Given, $$f(0)=f(1)=0$$

$$\begin{aligned} & \text { And } f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x, x \in[0,1] \\ & \Rightarrow e^{-x} f^{\prime \prime}(x)-e^{-x} f^{\prime}(x)-e^{-x} f^{\prime}(x)+e^{-x} f(x) \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)\right)^{\prime}-\left(e^{-x} f^{\prime}(x)\right)-\left(e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)-e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f(x)\right)^{\prime \prime} \geq 1 \end{aligned}$$

Let $$g(x)=e^{-x} f(x)$$

Now, $$g(0)=e^{-0} f(0)=0$$ and $$g(1)=e^{-1} f(1)=0$$

Here, $$g^{\prime \prime}(x)=\left(e^{-x} f(x)\right)^{\prime \prime} \geq 1>0$$

$$\therefore$$ Concavity of $$g(x)$$ is up and $$g(0)=g(1)=0$$

$$\begin{aligned} & \Rightarrow g(x)=e^{-x} f(x)<0 \forall x \in(0,1) \\ & \Rightarrow f(x)<0 \end{aligned}$$

Hints:

(i) $$f^{\prime \prime}(x)>0$$ implies the concavity of $$f(x)$$ is upward and $$f^{\prime \prime}(x)<0$$ implies the concavity of $$f(x)$$ is downward.

(ii) $$f^{\prime \prime}(x)>0 \quad \forall x \in(a, b)$$ and $$f(a)=f(b)=0$$ implies that the concavity of the function is upward and $$f(x)$$ lies below the $x$-axis in the interval $$x \in(a, b)$$