Question

If $f\left(x\right)=\left\{\begin{array}{ll}a{x}^2-b,& \mathrm{if}\left|x\right|<1\\ \frac{1}{\left|x\right|},& \mathrm{if}\left|x\right|\ge 1\end{array}\right.$ is differentiable at x = 1, find a, b.

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{l}a{x}^2+b,\left|x\right|<1\\ \frac{1}{\left|x\right|},\left|x\right|\ge 1\end{array}\right.$
$\Rightarrow f\left(x\right)=\left\{\begin{array}{l}-\frac{1}{x},x<-1\\ \begin{array}{l}a{x}^2-b,-1<x<1\\ \frac{1}{x},x\ge 1\end{array}\end{array}\right.$

It is given that the given function is differentiable at x = 1.

We know every differentiable function is continuous. Therefore it is continuous at x=1. Then,

$$\begin{gathered} \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right) \\ \Rightarrow \underset{x\to 1}{\lim }a{x}^2-b=\underset{x\to 1}{\lim }\frac{1}{x} \\ \Rightarrow a-b=1...\left(\mathrm{i}\right) \end{gathered}$$


It is also differentiable at x=1. Therefore,

(LHD at x = 1) = (RHD at x = 1)

$$\begin{gathered} \Rightarrow \underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ \Rightarrow \underset{x\to 1}{\lim }\frac{a{x}^2-b-1}{x-1}=\underset{x\to 1}{\lim }\frac{{\displaystyle \frac{1}{x}}-1}{x-1} \\ \Rightarrow \underset{x\to 1}{\lim }\frac{a{x}^2+1-a-1}{x-1}=\underset{x\to 1}{\lim }\frac{-(x-1)}{x-1}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow \underset{x\to 1}{\lim }a(x+1)=\underset{x\to 1}{\lim }-1 \\ \Rightarrow 2a=-1 \\ \Rightarrow a=-\frac{1}{2} \end{gathered}$$
From (i), we have:
$$\begin{gathered} a-b=1 \\ \Rightarrow -\frac{1}{2}-b=1 \\ \Rightarrow b=-\frac{3}{2} \end{gathered}$$

Hence, when $a=-\frac{1}{2}$ and $b=-\frac{3}{2}$ the function is differentiable at x = 1.