Question

The least possible value of $k$ for which the function $f\left(x\right)=x^2+kx+1$ may be increasing on $[1,2]$ is

• A. $2$
• B. $-2$
• C. $0$
• D. none of these

Answer 1

The correct option is B $-2$

We have. $f\left(x\right)=x^2+kx+1$

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

Also, $f^{\prime \prime }\left(x\right)=2$

Now, $f^{\prime \prime }\left(x\right)=2,\forall x\in [1,2]$

$\Rightarrow f^{\prime \prime }\left(x\right)>0,\forall x\in [1,2]$

$\Rightarrow f^{\prime }\left(x\right)$ is an increasing function in the interval $[1,2].$

$\Rightarrow f^{\prime }\left(1\right)$ is the least value of $f^{\prime }\left(x\right)$ on $[1,2]$

But $f^{\prime }\left(x\right)>0\forall x\in [1,2]$ $[\because f\left(x\right)$is increasing on $[1,2]]$

$\therefore f^{\prime }\left(1\right)>0,\forall x\in [1,2]\Rightarrow k>-2$

Thus, the least value of $k$ is $-2$