Find the least value of $a$ such that the function $f$ given by $f(x) = {x^2} + ax + 1$ is strictly increasing on $(1,\;2)$.

We have, $f(x) = {x^2} + nx + 1$ …(i)
$\Rightarrow f'(x) = 2x + a$
If $1 < x < 2 \Rightarrow 2 < 2x < 4 \Rightarrow 2 + a < 2x + a < 4 + a$
$\Rightarrow 2 + a < f'(x) < 4 + a$

Now, $f(x)$ is strictly increasing on $(1,\;2)$ if and only if $f'(x) > 0$

$\Rightarrow 2 + a \ge 0 \Rightarrow a \ge - 2$

Therefore the required least value of $a$ is $- 2.$