Question

If $f(x) = |\cos \, x - \sin |x $, then $f' \left( \frac{\pi}{6}\right)$ is equal to

• A. $- \frac{1}{2} ( 1 + \sqrt{3} ) $
• B. $\frac{1}{2} ( 1 + \sqrt{3} ) $
• C. $- \frac{1}{2} ( 1 - \sqrt{3} ) $
• D. $ \frac{1}{2} ( 1 - \sqrt{3} ) $

Answer 1

Answer: (A)

At $x =\frac{\pi}{6}, \cos x -\sin x >\,0 $
$\therefore f\left(x\right)=\cos \,x -\sin \,x$
$ \Rightarrow f'\left(x\right)=-\sin x -\cos \,x $
$\therefore f'\left(\frac{\pi}{6}\right) = - \frac{1}{2} - \frac{\sqrt{3}}{2}$
$ = - \frac{1}{2} \left(1+ \sqrt{3}\right) $