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?

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

(A) $\mathrm{f}^{\prime}(0)=-\frac{\pi}{2}$

(B) f is not differentiable at $\mathrm{x}=0$

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

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Continuity and Differentiability

Let

f (x) = x cos^{- 1} (- sin ∣ x ∣), x \in [- \frac{π}{2}, \frac{π}{2}]

, then which of the following is true?

(A) $f^{'} (0) = - \frac{π}{2}$
(B)f is not differentiable at $x = 0$
(C) $f$ ' is decreasing in $(- \frac{π}{2}, 0)$ and increasing in $(0, \frac{π}{2})$ ✓ Correct
(D) $f^{'}$ is increasing in $(- \frac{π}{2}, 0)$ and decreasing in $(0, \frac{π}{2})$

Answer: (C)

The inner inverse cosine simplifies cleanly, since for

∣ x ∣ \leq \frac{π}{2}

one has

cos^{- 1} (- sin ∣ x ∣) = π - cos^{- 1} (sin ∣ x ∣) = π - (\frac{π}{2} - ∣ x ∣) = \frac{π}{2} + ∣ x ∣

. Hence

f (x) = x (\frac{π}{2} + ∣ x ∣)

, equal to

\frac{π}{2} x + x^{2}

for

x \geq 0

and

\frac{π}{2} x - x^{2}

for

x < 0

. Differentiating gives

f^{'} (x) = \frac{π}{2} + 2 x

on the right and

\frac{π}{2} - 2 x

on the left; both one-sided slopes equal

\frac{π}{2}

at the origin, so

f

is differentiable there with

f^{'} (0) = \frac{π}{2}

. On

(0, \frac{π}{2})

the slope

\frac{π}{2} + 2 x

grows with

x

, so

f^{'}

is increasing, whereas on

(- \frac{π}{2}, 0)

the slope

\frac{π}{2} - 2 x

falls as

x

rises toward zero, so

f^{'}

is decreasing.

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option