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UPSC NDA 2 2017 Mathematics Solved Paper

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Question : 66 of 120

Marks: +1, -0

If

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

\frac{{{\rm{dy}}}}{{{\rm{dx}}}}

is equal to

0
1
$\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}$
$\frac{{{\rm{x}} + 1}}{{{\rm{x}} - 1}}$

Solution:

Concept:

Use the identities

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

and

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

.

Explanation:

Start with

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

.

Rewrite the first term using

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

.

Here

A = \frac{x+1}{x-1}

, so

\frac{1}{A} = \frac{x-1}{x+1}

.

Thus

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

.

Now

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

.

Let

u = \frac{x-1}{x+1}

. Then

y = \cos^{-1}u + \sin^{-1}u

.

For any

u

in the domain

[-1,1]

, we have

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

.

Therefore

y = \frac{\pi}{2}

, a constant.

Differentiate:

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) = 0

.

Answer:

0

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