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  • Let <img alt="f(x)=x\: \cos ^{-1}\left ( -\sin \left | x \right | \right ),x\: \epsilon \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]," src="https://learn.careers360.com/latex-image/?f%28x%29%3Dx%5C%3A%20%5Ccos%20%5E%7B-1%7D%5Cleft%20%28%20-%5Csin%20

Let f(x)=x\: \cos ^{-1}\left ( -\sin \left | x \right | \right ),x\: \epsilon \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ], then which of the following is true ?
   
Option: 1 f'(0)=-\frac{\pi }{2},
Option: 2 f' is decreasing in \left ( -\frac{\pi }{2},0 \right ) and increasing in \left (0,\frac{\pi }{2} \right ),
Option: 3 f is not differentiable at x=0  ,
Option: 4 f' is increasing in \left ( -\frac{\pi }{2},0 \right ) and decreasing in \left (0,\frac{\pi }{2} \right ),
 

Answers (1)

best_answer

 

 

Application of Monotonicity (Part 1 ) -

-

\begin{array}{l}{f^{\prime}(x)=x\left(\pi-\cos ^{-1}(\sin |x|)\right)} \\ {=x\left(\pi-\left(\frac{\pi}{2}-\sin ^{-1}(\sin |x|)\right)\right)} \\ {=x\left(\frac{\pi}{2}+|x|\right)}\end{array}

\begin{array}{ll}{f(x)=} {\left\{\begin{array}{ll}{x\left(\frac{\pi}{2}+x\right)} & {x \geq 0} \\ {x\left(\frac{\pi}{2}-x\right)} & {x<0}\end{array}\right.} \\ {f^{\prime}(x)=\left\{\begin{array}{ll}{\frac{\pi}{2}+2 x} & {x \geq 0} \\ {\frac{\pi}{2}-2 x} & {x<0}\end{array}\right.}\end{array}

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

Correct Option (2)

Posted by

Kuldeep Maurya

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