| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Moderate -0.3 This is a straightforward composite function question requiring basic substitution and algebraic manipulation. Part (a) is direct evaluation, and part (b) involves forming the composite gf(x), setting it equal to 6, and solving a simple linear equation. While it tests understanding of function composition, it requires only routine techniques with no conceptual challenges or multi-step problem-solving. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(= f\left(\frac{1}{2}\right) = -\frac{5}{8}\) | M1 A1 | |
| (b) \(g(x) = \frac{2}{(3x - 4) + 3} = \frac{2}{3x - 1}\) | M1 A1 | |
| \(\therefore \frac{2}{3x - 1} = 6\) | M1 | |
| \(2 = 6(3x - 1)\) | M1 | |
| \(x = \frac{2}{9}\) | A1 | (6 marks) |
**(a)** $= f\left(\frac{1}{2}\right) = -\frac{5}{8}$ | M1 A1 |
**(b)** $g(x) = \frac{2}{(3x - 4) + 3} = \frac{2}{3x - 1}$ | M1 A1 |
$\therefore \frac{2}{3x - 1} = 6$ | M1 |
$2 = 6(3x - 1)$ | M1 |
$x = \frac{2}{9}$ | A1 | (6 marks)\begin{enumerate}
\item The functions $f$ and $g$ are defined by
\end{enumerate}
$$\begin{aligned}
& \mathrm { f } : x \rightarrow 3 x - 4 , \quad x \in \mathbb { R } , \\
& \mathrm {~g} : x \rightarrow \frac { 2 } { x + 3 } , \quad x \in \mathbb { R } , \quad x \neq - 3
\end{aligned}$$
(a) Evaluate fg(1).\\
(b) Solve the equation $\operatorname { gf } ( x ) = 6$.\\
\hfill \mbox{\textit{Edexcel C3 Q2 [6]}}