Question

Let $f:[0,1]\to R$ be a function. Suppose the function $f$ is twice differentiable, $f\left(0\right)=f\left(1\right)=0$ and satisfies $f^{\prime \prime }\left(x\right)-2f^{\prime }\left(x\right)+f\left(x\right)\ge e^x$, $x\in [0,1]$. Which of the following is true for $0<x<1$?

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

Answer 1

The correct option is D

$-\infty <f\left(x\right)<0$

$\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}+y\ge e^x{e}^{-x}\frac{d^{2}y}{d{x}^2}-2e^{-x}\frac{dy}{dx}+e^{-x}y\ge 1\frac{d}{dx}\left(\frac{dy}{dx}{e}^{-x}\right)-\frac{d}{dx}\left(y{e}^{-x}\right)\ge 1\frac{d^2}{d{x}^2}\left(y{e}^{-x}\right)\ge 1>0$
$y{e}^{-x}$ is concave upward