A) \[f'(2)=f'(3)\]
B) \[f'(2)=0\]
C) \[f'(1/2)=16/5\]
D) All of these
Correct Answer: C
Solution :
We know that, \[{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)=\left\{ \begin{matrix} 2{{\tan }^{-1}}x, & if-1\le x\le 1 \\ \pi -2{{\tan }^{-1}}x, & if\,\,x>1 \\ -\pi -2{{\tan }^{-1}}x, & if\,\,x<-1 \\ \end{matrix} \right.\] \[\therefore \] \[f(x)=\left\{ \begin{matrix} 4{{\tan }^{-1}}x, & if\,\,-1\le x\le 1 \\ \pi & if\,\,x>1 \\ -\pi , & if\,\,x<-1 \\ \end{matrix} \right.\] \[\Rightarrow \] \[f'(x)=\left\{ \begin{matrix} \frac{4}{1+{{x}^{2}}}, & if\,\,-1<x<1 \\ 0, & if\,\,|x|\,\,>1 \\ \end{matrix} \right.\] \[\Rightarrow \] \[f'(2)=f'(3)=0\]and\[f'\left( \frac{1}{2} \right)=\frac{4}{1+{{\left( \frac{1}{2} \right)}^{2}}}=\frac{16}{5}\]
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