If {f}(x)={1+ ²≤ft(x²)}, then {f}^{}≤ft({{π}}{2}) is

Correct Answer is : −π6-{π/6}−6π

Correct Answer is : −π6-{π/6}−6π

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MHTCET 26 Apr 2025 Shift 1 Paper

Question : 112 of 150

Marks: +1, -0

\mathrm{f}(x)=\sqrt{1+\cos ^{2}\left(x^{2}\right)}

\mathrm{f}^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)

Solution:

Concept:

Differentiate the given composite function using the chain rule, then substitute the given value of

x

Explanation:

We have

f(x)=\sqrt{1+\cos^2(x^2)}

Differentiate with respect to

x

f'(x)=\frac{1}{2\sqrt{1+\cos^2(x^2)}}\cdot 2\cos(x^2)\cdot(-\sin(x^2))\cdot 2x

Simplify using

2\sin\theta\cos\theta=\sin 2\theta

f'(x)=\frac{-x\sin(2x^2)}{\sqrt{1+\cos^2(x^2)}}

Now substitute

x=\frac{\sqrt{\pi}}{2}

x^2=\frac{\pi}{4},\quad 2x^2=\frac{\pi}{2},\quad \sin\left(\frac{\pi}{2}\right)=1

Also,

\cos^2\left(\frac{\pi}{4}\right)=\frac{1}{2}

, so:

f'\left(\frac{\sqrt{\pi}}{2}\right)=\frac{-\frac{\sqrt{\pi}}{2}\cdot 1}{\sqrt{1+\frac{1}{2}}}=-\frac{\frac{\sqrt{\pi}}{2}}{\sqrt{\frac{3}{2}}}=-\sqrt{\frac{\pi}{6}}

Answer:

f'\left(\frac{\sqrt{\pi}}{2}\right)=-\sqrt{\frac{\pi}{6}}

Hence, the correct option is B.

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