| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 This is a straightforward inverse function question requiring algebraic manipulation to make x the subject (swap x and y, cross-multiply, rearrange). Part (b) is simple algebraic verification, and part (c) tests basic understanding that range of f must match domain of f for composition. Slightly above routine due to the rational function form, but still a standard P1 exercise with no novel insight 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 |
| \(y = \frac{x^2-4}{x^2+4}\) leading to \((x^2+4)y = (x^2-4)\) leading to \(x^2y + 4y = x^2 - 4\) | *M1 | For clearing denominator and expanding brackets. If swap variables first, look for \(y^2x + 4x = y^2 - 4\) |
| \(x^2y - x^2 = -4y - 4\) leading to \(x^2(1-y) = 4y+4\) leading to \(x^2 = \ldots\) | DM1 | For making \(x^2\) the subject. If swap variables first, look for \(y^2(1-x) = 4x+4 \Rightarrow y^2 = \ldots\) |
| \(x^2 = \frac{4y+4}{1-y}\) leading to \(x = \sqrt{\frac{4y+4}{1-y}}\) leading to \([f^{-1}(x)] = \sqrt{\frac{4x+4}{1-x}}\) | A1 | OE e.g. \(\sqrt{\frac{-4x-4}{x-1}}\) without \(\pm\) in final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \frac{y^2-4}{y^2+4}\) leading to \(x = 1 - \frac{8}{y^2+4}\) leading to \(x-1 = \frac{-8}{y^2+4}\) | *M1 | For division and reaching \(x-1 = \ldots\) (or \(y-1 = \ldots\)) |
| \(y^2+4 = \frac{-8}{x-1}\) leading to \(y^2 = \frac{-8}{x-1} - 4\) | DM1 | For making \(y^2\) (or \(x^2\)) the subject |
| \([y=][f^{-1}(x)] = \sqrt{\frac{-8}{x-1}-4}\) | A1 | OE without \(\pm\) in final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - \frac{8}{x^2+4} = \frac{x^2+4}{x^2+4} - \frac{8}{x^2+4} = \left[\frac{x^2+4-8}{x^2+4}\right] = \frac{x^2-4}{x^2+4}\) | M1 A1 | Using common denominator or division to reach 1. Remainder \(-8\). WWW |
| \(0 < f(x) < 1\) | B1 B1 | B1 for each correct inequality. B0 if contradictory statement seen. Accept \(f(x)>0\), \(f(x)<1\); \(1>f(x)>0\); \((0,1)\). SC B1 for \(0 \leqslant f(x) \leqslant 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Because the range of f does not include the whole of the domain of f (or any of it) | B1 | Accept an answer that includes an example outside the domain of f, e.g. \(f(4) = \frac{12}{20}\). Must refer to the domain or \(> 2\). Need not explicitly use the term 'domain' but must not refer just to the range |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{x^2-4}{x^2+4}$ leading to $(x^2+4)y = (x^2-4)$ leading to $x^2y + 4y = x^2 - 4$ | *M1 | For clearing denominator and expanding brackets. If swap variables first, look for $y^2x + 4x = y^2 - 4$ |
| $x^2y - x^2 = -4y - 4$ leading to $x^2(1-y) = 4y+4$ leading to $x^2 = \ldots$ | DM1 | For making $x^2$ the subject. If swap variables first, look for $y^2(1-x) = 4x+4 \Rightarrow y^2 = \ldots$ |
| $x^2 = \frac{4y+4}{1-y}$ leading to $x = \sqrt{\frac{4y+4}{1-y}}$ leading to $[f^{-1}(x)] = \sqrt{\frac{4x+4}{1-x}}$ | A1 | OE e.g. $\sqrt{\frac{-4x-4}{x-1}}$ without $\pm$ in final answer |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \frac{y^2-4}{y^2+4}$ leading to $x = 1 - \frac{8}{y^2+4}$ leading to $x-1 = \frac{-8}{y^2+4}$ | *M1 | For division and reaching $x-1 = \ldots$ (or $y-1 = \ldots$) |
| $y^2+4 = \frac{-8}{x-1}$ leading to $y^2 = \frac{-8}{x-1} - 4$ | DM1 | For making $y^2$ (or $x^2$) the subject |
| $[y=][f^{-1}(x)] = \sqrt{\frac{-8}{x-1}-4}$ | A1 | OE without $\pm$ in final answer |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - \frac{8}{x^2+4} = \frac{x^2+4}{x^2+4} - \frac{8}{x^2+4} = \left[\frac{x^2+4-8}{x^2+4}\right] = \frac{x^2-4}{x^2+4}$ | M1 A1 | Using common denominator or division to reach 1. Remainder $-8$. WWW |
| $0 < f(x) < 1$ | B1 B1 | B1 for each correct inequality. B0 if contradictory statement seen. Accept $f(x)>0$, $f(x)<1$; $1>f(x)>0$; $(0,1)$. **SC B1** for $0 \leqslant f(x) \leqslant 1$ |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Because the range of f does not include the whole of the domain of f (or any of it) | B1 | Accept an answer that includes an example outside the domain of f, e.g. $f(4) = \frac{12}{20}$. Must refer to the domain or $> 2$. Need not explicitly use the term 'domain' but must not refer just to the range |
---6 The function $f$ is defined as follows:
$$\mathrm { f } ( x ) = \frac { x ^ { 2 } - 4 } { x ^ { 2 } + 4 } \quad \text { for } x > 2$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
\item Show that $1 - \frac { 8 } { x ^ { 2 } + 4 }$ can be expressed as $\frac { x ^ { 2 } - 4 } { x ^ { 2 } + 4 }$ and hence state the range of f .
\item Explain why the composite function ff cannot be formed.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q6 [8]}}