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 $x=\frac{1}{4}$, 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 $f^{\prime }\left(x\right)<f\left(x\right)$, $0<x<\frac{1}{4}$

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

At $x=\frac{1}{4},h\left(x\right)$ attains minima. So, $h\left(x\right)$ is increasing for $x>\frac{1}{4}$ and decreasing for $x<\frac{1}{4}$
So, $f^{\prime }\left(x\right)<f\left(x\right)$, $0<x<\frac{1}{4}$