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MHTCET 22 Apr 2025 Shift 2 Paper

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Question : 133 of 150

Marks: +1, -0

The derivative of

\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)

\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)

x=0

is

$\frac{1}{8}$
$\frac{1}{4}$
$\frac{1}{2}$
1

Solution:

Concept:

Use a suitable trigonometric substitution for

x

to simplify both functions before differentiating.

Explanation:

Let

y = \tan^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)

and put

x = \tan\theta

.

Then

y = \tan^{-1}\left(\frac{\sec\theta - 1}{\tan\theta}\right) = \tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)

.

Now

\frac{1-\cos\theta}{\sin\theta} = \tan\frac{\theta}{2}

, so

y = \tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2}

.

Hence

\frac{dy}{d\theta} = \frac{1}{2}

.

Let

z = \tan^{-1}\left(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}}\right)

and put

x = \sin\theta

.

Then

z = \tan^{-1}\left(\frac{2\sin\theta\cos\theta}{1-2\sin^{2}\theta}\right) = \tan^{-1}\left(\frac{\sin2\theta}{\cos2\theta}\right) = \tan^{-1}(\tan2\theta) = 2\theta

.

Thus

\frac{dz}{d\theta} = 2

.

At

x = 0

, we have

\theta = 0

, and the required derivative is

\frac{dy}{dz} = \frac{dy/d\theta}{dz/d\theta} = \frac{1/2}{2} = \frac{1}{4}

.

Answer:

\frac{1}{4}

, i.e. option B.

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