The greatest possible value of 'a' such that the function \(f(x) = x^2+ax + 1\) is always decreasing in the interval [1, 2], is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → -4
f is decreasing on [1, 2] when \(f'(x) \le 0\) at every point of the interval:
\(f'(x) = 2x + a\)
\(2x + a \le 0\) for all \(x \in [1, 2]\)
\(a \le -2x\) for all \(x \in [1, 2]\)
The condition must hold even where \(-2x\) is smallest, that is at x = 2:
\(a \le -4\)
So the greatest value is a = −4. With a = −4, \(f'(x) = 2x - 4 \lt 0\) for \(1 \le x \lt 2\) and is 0 only at the end point x = 2, so f is decreasing on [1, 2].
Why the other options are incorrect
Option (1) -2 — Incorrect. It meets the condition only at x = 1, where \(f'(1) = 0\). For \(1 \lt x \le 2\), \(f'(x) = 2x - 2 \gt 0\), so f increases there.
Option (3) 2 — Incorrect. \(f'(x) = 2x + 2 \gt 0\) on [1, 2], so f is increasing.
Option (4) 4 — Incorrect. \(f'(x) = 2x + 4 \gt 0\) on [1, 2], so f is increasing.