Question

Let $f\left(x\right)$ be a non-negative differentiable function on $[0,\infty )$ such that $f\left(0\right)=0$ and $f^{\prime }\left(x\right)\le 2f\left(x\right)$ for all $x>0$. Then, on $[0,\infty )$

• A. $f\left(x\right)$ is always a constant function
• B. $f\left(x\right)$ is strictly increasing
• C. $f\left(x\right)$ is strictly decreasing
• D. $f^{\prime }\left(x\right)$ changes sign

Answer 1

The correct option is A $f\left(x\right)$ is always a constant function

$f^{\prime }\left(x\right)\le 2f\left(x\right)\Rightarrow f^{\prime }\left(x\right)e^{-2x}\le 2f\left(x\right)e^{-2x}\Rightarrow \frac{d}{dx}\left(f\right(x\left)e^{-2x}\right)\le 0$
Let $g\left(x\right)=f\left(x\right)e^{-2x}$ then,
$g^{\prime }\left(x\right)\le 0\Rightarrow g\left(x\right)$ is non increasing function.
$\therefore g\left(x\right)\le g\left(0\right)\Rightarrow f\left(x\right)e^{-2x}\le f\left(0\right)\Rightarrow f\left(x\right)e^{-2x}\le 0$
We know that $e^{-2x}>0$
So only possible solution will be $f\left(x\right)=0$