Let f(x) = 2 ^ - 1 x + 4 ^ - 1 x - 3 x ^ 2 - 2x + 10, x ∈ [- 1, 1]....

Given $f\left(x\right) = 2\cos^{- 1}x + 4\cot^{- 1}x - 3{x}^{2} - 2x + 10$

${f}^{'}\left(x\right) = \dfrac{- 2}{\sqrt{1 - {x}^{2}}} - \dfrac{4}{1 + {x}^{2}} - 6x - 2$

$= - 2\left[\dfrac{1}{\sqrt{1 - {x}^{2}}} + \dfrac{2}{1 + {x}^{2}} + 3x + 1\right]$

${f}^{'}\left(x\right) < 0 ⇒ f\left(x\right) is a decreasing function</p><p>Now$f\left(1\right)$= \pi + 5 and$f\left(- 1\right)$= 5\pi + 5$

So, range: \left[a, b\right] ≡ $\left[\pi + 5, 5\pi + 5\right]</p><p>$a = \pi + 5, b = 5\pi + 5 ⇒ 4a - b = 11 - \pi$</p>$