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

For a function to be increasing, its derivative must be positive.

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

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

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

This means:

$2x + b \geq 0$ for all $x \in [1,2]$

Since $f'(x) = 2x + b$ is an increasing function of $x$, the minimum value of $f'(x)$ on $[1,2]$ occurs at $x = 1$.

At $x = 1$, we need:

$f'(1) \geq 0$

$2(1) + b \geq 0$

$2 + b \geq 0$

$b \geq -2$

Since we want the least value of $b$, and $b$ must be greater than or equal to $-2$, the least value is $b = -2$.

To verify: When $b = -2$, $f'(x) = 2x - 2$

At $x = 1$: $f'(1) = 0$ (just starts increasing)

At $x = 2$: $f'(2) = 2 > 0$ (still increasing)

Therefore, the least value of $b$ is $-2$.