# Solve this

Question:

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

Solution:

Given: $f(x)= \begin{cases}a x^{2}+b, & |x|<1 \\ \frac{1}{|x|}, & |x| \geq 1\end{cases}$

$\Rightarrow f(x)= \begin{cases}-\frac{1}{x}, & x<-1 \\ a x^{2}-b, & -1 It is given that the given function is differentiable at x = 1. We know every differentiable function is continuous. Therefore it is continuous at x=1. Then,$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)\Rightarrow \lim _{x \rightarrow 1} a x^{2}-b=\lim _{x \rightarrow 1} \frac{1}{x}\Rightarrow a-b=1$....(1) It is also differentiable at x=1. Therefore, (LHD at x = 1) = (RHD at x = 1)$\Rightarrow \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}+1-a-1}{x-1}=\lim _{x \rightarrow 1} \frac{-(x-1)}{x-1}[$Using (i) ]$\Rightarrow \lim _{x \rightarrow 1} a(x+1)=\lim _{x \rightarrow 1}-1\Rightarrow 2 a=-1\Rightarrow a=-\frac{1}{2}$From (i), we have:$a-b=1\Rightarrow-\frac{1}{2}-b=1\Rightarrow b=-\frac{3}{2}$Hence, when$a=-\frac{1}{2}$and$b=-\frac{3}{2}$the function is differentiable at$x=1\$.