Here is the problem statement:
Let $f(x)$ is a quadaratic expression with positive integral coefficients such that for every $\alpha, \beta\; \epsilon\; \Re$, $\beta>\alpha$, $\int_\alpha^\beta f(x) dx>0$. Let $g(t)=f''(t)f(t)$ and $g(0)=12$, then
(a)$16$ such quadaratics are possible (b) $f(x)=0$ has either no real root or distinct roots
(c) minimum value of $f(1)$ is $6$ (d) maximum value of $f(1)$ is $11$
From the condition $\int_\alpha^\beta f(x) dx>0$, it is clear that $f(x)$ must always be positive or zero meaning that $f(x)=0$ will have either identical roots or no roots at all. So, option (b) can be eliminated.
Now let $f(x)=ax^2 + bx+c$. So $f''(x)=2a$ (obviously $>0$).
Therefore, $g(t)=2a(ax^2+bx+c) \implies 2ac=12 \implies ac=6$
Also the minimum value for $b$ is $0$ amd the maximum can be found out in this way :
Since $f(x)=0$ has either no or equal roots, then
$$ b^2-4ac\le 0
\implies b^2\le 4ac \implies b^2\le 24\implies b^2\le 16\implies b\le4$$
Therefore max value of b is 4. Now this means that there should be $5 \times 4=20$ quadratics possible ($5$ for $b$ and $4$ for the combination of $a$ and $c$). But the option (a) is given to be true. How then is it possible? Please provide a hint. Moreover, please also provide a hint for finding the maximum and minimum values of $f(1)$.
Positive integer coefficients means $b\neq0$ so there are only $\textbf{4}$ possibilities for $b$.
For $f(1)$ I would work through an enumeration of the possible values of $a$, $b$ and $c$ and find which set gives the smallest value.