Let $f(x)= \begin{cases}\frac{1}{|x|} & \text { for }|x| \geq 1 \\ a x^{2}+b & \text { for }|x|<1\end{cases}$
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
(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, $b=\frac{3}{2}$ $\therefore a=\frac{-1}{2}, b=\frac{3}{2}$