Let `F (x) = (f (x ))^(2) + (f' (x ))^(2), F (0) =6,` whtere f (x) is a thrice differentiable function such that `|f (x) || le 1 AA x in [-1, 1], ` then choose the correct statement (s)

Options

A. there is atleast one point in each of the intervals `(-1,0) and (0,1) ` where `|f' (x) le 2`
B. there is atleast one point in each of the intervals `(-1,0) and (0,1) ` where `F (x) le 5`
C. there is no poin tof local maxima of `F(x) ` in `(-1,1)`
D. for some `c in (-1,1) , F (c ) ge 6, F'(c ) =0 and f ''(c ) le 0`

Answer

To solve the given problem, we start with the function \( F(x) = (f(x))^2 + (f'(x))^2 \) where \( F(0) = 6 \) and \( |f(x)| \leq 1 \) for \( x \in [-1, 1] \). We need to analyze the statements provided based on this information.

### Step 1: Analyze the function \( F(x) \)

Given:
\[ F(x) = (f(x))^2 + (f'(x))^2 \]
At \( x = 0 \):
\[ F(0) = (f(0))^2 + (f'(0))^2 = 6 \]
Since \( |f(0)| \leq 1 \), we have \( (f(0))^2 \leq 1 \). Therefore, \( (f'(0))^2 \) must account for the remaining value to make the sum equal to 6.

### Step 2: Determine the implications for \( f'(0) \)

From \( (f(0))^2 + (f'(0))^2 = 6 \):
- Let \( (f(0))^2 = a \) where \( a \leq 1 \).
- Then \( (f'(0))^2 = 6 - a \).

Since \( a \leq 1 \), we have:
\[ (f'(0))^2 = 6 - a \geq 6 - 1 = 5 \]
Thus, \( |f'(0)| \geq \sqrt{5} \).

### Step 3: Apply the Mean Value Theorem

Since \( f(x) \) is thrice differentiable, we can apply the Mean Value Theorem. For any interval, say \( [-1, 1] \):
\[ f'(c) = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{f(1) - f(-1)}{2} \]
Given \( |f(x)| \leq 1 \), we know \( |f(1)| \leq 1 \) and \( |f(-1)| \leq 1 \). Thus:
\[ |f(1) - f(-1)| \leq |f(1)| + |f(-1)| \leq 1 + 1 = 2 \]
This implies:
\[ |f'(c)| \leq 1 \]

### Step 4: Evaluate the statements

1. **Statement A**: There is at least one point in the interval \((-1, 0)\) and \((0, 1)\) where \(|f'(x)| \leq 2\).

Since we found \( |f'(c)| \leq 1 \) for some \( c \) in both intervals, this statement is **correct**.

2. **Statement B**: There is at least one point in each of the intervals \((-1, 0)\) and \((0, 1)\) where \( F(x) \leq 5 \).

Since \( F(x) = (f(x))^2 + (f'(x))^2 \) and we have shown that \( (f(x))^2 \leq 1 \) and \( (f'(x))^2 \leq 4 \) (from \( |f'(x)| \leq 2 \)), we conclude \( F(x) \leq 5 \). Thus, this statement is also **correct**.

3. **Statement C**: There is no point of local maxima of \( f(x) \) in \((-1, 1)\).

Since \( f(0) = 6 \) and \( F(x) \leq 5 \) in the intervals, this implies that \( f(x) \) must have a local maximum at \( x = 0 \). Therefore, this statement is **incorrect**.

4. **Statement D**: For some \( c \in (-1, 1) \), \( f(c) > 6 \) and \( f'(c) = 0 \) and \( f''(c) < 0 \).

Since \( f(0) = 6 \) and there exists a local maximum, this statement is **correct**.

### Conclusion

The correct statements are A, B, and D.