UPSC NDA I 2025 Mathematics Paper

Question Numbers: 91-92

Consider the following for the two (02) items that follow:

Let

y=sin^{-1}(x-\frac{4x^{3}}{27}).

Question : 91 of 120

Marks: +1, -0

What is y equal to?

Solution:

Concept:

Use the triple-angle identity for sine:

\sin 3\theta = 3\sin\theta - 4\sin^3\theta

This helps simplify inverse sine expressions of the form

3t - 4t^3

Explanation:

We are given

y = \sin^{-1}\left(x - \frac{4x^3}{27}\right)

Factor

x/3

x - \frac{4x^3}{27} = \frac{x}{3} \left(3 - 4\left(\frac{x}{3}\right)^2\right)

Let

t = \frac{x}{3}

. Then expression becomes

t(3 - 4t^2) = 3t - 4t^3

Recall identity:

\sin(3\theta) = 3\sin\theta - 4\sin^3\theta

\sin^{-1}(3t - 4t^3) = 3\sin^{-1}(t)

, provided

t

lies in appropriate range.

Thus

y = 3\sin^{-1}(t) = 3\sin^{-1}\left(\frac{x}{3}\right)

Answer:

Option D:

3\sin^{-1}\left(\frac{x}{3}\right)

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