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

The Correct Option is B

For the quadratic function $$f(x) = ax^2 + bx + c$$, where $$a, b, c$$ are non-zero real numbers and $$b^2 < 4ac$$, we seek the set $$S$$ of integers $$m$$ for which $$f(m) < 0$$. The condition $$b^2 < 4ac$$ implies that the discriminant $$\Delta = b^2 - 4ac$$ is negative ($$\Delta < 0$$). This means the quadratic equation $$ax^2 + bx + c = 0$$ has two distinct complex roots, and the parabola representing $$f(x)$$ does not intersect the x-axis.

The parabola opens upwards if $$a > 0$$ and downwards if $$a < 0$$. Since it does not intersect the x-axis, $$f(x)$$ is either always positive (if $$a > 0$$) or always negative (if $$a < 0$$) for all real $$x$$.

Consequently, if $$a > 0$$, the parabola is entirely above the x-axis, meaning $$f(m) > 0$$ for all integers $$m$$. In this case, the set $$S$$ of integers $$m$$ such that $$f(m) < 0$$ is the empty set.

If $$a < 0$$, the parabola is entirely below the x-axis, meaning $$f(m) < 0$$ for all integers $$m$$. In this case, the set $$S$$ is the set of all integers.

Therefore, the set $$S$$ is either the empty set or the set of all integers, depending on the sign of $$a$$.