A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[-\frac{\pi }{2}\]
D) \[-\frac{\pi }{6}\]
Correct Answer: A
Solution :
Key Idea: For maximum or minimum, the second derivative of that function is negative or positive. Given that, \[y=a(1-\cos x)\] On differentiating w.r.t. x, we get \[\frac{dy}{dx}=a\sin x\] For maxima or minima, put\[\frac{dy}{dx}=0\] \[\Rightarrow \] \[a\sin x=0\Rightarrow x=0,\,\,\pi \] On again differentiating, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=a\cos x\] At\[x=0\], \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=a>0,\]minima At\[x=\pi \], \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-a<0,\]maxima \[\therefore \]Given function is maximum at\[x=\pi \].
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