| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a standard C3 composite and inverse functions question with routine techniques: evaluating a composition, finding an inverse of a logarithmic function, sketching a modulus graph, and solving a modulus equation. All parts are textbook exercises requiring direct application of learned methods with no novel problem-solving or insight needed. Slightly easier than average due to straightforward structure. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Finding \(g(4) = k\) and \(f(k) = ....\) or \(fg(x) = \ln\left(\frac{4}{x-3}-1\right)\) | M1 | |
| \([f(2) = \ln(2x^2 - 1)\) \(fg(4) = \ln(4-1)]\) \(= \ln 3\) | A1 | (2 marks) |
| (b) \(y = \ln(2x - 1) \Rightarrow e^y = 2x - 1\) or \(e^y = 2y - 1\) | M1, A1 | |
| \(f^{-1}(x) = \frac{1}{2}(e^x + 1)\) | A1 | |
| Allow \(y = \frac{1}{2}(e^x + 1)\) | ||
| Domain \(x \in \mathbb{R}\) | B1 | |
| [Allow \(\mathbb{R}\), all reals, \((-\infty, \infty)\)] | independent | (4 marks) |
| (c) | B1 | |
| Shape, and x-axis should appear to be asymptote. Equation \(x = 3\) needed, may see in diagram (ignore others). Intercept \((0, \frac{2}{3})\) no other; accept \(y = \frac{2}{3}\) (0.67) or on graph | B1 ind. | (3 marks) |
| (d) \(\frac{2}{x-3} = 3\) \(\Rightarrow x = 3\frac{2}{3}\) or exact equiv. | B1 | |
| \(\frac{2}{x-3} = -3\), \(\Rightarrow x = 2\frac{1}{3}\) or exact equiv. | M1, A1 | (3 marks) |
| Answer | Marks |
|---|---|
| Alt: Squaring to quadratic (\(9x^2 - 54x + 77 = 0\)) and solving M1; B1A1 | (12 marks) |
**(a)** Finding $g(4) = k$ and $f(k) = ....$ or $fg(x) = \ln\left(\frac{4}{x-3}-1\right)$ | M1 | |
$[f(2) = \ln(2x^2 - 1)$ $fg(4) = \ln(4-1)]$ $= \ln 3$ | A1 | (2 marks) |
**(b)** $y = \ln(2x - 1) \Rightarrow e^y = 2x - 1$ or $e^y = 2y - 1$ | M1, A1 | |
$f^{-1}(x) = \frac{1}{2}(e^x + 1)$ | A1 | |
Allow $y = \frac{1}{2}(e^x + 1)$ | | |
Domain $x \in \mathbb{R}$ | B1 | |
[Allow $\mathbb{R}$, all reals, $(-\infty, \infty)$] | independent | (4 marks) |
**(c)** | B1 | |
Shape, and x-axis should appear to be asymptote. **Equation $x = 3$ needed**, may see in diagram (ignore others). Intercept $(0, \frac{2}{3})$ no other; accept $y = \frac{2}{3}$ (0.67) or on graph | B1 ind. | (3 marks) |
**(d)** $\frac{2}{x-3} = 3$ $\Rightarrow x = 3\frac{2}{3}$ or exact equiv. | B1 | |
$\frac{2}{x-3} = -3$, $\Rightarrow x = 2\frac{1}{3}$ or exact equiv. | M1, A1 | (3 marks) |
**Note:** $2 = 3(x + 3)$ or $2 = 3(-x - 3)$ o.e. is M0A0
**Alt:** Squaring to quadratic ($9x^2 - 54x + 77 = 0$) and solving M1; B1A1 | (12 marks) |
---5. The functions $f$ and $g$ are defined by
$$\begin{array} { l l }
\mathrm { f } : x \mapsto \ln ( 2 x - 1 ) , & x \in \mathbb { R } , x > \frac { 1 } { 2 } \\
\mathrm {~g} : x \mapsto \frac { 2 } { x - 3 } , & x \in \mathbb { R } , x \neq 3
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of fg(4).
\item Find the inverse function $\mathrm { f } ^ { - 1 } ( x )$, stating its domain.
\item Sketch the graph of $y = | \mathrm { g } ( x ) |$. Indicate clearly the equation of the vertical asymptote and the coordinates of the point at which the graph crosses the $y$-axis.
\item Find the exact values of $x$ for which $\left| \frac { 2 } { x - 3 } \right| = 3$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2007 Q5 [12]}}