Question

If the equation $x^3-9x^2+24x+k=0$ has exactly one root in $(2,4)$, then $k$ lies in the interval

• A. $(-20,-16)$
• B. $($16, $\infty )$
• C. $(-\infty ,-20)$
• D. None of these

Answer 1

The correct option is A $(-20,-16)$

Let $f\left(x\right)=x^3-9x^2+24x+k=0$

If $f$ has exactly one root in $(a,b)$, then $f\left(a\right)\cdot f\left(b\right)<0$

For $(a,b)\equiv (2,4)$, we can say
$\Rightarrow f\left(2\right)\cdot f\left(4\right)<0$
$\Rightarrow (8-9(4)+24(2)+k)(64-9(16)+24(4)+k)<0$
$\Rightarrow (k+20)(k+16)<0$
$\Rightarrow k\in (-20,-16)$
$f^{\prime }\left(x\right)=3(x-2)(x-4)$
Hence in the interval $(2,4),f^{\prime }\left(x\right)$ is negative.

Hence, it is monotonic decreasing in that interval and can have only one real root.

Hence, option A.