Question

Let $f\left(x\right)=\{\begin{array}{c}x+a;x<0\\ |x-1|;x\ge 0\end{array}$ and $g\left(x\right)=\{\begin{array}{c}x+1;ifx<0\\ {(x-1)}^2+b;x\ge 0\end{array}$
If $\text{gof}$ is continuous $(a>0)$, then

• A. $a=2,b=0$
• B. $a=2,b=1$
• C. $a=1,b=0$
• D. $a=1,b=1$

Answer 1

The correct option is C $a=1,b=0$

Given: $f\left(x\right)=\{\begin{array}{c}x+a;x<0\\ |x-1|;x\ge 0\end{array}g\left(x\right)=\{\begin{array}{c}x+1;x<0\\ {\left(x--1\right)}^2+b;x\ge 0\end{array}$

To find: a & b if gof is continuous

Sol: $g\left(f\left(x\right)\right)=\{\begin{array}{c}f\left(x\right)+1;f\left(x\right)<0\\ {\left\{f\left(x\right)\right\}}^2+b;f\left(x\right)\ge 0\end{array}$

$g\left(f\left(x\right)\right)$ is continuous only when$f\left(x\right)$ is continuous $⟹a=1$

$⟹g\left(f\left(x\right)\right)=\{\begin{array}{c}f\left(x\right)+1;x<-a\\ {\{f\left(x\right)-1\}}^2+b;x\ge a\end{array}$

$⟹g\left(f\left(x\right)\right)=\{\begin{array}{c}(x+1)+1=(x+2)x<-1\\ {\{(x+1)-1\}}^2+b=x^2+b-1\le x<0\\ {\{(1-x)-1\}}^2+b=x^2+b0\le x<1\\ {\{(x-1)-1\}}^2+b={(x-1)}^2+bx\ge 1\end{array}$

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

$⟹1=1+b⟹b=0$

Hence, the correct answer is $a=1$ and $b=0$