CAT 2022 Slot 1 · Quant · Question 5

The discriminant of f is {b}^{2}-4ac, and the condition {b}^{2}<4ac makes it negative.

A negative discriminant means the quadratic has no real roots.

With no real roots, the parabola never touches the x-axis, so f\left( x \right) never equals zero for any real x.

A continuous function that is never zero cannot change sign, so f\left( x \right) keeps one sign throughout.

That sign is the sign of the leading coefficient a.

If a>0 the parabola opens upwards and sits entirely above the axis, so f\left( m \right)<0 holds for no integer at all, making S empty.

If a<0 the parabola opens downwards and sits entirely below the axis, so f\left( m \right)<0 holds for every integer, making S the set of all integers.

Both cases are genuinely possible, and nothing in the question fixes the sign of a.

So S is either the empty set or the set of all integers — the second option.

Shortcut. Option 4 can be dismissed on symmetry alone: nothing in f distinguishes positive integers from negative ones, so no condition could single them out. That leaves only the option covering both sign cases.