| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| 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 routine algebraic manipulation. Part (a) involves substituting g(x) into f, setting equal to x, and solving a simple quadratic. Part (b) uses the result from (a) with minimal additional insight. Both parts are standard C3 exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(fg(x) = \frac{28}{x-2} - 1\) \(\left(= \frac{30-x}{x-2}\right)\) | M1 | Sets or implies \(fg(x) = \frac{28}{x-2} - 1\). Accept \(fg(x) = 7\left(\frac{4}{x-2}\right) - 1\) followed by \(fg(x) = \frac{7 \times 4}{x-2} - 1\). Note \(fg(x) = 7\left(\frac{4}{x-2}\right) - 1 = \frac{28}{7(x-2)} - 1\) is M0 |
| Sets \(fg(x) = x \Rightarrow \frac{28}{x-2} - 1 = x \Rightarrow 28 = (x+1)(x-2) \Rightarrow x^2 - x - 30 = 0\) | M1 | Sets up a 3TQ \((= 0)\) from an attempt at \(fg(x) = x\) or \(g(x) = f^{-1}(x)\) |
| \(\Rightarrow (x-6)(x+5) = 0\) | dM1 | Method of solving 3TQ \((= 0)\) to find at least one value for \(x\). Dependent on previous M |
| \(\Rightarrow x = 6,\ x = -5\) | A1 | Both \(x = 6\) and \(x = -5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 6\) | B1 ft | For \(a = 6\) but may follow through on largest solution from part (a) provided more than one answer found. Accept 6, \(a = 6\) and even \(x = 6\). Do not award marks for part (a) for work in part (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(fg(x) = x \Rightarrow g(x) = f^{-1}(x)\); \(\frac{4}{x-2} = \frac{x+1}{7}\) | M1 | Alternatively sets \(g(x) = f^{-1}(x)\) where \(f^{-1}(x) = \frac{x \pm 1}{7}\) |
| \(\Rightarrow x^2 - x - 30 = 0 \Rightarrow (x-6)(x+5) = 0\) | M1 | |
| \(\Rightarrow x = 6,\ x = -5\) | dM1 A1 |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $fg(x) = \frac{28}{x-2} - 1$ $\left(= \frac{30-x}{x-2}\right)$ | M1 | Sets or implies $fg(x) = \frac{28}{x-2} - 1$. Accept $fg(x) = 7\left(\frac{4}{x-2}\right) - 1$ followed by $fg(x) = \frac{7 \times 4}{x-2} - 1$. Note $fg(x) = 7\left(\frac{4}{x-2}\right) - 1 = \frac{28}{7(x-2)} - 1$ is M0 |
| Sets $fg(x) = x \Rightarrow \frac{28}{x-2} - 1 = x \Rightarrow 28 = (x+1)(x-2) \Rightarrow x^2 - x - 30 = 0$ | M1 | Sets up a 3TQ $(= 0)$ from an attempt at $fg(x) = x$ or $g(x) = f^{-1}(x)$ |
| $\Rightarrow (x-6)(x+5) = 0$ | dM1 | Method of solving 3TQ $(= 0)$ to find at least one value for $x$. Dependent on previous M |
| $\Rightarrow x = 6,\ x = -5$ | A1 | Both $x = 6$ and $x = -5$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 6$ | B1 ft | For $a = 6$ but may follow through on largest solution from part (a) provided more than one answer found. Accept 6, $a = 6$ and even $x = 6$. Do not award marks for part (a) for work in part (b) |
## Alt 1(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $fg(x) = x \Rightarrow g(x) = f^{-1}(x)$; $\frac{4}{x-2} = \frac{x+1}{7}$ | M1 | Alternatively sets $g(x) = f^{-1}(x)$ where $f^{-1}(x) = \frac{x \pm 1}{7}$ |
| $\Rightarrow x^2 - x - 30 = 0 \Rightarrow (x-6)(x+5) = 0$ | M1 | |
| $\Rightarrow x = 6,\ x = -5$ | dM1 A1 | |
---\begin{enumerate}
\item The functions $f$ and $g$ are defined by
\end{enumerate}
$$\begin{aligned}
& \mathrm { f } : x \rightarrow 7 x - 1 , \quad x \in \mathbb { R } \\
& \mathrm {~g} : x \rightarrow \frac { 4 } { x - 2 } , \quad x \neq 2 , x \in \mathbb { R }
\end{aligned}$$
(a) Solve the equation $\operatorname { fg } ( x ) = x$\\
(b) Hence, or otherwise, find the largest value of $a$ such that $\mathrm { g } ( a ) = \mathrm { f } ^ { - 1 } ( a )$\\
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel C3 2016 Q1 [5]}}