Question

If $f\left(x\right)=x^3-6x^2+9x+k=0$ has one root in $(1,3)$ then range of $k$ is

• A. $R$
• B. $(-\infty ,0)$
• C. $(-4,0)$
• D. $(0,\infty )$

Answer 1

The correct option is C $(-4,0)$

Given:-

$f\left(x\right)=x^3-6x^2+9x+k=0$
Using Intermediate value theorem,
As $f\left(x\right)$ is continuous function in $(1,3).$ $f\left(x\right)$ will have one root in $(1,3)$ iff $f\left(1\right)\cdot f\left(3\right)<0$
$\Rightarrow (1-6+9+k)(27-54+27+k)<0(4+k)\left(k\right)<0$

$\Rightarrow k\in (-4,0)$

Hence, Range of $k$ is $(-4,0)$