Question

Let $f(x)=\{\begin{array}{cc}x+a,& \text{ if }x<0\\ âˆ\pounds x−1âˆ\pounds ,\text{ if }xâ‰\yen 0& \end{array}$ and
$g(x)=\{\begin{array}{cc}x+1,& \text{ if }x<0\\ (x−1{)}^2+b,& \text{ if }xâ‰\yen 0\end{array}$
where $a$ and $b$ are non-negative real numbers. If the composite function $gof(x)$ is continuous for all real $x,$ then the values of $a+b$ is _____.

Answer 1

Answer: 1

$\begin{array}{rl}gof(x)& =\{\begin{array}{cc}f(x)+1,& \text{ if }{f}^{(x)}<0\\ {\left\{f^{(x)}−1\right\}}^2+b,& \text{ if }f(x)â‰\yen 0\end{array}\\ & =â\begin{array}{cc}x+a+1,& \text{ if }x<−a\\ (x+a−1{)}^2+b,& \text{ if }−a≤x<0\\ (âˆ\pounds x−1âˆ\pounds −1{)}^2+b,& \text{ if }xâ‰\yen 0\end{array}\end{array}$
As $gof(x)$ is continuous at $x=−a$ ,
$L.H.L=1$ and $R.H.L=1+b$
$∴$ $b=0$
$gof\left(−a\right)=gof\left(−a^{+}\right)=gof\left(−a^{−}\right)$
$⇒1+b=1+b=1⇒b=0$
Also, $gof(x)$ is continuous at $x=0$ .
$⇒gof\left(0\right)=gof\left(0^{+}\right)=gof\left(0^{−}\right)$
$⇒b={\left(a−1\right)}^2+b$
$⇒a=1$
$⇒a+b=1$