Question

Let $f:R\to R$ and $f:R\to R$ be defined by $f\left(x\right)=x+1$, $g\left(x\right)=x^2-2$ and $g\circ f$: $[-1,\infty )\to [-2,\infty )$. Then the range of ${(g\circ f)}^{-1}$ for $x\in [-2,-1]$ is

• A. $[-1,0]$
• B. $[0,1]$
• C. $[-2,-1]$
• D. $[0,2]$

Answer 1

The correct option is A $[-1,0]$

$g\circ f=y=(x+1{)}^2-2$

$\pm \sqrt{y+2}-1=x=(g\circ f{)}^{-1}(y)$

Now $xϵ[-1,\infty )$

$\therefore x=\sqrt{y+2}-1$

$\therefore g\circ f^{-1}\left(x\right)=\sqrt{x+2}-1$

$xϵ[-2,-1]$

$\therefore g\circ f^{-1}\left(x\right)ϵ[\sqrt{-2+2-1},\sqrt{2-1-1}]$

$ϵ[-1,0]$