IIM B UG (B.Sc-Hons.)- Sample Paper 2026 · Quant & DI · Question 3

IIM B UG (B.Sc-Hons.)- Sample Paper 2026 question on Inverse Trigonometric Functions, with the answer and a step-by-step solution.

Inverse Trigonometric FunctionsTrigonometryMultiple choiceMedium
For x > 1, using principal values of inverse trigonometric functions, si{n}^{-1}\left( \frac{2x}{1 +{x}^{2}}\right) equals:
  1. A
    \pi - 2ta{n}^{-1}x
  2. B
    \frac{\pi}{2}- 2ta{n}^{-1}x
  3. C
    2ta{n}^{-1}x
  4. D
    2ta{n}^{-1}x - \pi
Pick an option, then check.
Answer & solution

Answer: Option A

Put x = tan θ, so that θ = tan⁻¹x. Then 2x/(1 + x²) = sin 2θ, and the expression becomes sin⁻¹(sin 2θ).

The principal value of sin⁻¹ lies in [−π/2, π/2], so sin⁻¹(sin 2θ) equals 2θ only when 2θ lies in that range, that is when θ ≤ π/4, i.e. x ≤ 1.

For x > 1 we have θ > π/4, so 2θ > π/2 and 2θ falls outside the principal range.

In that case sin⁻¹(sin 2θ) = π − 2θ.

So for x > 1 the expression equals π − 2tan⁻¹x.

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