JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Quadratic Equations and Inequalities

We need $\frac{a{x}^2+2(a+1)x+9a+4}{x^2-8x+32}<0$ for all real $x$.

The denominator $x^2-8x+32=(x-4{)}^2+16>0$ always, so the inequality requires the numerator $a{x}^2+2(a+1)x+9a+4<0$ for all $x$, which needs $a<0$ (parabola opens downward) together with a negative discriminant.

But $S$ is defined as the set of positive integral values of $a$, and $a<0$ is required — no positive integer can satisfy this. So $S=\varnothing$, and the number of elements in $S$ is $0$.