Question

If $f\left(x\right)=\{\begin{array}{cc}a{x}^2-b,& |x|<1\\ -\frac{1}{|x|},& |x|\ge 1\end{array}$ is differentiable at $x=1,$ then the value of $a+b$ is

• A. $0$
• B. $-1$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

$f\left(x\right)=\{\begin{array}{cc}a{x}^2-b,& |x|<1\\ -\frac{1}{|x|},& |x|\ge 1\end{array}\Rightarrow f\left(x\right)=\{\begin{array}{cc}a{x}^2-b,& |x|<1\\ -\frac{1}{|x|},& x\ge 1\text{or}x\le -1\end{array}\Rightarrow f\left(x\right)=\{\begin{array}{cc}a{x}^2-b,& -1<x<1\\ \frac{1}{x},& x\le -1\\ -\frac{1}{x},& x\ge 1\end{array}{f}^{\prime }\left(x\right)=\{\begin{array}{cc}2ax,& -1<x<1\\ \frac{-1}{x^2},& x\le -1\\ \frac{1}{x^2},& x\ge 1\end{array}$

$f\left(x\right)$ is continuous at $x=1.$
So, $f\left(1^{-}\right)=f\left(1^{+}\right)$
$\Rightarrow a-b=-1\dots \left(1\right)$

Given, $f\left(x\right)$ is differentiable at $x=1$
So, $f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)$
$\Rightarrow 2a\left(1\right)=\frac{1}{1^2}\Rightarrow a=\frac{1}{2}$
From equation $\left(1\right),$
$b=\frac{1}{2}+1=\frac{3}{2}$
$\therefore a+b=\frac{1}{2}+\frac{3}{2}=2$