For $g\left(x\right)$ to be continuous and differentiable $\forall x\in R, f\left(x\right)$ must be continuous and differentiable $\forall x\in R$
Since, $f\left(x\right)$ is continuous for all $x,$ therefore, it is continuous at $x=1$ also.
$\therefore f\left(1\right)=\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(1-h\right)=1=\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\left[a{\left(1-h\right)}^2+b\right]$
$\Rightarrow a+b=1$ … $\left(1\right)$
Also, $f\left(x\right)$ is differentiable at $x=1\Rightarrow f^{\text{'}}\left(1^{-}\right)=f^{\text{'}}\left(1^{+}\right)$
$\Rightarrow \underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{f\left(1-h\right)-f\left(1\right)}{-h}=\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{f\left(1+h\right)-f\left(1\right)}{h}$
$\Rightarrow \underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{a{\left(1-h\right)}^2+b-1}{-h}=\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\frac{1}{\left|1+h\right|}-1}{h}$
$\Rightarrow \underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\left(a+b-1\right)+\left(h^2-2h\right)a}{-h}=\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1-1-h}{h\left(1+h\right)}$
$\Rightarrow 2a=-1$ $\left(\mathrm{U}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g }\mathrm{}\mathrm{}a+b=1\right)$
$\therefore a=\frac{-1}{2}$
Hence, $a+b=1\Rightarrow b=1-a=\frac{3}{2}.$