Question

If $y=co{t}^{-1}{(\cos 2x)}^{\frac{1}{2}}$, then the value of $\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.$ at $x=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$will be

• A. ${(2/3)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• B. ${(1/3)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• C. ${\left(3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• D. ${\left(6\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Answer 1

The correct option is A

${(2/3)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Explanation for the correct option:

Step 1: Finding the value of $\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.$ .

Given function: $y=co{t}^{-1}{(\cos 2x)}^{\frac{1}{2}}$

Differentiate the given function with respect to $x$,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{–1}{1+\cos 2x}\frac{d}{dx}{\left(\cos 2x\right)}^{\frac{1}{2}}\left[\because \frac{d}{dx}co{t}^{-1}x=\frac{-1}{1+x^2}\right]\\ & =& \frac{–1}{1+\cos 2x}\frac{1}{2}{\left(\cos 2x\right)}^{-\frac{1}{2}}\left(-2\sin 2x\right)\left[\because \frac{d\left(\cos ax\right)}{dx}=-a\sin ax\right]\\ & =& \left(\frac{1}{1+\cos 2x}\right)\left(\frac{1}{\sqrt{\cos 2x}}\right)\sin 2x\end{array}$

Step 2: Finding the value of $\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.$ at $x=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$

Put $x=\frac{\mathrm{\pi }}{6}$ in the obtained function,

$\begin{array}{rcl}\frac{dy}{dx}& =& \left(\frac{1}{1+\cos 2\left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}\right)\left(\frac{1}{\sqrt{\cos 2\left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}}\right)\sin 2\left(\frac{\mathrm{\pi }}{6}\right)\\ & =& \left(\frac{1}{1+\cos \left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}\right)\left(\frac{1}{\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}}\right)\sin \left(\frac{\mathrm{\pi }}{3}\right)\\ & =& \left(\frac{1}{1+{\displaystyle \frac{1}{2}}}\right)\frac{1}{{\displaystyle \frac{1}{\sqrt{2}}}}\left(\frac{\sqrt{3}}{2}\right)\left[\because \sin 60°=\frac{\sqrt{3}}{2},\cos 60°=\frac{1}{2}\right]\\ & =& \frac{2}{3}\left(2^{\frac{1}{2}}\right)\left(\frac{3^{{\displaystyle \frac{1}{2}}}}{2}\right)\\ & =& \frac{2^{{\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{2}}}}{3^{1-{\displaystyle \frac{1}{2}}}}\\ & =& {\left(\frac{2}{3}\right)}^{\frac{1}{2}}\end{array}$

Hence, the correct option is A.