Figure 2 shows a sketch of part of the graph $y = f(x)$, where $f(x) = 2/3 - |x| + 5, \, x > 0$ - Edexcel - A-Level Maths Pure - Question 12 - 2017 - Paper 2

To solve the equation $f(x) = \frac{1}{2}x + 30$, we first substitute the function:

$2/3 - |x| + 5 = \frac{1}{2}x + 30$

This simplifies to:

$- |x| + \frac{17}{3} = \frac{1}{2}x + 30$

Next, we isolate the absolute value term:

$- |x| = \frac{1}{2}x + 30 - \frac{17}{3}$

Perform the calculations for the right side:

$- |x| = \frac{1}{2}x + \frac{90-17}{3}$

Multiply through by 3 to eliminate fractions:

$-3|x| = 3 \cdot \frac{1}{2}x + 73$

This leads to working through both cases of the absolute value and solving for x, concluding with:

$x = \frac{62}{3}$

For the equation $f(x) = k$ to have two distinct roots, the line $y = k$ must intersect the graph of $f(x)$ in two distinct places. From our earlier analysis: Since $f(x)$ achieves a maximum of $\frac{17}{3}$ (which is approximately 5.67) when $x = 0$ and extends to $-\,infty$ as $x$ increases, the range of $k$ must satisfy:

$5 < k < 11$

Thus, the set of possible values for $k$ is:

$\{ k \in \mathbb{R} \mid 5 < k < \frac{17}{3}\}$