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 D $-2$

Given,

$x^2+ax+1\to [1,2]$

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

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

Given, $f^{\prime }\left(x\right)=2x+a>0\to [1,2]$

$2\left(1\right)+a>0,2\left(2\right)+a>0$

$2+a>0,4+a>0$

$a>-2,a>-4$

Therefore the least value is $-2$