Dig the Range

Note:- Using definition of inverse functions domain of f(x) is $[-2,2]$ .

Use $\color{#D61F06}{\cos^{-1}\dfrac x2=\dfrac{\pi}2-\sin^{-1}\dfrac x2}$ and then expand using identity $(a-b)^2=a^2-2ab+b^2$ so that most of the terms would cancel and equation would be reduced to a quadratic in x!! $f(x)=\dfrac{\pi^2}{4}+\dfrac{\pi^2}{12}\underbrace{(x^2+6x+8)}_{(x+3)^2-1}$ $=\dfrac{\pi^2}{6}+\dfrac{\pi^2}{12}(x+3)^2$

Using calculus we can see $(x+3)^2$ is increasing in $[-2,2]$ . Hence:-

$b\pi^2=f(x)_{\text{max}}=f(2)=\color{#D61F06}{\dfrac{\pi^2}6+\dfrac{25\pi^2}{12}}$

$a\pi^2=f(x)_{\text{min}}=f(-2)=\color{#20A900}{\dfrac{\pi^2}{6}+\dfrac{\pi^2}{12}}$

Hence,

$2\left( \color{#D61F06}{\dfrac{1}6+\dfrac{25}{12}}+\color{#20A900}{\dfrac{1}{6}+\dfrac{1}{12}}\right)=\large\boxed 5$

Let $\begin{cases} \alpha = \sin^{-1} \dfrac{x}{2} & \implies \sin \alpha = \dfrac{x}{2} & \implies x = 2 \sin \alpha \\ \beta = \cos^{-1} \dfrac{x}{2} & \implies \cos \beta = \dfrac{x}{2} \end{cases}$

$\begin{aligned} \implies \sin \alpha & = \cos \beta \\ & = \sin \left(\frac{\pi}{2} - \beta \right) \\ \implies \alpha & = \frac{\pi}{2} - \beta \\ \beta & = \frac{\pi}{2} - \alpha \end{aligned}$

Now consider,

$\begin{aligned} f(x) & = \left(\cos^{-1}\frac{x}{2}\right)^2+\pi \sin^{-1}\frac{x}{2}-\left(\sin^{-1}\frac{x}{2}\right)^2+\frac{\pi^2}{12}(x^2+6 x+8) \\ & = \beta^2 + \pi \alpha - \alpha^2 + \frac{\pi^2}{12} (4\sin^2 \alpha + 12\sin \alpha + 8) \\ & = \left(\frac{\pi}{2}-\alpha \right)^2 + \pi \alpha - \alpha^2 + \frac{\pi^2}{3} (\sin^2 \alpha + 3\sin \alpha + 2) \\ & = \frac{\pi ^2}{4}-\pi \alpha + \alpha^2 + \pi \alpha - \alpha^2 + \frac{\pi^2}{3} (\sin \alpha + 1)(\sin \alpha + 2) \\ \implies f(x) & = \frac{\pi ^2}{4} + \frac{\pi^2}{3} (\sin \alpha + 1)(\sin \alpha + 2) \end{aligned}$

We note that $f(x)$ is maximum and minimum when $\sin \alpha$ is maximum and minimum or $\sin \alpha = 1$ and $\sin \alpha = -1$ respectively. Therefore,

$\begin{aligned} f_{max}(x) & =\frac{\pi ^2}{4} + \frac{\pi^2}{3} (1 + 1)(1 + 2) = \frac{9\pi^2}{4} \\ f_{min}(x) & =\frac{\pi ^2}{4} + \frac{\pi^2}{3} (-1 + 1)(-1 + 2) = \frac{\pi^2}{4} \end{aligned}$

$\implies 2(a+b) = 2\left(\dfrac{9}{4} + \dfrac{1}{4}\right) = \boxed{5}$