Question

$\text{If}f\left(x\right)=|\cos x-\sin x|$, 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

The correct option is A $-\frac{1}{2}(1+\sqrt{3})$

$f\left(x\right)=\cos x-\sin x\text{for}x\in [0,\frac{\pi }{4}]$
$\text{Then},f^{\prime }\left(x\right)=-\sin x-\cos x$
$f^{\prime }\left(\frac{\pi }{6}\right)=-\frac{1}{2}-\frac{\sqrt{3}}{2}=-\frac{1}{2}(1+\sqrt{3})$