Show that there exists $a\in[0,1]$ such that $\frac{f''(a)}{f(a)}>0$ I would appreciate if somebody could help me with the following problem:
Given that :
(1) $f(x)$ is a twice differential function and $f''(x)$ is continuous on $\mathbb{R},$
(2) $\frac{f(1)}{f(0)}=\frac{f'(1)}{f'(0)}>0,$ and,
(3) $f(x)f'(x)\neq 0. \; (0\leq x\leq 1).$ 
Show that:
there exists $a\in[0,1]$ such that
$$\frac{f''(a)}{f(a)}>0$$
I tried to consider MVT .But not getting the result!
 A: Consider the function $$g(x)=\frac{f'(x)}{f(x)}$$ Condition 2. implies that $$\frac{f(1)}{f(0)}=\frac{f'(1)}{f'(0)} \implies \frac{f'(0)}{f(0)}=\frac{f'(1)}{f(1)}\implies g(0)=g(1)>0$$ According to the MVT there exists $a \in (0,1)$ such that $$g'(a)=\frac{g(1)-g(0)}{1-0}=0 \iff \frac{f''(a)f(a)-(f'(a)^2)}{f(a)^2}=0 \iff \frac{f''(a)}{f(a)}=\left(\frac{f'(a)}{f(a)}\right)^2>0$$ (and not $\ge0$ due to condition 3).
A: We have that 
$$ \frac{f'(1)}{f(1)} = \frac{f'(0)}{f(0)} $$
from (2). Hence, by the mean value theorem, there is some $a \in (0,1)$ such that
\begin{align*}
  0 &= \left(\frac{f'}f\right)(a)\\
    &= \frac{f''f - (f')^2}{f^2}(a)\\
\iff f''(a)f(a) &= f'(a)^2\\
\implies f''(a)f(a) &> 0 \qquad\quad \text{as $f'(a) \ne 0$ by (3)}
\end{align*}
Dividing by $f(a)^2 > 0$ (again by (3)), we get
$$ \frac{f''(a)}{f(a)} > 0 $$
A: By $(3)$ you get $f(x)\neq 0$ on $[0,1]$. So if $f(0)>0$ then $f(x)>0$ in $[0,1]$. 
Similarly, by $(3)$ if $f'(0)>0$ then $f'(x)>0$ in $[0,1]$, and if $f'(0)<0$ then $f'(0)<0$ in $[0,1]$.
Replacing $f$ with $-f$  does not change all the ratios in the hypothesis and the thesis, so you may assume $f>0$.
The claim then reduce to show that there is $a\in[0,1]$ such that $f''(a)>0$. 
By contradiction, supppose $f''(x)<0$ on $[0,1]$. Then $f'$ is monotone strictly decreasing. Thus $f'(1)<f'(0)$. If $f'>0$, this implies that $f'(1)/f'(0)<1$. By $(2)$ we have $f(1)/f(0)<1$. But this implies that there is $b\in[0,1]$ s.t. $f'(x)<0$, a contradiction.
Similarly, if $f'(x)<0$ then $f'(1)<f'(0)$ implies $f'(1)/f'(0)<1$, and as before we conclude that there must be $a\in[0,1]$ with $f'(a)>0$. Contradiction.
Therefore the assumption $f''(x)<0$ cannot be satisfied.
