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UPSC NDA II 2024 Mathematics Paper

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Question : 21 of 120

Marks: +1, -0

If

4 \sin^{-1}x + \cos^{-1}x = \pi

\sin^{-1}x + 4 \cos^{-1}x

equal to?

$\frac{\pi}{2}$
π
$\frac{3\pi}{2}$
$2\pi$

Solution:

Concept:

Use the identity

\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}

to relate the given and required expressions.

Explanation:

Let

a = \sin^{-1}x

and

b = \cos^{-1}x

. Then

a + b = \frac{\pi}{2}

.

Given:

4a + b = \pi

.

From

a + b = \frac{\pi}{2}

, we get

b = \frac{\pi}{2} - a

.

Substitute into

4a + b = \pi

:

4a + \left(\frac{\pi}{2} - a\right) = \pi

.

Simplify:

3a + \frac{\pi}{2} = \pi \implies 3a = \frac{\pi}{2} \implies a = \frac{\pi}{6}

.

Then

b = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}

.

Now compute

\sin^{-1}x + 4\cos^{-1}x = a + 4b = \frac{\pi}{6} + 4 \times \frac{\pi}{3} = \frac{\pi}{6} + \frac{4\pi}{3} = \frac{\pi}{6} + \frac{8\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2}

.

Answer:

\frac{3\pi}{2}

(Option C)

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