| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a straightforward inverse function question with standard algebraic manipulation. Part (a) requires simple substitution (ff(5) means f(f(5))), and part (b) involves the routine procedure of swapping x and y then rearranging—a technique practiced extensively at this level. The rational function form is standard for P1, making this 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 |
| \(f(5)=[2]\) and \(f(\textit{their}\,2)=[5]\) OR \(ff(5)=\left[\frac{2+3}{2-1}\right]\) OR \(\frac{\frac{x+3}{x-1}+3}{\frac{x+3}{x-1}-1}\) and attempt to substitute \(x=5\) | M1 | Clear evidence of applying f twice with \(x=5\) |
| \(5\) | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{x+3}{x-1}=y \Rightarrow x+3=xy-y\) OR \(\frac{y+3}{y-1}=x \Rightarrow y+3=xy-x\) | *M1 | Setting \(f(x)=y\) or swapping \(x\) and \(y\), clearing fractions and expanding brackets. Allow \(\pm\) sign errors |
| \(xy-x=y+3 \Rightarrow x=\frac{y+3}{y-1}\) OE OR \(y+3=xy-x \Rightarrow y=\left[\frac{x+3}{x-1}\right]\) OE | DM1 | Finding \(x\) or \(y=\,\). Allow \(\pm\) sign errors |
| \([f^{-1}(x)\text{ or }y]=\dfrac{x+3}{x-1}\) | A1 | OE e.g. \(1+\frac{4}{x-1}\) etc. Must be a function of \(x\), cannot be \(x=\,\) |
| Total | 3 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(5)=[2]$ and $f(\textit{their}\,2)=[5]$ OR $ff(5)=\left[\frac{2+3}{2-1}\right]$ OR $\frac{\frac{x+3}{x-1}+3}{\frac{x+3}{x-1}-1}$ and attempt to substitute $x=5$ | M1 | Clear evidence of applying f twice with $x=5$ |
| $5$ | A1 | |
| **Total** | **2** | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{x+3}{x-1}=y \Rightarrow x+3=xy-y$ OR $\frac{y+3}{y-1}=x \Rightarrow y+3=xy-x$ | *M1 | Setting $f(x)=y$ or swapping $x$ and $y$, clearing fractions and expanding brackets. Allow $\pm$ sign errors |
| $xy-x=y+3 \Rightarrow x=\frac{y+3}{y-1}$ OE OR $y+3=xy-x \Rightarrow y=\left[\frac{x+3}{x-1}\right]$ OE | DM1 | Finding $x$ or $y=\,$. Allow $\pm$ sign errors |
| $[f^{-1}(x)\text{ or }y]=\dfrac{x+3}{x-1}$ | A1 | OE e.g. $1+\frac{4}{x-1}$ etc. Must be a function of $x$, cannot be $x=\,$ |
| **Total** | **3** | |3 The function f is defined as follows:
$$\mathrm { f } ( x ) = \frac { x + 3 } { x - 1 } \text { for } x > 1$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\mathrm { ff } ( 5 )$.
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q3 [5]}}