Question

Find the points of local maxima and local minima respectively for the function $f\left(x\right) = sin\,2x-x$, where $-\frac{\pi}{2} \le x \le\frac{\pi }{2}$

• A. $-\pi /6$, $\pi/6$
• B. $\pi /3$, $-\pi/3$
• C. $-\pi /3$, $\pi/3$
• D. $\pi /6$, $-\pi/6$

Answer 1

The Correct Option is D

We have, $f(x) = sin\, 2x - x$ $\Rightarrow f'\left(x\right) = 2\, cos \,2x - 1$ For local maximum or minimum, we must have $f' \left(x\right) = 0$ $\Rightarrow 2cos \,2x - 1 = 0$ $\Rightarrow cos\, 2x =\frac{1}{2}$ $\Rightarrow 2x = -\frac{\pi}{3}$ or, $2x = \frac{\pi}{3}$ $\left[\because -\frac{\pi }{2} \le x \le \frac{\pi }{2} \therefore -\pi \le 2x \le \pi\right]$ $\Rightarrow = - \frac{\pi }{6}$ or, $x = \frac{\pi }{6}$ Thus, $x = -\frac{\pi }{6}$ and $x = \frac{\pi }{6}$ are possible points of local maxima or minima. Now, we test the function at each of these points, We have, $f''\left(x\right) = - 4 \,sin\, 2x$ At $x = - \pi/6$ : We have, $f''\left(-\frac{\pi }{6}\right) = -4\,sin \left(-\frac{\pi }{3}\right)$ $ = -4\times\frac{-\sqrt{3}}{2} = 2\sqrt{3} > 0$ So, $x = -\frac{\pi }{6}$ is a point of local minimum. At $x = \frac{\pi }{6}$ : We have, $f''\left(\frac{\pi }{6}\right) = -4\,sin \frac{\pi }{3}$ $ = -4\times \left(\frac{-\sqrt{3}}{2}\right) = -2\sqrt{3} < 0$ So, $x = \frac{\pi }{6}$ is a point of local maximum.