A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) None of these
Correct Answer: B
Solution :
[b] \[\because f\left( x \right)=log\left( sinx \right)\] By Lagrange theorem, \[f\left( x \right)=logsinx\] is (i) continuous in \[\left[ \frac{\pi }{6},\frac{5\pi }{6} \right]\] (ii) differentiable in \[\left[ \frac{\pi }{6},\frac{5\pi }{6} \right]\] Then \[\exists c\in \left( \frac{\pi }{6},\frac{5\pi }{6} \right)\] such that \[f'(\mathbf{c})=\frac{f(b)-f(a)}{b-a}\therefore f(x)=\log \sin x\] \[f'(c)=\cot c\] Now, \[cotc=\frac{\log \sin \frac{5\pi }{6}-\log \sin \frac{\pi }{6}}{\frac{5\pi }{6}-\frac{\pi }{6}}\] \[\Rightarrow cotc=\frac{\log \left( \frac{\sin \frac{5\pi }{6}}{\sin \frac{5\pi }{6}} \right)}{4\pi /6}=\log \frac{\left( \frac{\sin \left( \frac{\pi }{2}+\frac{\pi }{3} \right)}{\sin \left( \frac{\pi }{2}-\frac{\pi }{3} \right)} \right)}{2\pi /3}\] \[\cot c=\frac{\log \left( \frac{\cos \frac{\pi }{3}}{\cos \frac{\pi }{3}} \right)}{2\pi /3}\] \[\Rightarrow cotc=\frac{0}{2\pi /3}=0\text{ }\Rightarrow \cot c=cot\frac{\pi }{2}~\Rightarrow c=\frac{\pi }{2}\] Hence, option [b] is correct.
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