| 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 | Standard +0.2 This is a straightforward composite function question requiring basic substitution and solving a simple equation. Part (a) is direct evaluation, and part (b) involves composing functions then solving a linear equation after simplification—all standard C3 techniques with no novel problem-solving required, making it easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks |
|---|---|
| (a) \(f(2) = -2\) | M1 A1 |
| (b) \(g(x) = g(2-x^2) = \frac{3(2-x^2)}{2(2-x^2)-1} = \frac{6-3x^2}{3-2x^2}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 = \frac{9}{4}\), \(x = \pm\frac{3}{2}\) | M1 A1 | (6) |
**(a)** $f(2) = -2$ | M1 A1
**(b)** $g(x) = g(2-x^2) = \frac{3(2-x^2)}{2(2-x^2)-1} = \frac{6-3x^2}{3-2x^2}$ | M1 A1
$\frac{6-3x^2}{3-2x^2} = \frac{1}{2}$, so $2(6-3x^2) = 3-2x^2$
$x^2 = \frac{9}{4}$, $x = \pm\frac{3}{2}$ | M1 A1 | **(6)**\begin{enumerate}
\item The functions $f$ and $g$ are defined by
\end{enumerate}
$$\begin{aligned}
& f : x \rightarrow 2 - x ^ { 2 } , \quad x \in \mathbb { R } , \\
& g : x \rightarrow \frac { 3 x } { 2 x - 1 } , \quad x \in \mathbb { R } , \quad x \neq \frac { 1 } { 2 }
\end{aligned}$$
(a) Evaluate fg(2).\\
(b) Solve the equation $\operatorname { gf } ( x ) = \frac { 1 } { 2 }$.\\
\hfill \mbox{\textit{Edexcel C3 Q1 [6]}}