Let $f(x)=\left\{\begin{array}{c}\frac{1}{|x|}~~~~~, \text { if }|x|>2 \\ a+b x^2, \text { if }|x| \leq 2\end{array}\right.$, then f(x) is differentiable at x = -2 for

$a=\frac{3}{4}, b=-\frac{1}{16}$

We have,

$f(x)=\left\{\begin{aligned} -\frac{1}{x}~~~~~, & \text { if } x<-2 \\ a+b x^2, & \text { if }-2 \leq x \leq 2 \\ \frac{1}{x}~~~~~, & \text { if } x>2 \end{aligned}\right.$

Since f(x) is differentiable at x = -2. So, it is continuous there at

∴  $\lim\limits_{x \rightarrow-2^{-}} f(x)=\lim\limits_{x \rightarrow-2^{+}} f(x)=f(-2)$

$\Rightarrow \lim\limits_{x \rightarrow-2^{-}} f(x)=\lim\limits_{x \rightarrow-2^{+}} a+b x^2=a+b(-2)^2$

$\Rightarrow \frac{1}{2}=a+4 b$                .......(i)

As f(x) is differentiable at x = -2. Therefore,

$\lim\limits_{x \rightarrow-2^{-}} \frac{f(x)-f(-2)}{x-(-2)}=\lim\limits_{x \rightarrow-2^{+}} \frac{f(x)-f(-2)}{x-(-2)}$

$\Rightarrow \lim\limits_{x \rightarrow-2^{-}} \frac{-\frac{1}{x}-(a+4 b)}{x+2}=\lim\limits_{x \rightarrow-2^{+}} \frac{\left(a+b x^2\right)-(a+4 b)}{x+2}$

$\Rightarrow \lim\limits_{x \rightarrow-2^{-}} \frac{-\frac{1}{x}-\frac{1}{2}}{x+2}=b \lim\limits_{x \rightarrow-2^{+}} \frac{x^2-4}{x+2}$

$\Rightarrow \lim\limits_{x \rightarrow-2^{-}}-\frac{1}{2 x}=b \lim\limits_{x \rightarrow-2^{+}}(x-2)$

$\Rightarrow \frac{1}{4}=b(-4) \Rightarrow b=-\frac{1}{16}$               ........(ii)

Solving (i) and (ii), we get $a=\frac{3}{4}$ and $b=-\frac{1}{16}$