If the function f(x) = \ cases (1)/(| x |), & | x | ≥ 2 a x ^ 2 +...

Given,

$f$\left(x\right) = \left\{$$\begin{array}{ll}\dfrac{1}{x}, x \ge 2 \\ a{x}^{2} + 2b, - 2 < x < 2 \\ - \dfrac{1}{x}, x \le - 2\end{array}$$\right)

Also given $f\left(x\right)$ is differentiable on $R$

So, finding LHD and RHD at $x = 2$ we get,

$- \dfrac{1}{{x}^{2}} = 2ax$

$⇒ \dfrac{- 1}{4} = 4a$

$⇒ a = \dfrac{- 1}{16}$

Now, for function to be continuous at $x = 2$ we get,

$\dfrac{1}{2} = 4a + 2b$

$⇒ \dfrac{1}{2} = \dfrac{- 4}{16} + 2b$

$⇒ b = \dfrac{3}{8}$

Hence, $48\left(a + b\right) = 48\left(\dfrac{3}{8} - \dfrac{1}{16}\right)$

$⇒ 48\left(a + b\right) = 48\left(\dfrac{5}{16}\right) = 15$