Question

Let $f\left(x\right)=\{\begin{array}{cc}2-x,& x\le 0\\ x+1,& x>0\end{array}$ and $g\left(x\right)=\{\begin{array}{cc}x+3,& x<1\\ x^2-2x-2,& 1\le x<2\\ x-5,& x\ge 2\end{array}$. If $g\left(f\right(x\left)\right)$ is continuous at $x=0$, then the value of $g\left(f\right(0\left)\right)$ is

• A. $0$
• B. $-1$
• C. $-3$
• D. $3$

Answer 1

The correct option is C $-3$

$L.H.L.=\underset{x\to 0^{-}}{lim}g\left(f\right(x\left)\right)=\underset{x\to 0^{-}}{lim}g(2-x)$
$=\underset{h\to 0}{lim}g(2-(0-h\left)\right)$
$=\underset{h\to 0}{lim}g(2+h))$
$=\underset{h\to 0}{lim}(2+h-5)$
$=\underset{h\to 0}{lim}(h-3)$
$=-3$

$R.H.L.=\underset{x\to 0^{+}}{lim}g\left(f\right(x\left)\right)=\underset{x\to 0^{+}}{lim}g(x+1)$
$=\underset{h\to 0}{lim}g\left(\right(0+h)+1)$
$=\underset{h\to 0}{lim}g(1+h))$
$=\underset{h\to 0}{lim}\left[\right(1+h{)}^2-2(1+h)-2]$
$=1-2-2$
$=-3$

Since $g\left(f\right(x\left)\right)$ is continuous at $x=0,$
$\Rightarrow L.H.L.=R.H.L.=g\left(f\right(0\left)\right)=-3$
$\therefore$ $g\left(f\right(0\left)\right)=-3$