Question

If $f\left(x\right)$ is a non-negative continuous function for all $x\ge 1$ such that $f^{\prime }\left(x\right)\le pf\left(x\right)$, where $p>0$ and $f\left(1\right)=0$, then $[f\left(\sqrt{e}\right)+f\left(\sqrt{\pi }\right)]$ is equal to

• A. $0$
• B. Negative
• C. Positive
• D. None of these

Answer 1

The correct option is A $0$

Given, $f^{\prime }\left(x\right)\le pf\left(x\right)$

Thus $f^{\prime }\left(x\right)-pf\left(x\right)\le 0$
$\Rightarrow \frac{d}{dx}{e}^{-px}f\left(x\right)\le 0$
Let $g\left(x\right)=e^{-px}f\left(x\right)$
$g^{\prime }\left(x\right)\le 0,\forall x\ge 1$
$\Rightarrow g\left(x\right)\le g(\le 1),\forall x\ge 1$
$\Rightarrow g\left(x\right)\le 0$
$\Rightarrow f\left(x\right)\le 0$
But given $f\left(x\right)\ge 0$
Therefore, $f\left(x\right)\le =0,\forall x\ge 1$