Question

If $f(x)=|\cos x-\sin |x$, then $f^{\prime }\left(\frac{\pi }{6}\right)$ 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 $x=\frac{\pi }{6},\cos x-\sin x>0$
$\therefore f\left(x\right)=\cos x-\sin x$
$\Rightarrow f^{\prime }\left(x\right)=-\sin x-\cos x$
$\therefore f^{\prime }\left(\frac{\pi }{6}\right)=-\frac{1}{2}-\frac{\sqrt{3}}{2}$
$=-\frac{1}{2}\left(1+\sqrt{3}\right)$