A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{2\pi }{3}\]
D) None of these
Correct Answer: B
Solution :
\[f\left( \frac{5\pi }{6} \right)=\log \sin \left( \frac{5\pi }{6} \right)=\log \sin \frac{\pi }{6}=\log \frac{1}{2}\] \[=-\log 2\] \[f\left( \frac{\pi }{6} \right)=\log \sin \left( \frac{\pi }{6} \right)=-\log 2\] Now, \[f(x)=\frac{1}{\sin x}\cos x=\cot x\] By LMV theorem, \[\frac{f\left( \frac{5\pi }{6} \right)-f\left( \frac{\pi }{6} \right)}{\left( \frac{5\pi }{6} \right)-\left( \frac{\pi }{6} \right)}=\cot C\] \[\Rightarrow \] \[\cot C=0\] \[\Rightarrow \] \[C=\frac{\pi }{2}\in \left( \frac{\pi }{6},\frac{5\pi }{6} \right)\]You need to login to perform this action.
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