Question

Let f:[0,1]\(\rightarrow\)R be a function. suppose the function f is twice differentiable,f(0)=0=f(1) and satisfies f''(x)-2f'(x)+f(x)\(\geq\)ex,x\(\in\)[0,1], which of the following is true?
 

• A. 0<f(x)<\(\infty\)
• B. \(-\frac{1}{2}<f(x)<\frac{1}{2}\)
• C. \(-\frac{1}{4}<f(x)<1\)
• D. \(-\infty<f(x)<0\)

Answer 1

The Correct Option is D

 To solve the given problem, we need to analyze the function \( f:[0,1]\rightarrow \mathbb{R} \) which is twice differentiable, satisfies \( f(0)=0=f(1) \), and the inequality \( f''(x) - 2f'(x) + f(x) \geq e^x \) for \( x \in [0,1] \).

The inequality can be rewritten as:

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

This is a second-order linear differential inequality. We know that the left-hand side is related to the homogeneous solution of the equation:

\(y'' - 2y' + y = 0\)

The characteristic equation is:

\(m^2 - 2m + 1 = 0\)

which simplifies to:

\((m-1)^2 = 0\)

This gives a repeated root \( m = 1 \). Therefore, the general solution of the homogeneous equation is:

\(y_h(x) = (A + Bx)e^x\)

To analyze the non-homogeneous inequality, we notice that any particular solution must satisfy:

\(y_p(x) \geq e^x\)

The problem states \( f(0) = 0 \) and \( f(1) = 0 \), giving us boundary conditions for any particular solution. Considering these conditions, we articulate:

• Trying \( f(x) = -(A + Bx)e^x \) aligns with negative solutions because the inequality \( f''(x) - 2f'(x) + f(x) \geq e^x \) allows negative values on the left-hand side if \( g(x) = e^x \) is subtracted.

Let us check the options:

• Option 1: \( 0 < f(x) < \infty \) does not satisfy \( f(0) = 0 \) due to the positivity range constraint.
• Option 2: \( -\frac{1}{2} < f(x) < \frac{1}{2} \) and Option 3: \( -\frac{1}{4} < f(x) < 1 \) do not inherently satisfy \( f(0) = 0 \) and the possible values for \( f(x) \).
• Option 4: \( -\infty < f(x) < 0 \) is the only viable range according to constraints \( f(0) = 0 \), \( f(x) \) being possibly negative with respect to function behavior.

Thus, the correct answer is:

\(-\infty < f(x) < 0\)