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 $f"\left(x\right)-2f^{\prime }\left(x\right)+f\left(x\right)\ge e^x,x\in [0,1]$
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 $-\infty <f\left(x\right)<0$

We have $f^{\prime \prime }\left(x\right)-2f^{\prime }\left(x\right)+f\left(x\right)\ge e^x$
$\Rightarrow \left[f^{\prime \prime }\right(x)-f^{\prime }(x\left)\right]-\left[f^{\prime }\right(x)-f(x\left)\right]\ge e^x\Rightarrow \left[e^{-x}{f}^{\prime \prime }\right(x)-e^{-x}{f}^{\prime }(x\left)\right]-\left[e^{-x}{f}^{\prime }\right(x)-e^{-x}f(x\left)\right]\ge 1\Rightarrow \frac{d}{dx}\left[e^{-x}{f}^{\prime }\right(x\left)\right]-\frac{d}{dx}\left[e^{-x}f\right(x\left)\right]\ge 1\Rightarrow \frac{d}{dx}\left[e^{-x}{f}^{\prime }\right(x)-e^{-x}f(x\left)\right]\ge 1\Rightarrow \frac{d}{dx}\left[\frac{d}{dx}\right(e^{-x}f\left(x\right)\left)\right]\ge 1$
Let $g\left(x\right)=e^{-x}f\left(x\right)$
Then we have $g^{\prime \prime }\left(x\right)\ge 1>0$
So $g$ is concave upward.
Also $g\left(0\right)=g\left(1\right)=0$
$g\left(x\right)<0\forall xϵ(0,1)\Rightarrow e^{-x}f\left(x\right)<0\Rightarrow f\left(x\right)<0,\forall xϵ(0,1)$