$$f(x) = \left\{ {\matrix{ {{1 \over {|x|}},} & {|x| \ge 1} \cr {a{x^2} + b,} & {|x| < 1} \cr } } \right.$$

$$ = \left\{ {\matrix{ { - {1 \over x};} & {x \le - 1} \cr {a{x^2} + b;} & { - 1 < x < 1} \cr {{1 \over x};} & {x \ge 1} \cr } } \right.$$

As f(x) is differentiable so it is also continuous,

at x = 1,

$$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x)$$

$$ \Rightarrow a + b = {1 \over 1}$$

$$ \Rightarrow a + b = 1$$ ...... (1)

As f(x) is differentiable, so at x = 1

L.H.D. = R.H.D.

$$ \Rightarrow 2ax = - {1 \over {{x^2}}}$$

$$ \Rightarrow 2a = - 1$$

$$ \Rightarrow a = - {1 \over 2}$$

From (1), $$b = {3 \over 2}$$