Prerna Jee Advanced 2013 maths- Paper 2

JEE Advanced 2013 (02Jun13) Question & Solutions PaperII 1 www. prernaclasses.com
Mathematics Paper II Jee Advance 2013
R
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PART III MATHEMATICS
SECTION   1 : (One or more options correct Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B),
(C) and (D), out of which ONE or MORE are correct.
41. In a triangle PQR, P is the largest angle and cos P = 1 / 3. Further the incircle of the triangle
touches the sides PQ, QR and RPat N, L and M respectively, such that the lengths of PN, QL
and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle
is(are)
(A) 16 (B) 18 (C) 24 (D) 22
41. (BD) PM = 2k – 2, RL = 2k + 2, QN = 2k
PQ = 4k – 2, QR = 4k + 2, RP = 4k
k k
k k k
P
4 . ) 2 4 ( 2
) 2 4 ( 16 ) 2 4 (
3
1
cos
2 2 2
-
+-+-
== Þ k = 0, 5
For k = 0, PQ is negative.
For k = 5, sides 18, 20, 22 units.
42. Two lines L 1 : x = 5,
2 3 -
=
a-
z y
and L 2 : x = a,
a-
=
- 2 1
z y
are coplanar. Then a can take
value(s)
(A) 1 (B) 2 (C) 3 (D) 4
42. (AD) Point on first line can be considered as (5, 3 – a, – 2)
Point on second line :  (a, – 1, 2 – a)
0
2 1 0
2 3 0
4 4 5
=
a--
-a-
-aa-a-
Þ 0
2 1 0
2 1 0
4 0 5
=
a-a-
-a-
-aa-
Þ 0
2 1 0
2 1 0
4 0 5
) 1 ( =
a-
-
-aa-
a- Þ (1 – a) [(5 – a) (4 – a)] = 0
Þ a = 1, 4, 5.

43. Circle(s) touching xaxis at a distance 3 from the origin and having an intercept of length 2Ö7 on
yaxis is (are)
(A) x 2 + y 2 – 6x + 8y + 9 = 0 (B) x 2 + y 2 – 6x + 7y + 9 = 0
(C) x 2 + y 2 – 6x – 8y + 9 = 0 (D) x 2 + y 2 – 6x – 7y + 9 = 0
43. (AC) Centre º (– g, – f)
g = ± 3 and f 2 – c = 7
Now, g 2 – c = 0 Þ c = 9 and f = ± 4
Circle is x 2 + y 2 ± 6x ± 8y + 9 = 0.
44. For a Î R (the set of all real numbers), a ¹ – 1,
60
1
)] ( ..... ) 2 ( ) 1 [( ) 1 (
) ..... 2 1 (
1
=
+++++++
+++
-¥® n na na na n
n
Lim
a
a a a
n
. Then a =
(A) 5 (B) 7 (C) – 15 / 2 (D) – 17 / 2
44. (BD) Limit =
ò
) 2 / 1 ( ) 1 (
2
) 1 (
) / 1 ( ) / (
1
0
1
1
2
1
+
=
+
ú
û
ù
ê
ë
é +
+
+
-
=
¥®
å
a
dx x
n
n n n
a n
n n r
Lim
a
a
a
n
r
a
n
Þ 60 / 1
2
1 2
1
1
=
+
+
a
a Þ (2a + 1) (a + 1) = 120
Þ 2a 2 + 3a – 119 = 0 Þ 7 or, 2 / 17
4
31 3
-=
±-
=a .
45. The function f (x) = 2 | x | + | x + 2 | – | | x + 2 | – 2 | x | | has a local minimum or a local maximum
at x =
(A) – 2 (B) – 2 / 3 (C) 2 (D) 2 / 3
45. (AB) f (x) = 2 | x | + | x + 2 | – | | x + 2 | – 2 | x | |
Redefining the function considering the critical points.
f (x) = – 2x – 4, x < – 2
2x + 4, – 2 £ x < – 2 / 3
– 4x, – 2 / 3 £ x < 0
4x, 0 £ x < 2
2x + 4, x ³ 2
f ' (x) = – 2, x < – 2
2, – 2 £ x < – 2 / 3
– 4, – 2 / 3 £ x < 0
4, 0 £ x < 2
2, x ³ 2
f ' (x) changing sign at x = – 2, – 2 / 3 and 0
Answer (A) and (B).

46. Let wbe a complex cube root of unity with w¹ 1 and P = [p ij ] be a n × n matrix with p ij = wi+ j . Then
P 2 ¹ 0, when n =
(A) 57 (B) 55 (C) 58 (D) 56
46. (BCD) Suppose C = P 2
åå
=
+
=
ww==
n
k
k j i
n
k
kj ik ij p p C
1
2
1
. [Q p ik = wi + k , p kj = wk + j ]
2
2
2
1
1
. w
ïþ
ï
ý
ü
ïî
ï
í
ì
w-
w-
w= +
n
j i
which equals to zero " i, j iff 2n = 3k, k Î I
which is correct only in option (A) but it is given that P 2 ¹ 0. Hence, n can take 55, 58, 56 as
its values.
47. If 3 x = 4 x – 1 , then x=
(A)
1 2 log 2
2 log 2
3
3
-
(B) 3 log 2
2
2- (C) 3 log 1
1
4- (D) 1 3 log 2
3 log 2
2
2
-
47. (ABC) (3 / 4) x = 1 / 4 Þ Taking log with base 2
Þ
3 log 1
1
3 log 2
2
4 2 -
=
-
=x Þ (BC)
Taking log with base 3 Þ
1 2 log 2
2 log 2
3
3
-
=x Þ (A)
48. Let
2
3 i
w
+
= and P = {w n : n = 1, 2, 3, ......}. Further H 1 = {z Î C : Re z> 1 / 2} and H 2 = {z Î C
: Re z < – 1 / 2}, where C is the set of all complex numbers. If z 1 Î P Ç H 1 , z 2 Î P Ç H 2 and O
represents the origin, then Ðz 1 O z 2 =
(A) p / 2 (B) p / 6 (C) 2p / 3 (D) 5p / 6
48. (CD)
e +i5p / 6
e-i5p / 6
e ip / 6
e-ip / 6
-1/2 1/2
1-1
6 /
2
3 p
=
+
= i
e
i
w P = e inp / 6
As z 1 Î P Ç H 1 Þ z 1 = 1, e ip / 6 , e – ip / 6
As z 2 Î P Ç H 2 Þ z 2 = – 1, e i5p / 6 , e – i5p / 6
Ðz 1 O z 2 = 2p / 3 where z 1 = e ip / 6 , z 2 = e i5p / 6
Ðz 1 O z 2 = 5p / 6 where z 1 = 1, z 2 = e i5p / 6
Taking suitable combination we get options (C) and (D) correct.

SECTION 2 (Paragraph Type)
This section contains 4 paragraphs each describing theory, experiment, data etc. Eight
questions relate to four paragraphs with two questions on each paragraph. Each question on
a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).
Paragraph for Questions 49 and 50
Let PQ be a focal chord of the parabola y 2 = 4ax. The tangents to the parabola at P and Q meet
at a point lying on the line y = 2x + a, a > 0.
49. If chord PQ subtends an angle q at the vertex of y 2 = 4ax, then tan q =
(A) 7
3
2
(B) 7
3
2
- (C) 5
3
2
(D) 5
3
2
-
50. Length of chord PQ is
(A) 7a (B) 5a (C) 2a (D) 3a
Sol. 49. (D) 50. (B)
Since PQ is a focal chord Þ t 1 t 2 = – 1, where P(at 1
2 , 2at 1 ) and Q(at 2
2 , 2at 2 )
Also, these tangents meet in directrix, hence their point of intersection is (– a, – a).
t 1 + t 2 = – 1 Þ | t 1 – t 2 | = Ö5
3 / 5 2
4 1
) / 2 ( ) / 2 (
tan 2 1 -=
-
-
=q
t t
, since vertex is on the right side of directrix, hence angle
must be obtuse.
And length a t t t t a PQ 5 4 ) ( 2
2 1 2 1 =++-= .
Paragraph for question 51 and 52
Let f : [0,1] ® R (the set of all real numbers) be a function. Suppose the function f is twice
differentiable, f (0) = f (1) = 0 and satisfies f '' (x) – 2f '(x) + f(x) ³ e x , x Î [0, 1].
51. If the function e –x f(x) assumes its minimum in the interval [0,1] at x=
4
1
, which of the following is
true?
(A) f '(x) < f (x) ,
4
3
4
1
<< x (B) f '(x) > f (x) ,
4
1
0 << x
(B) f ' (x) < f (x),
4
1
0 << x (D) f ' (x) < f (x), 1
4
3
<< x
52. Which of the following is true for 0 < x < 1 ?
(A) 0 < f (x) < ¥ (B)
2
1
) (
2
1
<<- x f
(C) 1 ) (
4
1
<<- x f (D) 0 ) ( <<¥- x f

Sol. 51. (C) 52. (D)
51. Let g(x) = e – x f (x)
Since f '' (x) – 2 f ' (x) + f (x) ³ e x Þ e – x {f '' (x) – 2 f ' (x) + f (x)} ³ 1
Þ g'' (x) ³ 1 Q g'' (x) > 0
Q g(x) has a min. at x = 1 / 4.
Þ g' (x) changes sign from (–)ve to (+)ve around x = 1 / 4.
Þ e – x {f ' (x) – f (x)} < 0 for 0 < x < 1 / 4 Þ f ' (x) < f (x) for 0 < x < 1 / 4.
52. f (x) = e x g(x)
g(0) = g(1) = 0 and g(x) is negative for all x Î [0, 1]
Þ f (x) < 0 for 0 < x < 1.
Paragraph for question 53 and 54
A box B 1 contains 1 white ball, 3 red balls and 2 black balls. Another box B 2 contains 2 white
balls, 3 red balls and 4 black balls. A third box B 3 contains 3 white balls, 4 red balls and 5 black
balls.
53. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is
white and the other is red, the probability that these 2 balls are drawn from box B 2 is
(A)
181
116
(B)
181
126
(C)
181
65
(D)
181
55
54. If 1 ball is drawn from each of the boxes B 1 , B 2 and B 3 , the probability that all 3 drawn balls are
of the same colour is
(A)
648
82
(B)
648
90
(C)
648
558
(D)
648
566
53. (D) 54. (A)
53. Let E = event that one ball is white and the other red.
Then
181
55
2
12
1
4
1
3
2
9
1
3
1
2
2
6
1
3
1
1
2
9
1
3
1
2
2 =
´
+
´
+
´
´
=÷
ø
ö
ç
è
æ
C
C C
C
C C
C
C C
C
C C
E
B
P .
54. Required probability = P(W 1 ) × P(W 2 ) × P(W 3 ) + P(R 1 ) × P(R 2 ) × P(R 3 ) + P(B 1 ) × P(B 2 ) × P(B 3 )
= (1 / 6) × (2 / 9) × (3 / 12) + (3 / 6) × (3 / 9) × (4 / 12) + (2 / 6) × (4 / 9) × (5 / 12)
= 82 / 648.
ParParagraph for question 55 and 56agraph for question 53 and 54
Let S = S 1Ç S 2 Ç S 3 , where
S 1 = { z Î C : |z |< 4}, S 2 =
ïþ
ï
ý
ü
ïî
ï
í
ì
>ú
û
ù
ê
ë
é
-
+-
Î 0
3 1
3 1
Im :
i
i z
C z and
S 3 = {z Î C : Re z > 0}
55. =--
Î
z i
S Z
3 1 min
(A)
2
3 2 -
(B)
2
3 2 +
(C)
2
3 3 -
(D)
2
3 3 +

56. Area of S =
(A)
3
10π
(B)
3
20π
(C)
3
16π
(D)
3
32π
Sol. 55. (C) 56. (B)
Let z = x + iy, S 1 : x 2 + y 2 < 16
Now, 0
3 1
) 3 ( ) 1 (
Im >÷
÷
ø
ö
ç
ç
è
æ
-
++-
i
y i x
Þ S 2 : Ö3x + y > 0
S 3 : x > 0
60°
(0, 4)
P
(1, -3)
(4, 0)
S
Ö3x + y = 0
55. min | 1 – 3i – z | = min | z – 1 + 3i |
= perpendicular distance of the point (1, – 3) from the straight line Ö3x + y = 0
2
3 3
2
3 3 -
=
-
=
56. Area of S = (1 / 4) p × (4) 2 + (1 / 6) p × (4) 2 = 20p / 3.

SECTION 3 (Matching list Type)
This section contains 4 multiple choice questions. Each question has matching lists. The codes
for the lists have choices. (A), (B), (C) and (D)  out of which ONLY ONE is correct.
57. Match List – I with List II and select the correct answer using the code given below the lists :
List I List II
P. Volume of parallelepiped determined by vectors a
r
, b
r
and c
r
is 2. 1. 100
Then the volume of the parallelepiped determined by vectors.
) ( ) ( 3 ), ( 2 a c and c b b a
rrrrrr
´´´
Q. Volume of parallelepiped determined by vector a
r
, b
r
and c
r
is 5. 2. 30
Then the volume of the parallepiped determined by vectors
) ( 2 ) ( ), ( 3 a c and c b b a
rrrrrr
+++ is
R.  Area of a triangle with adjacent sides determined by vectors a
r
and 3. 24
b
r
is 20. Then the area of the triangle with adjacent sides determined
by vectors ) ( ) 3 2 ( b a and b a
rrrr
-+ is
S.  Area of a parallelogram with adjacent sides detemined by vectors 4. 60
a
r
and b
r
is 30. Then the area of the parallelogram with adjacent
sides determined by vectors ) ( b a
rr
+ and a
r
is
Codes
P Q R S
(A) 4 2 3 1
(B) 2 3 1 4
(C) 3 4 1 2
(D) 1 4 3 2
57. (C) P : 24 4 6 ] [ 6 )} ( ) ( 3 { . ) ( 2 2
=´==´´´´ c b a a c c b b a
rrrrrrrrr
Q : ))} ( 2 ( ) {( . ) ( 3 a c c b b a
rrrrrr
+´++
= 60 ] [ 12 } 0 { . ) ( 6 ==´+´++´+ abc a c a b c b b a
rr
R : (1 / 2) | a × b | = 20
| a × b | = 40
| } 3 2 { | ) 2 / 1 ( | )} ( ) 3 2 { | ) 2 / 1 ( b a b a b a b a
rrrrrrrr
´-+-=-´+=D
100 40 ) 2 / 5 ( | | ) 2 / 5 ( =´=´= b a
rr
S : 30 | | =´ b a
rr
30 | ) ( | =´+ a b a
rrr
.

58. Consider the lines L 1 :
1
3
1 2
1 +
=
-
=
- z y x
, L 2 :
2
3
1
3
1
4 +
=
+
=
- z y x
and the planes
P 1 : 7x + y + 2z = 3, P 2 : 3x + 5y – 6z = 4. Let ax + by + cz = d be the equation of the plane
passing through the point of intersection of lines L 1 and L 2 . and perpendicular to planes P 1 and
P 2 .
Match List I with List II and select the correct answer using the code given below the lists :
List I List II
P. a = 1. 13
Q. b = 2. – 3
R. c = 3. 1
S. d = 4. – 2
Codes
P Q R S
(A) 3 2 4 1
(B) 1 3 4 2
(C) 3 2 1 4
(D) 2 4 1 3
58. (A) Point of intersection of L 1 : k
z y x
=
+
=
-
=
-
1
3
1 2
1
(let)
L 2 :
2
3
1
3
1
4 +
=
+
=
- z y x
= m (let)
then 2k + 1 = m + 4 ..........(1)
– k = m – 3 ........ (2)
Solving, k = 2
So, point of intersection is (5, – 2, – 1)
Q ax + by + cz = d is perpendicular to both plane P 1 and P 2
So, normal vector to the required plane k j i
j k i
n ˆ 32 ˆ 48 ˆ 16
6 5 3
2 1 7
ˆ ˆ ˆ
++-=
-
=
r
So, Equation of plane is (x – 5) (– 16) + (y + 2) 48 + (z + 1) 32 = 0
Þ x – 3y – 2z = 13
So a = 1, b = – 3, c = – 2, d = 13.

59. Match List I with List II and select the correct answer using the code given below the lists :
List I List II
P.
1/2
1
1 1
2 ÷
÷
ø
ö
ç
ç
è
æ
+÷÷
ø
ö
çç
è
æ
+
+
--
--
4
1
) tan(sin ) cot(sin
) sin(tan ) cos(tan 1
y
y y
y y y
y
takes value 1.
3
5
2
1
Q. If cos x + cos y + cos z = 0 sin x + sin y + sin z then 2. 2
possible value of cos
2
y x -
is
R. ÷
ø
ö
ç
è
æ
- x
π
4
cos cos 2x + sin x sin 2x secx = cos x sin 2x sec x + 3.
2
1
÷
ø
ö
ç
è
æ
+ x
π
4
cos cos 2x then possible value of sec x is
S. If ( ) , 0 )), 6 ( 1 sin(tan 1 sin cot ¹-=--
x x x 2 1 4. 1
then possible value of x is
Codes
P Q R S
(A) 4 3 1 2
(B) 4 3 2 1
(C) 3 4 2 1
(D) 3 4 1 2
59. (B) (P)
2 / 1
4
2
2
1
2
1
2
1
2
1
2
1
tan tan
1
cot cot
1
sin sin
1
1
cos cos
1
÷÷
÷
÷
÷
÷
÷
÷
ø
ö
çç
ç
ç
ç
ç
ç
ç
è
æ
+
÷÷
÷
÷
÷
÷
÷
ø
ö
çç
ç
ç
ç
ç
ç
è
æ
÷÷
÷
ø
ö
çç
ç
è
æ
-
+
÷÷
÷
ø
ö
çç
ç
è
æ -
÷÷
÷
ø
ö
çç
ç
è
æ
+
+
÷÷
÷
ø
ö
çç
ç
è
æ
+
--
--
y
y
y
y
y
y
y
y
y
y
Þ
2 / 1
4
2
4
2
1
1
÷
÷
ø
ö
ç
ç
è
æ
+÷
ø
ö
ç
è
æ - y y y
y
= 1
(Q) cos 2 x + cos 2 y + 2 cos x . cos y = cos 2 z
sin 2 x + sin 2 y + 2 sin x . sin y = sin 2 z
Adding we have 1 + cos (x – y) = 1 / 2
Þ cos 2 4 / 1
2
=
- y x
Þ cos 2 / 1
2
=
- y x
(R) cos 2x {cos (p / 4 – x) – cos (p / 4 + x)} = sin 2x . sec x (cos x – sin x)
Þ cos 2x {Ö2 sin x} = sin 2x . sec x (cos x – sin x)
Þ ) tan 1 (
2
1
sin cos
sin 2
2 sin
2 cos
sec x
x x
x
x
x
x +=
-
´=
Put tan x = 1, we have sec x = Ö2.

(S) ( ) , 0 )), 6 ( 1 sin(tan 1 sin cot ¹-=--
x x x 2 1
÷
÷
ø
ö
ç
ç
è
æ
+
=
÷
÷
ø
ö
ç
ç
è
æ
-
--
2
1
2
6 1
6
sin sin
1
| |
cot cot
x
x
x
x 1
Þ x 2 + 6x 4 = 6x 2 – 6x 4 Þ x =
3
5
2
1
.
60. A line L : y = mx + 3 meets y axis at E(0,3) and the arc of the parabola y 2 = 16x,
0 £ y £ 6 at the point F(x 0 , y 0 ). The tangent to the parabola at F(x 0 , y 0 ) intersects the yaxis at
G(0, y). The slope m of the line L is chosen such that the area of the triangle EFG has a local
maximum.
Match List I with List II and select the correct answer using the code given below the lists :
List I List II
P m = 1.
2
1
Q Maximum area DEFG is / DEFG 2.   4
R. y 0 = 3.   2
S. y 1 = 4.   1
Codes
P Q R S
(A) 4 1 2 3
(B) 3 4 1 2
(C) 1 3 2 4
(D) 1 3 4 2
60. (A) Equation of tangent at (x 0 , y 0 )
y y 0 = 8x + 8x 0
Put x = 0, y 1 = 8x 0 / y 0
Þ G = (0, 8x 0 / y 0 )
D = 1 / 2 | 3 – y 1 | x 0 = 1 / 2 (3 – 8x 0 / y 0 ) x 0
D(y 0 ) = 3 / 32 y 0
2 – y 0
3 / 64 (Q y 0
2 = 16x 0 )
D' (y 0 ) = 3 / 16 y 0 (1 – y 0 / 4)
D' (y 0 ) = 0
Þ y 0 = 4, y 1 = 2, m = 1, D = 1 / 2.

Prerna Jee Advanced 2013 maths- Paper 2

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