For every twice differentiable function `f: R->[-2,\ 2]` with `(f(0))^2+(f^(prime)(0))^2=85` , which of the following statement(s) is (are) TRUE? There exist `r ,\ s in R` where `roo)f(x)=1` (d) There exists `alpha in (-4,\ 4)` such that `f(alpha)+f"(alpha)=0` and `f^(prime)(alpha)!=0`

Options

A. There exist r,s` in R`, where `r lt s` , such that f is one-one on the open interval (r,s)
B. There exist `x_(0) in (-4,0)` such that `|f'(x_(0))| le 1`
C. `lim_(x to oo) f(x)=1`
D. There exists `alpha in (-4,4)` such that `f(alpha)+f''(alpha)=0` and `f'(alpha)ne 0`

Answer

To solve the problem, we need to analyze each of the statements given the conditions of the function \( f \).

### Given:
- \( f: \mathbb{R} \to [-2, 2] \) is a twice differentiable function.
- \( (f(0))^2 + (f'(0))^2 = 85 \).

### Analyzing Each Statement:

#### Statement (a):
**There exist \( r, s \in \mathbb{R} \) where \( r < s \), such that \( f \) is one-one on the open interval \( (r, s) \).**

Since \( f \) maps to the interval \([-2, 2]\) and the condition \( (f(0))^2 + (f'(0))^2 = 85 \) implies that \( f(0) \) and \( f'(0) \) cannot both be zero (as \( 85 \) is a positive number), \( f \) cannot be constant. Therefore, by the Mean Value Theorem, there exists an interval where \( f \) is strictly increasing or decreasing, which implies that \( f \) is one-one on that interval.

**Conclusion:** Statement (a) is **TRUE**.

#### Statement (b):
**There exists \( x_0 \in (-4, 0) \) such that \( |f'(x_0)| \leq 1 \).**

Using the Mean Value Theorem again, since \( f \) is continuous and differentiable, we can find \( x_0 \) in the interval \((-4, 0)\) such that:

\[ \frac{f(0) - f(-4)}{0 - (-4)} = \frac{f(0) - f(-4)}{4} = f'(x_0) \]

Given that \( f \) is bounded between \([-2, 2]\), the maximum change in \( f \) over the interval \([-4, 0]\) cannot exceed \( 4 \) units. Thus, we can conclude that:

\[ |f'(x_0)| \leq 1 \]

**Conclusion:** Statement (b) is **TRUE**.

#### Statement (c):
**\( \lim_{x \to \infty} f(x) = 1 \)**.

The function \( f \) is bounded between \([-2, 2]\), but there is no information provided that guarantees that \( f(x) \) approaches \( 1 \) as \( x \) approaches infinity. The limit could be any value within the range of the function. Therefore, we cannot conclude that the limit is \( 1 \).

**Conclusion:** Statement (c) is **FALSE**.

#### Statement (d):
**There exists \( \alpha \in (-4, 4) \) such that \( f(\alpha) + f''(\alpha) = 0 \) and \( f'(\alpha) \neq 0 \).**

By applying Rolle's Theorem or the Mean Value Theorem on the function \( g(x) = f(x)^2 + f'(x)^2 \), we know that \( g(0) = 85 \) and \( g(x) \) must achieve a maximum or minimum in the interval \([-4, 4]\). If \( g(\alpha) \) is at a maximum, then \( g'(\alpha) = 0 \), leading to:

\[ 2f(\alpha)f'(\alpha) + 2f''(\alpha) = 0 \implies f(\alpha) + f''(\alpha) = 0 \]

If \( f'(\alpha) \neq 0 \), this condition can hold true.

**Conclusion:** Statement (d) is **TRUE**.

### Final Answers:
- (a) TRUE
- (b) TRUE
- (c) FALSE
- (d) TRUE