If \[f(x)=2x+\left| x \right|,g(x)=\frac{1}{3}(2x-\left| x \right|)\] and \[h(x)=f(g(x)),\] then domain of \[{{\sin }^{-1}}\]\[\underbrace{(h(h(h(h...h(x)...))))}_{n\,times}\] is

\[\left[ -1,-\frac{1}{2} \right]\cup \left[ \frac{1}{2},1 \right]\]

\[\left[ -1,-\frac{1}{2} \right]\]

\[\left[ \frac{1}{2},1 \right]\]

If \[f(x)=2x+\left| x \right|,g(x)=\frac{1}{3}(2x-\left| x \right|)\]

1.	If \[f(x)=2x+\left\| x \right\|,g(x)=\frac{1}{3}(2x-\left\| x \right\|)\] and \[h(x)=f(g(x)),\] then domain of \[{{\sin }^{-1}}\]\[\underbrace{(h(h(h(h...h(x)...))))}_{n\,times}\] is
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.	\[\left[ \frac{1}{2},1 \right]\]
Answer» B. \[\left[ -1,-\frac{1}{2} \right]\cup \left[ \frac{1}{2},1 \right]\]

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If \[f(x)=2x+\left| x \right|,g(x)=\frac{1}{3}(2x-\left| x \right|)\] and \[h(x)=f(g(x)),\] then domain of \[{{\sin }^{-1}}\]\[\underbrace{(h(h(h(h...h(x)...))))}_{n\,times}\] is

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