Question

Let $f\left(x\right)=x^3-12x$ be function such that the equation $|f\left(|x|\right)|=n(n\in N)$ has exactly $6$ distinct real roots then number of possible values of $n$ are :

• A. $15$
• B. $16$
• C. $17$
• D. $14$

Answer 1

The correct option is A $15$

$f\left(x\right)=x^3-12x=x(x^2-12)$

The roots can be obtained by $x(x^2-12)=0$

$\Rightarrow x=0,x=\pm \sqrt{12}=\pm 2\sqrt{3}$

$\left|f\left(\left|x\right|\right)\right|=n$(given)

$\therefore f^{\prime }\left(x\right)=3x^2-12=0$

$\Rightarrow 3x^2=12$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

$\therefore$ At $x=2$
$f\left(2\right)=2(2^2-12)=2(4-12)=-16$

Thus, the slope is $(2,-16)$

$\therefore$ from the graph, we have $n\in (0,16)$

There are $16$ solutions

Since $n\in N$ we have $n$ takes the values

$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

Hence the number of possible values of $n$ are $15$