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

Accepted Answer

B. Given $f(x) = ax^2 + bx + c$ with discriminant $D = b^2 - 4ac < 0$. Since $D < 0$, $f(x)$ has no real roots, which means $f(x)$ always retains the same sign as $a$ for all real $x$. - If $a > 0$, $f(x) > 0$ for all real $x$. Therefore, $f(m) < 0$ is never true for any integer $m$. Thus $S = \emptyset$ (empty set). - If $a < 0$, $f(x) < 0$ for all real $x$. Therefore, $f(m) < 0$ is true for all integers $m$. Thus $S$ is the set of all integers. Hence, $S$ must necessarily be either the empty set or the set of all integers.

Set of Integers Satisfying Quadratic Inequality

Explanation

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