The function f is defined by $f: x \mapsto |2x - 5|, \; x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 5

To solve the equation $|2x - 5| = 15 + x$, we will consider two cases based on the absolute value:

Case 1: $2x - 5 = 15 + x$

• Rearranging gives: $2x - x = 15 + 5 \Rightarrow x = 20$

Case 2: $-(2x - 5) = 15 + x$

• This simplifies to: $5 - 2x = 15 + x \Rightarrow 5 - 15 = 3x \Rightarrow -10 = 3x \Rightarrow x = -\frac{10}{3}$

Thus, the solutions are $x = 20$ and $x = -\frac{10}{3}$.

To find $fg(2)$, we first compute $g(2)$:

1. Plugging $x = 2$ into the function $g(x)$: $g(2) = 2^{2} - 4(2) + 1 = 4 - 8 + 1 = -3$

2. Now we need to evaluate $f(g(2)) = f(-3)$: $f(-3) = |2(-3) - 5| = |-6 - 5| = | -11 | = 11$

Thus, $fg(2) = f(g(2)) = 11.$

The function $g(x) = x^{2} - 4x + 1$ is a quadratic function. To find the range, we need to analyze its vertex and the behavior within the interval $0 \leq x \leq 5$.

1. Calculate the vertex using the formula $x = -\frac{b}{2a}$:

   • Here, $a = 1$, $b = -4$: $x_{vertex} = -\frac{-4}{2 imes 1} = 2$

2. Evaluate $g(2)$: $g(2) = 2^{2} - 4(2) + 1 = -3$

3. Find the values of $g(x)$ at the endpoints:

   • $g(0) = 0^{2} - 4(0) + 1 = 1$
   • $g(5) = 5^{2} - 4(5) + 1 = 6$

4. The values at $x = 0, 2, 5$ gives:

   • Minimum at $g(2) = -3$ and maximum at $g(5) = 6$. Thus, the range of $g(x)$ is: $[-3, 6]$