Figure 2 shows part of the graph with equation $y = f(x)$, where $f(x) = 2|5 - x| + 3, \, x > 0$ Given that the equation $f(x) = k$, where $k$ is a constant, has exactly one root, (a) state the set of possible values of $k$; (b) Solve the equation $f(x) = \frac{1}{2} + 10$ - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 5

To solve for $f(x) = \frac{1}{2} + 10 = \frac{21}{2}$:

1. Set up the equation:

   $2|5 - x| + 3 = \frac{21}{2}$

2. Rearrange to isolate the absolute value:

   $2|5 - x| = \frac{21}{2} - 3 = \frac{21}{2} - \frac{6}{2} = \frac{15}{2}$

   $|5 - x| = \frac{15}{4}$

3. Solve for the two cases:

   • Case 1: $5 - x = \frac{15}{4}$

     $x = 5 - \frac{15}{4} = \frac{20 - 15}{4} = \frac{5}{4}$

   • Case 2: $5 - x = -\frac{15}{4}$

     $x = 5 + \frac{15}{4} = \frac{20 + 15}{4} = \frac{35}{4}$

The solutions are:

$x = \frac{5}{4} \text{ and } x = \frac{35}{4}$

The transformation $y = 4f(x - 1)$ will shift the graph of $f(x)$ to the right by 1 unit and vertically stretch it by a factor of 4. The vertex of the original function $f(x)$ is at $(5, 3)$. After the transformation:

1. The new x-coordinate is:

   $p = 5 + 1 = 6$

2. The new y-coordinate is:

   $q = 4 \times 3 = 12$

Thus, the values of $p$ and $q$ are:

$(p, q) = (6, 12)$