Question

Let $f\left(x\right)=\left\{\begin{array}{ll}\frac{1}{\left|x\right|}& \mathrm{for}\left|x\right|\ge 1\\ a{x}^2+b& \mathrm{for}\left|x\right|<1\end{array}\right.$

If f (x) is continuous and differentiable at any point, then
(a) $a=\frac{1}{2},b=-\frac{3}{2}$
(b) $a=-\frac{1}{2},b=\frac{3}{2}$
(c) a = 1, b = − 1
(d) none of these

Answer 1

(b) $a=-\frac{1}{2},b=\frac{3}{2}$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=\left\{\begin{array}{l}\begin{array}{l}\frac{-1}{x},x\le -1\\ a{x}^2+b,-1<x<1\end{array}\\ \frac{1}{x},x\ge 1\end{array}\right. \\ \mathrm{Given}:f\left(x\right)\mathrm{is}\mathrm{differentiable}\mathrm{and}\mathrm{continuous}\mathrm{at}\mathrm{every}\mathrm{point}. \\ \mathrm{Consider}\mathrm{a}\mathrm{point}x=1 \\ \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right) \\ \Rightarrow \underset{x\to 1^{-}}{\lim }\left(a{x}^2+b\right)=\underset{x\to 1^{+}}{\lim }\frac{1}{x} \\ \Rightarrow a+b=1...\left(\mathrm{i}\right) \\ \mathrm{It}\mathrm{is}\mathrm{also}\mathrm{differentiable}\mathrm{at}x=1 \\ \underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ \Rightarrow \underset{x\to 1^{-}}{\lim }\frac{a{x}^2+b-1}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{{\displaystyle \frac{1}{x}-1}}{x-1} \\ \Rightarrow \underset{x\to 1^{-}}{\lim }\frac{a{x}^2-a}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{1-x}{\left(x-1\right)x}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow \underset{x\to 1^{-}}{\lim }a\left(x+1\right)=\underset{x\to 1^{+}}{\lim }\left(-x\right) \\ \Rightarrow 2a=-1 \\ \Rightarrow a=\frac{-1}{2} \\ \mathrm{Plugging}a=\frac{-1}{2}\mathrm{in}\left(\mathrm{i}\right)\mathrm{we}\mathrm{get}, \\ b=\frac{3}{2} \\ \therefore a=\frac{-1}{2},b=\frac{3}{2} \end{gathered}$$