A) \[2x-\frac{\pi }{3}\]
B) \[\frac{\pi }{3}-x\]
C) \[\frac{\pi }{6}-x\]
D) \[x-\frac{\pi }{6}\]
Correct Answer: D
Solution :
Consider \[{{\cot }^{-1}}\left( \frac{\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x}{\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\sin x} \right)\] \[={{\cot }^{-1}}\left( \frac{\sin \left( x+\frac{\pi }{3} \right)}{\cos \left( x+\frac{\pi }{3} \right)} \right)\] \[={{\cot }^{-1}}\left( \tan \left( x+\frac{\pi }{3} \right) \right)=\frac{\pi }{2}-{{\tan }^{-1}}\left( \tan \left( x+\frac{\pi }{3} \right) \right)\] \[\left\{ \begin{align} & \left\{ \frac{\pi }{2}-\left( x+\frac{\pi }{3} \right) \right.=\left( \frac{\pi }{6}-x \right);0<x<\frac{\pi }{6} \\ & \frac{\pi }{2}-\left( \left( x-\frac{\pi }{3} \right)-\pi \right)=\left( \frac{7\pi }{6}-x \right);\frac{\pi }{6}<x<\frac{\pi }{2} \\ \end{align} \right.\] \[\therefore \]\[2y=\left\{ \begin{align} & {{\left( \frac{\pi }{6}-x \right)}^{2}};0<x<\frac{\pi }{6} \\ & {{\left( \frac{7\pi }{6}-x \right)}^{2}};\frac{\pi }{6}<x<\frac{\pi }{2} \\ \end{align} \right.\] \[\therefore \]\[2\frac{dy}{dx}=\left\{ \begin{align} & 2\left( \frac{\pi }{6}-x \right).(-1);0<x<\frac{\pi }{6} \\ & 2\left( \frac{7\pi }{6}-x \right).(-1);\frac{\pi }{6}<x<\frac{\pi }{2} \\ \end{align} \right.\]
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