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 $f^{\prime \prime }\left(x\right)-2f^{\prime }\left(x\right)+f\left(x\right)\ge e^x,x\in [0,1]$
If the function $e^{-x}f\left(x\right)$ assumes its minimum in the interval [0, 1] at $=\frac{1}{4}$, which of the following is true?

• A. $f^{\prime }\left(x\right)<f\left(x\right),\frac{1}{4}<x<\frac{3}{4}$
• B. $f^{\prime }\left(x\right)<f\left(x\right),\frac{1}{4}<x<\frac{1}{4}$
• C. $f^{\prime }\left(x\right)<f\left(x\right),0<x<\frac{1}{4}$
• D. $f^{\prime }\left(x\right)<f\left(x\right),\frac{3}{4}<x<1$

Answer 1

The correct option is C $f^{\prime }\left(x\right)<f\left(x\right),0<x<\frac{1}{4}$

$g\left(x\right)=e^{-x}f\left(x\right)$
$\Rightarrow g^{\prime }\left(x\right)=e^{-x}{f}^{\prime }\left(x\right)-e^{-x}f\left(x\right)=e^{-x}f\left(x\right)-f\left(x\right)$
As $x=\frac{1}{4}$ is point of local minima in [0, 1]
$\therefore g^{\prime }\left(x\right)<0\text{for}xϵ(0,\frac{1}{4})$
and $g^{\prime }\left(x\right)>0\text{for}xϵ(\frac{1}{4},1)$
$\therefore In(0,\frac{1}{4}),g^{\prime }\left(x\right)<0$
$\Rightarrow e^{-x}\left(f^{\prime }\right(x)-f(x\left)\right)<0\Rightarrow f^{\prime }\left(x\right)<f\left(x\right)$