Let f(x) = x² + ax + b and g(x) = f(x + 1) – f(x – 1). If f(x) ≥ 0 for all real x, and g(20) = 72, then the smallest possible value of b is
Explanation:
Given, g(x) = f(x + 1) – f(x – 1).
f(x) = x² + ax + b
∴ g(x) = (x + 1)² + a(x + 1) + b – [(x - 1)² + a(x - 1) + b]
⇒ g(x) = x2 + 2x + 1 + ax + a + b - x2 + 2x - 1 - ax + a – b
⇒ g(x) = 4x + 2a
Given, g(20) = 72
⇒ 4 × 20 + 2a = 72
⇒ a = -4
Also given, f(x) ≥ 0
⇒ x² - 4x + b ≥ 0
This is possible when discriminant is less than or equal to 0.
⇒ 16 – 4b ≤ 0
⇒ b ≥ 4
∴ Least possible value of b is 4.
Hence, option (c).
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