Question

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

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

Define a function $g\left(x\right)=e^{-x}f\left(x\right)$
$\Rightarrow g^{\prime }\left(x\right)=e^{-x}\left(f^{\prime }\right(x)-f(x\left)\right)$
$\Rightarrow g^{\prime \prime }\left(x\right)=e^{-x}\left[f^{\prime \prime }\right(x)-2f^{\prime }(x)+f(x\left)\right]$
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 $g^{\prime \prime }\left(x\right)\ge 1>0$
So $g^{\prime }\left(x\right)$ is increasing.
So, for $x<1/4,g^{\prime }\left(x\right)<g^{\prime }\left(1/4\right)$
$\Rightarrow g^{\prime }\left(x\right)<0$
$\Rightarrow \left(f^{\prime }\right(x)-f(x\left)\right)e^{-x}<0$
$\Rightarrow f^{\prime }\left(x\right)<f\left(x\right)$ in $(0,1/4)$ .