A) \[[-1,1]\]
B) \[\left[ -1,-\frac{1}{2} \right]\cup \left[ \frac{1}{2},1 \right]\]
C) \[\left[ -1,-\frac{1}{2} \right]\]
D) \[\left[ \frac{1}{2},1 \right]\]
Correct Answer: A
Solution :
[a] Since, f(x)= \[\left\{ \begin{matrix} 2x+x.x\ge 0 \\ 2x-x,x<0 \\ \end{matrix}=\left\{ \begin{matrix} 3x,x\ge 0 \\ x,x<0 \\ \end{matrix} \right. \right.\] and \[g(x)=\frac{1}{3}\left\{ \begin{matrix} 2x-x,x\ge 0 \\ 2x+x,x<0 \\ \end{matrix} \right.=\left\{ \begin{matrix} \frac{x}{3},x\ge 0 \\ x,x<0 \\ \end{matrix} \right.\] \[\therefore f(g(x))=\left\{ \begin{matrix} 3\left( \frac{x}{3} \right),x\ge 0 \\ x,x<0 \\ \end{matrix} \right.\] \[\therefore f(g(x))=x\forall x\in R\therefore h(x)=x\] \[\Rightarrow {{\sin }^{-1}}(h(h(h...(h(x)..)))=si{{n}^{-1}}x\] Thus, domain of \[{{\sin }^{-1}}(h(h(h(h..h(x)...))))is[-1,1].\]
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