Question

Let a,b,c be non-zero real numbers such that b2<4ac, and f(x)=ax2+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 setthe 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.