If a function f(x) = x² +bx+1 is increasing in the interval [1,2], then the least value of b is:

The correct option is -2.

To find the least value of $b$ such that the function is increasing in the interval $[1,2]$, we can use the concept of derivatives.

Step 1: Find the derivative of the function
The given function is:
$$f\left(x\right)=x^2+bx+1$$
Differentiating both sides with respect to $x$, we get:
$$f^{\prime }\left(x\right)=2x+b$$

Step 2: Apply the condition for an increasing function
For the function $f\left(x\right)$ to be increasing in the interval $[1,2]$, its derivative must be non-negative for all $x$ in that interval:
$$f^{\prime }\left(x\right)\ge 0\text{for all}x\in [1,2]$$
Substituting the derivative, we get:
$$2x+b\ge 0$$
Rearranging the inequality for $b$:
$$b\ge -2x$$

Step 3: Find the least value of b
Since $b\ge -2x$ must hold true for all $x$ in the interval $[1,2]$, $b$ must be greater than or equal to the maximum value of $-2x$ on this interval.
Let us evaluate $-2x$ at the boundaries of the interval:
At $x=1$:
$$-2\left(1\right)=-2$$
At $x=2$:
$$-2\left(2\right)=-4$$
The maximum value of $-2x$ on the interval $[1,2]$ is $-2$.
Therefore, to satisfy $b\ge -2x$ for all $x\in [1,2]$, we must have:
$$b\ge -2$$
Thus, the least value of $b$ is $-2$.