If f(x) is a polynomial of degree three with leading coefficient 1 and $f(1)=1$, $f(2)=4$, $f(3)=9$ then prove $f(x)=0$ has a root in interval $(0,1)$ 
If f(x) is a polynomial of degree three with leading coefficient 1 such that $f(1)=1$, $f(2)=4$, $f(3)=9$then prove $f(x)=0$ has a root in interval $(0,1)$.     

This is a reframed version of "more than one option correct type" questions. I could identify all the other answers but this one got left out.    
My Attempt:
From the information given in the question, the cubic equation is $$x^3-5x^2+11x-6=0$$     
Now I don't know how to prove the fact that one of the roots lies in the interval $(0,1)$. As it is a cubic equation I can't even find the roots directly to prove this.

You have that:
$$f(0) = c,\ f(1)=1,$$
Since you have shown that $c<0$, and since $f$ is continuous, by the intermediate value theorem it must pass through zero at some point in $[0,1]$.
