| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a standard multi-part question on functions requiring routine techniques: finding range from a rational function, finding inverse by swapping and rearranging, and composing functions. All parts follow textbook procedures 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 | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([y] < 2\) OR \([f(x)] < 2\) | B1 | OE e.g. \(f < 2, (-\infty, 2), -\infty < f[x] < 2\). Do not accept \(x < 2\) or \(f(x) \leq 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 2 - \frac{5}{x+2}\) leading to \(y(x+2) = 2(x+2) - 5\) leading to \(xy + 2y = 2x - 1\) | M1 | or \(\frac{5}{x+2} = 2 - y\) (allow sign errors) |
| \(2y + 1 = 2x - xy\) leading to \(2y + 1 = x(2-y)\) | DM1 | or \(\frac{5}{2-y} = x + 2\) (allow sign errors) |
| \(x = \frac{2y+1}{2-y} \rightarrow f^{-1}(x) = \frac{2x+1}{2-x}\) | A1 | OE or \(y = \frac{5}{2-x} - 2\) |
| Domain is \(x < 2\) | B1 FT | FT on the numerical part of their range from part (a), including \(x \neq 2\) not penalized. No FT for \(x \in \mathcal{R}, x = k, x \neq k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(fg(x) = 2 - \frac{5}{x + 3 + 2}\) | B1 | |
| \(= \frac{2(x+5)-5}{x+5}\) or \(\frac{2(x+5)}{x+5} - \frac{5}{x+5}\) | M1 | Use of their common denominator |
| \(= \frac{2x+5}{x+5}\) | A1 |
## Question 7:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[y] < 2$ OR $[f(x)] < 2$ | B1 | OE e.g. $f < 2, (-\infty, 2), -\infty < f[x] < 2$. Do not accept $x < 2$ or $f(x) \leq 2$ |
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 2 - \frac{5}{x+2}$ leading to $y(x+2) = 2(x+2) - 5$ leading to $xy + 2y = 2x - 1$ | M1 | or $\frac{5}{x+2} = 2 - y$ (allow sign errors) |
| $2y + 1 = 2x - xy$ leading to $2y + 1 = x(2-y)$ | DM1 | or $\frac{5}{2-y} = x + 2$ (allow sign errors) |
| $x = \frac{2y+1}{2-y} \rightarrow f^{-1}(x) = \frac{2x+1}{2-x}$ | A1 | OE or $y = \frac{5}{2-x} - 2$ |
| Domain is $x < 2$ | B1 FT | FT on the numerical part of their range from part (a), including $x \neq 2$ not penalized. No FT for $x \in \mathcal{R}, x = k, x \neq k$ |
**Part (c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $fg(x) = 2 - \frac{5}{x + 3 + 2}$ | B1 | |
| $= \frac{2(x+5)-5}{x+5}$ or $\frac{2(x+5)}{x+5} - \frac{5}{x+5}$ | M1 | Use of their common denominator |
| $= \frac{2x+5}{x+5}$ | A1 | |7 The function f is defined by $\mathrm { f } ( x ) = 2 - \frac { 5 } { x + 2 }$ 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 $\mathrm { g } ( x ) = x + 3$ for $x > 0$.
\item Obtain an expression for $\operatorname { fg } ( x )$ giving your answer in the form $\frac { a x + b } { c x + d }$, where $a , b , c$ and $d$ are integers.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q7 [8]}}