A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[-\frac{\pi }{2}\]
D) \[-\frac{\pi }{6}\]
E) \[\frac{\pi }{3}\]
Correct Answer: A
Solution :
\[\because \] \[y=a(1-\cos x)\] On differentiating w.r.t\[x,\]we get \[\frac{dy}{dx}=a\sin x\] Again differentiating w. r. t.\[x,\]we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=a\cos x\] Put\[\frac{dy}{dx}=0\]for maxima or minima \[\Rightarrow \] \[\sin x=0\Rightarrow x=\pi \] \[\therefore \] \[{{\left( \frac{{{d}^{2}}u}{d{{x}^{2}}} \right)}_{x=\pi }}=a\cos \pi =-a\] \[\therefore \]Function is maximum at\[x=\pi \].
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