Question

Let $f:\left[0,1\right]\to R$ (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$f"\left(x\right)-2f\text{'}\left(x\right)+f\left(x\right)\ge e^x$,$x\in \left[0,1\right]$

• 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

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

Explanation for the correct option:

Determining the range of $f\left(x\right)$

Consider the given data as

$f"\left(x\right)-2f\text{'}\left(x\right)+f\left(x\right)\ge e^x$

$e^x>0$ and divided by the $e^x$

$$\begin{gathered} f"\left(x\right)e^{-x}-2f\text{'}\left(x\right)e^{-x}+f\left(x\right)e^{-x}\ge 1 \\ \Rightarrow f"\left(x\right)e^{-x}-f\text{'}\left(x\right)e^{-x}-f\text{'}\left(x\right)e^{-x}+f\left(x\right)e^{-x}\ge 1 \end{gathered}$$

Differentiate the above Equation by parts

$$\begin{gathered} \frac{d}{dx}\left(f\text{'}\left(x\right)e^{-x}\right)-\frac{d}{dx}\left(f\left(x\right)e^{-x}\right)\ge 1 \\ \Rightarrow \frac{d}{dx}\left(\left(f\text{'}\left(x\right)e^{-x}\right)-\left(f\left(x\right)e^{-x}\right)\right)\ge 1 \\ \Rightarrow \frac{d}{dx}\left(\frac{d}{dx}\left(f\left(x\right)e^{-x}\right)\right)\ge 1 \\ \Rightarrow \frac{d^2}{d{x}^2}\left(f\left(x\right)e^{-x}\right)\ge 1 \end{gathered}$$

Let, $\varphi \left(x\right)=e^{-x}f\left(x\right)$

$$\begin{gathered} \frac{d^2}{d{x}^2}\varphi \left(x\right)\ge 1 \\ \varphi "\left(x\right)>1 \end{gathered}$$

It is the concave upward as the derivative is positive.

$$\begin{gathered} \varphi \left(0\right)=0=\varphi \left(1\right) \\ and\varphi \left(x\right)isconceveupward \end{gathered}$$

$\varphi \left(x\right)<0\Rightarrow f\left(x\right)<0$

Hence, the correct answer is option D.