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प्रश्न
`f(x)=sin2x-x, -pi/2<=x<=pi/2`
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उत्तर
\[\text { Given }: \hspace{0.167em} f\left( x \right) = \sin 2x - x\]
\[ \Rightarrow f'\left( x \right) = 2 \cos 2x - 1\]
\[\text { For a local maximum or a local minimum, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 2 \cos 2x - 1 = 0\]
\[ \Rightarrow \cos 2x = \frac{1}{2}\]
\[ \Rightarrow x = \frac{- \pi}{6} or \frac{\pi}{6}\]
Since f'(x) changes from positive to negative when x increases through \[\frac{\pi}{6}\] x = \[\frac{\pi}{6}\] is the point of local maxima.
The local minimum value of f (x) at x = \[- \frac{\pi}{6}\] is given by \[\sin \left( \frac{- \pi}{3} \right) + \frac{\pi}{6} = \frac{\pi}{6} - \frac{\sqrt{3}}{2}\]
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