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 $a, b$.
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 $x=1$.
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