I must demonstrate that there exists $c \in (a,b)$ where $a,b \ne 0$ such that $f'(c)=c\!\cdot\!f(c)$ if the Rolle's theorem's third condition, which is $f(a)=f(b)$, is replaced by $\dfrac{f(a)}{a}=\dfrac {f(b)}{b}$. Please suggest how I should proceed.
EDIT: Attaching the snap of the question.
(The answer is option c)
EDIT1: The answer key must be incorrect. The obvious answer should be option d) as evident from the counter examples in the answers and comments below.
If $\;0<a<b\;$ or $\;a<b<0\;$, we can take into consideration the following function:
$g:[a,b]\to\Bbb R\;$ defined as
$g(x)=\dfrac{f(x)}x\;\;$ for any $\;x\in[a,b]\,.$
The function $\;g(x)\;$ satisfies all the hypothesis of Rolle's theorem, hence there exists $\;\xi\in(a,b)\;$ such that $\;g’(\xi)=0\,.$
Since $\;g’(\xi)=\dfrac{\xi f’(\xi)-f(\xi)}{\xi^2}\;,\;$ it follows that
$\xi f’(\xi)=f(\xi)\,.$
It means that actually the correct answer is the option (D).
Addendum :
If $\;a<0<b\;,\;$ there are functions $\;f:[a,b]\to\Bbb R\;$ continuous on $\,[a,b]\,,\,$ differentiable on $\,(a,b)\,$ such that $\,\dfrac{f(a)}a=\dfrac{f(b)}b\,,\,$ but the option (C) is not satisfied.
For example :
$f:[-1,1]\to\Bbb R\;$ defined as
$f(x)=x^2+x-1\;\;$ for any $\;x\in[-1,1]\,.$
This function $\,f(x)\,$ is continuous on $\,[-1,1]\,,\,$ differentiable on $\,(-1,1)\,,\;$ moreover
$\dfrac{f(-1)}{-1}=\dfrac{f(1)}1\;,\;\;$ but
$xf’(x)=2x^2+x>f(x)\;\;$ for any $\;x\in(-1,1)\,.$
Consequently, there does not exist any $\;\xi\in(-1,1)\;$ such that $\;\xi f’(\xi)=f(\xi)\,.$
Another addendum :
The option (C) cannot be correct, indeed there exists the following counterexample:
$f:[2,3]\to\Bbb R\;\;$ defined as $\;f(x)=x\;$ for any $\,x\in[2,3]\,.$
This function $\,f(x)\,$ is continuous on $\,[2,3]\,,\,$ differentiable on $\,(2,3)\,,\,$ moreover
$\dfrac{f(2)}2=\dfrac{f(3)}3\;\;$ but
$\dfrac{f’(x)}{f(x)}=\dfrac1x<\dfrac12<2<x\;\;$ for any $\,x\in(2,3)\,.$
Consequently, there does not exist any $\;\xi\in(2,3)\;$ such that
$\dfrac{f’(\xi)}{f(\xi)}=\xi\,.$