If the minimum value of a is $-\frac{k}{2}$, such that the function $f(x) = x^2 + ax + 5$ is increasing in [1, 2]. Then value of $k$ is

4

The correct answer is Option (3) → 4

Given: $f(x) = x^2 + ax + 5$

The function is increasing on $[1, 2]$.

To ensure a function is increasing over an interval, its derivative must be non-negative over that interval.

Step 1: Compute derivative

$f'(x) = \frac{d}{dx}(x^2 + ax + 5) = 2x + a$

For $f(x)$ to be increasing on $[1, 2]$, require:

$f'(x) \geq 0$ for all $x \in [1, 2]$

The minimum value of $f'(x)$ on $[1, 2]$ occurs at $x = 1$:

$f'(1) = 2(1) + a = 2 + a \geq 0 \Rightarrow a \geq -2$

So the minimum possible value of $a$ is $-2$.

Step 2: Compare with the given condition

Minimum value of $a = -\frac{k}{2}$

Equating:

$-\frac{k}{2} = -2 \Rightarrow k = 4$