| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on inverse functions and composition requiring standard algebraic manipulation. Finding the inverse involves routine rearrangement, the range follows directly from the domain restriction, and the composition is simple substitution. Slightly easier than average as it's purely procedural with no conceptual challenges or novel problem-solving required. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Range is \([y] > 1\) | B1 | Allow f, f\((x)\), \((1,\infty)\), etc |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{3}{x-2} + 1\) leading to \(y - 1 = \frac{3}{x-2}\) leading to \((x-2)(y-1) = 3\) | M1 | Clearing the fraction |
| \(x - 2 = \frac{3}{y-1}\) | DM1 | Reaching a stage which only requires one further operation |
| \(x = \frac{3}{y-1} + 2\) leading to \(f^{-1}(x) = \frac{3}{x-1} + 2\) | A1 | OE. Slightly longer routes lead to \(f^{-1}(x) = \frac{2x+1}{x-1}\) |
| [Domain is] \(x > 1\) | B1FT | Must use x FT *their* (a), must be a range |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{gf}(x) = 2\left(\frac{3}{x-2}+1\right) - 2\) or \(2\left(\frac{x+1}{x-2}\right) - 2\) | M1 | Substitute f\((x)\) into g\((x)\) |
| \(\frac{6}{x-2}\) | A1 | |
| 2 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Range is $[y] > 1$ | B1 | Allow f, f$(x)$, $(1,\infty)$, etc |
| | **1** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{3}{x-2} + 1$ leading to $y - 1 = \frac{3}{x-2}$ leading to $(x-2)(y-1) = 3$ | M1 | Clearing the fraction |
| $x - 2 = \frac{3}{y-1}$ | DM1 | Reaching a stage which only requires one further operation |
| $x = \frac{3}{y-1} + 2$ leading to $f^{-1}(x) = \frac{3}{x-1} + 2$ | A1 | OE. Slightly longer routes lead to $f^{-1}(x) = \frac{2x+1}{x-1}$ |
| [Domain is] $x > 1$ | B1FT | Must use x FT *their* **(a)**, must be a range |
| | **4** | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{gf}(x) = 2\left(\frac{3}{x-2}+1\right) - 2$ or $2\left(\frac{x+1}{x-2}\right) - 2$ | M1 | Substitute f$(x)$ into g$(x)$ |
| $\frac{6}{x-2}$ | A1 | |
| | **2** | |7 The function f is defined by $\mathrm { f } ( x ) = 1 + \frac { 3 } { x - 2 }$ for $x > 2$.
\begin{enumerate}[label=(\alph*)]
\item State the range of f.
\item Obtain an expression for $\mathrm { f } ^ { - 1 } ( x )$ and state the domain of $\mathrm { f } ^ { - 1 }$.\\
The function g is defined by $\mathrm { g } ( x ) = 2 x - 2$ for $x > 0$.
\item Obtain a simplified expression for $\mathrm { gf } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q7 [7]}}