The second degree polynomial $f(x)$ satisfying $f(0) = 0$, $f(1) = 1$ and $f'(x)>0$ for all $x\in[0,1]$ The second degree polynomial $f(x)$ satisfying $f(0) = 0$, $f(1) = 1$ and $f'(x)>0$ for all $x\in[0,1]$
Then which of the following is right?
Options
(a) $ax+(1-a)x^2$, $a\in \mathbb{R}$
(b) $ax+(1-a)x^2$, $a\in (0,2)$
(c) no such polynomial exists.
My try
Let $f(x) = ax^2+bx+c$. Now let $x=0$; we get $f(0)=0=c\implies c=0$
Again let $x=1$; we get $f(1) = 1= a+b+c=a+b+0\implies a+b=1$
So $f(x)=ax^2+bx=(1-b)x^2+bx\implies f(x)=(1-b)x^2+bx$.
Now $f'(x)>0\implies 2ax+b>0$ for all $x\in [0,1]$
Now I did not understand how can I solve it.
 A: So you found out that $f(x) = (1 - b)x^2 + bx$
Then $f'(x) = (2 - 2b)x + b$. This we know is a linear equation.
So, if this line has a positive gradient, then $2 - 2b > 0 \Rightarrow 1 > b$ We also want the points $f'(0)$ and $f'(1)$ to both be greater than 0.
$f'(0) > 0 \Rightarrow b > 0, f'(1) > 0 \Rightarrow 2 - b > 0 \Rightarrow 2 > b$
So for this case $b \in (0,1)$
If the line has a negative gradient, then $2 - 2b < 0 \Rightarrow 1 < b$
As before $b > 0$ and $2 > b$, and for this case $b \in (1,2)$
If the gradient is $0$, then $b = 1$.
Therefore, the answer is option (b).
A: I think there are typos in $(a)(b)$.
I'm going to write an answer under the condition that $(a)$ is $ax^2+(1-a)x,a\in\mathbb R$, and that $(b)$ is $ax^2+(1-a)x, a\in\mathbb (0,2)$.
The answer is $(b)$. If you just want to know which is correct, what you need is just to check the following two cases :
The $a=1$ case is true. And the $a=3$ case is not true. Hence, we know the answer is $(b)$. 
A: Even if you have typo's, as reported by mathlove, if
f(x) = (1-b) x^2 + b x
then
f'(x) = 2 (1-b) x + b
which is required to be positive for all x (0 < x < 1). Since f'(0)=b and f'(1)=2-b, both terms must be positive then 0 < b < 2. Suppose b=3/2; then f(0)=0, f(1)=1,f'(x)=x/2+3/4 which is always positive.    
Now replace "b" by "a" to be conform with problem notations.
