Question

The least value of $k$ for which the function $x^2+kx+1$ is an increasing function in the interval $1<x<2$ is

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

Answer 1

The correct option is D

$-2$

Calculate the value of $k$ based on given information

The given expression is $x^2+kx+1$ where $1<x<2$

Say, $f\left(x\right)=x^2+kx+1$

On differentiating the above with respect to $x$, we get,

$f\text{'}\left(x\right)=2x+k$ $...\left(\mathrm{i}\right)$ $\left[\because \frac{d}{dx}{x}^n=n{x}^{n-1}\right]$

Now, $f\left(x\right)$ is strictly increasing on the interval $(1,2)$ if,

$f\text{'}\left(x\right)>0$ for $x\in (1,2)$

$\Rightarrow$ $2x+k>0$ for $x\in (1,2)$ [using equation $\left(\mathrm{i}\right)$]

$\Rightarrow$ $k>-2x$ for $x\in (1,2)$ $...\left(\mathrm{ii}\right)$

Now, since $x\in (1,2)$

Or, $1<x<2$

$\Rightarrow$ $2<2x<4$

$\Rightarrow$ $-2>-2x>-4$

Or, $-4<-2x<-2$ $...\left(\mathrm{iii}\right)$

Using inequalities $\left(\mathrm{ii}\right)$ and $\left(\mathrm{iii}\right)$, it can be concluded that,

$k\ge -2$

Then, the least value of $k$ is $-2$.

Hence, option (D) is the correct option.