JEE Mathematics Inverse Trigonometric Functions MCQs Set 14

Answer: (c) \( [-2,-1]\cup[1,2] \)

Explanation:
1. We need −1 ≤ log₂(x²/2) ≤ 1, so 1/2 ≤ x²/2 ≤ 2, i.e. 1 ≤ x² ≤ 4.
2. This gives x ∈ [−2, −1] ∪ [1, 2].

Teacher's Note:
1. Convert the logarithmic bounds to bounds on x².

Answer: (c) \( \frac{2b}{a} \)

Explanation:
1. Let θ = cos⁻¹(a/b). tan(π/4 + θ/2) + tan(π/4 − θ/2) = (1 + t)/(1 − t) + (1 − t)/(1 + t) with t = tan(θ/2).
2. This equals 2(1 + t²)/(1 − t²) = 2/cos θ = 2b/a.

Teacher's Note:
1. Put t = tan(θ/2) and add the fractions.

Answer: (b) \( \frac{\pi-x}{2} \)

Explanation:
1. √(1 ± sin x) = |cos(x/2) ± sin(x/2)|. For x/2 ∈ [π/4, π/2], sin(x/2) ≥ cos(x/2), so √(1 + sin x) = sin(x/2) + cos(x/2) and √(1 − sin x) = sin(x/2) − cos(x/2).
2. The fraction is 2 sin(x/2)/(2 cos(x/2)) = tan(x/2).
3. cot⁻¹(tan(x/2)) = π/2 − x/2 = (π − x)/2, which lies in (0, π/4], within the principal range.

Teacher's Note:
1. Write 1 ± sin x as a perfect square and mind the sign of cos(x/2) − sin(x/2).

Answer: (b) \( \frac{1}{2} \)

Explanation:
1. tan⁻¹A − tan⁻¹B = tan⁻¹[(A − B)/(1 + AB)] with A − B = 1/(x + 2) and 1 + AB = 1 + x(x + 1)/(x + 2)².
2. As x → ∞, 1 + AB → 2, so the bracket ~ 1/(2x).
3. Multiplying by x gives the limit 1/2.

Teacher's Note:
1. Combine the two inverse tangents first.

Answer: (a) \( \tan^{-1}(n+1)-\frac\pi4 \)

Explanation:
1. The r-th term is cosec⁻¹√((r² + 1)(r² + 2r + 2)) = tan⁻¹(1/(r² + r + 1)) = tan⁻¹(r + 1) − tan⁻¹r.
2. The sum telescopes to tan⁻¹(n + 1) − tan⁻¹1 = tan⁻¹(n + 1) − π/4.

Teacher's Note:
1. Check the first term: cosec⁻¹√10 = tan⁻¹(1/3).

Answer: (b) \( \sum\cot x\cot y=1 \)

Explanation:
1. With s = a + b + c: cot x cot y = √(bc/(a s)) · √(ca/(b s)) = c/s, and similarly for the other pairs.
2. The sum is (a + b + c)/s = 1.
3. So Σ cot x cot y = 1 (and then Σ x = π, not π/2; Σ tan x tan y = s(1/a + 1/b + 1/c) ≠ 1).

Teacher's Note:
1. Compute cot x cot y for each pair.

Answer: (d) x² + y² + z² + 2xyz = 1

Explanation:
1. Let A = cos⁻¹x, B = cos⁻¹y, C = cos⁻¹z with A + B + C = π.
2. For angles with A + B + C = π, cos²A + cos²B + cos²C + 2 cos A cos B cos C = 1.
3. So x² + y² + z² + 2xyz = 1.

Teacher's Note:
1. Use the identity for angles of a triangle.

Answer: (c) 3

Explanation:
1. Each inverse cosine is at most π, so the sum is 3π only if each equals π.
2. Then λ = μ = γ = −1.
3. λμ + μγ + γλ = 1 + 1 + 1 = 3.

Teacher's Note:
1. The maximum of cos⁻¹ is π.

Answer: (c) x = 0

Explanation:
1. The sum of two inverse tangents is π/2 when the product of the arguments is 1 (both positive).
2. (1 + x)(1 − x) = 1 gives x² = 0, so x = 0.
3. Check: tan⁻¹1 + tan⁻¹1 = π/2.

Teacher's Note:
1. Use tan⁻¹a + tan⁻¹b = π/2 when ab = 1.

Answer: (a) \( \sqrt{\frac{x^2+1}{x^2+2}} \)

Explanation:
1. cos(tan⁻¹x) = 1/√(1 + x²) = c.
2. sin(cot⁻¹c) = 1/√(1 + c²) = 1/√(1 + 1/(1 + x²)) = √((x² + 1)/(x² + 2)).

Teacher's Note:
1. Evaluate from the inside out using right triangles.

Answer: (c) {−1, 1, 0}

Explanation:
1. Let x = tan t with sin⁻¹x = 2t in the range where it is valid; then x = sin 2t = 2x/(1 + x²).
2. So x = 0 or 1 + x² = 2, i.e. x = ±1 (x ∈ [−1, 1]).
3. Check x = 1: sin⁻¹1 = π/2 = 2 tan⁻¹1; similarly for x = −1 and x = 0.

Teacher's Note:
1. Take the sine of both sides.

Answer: (a) \( \frac{ab}{\sqrt{a^2-1}+\sqrt{b^2-1}} \)

Explanation:
1. Take the sine of both sides: 1/x = (1/a)√(1 − 1/b²) + (1/b)√(1 − 1/a²).
2. This simplifies to 1/x = (√(b² − 1) + √(a² − 1))/(ab).
3. So x = ab/(√(a² − 1) + √(b² − 1)).

Teacher's Note:
1. Use sin(A + B) = sin A cos B + cos A sin B.

Answer: (b) 2n

Explanation:
1. cos⁻¹ is never negative, so a zero sum needs every term to be 0.
2. Then each x_i = 1 and the sum is 2n.

Teacher's Note:
1. A sum of non-negative terms is zero only if each term is zero.

Answer: (c) \( [\frac12,1] \)

Explanation:
1. Put x = cos A (A ∈ [0, π]). The second argument is cos A · (1/2) + sin A · (√3/2) = cos(A − π/3).
2. So the left side is A + |A − π/3|, which equals π/3 only if A ≤ π/3.
3. A ≤ π/3 means x = cos A ∈ [1/2, 1].

Teacher's Note:
1. Write the second argument as cos(A − π/3).

Answer: (d) 2abc

Explanation:
1. Let a = sin A, b = sin B, c = sin C with A + B + C = π.
2. The expression is sin A cos A + sin B cos B + sin C cos C = (1/2)(sin 2A + sin 2B + sin 2C).
3. For a triangle sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C, so the value is 2abc.

Teacher's Note:
1. Use the triangle identity for sin 2A + sin 2B + sin 2C.

Answer: (c) \( \pi \)

Explanation:
1. Evaluating in radians: 3 tan⁻¹(1/2) ≈ 1.3909, 2 tan⁻¹(1/5) ≈ 0.3948, sin⁻¹(142/(65√5)) ≈ 1.3559.
2. The three add to 3.14159 = π (verified to 10 decimal places).

Teacher's Note:
1. Evaluate the angles in radians and add.

Answer: (a) \( \left(0,\frac{\pi^2}{4}\right] \)

Explanation:
1. Let θ = cot⁻¹x ∈ (0, π); then cot⁻¹(−x) = π − θ.
2. y = θ(π − θ), a downward parabola in θ with maximum π²/4 at θ = π/2 and values tending to 0 at the ends.
3. The range is (0, π²/4].

Teacher's Note:
1. Use cot⁻¹(−x) = π − cot⁻¹x.

Answer: (c) \( \tan^{-1}(\sqrt n) \)

Explanation:
1. Each term equals tan⁻¹√r − tan⁻¹√(r − 1) (check r = 2: both sides ≈ 0.1699).
2. The sum telescopes to tan⁻¹√n − tan⁻¹0 = tan⁻¹√n.

Teacher's Note:
1. Express each term as a difference of two inverse tangents.