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Solution for F(X) = Sin 2x − X, − π 2 < < X < π 2 . - CBSE (Commerce) Class 12 - Mathematics

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Question

f(x) = sin 2x \[-\] x, \[- \frac{\pi}{2} < \frac{<}{}x\frac{<}{}\frac{\pi}{2}\] .

Solution

\[\text { Given }: \hspace{0.167em} f\left( x \right) = \sin 2x - x\]

\[ \Rightarrow f'\left( x \right) = 2 \cos 2x - 1\]

\[\text { For a local maximum or a local minimum, we must have }\]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 2 \cos 2x - 1 = 0\]

\[ \Rightarrow \cos 2x = \frac{1}{2}\]

\[ \Rightarrow x = \frac{- \pi}{6} or \frac{\pi}{6}\]

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 maximum value of  f (x) at x = \[\frac{\pi}{6}\] is given by \[\sin \left( \frac{\pi}{3} \right) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6}\] 
Sincef '(x) changes from negative to positive when x increases through
\[- \frac{\pi}{6}\] x = \[- \frac{\pi}{6}\]  is the point of local minima.
The local minimum value of  f (x)  at x = \[- \frac{\pi}{6}\] is given by
 
\[\sin \left( \frac{- \pi}{3} \right) + \frac{\pi}{6} = \frac{\pi}{6} - \frac{\sqrt{3}}{2}\]
  Is there an error in this question or solution?

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Solution for question: F(X) = Sin 2x − X, − π 2 < < X < π 2 . concept: Graph of Maxima and Minima. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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