REVISION ASSIGNMENT
JEE (Main+Advanced) 2024
ENTHUSIAST COURSE
Revision Assignment # 01
Mathematics
Topic : Quadratic Equation/FOA/Set Theory and Number System
ONE OR MORE THAN ONE
1.
If 𝛼 and 𝛽 are the roots of the equation 𝑥 2 − 𝑥 + 1 = 0, then 𝛼 2009 + 𝛽 2009 is greater than or equal
to
(A) –2
2.
(B) –1
(C) 1
(D) 2
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 :
(A) 𝑝(2) = 19
3.
4.
(B) 𝑝(−2) = 19
(C) 𝑝(−2) = 11
(D) 𝑝(2) = 15
If 𝑏 > 𝑎 and 𝛼, 𝛽(𝛼 < 𝛽) are the roots of the equation, (𝑥 − 𝑎)(𝑥 − 𝑏) − 1 = 0, then (A) 𝛼  [𝑏, )
(B) 𝛼  (–, 𝑎)
(C) 𝛽  (b, )
(D) 𝛼  [𝑎, 𝑏] & 𝛽  [𝑎, 𝑏]
If 𝛼, 𝛽 are the roots of 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, and 𝛼 + ℎ,𝛽 + ℎ are the roots of 𝑝𝑥 2 + 𝑞𝑥 + 𝑟 = 0,
(where ℎ  0 ), then
(A)
5.
b c
a
= =
q r
p
(B) ℎ =
1 b q
 − 
2 a p
(C) ℎ =
1 b q
 + 
2 a p
(D)
b2 − 4ac q2 − 4 pr
=
p2
a2
For which of the following graphs of the quadratic expression 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, the product 𝑎𝑏𝑐
is negative?
𝑦
0
(A)
6.
𝑦
𝑥 (B)
𝑦
𝑦
𝑥
0
(C)
0
𝑥
(D)
0
𝑥
Let 𝑆 be the set of all non-zero numbers 𝛼 such that the quadratic equation 𝛼𝑥 2 − 𝑥 + 𝛼 = 0 has
two distinct real roots 𝑥1 and 𝑥2 satisfying the inequality |𝑥1 − 𝑥2 | < 1. Which of the following
intervals is(are) a subset(s) of 𝑆 ?
 1
1 
(A)  − , −

5
 2
7.
 1

, 0
(B)  −
5 


1 
(C)  0,

5

 1 1
, 
(D) 
 5 2
The value(s) of m for which the equation (1 + 𝑚2 )𝑥 2 − 2(1 + 3𝑚)𝑥 + (1 + 8𝑚) = 0 has no real
root is/are :
(A)
1
2
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8.
If 𝑎 and 𝑏 are non-zero real numbers and 𝛼, 𝛽 the roots of 𝑥 2 + 𝑎𝑥 + 𝑏 = 0, then
(A) 𝛼 2 , 𝛽 2 are the roots of 𝑥 2 − (2𝑏 − 𝑎2 )𝑥 + 𝑎2 = 0
1 1
(B) , are the roots of 𝑏𝑥 2 + 𝑎𝑥 + 1 = 0
(C)
,
are the roots of 𝑏𝑥 2 + (2𝑏 − 𝑎2 )𝑥 + 𝑏 = 0
(D) (𝛼 − 1), (𝛽 − 1) are the roots of the equation 𝑥 2 + 𝑥(𝑎 + 2) + 1 + 𝑎 + 𝑏 = 0
9.
𝑥 2 + 𝑥 + 1 is a factor of 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 = 0, then the real root of above equation is (𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝑅)
𝑑
(𝑏−𝑎)
(𝑎 – 𝑏)
𝑑
(A) − 𝑎
(B)
(C)
(D)
𝑎
𝑎
𝑎
10.
Let 𝑝 and 𝑞 be real numbers such that 𝑝  0, 𝑝3  𝑞 and 𝑝3  –𝑞. If  and  are nonzero complex
numbers satisfying  +  = – 𝑝 and 3 +  3 = 𝑞, then a quadratic equation having
and
as its
roots is
(A) (𝑝3 + 𝑞)𝑥 2 − (𝑝3 + 2𝑞)𝑥 + (𝑝3 + 𝑞) = 0
(B) (𝑝3 + 𝑞)𝑥 2 − (𝑝3 − 2𝑞)𝑥 + (𝑝3 + 𝑞) = 0
(C) (𝑝3 − 𝑞)𝑥 2 − (5𝑝3 − 2𝑞)𝑥 + (𝑝3 − 𝑞) = 0
(D) (𝑝3 − 𝑞)𝑥 2 − (5𝑝3 + 2𝑞)𝑥 + (𝑝3 − 𝑞) = 0
COMPREHENSION
11.
12.
13.
Paragraph for Questions 11 and 13
Consider the expression 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, 𝑎  0 and 𝑎, 𝑏, 𝑐  ℝ then the graph between 𝑥, 𝑦 is
always a parabola. If 𝑎 > 0 then the shape of the parabola is concave upward and if 𝑎 < 0 then the
shape of the parabola is concave down ward. If 𝑦 > 0 or 𝑦 < 0 then discriminant 𝐷 < 0.
Let 𝑥 2 + 2𝑎𝑥 + 10 − 3𝑎 > 0 for every real value of 𝑥, then –
(A) 𝑎 > 5
(B) 𝑎 <– 5
(C) – 5 < 𝑎 < 2
(D) 2 < 𝑎 < 5
The value of 𝑥 2 + 2𝑏𝑥 + 𝑐 is positive if –
(A) 𝑏 2 − 4𝑐 > 0
(B) 𝑏 2 − 4𝑐 < 0
(C) 𝑐 2 < 𝑏
(D) 𝑏 2 < 𝑐
The diagram shows the graph of 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 then –
𝑦
(𝑥2 , 0)
(A) 𝑎 < 0
[2]
(B) 𝑐 < 0
𝑥
(𝑥1 , 0)
(C) 𝑏 2 − 4𝑎𝑐 < 0
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(D) 𝑏 2 − 4𝑎𝑐 = 0
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Paragraph for Questions 14 and 15
Let 𝑝, 𝑞 be integers and let 𝛼, 𝛽 be the roots of the equation, 𝑥 2 − 𝑥 − 1 = 0, where 𝛼 ≠ 𝛽. For
𝑛 = 0,1,2, . . . ., let 𝑎𝑛 = 𝑝𝛼 𝑛 + 𝑞𝛽 𝑛 .
FACT : If 𝑎 and 𝑏 are rational numbers and a + b 5 = 0 , then 𝑎 = 0 = 𝑏.
14.
If 𝑎4 = 28, then 𝑝 + 2𝑞 =
(A) 14
15.
(B) 7
(C) 12
(D 21
(B) 𝑎11 – 𝑎10
(C) 𝑎11 + 𝑎10
(D) 𝑎11 + 2𝑎10
𝑎12 =
(A) 2𝑎11 + 𝑎10
MATCHING LIST
16.
(P)
List– 
List– 
If 𝛼, 𝛼 + 4 are two roots of 𝑥 2 − 8 𝑥 + 𝑘 = 0,
(1)
2
(2)
3
(3)
12
(4)
10
then possible value of 𝑘 is
(Q)
Number of real roots of equation 𝑥 2 − 5|𝑥| + 6 = 0
are '𝑛', then value of
(R)
n
is
2
If 3 − 𝑖 is a root of 𝑥 2 + 𝑎𝑥 + 𝑏 = 0 (𝑎, 𝑏  ℝ),
then 𝑏 is
(S)
If both roots of 𝑥 2 − 2𝑘𝑥 + 𝑘 2 + 𝑘 − 5 = 0
are less than 5, then '𝑘' may be equal to
(A) (P) → (3), (Q) → (1), (R) → (4), (S) → (2)
(B) (P) → (1), (Q) → (2), (R) → (3), (S) → (4)
(C) (P) → (3), (Q) → (2), (R) → (3), (S) → (2)
(D) (P) → (4), (Q) → (3), (R) → (2), (S) → (1)
MATCH THE COLUMN
17.
Let f ( x ) =
x − 6x + 5
x2 − 5x + 6
2
Match the expressions/statements in Column I with expressions/statements in Column II.
Column-I
Column-II
(A)
If −1 < 𝑥 < 1, then 𝑓(𝑥) satisfies
(p)
0 < 𝑓(𝑥) < 1
(B)
If 1 < 𝑥 < 2, then 𝑓(𝑥) satisfies
(q)
𝑓(𝑥) < 0
(C)
If 3 < 𝑥 < 5, then 𝑓(𝑥) satisfies
(r)
𝑓(𝑥) > 0
(D)
If 𝑥 > 5, then 𝑓(𝑥) satisfies
(s)
𝑓(𝑥) < 1
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NUMERICAL
18.
The sum of all real roots of the equation |𝑥 – 2|2 + |𝑥 – 2|– 2 = 0 is ...
19.
If 𝑥 + 𝑦 + 𝑧 = 12 & 𝑥 2 + 𝑦 2 + 𝑧 2 = 96 and
20.
If x = 4 − 2 3 and y = 9 − 4 5 then the value of
1 1 1
x3 + y3 + z3
+ + = 36 . Find the value of
x y z
4
(
5x − 3 y ) is equal to ( a − b c ) where 𝑎, 𝑏, 𝑐
2
are coprime numbers then 𝑎 + 𝑏 + 𝑐 is equal to (where '𝑐' is an odd integer)
21.

6

6
 3− 2+
Value of 
6

3− 2+
( 3 − 2) +



2



 is




22.
The smallest value of 𝑘, for which both the roots of the equation, 𝑥 2 − 8𝑘𝑥 + 16(𝑘 2 − 𝑘 + 1) = 0
are real, distinct and have values at least 4, is
23.
For 𝑥   , then number of real roots of the equation 3𝑥 2 − 4|𝑥 2 − 1| + 𝑥 − 1 = 0 is ____.
24.

2
Let S =  : log 2 (9

−4
5
+ 13) − log 2  .32
2
−4
 
+ 1  = 2 . Then the maximum value of  for which the
 
2


equation x − 2    x + ( + 1)2 = 0 has real roots, is ____ .
s
 s 
2
25.
We call '𝑝' a good number if the inequality
2x2 + 2x + 3
≤ 𝑝 is satisfied for any real 𝑥. Find the
x2 + x + 1
smallest integral good number.
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ANSWER KEY
Que.
1
2
3
4
5
6
7
8
9
10
Ans.
A,B,C
B,D
B,C
B,D
A,B,C,D
A,D
C,D
B,C,D
A,D
B
Que.
11
12
13
14
15
16
17
Ans.
C
D
B
C
C
A
A→p, r, s; B→q, s; C→q, s; D→p, r, s
Que.
18
19
20
21
22
23
24
25
Ans.
4
216.50
36
2
2
4
25
4
SOLUTIONS REVISION ASSIGNMENT-01
1.
Ans. (A,B,C)
,
will satisfy the quadratic equation
2

− +1 =0
= −1
2
On squaring
( ) =(
2
2
(

2
=
4

− 1)
2
−2 +1

− 1) − 2 + 1
2

= −1
=−
4
2
=− .
6
=−
.

2
(
− 1) = −
2
+
= − ( − 1) + = 1
2009
=
( )
6
 (1 )
334
.
334
4
.
5
.
− .
−
2
=1−
Similarly,
2009
+
2009
2009
=1−
= 2− ( +
)

= 2 − (1)
+ = 1
=1
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2.
Ans. (B,D)
Let 𝑝(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑝(0) = 𝑎(0)2 + 𝑏(0) + 𝑐 = 1
𝑐 =1
𝑝(1) = 4
 𝑎(1)2 + 𝑏(1) + 1 = 4
𝑎+𝑏 =3
…(i)
𝑝(−1) = 6
 𝑎(−1)2 + 𝑏(−1) + 1 = 6
𝑎−𝑏 =5
…(ii)
Solving both equations, we get,
𝑎 = 4, 𝑏 = −1
𝑝(𝑥) = 4𝑥 2 − 𝑥 + 1
𝑃(2) = 4(2)2 − 2 + 1 = 15
𝑝(−2) = 4(−2)2 − (−2) + 1 = 19
3.
Ans. (B,C)
Let 𝑝(𝑥) = (𝑥 − 𝑎)(𝑥 − 𝑏) − 1
𝑝(𝑎) = (𝑎 − 𝑎)(𝑎 − 𝑏) − 1
= −1
𝑝(𝑏) = −1
Possible graph for the above expression will be 𝑝(𝑎) and 𝑝(𝑏) both are negative and leading
coefficient of quadratic is +𝑣𝑒 𝑎 and 𝑏 both will be between both the roots
𝑎
𝑏
𝛼
𝛽
Looking at the graph
𝛼 ∈ (−∞, 𝑎) and 𝛽 ∈ (𝑏, ∞)
4.
Ans. (B,D)
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
𝛼, 𝛽 are roots
So,
+ =
=
[6]
c
a
−b
a
…(i)
…(ii)
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(
−
)
=

(
+
)
2
2
2
+
2
−2
−4
2
c
 −b 
   −4
a
 a 
b2 − 4ac
a2

…(iii)
𝑝𝑥 2 + 𝑞𝑥 + 𝑟 = 0
𝛼 + ℎ, 𝛽 + ℎ are roots

(
+ h) + ( + h) =
+ + 2h =

−q
r
, ( + h )( + h ) =
p
p
−q
p
−b
−q
1 b q 
+ 2h =
h=  − 
a
p
2 a p 

 ( + h ) − ( + h )  = ( ( + h ) + ( + h ) ) − 4 ( + h )( + h )
2

(
−
2
2
)
2
 −q 
r q2 − 4 pr
=  −4 =
p
p2
 p 
…(iv)
Divide (iii) and (iv)
5.
 b2 − 4ac 


a2 

= 2
 q − 4 pr 


p2 


(
(

q2 − 4 pr b2 − 4ac
=
p2
a2
−
−
)

)
2
Ans. (A,B,C,D)
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
(A)
𝑦
0
𝑥
𝑎 < 0, 𝑏 < 0 < 𝑐 < 0
 𝑎𝑏𝑐 < 0
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()
𝑦
𝑥
0
𝑎 < 0, 𝑏 > 0, 𝑐 > 0
 𝑎𝑏𝑐 < 0
(C)
𝑦
𝑥
0
𝑎 > 0, 𝑏 < 0, 𝑐 > 0
 𝑎𝑏𝑐 < 0
(D)
𝑦
𝑥
0
𝑎 < 0, 𝑏 < 0, 𝑐 < 0
 𝑎𝑏𝑐 < 0
6.
Ans. (A,D)
𝛼𝑥 2 − 𝑥 + 𝛼 = 0
𝐷 = 1– 4𝛼 2
distinct real roots 𝐷 > 0

 1 1
 − , 
 2 2
...(i)
given |𝑥1 − 𝑥2 | < 1

1−4
1 1
2
1
 1 − 4𝛼 2 < 𝛼 2



−1   1
  − ,
, 

5  5


...(ii)
from (i) & (ii)
 −1 −1   1 1 
 ,
, 

5   5 2
 2
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7.
Ans. (C,D)
For no real roots discriminant must be less than zero
𝐷 = 𝑏 2 − 4𝑎𝑐 < 0
𝑎 = 1 + 𝑚2
𝑏 = −2(1 + 3𝑚)
𝑐 = 1 + 8𝑚
 [−2(1 + 3𝑚)]2 − 4(1 + 𝑚2 )(1 + 8𝑚) < 0
 (1 + 3𝑚)2 − (1 + 8𝑚 + 𝑚2 + 8𝑚3 ) < 0
 1 + 9𝑚2 + 6𝑚 − 1 − 8𝑚 − 𝑚2 − 8𝑚3 < 0
 8𝑚2 − 8𝑚3 − 2𝑚 < 0
 −2𝑚(4𝑚2 − 4𝑚 + 1) < 0
 −2𝑚(2𝑚 − 1)2 < 0
1
2
 𝑚 > 0 and m 
8.
Ans. (B,C,D)
𝑥 2 + 𝑎𝑥 + 𝑏 = 0
𝛼 + 𝛽 = −𝑎
𝛼𝛽 = 𝑏
(A)
If 𝛼 2 , 𝛽 2 are roots
 sum of roots = 𝛼 2 + 𝛽 2
 (𝛼 + 𝛽)2 − 2𝛼𝛽
 (−𝑎)2 − 2𝑏
 𝑎2 − 2𝑏
Product of roots = 𝛼 2 𝛽2
= (𝛼𝛽)2 = 𝑏 2
Quadratic equation
 𝑥 2 −(sum of roots)𝑥 + product of roots = 0
 𝑥 2 − (𝑎2 − 2𝑏)𝑥 + 𝑏 2 = 0
(B)
sum of roots =
=
+
=
1
+
1
−a
b
1 1
1
1
 =
=
b
Quadratic equation
1
 a
 x2 −  −  x + = 0
b
 b
2
 𝑏𝑥 + 𝑎𝑥 + 1 = 0
Product of roots =
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(C)
sum of roots 
2




(
+
2
+
)
2
( −a )
2
+
−2
− 2b
b
a − 2b
b
2
Product or roots =

=1
Quadratic equation
 a2 − 2b 
 x2 − 
 x +1 = 0
 b 
(
)
 bx2 + 2b − a2 x + b = 0
(D)
sum of roots = (𝛼 − 1) + (𝛽 − 1)
𝛼+𝛽−2
 −𝑎 − 2 = −(𝑎 + 2)
Product of roots = (𝛼 − 1)(𝛽 − 1)
 𝛼𝛽 − (𝛼 + 𝛽) + 1
𝑏+𝑎+1
Quadratic equation
 𝑥 2 − (−(𝑎 + 2))𝑥 + 𝑏 + 𝑎 + 1 = 0
 𝑥 2 + (𝑎 + 2)𝑥 + 𝑏 + 𝑎 + 1 = 0
9.
Ans. (A,D)
𝑥 2 + 𝑥 + 1 is a factor of 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 = 0
So, let 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 = (𝑎𝑥 + 𝑑)(𝑥 2 + 𝑥 + 1)
 (𝑎𝑥 + 𝑑)(𝑥 2 + 𝑥 + 1) = 0
 x=−
d
−1  3 i
or x =
= w , 𝑤2
a
2
Now, sum of roots ; −
[ 10 ]
 −
d
b
−1 = −
a
a
 −
d
b a −b
=1− =
a
a
a
d
b
+ w + w2 = −
a
a
(
𝑤 + 𝑤 2 + 1 = 0)
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10.
Ans. (B)
3
+
3
=q
 ( + )3 − 3

=
( + )= q
p3 + q
3p
sum of the roots =
+
=
2
+
2
 p3 + q 
p2 − 2 

3 p  p3 − 2q

=
= 3
p3 + q
p +q
3p
Product of the roots = 1.
Required equation is
(𝑝3 + 𝑞)𝑥 2 − (𝑝3 − 2𝑞)𝑥 + (𝑝3 + 𝑞) = 0
11.
Ans. (C)
𝑥 2 + 2𝑎𝑥 + 10 − 3𝑎 > 0 for every real value of 𝑥,
So, discriminant 𝐷 < 0
 (2𝑎)2 − 4 ⋅ (10 − 3𝑎) < 0
 4𝑎2 − 4(10 − 3𝑎) < 0
 𝑎2 + 3𝑎 − 10 < 0
 (𝑎 + 5)(𝑎 − 2) < 0
 −5 < 𝑎 < 2.
12.
Ans. (D)
𝑥 2 + 2𝑏𝑥 + 𝑐 > 0 then 𝐷 < 0
 (2𝑏)2 − 4𝑐 < 0
 4𝑏 2 − 4𝑐 < 0
 𝑏 2 < 𝑐.
13.
Ans. (B)
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑦
(𝑥2 , 0)
𝑥
(𝑥1 , 0)
Here, 𝑎 > 0, 𝐷 > 0
( two real roots)
for 𝑥 = 0; 𝑦 = 𝑐 which is negative.
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14.
Ans. (C)
2 =  + 1  4 = 3 + 2
 𝑎4 = 28  𝑝4 + 𝑞4 = 𝑝(3 + 2) + 𝑞(3 + 2) = 28
 𝑝(3 + 2) + 𝑞(3 – 3 + 2) = 28
 (3𝑝 – 3𝑞) + 2𝑝 + 5𝑞 = 28
(as   Qc)
 𝑝 = 𝑞, 2𝑝 + 5𝑞 = 28  𝑝 = 𝑞 = 4
 𝑝 + 2𝑞 = 12
Ans : C
15.
Ans. (C)
2 =  + 1  n = n – 1 + n – 2
 𝑝n + 𝑞n = 𝑝(n – 1 + n – 2) + q( n – 1 + n – 2)
𝑎n = 𝑎n – 1 + 𝑎n – 2
 𝑎12 = 𝑎11 + 𝑎10
16.
Ans. (A)
(P)
,  + 4 are two roots of 𝑥 2 − 8𝑥 + 𝑘 = 0,
 Sum of roots;  +  + 4 = 8
=2
Now, product of roots; 𝑘 = 𝛼 ⋅ (𝛼 + 4)
 𝑘 = 2(2 + 4)
 𝑘 = 12
(Q)
𝑥 2 − 5|𝑥| + 6 = 0
 (|𝑥|)2 − 5|𝑥| + 6 = 0
 (|𝑥| − 3)(|𝑥| − 2) = 0
 |𝑥| = 3 |𝑥| = 2
 𝑥 = ±3 𝑥 = ±2
n
So, 𝑛 = 4  = 2 .
2
2
(R)
𝑥 + 𝑎𝑥 + 𝑏 = 0
As 3 − 𝑖 is a root, 3 + 𝑖 is also a root.
 sum of roots; −𝑎 = 6  𝑎 = −6
 product of roots; 𝑏 = (3 − 𝑖)(3 + 𝑖)
 𝑏 = 9 + 1 = 10
(S)
Both roots of 𝑥 2 − 2𝑘𝑥 + 𝑘 2 + 𝑘 − 5 = 0 are less than 5,
So, sum of roots 2𝑘 < 5 + 5  𝑘 < 5
Discriminant, 𝐷 ≥ 0
 (−2𝑘)2 − 4(𝑘 2 + 𝑘 − 5) ≥ 0
 4𝑘 2 − 4(𝑘 2 + 𝑘 − 5) ≥ 0
 −𝑘 + 5 ≥ 0
𝑘≤5
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…(1)
…(2)
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𝑓(5) > 0  (5)2 − 2𝑘(5) + 𝑘 2 + 𝑘 − 5 > 0
 𝑘 2 − 9𝑘 + 20 > 0
 (𝑘 − 4)(𝑘 − 5) > 0
 𝑘 < 4 𝑘 > 5
…(3)
From (1), (2) and (3);
𝑘<4
So, 𝑘 = 3 possible, 𝑘 = 2 possible.
17.
𝑦
𝑓(5)
5
𝑥
Ans. A → p, r, s ; B → q, s ; C → q, s ; D → p, r, s
(
)
2
x2 − 6 x + 5 x − 5x + 6 − ( x + 1 )
f ( x) = 2
=
x − 5x + 6
x2 − 5x + 6
x +1
=1−
( x − 2)( x − 3)
for −1 < 𝑥 < 1,1 < 𝑥 < 2,3 < 𝑥 < 5 and 𝑥 > 5 ;
x +1
x +1
0  1−
1
( x − 2)( x − 3)
( x − 2)( x − 3)
 𝑓(𝑥) < 1
Now,
f ( x) =
x2 − 6 x + 5 ( x − 1)( x − 5)
=
x2 − 5 x + 6 ( x − 2)( x − 3)
+
–
1
(A)
(B)
(C)
(D)
18.
+
2
–
3
+
5
If −1 < 𝑥 < 1 ; 𝑓(𝑥) > 0
Hence, also 0 < 𝑓(𝑥) < 1
If 1 < 𝑥 < 2 ; 𝑓(𝑥) < 0
If 3 < 𝑥 < 5 ; 𝑓(𝑥) < 0
If 𝑥 > 5 ; 𝑓(𝑥) > 0
Hence, also, 0 < 𝑓(𝑥) < 1.
Ans. (4)
|𝑥 – 2|2 + |𝑥 – 2|– 2 = 0
Case – I
𝑥>2
 (𝑥 − 2)2 + (𝑥 − 2) − 2 = 0
 𝑥 2 − 4𝑥 + 4 + 𝑥 − 2 − 2 = 0
 𝑥 2 − 3𝑥 = 0
 𝑥(𝑥 − 3) = 0
 𝑥 = 0,3
𝑥=3
∵𝑥>2
Case – II
𝑥<2
 (𝑥 − 2)2 − (𝑥 − 2) − 2 = 0
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 𝑥 4 + 4𝑥 + 4 − 𝑥 + 2 − 2 = 0
 𝑥 2 − 5𝑥 + 4 = 0
 (𝑥 − 4)(𝑥 − 1) = 0
 𝑥 = 1,4
𝑥=1
∵𝑥<2
Sum of all real roots = 4
19.
Ans. (216.50)
𝑥 + 𝑦 + 𝑧 = 12
 𝑥 2 + 𝑦 2 + 𝑧 2 = 96
1
x
 +
1 1
+ = 36
y z
 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 = 36𝑥𝑦𝑧
 (𝑥 + 𝑦 + 𝑧)2 = 144
 𝑥 2 + 𝑦 2 + 𝑧 2 + 2(𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥) = 144
 96 + 2(𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥) = 144
 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 = 24
 (𝑥 3 + 𝑦 3 + 𝑧 3 ) − 3𝑥𝑦𝑧 = (𝑥 + 𝑦 + 𝑧)(𝑥 2 + 𝑦 2 + 𝑧 2 − 𝑥𝑦 − 𝑦𝑧 − 𝑧𝑥)
 (𝑥 3 + 𝑦 3 + 𝑧 3 ) − 3𝑥𝑦𝑧 = (12)(96 − 24)
 (𝑥 3 + 𝑦 3 + 𝑧 3 ) = 12 × 72 + 3𝑥𝑦𝑧
 24 
= 12 × 72 + 3 ×  
 36 
= 12 × 72 + 2
= 866
866 433
=
= 216.50

4
2
20.
Ans. (36)
x=
( 3 ) + 1 − 2
2
2
3 1
 𝑥 = √3 − 1
y=
( 5)
2
+ 22 − 2  5  2
 𝑦 = (√5 − 2)

(
(
) (
2
5x − 3 y =
= 2 3− 5
15 − 5 − 15 + 2 3
)
2
)
2
= 12 + 5 − 4√15
= 17 − 4√15
 𝑎 = 17, 𝑏 = 4, 𝑐 = 15
 𝑎 + 𝑏 + 𝑐 = 36
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21.
Ans. (2)
x=
(
6
)
3− 2 +x
 𝑥 2 + (√3 − √2)𝑥 − √6 = 0
 𝑥(𝑥 + √3) − √2(𝑥 + √3) = 0
 (𝑥 − √2)(𝑥 + √3) = 0
 𝑥 = √2, 𝑥 = −√3
But 𝑥 cannot be negative.
 𝑥 = √2
 𝑥2 = 2
22.
Ans. (2)
or
4
4
𝑓(𝑥) = 𝑥 2 − 8𝑘𝑥 + 16(𝑘 2 − 𝑘 + 1) = 0
roots are real, 𝐷 > 0
64𝑘 2 − 64(𝑘 2 − 𝑘 + 1) > 0
𝑘 > 1  k (1,  )
at both the roots  4
 f (4)  0
k 2 − 3k + 2  0


  −b
k 1

 2a  4
 k [2, ) least value of 𝑘 = 2
23.
Ans. (4)
3𝑥 2 + 𝑥 − 1 = 4|𝑥 2 − 1|
If 𝑥  [–1, 1],
3𝑥 2 + 𝑥 − 1 =– 4𝑥 2 + 4  7𝑥 2 + 𝑥 − 5 = 0
say 𝑓(𝑥) = 7𝑥 2 + 𝑥 − 5
0
–1
1
𝑓(1) = 3; 𝑓(– 1) = 1; 𝑓(0) = – 1
[Two Roots]
If 𝑥  (–, –1]  [1, )
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3𝑥 2 + 𝑥 − 1 = 4𝑥 2 − 4  𝑥 2 − 𝑥 − 3 = 0
Say 𝑔(𝑥) = 𝑥 2 − 𝑥 − 3
–1
1
𝑔(– 1) = – 1; 𝑔(1) = – 3
[Two Roots]
So total 4 roots.
24.
Ans. (25)

5
 
S =  : log 2 92−4 + 13 − log 2  32−4 + 1  = 2
2
 

(
Now,
)


2 92−4 + 13
 92−4 + 13 
=4
log 2 
 = 2 or
2−4
5
5
3
+
2
2− 4
 3
+1 
2

(
(
(
)
)
)
 92–4 + 13 = 10  (3)2–4 + 4
 92(–2) + 13 = 10  9(–2) + 4
Let
𝑦 = 9𝛼−2
 𝑦 2 + 13 = 10𝑦 + 4
 𝑦 2 − 10𝑦 + 9 = 0
 (𝑦 − 9)(𝑦 − 1) = 0
 𝑦 = 1, 9
when 𝑦 = 9
𝑦=1
9–2
9–2 = 90
= 91
–2=1
=2
=3
Now,
2


2
 x − 2     x +  (  + 1)  = 0
s
 s 
2
 𝑥 2 − (2 × 25)𝑥 + 25𝛽 = 0
 𝑥 2 − 50𝑥 + 25𝛽 = 0
for real roots 𝐷 ≥ 0
𝑏 2 − 4𝑎𝑐 ≥ 0  2500 – 4(25)  0
  25  max = 25.
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25.
Ans. (4)
2x2 + 2 x + 3
p
x2 + x + 1
Let f ( x ) =
= 2+
2x2 + 2x + 3
x2 + x + 1
1
x + x +1
2
 2
 3 
 x + x + 1  ,  
 4 

 10 
f ( x )   2, 
 3
Hence,
2x2 + 2 x + 3
p
x2 + x + 1
 10 
p   ,    𝑝 is a good no.
3

Smallest integral 𝑝 = 4
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