The least value of a for which the function \(f(x) = 2x^2 + ax + 1\) is strictly increasing on the interval [1, 3] is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-4\)
Differentiating:
\(f'(x) = 4x + a\)
f’(x) is smallest at the left end x = 1 of the interval, where it equals 4 + a. So f is strictly increasing on [1, 3] when
\(4 + a \ge 0\)
\(a \ge -4\)
For a = −4, f’(x) = 4x − 4 is zero only at x = 1 and positive on (1, 3], so f is still strictly increasing. For a < −4, f’(x) < 0 just to the right of x = 1, so f decreases there.
Hence the least value of a is −4.
Why the other options are incorrect
Option (1) \(-12\) — Incorrect. This uses the right end x = 3; with a = −12, f’(x) = 4x − 12 < 0 on [1, 3), so f decreases.
Option (3) \(-2\) — Incorrect. f is increasing for a = −2, but −4 is smaller and also works.
Option (4) 4 — Incorrect. f is increasing for a = 4, but it is not the least such value.