Which function satisfies $: \mathbb R → [−2, 2]$ with $((0))^ 2 + ( ′ (0))^ 2 = 85$,also $f$ is twice differentiable? 
Let $f:\mathbb R\to[−2,2]\,$ be a twice differentiable function 
  with $$\big((0)\big)^2+\big(′(0)\big)^2=85.$$ 
  Which of the following statements are necessarily TRUE?
(A) There exist ,  ∈ ℝ, where  < , such that  is one-one on
  the open interval (, )
(B) There exists $_0\in(−4, 0),\,$ such that $|f'(_0)|\le 1$
(C) $\lim_{x\to\infty}f(x)=1$
(D) There exists $\,\alpha\in(−4,4),\,$ such that 
  $\,f(\alpha) + f''(\alpha)=0$ and $f'(\alpha)\ne 0$

Solution:I tried it by using the function  $f'(x)=x^3+2x+9$,as it is satisfying $((0))^
2 + (
′
(0))^
2
= 85$. On checking all the 4 options on the chosen function ,i'm getting $A,D$ as correct options.
My query is that  the chosen function is not satisfying the definition of function(every element  of the domain has a unique image in codomain) as $f(3)=43$ is not in $[-2,2]$. 
Please provide a function which staisfies the definition in the problem.

Solution for first two. (C) is shown false by the counterexample of lulu $f(x)=\sin (\sqrt{85}x$), so that leaves (D) which has answers here: For every twice differentiable function $f : \bf R \rightarrow [–2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statements are TRUE? .
(A) True. Since $f\in C^2$, $f'\in C^1$, and $|f'(0)|^2=85 - |f(0)|^2 \ge 85 - 4 = 81>0$. So $f'(0)>9$ or $f'(0)<-9$. In either case, by continuity of $f'$, $f'(x)$ has a fixed sign on a small neighbourhood of $0$, and therefore $f$ is injective restricted to this neighbourhood.
(B) True. Suppose not. Then $|f'(x)|> 1$ on $(-4,0)$ . Then we have
$$f(0) = f(-4)+ \int_{-4}^0 f'(s)ds$$
By continuity, $f'$ has a fixed sign. If $f'<0$, then
$$ f(-4) = f(0)-\int_{-4}^0 f'(s)ds = f(0) + \int_{-4}^0 |f'(s)|ds > -2 + 4 =2 $$
contradicting the assumption on the range of $f$. Similarly, if $f'>0$, then we see that $f(-4) < -2$, which cannot happen.
