Question

If $f\left(x\right)=\sqrt{1+co{s}^2\left(x^2\right)}$, then the value of $f^{\prime }\left(\frac{\sqrt{\pi }}{2}\right)$

• A. $\frac{\sqrt{\pi }}{6}$
• B. $-\sqrt{\frac{\pi }{6}}$
• C. $\frac{1}{\sqrt{6}}$
• D. $\frac{\pi }{\sqrt{6}}$

Answer 1

Answer: (B)

We have, $f\left(x\right)=\sqrt{1+co{s}^2\left(x^2\right)}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2}\times \frac{-2sin{x}^{2}cos{x}^2}{\sqrt{1+co{s}^2{x}^2}}\left(2x\right)$
$=\frac{-sin2x^2}{\sqrt{1+co{s}^2{x}^2}}(x)$
$\therefore f^{\prime }\left(\frac{\sqrt{\pi }}{2}\right)$ $=-\frac{\sqrt{\pi }}{2}\cdot \frac{sin2\left(\frac{\pi }{4}\right)}{\sqrt{1+co{s}^2\left(\frac{\pi }{4}\right)}}$
$=-\sqrt{\frac{\pi }{6}}$