Functions Question 548
Question: If $ f(x)=2x+| x |,g(x)=\frac{1}{3}(2x-| x |) $ and $ h(x)=f(g(x)), $ then domain of $ {{\sin }^{-1}} $ $ \underbrace{(h(h(h(h…h(x)…))))}_{n,times} $ is
Options:
A) $ [-1,1] $
B) $ [ -1,-\frac{1}{2} ]\cup [ \frac{1}{2},1 ] $
C) $ [ -1,-\frac{1}{2} ] $
D) $ [ \frac{1}{2},1 ] $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Since, f(x)= $ \begin{cases} 2x+x.x\ge 0 \\ 2x-x,x<0 \\ \end{cases} = \begin{cases} 3x,x\ge 0 \\ x,x<0 \\ \end{cases} . . $ and $ g(x)=\frac{1}{3} \begin{cases} 2x-x,x\ge 0 \\ 2x+x,x<0 \\ \end{cases} .= \begin{cases} \frac{x}{3},x\ge 0 \\ x,x<0 \\ \end{cases} . $
$ \therefore f(g(x))= \begin{cases} 3( \frac{x}{3} ),x\ge 0 \\ x,x<0 \\ \end{cases} . $
$ \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]. $
BETA
