View source for Solved Question Papers - IIT JEE/MathsQP2009

=Paper 1=
==SECTION – I==
Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which
ONLY ONE is correct.
===Q21===

*(A) <math>1/4\,\!</math>
*(B) <math>-1/4\,\!</math>
*(C) <math>1/8\,\!</math>
*(D) <math>-1/8\,\!</math>
{{HideNseek|title=Answer|text=Key (A)
{{HideNseek|title=Solution|text=<math>Sol.: \overrightarrow{P Q} \
= \hat i\ (-2 -3 μ) + \hat j\ (μ - 3) + \hat k\ (5μ - 4)
\overrightarrow{P Q} \ is parallel to x - 4y + 3z = 1
⇒ 1 (-2 -3μ) - 4 (μ - 3) + 3(5μ - 4) = 0
⇒ μ = 1/4</math>}}
}}

===*Q22===
Tangents drawn from the point P(1, 8) to the circle
x2 + y2 - 6x - 4y - 11 = 0
touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is
(A) x2 + y2 + 4x - 6y + 19 = 0 (B) x2 + y2 - 4x - 10y + 19 = 0
(C) x2 + y2 - 2x + 6y - 29 = 0 (D) x2 + y2 - 6x - 4y + 19 = 0
Key (B)
Sol.: (x - 1) (x - 3) + (y - 8) (y - 2) = 0
x2 + y2 - 4x - 10y + 19 = 0
P
(1, 8) (3, 2)
A
B

===Q23===
Let f be a non-negative function defined on the interval [0, 1]. If
x
2
0
∫ 1− (f ′(t)) dt =
x
0
∫f (t)dt , 0 ≤ x ≤ 1, and f(0) = 0, then
(A) f 1 1
2 2
⎛ ⎞< ⎜ ⎟
⎝ ⎠
and f 1 1
3 3
⎛ ⎞> ⎜ ⎟
⎝ ⎠
(B) f 1 1
2 2
⎛ ⎞> ⎜ ⎟
⎝ ⎠
and f 1 1
3 3
⎛ ⎞> ⎜ ⎟
⎝ ⎠
(C) f 1 1
2 2
⎛ ⎞< ⎜ ⎟
⎝ ⎠
and f 1 1
3 3
⎛ ⎞< ⎜ ⎟
⎝ ⎠
(D) f 1 1
2 2
⎛ ⎞> ⎜ ⎟
⎝ ⎠
and f 1 1
3 3
⎛ ⎞< ⎜ ⎟
⎝ ⎠
Key. (C)
Sol.: 1−(f ′(x))2 = f(x),
1 - (f′(x))2 = (f(x))2
(f′(x))2 = 1 - (f(x))2
Let y = f(x)
dy f (x)
dx
= ′
dy 1 y2
dx
⎛ ⎞= ± − ⎜ ⎟
⎝ ⎠
2
dy dx
1 y
=
± − ∫ ∫
sin-1y = ±(x) + c
f(0) = 0 ⇒ c = 0
⇒ f(x) = sinx
(∴ f(x) is non-negative)
as sinx < x ∀ x > 0
⇒ f 1
2
⎛ ⎞
⎜ ⎟
⎝ ⎠
= sin 1
2
< 1
2
and f 1 1
3 3
⎛ ⎞< ⎜ ⎟
⎝ ⎠
===*Q24===
Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whose
vertices are the roots of the equation
z z3 + zz3 = 350 is
(A) 48 (B) 32
(C) 40 (D) 80
Key (A)
Sol.: Z Z 3 + Z Z3 = 350
Let Z = x + iy (x, y ∈Z)
(x2 + y2) (x2 - y2) = 175
⇒ x2 + y2 = 25 |x2 - y2 = 7
⇒ x = ± 4, y = ± 3
(-4, 3) (4, 3)
A(-4,-3) (4, -3)
6
8
⇒ area = 48 sq. units.

===*Q25===
The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse
x2 + 9y2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and
the origin O is
(A) 31
10
(B) 29
10
(C) 21
10
(D) 27
10
Key (D)
Sol.:
(0, 0) A(3, 0)
x + 3y = 3
•
12 , 9
5 5
⎛ − ⎞ ⎜ ⎟
⎝ ⎠
M
Area of Δle OAM = 1 27
2 5
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 27
10
IIT-JEE2009-Code-0-Questions and Solutions-Paper-I and II
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===Q26===
If a, b, c
􀁇 􀁇 􀁇 and d
􀁇
are unit vectors such that (a×b).(c×d)=
􀁇 􀁇 􀁇 􀁇 1 and a􀁇 .c􀁇 = 1
2
, then
(A) a,b, c
􀁇 􀁇 􀁇 are non-coplanar (B) b, c,d
􀁇 􀁇 􀁇 are non-coplanar
(C) b, d
􀁇 􀁇
are non-parallel (D) a,d
􀁇
􀁇 are parallel and b, c
􀁇 􀁇 are parallel
Key (b)
Sol.: a ×b = | a
( a ×b).(c ×d)=1
⇒ sinα . sinβ 1 2 (nˆ .nˆ ) = 1 where θ is the angle between 1 2 n􀁇 &n􀁇
⇒ sinα sinβ cosθ = 1
⇒ sinα = 1, sinβ = 1 and cosθ = 1
⇒ α = π/2, β = π/2 and θ = 0
Now, a . c 1
2
= ⇒ cosγ = 1/2 ⇒ γ = π/3
As a ×b || c×d , a, b, c, d are coplanar
There are two possibilities as shown
b 􀁇
c 􀁇
a􀁇
d 􀁇
or
b 􀁇
d 􀁇
a􀁇
c 􀁇
60°
60°
So option (C) is correct
===*Q27===
Let z = cosθ + isinθ. Then the value of
15
m 1
Im
= Σ
(z2m-1) at θ = 2° is
(A) 1
sin 2°
(B) 1
3sin 2°
(C) 1
2sin 2°
(D) 1
4sin 2°
Key (D)
Sol.
15
m 1
Im
= Σ
(Z2m-1) = Im
15
2m 1
m 1
Z −
= Σ
= Im
2 15
2
Z(1 (Z ) )
1 Z
⎡ − ⎤
⎢ − ⎥ ⎣ ⎦
= Im
1 Z30
1 Z
Z
⎡ ⎤
⎢ − ⎥
⎢ ⎥
⎢ − ⎥
⎣ ⎦
= Im 1 cos30 isin 30
2isin
⎡ − θ − θ⎤
⎢⎣ − θ ⎥⎦
= 1 cos30
2sin
− θ
θ
= 1 cos 60
2sin 2
− °
°
= 1
4sin 2°
===*Q28===The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2
and 3 only, is
(A) 55 (B) 66
(C) 77 (D) 88
IIT-JEE2009-Code-0-Questions and Solutions-Paper-I and II
12
Key (C)
Sol.: Case I: digits used 1, 1, 1, 1, 1, 3, 2
Number of integers formed = 7!
5!
= 42
Case II: digits used : 1, 1, 1, 1, 2, 2, 2
Number of integers formed = 7!
3! 4!
= 35
Total number of integers formed = 77.

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