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
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
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
Options
Answer: either the empty set or the set of all integers
Explanation:
Since b² - 4ac < 0, the quadratic f(x) = ax² + bx + c has no real roots, so its sign is constant for all real x. If a > 0, then f(x) > 0 for all x ⇒ S = ∅ (empty set). If a < 0, then f(x) < 0 for all x ⇒ S = all integers. ∴ The set S must necessarily be either the empty set or the set of all integers.
Explanation:
Since b² - 4ac < 0, the quadratic f(x) = ax² + bx + c has no real roots, so its sign is constant for all real x. If a > 0, then f(x) > 0 for all x ⇒ S = ∅ (empty set). If a < 0, then f(x) < 0 for all x ⇒ S = all integers. ∴ The set S must necessarily be either the empty set or the set of all integers.
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