Let f(x) = x² − 6x + 11. If a and b are the two values of x for which f(x) = 6, what is the value of f(a + b)?
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Answer & Explanation
Correct answer
D
Method 1 (sum of roots then evaluate): f(x) = 6 gives x² − 6x + 11 = 6, i.e. x² − 6x + 5 = 0, whose roots sum to 6 by the coefficient relation a + b = 6; then f(6) = 36 − 36 + 11 = 11. Method 2 (factor then add): x² − 6x + 5 = (x − 1)(x − 5), so a = 1 and b = 5, a + b = 6, and f(6) = 6² − 6(6) + 11 = 11.
Choice 2 is the minimum value of f, reached by completing the square as f(x) = (x − 3)² + 2 (least value 2 at x = 3), not f(a + b). Choice 5 drops the constant. Choice 6 echoes the target. Choice 36 stops at the square.