Question

Let $g\left(x\right)=\{\begin{array}{cc}2x+{\tan }^{-1}x+a,& -\infty <x\le 0\\ x^3+x^2+bx,& 0<x<\infty \end{array}.$
If $g\left(x\right)$ is differentiable for all $x\in (-\infty ,\infty ),$ then $(a^2+b^2)$ is equal to

Answer 1

$g\left(x\right)=\{\begin{array}{cc}2x+{\tan }^{-1}x+a,& -\infty <x\le 0\\ x^3+x^2+bx,& 0<x<\infty \end{array}$

$g\left(x\right)$ should be continuous at $x=0$
L.H.L. $=a,$ R.H.L. $=0$
$\therefore a=0$

Differentiability at $x=0$
$g^{\prime }\left(x\right)=\{\begin{array}{cc}2+\frac{1}{1+x^2},& -\infty <x\le 0\\ 3x^2+2x+b,& 0<x<\infty \end{array}$

L.H.D. $=3,$ R.H.D. $=b$
$\therefore b=3$

Hence, $a^2+b^2=0+9=9$