Let f(x) = x2 + 3ax + 4, x ∈ [−1, 6]. The range of a for which f(x) is a strictly decreasing function is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \((-\infty, -4]\)
\( f'(x) = 2x + 3a \)
For f to be strictly decreasing on [−1, 6], f’(x) must be negative inside the interval (it may be zero at an end point only).
f’(x) = 2x + 3a increases with x, so its largest value on the interval is at x = 6:
\( f'(6) = 12 + 3a \le 0 \)
\( a \le -4 \)
At a = −4 itself, f’(x) = 2x − 12 is negative for every x < 6 and zero only at the end point x = 6, so f is still strictly decreasing on [−1, 6]. For a > −4, f’(x) > 0 near x = 6 and f increases there.
So the range of a is \((-\infty, -4]\).