Question

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

• A. the set of all integers
• B. either the empty set or the set of all integers
• C. the empty set the set of all positive integers
• D. the set of all positive integers

Answer 1

Answer: (B)

either the empty set or the set of all integers

Answer 2

Given the quadratic function $f(x) = ax^2 + bx + c$ where $a, b,$ and $c$ are non-zero real numbers such that $b^2 < 4ac$, we want to find the set $S$ of all integers $m$ such that $f(m) < 0$.
The quadratic function $f(x)$ represents a parabola. The discriminant $\Delta$ of the quadratic equation
$ax^2 + bx + c = 0$ is given by
$\Delta = b^2 - 4ac$.
Since $b^2 < 4ac$, the discriminant is negative ($\Delta < 0$), which means that the quadratic equation has two distinct complex roots.
The vertex of the parabola is given by the point $(h, k)$ where $h = -\frac{b}{2a}$ and $k = f(h)$.
In this case, since $a$ is non-zero, the parabola opens upwards if $a > 0$ and downwards if $a < 0$.
Since the parabola opens upwards or downwards and the discriminant is negative, the parabola does not intersect the x-axis. This implies that the function $f(x)$ is either entirely above the x-axis (if $a > 0$) or entirely below the x-axis (if $a < 0$).

Now, we want to find the set $S$ of all integers $m$ such that $f(m) < 0$.
Depending on the sign of $a$, the parabola is either above or below the x-axis.
If the parabola is above the x-axis, there will be no integer $m$ for which $f(m) < 0\, (f(m)$ will always be positive or zero).
If the parabola is below the x-axis, then $f(m) < 0$ for all integers $m$.

Therefore, the set $S$ must be either the empty set (if $a > 0$ and the parabola opens upwards) or the set of all integers (if $a < 0$ and the parabola opens downwards), depending on the sign of $a$.
Hence, the correct answer is: either the empty set or the set of all integers.