Question

range of k if ||x^2 - 6x + 8|- 12| = k has no real solution

Answer 1

(-\infty, 0)

Answer 2

Let $f(x) = ||x^2 - 6x + 8|- 12|$. We need to find the range of $f(x)$.

1. $y_1 = x^2 - 6x + 8 = (x-3)^2 - 1$. Range of $y_1$ is $[-1, \infty)$.
2. $y_2 = |x^2 - 6x + 8|$. Reflecting the negative part of $y_1$ (which is $[-1,0]$) results in $[0,1]$. For $y_1 > 0$, $y_2=y_1$. Thus, the range of $y_2$ is $[0, \infty)$.
3. $y_3 = |x^2 - 6x + 8| - 12$. Shifting $y_2$ down by 12 units. Range of $y_3$ is $[0-12, \infty-12) = [-12, \infty)$.
4. $y_4 = ||x^2 - 6x + 8|- 12|$. Reflecting the negative part of $y_3$. The minimum value of $y_3$ is -12, which reflects to 12. The points where $y_3=0$ become the minimum values of $y_4$.

$|x^2 - 6x + 8| = 12 \implies x^2 - 6x + 8 = 12$ or $x^2 - 6x + 8 = -12$.

$x^2 - 6x - 4 = 0 \implies x = 3 \pm \sqrt{13}$ (real roots, $y_4=0$).

$x^2 - 6x + 20 = 0$ (no real roots, $D<0$).

Since $y_4$ has real roots where it is 0 and tends to $\infty$, its range is $[0, \infty)$.

For the equation $f(x) = k$ to have no real solution, $k$ must be outside the range of $f(x)$.

Thus, $k < 0$.