4
\
Þ
x4 3
y×x = +
4 4
69.
\
3
x2
+
y=
4 4 x2
2x 2
+ c ...(2)
sin x
y(x) =
Q
æp ö
eq. (2) passes through ç ,0 ÷
è2 ø
p2
+C
2
p2
2
Now, put the value of C in (1)
ÞC= -
= e - ln y =
solution is
1
y
x
1
= 3 y . dy
y ò
y
x
= 3y + c
y
which passes through (1, 1)
\ 1 = 3 + c Þ c = –2
\ solution becomes
Þ x = 3y2 – 2y
(b) Consider the given differential equation the
sinxdy + ycosxdx = 4xdx
Þ d(y.sinx) = 4xdx
Integrate both sides
Þ y.sinx = 2x2 + C ...(1)
Þ
1
- ò dy
y
Þ
æ 1ö 1
49
yç ÷ = + 3 =
è 2 ø 16
16
Þ0=
dx x
dx x
- = 3y
= + 3y Þ
dy y
dy y
if = e
2
\
e2 – 1 e2 – 1 2 –2
e2 – 1
Þ
= c e Þ C2 =
2
2
2
(b) y dx – x dy – 3y2 dy = 0
Then, from equation (1)
\
dy
+ 2 y = 0 Þ y = c e –2x
2
dx
2
æ e 2 – 1 ö –2 x
æ3ö e –1
\ y (x) çç
÷÷ e Þ y ç ÷ =
è2ø
2e3
è 2 ø
y × x2 =
Q
dy
1
+ 2 y = 1Þ y = + C1e –2 x
dx
2
72.
æ 1 1ö
which also passes through ç - , ÷
è 3 3ø
dy y
tan x
+ sec x =
(d)
dx 2
2y
dy
2y
+ y2 sec x = tan x
dx
dt
dy
Put y2 = t Þ 2y
=
dx
dx
dt
+ t sec x = tan x
dx
sec xdx
In(sec x + tan x )
I.f = e ò
= sec x + tan x
=e
Downloaded from @Freebooksforjeeneet
M-457
Differential Equations
- cosec2x
Now, integrating factor (I.F) = e ò
dt
(sec x + tan x) + t sec x (sec x + tan x)
dx
= tan x(sec x + tan x)
or, I.F = e
ò d (t (sec x + tan x)) = ò tan x (sec x + tan x) dx
73.
x
x
Þ y2 = 1 –
sec x + tan x
sec x + tan x
(a) Given,
log(log x)
=e
y. logx = ò 2 log xdx + c
I.F. = e
(c) Let
\ y cot x =
76.
2sin x cos x
dx + c
cos x
77.
y (sec x) = –2cos x + c
...(1)
Given y(0) = 1
\ put x = 0 and y = 1, we get
1 (sec 0) = – 2 cos 0 + c
Þc=1+2Þc=3
\ from eqn (1), we have
y sec x = –2 cos x + 3
...(2)
To find y (p), put x = p in eqn (2), we get
y (secp) = – 2 cos p + 3
y = – 2 (–1) (–1) + 3 (–1) = –2 – 3 = – 5
æ dy
ö
(d) Given, sin 2 x ç - tan x ÷ - y = 0
è dx
ø
dy
y
+ tan x
=
dx
sin 2 x
dy
or,
- y cosec2 x = tan x
dx
(d) Given differential equation is
\ Solution is
4x2
´1 + x 2 + C
y(1 + x2) = ò
1 + x2
4 x3
+C
Þ y(1 + x2) =
3
Þ Required curve is
3y (1 + x2) = 4x3 (Q C = 0)
y (sec x) = 2ò sin x dx + c
75.
cot x dx + c
This is linear diff. equation
2x
dx
ò
2
+
1
x2
I.F = e
= elog (1+ x ) = 1 + x 2
tan x dx
ò
ò tan x .
ò 1.dx + c
dy
+ 2 xy = 4 x 2
dx
dy æ 2 x ö
4 x2
+ç
y
=
Þ
dx è 1 + x2 ÷ø
1 + x2
= e - log cos x = sec x
I.F = eò
Required solution is
y (sec x) =
) -1
(1 + x 2 )
dy
+ y tan x = sin 2x
dx
ò sin 2 x sec x dx + c
tan x
\ y cot x = x + c
= log x
y (sec x) =
(
ò Q(I.F.) dx + c
\ y cot x =
y logx = 2[x log x – x] + c
Put x = 1, y.0 = –2 + c
c=2
Put x = e
y loge = 2e(log e – 1) + c
y(e) = c = 2
74.
log
1
y(I. F.) =
dy æ 1 ö
+
y=2
dx çè x log x ÷ø
1
ò x log x dx
= e
= cot x
tan x
Now, general solution of eq. (1) is written as
=
t (sec x + tan x) = sec x + tan x – x
t=1–
1
- log|tan x|
2
or,
(b) Given differential equation is ( x 2 - 1)
dy
x
2x
+
.y = 2
dx x 2 - 1
x -1
This is in linear form.
Þ
Integrating factor = ò
78.
2x
dt where t = x2 – 1
e t
dx = ò
e x -1
= elog t = x2 – 1
Hence, required integ