SECTION I: Single Correct Answer Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.  Let a₁, a₂, a₃, ... be in harmonic progression with a₁ = 5 and a₂₀ = 25. The least positive integer n for which an < 0
(A) 22
(B) 23
(C) 24
(D) 25  Sol. (D)
a₁, a₂, a₃, ... are in H.P.
⇒ 1/a₁, 1/a₂, 1/a₃, ... are in A.P.
⇒ 1/an = 1/a₁ + (n - 1)d < 0, where (1/25 - 5/25) / 19 = d = -4 / (19 × 25)
⇒ 1/5 + (n - 1)(-4 / (19 × 25)) < 0
4(n - 1) / (19 × 5) > 1
n - 1 > (19 × 5) / 4
n > (19 × 5) / 4 + 1 ⇒ n ≥ 25.  The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x - y + z = 3 and at a distance 2/√3 from the point (3, 1, -1) is
(A) 5x - 11y + z = 17
(B) √2x + y = 3√2 - 1
(C) x + y + z = √3
(D) x - √2y = 1 - √2  Sol. (A)
Equation of required plane is
P ≡ (x + 2y + 3z - 2) + λ(x - y + z - 3) = 0
⇒ (1 + λ)x + (2 - λ)y + (3 + λ)z - (2 + 3λ) = 0
Its distance from (3, 1, -1) is 2/√3
⇒ 2/√3 = |3(1 + λ) + (2 - λ) - (3 + λ) - (2 + 3λ)| / √((λ + 1)² + (2 - λ)² + (3 + λ)²)
⇒ 4/3 = (-2λ)² / (3λ² + 4λ + 14) ⇒ 3λ² + 4λ + 14 = 3λ²
⇒ λ = -7/2 ⇒ -(5/2)x + (11/2)y - z/2 + 17/2 = 0
-5x + 11y - z + 17 = 0.  IITJEE-2011-Paper 2-CPM21  Let PQR be a triangle of area Δ with a = 2, b = 7/2 and c = 5/2, where a, b, and c are the lengths of the sides of the triangle opposite to the angles at P, Q and R respectively. Then (2 sin P - sin 2P) / (2 sin P + sin 2P) equals
(A) 3 / (4Δ)
(B) 45 / (4Δ)
(C) (3 / (4Δ))²
(D) (45 / (4Δ))²  Sol. (C)
(2 sin P - 2 sin P cos P) / (2 sin P + 2 sin P cos P) = (1 - cos P) / (1 + cos P) = (2 sin²(P/2)) / (2 cos²(P/2)) = tan²(P/2)
= (s - b)(s - c) / (s(s - a))
= ((s - b)(s - c))² / Δ² = (((1/2)(3/2)) / Δ)² = (3 / (4Δ))²            P
         / \
        / P \
c = 5/2 /     \ b = 7/2//Q───────────R
a = 2  If a⃗ and b⃗ are vectors such that |a⃗ + b⃗| = √29 and a⃗ × (2î + 3ĵ + 4k̂) = (2î + 3ĵ + 4k̂) × b⃗, then a possible value of (a⃗ + b⃗) · (-7î + 2ĵ + 3k̂) is
(A) 0
(B) 3
(C) 4
(D) 8  Sol. (C)
a⃗ × (2î + 3ĵ + 4k̂) = (2î + 3ĵ + 4k̂) × b⃗
(a⃗ + b⃗) × (2î + 3ĵ + 4k̂) = 0⃗
⇒ a⃗ + b⃗ = ±(2î + 3ĵ + 4k̂) (as |a⃗ + b⃗| = √29)
⇒ (a⃗ + b⃗) · (-7î + 2ĵ + 3k̂)
= ±(-14 + 6 + 12) = ±4.  If P is a 3 × 3 matrix such that Pᵀ = 2P + I, where Pᵀ is the transpose of P and I is the 3 × 3 identity matrix, then there exists a column matrix X = [x y z]ᵀ ≠ [0 0 0]ᵀ such that
(A) PX = [0 0 0]ᵀ
(B) PX = X
(C) PX = 2X
(D) PX = -X  Sol. (D)
Given Pᵀ = 2P + I
⇒ P = 2Pᵀ + I = 2(2P + I) + I
⇒ P + I = 0
⇒ PX + X = 0
PX = -X.  IITJEE-2012-Paper 2-PCM-22  Let α(a) and β(a) be the roots of the equation (∛(1 + a) - 1)x² + (√(1 + a) - 1)x + (⁶√(1 + a) - 1) = 0 where a > -1. Then lim_{a→0⁺} α(a) and lim_{a→0⁺} β(a) are
(A) -5/2 and 1
(B) -1/2 and -1
(C) -7/2 and 2
(D) -9/2 and 3  Sol. (B)
Let 1 + a = y
⇒ (y^(1/3) - 1)x² + (y^(1/2) - 1)x + y^(1/6) - 1 = 0
⇒ ((y^(1/3) - 1) / (y - 1))x² + ((y^(1/2) - 1) / (y - 1))x + (y^(1/6) - 1) / (y - 1) = 0
Now taking lim on both the sides
⇒ (1/3)x² + (1/2)x + 1/6 = 0
⇒ 2x² + 3x + 1 = 0
x = -1, -1/2.  Four fair dice D₁, D₂, D₃ and D₄, each having six faces numbered 1, 2, 3, 4, 5, and 6, are rolled simultaneously. The probability that D₄ shows a number appearing on one of D₁, D₂ and D₃ is
(A) 91/216
(B) 108/216
(C) 125/216
(D) 127/216  Sol. (A)
Required probability = 1 - (6 · 5³) / 6⁴ = 1 - 125/216 = 91/216  The value of the integral ∫_{-π/2}^{π/2} (x² + ln((π + x)/(π - x))) cos x dx is
(A) 0
(B) π²/2 - 4
(C) π²/2 + 4
(D) π²/2  Sol. (B)
∫{-π/2}^{π/2} {x² + ln((π + x)/(π - x))} cos x dx
= ∫{-π/2}^{π/2} x² cos x dx + ∫_{-π/2}^{π/2} ln((π + x)/(π - x)) cos x dx
= 2 ∫₀^{π/2} x² cos x dx
= 2 [x² sin x + 2x cos x - 2 sin x]₀^{π/2}
= 2 [π²/4 - 2] = π²/2 - 4.  IITJEE-2011-Paper 2-CPM23  SECTION II: Paragraph Type
This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.  Paragraph for Questions 49 and 50
A tangent PT is drawn to the circle x² + y² = 4 at the point P(√3, 1). A straight line L, perpendicular to PT is a tangent to the circle (x - 3)² + y² = 1.  A possible equation of L is
(A) x - √3y = 1
(B) x + √3y = 1
(C) x - √3y = -1
(D) x + √3y = 5  Sol. (A)
Equation of tangent at P(√3, 1)
√3x + y = 4
Slope of line perpendicular to above tangent is 1/√3
So equation of tangents with slope 1/√3 to (x - 3)² + y² = 1 will be
y = (1/√3)(x - 3) ± 1√(1 + 1/3)
√3y = x - 3 ± (2)
√3y = x - 1 or √3y = x - 5.  A common tangent of the two circles is
(A) x = 4
(B) y = 2
(C) x + √3y = 4
(D) x + 2√2y = 6  Sol. (D)
Point of intersection of direct common tangents is (6, 0)  \                                             /\       ╭─────────╮                         /\    ╭─╯         ╰─╮                      /\  ╭╯  x²+y²=4    ╰╮  ╭─────╮           /\ │              │ ╭╯(x-3)²╰╮         /────────┼┼──────●───────┼─┼──+y²=1─┼────────●───────/ │              │ ╰╮      ╭╯         \ R(6, 0)/  ╰╮             ╭╯  ╰─────╯/    ╰─╮         ╭─╯/       ╰─────────╯/                                               so let the equation of common tangent be
y - 0 = m(x - 6)
as it touches x² + y² = 4
⇒ |(0 - 0 + 6m) / √(1 + m²)| = 2  IITJEE-2012-Paper 2-PCM-24  9m² = 1 + m²
m = ± 1 / (2√2)
So equation of common tangent
y = (1 / (2√2))(x - 6), y = -(1 / (2√2))(x - 6) and also x = 2.  Paragraph for Questions 51 and 52
Let f(x) = (1 - x)² sin²x + x² for all x ∈ ℝ and let g(x) = ∫₁ˣ ((2(t - 1))/(t + 1) - ln t) f(t) dt for all x ∈ (1, ∞).  Consider the statements:
P: There exists some x ∈ ℝ such that f(x) + 2x = 2(1 + x²)
Q: There exists some x ∈ ℝ such that 2f(x) + 1 = 2x(1 + x)
Then
(A) both P and Q are true
(B) P is true and Q is false
(C) P is false and Q is true
(D) both P and Q are false  Sol. (C)
f(x) = (1 - x)² sin²x + x²  ∀ x ∈ ℝ
g(x) = ∫₁ˣ ((2(t - 1))/(t + 1) - ln t) f(t) dt  ∀ x ∈ (1, ∞)
For statement P:
f(x) + 2x = 2(1 + x²)
(1 - x)² sin²x + x² + 2x = 2 + 2x²   ...(i)
(1 - x)² sin²x = x² - 2x + 2 = (x - 1)² + 1
(1 - x)² (sin²x - 1) = 1
-(1 - x)² cos²x = 1
(1 - x)² cos²x = -1
So equation (i) will not have real solution.
So, P is wrong.
For statement Q:
2(1 - x)² sin²x + 2x² + 1 = 2x + 2x²   ...(ii)
2(1 - x)² sin²x = 2x - 1
2 sin²x = (2x - 1) / (1 - x)²
Let h(x) = (2x - 1) / (1 - x)² - 2 sin²x
Clearly h(0) = -ve, lim_{x→1⁻} h(x) = +∞
So by IVT, equation (ii) will have solution.
So, Q is correct.  Which of the following is true?
(A) g is increasing on (1, ∞)
(B) g is decreasing on (1, ∞)
(C) g is increasing on (1, 2) and decreasing on (2, ∞)
(D) g is decreasing on (1, 2) and increasing on (2, ∞)  Sol. (B)
g'(x) = ((2(x - 1))/(x + 1) - ln x) f(x). For x ∈ (1, ∞), f(x) > 0
Let h(x) = ((2(x - 1))/(x + 1) - ln x) ⇒ h'(x) = 4/(x + 1)² - 1/x = -(x - 1)² / ((x + 1)² x) < 0
Also h(1) = 0 so, h(x) < 0 ∀ x > 1
⇒ g(x) is decreasing on (1, ∞).  IITJEE-2011-Paper 2-CPM-25  Paragraph for Questions 53 and 54
Let an denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let bn = the number of such n-digit integers ending with digit 1 and cn = the number of such n-digit integers ending with digit 0.  The value of b₆ is
(A) 7
(B) 8
(C) 9
(D) 11  Sol. (B)
an = bn + cn
bn = an-1
cn = an-2 ⇒ an = an-1 + an-2
As a₁ = 1, a₂ = 2, a₃ = 3, a₄ = 5, a₅ = 8 ⇒ b₆ = 8.  Which of the following is correct?
(A) a₁₇ = a₁₆ + a₁₅
(B) c₁₇ ≠ c₁₆ + c₁₅
(C) b₁₇ ≠ b₁₆ + c₁₆
(D) a₁₇ = c₁₇ + b₁₆  Sol. (A)
As an = an-1 + an-2
for n = 17
⇒ a₁₇ = a₁₆ + a₁₅.  SECTION III: Multiple Correct Answer(s) Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.  For every integer n, let an and bn be real numbers. Let function f : ℝ → ℝ be given by
f(x) = { an + sin πx, for x ∈ [2n, 2n + 1]
{ bn + cos πx, for x ∈ (2n - 1, 2n)
for all integers n. If f is continuous, then which of the following hold(s) for all n?
(A) an-1 - bn-1 = 0
(B) an - bn = 1
(C) an - bn+1 = 1
(D) an-1 - bn = -1  Sol. (B, D)
At x = 2n
L.H.L. = lim_{h→0} (bn + cos π(2n - h)) = bn + 1
R.H.L. = lim_{h→0} (an + sin π(2n + h)) = an
f(2n) = an
For continuity bn + 1 = an ⇒ an - bn = 1
At x = 2n + 1
L.H.L. = lim_{h→0} (an + sin π(2n + 1 - h)) = an
R.H.L. = lim_{h→0} (bn+1 + cos(π(2n + 1 + h))) = bn+1 - 1
f(2n + 1) = an
For continuity
an = bn+1 - 1
an-1 - bn = -1.  IITJEE-2012-Paper 2-PCM-26  If the straight lines (x - 1)/2 = (y + 1)/k = z/2 and (x + 1)/5 = (y + 1)/2 = z/k are coplanar, then the plane(s) containing these two lines is(are)
(A) y + 2z = -1
(B) y + z = -1
(C) y - z = -1
(D) y - 2z = -1  Sol. (B, C)
For given lines to be coplanar, we get
| 2  k  2 |
| 5  2  k | = 0 ⇒ k² = 4, k = ±2
| 2  0  0 |
For k = 2, obviously the plane y + 1 = z is common in both lines
For k = -2, family of plane containing first line is x + y + λ(x - z - 1) = 0
Point (-1, -1, 0) must satisfy it
-2 + λ(-2) = 0 ⇒ λ = -1
⇒ y + z + 1 = 0.  If the adjoint of a 3 × 3 matrix P is
[ 1  4  4 ]
[ 2  1  7 ]
[ 1  1  3 ]
then the possible value(s) of the determinant of P is (are)
(A) -2
(B) -1
(C) 1
(D) 2  Sol. (A, D)
|Adj P| = |P|² as (|Adj(P)| = |P|ⁿ⁻¹)
Since |Adj P| = 1(3 - 7) - 4(6 - 7) + 4(2 - 1)
= 4
|P| = 2 or -2.  Let f : (-1, 1) → ℝ be such that f(cos 4θ) = 2 / (2 - sec²θ) for θ ∈ (0, π/4) ∪ (π/4, π/2). Then the value(s) of f(1/3) is (are)
(A) 1 - √(3/2)
(B) 1 + √(3/2)
(C) 1 - √(2/3)
(D) 1 + √(2/3)  Sol. (A, B)
For θ ∈ (0, π/4) ∪ (π/4, π/2)
Let cos 4θ = 1/3
f(cos 4θ) = 2 / (2 - sec²θ)
⇒ cos 2θ = ±√((1 + cos 4θ)/2) = ±√(2/3)
f(1/3) = 2 / (2 - sec²θ) = (2 cos²θ) / (2 cos²θ - 1) = 1 + 1 / cos 2θ
f(1/3) = 1 - √(3/2) or 1 + √(3/2)  Let X and Y be two events such that P(X|Y) = 1/2, P(Y|X) = 1/3 and P(X ∩ Y) = 1/6. Which of the following is (are) correct?
(A) P(X ∪ Y) = 2/3
(B) X and Y are independent
(C) X and Y are not independent
(D) P(Xᶜ ∩ Y) = 1/3  Sol. (A, B)
P(X/Y) = P(X ∩ Y) / P(Y) = 1/2 and P(X ∩ Y) / P(X) = 1/3
P(X ∩ Y) = 1/6 ⇒ P(Y) = 1/3 and P(X) = 1/2
Clearly, X and Y are independent
Also, P(X ∪ Y) = 1/2 + 1/3 - 1/6 = 2/3.  IITJEE-2011-Paper 2-CPM-27  If f(x) = ∫₀ˣ e^{t²} (t - 2)(t - 3) dt for all x ∈ (0, ∞), then
(A) f has a local maximum at x = 2
(B) f' is decreasing on (2, 3)
(C) there exists some c ∈ (0, ∞) such that f''(c) = 0
(D) f has a local minimum at x = 3  

Sol. (A, B, C, D)
f'(x) = e^{x²} (x - 2)(x - 3)
Clearly, maxima at x = 2, minima at x = 3 and decreasing in x ∈ (2, 3).
f'(x) = 0 for x = 2 and x = 3
(Rolle's theorem)
so there exist c ∈ (2, 3) for which f''(c) = 0

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