Let $f:[0,1] \rightarrow R$ be a function. Suppose the function $f$ is twice differentiable with $f(0)=f(1)=0$ and satisfies $f''(x)-2 f^{\prime}(x)+f(x) \geq e^x$ for all $x \in[0,1]$. Which of the following is true for $x \in(0,1)$ ? 

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

We have,

$f''(x)-2 f'(x)+f(x) \geq e^x$ for all $x \in[0,1]$

$\Rightarrow e^{-x} f''(x)-2 f'(x) e^{-x}+f(x) e^{-x} \geq 1$ for all $x \in[0,1]$

$\Rightarrow \left\{e^{-x} f''(x)-e^{-x} f'(x)\right\}-\left\{-e^{-x} f(x)+f'(x) e^{-x}\right\} \geq 1$ for all $x \in[0,1]$

$\Rightarrow \frac{d}{d x}\left(f'(x) e^{-x}\right)-\frac{d}{d x}\left(f(x) e^{-x}\right) \geq 1$ for all $x \in[0,1]$

$\Rightarrow \frac{d}{d x}\left(f'(x) e^{-x}-f(x) e^{-x}\right) \geq 1$ for all $x \in[0,1]$

$\Rightarrow \frac{d}{d x}\left\{\frac{d}{d x} f(x) e^{-x}\right\} \geq 1$ for all $x \in[0,1]$

$\Rightarrow \frac{d^2}{d x^2}\left(f(x) e^{-x}\right) \geq 1$ for all $x \in[0,1]$

$\Rightarrow \frac{d^2}{d x^2}(\phi(x)) \geq 1$ for all $x \in[0,1]$, where $\phi(x)=f(x) e^{-x}$

$\Rightarrow \phi(x)$ is concave upward on [0, 1]

It is given that $f(0)=f(1)=0$. Therefore, $\phi(0)=f(0)=0$ and $\phi(1)=f(1) e^{-1}=0$.

Thus, $\phi(x)$ is concave upward on $[0,1]$ such that $\phi(0)=\phi(1)=0$. Therefore,

$\phi(x)<0$ for all $x \in(0,1) \Rightarrow f(x) e^{-x}<0$ for all $x \in(0,1)$

$\Rightarrow f(x)<0$ for all $x \in(0,1) \Rightarrow-\infty<f(x)<0$ for all $x \in(0,1)$