A) \[f'(0)=-\frac{\pi }{2}\]
B) \[f'\] is decreasing in \[\left( -\frac{\pi }{2},0 \right)\] and increasing in \[\left( 0,\frac{\pi }{2} \right)\]
C) \[f\] is not differentiable at x=0
D) \[f'\] is increasing in \[\left( -\frac{\pi }{2},0 \right)\] and decreasing in \[\left( 0,\frac{\pi }{2} \right)\]
Correct Answer: B
Solution :
[b] \[f(x)=\frac{\pi x}{2}-x{{\sin }^{-1}}(sin-\left| x \right|)\] \[=\frac{\pi x}{2}+x\left| x \right|\] \[\therefore f'(x)=\frac{\pi }{2}+2x\] \[x\ge 0\] \[\frac{\pi }{2}-2x\] \[x<0\] \[\therefore f''(x)=2\] \[x\ge 0\] \[-\,2\] \[x<0\] \[\therefore f'(x)\] is decreasing in \[x\in \left( -\frac{\pi }{2},0 \right)\] and increasing in \[x\in \left( 0,\frac{\pi }{2} \right)\]
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