Q56 / 170 · Manipal (MET) 2016 · Mathematics
If $f(x)$ is a non-negative continuous function for all $x \geq 1$ such that $f^{\prime}(x) \leq p f(x)$, where $p\gt0$ and $f(1)=0$, then $[f(\sqrt{e})+f(\sqrt{\pi})]$ is equal to
- Correct answerA$0$
- Bnegative
- Cpositive
- DNone of these
Solution
$f^{\prime}(x)-p f(x) \leq 0$
$\Rightarrow \quad \frac{d}{d x} e^{-p x} f(x) \leq 0$
Let $\quad g(x)=e^{-p x} f(x)$
$g^{\prime}(x) \leq 0, \forall x \geq 1$
$\begin{array}{ll}\Rightarrow & g(x) \leq g(1), \forall x \geq 1 \\ \Rightarrow & g(x) \leq 0 \\ \Rightarrow & f(x) \leq 0\end{array}$
$\begin{array}{ll}\text { But given, } & f(x) \geq 0 \\ \therefore & f(x)=0, \forall x \geq 1\end{array}$
Hence, the correct answer is option: 'A'.