Question

ƒ(x) can't be constant throughout the domain. Hence we can find x $\in$(r, s) such that ƒ(x) is one- one option (A) is true
Option (B) :$\left|f\text{'}\left(x_o\right)\right|=\left|\frac{f\left(o\right)-f(-4)}{4}\right|\le 1$   (LMVT)
Option (C) :ƒ(x) = sin$\left(\sqrt{85x}\right)$ satisfies given condition
but$\underset{x\to \infty }{lim \sin }\left(\sqrt{85}\right)$D.N.E.
$\Rightarrow$Incorrect
Option (D) : g(x) = ƒ²(x) + (ƒ'(x))²
|ƒ'(x₁) $\le$ 1 (by LMVT)
|ƒ(x₁)| $\le$2 (given)
$\alpha$g(x₁) $\le$5 $\exists$x ₁$\in$(-4,0)
Similarly g(x₂)$\le$  5      $\exists$ 2x $\in$(0,4)
g(0) = 85            $\Rightarrow$ g(x) has maxima in (x₁, x₂) say at $\alpha$.
g'($\alpha$) = 0 & g($\alpha$) > 85
2ƒ'($\alpha$) (ƒ($\alpha$) + ƒ''($\alpha$)) = 0
If ƒ'($\alpha$) = 0      $\Rightarrow$  g($\alpha$) = ƒ²($\alpha$) $\ge$85 Not possible
$\Rightarrow$ ƒ($\alpha$) + ƒ''($\alpha$) = 0          $\exists \alpha \in$(x₁ ,x₂ )$\in$(-4,4)
option (D) correct