Miscellaneous Questions
1.
UNIT
7
Answer (3)
We have,
⇒
x
1
φ( x ) =
 {4t
x2
2
− 2φ′(t )}dt
4
x
x 2 φ( x ) =  {4t 2 − 2φ′(t )}dt
4
Differentiating both sides w.r.t. x we get
x2φ′(x) + 2xφ(x) = 4x 2 – 2φ′(x)
⇒ Putting x = 4, we get
16φ′(4) + 8φ(4) = 64 – 2φ′(4)
18φ′(4) = 64 ⇒ φ′(4) =
2.
32
( φ(4) = 0)
9
Answer (4)
25
1 2 3 4 th 5 6
For '25' to bear 4th place,
Required no. of ways = 25C3 × 24C2 =
3.
25C
22 ×
24C
22
Answer (2)
Let f ( x ) =
sin[cos x ]
1 + [cos x ]
We find that
sin[cos(0 − h )]
lim f ( x ) = lim
h →0 1 + [cos( 0 − h )]
x →0 −
=
sin[cos h] sin 0
;
=
1 + [cos h] 1 + 0 [cos h] = 0 in the label of 0
=0
and lim f ( x ) = lim
sin[cos( 0 + h )]
sin[cos h]
= lim
+ h )] h →0 1 + [cos h]
h →0 1 + [cos(0
x →0 +
=
sin 0
, as [cos h] = 0 in the neighbourhood of 0
1+ 0
=0
Then lim
sin[cos x ]
x →0 1 + [cos x ]
=0
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4.
Miscellaneous Questions
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Answer (3)
4!
=3
Number of groups of girls possible = ( ) 2
2! .2!
_B_B_B_B_B_
total number of ways = 3 × 6P2 × 2! × 2! × 5! = 60 × 6!
5.
Answer (4)
Number of selecting 5 boxes
= coefficient of x5 in (4C1x + 4C2x2 + 4C3x3 + 4C4x4) × (2C1x + 2C2x2)2
= coefficient of x2 in (4 + 6x + 4x2 + x3) (2 + x)2
= coefficient of x2 in ( 4 + 6x + 4x2 + x3) (4 + 4x + x2)
= 4 + 24 + 16 = 44
∴ Number of arrangements of letters = 44 ×
5!
3!
= 44 × 20 = 880
6.
Answer (1)
Case-I
when 4 bowler's were selected,
6C
4
× 3C2 × 6C4 + 6C4 × 3C3 × 6C3 = 975
Case-II
when 5 bowler's were selected,
6C
5
× 3C2 × 6C3 + 6C5 × 3C3 × 6C2 = 450
Case-III
When 6 bowler's were selected,
6C
6
× 3C2 × 6C2 + 6C6 × 3C3 × 6C1 = 51
Total number of ways = 1476
7.
Answer (3)
Number of selecting six boxes
= coefficient of x6 in (4C1x + 4C2x2 + 4C3x3 + 4C4x4) × (2C1x + 2C2x2)2
= coefficient of x3 in (4 + 6x + 4x2 + x3) (2 + x)2
= coefficient of x3 in ( 4 + 6x + 4x2 + x3) (4 + 4x + x2)
= 6 + 16 + 4 = 26
∴ Number of arrangement of letters = 26 × 6!
3!
= 26 × 120 = 26 × 5!
Second Method
Number of ways of selecting 6 boxes out of 8 boxes = 8C6 = 28
Number of ways in which a row can be empty = 2
Number of ways in which no row remains empty = 28 – 2 = 26
Hence, number of ways of putting the letters = 26 ×
6!
= 26 × 5!
3!
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8.
Miscellaneous Questions
393
Answer (3)
From the adjoining figure it is clear that
π π
cos x ≤ sin x, ∀x ∈  , 
4 2
 π
and sinx ≤ cosx for x ∈ 0, .
 4
Hence the required area
π/4

=
O
π/2
(cos x − sin x )dx +

(sin x − cos x )dx
π
4
π
2
π
π/ 4
0
π
π
2

4 
= sin x + cos x  +  – cos x − sin x 
π

0 
4
= ( 2 − 1) + [ −1 + 2 ] = 2( 2 − 1) sq. units
8a.
Answer (1)
(AIEEE 2010)
π
2
O
3π
2
π
Required area
=
π /4

(cos x − sin x )dx +
0
5 π /4

(sin x − cos x )dx +
π /4
3 π /2

(cos x − sin x )dx
5 π /4
= (4 2 − 2) sq. units
9.
Answer (4)
We have,
     
  
(a + b ).p + (b + c ).q + (c + a ).r
 
 
 
 
 
 
1
1
1
= (a + b ).    (b × c ) + (b + c ).    (c × a ) + (c + a ).    (a × b )
[a b c ]
[a b c ]
[a b c ]
=1+1+1=3
10. Answer (2)

 
 
 
r = x ( a × b) + y ( b × c ) + z ( c × a)
 

 r . a = y a b c 
... (i)
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similarly
 

 r . b = z a b c 
... (ii)
 

 r . c = x a b c 
... (iii)
adding (i), (ii), (iii)
   

r . ( a + b + c ) = [ x + y + z ] a b c 
   
 x + y + z = r .( a + b + c )
11. Answer (2)
1
Δ= 5
2
6
3
7 =0
9 10 11
4
Δ1 = 8
2
6
3
7 =0
12 10 11
similarly Δ2 = Δ3 = 0
Hence planes will form a common line.
12. Answer (4)
If point lies between the planes
then (a + a + a – 3) (a + a + a – 6) < 0
⇒ 3(a – 1) (3) (a – 2) < 0
(a – 1) (a – 2) < 0
⇒ a ∈ (1, 2)
13. Answer (2)
The vertices of the triangle formed by the lines are given by A(0, 6), B(2 3 , 0) and C( −2 3 , 0)
⇒ Clearly AB = BC = CA = 4 3
Y
A(0, 6)
The centroid of triangle ABC is (0, 2)
⇒ AG = 4, OG = 2
G
Hence the required equation of the circumcircle is
(x – 0)2 + (y – 2)2 = 42 ⇒ x2 + y2 – 4y = 12
C(–2 3, 0)
O
B(2 3, 0)
X
14. Answer (1)
Let p = probability of getting a tails in a single trial =
1
2
n = number of trials = 2008
x = number of trails in 2008 trials
we have P(x = r) =
r
2008C
 1  1
= 2008Cr    
2 2
r
p r q 2008 – r.
2008 − r
 1
= 2008Cr  
2
2008
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Now,
P(x = odd) = P(x = 1) + P(x = 3) + P(x = 5) +..........+ P(x = 2007)
 1
= 2008C1 
2
 1
= 
 2
=
 1
+ 2008C3  
2
2008
 1
+ 2008C5  
2
2008
 1
+ ....... + 2008C2007  
2
2008
2008
1
2
2008
2008
.( 2008C1 + 2008 C3 + ....... + 2008 C2007 )
.2 2007 =
1
2
15. Answer (3)
Comparing the given equation
x2 + 4y2 + 2x + 16y + 13 = 0
with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
we get a = 1, h = 0, b = 4, g = 1, f = 8, c = 13
⇒ abc + 2fgh – af2 – bg2 – ch2 = –16 ≠ 0
⇒ The given equation does not represent a pair of straight lines
we find that h2 – ab = 0 – 4 < 0
⇒ The given equation represents an ellipse.
16. Answer (1)
The given equation can be written as
(x – 1)2 + (y – 1)2 + (z – 1)2 + (t – 1)2 = 0
⇒ x=y=z=t=1
Hence the required product = 1
17. Answer (2)
32x + 1 + 32 = 3x + 3 + 3x
⇒ 3 . y2 + 9 = 27 . y + y
(where y = 3x )
⇒ 3y2 – 28y + 9 = 0
⇒ 3y2 – 27y – y + 9 = 0
⇒ (3y – 1) (y – 9) = 0
⇒
y=
1
x
,9=3
3
⇒ x = 2, –1
∴ Number of solutions = 2
18. Answer (3)
We have,
s(n) – s′(n) = (3 + 8 + 15 + ..............) – (6 + 11 + 18 + .......)
= –(3 + 3 + 3 +........ to n term) = –3n
⇒ s(111) – s′(111) = –333
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19. Answer (1)
We have,
n + 1 = 2008! + 2
n + 2 = 2008! + 3
n + 3 = 2008! + 4
.............................
n + 2007 = 2008! + 2008
Thus 2 divides n + 1, 3 divides n + 2, 4 divides n + 3, ........... and finally 2008 divides n + 2007.
Note that 2008! is divisible by each of 2, 3, 4........ and 2008. We have then shown that all the numbers in
this list are composite, establishing that there are no primes in the list.
20. Answer (4)
21. Answer (1)
Let z be the variable point P, –i and i be respectively A and B as two fixed points. The equation |z – i| – |z + i| = k
reduces to PA – PB = k
The locus of P under suitable condition on k can be hyperbola having foci as A and B. By triangle inequality
|PA – PB| < AB (the difference of two sides being less than the third side). Thus | k | < |–i – i| = 2
⇒ 0 < |k| < 2
22. Answer (2)
z1 – z2 = (z3 – z2) e
i
z1
π
2
z1 – z2 = (z3 – z2)
... (i)
on squaring,
z2
z3
z12 + z22 – 2z1z2 = –1(z32 + z22 – 2z3z2)
⇒ 2z22 = 2z1z2 + 2z3z2 – z32 – z12
⇒ z12 + z32 + 2z22 = 2z2(z1 + z3)
23. Answer (4)
z=
b+ic
1+ a
iz =
−c + i b
1
1+ a

=
1+ a
i z −c + i b
1+ i z 1+ a − c + i b
=
1 − iz 1 + a + c − i b
⇒
[a2 + b2 = 1 – c2 and (a + ib) (a – ib) = (1 + c) (1 – c)]
1+ i z a + i b + 1− c a + i b
=
=
1− i z 1+ c + a − i b
1+ c
24. Answer (3)
–bz2 = az1 + cz3
z2 =
az1 + cz3 az1 + cz3
=
−b
a+c
⇒ z1, z2 and z3 are collinear.
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Miscellaneous Questions
397
25. Answer (4)
We have,
aaa + aa + a + a + a = 10k
⇒ aaa + aa + 3a = 10k
⇒ a(111 + 11 + 3) = 10k
⇒ a125 = 10k
As 125 must divide 10k, then the minimum value of k is 3 for
k = 3, a = 8
k = 4, a = 80
k = 5, a = 800 (as a is single digit number ∴ a ≠ 80, 800)
26. Answer (1)
The number divisible by 30 and 35 both is also divisible by their LCM i.e., 210. The least and the greatest
4 digit number divisible by 210 are 1050 and 9870 respectively.
The number of numbers divisible by 210
=
9870 − 1050
8820
+1=
+ 1 = 43
210
210
Similarly the number of 4-digit numbers divisible by 420 =
9660 − 1260
+ 1 = 21. Hence the number of numbers
420
divisible by 210 but not divisible by 420 = 43 – 21 = 22
27. Answer (3)
Solving the equation y = mx + 9 and x + 4y = 81 we get,
81 − x
= mx + 9 ⇒
4
⇒
81 – x = 4mx + 36
x =
45
4m + 1
As m is a positive integer, so is 4m + 1 and then so is x. The above equation is to be read as 45 being
the product of two positive integers. 4m + 1 can be 5, 9, 45. So m can be 1, 2, 11. With these values of
m, both x and y turn out to be integers.
28. Answer (4)
There were 6 clear mornings mean there were n – 6 rainy morning. Similarly, there were 5 clear afternoon
mean there were n – 5 rainy afternoons.
When it rained in the afternoon it was clear in the morning implies that there is no day when it rained both
in the afternoon and in the morning. The number of rainy days = the number of rainy afternoons + the number
of rainy mornings.
= n – 5 + n – 6 = 2n – 11
But 2n – 11 = 7 (given)
⇒ 2n = 18 ⇒ n = 9
29. Answer (1)
We have,
2x2 + 3y2 – 4x – 12y + 20
= 2(x2 – 2x) + 3(y2 – 4y) + 20
= 2(x2 – 2x + 1) + 3(y2 – 4y + 4) + 6
= 2(x – 1)2 + 3(y – 2)2 + 6 ≥ 6
The least value occurs at x = 1 and y = 2.
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30. Answer (4)
We have,
A3 = B3 & A2B = B2A
⇒ A3 – A2B = B3 – B2A
⇒ A2(A – B) = B2(B – A)
⇒ (A2 + B2)(A – B) = 0
⇒ det(A2 + B2) det (A – B) = 0
⇒ Either det(A2 + B2) = 0 or det (A – B) = 0
31. Answer (4)
We have,
f (g ( x )) = f (1 − x ) = sin(1 − x ) whose domain [0, ∞) and range [–1, 1]
g (f ( x )) = 1 − f ( x ) = 1 − sin x whose domain = 2kπ ≤ x ≤ 2kπ + π, k ∈ z and range [0, 1]
(fof)(x) = f(f(x)) = f(sinx) = sin(sinx) whose domain = R and range [–sin 1, sin 1]
(gog)(x) = g(g(x)) = 1 − g ( x ) = 1 − 1 − x whose domain is 0 ≤ x ≤ 1 and range = [0, 1]
Clearly gof and gog have same range.
32. Answer (2)
1
f(– x) ⇒
2
1
3
4
3
4
1
|f(– x)| ⇒
–|f(– x)| ⇒
2
1
0
1
2
3
4
1
2
3
4
–1
1
1– |f(– x)| ⇒
33. Answer (3)
Obviously
34. Answer (1)
 [ sin x ] = 1 at sin x = 1
35. Answer (2)
The given function f ( x ) = cos −1(log2 ( x 2 + 3 x + 4)) is well defined iff –1 ≤ log2(x2 + 3x + 4) ≤ 1
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⇒
Miscellaneous Questions
399
1
≤ x 2 + 3x + 4 ≤ 2
2
⇒ x ∈ [–2, –1]
36. Answer (4)
| x |
log[ x ] 
 is well defined when x > 0, [x] > 0 and [x] ≠ 1
 x 
⇒
x ∈ [2, ∞ ) and x > 0,
|x|
|x|
= 1  log[ x ]
=0
x
x
π
⇒ range of the given function =  
2
π
The domain and range of the given function are [2, ∞) and   respectively.
2
37. Answer (3)
We have,
P ( A)
 A  P ( A ∩ ( A ∪ B ))
P
=
=
P(A ∪ B)
P ( A ) + P (B )
 A∪B
37a.
Answer (3)
(AIEEE 2011)
We have
C⊂D
⇒
C∩D=C
⇒
 C  P (C ∩ D) P (C )
P  =
=
≥ P (C )
P (D )
P (D )
D
as 0 < P(D) ≤ 1
38. Answer (2)
n(s) = 26 = 64. Number of ways of getting atleast 4 heads = 6C4 + 6C5 + 6C6 and 5 heads = 6C5 + 6C6.
( 6C5 + 6 C6 )
Required probability =
7
26
=
6
( C 4 + C5 + C 6 ) 22
6
6
26
39. Answer (4)
The modulus of a function is always non-negative and hence LHS can't be zero.
40. Answer (2)
We have,
cos103x = 1 + sin103x
Since cos103x never exceeds 1 while 1 + sin103x exceeds 1 unless sinx ∈ [–1, 0].
⇒ x = 0,
−π
are the solution of the given equation.
2
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41. Answer (2)
We have,
|f(x) + g(x)| ≤ |f(x)| + |g(x)| and equality holds when f(x) and g(x) are both non-negative or negative
i.e., f(x)g(x) ≥ 0
⇒ tanx secx ≥ 0
sin x
⇒
cos 2 x
≥ 0  sin x ≥ 0 but cos x ≠ 0
π
 π π 
x ∈ [0, π] ~    x ∈ 0,  ∪  , π
2
 2 2 
⇒
42. Answer (2)
Applying A.M-G.M. inequality we get 2 sin x + 2cos x ≥ 2 2 sin x +cos x
1−
1
⇒ 2 sin x + 2cos x ≥ 2
2
equality holds when 2sinx = 2cosx
⇒ sinx = cosx
⇒ tanx = 1 ⇒ x = nπ +
π
4
43. Answer (2)
We have,
sinθ, sinφ, cosθ are in G.P.
⇒ 2sin2φ = 2sinθcosθ
⇒ 1 – cos2φ = sin2θ
⇒ cos2φ = 1 – sin2θ ≥ 0
The discriminant of the quadratic equation x2 + 2xcotφ + 1 = 0
is given by
4cot2φ – 4
= 4(cot2φ – 1)
 cos2 φ − sin2 φ 

= 4

sin2 φ


= 4cos2φ.cosec2φ ≥ 0
⇒ roots are always real
44. Answer (1)
We have,
 x − 6, x ≥ 6
f ( x ) =| x − 6 |= 
 6 − x, x < 0
∴ g(x) = f(f(x)) = f(x – 6) = x – 12 for x ≥ 18
⇒ g′(x) = 1
45. Answer (1)
We have,
n
k=
e
x(a− x )
dx
a −n
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let us put a – x = t so that
n

I=
xe x ( a − x )dx =
a −n
=−
a−n
 (a − t ) e
t ( a −t )
( −dt )
n
a −n

(a − t )et (a −t )dt =
n

aet (a −t )dt −
a −n
n
⇒ 2I = ak  I =
n

tet ( a −t )dt
a −n
ak
2
46. Answer (4)
a

f ( x ) , if f ( x ) even
 f ( x ) dx = 
−a
0, if f ( x ) old
2
 [ x ] dx
2
=2
0
= 2 ( 2 − 1) + 2 ( 3 − 2 ) + 3 ( 2 − 3 ) 
= 10 − 2 2 − 2 3
so, [ I ] = 3
47. Answer (3)
π
I=
dx
 5 − 3 cos x
0
π
10


so 2I = 
dx
 25 − 9 cos2 x 

0
π
10


dx
= 
 25 − 9 + 9 sin2 x 

0
π
so I = 5
cosec 2 x
 (16 cot
2
0
x + 25)
1
 4 cot x 
= − tan−1 
 5 
4
dx
π
=
0
π
4
48. Answer (4)
dI x 4 − 5 x 2 + 4
=
.(2x ) = 0
2
dx
2 + ex
⇒ x = 0, ± 1, ± 2
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49. Answer (4)
We have,
f(x + y) = f(x) + f(y)
⇒ f(2x) = f(x) + f(x) = 2f(x)
x
 2f ( x )dx
⇒ π(x) =
0
⇒ π′(x) = 2f(x)
⇒ π′(3) = 2f(3) = 2[f(2) + f(1)]
= 2[f(1) + f(1) + f(1)] = 12
50. Answer (4)
We have,
2(sec2θ – 1)(sec2φ – 1)(sec2ψ – 1) + (sec2θ – 1)(sec2φ – 1) + (sec2φ – 1)(sec2ψ – 1) + (sec2θ – 1)(sec2ψ – 1) = 1
⇒ 2sec 2 θsec 2 φsec 2 ψ – 2sec 2 θsec 2 φ – 2sec 2 φsec 2 ψ – 2sec 2 θ sec 2 ψ + 2sec 2φ + 2sec 2 θ + 2sec 2 ψ
–2 + sec2θsec2φ – sec2θ – sec2φ + 1 + sec2φsec2ψ –sec2φ – sec2ψ + 1 + sec2θsec2ψ – sec2θ – sec2φ + 1 = 1
⇒ 2sec2θsec2φsec2ψ = sec2θsec2φ + sec2φsec2ψ + sec2ψsec2θ
⇒ 2 = cos2θ + cos2φ + cos2ψ
⇒ 2 = 1 – sin2θ + 1 – sin2φ + 1 – sin2ψ
⇒ sin2θ + sin2φ + sin2ψ = 1
51. Answer (3)
We have,
24 = 3 × 2 × 2 × 2 = (2 + 1) ( 1 + 1) (1 + 1) (1 + 1)
Hence the prime factor are 2, 2, 3, 5, 7 and corresponding number = 2 × 2 × 3 × 5 × 7 = 420.
But 7 > 2 × 3.
Hence replace factor 7 by 2 × 3
∴ break 24 = 4 × 3 × 2 = (3 + 1) (2 + 1) (1 + 1) and corresponding number = (2 × 2 × 2) (3 × 3) × 5 = 360
52. Answer (3)
We have,
a + b + c = 0 ⇒ a3 + b3 + c3 – 3abc = 0 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
⇒ a2 + b2 + c2 = ab + bc + ca
We find that,
a2
2a 2 + bc
+
b2
2b 2 + ca
+
c2
2c 2 + ab
=
=
a2
2a 2 + bc
− a 2 − bc
− 1+
−
b2
2b 2 + ca
b 2 + ca
−
− 1+
c2
2c 2 + ab
− 1+ 3
c 2 + ab
+ 3 , which will be reduced to 1.
2a 2 + bc 2b 2 + ca 2c 2 + ab
Alternatively we can take a = 1, b = 2, c = –3 so that a + b + c = 0.
53. Answer (2)
We have,
2
e −( x −1) has maximum value 1 at x = 1.
0
1
2
⇒ a – 1 = 0, a + 1 = 2 gives a = 1 which is mid point of interval [a – 1, a + 1], hence a = 1
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403
54. Answer (3)
Let (x1, y1) be a point of contact of tangents from origin to curve.
π

y = cos  x −  = sin x

2

dy
= cos x
dx
so tangent is y = (cos x1)x
(x1, y1) is on tangent and curve so
y1 = cos x1.x and y1 = sin x1
Now locus of (x1, y1) ⇒ x2 y2 = x2 – y2
55. Answer (3)
a
+ 2b x + 1 = 0
x
y' =
at x = 1, 2
a + 2b + 1 = 0 

a
+ 4b + 1 = 0 

2
( a, b) ≡  −
2
1
,− 
3 6
which lies on 3x2 – 12y2 = 1
56. Answer (4)
Consider y = x1/x for x > 0
1
dy  1 − ln x  x
=
 x > 0 for x < e
dx  x 2 
so function is increasing for x ∈ (0, e)
and decreasing for x ∈ ( e, ∞ )
1
1
∴ ee > π π
57. Answer (4)
We have,
3(x2 + y2 + z2 + 1) = 2(x + y + z + xy + yz + zx)
⇒ (x – 1)2 + (y – 1)2 + (z – 1)2 + (x – y)2 + (y – z)2 + (z – x)2 = 0
⇒ x=y=z=1
Here it should be noted that when x = y = z, x, y, z are in A.P., G.P. and H.P. but x = y = z = 1 is the most
suitable option.
58. Answer (2)
Let R be the radius of the bigger circle.
∠AOB =
2π
n
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∠AOC =
Success Achiever (Solutions)
π
n
O
OA = R – r = OB
r
 π  AC
=
In Δ OAC, sin  =
 n  OA R − r
A
C
B
π
⇒ R − r = r cosec  
n
π
π
⇒ R = r + r cosec   = r 1 + cosec 
n
n
 


59. Answer (4)
Since the required number of 6 digit is divisible by 4, lets two digit will be 12 or 32
L
T Th H T U
1, 2, 3, 3, 2
3!
2!
1, 1, 2, 3, 3
3!
2!
Total number of ways = 2 ×
3!
3!
+ 3 × = 6 + 9 = 15
2!
2!
60. Answer (3)
We have,
tf ′(t) = f(t)
⇒
f ′(t ) 1
=
f (t ) t
⇒ ln|f(t)| = ln| t | + ln|k |
⇒ f(t) = k.t = kt
⇒ f(x) = kx, x ∈ R
⇒ f(p) = kp & f(q) = kq
⇒
60a.
f ( p ) f (q )
=
=k
p
q
Answer (3)
[JEE (Main)-2014]
dp(t ) 1
= p(t ) − 200
dt
2
1
d ( p(t ))

 p(t ) − 200
2
=  dt
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 p(t )

2log 
− 200 = t + c
 2

t
p(t )
− 200 = e 2 k
2
Using given condition p(t) = 400 – 300 et/2
61. Answer (2)
We have,
x
det( A) = abc = 7 x .7 7 .7 7
7
77
x
7x
x
.7 7 .7 x dx
Put 77
7x
=z
7x
dz
x
7 7 ..7 7 .7 x dx =
(ln 7)3
=
dz
 (ln 7)
3
=
7x
z
7
+k = 7
+k
3
(ln 7)
(ln 7)3
62. Answer (3)
We have,
1
2
1
(1 + e n + e n + ........ + e
n →∞ n
lim
n −1
n )
n −1
r
1
1
e n = e x dx = e − 1
n →∞ n
r =0
= lim


0
63. Answer (1)
1
Let y = (1 + x ) x
ln y =
 x x2 
1
log (1 + x ) = 1 − +
...
x
 2 3 
 x x2

− +
+...
2 3

so y = e . e 
2
  x x2

 1  x x3 
+ ... +  − + ... + ...
so = e 1 +  − +
 2 ! 2 3 
  2 3

so
y −e
1
→− e
x
2
64. Answer (2)
Usex – 1 < [x] ≤ x
for x, 2x, 3x,....nx and then add them we have
n ( n + 1)
.x − n <
2
Now divide by
so limit =
n
 [r x ] ≤
r =1
n2
n ( n + 1)
x
2
take limit
x
2
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65. Answer (3)
π − cos−1 x
lim
x +1
x →−1+
= lim
cos−1 ( − x )
x +1
x →−1+
= lim
cos −1 (1 − h )
h
h→0
= lim
sin−1
(
1
×
π + cos−1 x
1


×
 π + π 
×
2 π
h 2−h
h 2−h
h→0
1
)×
2 − h ×1
2 π
=
2
2 π
66. Answer (3)
Let us plot the graph of the curves given by y = x2 and y = 2x
y = x2
y=2
x
Since we have two points of intersection of the curves y = 2x and y = x2, hence the given equation x2 = 2 x
has exactly three roots.
67. Answer (3)
We have,
n2 + k + 1 − n2 + k =
67a.
1
2
2
n + k +1+ n + k
<
1
1
=
n + n 2n
Answer (2)
(AIEEE 2011)
 1− cos2( x − 2) 
lim 

x −2


x →2 
= lim
x →2
2 |sin( x − 2)|
x −2
which doesn't exist as L.H.L. = − 2 whereas
R.H.L. = 2
68. Answer (2)
We have,
loge ( x + 1) loge ( x + 2)
loge x m
.
....
x →∞
loge x
loge ( x + 1) loge ( x m − 1)
lim log x ( x + 1) log( x +1) ( x + 2).log( x + 2 ) ( x + 3).... log( x m −1) ( x m ) = lim
x →∞
= lim
x →∞
m loge x
=m
loge x
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69. Answer (1)
 e − sin h − 1 1
f ( 0 − h ) = lim a 
=a
h → 0  ( − sin h ) 
 cos h
f (0 + h) = −
so a = −
1
b
1
 ab = −1
b
so a = 1 and b = –1 or a = –1 and b = 1
70. Answer (1)
f ( 0 − h ) = lim (1 − sin h )
f (0 + h) = e
 x2 + 2 x 


sin x 
a
= ea
− sin h
= e2
so b = 1, a = 2
71. Answer (2)
f ( x ) = lim ( sin x )
2n
n →∞
0,
=
1,
sin x < 1
sin x = 1
 π 
 5π 
so points  − , 1 and  , 1  g ( x ) = 1
2
2
Now intersection points are (x, 1) where
π π 3 π 5π
x=− , ,
,
2 2 2 2
72. Answer (3)
Function can be written as
 0 ; x < 0
f (x) =  2
2 x ; x ≥ 0
2h2 − 0
=0
h→0
h
so, f ' ( 0 + 0) = lim
f ' ( 0 − 0 ) = lim
h→0
0−0
= lim = 0
h→0
−h
so function is derivable at x = 0
when x = 0 then
2t – | t | = 0
If t ≥ 0 and 2t – t = 0
t=0
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73. Answer (2)
2>
2π
2
π
2
π
We have, tan–1(tan2) = (2 – π)
⇒ cos(tan4 (tan2)) = cos(2 – π) = cos(π – 2) = –cos2 = cos(π + 2)
74. Answer (4)
75. Answer (4)
We know
cosec–1x is defined for x ≥ 1 or x ≤ 1 and –1 ≤ cosx ≤ 1
⇒ cosec–1(cosx) is defined only when cosx = ± 1
⇒ x = nπ, n ∈ Z
76. Answer (4)
We have,
 1
 1
a = tan −1 1 − cos −1 −  + sin −1 
 2
2
=
π
π π
π
−π+ + = −
4
3 6
4
1 3
1
and b = cos cos −1  =
8 4
2
tan a = − tan
π
–4
4 3
= −1 and
× b = − × = −1
4
3
3 4
∴
4
b
3
tan a = −
77. Answer (2)
From figure
tan 75º =
x=
C
50
x
50 ft
50
tan 75 º
= 50 (2 − 3 )
tan 15 º =
75º
O
x
A
15º
d
B
...... (i)
50
x +d
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⇒
x+d=
Miscellaneous Questions
50
= 50 (2 + 3 )
tan15º
409
....... (ii)
Eq. (ii) – Eq. (i)
d = 100 3
78. Answer (1)
B
Let the equation of the plane
Q(1, 0, –3)
be x – y – z = 9
and the straight line AB is
A
P
x−2 y +2 z−6
=
=
2
3
−6
Let the point (1, 0, –3) be Q. Through Q and the equation of the line passing parallel to AB may be written as
x −1 y − 0 z + 3
=
=
2
3
−6
Any point on the line may be taken as (1 + 2r, 3r, –3 – 6r) which will lie in the plane if
1 + 2r – 3r + 3 + 6r = 9
⇒ r=1
⇒ P(3, 3, –9)
Required distance = PQ = (3 − 1)2 + (3 − 0 )2 + ( −9 + 3)2
= 7 units
79. Answer (4)
The normal to the plane and the straight line are perpendicular to each other. Hence the given Straight line
is parallel to the plane.
80. Answer (1)
We have,
(–a, 0)
a2 =
2
a + 4b
4
(a, 0)
2
⇒ 4a2 = a2 + 4b2
⇒ 3a2 = 4(a2) (1 – e2)
⇒ 4e2 = 4 – 3 = 1
2
⇒ e =
1
1
e=
4
2
81. Answer (4)
The chord of contact of the point (4, 4) w.r.t. the circle
x2 + y2 – 2x – 2y – 7 = 0 is 4x + 4y –1(x + 4) –1(y + 4) = 0
3x + 3y – 15 = 0
⇒ x + y –5 = 0
p = the length of perpendicular drawn from (1, 1) to the chord x + y – 5 = 0 is
3
2
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BD 2 = CB 2 − CD 2 = 9 −
Success Achiever (Solutions)
9 9
=
2 2
3
P
C
3
⇒ BD =
(4, 4)
(1, 1)
2
⇒ The length of the chord AB
= 2×
3
2
= 3 2 units
82. Answer (1)
Since 1 lies between the roots of x2 – kx – k + 3 = 0
hence k2 + 4(k – 3) > 0 and 1 – k – k + 3 < 0
⇒ k2 + 4k – 12 > 0 and 2k – 4 > 0
⇒ (k + 6) (k – 2) > 0 and k > 2
⇒ k > 2 or k < –6 and k > 2
⇒ k ∈ (2, ∞)
83. Answer (4)
We have,
sin–1(sin 4) = sin–1(sin(π – 4)) = π – 4
Discriminant = p2 – 4(π – 4) = p2 + 4(4 – π) = positive always.
Thus x2 – px + sin–1(sin4) is positive when x lies outside the roots and it is negative when x lies between
roots.
⇒ x2 – px + (sin–1(sin 4) can't be always positive ∀ x ∈ R ⇒ p ∈ φ
84. Answer (1)
We have,
cos3A + cos3B + cos3C = 3cosAcosBcosC
⇒ cos3A + cos3B + cos3C – 3cosAcosBcosC = 0
⇒ (cosA + cosB + cosC)(cos2A + cos2B + cos2C – cosAcosB – cosBcosC – cosCcosA) = 0
⇒
1
(cos A + cos B + cos C )[(cos A − cos B )2 + (cos B − cos C )2 + (cos C − cos A)2 ] = 0
2
A
B
C
⇒ 1 + 4 sin sin sin [(cos A − cos B ) 2 + (cos B − cos C ) 2 + (cos C − cos A )2 ] = 0
2
2
2

⇒ cos A = cos B = cos C as 1 + 4 sin
A
B
C
sin sin ≠ 0
2
2
2
⇒ A=B=C
⇒ ΔABC is an equilateral triangle.
85. Answer (1)
Let z = x + iy so that the given condition reduces to (x + 5)2 + y2 – (x – 5)2 – y2 = 10
⇒ 2x = 10 ⇒ x = 5.
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86. Answer (2)
The curve y =
1
x2
2 −|x|
intersects the curve y = e
in four points.
y
y
1
2
x
2 −|x|
y=e
x
87. Answer (3)
We have,
φ′( x ) =
x
x2
−
sin x
1 − cos x 2(1 − cos x )2
which is > 0 for 0 < x < 1 and ψ ′(x) < 0 for 0 < x < 1
88. Answer (1)
We have,
i=
2i 1 + 2i − 1 (1 + i )2
=
=
2
2
2
⇒ in =
(1 + i )2n
i n + i −n =
2n
2n
(1 + i )
n
=
+
(1 + i )2n (1 − i )2n
=
(1 − i )2n .2n
2 2n
2n (1 − i )2n
=
2n
(1 − i )2n
2n
(1 − i )2n
89. Answer (1)
We have,
x
y
+
= x cos 2α + y sin 2α
sec 2α cosec 2α
=x
1 − tan 2 α
1 + tan 2 α
+y
2 tan α
1 + tan 2 α
y2
y
2.
2
x
x
=x
+ y.
y2
y2
1+ 2
1+ 2
x
x
1−
= x.
=x
x2 − y 2
2
x +y
2
+
(x 2 + y 2 )
(x 2 + y 2 )
2xy 2
x2 + y 2
=x
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90. Answer (4)
We have,
log 1 (log 4 ( x 2 − 5)) > 0
3
 1
2
⇒ log 4 ( x − 5) <  
3
o
and log 4 ( x 2 − 5 ) > 0
⇒ x2 – 5 < 4 and x ≠ − 6 , 6
⇒ x2 < 9
⇒ − 3 < x < 3 and x ≠ − 6 ,
6
x 2 − 5 > 0  x > 5 and x < − 5
⇒
x ∈ ( −3, − 6 ) ∪ ( 6 , 3 )
91. Answer (1)
nπ + p

nπ + p
nπ

| cos x | dx = | cos x | dx +
0
 | cos x | dx
nπ
0
p
π/2

= 2n + | cos x | dx = 2n +
0

cos x dx −
0
p
 cos x dx
π/2
= 2n + 2 – sinp
92. Answer (3)
We have,
0 if x ∈ Z
[ x ] + [− x ] = 
- 1 if x ∉ Z
thus,
q

q

q

q

[ x ]dx + [ − x ]dx = ([ x ] + [ − x ]) dx = ( −dx )
p
p
p
p
= –(q – p) = p – q
92a.
Answer (3)
(AIEEE 2009)
π
I =  [cot x ]dx
0
π
I =  [cot( π − x )]dx
0
π
2I =  ([cot x ] + [ − cot x ])dx
0
π
2I =  ( −1)dx = −π
0
I=−
π
2
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93. Answer (3)
We have,
f ′(sin2x) = cos2x = 1 – sin2x
⇒ f ′( x ) = 1 − x  f ( x ) = x −
 f (1) = 1 −
 k=
∴
f (x) = x −
x2
+k
2
1
1
+k = k + =1
2
2
1
2
x2 1
+
2 2
94. Answer (1)
Since f(x) is an even function
⇒ f(x) = f(–x)
⇒ f ′ (x) = –f ′ (–x)
⇒ f ′′(x) = f ′′(–x)
1004

⇒
1004
( x 3 f ( x ) + xf ′′( x ) + 2)dx =
−1004

( x 3 f ( x ) + xf ′′( x ))dx +
−1004
1004
 1dx
−1004
= 0 + 1 × (1004 + 1004)
= 2008
95. Answer (3)
Let x + ex = t then
I=
dt
 (t − 1) t
2
96. Answer (1)
1
2
1 −3
x dx = dt
3
Let x 3 = t ,
I=
3t 2
 t (t − 1) dt
97. Answer (3)
Let tan θ = x then,
I=
=

cos7 x − cos8 x
dx
1 + 2cos5 x

15
x
x sin
2
2 dx
5x 

1 + 2  2cos2
− 1


2
2 sin
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2 sin
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5x 
5x 
x
3 − 4 sin2
sin d x
2 
2 
2

2 5x 
 3 − 4 sin

2 

=

=
 (cos 2x − cos3 x ) dx
=
sin2 x sin3 x
−
+C
2
3
where x = tan θ
98. Answer (1)
The equation of the normal at (x, y) is
(Y − y )
dy
+ ( X − x ) = 0 whose intercept on
dx
dy 


x-axis is  x + y
dx 

dy 

Hence x  x + y
 = 2( x 2 + y 2 )
dx 

dy x 2 + 2y 2
=
putting y = ux
dx
xy
so that,
⇒ ln
2u
1+ u 2
(1 + u 2 )
x
2
du =
2dx
x
= ln c 
x2 + y 2
x4
=c
⇒ x2 + y2 = cx4
99. Answer (4)
When sinx is positive, the given equation has no solution.
When sinx is negative, then the given equation has solution for sinx = –1. Due to periodic nature of sinx, the
given equation has infinitely many solutions.
100. Answer (3)
101. Answer (4)
102. Answer (2)
103. Answer (3)
We have
 
 
 
(a + b ) × (a − b ) = 2b × a
 
 
 
 
⇒ {(a + b ) × (a − b )} 2 = 2(b × a ) ⋅ 2(b × a )


π
= 4 | a |2 | b |2 sin n 2
4


= 2 | a |2 | b |2
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104. Answer (1)
Squaring and adding the given results
we get,
u2 + v2 = 25 + 49 = 74
⇒ 2uu + 2vv = 0
⇒ uu + vv = 0
⇒
u
v
=−

v
u
⇒ Again differentiating it w.r.t. x we get,
vu − uv
uv − vu vu − uv
=−
=
2
v
u2
u2
  − uv
  =
uv
=
u2
(vu − uv )
v 2 
v

 −v ⋅ ⋅ v − uv 
u

u2 
=−
=
v 2
v 3  v 2 + u 2 


u2  u

−74(v )3
u3
105. Answer (4)
We have,
  
1
a × (b × c ) = b
2
    1
⇒ (a.c )b − (a.b )c = b
2
   1    
⇒  a.c − b − (a.b )c = 0
2

 
   1
⇒  a.c − b = (a.b )c
2

 
 1 
⇒ a.c = & a.b = 0 as b & c
2
are non-parallel unit vectors.


⇒ a perpendicular b
106. Answer (1)
We have,
The required ratio = −
 4 + 3 − 10 + 1  2
(ax1 + by 1 + cz1 + d )
= −
=
(ax 2 + by 2 + cz2 + d )
 6 + 12 − 6 + 1  13
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107. Answer (4)
−1  1 
−1
Angle between two diagonals of a cube = cos   = tan ( 2 2 )
3
108. Answer (1)
Let
1
1
1
x y z
+ + = 1 be the equation of the plane, then 2 + 2 + 2 = 1
a b c
a
b
c
a b c 
The centroid of the ΔABC is given by  , , 
3 3 3
⇒
1
a
 
3
2
+
1
b
 
3
+
2
1
c 
 
3
2
C(0, 0, c)
=9
0)
A(a, 0,
B(0,
b,
0)
a b c 
1
1
1
⇒  , ,  lies on 2 + 2 + 2 = 9
3
3
3


x
y
z
∴ k = 9
109. Answer (1)
The point (4, 2, k) lies on the plane
2x – 4y + z = 7
⇒ 8–8+k=7
⇒ k = 7
109a. Answer (1)
(AIEEE 2009)
The point (2, 1, –2) is on the plane x + 3y – αz + β = 0
Hence
2 + 3 + 2α + β = 0
2α + β = –5
Also
... (i)
1(3) + 3(–5) + –α(2) = 0
3 – 15 – 2α = 0
2α = –12
α = –6
Put α = –6 in (i)
β = 12 – 5 = 7
∴
(α, β) ≡ (–6, 7)
110. Answer (4)
Let sin–1x = θ ⇒ sinθ = x
We have,
−1≤ x ≤ 0 
and so,
−π
≤ sin −1 x ≤ 0
2
−π
≤θ≤0
2
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π
≥ −θ ≥ 0
2
⇒ cos(−θ) = 1 − x 2
⇒ − θ = cos −1( 1 − x 2 )
⇒ θ = − cos −1( 1 − x 2 )
111. Answer (1)
The given system of equation may reduce to
3logex + 2logey = 4 + logez
4logex + 3logez = 5 + 2logey
and 2logey + 6logez = 6 + 2logex
We observe that the determinant of the coefficient of logex, logey, logez
2 −1
− 2 3 = −108 ≠ 0
3
= 4
−2
2
6
Hence the given system of equation has unique solution.
112. Answer (3)
1
3f ′( x )
loge (1 + 3f ( x ))
1
3
3
1 + 3f ( x )
lim
= lim
= .
=
x →a
x →a
2f ( x )
2f ′( x )
2 1 + 3f (a ) 2
113. Answer (1)
π

= sinx + sin  x − 
6

Y
= sinx + sinx
1
3
− cos x.
2
2

3  1

= sinx 1 + 2  − 2 cos x


2
Ymax =


1 + 3  + 1


2
4


3
1
+ 3+
4
4
=
1+
=
2+ 3
≈
2 + 1.73
≈
3 . 73
≈ 1.9
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≈ –1.9
d
= Ymax–Ymin ≈ 3.8
[d] = 3
114. Answer (4)
 nπ

sin
+ θ  = ( −1)
 2

n −1
2 cos θ
if x is odd
n
 nπ

sin
+ θ  = ( −1) 2 sin θ if n is even.
 2

1 + tan 2 θ = sec 2 θ , if θ ≠ (2n + 1)
π
2
115. Answer (2)
using A.M. ≥ G.M.
Let y =
sin12 x + cos12 x
≥
2
(sin x. cos x )12
minimum value of y is obtained when numbers are equal  x =
 1 

Ymin = 
 2
12
12
 1 

+ 
 2
=
1
+
26
1
26
=
2
26
=
1
25
=
π
4
1
32
116. Answer (3)
We have
31.4 < 10 π
The number of θ satisfying the given condition is 10 viz 0, π, 2π.....9π, only 10 solutions exist for the given
equation.
117. Answer (1)
Common domain of the function is {–1, 1} as
for sin −1 x, cos −1 x
−1≤ x ≤ 1
for tan–1x, cot–1x
–∞ < x < ∞
for cosec–1x, sec–1x
x ∈ ( −∞,−1] ∪ [1, ∞ )
118. Answer (2)
By graph
y
π
–1
cos x
1
–π –1
2
cosx
O
1
π
x
2
Clearly only one solution exists.
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419
119. Answer (3)
sin −1 x1 =
π
 x1 = 1
2
sin −1 2 x 2 =
π
1
 x2 =
2
2
sin −1 3 x 3 =
1
π
 x3 =
2
3
tan −1 x1 + tan −1 x 2 + tan −1 x 3
=
1
1
π
+ tan −1 + tan −1
4
2
3
 1 1
5 
 +3 π
π
−1 2
−1
6 = π + π = π
+ tan 
 = + tan 
= 4
1
5

 4 4 2
 1 −  4
 6
6 

120. Answer (2)
Option (1) and (4) are not valid in right triangle and cosA + cosB + cosC = 1 + 4 sin
option (2) is correct.
A
B
C
sin sin
hence
2
2
2
121. Answer (3)
x
10 m
45°
sin 45° =
10
x
x = 10 2 m
122. Answer (4)
at k = 0, z → Point
at k ≠ 1, k > 0, z → circle
at k = 1, z → straight line
Hyperbola is never formed.
123. Answer (1)
a=0
b = ( −1)100 −1 = ( −1)99 = −1
a + b = 0 − 1 = −1
124. Answer (3)
( | z–z1|+| z–z2| ) min = | z1 – z2 |
In this case the locus of z is the line segment joining z1 and z2 .
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125. Answer (3)
If f(x) = ax3 + bx2 + cx+d = 0
has only one real roots then its derivative has no real root
i.e., f ′( x ) = 3ax 2 + 2bx + c = 0 has
no real root
if (2b )2 − 4 × 3a × c < 0
⇒ b2 – 3ac < 0
b 2 < 3ac
126. Answer (4)
Let α = p + iq  | α | = p 2 + q 2
β = piq  | β | = p 2 + q 2
αβ = p 2 + q 2 > 0, arg (αβ ) = 0 = arg (α ) = –arg (β) also α = β
hence αβ < 0, is not true.
127. Answer (4)
1.f(α) = –2 < 0
1.f(β) = –2 < 0
hence α , β lies between the roots, hence (4) is correct.
128. Answer (1)
Let A.P is a, a+d, a + 2d, ……
Sn =
n
d
[2a + (n − 1)d ] = na + n(n − 1) = quadratic expression in n.
2
2
129. Answer (2)
Let number are a and b.
A.M. = A =
a+b
2
G.M. = G =
ab
H.M. = H =
2ab
a+b
Clearly G2 = A H
130. Answer (3)
Using A.M. ≥ G.M.
1
a1 + a2 + …an −1 + 3an
≥, (a1.a2 …an −1.3 an ) n
n
1
a1 + a2 + … + …3an ≥ n(3c ) n
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131. Answer (4)
Let the number are a and b.
A.M. = A, G.M. = G, H.M. = H
G2 = AH
 G2 = A.
G
because G = 2H
2
2G2 = AG
2G = A
A
=2
G
a
=
b
A + A2 − G 2
A − A2 − G 2
or
A − A2 − G 2
A + A2 − G 2
A + A2 − G 2
2G + 4G 2 − G 2
a
=
Let
=
b
A − A2 − G 2
2G − 4G 2 − G 2
=
Also,
2+ 3
2− 3
=7+4 3
b 2− 3
=
=7−4 3
a 2+ 3
132. Answer (2)
cos x −
cos 2 x cos 3 x cos 4 x
+
−
…
2
3
4
x
x


log(1 + cos x ) = log 2 cos 2  = log 2 + 2 log cos 
2
2


133. Answer (1)
Let the polygon has x sides.
∴ No. of diagonals =
x ( x − 3)
= 35 = x (x–3) = 70
2
⇒ x 2 − 3 x − 70 = 0
⇒ (x – 10) (x + 7) = 0 ⇒ x = 10
134. Answer (3)
Sum of coefficients = a = (5 – 3)10 = 210
Sum of binomial coefficients = b = 210 ⇒
a
b
=1 ⇒
=1
b
a
135. Answer (1)
Total term =
n + 4 – 1C
4–1
=
n + 3C
3
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136. Answer (1)
Adding first and last term, second and second last term and so on, we get, using C0= Cn, C1 = Cn – 1…
(C0 + C1 + C 2 + C n −1 + C n ) + (C0 + C1 + C 2 + Cn −1 + Cn ) + … to
n
terms as n is even.
2
n
hence sum = 2 n + 2 n + 2 n + … upto
term
2
sum = 2 n.
n
= n.2 n −1
2
137. Answer (3)
Total number of such triangles = number of sides of polygon.
138. Answer (3)
n(S) = 10!
n(E) =
10 !
3!
10! 1 1
= =
3! 6
p(E) = 3!
10!
139. Answer (3)
Total probability =
12 3 1
 + =
25 5 2
1 3
×
2 5 =3
∴ Required probability =
1
5
×1
2
140. Answer (3)
12
8
4
Required number of ways = C4 × C4 × C4 =
12!
(4!)3
141. Answer (2)
Let the line is ax + by + c = 0 … (i)
pi =
ai + bi 2 + c
a2 + b2
Σpi = 0 = aΣi + bΣi 2 + nc = 0
⇒ a
Σi
Σi 2
+b
+c =0
n
n
... (ii)
Comparing (i) and (ii)
x=
Σi
Σi 2
, y=
n
n
⇒ x=
n +1
(n + 1)( 2n + 1)
, y=
2
6
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142. Answer (2)
 0 + 1+ 5 0 + 1+ 8 
,
 = ( 2, 3 )
Centroid of Δ ABC = centroid of Δ A1B1C1 = centroid of Δ A2B2C2 = 
3
3


143. Answer (3)
Let a = b = 2, c = –2
ax 2 + 3 y 2 + c = 2 x 2 + 2y 2 − 2 = 0
x 2 + y 2 = 1; circle
x 2 − 4 xy + 5y 2 = 0
 ( x − y )( x − 5 y ) = 0 two lines.
144. Answer (3)
=
{
1 + x 2  x cos(cot −1 x ) + sin(cot −1 x

} − 1
2
1
2

1
x
 x cos . cos −1
+ sin sin −1

2
1+ x
1+ x 2

=
1+ x 2
=
2
 2
2
 x + 1
 = x 1+ x 2
1 + x 2 
−
1

1 + x 2 



1
2
2


−
1




1
145. Answer (1)
a1 = 3, b1 = 4
a2 = 3, b1 = 4
a1a2+b1b2 = 9 – 16 = –7 < 0
Hence equation of acute angle bisector is
3 x + 4 y + 5 3 x − 4 y + 10
=
5
5
 3x + 4y + 5 = 3x – 4y + 10
 8y – 5 = 0
y=
5
8
146. Answer (3)
old line
x y
+ =1
3 4
new line
x y
+ =1
4 a
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The distance of the line from (0,0) is fixed, hence
1
1
=
1 1
1
1
+
+ 2
9 16 16 a
=
1
a
2
=
1
 a2 = 9
9
a= ± 3
147. Answer (4)
x2y + xy2 – xy = 0 ⇒ xy (x + y – 1) = 0
x = 0, y = 0, x + y = 1
Clearly orthocentre = (0, 0)
148. Answer (1)
y
(0, q)
D
(0, b) B
(0,0) A(a, 0)
C(p, 0)
x
O
If ABCD are concyclic then
OA.OC = OB.OD
 ap = bq 
a q
=
b p
149. Answer (1)
The point of intersection 2x + 3y – 5 = 0 and x + y – 2 = 0 is (1, 1), Let A ≡ (1, 1) and B ≡ (4, 5). The required line
will be perpendicular to AB.
slope of AB =
5 −1 4
=
4 −1 3
slope of the required line = −
3
4
equation of the line is y − 1 = −
3
( x − 1)
4
4y – 4 = –3x + 3
4y + 3x = 7
150. Answer (2)
If line (i) is parallel to (iii) then
−
2
k
2
=− k =
3
1
3
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If line (ii) is parallel to (iii)
1
k
then − = −  k = 1
1
1
sum of values =
2
5
+1=
3
3
151. Answer (3)
The equation of line mirror will the equation of angle bisectors of image line and object line.
3x + 4y + 5
5 x + 12y + 13
=±
5
13
⇒ 39x + 52y + 65 = ± (25x + 60y + 65)
taking positive sign we get
39x + 52y + 65 = 25x + 60y + 65
⇒ 14x – 8y = 0
...(i)
7x – 4y = 0
taking (–) we get
39x + 52y + 65 = – (25x + 60y + 65)
⇒ 39x + 25x + 52y + 60y +130 = 0
⇒ 64x + 112y + 130 = 0
⇒ 32x + 56y + 65 = 0
152. Answer (1)
x3 + kx2y + 2xy2 + ky3 = 0
2
y
y
y
⇒ 1 + k   + 2   + k  
x
x
x
3
But the equation of line passes through (0, 0) and having slope m is y = mx or m =
y
y
, we may use = m in the
x
x
above equation
⇒ km3 + 2m2 + km + 1 = 0
...(A)
Let the slopes of the lines represented by the equation (A)
are m1, m2, m3
⇒ m1 + m2 + m3 = −
2
k
m1m2 + m2m3 + m3m1 =
m1m2m3 = −
1
k
... (i)
k
=1
k
... (ii)
... (iii)
If m1m2 = –1, then by (iii)
− m3 = −
1
1
 m3 =
k
k
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as we know that m3 is a root of the equation (A)
3
2
 1
 1
 1
⇒ k   + 2   + k   + 1 = 0
k
k
k
1
+
k2
⇒
2
k2
+ 1+ 1 = 0
3
3
+ 2 = 0  k2 = − ,
2
2
k
hence no real value of k exists.
153. Answer (4)
Equation of chord of contact AB is
5x + 6y −
x+5 y +6
−
=0
2
2
10x + 12y – x – 5 – y – 6 = 0
9x + 11y – 11 = 0
A
P(5, 6)
B
Equation of circle passing through P, A and B is
x2 + y2 – x – y + λ(9x + 11y – 11) = 0
But it passes through (5, 6) hence
25 + 36 – 5 – 6 + λ(9 × 5 + 11 × 6 – 11) = 0
λ=−
1
2
Hence the equation of the circle becomes
x2 + y2 – x – y –
1
(9x + 11y – 11) = 0
2
2x2 + 2y2 – 2x – 2y – 9x – 11y + 11 = 0
⇒ 2x2 + 2y2 – 11x – 13y + 11 = 0
154. Answer (1)
Let C : x2 + y2 + 2gx + 2fy + C = 0,
Let –g = α,
–f = β ⇒ f = –β
C1 : x2 + y2 + 2g1x + 2f1y + k1 = 0
C2 : x2 + y2 + 2g2x + 2f2y + k2 = 0
If C is orthogonal to C1 and C2 then
2gg1 + 2ff1 = C + k1
...(i)
2gg2 + 2ff2 = C + k2
...(ii)
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by (i) and (ii)
2gg1 – 2gg2 + 2ff1 – 2ff2 = k1 – k2
2g(g1 – g2) + 2f(f1 – f2) = k1 – k2
Putting (g = – α) (f = –β), we get
–2α(g1 – g2) – 2β(f1 – f2) = k1 – k2
Locus of (α, β)
x(g1 – g2) + y(f1 – f2) = k2 – k1
g1 − g 2
slope of AB = − f − f = m1
1
2
f2 − f1
slope of C1C2 = g − g = m 2
2
1
⇒ m1m2 = –1,
Hence angle = 90° =
π
2
155. Answer (3)
d=
3 × 1+ 4 × 1+ 8
9 + 16
=
15
=3
5
Hence the length of latus rectum = 3 × 4 = 12
156. Answer (3)
This is standard definition of a parabola.
157. Answer (2)
y2 = 4x – (i)
x2 + y2 – 6x + 1 = 0 (ii)
solving (i) and (ii) for x
x2 + 4x – 6x + 1= 0
x2 – 2x + 1 = 0
x = 1, 1
Hence the curves touch each other at two points.
158. Answer (3)
The point of intersection of tangents at A and B is
P(t1t2, t1+t2) = (h, k)
The point of intersection of normals at A and B is
(2 + (t12 + t22 + t1t2) – t1t2(t1 + t2) = (x1, y1)
h = t1t2, k = t1 + t2
x1 = 2+(t12 + t22 + t1t2)
= 2 + ((t1 + t2)2 – t1t2)
x1 = 2+(k2 – h)
x 1 = 2 + k2 – h
similarly, y1 = –t1t2(t1 + t2)
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y1 = –hk
Hence point of intersection of normals is
(2 + k2 – h, –h k)
159. Answer (1)
Let the chord is y = mx + c
y
0
A
90°
x
y = mx + c
B
Combined equation of OA and OB is obtained by getting a homogeneous equation of second degree that is
given by
 y − mx 
y 2 = 4x 
 c 
cy2 = 4xy – 4mx2
⇒ cy2 + 4mx2 – 4xy = 0
As the angle between OA and OB is 90° hence
coefficient of x2 + coefficient of y2 = 0
⇒ c + 4m = 0
But y – mx – c = 0
...(i)
4m + c = 0
...(ii)
Comparing (i) and (ii)
y
mx
c
=−
=−
0
4m
c
⇒ y = 0, x = 4
fixed point is (4, 0)
160. Answer (4)
Let P ≡ (1, 1) and a, b, e be the semi-major axis, semi-minor axis and eccentricity respectively.
2
2
PS1 = ( 4 − 1) + ( 5 − 1) = 5
2
2
PS2 = ( 6 − 1) + (13 − 1) = 13
As we know that
PS1 + PS2 = 2a
⇒ 2a = 18 ⇒ a = 9
and S1S2 = 2ae
⇒
( 6 − 4)2 + (13 − 5)2 = 2ae
⇒
4 + 64 = 18 × e ⇒
68
=e ⇒
18
17
=e
9
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161. Answer (1)
Let the ellipse is
x2 y 2
+
=1
a2 b2
and middle point is (h, k)
and fixed point is (α, β)
Equation of chord is T = S1
⇒
xh yk
h2 k 2
+
−
1
=
+
−1
a2 b2
a 2 b2
But the chord passes through (α, β)
⇒
αh βk h 2 k 2
+
=
+
a 2 b2 a2 b2
x 2 y 2 αx βy
+
−
−
= 0 , which is clearly an ellipse.
a 2 b 2 a2 b2
⇒ locus of (h, k) is
162. Answer (2)
x2 y 2
−
=1
2
3
The equation of any tangent to the hyperbola is y = mx ± 2m 2 − 3
As the tangent passes through (1, 1)
⇒ 1 = m ± 2m 2 − 3
⇒ 1 − m = ± 2m 2 − 3
⇒ 1 + m2 – 2m = 2m2 – 3
⇒ m2 – 2m + 1 = 2m2 – 3
⇒ m2 + 2m – 4 = 0
⇒ m1 + m2 = –2
⇒ m1m2 = –4
If angle between the tangent θ.
then tan θ = ±
=±
m1 − m2
1 + m1m2
( m1 + m2 )2 − 4m1m2
1 + m1m2
=±
4 − 4 × ( −4 )
1− 4
=±
2 5
2 5
=±
−3
3
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If θ is acute then
 2 5
θ = tan−1 
 3 
163. Answer (1)
Let us take general point
P ≡ (3 sec θ, 2 tan θ)
Equation of tangent at P is
x ( 3sec θ) y ( 2tan θ)
−
=1
9
4
x sec θ tan θ
−
y =1
3
2
y
B
x
y
sec θ - tanθ =1
3
2
M
x
A
⇒
2 
 3


A≡
, 0 , B ≡  0,
 sec θ 
 tan θ 
Let middle point of AB is M ≡ (h, k)
h=
3
2 sec θ
sec θ =
k=
3
2h
... (i)
−2
−1
=
2 tan θ tan θ
⇒ tan θ =
−1
k
... (ii)
by (i) and (ii)
sec2θ – tan2θ = 1
⇒
9
4h
2
−
1
k2
=1
locus of (h, k),
9
1
−
=1
4x 2 y 2
164. Answer (4)
Equation of asymptotes of xy = c is x = 0 and y = 0
...(i)
Now we are given xy = 2x + 3y
xy – 2x – 3y + 6 = 6
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431
y(x – 3) – 2(x – 3) = 0
(x – 3) (y – 2) = 0
...(ii)
by (i) and (ii) we conclude that
Asymptotes are x = 3 and y = 2
165. Answer (2)
x2 + y2 = 1
required curves is
x2 + y2 – 12x – 16y + λ (x2 + y2 –1) = 0
at x = 1, y =1
1 + 1 – 12 – 16 + λ (1 + 1 – 1) = 0
– 26 + λ = 0
λ = 26.
Thus the equation of the required the equation of required circle is x2 + y2 –12x – 16y + 26 (x2 + y2 –1) = 0
27x2 + 27y2 – 12x –16y – 26 = 0
⇒ x2 + y2 –
12
16
26
x−
y−
=0
27
27
27
165a. Answer (1)
(AIEEE 2009)
x2 + y2 + 3x + 7y + 2p – 5 + λ(x2 + y2 + 2x + 2y – p2) = 0, λ ≠ –1 passes through point of intersection
of given circles.
Since it passes through (1, 1), hence
7 – 2p + λ(6 – p2) = 0
⇒
7 – 2p + 6λ – λp2 = 0
If
λ = –1, then 7 – 2p – 6 + p2 = 0
p2 – 2p + 1 = 0
p=1

λ ≠ –1 hence p ≠ 1
∴
All values of p are possible except p = 1
166. Answer (4)
C1 ≡ (0, 0)
C2 ≡ (3, 3)
C1C2 = 3 2
r1 = |sinθ|
r2 = |cosθ|
r1 + r2 =| sin θ | + | cos θ |=
(r1 + r2 )max = 2
Here, C1C2 > r1r2 and hence 4 common tangents are possible.
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167. Answer (2)
5×4
= 10
2
168. Answer (1)
5
C2 =
The length of tangent is
3 sin θ + 4 cos θ − 1
∴ maximum length =
9 + 16 − 1 = 2
Y
169. Answer (3)
AB = l, AR = lsinθ
BR = OA = lcosθ
PA = ARcos(90 – θ) = lsinθ.sinθ = lsin2θ
K = PS = PAsinθ =
P
T
(h, K)
lsin3θ
Similarly, PT = lcos3θ = h
2
2
2
(90 – θ )
2
2
 K 3  h 3
  +   = 1 x 3 + y 3 = l 3
 l 
l
O
S
θ
A
X
170. Answer (1)
We have,
a – b + b – a = 0 ⇒ coefficient of x 2 + coefficient of y 2 = 0, hence angle between the lines is = 90°
171. Answer (4)
Radical axis is perpendicular bisector of the segment joining (0, 0) and (2, 2) that is given by
radical axis
m=1
(0, 0)
(1, 1) (2, 2)
m1 = –1
radical axis is y – 1 = –1(x – 1)
y – 1 = –x + 1
x+y=2
172. Answer (4)
Because it is not fixed that the tangent and normal are at the same point.
173. Answer (1)
e1 = 1, e2 < 1, e3 > 1
minimum of [e1 + e2 + e3] = 2
174. Answer (3)
Director circle of the parabola is directrix = 3x + 4y + 8 = 0
Latus rectum = 2 × perpendicular distance of focus from directrix
= 2×
3 × 1+ 4 × 1+ 8
= 2×3 = 6
5
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175. Answer (1)
Semi-latus rectum = Harmonic mean of focal radii
2×3× 6
=4
3+6
Latus rectum = 2 × 4 = 8
176. Answer (3)
16 9
+ −1= 1> 0
16 9
E ( 4, 3) =
C(4, 3) = 16 + 9 – 9 = 16 > 0
P(4, 3) = 16 – 9 = 7 > 0
Hence point is outsides all P, C and E.
177. Answer (4)
The equation of normal at (x1, y1) of
x2
a2
−
y2
b2
= 1 is
a2 x b2 y
+
= a2 + b2
x1
y1
⇒
4.x
2 2
+
1.y
=5
1
2x + y = 5
178. Answer (2)
Obviously
178a. Answer (4)
[JEE (Main)-2017]
P ( A) + P (B ) − P ( A ∩ B ) =
1
4
P (B ) + P (C ) − P (B ∩ C ) =
1
4
P (C ) + P ( A) − P ( A ∩ C ) =
1
4
P ( A) + P (B ) + P (C ) − P ( A ∩ B ) − P (B ∩ C ) − P ( A ∩ C ) =
 P(A ∩ B ∩ C ) =
∴
3
8
1
16
P(A ∪ B ∪ C ) =
3 1
7
+
=
8 16 16
179. Answer (4)
x2 = x |y + 1| ⇒ x = 0 or |y + 1| = x
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180. Answer (4)
for reflexive → (b, b), (d, d)
for symmetric → (b, a), (c, a)
for transitive → (b, c), (c, b)
181. Answer (3)
f ( x + 1) + f ( x − 1) = 3f ( x )
...(i)
x→x+1
⇒ f ( x + 2) + f ( x ) = 3f ( x + 1) , substituting x + 1 for x
...(ii)
and f ( x ) + f ( x − 2) = 3f ( x − 1) , substituting x – 1 for x
...(iii)
2f ( x ) + f ( x + 2) + f ( x − 2) = 3 (f ( x + 1) + f ( x − 1)) [adding (ii) & (iii)]
⇒ 2f ( x ) + f ( x + 2) + f ( x − 2) = 3 . 3f ( x )
⇒ f(x + 2) + f(x – 2) = f(x)
...(iv)
f(x + 4) + f(x) = f(x + 2), substituting x + 2 for x
...(v)
Adding (iv) & (v)
f(x + 4) + f(x – 2) = 0
⇒ f(x + 6) + f(x) = 0, substituting x + 2 for x
...(vi)
x→x+6
⇒ f(x + 12) + f(x + 6) = 0
...(vii)
by (vi) & (vii)
f(x) = f(x + 12) period of f(x) = 12
182. Answer (1)
f(x) = 1 + {2x}
period of f(x) =
1
2
183. Answer (4)
If n(A) = n
n(B) = m
then number of one-one function from A to B is
mP
n
= 5P4 = 5 × 4 × 3 × 2 = 120
184. Answer (1)
  x   x  1
x x 1
h′( x ) = 2. u   .u′ . + 2v  v ′ .
2
2
2
2 2 2
    
x x
x x
= u  .u′  + v  .v ′ 
2
2
   
2 2
x x
x x
= u′′ .u′  − u′ u ′′ 
2
2
   
2 2
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using the given relation, we get
h′(x) = 0 ⇒ h(x) = constant
⇒ h(6) = h(12) = 6
185. Answer (3)
If f(x) and g(x) are mirror image of each other through the line y = x then they are inverse functions of each
other.
g(x) = f –1(x)
f(x) = 2x + 3
f(f–1(x)) = 2f–1(x) + 3
f –1( x ) =
x −3
2
186. Answer (1)
 3

 1
 x sin  x  + x 
 

lim 
x →−∞ 


1+ x 2


 1
 sin  x  1 



+ 
x
  1
  x 

1+ 0
=1
= lim 
 =
0 +1
1
x →−∞ 

+
1
 x 2

187. Answer (1)
Since f(x) is a continuous function and takes only rational values hence it is a constant function
5
⇒ f   = 3
2
188. Answer (4)
Clearly f(x) = 0, hence continuous and differentiable everywhere.
189. Answer (4)
e
 1+ 4 x 2  2
−1 2
lim 
x


x → 0  1+ 5 x 2
=e
=e
 1+ 4 x 2 −1−5 x 2
lim 
1+ 5 x 2
x → 0 

lim 
−2


x → 0  1+ 5 x 2 
 2
×
 x2

= e −2 =
1
e2
190. Answer (2)
y =| tan x |= − tan x at x =
3π
4
dy
= − sec 2 x
dx
dy
dy
3π
 3π 
= − sec 2 
at x =
 = −2
is
dx
dx
4
 4 
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191. Answer (4)
d
(tan x ) = sec 2 x > 0
dx
d
(x 3 ) = 3x 2 ≥ 0
dx
d
( x 2 ) = 2 x > 0 as well as < 0
dx
d
(cot x ) = −cosec 2 x < 0, hence
dx
It is a decreasing function in its domain.
192. Answer (4)
 dy 

Length of subnormal is  y
 dx  (1, 3 )
= |3 × 3| = 9
193. Answer (1)
e
=
x
(3 sin x − 4 sin 3 x ) + 3( 4 cos 3 x − 3 cos x )dx
e
x
(sin 3 x + 3 cos 3 x )dx = e x sin 3 x + C
194. Answer (3)
Period of | sin 2πx |=
π
1
=
2π 2
50
1
2
0
0
 | sin 2πx | dx = 100  | sin 2πx | dx
1
2

= 100 sin 2πx dx
0
1
 cos 2πx  2
= 100 −
2π  0

=
100
100
[1 + 1] =
2π
π
195. Answer (1)
g(x) = g(3 – 2x)
g′(x) = g′(3 – 2x))(–2)
g′(1) = g′(1)(–2)
⇒ 3g′(1) = 0
⇒ g′(1) = 0
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196. Answer (2)
f(x) = f(1 – x)
x→x+
1
2
 
1
1 

1

f  x +  = f 1 −  x +   = f  − x 
2
2
2





 
1

Let h( x ) = f  x +  sin x
2

1

h( − x ) = f  − x  sin( − x )
2

1

h( − x ) = − sin x f  − x + 
2

1

1

1
⇒ h( − x ) = − sin x f  x +  , as f  + x  = f  − x 
2
2
2






h(–x) = –h(x)
Hence h(x) is odd function.
1
2
1

 f  2 + x  sin x dx = 0
⇒
1
2
−
197. Answer (3)
π
I=
cos x
 1+ a
x
dx
...(i)
−π
π
I=
cos( − x )
 1+ a
−π
−x
π
dx =

−π
a x cos x
ax + 1
dx
...(ii)
Adding (i) & (ii),
π
2I =

cos x (1 + a x )
(1 + a x )
−π
π
dx =
 cos x d x
−π
π

= 2 cos xdx = 2[sin x ]0π = 0
0
⇒ I=0
198. Answer (3)
R is not symmetric because (2, 1) is not in the set. Hence R is not equivalence.
R is transitive as (1, 2) ∈ R, (2, 2) ∈ R ⇒ (1, 2) ∈ R
(1, 1) ∈ R, (1, 2) ∈ R ⇒ (1, 2) ∈ R
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199. Answer (4)
n((A × B) ∩ (B × A) = (n(A ∩ B))2 = 42 = 16
200. Answer (1)
y
y = |x|
1
–π
+π
2
y = |sinx|
x
π
because y = | x | and y = |sin x| meet only at x = 0, hence the area bounded by the curve = 0.
201. Answer (2)
y2 = 4ax
...(i)
x2 = 4ay
...(ii)
on solving
x2 = 4ay
x = 0, y = 0
x = 4a, y = 4a
2
(4
a, y = 4ax
4a
)
4a
Area bounded =
2 

 4ax − x dx = 16 a 2

3
4a 
0

202. Answer (3)
b4
 f ( x ) dx = e
b
sin b
0
Differentiating w.r.t. b, we get
f (b4).4b3 = ebsinb + cosb.e b
1
Let, b 4 = x  b = (x ) 4
⇒ f (x) =
1
3
4x 4
[e x
1/ 4
sin( x 1 / 4 ) + cos( x 1/ 4 )e x (1/ 4 ) ]
203. Answer (2)
P(x, y)
2A/3
A/3
x

3 y dx = xy
0
 3y = x
dy
+ y  y = kx 2
dx
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204. Answer (1)
Put x = r cos θ, y = r sin θ
r dr
=
r 2d θ
sin−1
a2 − r 2

r
dr
2
a −r2
= dθ
r
=θ+c
a
205. Answer (4)
p = e− x 
dy
= e− x
dx
⇒ y = – e–x + c
206. Answer (4)
We have,
xy = c 2 
xdy
+y =0
dx
dy
y
=−
dx
x
The differential equation of orthogonal trajectory is
−
dx
y
dx y
=− 
=
dy
x
dy x
xdx = ydy
⇒
x2 y 2
−
= c  x 2 − y 2 = c, Which is a hyperbola
2
2
207. Answer (4)
I =  ( x 2 y cos xydy + xy 2cos xy dx + x sin xydy + y sin xy dx )
=  ( x 2 y cos xydy + xy 2cos xy dx + x sin xydy + y sin xy dx )

= (( xdy + ydx )( xy cos xy + sin xy )

= ( xy . cos( xy )( xdy + ydx ) + sin( xy ).( xdy + ydx )

= d ( xy sin xy )
= x ysinx y + c, where c is a constant of integration.
208. Answer (3)
| A | = 23 | BCDB −1C −1D −1 |
= 23 | B | ⋅| C | ⋅| D | ⋅| B −1 | ⋅| C −1 | ⋅| D −1 |
= 23 | B | ⋅ | C | ⋅ | D | ⋅
1
1
1
⋅
⋅
|B | |C | | D |
=8
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209. Answer (3)
1
Δ1 = 9
3 5
11 13 = 0
17 19 21
7 3 5
Δ1 = 15 11 13 = 0
23 19 21
1
Δ2 = 9
7 5
1
15 13 = 0, Δ 3 = 9
17 23 21
3 7
11 15 = 0
17 19 23
Δ = Δ1 = Δ2 = Δ3 = 0 hence the equations have infinitely many solutions.
210. Answer (1)
a b c
b c a =0
c a b
on expanding
–(a3 + b3 + c3 – 3abc) = 0
a3 + b3 + c3 – 3abc = 0
(a + b + c)(a2 + b2 + c2 – ab – bc – ca) = 0
⇒ a+b+c=0
...(i)
also, ax + 2by + 3c = 0
...(ii)
comparing (i) & (ii)
ax 2by 3c
=
=
a
b
c
x 2y
=
=3
1
1
x=3
y=
3
2
 3
Fixed point  3, 
 2
211. Answer (3)
AB = A
⇒ (AB)A = A2
⇒ A(BA) = A2
(by associative law)
⇒ AB = A2
(by BA = B)
⇒ A = A2
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BA = B
⇒ (BA)B = B2
B(AB) = B2
⇒ BA = B 2
⇒ B = B2
212. Answer (4)
AA′ = I ⇒ | A | |A′| = 1 ⇒ |A2| = 1 ⇒ |A| = ± 1
|A| = ±1
sinx = ±1
π
x = nπ + ; n ∈ Z
2
213. Answer (1)
Equation of line y + 3 = 1 (x – 0) as it passes through the centre of circle
⇒ y=x–3
Applying condition of tangency
9 = a2 + b2
Area of ellipse A = πab
2
= πa 9 − a
dA
9
3
= 0  a2 =
a=
da
2
2
 b2 =
9
3
b=
2
2
Maximum area of ellipse
= π×
=
3
2
×
3
2
9π
2
214. Answer (4)
Since α, β are the roots of the equation
x2 – px + q = 0
∴ α + β = p and αβ = q
Now, (α1/4 + β1/4)4 = [(α1/4 + β1/4)2]2
= [α1/2 + β1/2 + 2(αβ)1/4]2
1/ 4
=  α + β + 2 αβ + 2( αβ) 


1/ 4
=  p + 2 q + 2(q ) 


2
2
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1/ 4
p+2 q
= p + 6 q + 4q
∴ α1/4 + β1/4
1/ 4
p+2 q
=  p + 6 q + 4q


1/ 4
215. Answer (3)
2z−
1 i
− = 2
2 2
1
 1+ i 
z−
 =
2
2


 1 1
⇒ A circle with centre  , 
2 2
1
radius =
2
2
2
1 
1
1

x − 2 + y − 2 = 2

 

Simplifying we get
x2 + y2 – x – y = 0
⇒ x(x – 1) + y(y – 1) = 0
ω=
…(i)
z − 1 x + iy − 1
=
z − i x + iy − i
=
( x − 1) + iy x − i ( y − 1)
×
x + i ( y − 1) x − i ( y − 1)
=
( x( x − 1) + y ( y − 1) + i ( yx − ( x − 1)( y − 1))
x 2 + ( y − 1)2
 yx − ( x − 1)( y − 1) 
ω= i

2
2
 x + ( y − 1)

⇒ ω is purely imaginary
⇒ ω+ω = 0
216. Answer (2)
The required number of ways is equal to the coefficient of x12 in
(x1 + x2 + x3 + …)4 (x3 + x4 + x5 + …)1
= Coefficient of x12 in x7(1 + x + x2 +…)5
1
= Coefficient of x5 in (1 − x )5
= Coefficient of x5 in (1 – x)–5
= 5 + 5 – 1C5 – 1 = 9C4
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443
217. Answer (3)
y2 = 8x, a = 2
Normal of slope 2
y = mx – 2am – am3
y = 2x – 24
…(i)
Tangent of slope 2
y = mx +
a
m
y = 2x + 1
…(ii)
Distance between equation (i) & (ii)
| 25 |
1+ 4
=5 5
218. Answer (1)
Let f(x) is satisfying LMVT in [0, x], then there exist C in [0, x] such that
f ′(C ) =
f ( x ) − f (0)
x −0
f ′(C ) =
f (x)
x
As |f ′(x)| ≤ a
|f ′(C)| ≤ a
⇒
f (x)
≤a
x
⇒ |f(x)| ≤ ax
As the interval is [0, 8]
|f(x)| ≤ 8a
219. Answer (3)
The four digits 3, 3, 5, 5 can be arranged at four even places in
8, 8, 8 can be arranged at the five odd places in
4!
= 6 ways and the remaining digits viz., 2, 2,
2!2!
5!
= 10 ways. Thus, the number of possible arrangements is
2!3!
(6) (10) = 60.
220. Answer (1)
 ab
A= 2
 −a
b2 

−ab 
0 0 
A2 = 

0 0 
A3 = 0
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A4 = 0
B=I+A
B100 = (I + A)100
= 100C0 I + 100C1 A + 100C2 A2 +…100C100A100
= I + 100A
(B100) = (I + 100A)
1 + 100ab
100b 2
−100a 2 1 − 100ab
|B100| =
= 1 – 1002 a2 b2 + 1002 a2 b2 = 1
|B|100 = 1
Independent of a, b.
221. Answer (2)
lim
x →0
x 3 z 2 − ( z − x )2
( 3 8 xz − 4 x 2 + 3 8 xz )4
= lim
x →0
( 3 x 3 8 z − 4 x + 3 8 z 3 x )4
x 4/3 3 2z − x
lim
=
=
x 3 2zx − x 2
x →0
x
4/3
 3 8z − 4 x + 3 8 z 


4
=
3
2z
2 3 8z 


4
1
223/3 z
222. Answer (2)
100  1

Let I =    f (r − 1 + x ) dx 
r =1  0

1
1
1
1
0
0
0
0
1
2
3
100
0
1
2
99
⇒ I =  f ( x ) dx +  f ( x + 1) dx +  f (2 + x ) dx + …… +  f (99 + x ) dx
⇒ I =  f ( x ) dx +  f ( x ) dx +  f ( x ) dx + …… +
 f ( x ) dx
100
∴
I=
 f ( x ) dx = a (given)
0
223. Answer (2)
9(x – 3)2 + 9(y – 4)2 = y2
⇒ 9(x – 3)2 + 8y2 – 72y + 144 = 0
9(x – 3)2 + 8(y2 – 9y) + 144 = 0
2

9  81
2
9(
x
−
3)
+
8
y
−
−  + 144 = 0


⇒
2 
4 

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2
9

9( x − 3)2 + 8  y −  = 162 − 144 = 18
2

2
9

8 y − 
2
9( x − 3)
2
+ 
=1
18
18
2
9

y − 2
( x − 3)
 =1
+
9
2
4
2
⇒
e2 = 1 −
∴
e=
2.4 1
=
9
9
1
3
224. Answer (3)
Given equation of lines
3 y 2 − 4 xy + 3 x 2 = 0
⇒
3 y 2 − 3 xy − xy + 3 x 2 = 0
⇒ ( 3y − x )( y − 3 x ) = 0
y=
x
, y = 3x
3
∴ ∠APO = 75°
⇒
In ∠AMP, sin75° =
AM
3
⇒ AM = 3sin75°
Now, length of chord of contact AB = 2AM
= 2(3sin75°) = 6sin75°
 3 + 1
( 3 + 1)
 = 3
= 6 
2
2
2


y = 3x
y=x
A
15
º
M
º
15
75º
3P
3
y=
x
3
B
O (0, 0)
225. Answer (1)
Since the lines are perpendicular
⇒ –12 + 2 + 2β = 0
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⇒ β=5
Since the lines intersect applying the condition of coplanarity
α−2 1
3
3
2
2 =0
−4
1
5
3(10 – 2) + (α – 2)(–8 – 25) + 1(3 + 8) = 0
α=
81
23
226. Answer (1)
Given that
|a1tanx + a2 tan2x + ... + antannx| ≤ |tanx|
tan x − 0
a1 tan x + a2 tan2 x + ... + an tan nx − 0
≤ lim
x →0
x −0
x −0
Apply lim
x →0
∴
∴
|a1 + 2a2 + 3a3 + .... + an.n| ≤ 1
max value of |a1 + 2a2 + ... + n.an| is K = 1
lim (sin x )1/x
 lim (1 + sin x )1/ x = e x→0
x →0
= e1 = e
227. Answer (2)
Let the equation of straight line is ax + by + c = 0 it is given that
1000

an + bn + c
n =1
a2 + b2
=0
1000
1000
1000
1
1
1
∴ a n + b n +
c = 0
 1000 × 1001 
 1000 × 1001 
a
 + b
 + 100c = 0

2


2

 1001 1001 
,

∴ the fixed point (α, β) = 
 2
2 
∴ α + β = 1001= 7 × 11 × 13
228. Answer (4)
Given
y = ax3 + bx2 + cx + d
y′ = 3ax2 + 2bx + c
y″ = 6ax + 2b
⇒ y″ – y′ – 2y = 4x2
6ax + 2b – 3ax2 – 2bx – c – 2ax3 – 2cx – 2d = 4x2
–2ax3 + (–3a – 2b)x2 + (6a – 2b – 2c)x + (2b – c – 2d) = 4x2
on comparing coefficient
a=0
−2b = 4 b = −c = −2
b = −2
c=2
2b – c – 2d = 0
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d=
2b − c
2
d=
−4 − 2
2
Miscellaneous Questions
d = −3
∴
 0 −2 
B=

2 −3 
 0 −2   0 −2  −4 6  
B2 = 
=


 2 −3   2 −3  −6 5  
Tr(C + CT) = 2Tr(C) = 2 × 1 = 2
229. Answer (1)
The locus of point of intersection of ⊥ tangents drawn to ellipse is director circle.
x2 y 2
+
= 1, equation of director
3
2
i.e. for
Circle is x2 + y2 = 5
Even for
x2 y 2
−
= 1, equation of director
9
4
Circle is x2 + y2 = 5
∴ θ = 90°
230. Answer (4)
Equation of tangent at (1, 2) to the parabola is T = 0.
y(2) = 2(x + 1) ⇒y = x + 1
Equation of directrixis x = –1
Let centre of the circle be (–1, b)
2
∴ 4 + (b − 2) =
| −1 − b + 1|
1+ 1
2[4 + (b – 2)2] = b2
 b=4
 r= 2 2
231. Answer (3)
As g is continuous function by definition
lim g ( x ) = g (1)
x →1
g (1) = lim g ( x )
x →1
∴
 n2 + n + 1 
g (1) = lim g  2

n →∞
 n + 2n + 2 
 n2 + 1 n + 1
n2 + n + 1 
= lim cos−1  2
⋅ 2
+ 2

n →∞
 n + 2 n + 1 n + 2n + 2 
= cos–1(1) = 0
g(1) = 0
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232. Answer (2)
Given f ( x ) =
 (x
f (x) =
(x
5
x 2 + 20
 ( x sin x + 5cos x )
5
2
⋅ dx
)
+ 20 x 3 cos x
 x5 
⋅
 dx
sin x + 5 x cos x )  cos x 
4
2
(II)
(I)
By applying integration by parts
−x
+ tan x + c
cos x [ x sin x + 5cos x ]
f (x) =
f (0) = 0 = c; f ( π) =
−π
−π
+0+0=
−1(0 + 5( −1)]
5
233. Answer (1)
Variance =
250
1
2
= 25
( x − xi ) =

10
N
Standard deviation =
25 = 5
σ
5
× 100 =
× 100 = 10%
x
50
C.V. =
234. Answer (3)
1
1
+
=1
2
By using e 2
( f (e ) )

f (e ) =
e
e2 − 1
 e ; if ' n ' is even
f (f (f ....f (e )))....) = 
 f (e ) ; if ' n ' is odd
n + times
∴

3
1
ede =
e2
2
3
1
=
9 1
− =4
2 2
235. Answer (2)
n
Sn =  r ⋅ nCr = n ⋅ 2n −1
r =1
⇒ n⋅2n – 1 must be 100 multiple
∴ n = 25, 50, 75 and 100
Hence number of values of ‘n’ = 4
(2) → x1 + x2 = 3 ⇒ # non negative integral solutions
= 3 + 2 – 1C2 – 1
=4
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236. Answer (1)
A5
A6
A4
O
A1
B1
B2
A3
A2
A1 = 1 − 3; = (1, − 3)
ΔOA1A2 is an equilateral Δ
⇒
Side of hexagon A1A2A3A4A5A6 is 2
Let B1, B2, B3, B4, B5, B6 be the mid points of A1A3, A2A4, A3A5, A4A6, A5A1, A6A2 respectively.
⇒ B1 B2 =
1
1
A2 A3 = (2) = 1
2
2
∴ Area of (B1B2B3B4B5) =
6 3 2 3 3
(1) =
sq. units
4
2
237. Answer (4)
Given that y = f(x) be a polynomial of degree 2 and symmetric about y-axis
∴ f(x) = ax2 + c (even function)
l , m, n → A : P
∴ f ′( x ) = 2ax
1 ∴ l , l + d, l + 2d
f ′(l + d )
=
f ′(l ) + f ′(l + 2d ) 2
238. Answer (1)
|[a b c]| = |a| |b| |c|
∴

= 2 × 2 × 3 = 12
  
  
  
a, b, c are mutually perpendicular vectors and [b c d ] = 0 ( b, c, d coplanar)
     
b ⋅b b ⋅c b ⋅d
  2
     
[b c d ] = c ⋅ b c ⋅ c c ⋅ d = 0
     
d ⋅b d ⋅c d ⋅d
 
3 15
c ⋅d = ±
2
   2
and | (a × c ) ⋅ d | = 9
  
  
| (a × c ) × d |2 = | 0 − (c ⋅ d )a |2

= 135

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