Question

The least integral value of $a$ such that the function $x^2+ax+1$ is strictly increasing on $[1,2]$ is

• A. $-1$
• B. $-3$
• C. $-4$
• D. $-2$

Answer 1

The correct option is A $-1$

$f\left(x\right)=x^2+ax+1$
$\Rightarrow f^{\prime }\left(x\right)=2x+a$
Since, $1\le x\le 2$
$\Rightarrow 2\le 2x\le 4$
$\Rightarrow a+2\le f^{\prime }\left(x\right)\le 4+a$
Since, f(x) is strictly increasing.
$\Rightarrow f^{\prime }\left(x\right)>0$
$\Rightarrow 2+a>0$
$\Rightarrow a>-2$
So, the least integral value of a is -1.