Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{f}(\mathrm{x})$ be a non-negative differentiable function on $[0, \infty)$ such that $\mathrm{f}(0)=0$ and $\mathrm{f}(\mathrm{x}) \leq 2 \mathrm{f}(\mathrm{x})$ for all $\mathrm{x}>0$. Then, on $[0, \infty)$
Options:
Solution:
1793 Upvotes
Verified Answer
The correct answer is:
$\mathrm{f}(\mathrm{x})$ is always a constant function
$\mathrm{f}^{\prime}(\mathrm{x}) \leq 2 \mathrm{f}(\mathrm{x})$
$\mathrm{f}^{\prime}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}} \leq 2 \mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}}$
$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{n}}\right) \leq 0$
$\mathrm{~g}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}}$ is non-Increasing function
$\mathrm{x} \geq 0$
$\mathrm{~g}(\mathrm{x}) \leq \mathrm{g}(0)$
$\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}} \leq \mathrm{f}(0) \mathrm{e}^{-0}$
$\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}} \leq 0$
$\mathrm{f}(\mathrm{x}) \leq 0 \quad$ but given $\mathrm{f}(\mathrm{x})$ is not negative
$\therefore \mathrm{f}(\mathrm{x})=0 \quad$ Constant function
$\mathrm{f}^{\prime}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}} \leq 2 \mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}}$
$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{n}}\right) \leq 0$
$\mathrm{~g}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}}$ is non-Increasing function
$\mathrm{x} \geq 0$
$\mathrm{~g}(\mathrm{x}) \leq \mathrm{g}(0)$
$\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}} \leq \mathrm{f}(0) \mathrm{e}^{-0}$
$\mathrm{f}(\mathrm{x}) \mathrm{e}^{-2 \mathrm{x}} \leq 0$
$\mathrm{f}(\mathrm{x}) \leq 0 \quad$ but given $\mathrm{f}(\mathrm{x})$ is not negative
$\therefore \mathrm{f}(\mathrm{x})=0 \quad$ Constant function
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.