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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 $x=1$.