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NDA 1 2025 Maths, question 91

Topic: Differential calculus

Passage · Q91 to Q92

Let y=sin⁡−1(x−4x327)y = \sin^{- 1}\left( x - \frac{4x^{3}}{27} \right).

Q91

What is yy equal to?

  1. (a) sin⁡−1x\sin^{- 1}x
  2. (b) sin⁡−1(x/3)\sin^{- 1}(x/3)
  3. (c) 3sin⁡−1x3\sin^{- 1}x
  4. (d) 3sin⁡−1(x/3)3\sin^{- 1}(x/3)Correct answer

Solution.

Answer: (d) 3sin⁡−1(x/3)3\sin^{-1}(x/3)

  1. Put x=3sin⁡θx = 3\sin\theta. Then x−4x327=3sin⁡θ−4sin⁡3θ=sin⁡3θx - \frac{4x^{3}}{27} = 3\sin\theta - 4\sin^{3}\theta = \sin 3\theta.
  2. So y=sin⁡−1(sin⁡3θ)=3θ=3sin⁡−1(x/3)y = \sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1}(x/3).

Trap: This holds only while 3θ3\theta stays within [−π/2,π/2][-\pi/2, \pi/2], that is ∣x∣≤3/2|x| \leq 3/2; outside that range the identity breaks, for example at x=3x = 3.