Figure 1 shows the sketch of a curve with equation $y = f(x)$, $x eq ext{R}$ - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 1

To solve $f(2x) = 0$, we need to set the argument of the function $2x$ equal to the x-value where the original function crosses the x-axis. The original function crosses the x-axis at $x = 5$. Therefore, setting $2x = 5$ gives us:

x = rac{5}{2} = 2.5.

The asymptote of the curve defined by $y = f(-x)$ remains the same as that of the original curve, since horizontal shifts do not affect the horizontal asymptotes. The original asymptote is given by the equation $y = 1$; thus, the asymptote for $y = f(-x)$ is also:

$y = 1.$

For the line $y = k$ to intersect the curve $y = f(x)$ at only one point, it must either be tangent to the curve or located at the maximum or minimum points. Since $f(x)$ has a maximum at $(2, 7)$ and approaches the horizontal asymptote at $y = 1$, the values for $k$ must be constrained between these two values:

$k < 1 ext{ or } k = 7.$