If {f}(x)={ ² x}{1+ x}+{ ² x}{1+ x}, then the value of {f}^{}≤ft(π/6) is equal to

Correct Answer is : −12-1/2−21

Correct Answer is : −12-1/2−21

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MHTCET 26 Apr 2025 Shift 1 Paper

Question : 148 of 150

Marks: +1, -0

\mathrm{f}(x)=\frac{\sin ^{2} x}{1+\cot x}+\frac{\cos ^{2} x}{1+\tan x}

\mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)

is equal to

Solution:

Concept:

Simplify

\mathrm{f}(x)

using trigonometric identities before differentiating.

Explanation:

First, rewrite each term using

\cot x=\frac{\cos x}{\sin x}

and

\tan x=\frac{\sin x}{\cos x}

\mathrm{f}(x)=\frac{\sin ^{2} x}{1+\frac{\cos x}{\sin x}}+\frac{\cos ^{2} x}{1+\frac{\sin x}{\cos x}}

=\frac{\sin ^{3} x}{\sin x+\cos x}+\frac{\cos ^{3} x}{\sin x+\cos x}

=\frac{\sin ^{3} x+\cos ^{3} x}{\sin x+\cos x}

Using

a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})

, we get:

\mathrm{f}(x)=\sin ^{2} x-\sin x\cos x+\cos ^{2} x

=1-\sin x\cos x

=1-\frac{\sin 2x}{2}

Differentiate with respect to

x

\mathrm{f}^{\prime}(x)=-\cos 2x

Now substitute

x=\frac{\pi}{6}

\mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)=-\cos\left(2\cdot\frac{\pi}{6}\right)=-\cos\frac{\pi}{3}=-\frac{1}{2}

Answer:

-\frac{1}{2}

(Option C)

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