The function $f(x) = x^2 + bx + 1$ is increasing in the interval [1, 2], then the least value of b is:

-2

The correct answer is Option (4) → -4

The function $f(x) = x^2 + bx + 1$ is increasing if $f'(x)≥0$ for all $x∈ [1, 2]$. Differentiating $f(x)$:

$f'(x)=2x+b$

For $f'(x)≥0$ in $[1,2]$, check the boundary points:

At $x=1$

$2(1)+b≥0⇒b≥-2$

At $x=2$

$2(2)+b≥0⇒b≥-4$

Thus, the least b satisfying both conditions is $b= -2$