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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.

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  • $\begingroup$ $f=g=0$ is a solution for all this equations, perhaps the question here is about non trivial solution ? $\endgroup$
    – Hamza
    Aug 24 at 9:37
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    $\begingroup$ I think all the options are correct. By the way, @Hamza, I don't think the question asks about existence of such functions. I think the question asks about the solutions of the given equations for any differentiable function $f$ and $g$ as defined. $\endgroup$
    – Aditya
    Aug 24 at 10:22
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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 $(0,1)$.

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