Find the least value of a so that the function $f(x) = x^2 + ax + 1$ is strictly increasing on [1, 2].

$a=−2$

The correct answer is Option (2) → $a=−2$

Given $f(x) = x^2 + ax + 1, x = [1, 2]$.

Differentiating w.r.t. x, we get $f'(x) = 2x + a$.

Now $x ∈ [1, 2] ⇒ 1≤x≤2 ⇒ 2≤2x≤4$

$⇒ 2+ a≤2x+a≤4+a⇒2+a≤f'(x) ≤ 4 + a$.

For $f(x)$ to be strictly increasing on [1, 2], $2+ a ≥ 0⇒a≥-2$.

Hence, the least value of $a = -2$.