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

$\begin{array}{1 1} a=2 \\ a=-2 \\ a=\pm 2 \\ None\;of\;the\;above \end{array}$

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Let $f(x)=x^2+ax+1$
Differentiating w.r.t $x$ we get,
$f'(x)=2x+a$
The function $f$ will be increasing in $(1,2)$ if $f'(x) > 0$ in $(1,2)$
$f'(x) > 0\Rightarrow 2x+a > 0$
$\Rightarrow 2x > a$
$\Rightarrow x > \large\frac{-a}{2}$
Step 2:
$x >\large\frac{-a}{2}$ when $x \in (1,2)$
$\Rightarrow x>\large\frac{-a}{2}$$(when 1 < x< 2) The least value of 'a' for f to be increasing on (1,2) is given by \large\frac{-a}{2}$$=1$
$\Rightarrow -a=2$
$a=-2$
Hence the value of $'a'$ is $-2$
answered Jul 10, 2013