If f x   = 1 x ; x   ≥ 1 a x 2 + b ; – 1 < x < 1 is differentiable ∀ x ,…

Given,

$f\left(x\right) =\left\{\begin{array}{ll}\frac{1}{\left|x\right|};& \left|x\right| \ge 1\\ a{x}^2+b;& –1<x<1\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{ll}-\frac{1}{x}& x\le -1\\ a{x}^2+b& -1<x<1\\ \frac{1}{x}& x>1\end{array}\right.$

Also given funciton at $x=1$ is continuous

So, $a.1+b=1 ......\left(1\right)$

Also at $x=1$ differentiable

$2ax=–\frac{1}{x^2}$

Now at $x=1$ $\Rightarrow 2a=–1$

So, $a=\frac{–1}{2}$

Now using equation $\left(1\right)$ we get, $b=\frac{3}{2}.$