Figure 2 shows a sketch of part of the graph $y = f(x)$, where $f(x) = 2/3 - |x| + 5, \quad x > 0$ (a) State the range of $f$ (b) Solve the equation $f(x) = \frac{1}{2}x + 30$ (c) Given that the equation $f(x) = k$, where $k$ is a constant, has two distinct roots, (e) state the set of possible values for $k$. - Edexcel - A-Level Maths Pure - Question 11 - 2017 - Paper 2

To solve the equation: $f(x) = \frac{1}{2}x + 30$ We begin with the expression for $f(x)$: $2/3 - |x| + 5 = \frac{1}{2}x + 30$ Combining like terms yields: $- |x| + 5 + \frac{2}{3} = \frac{1}{2}x + 30$ Next, multiply through by -1 to isolate $|x|$ on one side: $|x| - \frac{1}{2}x = 30 - 5 - \frac{2}{3}$ This simplifies to: $|x| = \frac{62}{3}$ To find the solution consider $x = \frac{62}{3}$. Since for $x > 0$, we have: $x = \frac{62}{3}$

For the equation $f(x) = k$ to have two distinct roots, the graph of $y = f(x)$ must intersect the line $y = k$ at two points.

From the analysis of the function, we find:

• The maximum value of $f(x)$ occurs when $x = 0$, yielding $f(0) = 5.666...$
• The function decreases to negative infinity as $x$ increases. Therefore, $k$ must satisfy the condition: $k < 5 ext{ and } k > -\infty$ Thus, the set of possible values for $k$ is: $\{ k : 5 < k < 11 \}$