Answer
See explanations.
Work Step by Step
Step 1. Based on the Mean Value Theorem, there exists at least one point $c$ in $(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$
Step 2. We need to show that point $c$ is unique for $f(x)=px^2+qx+r$. We have $f(a)=pa^2+qa+r$ and $f(b)=pb^2+qb+r$
Step 3. Thus $f'(c)=\frac{pb^2+qb+r-(pa^2+qa+r)}{b-a}=p(b+a)+q$
Step 4. On the other hand $f'(x)=2px+q$ and $f'(c)=2pc+q$
Step 5. Combining the above results, we get $2pc+q=p(b+a)+q$ and $c=(a+b)/2$ (where $p\ne0$).
Step 6. As the $a,b$ values are predefined, we conclude that the value $c$ is unique.
