PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for F(X) = Sin 2x − X, − π 2 < < X < π 2 . - PUC Karnataka Science 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?

#### Video TutorialsVIEW ALL [1]

Solution F(X) = Sin 2x − X, − π 2 < < X < π 2 . Concept: Graph of Maxima and Minima.
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