Logo
Jeeify
NVAMediumJEE 2025Inverse Trigonometric Functions

JEE Mathematics 2025 Question with Solution

Let S={x:cos⁡−1x=π+sin⁡−1x+sin⁡−1(2x+1)}S = \{ x : \cos^{-1}x = \pi + \sin^{-1}x + \sin^{-1}(2x+1) \}. Then ∑x∈S(2x−1)2\sum_{x \in S} (2x - 1)^2 is equal to:

Answer

Correct answer:16

Step-by-step solution

Standard Method

Given: S={x:cos⁡−1x=π+sin⁡−1x+sin⁡−1(2x+1)}S = \{ x : \cos^{-1}x = \pi + \sin^{-1}x + \sin^{-1}(2x+1) \}

Find: ∑x∈S(2x−1)2\sum_{x \in S} (2x - 1)^2

From the solution:

  1. Solve the equation cos⁡−1x=π+sin⁡−1x+sin⁡−1(2x+1)\cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1} (2x + 1).
  2. Find the possible values of xx that satisfy the equation.
  3. Calculate the sum ∑x∈S(2x−1)2\sum_{x \in S} (2x - 1)^2 for these values.

The provided solution concludes that the result is 1616.

Therefore, the required numerical value is 1616.

Common mistakes

  • Ignoring the domain restrictions of inverse trigonometric functions. Since sin⁡−1x\sin^{-1}x and sin⁡−1(2x+1)\sin^{-1}(2x+1) are defined only for inputs in [−1,1][-1,1], any candidate value of xx must satisfy these conditions before substitution.

  • Using incorrect identities between sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x. The standard relation is sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} for x∈[−1,1]x \in [-1,1], and missing this leads to an invalid simplification.

  • Treating inverse trigonometric functions like ordinary algebraic inverses without checking principal values. The ranges of sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x matter, so any transformed equation must remain consistent with those principal branches.

Practice more Inverse Trigonometric Functions questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Inverse Trigonometric Functions

Related questions