Figure 1 shows the sketch of a curve with equation $y = f(x), \, x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 1

To find the solution of $f(2x) = 0$, we need to determine when the function intersects the $x$-axis. Since $f(x)$ crosses the $x$-axis at $x = 5$, we set $2x = 5$. Thus, solving for $x$ yields:

$2x = 5 \implies x = \frac{5}{2} = 2.5.$

So the solution is $x = 2.5$.

The original curve has a horizontal asymptote at $y = 1$. The transformation to $y = f(-x)$ reflects the curve over the $y$-axis, but does not change the horizontal asymptote. Hence, the equation of the asymptote remains:

$y = 1.$

This shows that the behavior of the curve as it approaches infinity does not change.

For the line $y = k$ to intersect the curve at only one point, it must be tangent to the curve at that point. The curve has a maximum at $(2, 7)$. Hence:

• For $k < 1$, the line never intersects the curve.
• For $k = 1$, the line is tangent to the curve at the asymptote, which is valid for one point.
• For $k > 7$, the line does not intersect the curve.

Therefore, the possible values for $k$ are: $k \leq 1 \text{ or } k = 7.$

This means the set of possible values for $k$ is $(-\infty, 1] \cup \{7\}$.