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UPSC NDA Math Model Paper 2

Question : 92 of 120

Marks: +1, -0

y = \sec^{-1}\left(\frac{\sqrt{x}-1}{x+\sqrt{x}}\right)

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

\frac{dy}{dx} =

Solution:

Given:

y = \sec^{-1}\left(\frac{\sqrt{x}-1}{x+\sqrt{x}}\right)

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

Here we have to find dy/dx

As we know that,

\sec^{-1} x = \cos^{-1}\left(\frac{1}{x}\right)

\Rightarrow = \sec^{-1}\left(\frac{\sqrt{x}-1}{x+\sqrt{x}}\right)

= \cos^{-1}\left(\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)

\Rightarrow y = \cos^{-1}\left(\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)

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

As we know that,

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

⇒ y = π / 2

⇒ dy/dx = 0

Hence, Option D is the correct answer.

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