Question

If $f\left(x\right)=\sqrt{\left(1+{\cos }^2\left(x^2\right)\right)}$, then $f\text{'}\left(\frac{\sqrt{\mathrm{\pi }}}{2}\right)$ is

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

Answer 1

The correct option is B

$-\sqrt{\frac{\mathrm{\pi }}{6}}$

Explanation for the correct option:

Finding the value of $f\text{'}\left(\frac{\sqrt{\mathrm{\pi }}}{2}\right)$:

Given that,

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

Differentiate the given function with respect to $x$,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{{\displaystyle \frac{d\left({\cos }^2{x}^2\right)}{dx}}}{2\sqrt{1+{\cos }^2\left(x^2\right)}}\\ & =& \frac{-2\cos x^2\sin x^2\left(2x\right)}{2\sqrt{1+{\cos }^2\left(x^2\right)}}\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}\mathbf{\left(}\mathbf{cosx}\mathbf{\right)}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\mathbf{sinx}\mathbf{]}\\ & =& \frac{-x\sin 2x^2}{\sqrt{1+{\cos }^2\left(x^2\right)}}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{sin}\mathbf{x}\mathbf{cos}\mathbf{x}\mathbf{]}\end{array}$

Now, putting $x=\frac{\sqrt{\mathrm{\pi }}}{2}$ in the above differentiation, we get

$\begin{array}{rcl}f\text{'}\left(\frac{\sqrt{\mathrm{\pi }}}{2}\right)& =& \frac{-\sin 2{\left({\displaystyle \frac{\sqrt{\mathrm{\pi }}}{2}}\right)}^2\left({\displaystyle \frac{\sqrt{\mathrm{\pi }}}{2}}\right)}{\sqrt{1+{\cos }^2{\left({\displaystyle \frac{\sqrt{\mathrm{\pi }}}{2}}\right)}^2}}\\ & =& -\frac{\sqrt{\mathrm{\pi }}}{2}\left[\frac{\sin 2\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{\sqrt{1+{\cos }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}}\right]\\ & =& -\frac{\sqrt{\mathrm{\pi }}}{2}\left[\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{\sqrt{1+{\cos }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}}\right]\\ & =& -\frac{\sqrt{\mathrm{\pi }}}{2}\left[\frac{1}{\sqrt{1+{\displaystyle \frac{1}{2}}}}\right]\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{]}\\ & =& -\frac{\sqrt{\mathrm{\pi }}}{2}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)\\ & =& -\sqrt{\frac{\mathrm{\pi }}{6}}\mathbf{[}\mathbf{}\mathbf{by}\mathbf{}\mathbf{rationalizing}\mathbf{]}\mathbf{}\end{array}$

Hence, the correct option is B.