Question

If $f\left(x\right)=\left|\left|2sinx-1\right|-2cotx\right|$ , then the value of $f^{{}^{\prime }}\left(\frac{\pi }{3}\right)$ is equal to

• A. $0$
• B. $-\frac{5}{3}$
• C. $\frac{5}{3}$
• D. $\frac{8}{3}$

Answer 1

Answer: (C)

In the neighbourhood of $x=\frac{\pi }{3}$ , we know,
$\left(2sinx-1\right)>0$
$\Rightarrow f\left(x\right)=\left|\left(2sinx-1\right)-2cotx\right|$
Now, for the neighbourhood of $x=\frac{\pi }{3}$
$2sinx-1-2cotx=2\left(\frac{\sqrt{3}}{2}\right)-1-\frac{2}{\left(\sqrt{3}\right)}=\frac{3-2}{\sqrt{3}}-1$
$=\frac{1}{\sqrt{3}}-1<0$
$\therefore f\left(x\right)=\left(2cotx\right)-\left(2sinx-1\right)$
Now, $f^{{}^{\prime }}\left(x\right)=2\left(-cose{c}^{2}x\right)-2cosx$
$f^{{}^{\prime }}\left(\frac{\pi }{3}\right)=2{\left(-\frac{2}{\sqrt{3}}\right)}^2-2\left(\frac{1}{2}\right)$
$=2\frac{4}{3}-1=\frac{5}{3}$