Let $f(x)$ and $g(x)$ be differentiable for $0\leq x\leq 1$, such that $f(0)=0,g(0)=0,f(1)=6$. Let there exists a real number $c$ in $(0,1)$ such that $f'(c)=2g'(c)$, then prove that $g(1)=3$
It is solved in my reference as $$ F'(c)=0\implies\text{Apply Rolle's theorem}\\ F(x)=f(x)-2g(x)\\ F(0)=0\\ F(1)=f(1)-2g(1)=6-2.g(1)=0\\ g(1)=3 $$ But it seems the Rolle's theorem is applied in reverse here. How can we conclude that $F'(c)=0\implies F(0)=F(1)$ ?
The problem is wrong.
Consider $$f(x)=6x ; \, \; g(x)=x^6$$
Then $f(0)=0,g(0)=0, f(1)=6$ and $f'(c)=2g'(c)$ for $c =\frac{1}{\sqrt[5]{2}}$ but $g(1) \neq 3$.
The claim is true if $f'(c)=2g'(c)$ for all $c$.