QJEEM20Mac542: JEE Main 2020 Mathematics - Class 12 JEEM20Mac542 Solution: Let ƒ(x) = xcos –1 (–sin

Question

Verified Solution • 2025–26 Session moderate

Let ƒ(x) = xcos^–1(–sin|x|),

x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]

, then which of the following is true?

A ƒ' is decreasing in $\left( { - {\pi \over 2},0} \right)$ and increasing in $\left( {0,{\pi \over 2}} \right)$
B ƒ '(0) = ${ - {\pi \over 2}}$
C ƒ is not differentiable at x = 0
D ƒ' is increasing in $\left( { - {\pi \over 2},0} \right)$ and decreasing in $\left( {0,{\pi \over 2}} \right)$

Accepted Answer

We know, cos^-1(-x) =

\pi

- cos^-1x

\therefore

ƒ(x) = x(

\pi

- cos^–1(sin|x|))

= x(

\pi

-

{\pi \over 2}

+ sin^–1(sin|x|))

= x(

\pi

-

{\pi \over 2}

+ sin^–1(sin|x|))

= x

{\pi \over 2}

+ x|x|

\therefore

f(x) =

{\begin{matrix} x \frac{π}{2} - x^{2}, & x < 0 \\ x \frac{π}{2} + x^{2}, & x \geq 0 \end{matrix}

Now f'(x) =

{\begin{matrix} \frac{π}{2} - 2 x, & x < 0 \\ \frac{π}{2} + 2 x, & x \geq 0 \end{matrix}

and f''(x) =

{\begin{matrix} - 2, & x < 0 \\ 2, & x \geq 0 \end{matrix}

\therefore

ƒ' is decreasing in

\left( { - {\pi \over 2},0} \right)

and increasing in

\left( {0,{\pi \over 2}} \right)

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