JEE Mathematics 2024 Question with Solution

Given:

Find: The number of positive integral values of $a$.

First check the denominator:

Its discriminant is

Since the discriminant is negative and the coefficient of $x^2$ is positive, the denominator is always positive for all real $x$.

Therefore, for the fraction to be negative for all real $x$, the numerator must be negative for all real $x$:

Now consider the quadratic in $x$:

For a quadratic to be negative for all real $x$, its leading coefficient must be negative. Hence,

But the question asks for positive integral values of $a$. Hence no such value exists.

Therefore, the number of elements in $S$ is $0$. The correct option is B.

Given:

Find: How many positive integers $a$ satisfy the inequality.

Since

has discriminant

and leading coefficient $1 > 0$, we have

So the sign of the fraction is completely determined by the numerator:

It must be negative for every real $x$.

If $a > 0$, then as $|x|$ becomes large, the term $ax^2$ dominates and the quadratic becomes positive. Hence it cannot remain negative for all real $x$. Therefore the required condition forces

This contradicts the requirement that $a$ be a positive integer. Hence the set $S$ is empty.

Therefore, the number of elements in $S$ is $0$.