# Solve this

Question:

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

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}\$