Let $a, b, c$ be non-zero real numbers such that $b^2 < 4ac$, and $f(x) = ax^2 + bx + c$. If the set S consists of all integers m such that $f(m) < 0$, then the set S must necessarily be

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We have $f(x) = ax^2 + bx + c$ where $a, b, c$ are non-zero real numbers.

The key condition is $b^2 < 4ac$.

The discriminant of any quadratic equation $ax^2 + bx + c = 0$ is:

$\Delta = b^2 - 4ac$

Since $b^2 < 4ac$, we get:

$\Delta = b^2 - 4ac < 0$

When the discriminant is negative, the quadratic equation has no real roots. This means the parabola $y = f(x)$ never touches or crosses the x-axis.

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Since the parabola never crosses the x-axis, it must be entirely on one side of the x-axis. Which side depends on the sign of $a$:

If $a > 0$:

The parabola opens upward

Since it never crosses the x-axis, it's always above the x-axis

Therefore: $f(x) > 0$ for all real values of $x$

This means $f(m) > 0$ for all integers $m$

So the set $S = \{m : f(m) < 0\} = \emptyset$ (empty set)

If $a < 0$:

The parabola opens downward

Since it never crosses the x-axis, it's always below the x-axis

Therefore: $f(x) < 0$ for all real values of $x$

This means $f(m) < 0$ for all integers $m$

So the set $S = \{m : f(m) < 0\} = \mathbb{Z}$ (all integers)

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The problem asks what set $S$ "must necessarily be."

Since we don't know whether $a$ is positive or negative, we can't determine exactly which case applies. However, we can say definitively that:

$S$ must be either the empty set or the set of all integers.

There's no middle ground - the function is either always positive (giving us an empty set) or always negative (giving us all integers).

This is because when a quadratic has no real roots, it maintains the same sign (positive or negative) for all real numbers, including all integers.

Therefore, the answer is Option 2: S is either the set of all integers or the empty set.