For every twice differentiable function $f:ℝ\to \left[-2,2\right]$ with $\left(f{\left(0\right)}^2\right)+{\left(f^1\left(0\right)\right)}^2=85$  which of the  Following statement(s) is (are) TRUE?

ƒ(x) can't be constant throughout the domain. Hence we can find x  (r, s) such that ƒ(x) is one-one  

$\therefore$Option (A) is true

$\left|f^1\left(x_0\right)\right|=\left|\frac{f\left(0\right)-f\left(-4\right)}{4}\right|\le 1$

$\therefore$Option (B) is true

$f\left(x\right)=\sin \left(\sqrt{85x}\right)$ satisfies the given condition  but $\underset{x\to \infty }{\lim }\sin \left(\sqrt{85x}\right)$ but does not exist.

$\therefore$Option (C) is not true. 

Let $g\left(x\right)=f^2\left(x\right)+{\left(f^1\left(x\right)\right)}^2$

$Since \left|f^1\left(x_1\right)\right|\le 1$ and $\left|f\left(x_1\right)\right|\le 2$ it gives $\Rightarrow g\left(x_1\right)\le 5for some x_1\in \left(-4,0\right)$

$g\left(0\right)=85\Rightarrow g\left(x\right)\text{ has maxima in }\left(x_1,x_2\right)\text{ say at }\alpha$, $g^1\left(\alpha \right)=0$.

$\Rightarrow 2f^1\left(\alpha \right)\left(f\left(\alpha \right)+f^{11}\left(\alpha \right)\right)=0$

$\begin{array}{l}\text{ If }{f}^1\left(\alpha \right)=0\Rightarrow g\left(\alpha \right)=f^2\left(\alpha \right)=85 \text{not possible }\\ \Rightarrow f\left(\alpha \right)+f^{11}\left(\alpha \right)=0 for some \alpha \in \left(x_1,x_2\right)\in (-4,4)\end{array}$

$\therefore$Option (D) is correct