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Q.
If $f\left(x\right)=\left|\left|2 sin x - 1\right| - 2 cot x\right|$ , then the value of $f^{'}\left(\frac{\pi }{3}\right)$ is equal to
NTA AbhyasNTA Abhyas 2020
Solution:
In the neighbourhood of $x=\frac{\pi }{3}$ , we know,
$\left(2 sin x - 1\right)>0$
$\Rightarrow f\left(x\right)=\left|\left(2 sin x - 1\right) - 2 cot x\right|$
Now, for the neighbourhood of $x=\frac{\pi }{3}$
$2sin x-1-2cot x=2\left(\frac{\sqrt{3}}{2}\right)-1-\frac{2}{\left(\sqrt{3}\right)}=\frac{3 - 2}{\sqrt{3}}-1$
$=\frac{1}{\sqrt{3}}-1 < 0$
$\therefore f\left(x\right)=\left(2 cot x\right)-\left(2 sin x - 1\right)$
Now, $f^{'}\left(x\right)=2\left(- c o s e c^{2} x\right)-2cos x$
$f^{'}\left(\frac{\pi }{3}\right)=2\left(- \frac{2}{\sqrt{3}}\right)^{2}-2\left(\frac{1}{2}\right)$
$=2\frac{4}{3}-1=\frac{5}{3}$