Question

Let $f:[0,1]\to R$(the set of all real numbers) 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 $-\infty <f\left(x\right)<0$

The given inequality is expressed as,
$f^{\prime \prime }\left(x\right)-2f^{\prime }\left(x\right)+f\left(x\right)\ge e^x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+y\ge e^x$
$\Rightarrow e^{-x}\frac{d^{2}y}{d{x}^2}-2e^{-x}\frac{dy}{dx}+e^{-x}y\ge 1$
$\Rightarrow \frac{d^2\left(y{e}^{-x}\right)}{d{x}^2}\ge 1$
Thus, $y{e}^{-x}$ concave upward.
For, $0<x<1$, So,$-\infty <f\left(x\right)<0$.