QJEEAdv13Ma787e: JEE Advanced 2013 Mathematics - Class 12 JEEAdv13Ma787e Solution: Let (the set of all real numbers) be a function. Suppose the…

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 $x \in(a, b)$