<s>[INST] Question under quadratic equation Let $f(x)=x^2 + ax + b$ such that $f(2)\times f(3)= \frac{1}{2}$ and $1< f(2) + f(3)< 2$, then the equation $f(x)=1$ has $(a,b \in \mathbb{R})$

*

*Both roots real and distinct


*Both roots real and equal


*Non real roots


*Roots whose nature depends on value of a & b
Only one of the options is correct.
My attempt
I tried doing various manipulations to $f(x)$, I tried putting the values of $f(2)$ and $f(3)$ in the equations and inequality qiven above but that derived nothing. I tried thinking that either $f(2)$ or $f(3)$ must be less than 1 but came to a point where I disagreed myself because $f(2)$ and $f(3)$ can both have values less than 1. I doubt if this question is really tough or I can't  approach the question correctly.
 [/INST] We have $f(2)=2a+b+4$ and $f(3)=3a+b+9$. So, $a=f(3)-f(2)-5$ and $b=3f(2)-2f(3)+6$.
The discriminant of the equation $f(x)=1$ is
\begin{align*}
a^2-4(b-1)&=(f(3)-f(2)-5)^2-4(3f(2)-2f(3)+5)\\
&=(f(3)-f(2))^2-10(f(3)-f(2))+25-12f(2)+8f(3)-20\\
&=(f(3)+f(2))^2-4f(3)f(2)-2f(3)-2f(2)+5\\
&=(f(3)+f(2))^2-2f(3)-2f(2)+5-2\\
&=(f(3)+f(2)-1)^2+2\\
&>0
\end{align*}
The equation has two distinct real roots.
</s>