Question

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 $[1,2]$.

Answer 1

We have,

$f\left(x\right)=x^2+ax+1$

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

Now, function $f$ will be increasing in $(1,2)$, if $f^{\prime }\left(x\right)>0$ in $(1,2)$.

$\Rightarrow 2x+a>0$

$\Rightarrow 2x>-a$

$\Rightarrow x>\frac{-a}{2}$

Therefore, we have to find the least value of $a$ such that

$\Rightarrow x>\frac{-a}{2}$, when $x\in (1,2)$.

Thus, the least value of $a$ for $f$ to be increasing on $(1,2)$ is given by,

$\frac{-a}{2}=1\Rightarrow a=-2$

Hence, the required value of $a$ is $-2.$