Find the least value of a such that the function

We have,

$f(x)=x^{2}+a x+1$

$\therefore f^{\prime}(x)=2 x+a$

Now, function f is increasing on [1,2].

$\therefore f^{\prime}(x) \geq 0$ on $[1,2]$

Now, we have $1 \leqslant x \leqslant 2$

$\Rightarrow 2 \leqslant 2 x \leqslant 4$

$\Rightarrow 2+a \leqslant 2 x+a \leqslant 4+a$

$\Rightarrow 2+a \leqslant f^{\prime}(x) \leqslant 4+a$

Since $f^{\prime}(x) \geq 0$

$\Rightarrow 2+a \geq 0$

$\Rightarrow a \geq-2$

So, least value of $a$ is $-2$.