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

Define a function $g\left(x\right)=e^{-x}f\left(x\right)$

$\Rightarrow g^{\prime }\left(x\right)=e^{-x}\left(f^{\prime }\right(x)-f(x\left)\right)$

$\Rightarrow g"\left(x\right)=e^{-x}[f"(x)-2f^{\prime }(x)+f(x\left)\right]$

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

$\Rightarrow e^{-x}\left(f^{\prime \prime }\right(x)-2f^{\prime }(x)+f(x\left)\right)\ge 1$

Hence $g^{\prime \prime }\left(x\right)>1>0$

So, $g\left(x\right)$ is concave upward and $g\left(0\right)=g\left(1\right)=0$

Hence, $g\left(x\right)<0\forall x\in (O,1)$

$\Rightarrow e^{-x}f\left(x\right)<0$

$f\left(x\right)<0\forall x\in (O,1)$