Last modified on 24 September 2014, at 11:05

# Paper 1Edit

## SECTION – IEdit

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.

### Q21Edit

• (A) $1/4\,\!$
• (B) $-1/4\,\!$
• (C) $1/8\,\!$
• (D) $-1/8\,\!$
Answer

Key (A)

Solution

$P Q} \ = \hat i\ (-2 -3 μ) + \hat j\ (μ - 3) + \hat k\ (5μ - 4) \overrightarrow{P Q$

### *Q22Edit

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

### Q23Edit

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 ⎛ ⎞< ⎜ ⎟ ⎝ ⎠

### *Q24Edit

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.

### *Q25Edit

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 11

### Q26Edit

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

### *Q27Edit

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.