Let `f:[0,1]toR` (the set of all real numbers ) be a function. Suppose the function f is twice differentiable,f(0)=f(1)=0 and satisfies
`f''(x)-2f'(x)+f(x)gee^(x),x in [0,1]` Consider the statements.
I. There exists some `x in R` such that, `f(x)+2x=2(1+x^(2))`
(II) There exists some `x in R` such that, 2f(x)+1=2x(1+x)

Options

A. Both I and II are true
B. I is true and II is false
C. I is false and II is true
D. Both I and II are false

Answer

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

### Given:
1. The function \( f: [0, 1] \to \mathbb{R} \) is twice differentiable.
2. \( f(0) = 0 \) and \( f(1) = 0 \).
3. The differential inequality:
\[ f''(x) - 2f'(x) + f(x) \geq e^x \quad \text{for } x \in [0, 1]. \]

### Analyzing Statement I:
**Statement I:** There exists some \( x \in \mathbb{R} \) such that \( f(x) + 2x = 2(1 + x^2) \).

Rearranging gives:
\[ f(x) = 2(1 + x^2) - 2x = 2 + 2x^2 - 2x. \]
This simplifies to:
\[ f(x) = 2x^2 - 2x + 2. \]
Now, we need to check if this function can satisfy the boundary conditions \( f(0) = 0 \) and \( f(1) = 0 \):
- For \( x = 0 \):
\[ f(0) = 2(0)^2 - 2(0) + 2 = 2 \quad \text{(not equal to 0)}. \]
- For \( x = 1 \):
\[ f(1) = 2(1)^2 - 2(1) + 2 = 2 \quad \text{(not equal to 0)}. \]
Since \( f(0) \) and \( f(1) \) do not equal 0, Statement I is **false**.

### Analyzing Statement II:
**Statement II:** There exists some \( x \in \mathbb{R} \) such that \( 2f(x) + 1 = 2x(1 + x) \).

Rearranging gives:
\[ 2f(x) = 2x(1 + x) - 1 \quad \Rightarrow \quad f(x) = x(1 + x) - \frac{1}{2}. \]
This simplifies to:
\[ f(x) = x^2 + x - \frac{1}{2}. \]
Now, we need to check if this function can satisfy the boundary conditions \( f(0) = 0 \) and \( f(1) = 0 \):
- For \( x = 0 \):
\[ f(0) = 0^2 + 0 - \frac{1}{2} = -\frac{1}{2} \quad \text{(not equal to 0)}. \]
- For \( x = 1 \):
\[ f(1) = 1^2 + 1 - \frac{1}{2} = 1.5 \quad \text{(not equal to 0)}. \]
Since \( f(0) \) and \( f(1) \) do not equal 0, Statement II is also **false**.

### Conclusion:
Both statements are false. Thus, the correct answer is that both statements do not hold true under the given conditions.