Solving the equation of the type $g\left( x \right) = \int\limits_0^x {f\left( t \right)dt} $ Let $f:\left[ {0,1} \right] \to R$ be a differentiable function. Let $g\left( x \right) = \int\limits_0^x {f\left( t \right)dt} $ with $g\left( 1 \right) = 0$  . Which of these equations must have at least one solution for x in the interval $(0,1)$ ?
(A). $g\left( x \right) = f\left( x \right)$
(B). $xg\left( x \right) = \left( {1 - x} \right)f\left( x \right)$
(C). $f\left( x \right) = f'\left( x \right)g\left( x \right)$
(D). $f\left( x \right) = xg\left( x \right)$
This question has one or more than one correct
My approach is as follow
$g\left( 0 \right) = \int\limits_0^0 {f\left( t \right)dt}  = 0$
Given $g(1)=0$
On differentiating we get $g'\left( x \right) = f\left( x \right)$, hence $g'\left( c \right) = f\left( c \right)$ where $c\in (0,1)$. How we will use the Rolle's theorem for checking and verifying each option choice.
 A: I think all the options have at least one solution in $(0,1)$.
First of all, $g'(x)=f(x)$.
For option A,
Let $F(x)=e^{-x}g(x)$.
$\implies F(0)=F(1)=0$
$\implies \exists c \in (0,1)$ for which $F'(x)=0$
And $F'(x)=e^{-x}g'(x) -e^{-x}g(x)=e^{-x}(f(x)-g(x))=0$
$\implies f(x)-g(x)=0$ has a solution in $(0,1)$.
For option B,
Let $G(x)=e^{x+\log(1-x)}g(x)$
$G(0)=G(1)=0\implies G'(c)=0$ for some $c\in (0,1)$.
And $G'(x)=e^{x+\log(1-x)}g'(x)+e^{x+\log(1-x)}\left(1-\frac{1}{1-x}\right)g(x)=0$
$\implies e^{x+\log(1-x)}\left(f(x)-\frac{xg(x)}{1-x}\right)=0$
$\implies (1-x)f(x)=xg(x)$ has a solution in $(0,1)$.
For option C,
Let $H(x)=e^{-f(x)}g(x)$
So $H(0)=H(1)=0 \implies H'(c)=0$ for some $c\in (0,1)$
And $H'(x)=e^{-f(x)}f(x)-e^{-f(x)}f'(x)g(x)=0$
$\implies e^{-f(x)}(f(x)-f'(x)g(x))=0$
$\implies f(x)=f'(x)g(x)$ has a solution in $(0,1)$.
For option D,
Let $I(x)=e^{-\frac{x^2}{2}}g(x)$
$I(0)=I(1)=0 \implies I'(c)=0$ for some $c\in(0,1)$.
And $I'(x)=e^{-\frac{x^2}{2}}f(x)-xe^{-\frac{x^2}{2}}g(x)=0$
$\implies e^{-\frac{x^2}{2}}(f(x)-xg(x))=0$
$\implies f(x)=xg(x) $ has a solution in $(0,1)$.
