Question

Let $f\left(x\right)$ be a non-positive continuous function and $F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt\forall x\ge 0$ and $f\left(x\right)\ge cF\left(x\right)$ where $c>0$ and let $g:[0,\infty )\to R$ be a function such that $\frac{dg\left(x\right)}{dx}<g\left(x\right)\forall x>0$ and $g\left(0\right)=0.$
The total number of root(s) of the equation $f\left(x\right)=g\left(x\right)$ is/are

• A. $\infty$
• B. $1$
• C. $2$
• D. $0$

Answer 1

The correct option is B $1$

$f\left(x\right)\le 0$ and $F^{{}^{\prime }}\left(x\right)=f\left(x\right)$
or $f\left(x\right)\ge cF\left(x\right)$
or $F^{{}^{\prime }}\left(x\right)-cF\left(x\right)\ge 0$
or $e^{-cx}{F}^{{}^{\prime }}\left(x\right)-c{e}^{-cx}F\left(x\right)\ge 0$
or $\frac{d}{dx}\left(e^{-cx}F\right(x\left)\right)\ge 0$
Thus, $e^{-cx}F\left(x\right)$ is an increasing function
$\therefore e^{-cx}F\left(x\right)\ge e^{-c\left(0\right)}F\left(0\right)$
or $e^{-cx}F\left(x\right)\ge 0$
or $F\left(x\right)\ge 0$
or $f\left(x\right)\ge 0$ [as $f\left(x\right)\ge cF\left(x\right)\text{and c is positive}$]
$\therefore f\left(x\right)=0$
Also, $\left(\frac{dg\left(x\right)}{dx}\right)<g\left(x\right),\forall x>0$
or $e^{-x}\frac{d\left(g\right(x\left)\right)}{dx}-e^{-x}g\left(x\right)<0$
or $\frac{d}{dx}\left(e^{-x}g\right(x\left)\right)<0$
Thus, $e^{-x}g\left(x\right)$ is a decreasing function
$\therefore e^{-x}g\left(x\right)<e^{-\left(0\right)}g\left(0\right)$
or $g\left(x\right)<0$ [as $g\left(0\right)=0$]
Thus, $f\left(x\right)=g\left(x\right)$ has one solution, $x=0$