If the minimum value of \(a\) is \(-\displaystyle\frac{k}{2}\), such that the function \(f(x) = x^2 + ax + 5\) is increasing in [1, 2]. Then value of \(k\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
A differentiable function is increasing on an interval when \(f'(x) \ge 0\) throughout it.
\(f'(x) = 2x + a\)
Since 2x + a grows with x, its least value on [1, 2] is at x = 1:
\(f'(1) = 2 + a \ge 0\)
\(a \ge -2\)
At a = −2, \(f'(x) = 2x - 2\) is zero only at x = 1, so f is still increasing on [1, 2]. Hence the minimum value of a is −2, and
\(\displaystyle -\frac{k}{2} = -2\)
\(k = 4\)