Analysis of the given information

We are given a twice differentiable function $f: \mathbb{R} \to [-2, 2]$ such that ${(f(0))^2} + {(f'(0))^2} = 85$.

1. Boundedness of f: The range of $f$ is a subset of $[-2, 2]$, which means $|f(x)| \le 2$ for all $x \in \mathbb{R}$.
2. Information at x=0: We have $(f(0))^2 \le 2^2 = 4$. Substituting this into the given equation: $4 \ge (f(0))^2 = 85 - (f'(0))^2$ This implies $(f'(0))^2 \ge 85 - 4 = 81$. Taking the square root, we get $|f'(0)| \ge 9$.
3. Differentiability: The function $f$ is twice differentiable on $\mathbb{R}$. This implies that both $f(x)$ and $f'(x)$ are continuous functions.

Now, let's evaluate each statement.

Option A: There exist r, s $\in$ R, where r < s, such that f is one-one on the open interval (r, s)

1. We know that $|f'(0)| \ge 9$, which means $f'(0) \neq 0$. Let's assume $f'(0) > 0$ (the case $f'(0) < 0$ is analogous).
2. Since $f$ is twice differentiable, its first derivative $f'(x)$ is a continuous function.
3. By the property of continuous functions, if $f'(0) > 0$, then there exists an open interval around 0, say $(r, s)$, where $f'(x)$ maintains the same sign. Specifically, there exists a $\delta > 0$ such that for all $x \in (-\delta, \delta)$, $f'(x) > 0$.
4. A function with a strictly positive derivative on an open interval is strictly increasing on that interval.
5. A strictly increasing function is one-to-one (one-one).
6. Therefore, there exists an interval $(r, s) = (-\delta, \delta)$ on which $f$ is one-one.

Conclusion: Statement A is TRUE.

Option B: There exists x0 $\in$ ($-$ 4, 0) such that |f'(x0)| $\le$ 1

1. We can apply the Mean Value Theorem (MVT) to the function $f(x)$ on the interval $[-4, 0]$. Since $f$ is differentiable on $\mathbb{R}$, it is continuous on $[-4, 0]$ and differentiable on $(-4, 0)$.
2. According to MVT, there exists a point $x_0 \in (-4, 0)$ such that: $f'(x_0) = \frac{f(0) - f(-4)}{0 - (-4)} = \frac{f(0) - f(-4)}{4}$
3. We know that $|f(x)| \le 2$ for all $x$. Therefore, $-2 \le f(0) \le 2$ and $-2 \le f(-4) \le 2$.
4. The maximum value of the difference $f(0) - f(-4)$ is $2 - (-2) = 4$, and the minimum value is $-2 - 2 = -4$. So, $|f(0) - f(-4)| \le 4$.
5. Taking the absolute value of the expression for $f'(x_0)$: $|f'(x_0)| = \left| \frac{f(0) - f(-4)}{4} \right| = \frac{|f(0) - f(-4)|}{4}$
6. Using the bound from step 4: $|f'(x_0)| \le \frac{4}{4} = 1$
7. Thus, there exists an $x_0 \in (-4, 0)$ such that $|f'(x_0)| \le 1$.

Conclusion: Statement B is TRUE.

Option C: $\mathop {\lim }\limits_{x \to \infty } f(x) = 1$

1. This statement claims that the limit must be 1. Let's test this by trying to find a counterexample that satisfies the given conditions but does not have this limit.
2. Consider a function of the form $f(x) = A\cos(\omega x) + B\sin(\omega x)$. The range of this function is $[-\sqrt{A^2+B^2}, \sqrt{A^2+B^2}]$. To satisfy $|f(x)| \le 2$, we need $\sqrt{A^2+B^2} \le 2$.
3. Let's choose $A=1, B=1$. Then $\sqrt{1^2+1^2} = \sqrt{2} \le 2$. So, $f(x) = \cos(\omega x) + \sin(\omega x)$ is a candidate.
4. Let's check the condition at $x=0$. $f(0) = \cos(0) + \sin(0) = 1$. $f'(x) = -\omega\sin(\omega x) + \omega\cos(\omega x)$, so $f'(0) = \omega$.
5. The condition is $(f(0))^2 + (f'(0))^2 = 85$. $1^2 + \omega^2 = 85 \implies \omega^2 = 84 \implies \omega = \sqrt{84}$.
6. So, the function $f(x) = \cos(\sqrt{84}x) + \sin(\sqrt{84}x)$ satisfies all the given conditions: it's twice differentiable, its range is $[-\sqrt{2}, \sqrt{2}] \subseteq [-2, 2]$, and $(f(0))^2+(f'(0))^2=85$.
7. Now let's evaluate the limit: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} (\cos(\sqrt{84}x) + \sin(\sqrt{84}x))$. This limit does not exist because the function oscillates indefinitely.
8. Since we found a valid function for which the limit is not 1 (in fact, it doesn't exist), the statement is not always true.

Conclusion: Statement C is FALSE.

Option D: There exists $\alpha \in$($-$ 4, 4) such that f($\alpha$) + f"($\alpha$) = 0 and f'($\alpha$) $\neq$ 0

1. Let's define an auxiliary function $g(x) = (f(x))^2 + (f'(x))^2$. Since $f$ is twice differentiable, $g$ is differentiable.
2. The derivative of $g(x)$ is: $g'(x) = 2f(x)f'(x) + 2f'(x)f''(x) = 2f'(x)(f(x) + f''(x))$.
3. The problem asks to show that there's an $\alpha \in (-4, 4)$ where $f(\alpha) + f''(\alpha) = 0$ and $f'(\alpha) \neq 0$. This is equivalent to finding an $\alpha$ such that $g'(\alpha) = 0$ and $f'(\alpha) \neq 0$.
4. From the given information, $g(0) = (f(0))^2 + (f'(0))^2 = 85$.
5. From the analysis for Option B, we know there exists $x_0 \in (-4, 0)$ such that $|f'(x_0)| \le 1$. At this point, $g(x_0) = (f(x_0))^2 + (f'(x_0))^2 \le 2^2 + 1^2 = 5$.
6. Similarly, applying MVT to $f(x)$ on $[0, 4]$, there exists $x_1 \in (0, 4)$ such that $|f'(x_1)| \le 1$. At this point, $g(x_1) = (f(x_1))^2 + (f'(x_1))^2 \le 2^2 + 1^2 = 5$.
7. Consider the function $g(x)$ on the closed interval $[x_0, x_1]$. Since $g(x)$ is continuous on this interval, by the Extreme Value Theorem, it must attain a maximum value on $[x_0, x_1]$.
8. We have $g(x_0) \le 5$ and $g(x_1) \le 5$. However, $0 \in (x_0, x_1)$ and $g(0) = 85$. This means the maximum value of $g(x)$ on $[x_0, x_1]$ is at least 85, and it cannot occur at the endpoints $x_0$ or $x_1$.
9. Therefore, the maximum must occur at an interior point, say $\alpha \in (x_0, x_1)$. Since $(x_0, x_1) \subset (-4, 4)$, we have $\alpha \in (-4, 4)$.
10. By Fermat's Theorem, since $\alpha$ is an interior maximum, we must have $g'(\alpha) = 0$.
11. From $g'(\alpha) = 2f'(\alpha)(f(\alpha) + f''(\alpha)) = 0$, we conclude that either $f'(\alpha) = 0$ or $f(\alpha) + f''(\alpha) = 0$.
12. Let's test the possibility that $f'(\alpha) = 0$. If $f'(\alpha) = 0$, then $g(\alpha) = (f(\alpha))^2 + (f'(\alpha))^2 = (f(\alpha))^2$. Since $|f(x)| \le 2$, we have $(f(\alpha))^2 \le 4$. So, $g(\alpha) \le 4$.
13. But we know that $g(\alpha)$ is the maximum value on $[x_0, x_1]$, so $g(\alpha) \ge g(0) = 85$. This leads to the contradiction $85 \le 4$.
14. Therefore, our assumption that $f'(\alpha) = 0$ must be false. So, $f'(\alpha) \neq 0$.
15. Since $g'(\alpha) = 0$ and $f'(\alpha) \neq 0$, it must be that the other factor is zero: $f(\alpha) + f''(\alpha) = 0$.
16. We have found a point $\alpha \in (-4, 4)$ satisfying both conditions.

Conclusion: Statement D is TRUE.

Final Summary

• Statement A is TRUE.
• Statement B is TRUE.
• Statement C is FALSE.
• Statement D is TRUE.

The correct statements are A, B, and D.