Let `f:[0,1]rarrR` be a function. Suppose the function `f` is twice differentiable, `f(0)=f(1)=0` and satisfies `f\'\'(x)-2f\'(x)+f(x) ge e^x, x in [0,1]` Which of the following is true for `0 lt x lt 1 ?`

Options

A. (a) `0ltf(x)ltoo`
B. (b) `-(1)/(2)ltf(x)lt(1)/(2)`
C. (c) `-(1)/(4)ltf(x)lt1`
D. (d) `-ooltf(x)lt0`

Answer

To solve the problem, we start with the given differential inequality:

\[ f''(x) - 2f'(x) + f(x) \geq e^x, \quad x \in [0, 1] \]

### Step 1: Rewrite the inequality

We can rewrite the inequality as:

\[ f''(x) - 2f'(x) + f(x) - e^x \geq 0 \]

### Step 2: Define a new function

Let’s define a new function \( g(x) = f(x)e^{-x} \). We will compute the derivatives of \( g(x) \):

1. First derivative:
\[ g'(x) = f'(x)e^{-x} - f(x)e^{-x} \]
This simplifies to:
\[ g'(x) = e^{-x}(f'(x) - f(x)) \]

2. Second derivative:
\[ g''(x) = (f''(x)e^{-x} - f'(x)e^{-x}) - (f'(x)e^{-x} - f(x)e^{-x}) \]
Simplifying this gives:
\[ g''(x) = e^{-x}(f''(x) - 2f'(x) + f(x)) \]

### Step 3: Apply the inequality

From the original inequality, we have:

\[ g''(x) \geq e^{-x}e^x = 1 \]

This means:

\[ g''(x) \geq 1 \]

### Step 4: Integrate the inequality

Integrating \( g''(x) \geq 1 \) gives:

\[ g'(x) \geq x + C_1 \]

where \( C_1 \) is a constant of integration.

### Step 5: Integrate again

Integrating \( g'(x) \geq x + C_1 \):

\[ g(x) \geq \frac{x^2}{2} + C_1 x + C_2 \]

where \( C_2 \) is another constant of integration.

### Step 6: Analyze the boundary conditions

Given \( f(0) = 0 \) and \( f(1) = 0 \), we can evaluate \( g(0) \) and \( g(1) \):

1. At \( x = 0 \):
\[ g(0) = f(0)e^{0} = 0 \]

2. At \( x = 1 \):
\[ g(1) = f(1)e^{-1} = 0 \]

Since \( g(x) \) is greater than or equal to a quadratic function that opens upwards and has its values at both endpoints equal to zero, \( g(x) \) must be negative in the interval \( (0, 1) \).

### Step 7: Conclusion

Since \( g(x) < 0 \) for \( 0 < x < 1 \), we have:

\[ f(x)e^{-x} < 0 \implies f(x) < 0 \quad \text{for } 0 < x < 1 \]

Thus, we conclude that:

\[ f(x) \in (-\infty, 0) \quad \text{for } 0 < x < 1 \]

### Final Answer

The correct option is:

**d. \( f \) lies between negative infinity and 0.**