Let f(x) = x cos^{-1}(-sin |x|),x in left[-fr | vedclass

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(A) Given $f(x) = x \cos^{-1}(-\sin |x|)$. Since $\cos^{-1}(-u) = \pi - \cos^{-1}(u)$,we have $f(x) = x(\pi - \cos^{-1}(\sin |x|))$.Using $\cos^{-1}(\

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$f^{\prime}$ is decreasing in $\left(-\frac{\pi}{2}, 0\right)$ and increasing in $\left(0, \frac{\pi}{2}\right)$

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(A) Given $f(x) = x \cos^{-1}(-\sin |x|)$. Since $\cos^{-1}(-u) = \pi - \cos^{-1}(u)$,we have $f(x) = x(\pi - \cos^{-1}(\sin |x|))$.Using $\cos^{-1}(\sin |x|) = \cos^{-1}(\cos(\frac{\pi}{2} - |x|)) = \frac{\pi}{2} - |x|$ for $|x| \in [0, \frac{\pi}{2}]$,we get:$f(x) = x(\pi - (\frac{\pi}{2} - |x|)) = x(\frac{\pi}{2} + |x|)$.For $x \geq 0$,$f(x) = x(\frac{\pi}{2} + x) = \frac{\pi}{2}x + x^2$. Thus $f^{\prime}(x) = \frac{\pi}{2} + 2x$.For $x Checking differentiability at $x=0$: $LHD = \lim_{x 	o 0^-} (\frac{\pi}{2} - 2x) = \frac{\pi}{2}$ and $RHD = \lim_{x 	o 0^+} (\frac{\pi}{2} + 2x) = \frac{\pi}{2}$. Since $LHD = RHD$,$f$ is differentiable at $x=0$ and $f^{\prime}(0) = \frac{\pi}{2}$.Now,for $x \in (-\frac{\pi}{2}, 0)$,$f^{\prime\prime}(x) = -2 For $x \in (0, \frac{\pi}{2})$,$f^{\prime\prime}(x) = 2 > 0$,so $f^{\prime}$ is increasing.Thus,option $A$ is correct.

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

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