Let $f:[0,1] \rightarrow R$ be a twice differentiable function such that $f(0)=f(1)=0$ and $f''(x)-2 f'(x)+f(x) \geq e^x$ for all $x \in[0,1]$. If the function $f(x) e^{-x}$ assumes its minimum in the interval $[0,1]$ at $x=\frac{1}{4}$, which of the following is true?

$f'(x)<f(x)$ for $0<x<\frac{1}{4}$

Let $\phi(x)=e^{-x} f(x)$. We have $\phi(x)$ is convave upward on $[0,1]$. It is given that $\phi(x)$ attains a local minimum at $x=\frac{1}{4}$. Therefore,

$\phi'(x)<0$ for $0<x<\frac{1}{4}$ and $\phi'(x)>0$ for $\frac{1}{4}<x<1$

$\Rightarrow e^{-x}\left(f'(x)-f(x)\right)<0$ for $0<x<\frac{1}{4}$

and, $e^{-x}\left(f'(x)-f(x)\right)>0$ for $1 / 4<x<1$

$\Rightarrow f'(x)<f(x)$ for $0<x<\frac{1}{4}$ and $f'(x)>f(x)$ for $\frac{1}{4}<x<1$

Hence, option (c) is correct.