Find the least value of $a$ such that the function f given by $f\left(x\right) = x^2 +ax + 1$ is strictly increasing on $\left(1,2\right)$.
$f\left(x\right) = x^2 + ax + 1$
i.e. $f'\left(x\right) = 2x + a$
Since $f\left(x\right)$ is a strictly increasing function $\implies$ $f'\left(x\right) > 0$
i.e. $2 x + a > 0$ $\implies$ $x > -\dfrac{a}{2}$
i.e. To find $x > \dfrac{-a}{2}$ when x is in the interval $\left(1,2\right)$
i.e. To find $x > \dfrac{-a}{2}$ when $1 < x < 2$
Least value of x is 1.
$\therefore$ Least value of a is obtained when $x = 1$
$\therefore$ We have, $1 = \dfrac{-a}{2}$ $\implies$ $a = -2$
$\therefore$ The least value of $a = -2$.