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

Consider a function $g\left(x\right)=e^{2x}$

$g\left(0\right)=1$ and $g\left(x\right)>0$ for all $x$ in $(0,\infty )$. Also, $g^{\prime }\left(x\right)=2g\left(x\right)$.

Differentiating the function $\frac{f\left(x\right)}{g\left(x\right)}$,

$\Rightarrow \frac{d\frac{f\left(x\right)}{g\left(x\right)}}{dx}=\frac{g\left(x\right)f^{\prime }\left(x\right)-2f\left(x\right)g\left(x\right)}{g^2\left(x\right)}\le 0$ for all $x$ in the domain.

$\Rightarrow \frac{f\left(x\right)}{g\left(x\right)}$ is a decreasing function.

$\Rightarrow \frac{f\left(x\right)}{g\left(x\right)}\le \frac{f\left(0\right)}{g\left(0\right)}=0$

$\Rightarrow f\left(x\right)\le 0$

Since $f\left(x\right)$ is a non-negative function, $f\left(x\right)=0$ and is a constant function.