There is a polynomial $p(x)$ of degree two or less such that
$$p(0)=0, p(1)=1, p'(b)=2$$
$b$ can be any value between $[0,1]$ except one point lets call that $x_0$.
Find $x_0$ and find the polynomial for $b\ne x_0$
The solution gives me $x_0 = \frac{1}{2}$.
My first step is to let the polynomial be $f(x)$. Since, the polynomial, is of degree $2$ or less. Let it be $px^2 + qx + r = 0$.
Since, $f(0) = 0 \implies 0 + 0 + r = 0 \implies r=0$.
Then the next step should be $p(1)=1$ which leads to $p+q+r=0 \implies p=-q$ since $r=0$. What do I do next? Thanks.
Indeed, if $f(x)=px^2+qx+r$, then the conditions $f(0)=0$ and $f(1)=1$ imply $r=0$ and $q=1-p$, respectively.
Now we can reduce $f$ to the following form: $f(x)=px^2+(1-p)x$. Then $f'(x)=2px+(1-p)$, so the condition $f'(b)=2$ implies $$2pb+(1-p)=2\implies p(2b-1)=1\implies p=\frac{1}{2b-1}$$ So do you see what value of $b$ makes $p$ undefined?
Edit: the condition $f(1)=1$ actually implies $p+q=1\iff q=1-p$ (thanks to user296602 for pointing out the mistake). However, the value of $b$ which makes $p$ undefined is still the same, but the resulting polynomial will be different. My solution above is now correct.