Question

Let $F\left(x\right)$ be a non-negative continuous function and $F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(x\right)dx\forall x\ge 0.$ If for some $c>0,f\left(x\right)\le cF\left(x\right)$ for all $x\ge 0,$ then which of the following is/are always correct ?

• A. $F\left(x\right)+f\left(x\right)=1,x\ge 0$
• B. $F\left(x\right)=0,x\ge 0$
• C. $f\left(x\right)=0,x\ge 0$
• D. $F\left(x\right)-f\left(x\right)=1,x\ge 0$

Answer 1

Answer: (B), (C)

$f\left(x\right)=0,x\ge 0$

Given :
$F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(x\right)dx\Rightarrow F^{\prime }\left(x\right)=f\left(x\right)\cdots \left(i\right)$
Also, $f\left(x\right)\le cF\left(x\right)$
$\Rightarrow F^{\prime }\left(x\right)\le cF\left(x\right)\Rightarrow F^{\prime }\left(x\right)-cF\left(x\right)\le 0$
multiplying by $I.F.=e^{-cx},$ we get
$\frac{d}{dx}\left(e^{-cx}F\right(x\left)\right)\le 0$
So, $e^{-cx}F\left(x\right)$ is decreasing function
$\Rightarrow e^{-cx}F\left(x\right)\le 0,x\ge 0[\because F(0)=0]$
$\Rightarrow F\left(x\right)\le 0,$ but $F\left(x\right)$ is non-negative function
So, $F\left(x\right)=0,x\ge 0$
and so, $f\left(x\right)=0,x\ge 0$