| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.3 This is a straightforward composite function question requiring completing the square to find the range, then solving gf(3)=7 by substituting and solving a quadratic. All techniques are standard C3 material with no novel insight required, making it slightly 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 |
|---|---|
| \(g(x) = (x + a)^2 - a^2 + 2\) | M1 A1 |
| \(\therefore g(x) \geq 2 - a^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(g(f(3)) = g(1 - 3a) = (1 - 3a)^2 + 2a(1 - 3a) + 2\) | M1 | |
| \(\therefore 1 - 6a + 9a^2 + 2a - 6a^2 + 2 = 7, \quad 3a^2 - 4a - 4 = 0\) | A1 | |
| \((3a + 2)(a - 2) = 0\) | M1 | |
| \(a = -\frac{2}{3}, 2\) | A1 | (7) |
**(a)**
$g(x) = (x + a)^2 - a^2 + 2$ | M1 A1 |
$\therefore g(x) \geq 2 - a^2$ | A1 |
**(b)**
$g(f(3)) = g(1 - 3a) = (1 - 3a)^2 + 2a(1 - 3a) + 2$ | M1 |
$\therefore 1 - 6a + 9a^2 + 2a - 6a^2 + 2 = 7, \quad 3a^2 - 4a - 4 = 0$ | A1 |
$(3a + 2)(a - 2) = 0$ | M1 |
$a = -\frac{2}{3}, 2$ | A1 | (7)2. The functions $f$ and $g$ are defined by
$$\begin{aligned}
& \mathrm { f } : x \rightarrow 1 - a x , \quad x \in \mathbb { R } , \\
& \mathrm {~g} : x \rightarrow x ^ { 2 } + 2 a x + 2 , \quad x \in \mathbb { R } ,
\end{aligned}$$
where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the range of g in terms of $a$.
Given that $\operatorname { gf } ( 3 ) = 7$,
\item find the two possible values of $a$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q2 [7]}}