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MHTCET 11 May 2024 Shift 1 Paper

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Question : 107 of 150

Marks: +1, -0

If

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

\left(\frac{d y}{d x}\right)_{\text {at } x=1}=

2
$\frac{1}{2}$
$\frac{2}{3}$
-2

Solution:

Concept:

Simplify the given expression using logarithm properties and the identity

\frac{2\tan\theta}{1+\tan^2\theta}=\sin 2\theta

, then differentiate.

Explanation:

Given

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

.

Using

\log x^2 = 2\log x

, we get

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

.

Let

\log x = \tan\theta

.

Then

\frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin 2\theta

.

Hence

y = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}(\log x)

.

Differentiating with respect to

x

:

\frac{dy}{dx} = 2 \cdot \frac{1}{1+(\log x)^2} \cdot \frac{1}{x}

.

At

x = 1

, we have

\log 1 = 0

.

Therefore,

\frac{dy}{dx} = 2 \cdot \frac{1}{1+0} \cdot 1 = 2

.

Answer:

\left(\frac{dy}{dx}\right)_{x=1} = 2

, which is option A.

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