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Question

Let $$f(x) = x \cos^{-1} (-\sin |x|), \, x \in \left[ - \dfrac{\pi}{2} , \dfrac{\pi}{2}\right]$$, then which of the following is true?

A
$$f'$$ is decreasing in $$\left(-\dfrac{\pi}{2}, 0\right)$$ and increasing in $$\left(0, \dfrac{\pi}{2}\right)$$
B
$$f'$$ is increasing in $$\left(-\dfrac{\pi}{2}, 0\right)$$ and decreasing in $$\left(0, \dfrac{\pi}{2}\right)$$
C
$$f'(0) = -\dfrac{\pi}{2}$$
D
$$f$$ is not differentiable at $$x = 0$$
Solution
Verified by Toppr

Correct option is C. $$f'$$ is decreasing in $$\left(-\dfrac{\pi}{2}, 0\right)$$ and increasing in $$\left(0, \dfrac{\pi}{2}\right)$$
$$f(x) = x\cos^{-1}(-\sin (|x|))$$

$$=x (\pi - \cos^{-1}(\sin |x|))$$

$$= x(\pi - \dfrac{\pi}{2} + \sin^{-1}(\sin |x|))$$

$$=x\left(\dfrac{\pi}{2} + |x|\right)$$

$$\therefore f(x) = \begin{cases} x\left(\dfrac{\pi}{2} + x\right) & \dfrac{\pi}{2} > x \ge o \\ \\ x\left(\dfrac{\pi}{2} - x\right) & \dfrac{\pi}{2} < x < 0 \end{cases}$$

$$\therefore f'(x) = \begin{cases} \dfrac{\pi}{2} +2x & \dfrac{\pi}{2} > x \ge 0 \\ \\\dfrac{\pi}{2} - 2x & -\dfrac{\pi}{2} \le x < 0 \end{cases}$$

$$f'(x)$$ is increasing in $$\left(0, \dfrac{\pi}{2}\right)$$ and decreasing in $$\left(-\dfrac{\pi}{2}, 0\right)$$.

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