Properties of Twice Differentiable Bounded Function

To determine which of the given statement(s) is (are) true, we analyze each option systematically:

Analysis of Option (A):

Given $f: \mathbb{R} \to [-2, 2]$, we know that: $|f(0)| \le 2 \implies (f(0))^2 \le 4$

From the given condition: $(f(0))^2 + (f'(0))^2 = 85 \implies (f'(0))^2 = 85 - (f(0))^2 \ge 85 - 4 = 81$ $\implies |f'(0)| \ge 9 > 0$

Since $f$ is twice differentiable on $\mathbb{R}$, its derivative $f'$ is continuous on $\mathbb{R}$. Because $f'(0) \ne 0$, by the sign preservation property of continuous functions, there exists an open interval $(r, s)$ containing $0$ where $f'(x)$ maintains the same sign (strictly positive or strictly negative) for all $x \in (r, s)$.

Thus, $f$ is strictly monotonic on $(r, s)$, which implies that $f$ is one-one on $(r, s)$. Therefore, Option (A) is TRUE.

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Analysis of Option (B):

Applying the Mean Value Theorem (MVT) to the differentiable function $f$ on the interval $[-4, 0]$: There exists $x_0 \in (-4, 0)$ such that: $f'(x_0) = \frac{f(0) - f(-4)}{0 - (-4)} = \frac{f(0) - f(-4)}{4}$

Since $f(x) \in [-2, 2]$ for all $x \in \mathbb{R}$, we have: $|f(0) - f(-4)| \le |f(0)| + |f(-4)| \le 2 + 2 = 4$

Taking the absolute value on both sides: $|f'(x_0)| = \frac{|f(0) - f(-4)|}{4} \le \frac{4}{4} = 1$

Therefore, Option (B) is TRUE.

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Analysis of Option (C):

The function $f(x)$ is bounded and twice differentiable, but it is not required to converge to any specific limit as $x \to \infty$. For instance, a sinusoidal function like $f(x) = 2\sin(\omega x + \phi)$ can be constructed to satisfy the initial conditions, whose limit as $x \to \infty$ does not exist. Therefore, Option (C) is FALSE.

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Analysis of Option (D):

Consider the function $g: \mathbb{R} \to \mathbb{R}$ defined by: $g(x) = (f(x))^2 + (f'(x))^2$

Given that $g(0) = 85$.

1. By the Mean Value Theorem on $[-4, 0]$, there exists $x_1 \in (-4, 0)$ such that: $|f'(x_1)| = \frac{|f(0) - f(-4)|}{4} \le 1$ Since $|f(x_1)| \le 2$, we have: $g(x_1) = (f(x_1))^2 + (f'(x_1))^2 \le 2^2 + 1^2 = 5$

2. Similarly, by the Mean Value Theorem on $[0, 4]$, there exists $x_2 \in (0, 4)$ such that: $|f'(x_2)| = \frac{|f(4) - f(0)|}{4} \le 1$ Since $|f(x_2)| \le 2$, we have: $g(x_2) = (f(x_2))^2 + (f'(x_2))^2 \le 2^2 + 1^2 = 5$

Since $f$ is twice differentiable, $g(x)$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$. Since $g(0) = 85 > 5 \ge \max\{g(x_1), g(x_2)\}$, the absolute maximum of $g(x)$ on the closed interval $[x_1, x_2]$ must occur at an interior point $\alpha \in (x_1, x_2) \subset (-4, 4)$.

At this local extremum point $\alpha$, the derivative must vanish: $g'(\alpha) = 0$

Differentiating $g(x)$: $g'(x) = 2f(x)f'(x) + 2f'(x)f''(x) = 2f'(x)\left(f(x) + f''(x)\right)$

Hence: $2f'(\alpha)\left(f(\alpha) + f''(\alpha)\right) = 0$

If $f'(\alpha) = 0$, then: $g(\alpha) = (f(\alpha))^2 + (f'(\alpha))^2 = (f(\alpha))^2 \le 4$ However, $\alpha$ is the maximum of $g$ on $[x_1, x_2]$, so $g(\alpha) \ge g(0) = 85$, which contradicts $g(\alpha) \le 4$.

Thus, we must have: $f'(\alpha) \ne 0$

Consequently: $f(\alpha) + f''(\alpha) = 0 \quad \text{and} \quad f'(\alpha) \ne 0$

Therefore, Option (D) is TRUE.

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Conclusion:

The correct statements are A, B, and D.