Solve this

Solution:

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

We have,

$f(x)=\left\{\begin{array}{l}\frac{-1}{x}, \quad x \leq-1 \\ a x^{2}+b, \quad-1

Given: $f(x)$ is differentiable and continuous at every point.

Consider a point $x=1$

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$

$\Rightarrow \lim _{x \rightarrow 1^{-}}\left(a x^{2}+b\right)=\lim _{x \rightarrow 1^{+}} \frac{1}{x}$

$\Rightarrow \lim _{x \rightarrow 1^{-}}\left(a x^{2}+b\right)=\lim _{x \rightarrow 1^{+}} \frac{1}{x}$

$\Rightarrow a+b=1 \quad \ldots(\mathrm{i})$

It is also differentiable at $x=1$

$\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}$

$\Rightarrow \lim _{x \rightarrow 1^{-}} \frac{a x^{2}+b-1}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{\frac{1}{x}-1}{x-1}$

$\Rightarrow \lim _{x \rightarrow 1^{-}} \frac{a x^{2}-a}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{1-x}{(x-1) x}$         $\left[\begin{array}{ll}\text { Using } & \text { (i) }\end{array}\right]$

$\Rightarrow \lim _{x \rightarrow 1^{-}} a(x+1)=\lim _{x \rightarrow 1^{+}}(-x)$

$\Rightarrow 2 a=-1$

$\Rightarrow a=\frac{-1}{2}$

Plugging $a=\frac{-1}{2}$ in (i) we get,

$b=\frac{3}{2}$

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