| 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 | Standard +0.3 This is a multi-part question involving composite functions and finding an inverse, but all steps are routine. Part (a) requires substitution and solving simultaneous equations. Parts (b) and (c) involve standard inverse function techniques (finding range of f for the domain of f^{-1}, then algebraically rearranging). The rational function form is standard for P1 level, requiring no novel insight—just careful algebraic manipulation. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 + \dfrac{2a}{7-a} = \dfrac{5}{2} \Rightarrow \dfrac{2a}{7-a} = \dfrac{3}{2} \Rightarrow 7a = 21 \Rightarrow a = \ldots\) OR \(1 + \dfrac{2a}{7-a} = \dfrac{5}{2} \Rightarrow (7-a)+2a = \dfrac{5}{2}(7-a) \Rightarrow 7a=21 \Rightarrow a=\ldots\) | M1 | OE Substitute \(x=7\) then solve for \(a\) via legitimate mathematical steps. Condone sign errors only |
| \(a = 3\) | A1 | If M0, SC B1 for \(a=3\) with no working |
| \(f(5) = 1 + \dfrac{2(\textit{their}\,3)}{5-\textit{their}\,3} = 4 \Rightarrow 4b-2=4 \Rightarrow b=\ldots\) OR \(\text{gf}(5) = b\left(1+\dfrac{2(\textit{their}\,3)}{5-\textit{their}\,3}\right)-2 \Rightarrow 4b-2=4 \Rightarrow b=\ldots\) | M1 | Evaluate \(f(5)\), either separately or within \(\text{gf}\) then solve for \(b\) via legitimate mathematical steps. Condone sign errors only. FT *their* \(a\) value |
| \(b = \dfrac{3}{2}\) | A1 | OE e.g. \(\dfrac{6}{4}\), \(1.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x > 1\) | B1 | Accept \((1,\infty)\) or \(\{*: *>1\}\) where \(*\) is any variable. B0 for \(f^{-1}(x)>1\) or \(f(x)>1\) or \(y>1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER \(x - 1 = \dfrac{6}{y-3} \Rightarrow (y-3)(x-1)=6\) OR \(x = 1 + \dfrac{6}{y-3} \Rightarrow x(y-3)=(y-3)+6\) | \*M1 | OE \(y-1=\dfrac{6}{x-3} \Rightarrow (x-3)(y-1)=6\). OE \(y=1+\dfrac{6}{x-3} \Rightarrow y(x-3)=(x-3)+6\). Allow \*M1 for use of *their* 3 from (a) |
| \(y-3 = \dfrac{6}{x-1}\) or \(y(x-1)=3x+3\) | DM1 | OE \(x-3=\dfrac{6}{y-1}\) or \(x(y-1)=3y+3\). Allow DM1 for use of *their* 3 from (a) |
| \(\left[f^{-1}(x)\right] = 3 + \dfrac{6}{x-1}\) | A1 | OE Correct answer e.g. \(\dfrac{3x+3}{x-1}\) ISW. Must be in terms of \(x\). \*M1 DM1 possible for '\(a\)' used, but A0 so max 2/3 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 + \dfrac{2a}{7-a} = \dfrac{5}{2} \Rightarrow \dfrac{2a}{7-a} = \dfrac{3}{2} \Rightarrow 7a = 21 \Rightarrow a = \ldots$ OR $1 + \dfrac{2a}{7-a} = \dfrac{5}{2} \Rightarrow (7-a)+2a = \dfrac{5}{2}(7-a) \Rightarrow 7a=21 \Rightarrow a=\ldots$ | M1 | OE Substitute $x=7$ then solve for $a$ via legitimate mathematical steps. Condone sign errors only |
| $a = 3$ | A1 | If M0, SC B1 for $a=3$ with no working |
| $f(5) = 1 + \dfrac{2(\textit{their}\,3)}{5-\textit{their}\,3} = 4 \Rightarrow 4b-2=4 \Rightarrow b=\ldots$ OR $\text{gf}(5) = b\left(1+\dfrac{2(\textit{their}\,3)}{5-\textit{their}\,3}\right)-2 \Rightarrow 4b-2=4 \Rightarrow b=\ldots$ | M1 | Evaluate $f(5)$, either separately or within $\text{gf}$ then solve for $b$ via legitimate mathematical steps. Condone sign errors only. FT *their* $a$ value |
| $b = \dfrac{3}{2}$ | A1 | OE e.g. $\dfrac{6}{4}$, $1.5$ |
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x > 1$ | B1 | Accept $(1,\infty)$ or $\{*: *>1\}$ where $*$ is any variable. B0 for $f^{-1}(x)>1$ or $f(x)>1$ or $y>1$ |
---
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| EITHER $x - 1 = \dfrac{6}{y-3} \Rightarrow (y-3)(x-1)=6$ OR $x = 1 + \dfrac{6}{y-3} \Rightarrow x(y-3)=(y-3)+6$ | \*M1 | OE $y-1=\dfrac{6}{x-3} \Rightarrow (x-3)(y-1)=6$. OE $y=1+\dfrac{6}{x-3} \Rightarrow y(x-3)=(x-3)+6$. Allow \*M1 for use of *their* 3 from (a) |
| $y-3 = \dfrac{6}{x-1}$ or $y(x-1)=3x+3$ | DM1 | OE $x-3=\dfrac{6}{y-1}$ or $x(y-1)=3y+3$. Allow DM1 for use of *their* 3 from (a) |
| $\left[f^{-1}(x)\right] = 3 + \dfrac{6}{x-1}$ | A1 | OE Correct answer e.g. $\dfrac{3x+3}{x-1}$ ISW. Must be in terms of $x$. \*M1 DM1 possible for '$a$' used, but A0 so max 2/3 |
---8 The functions f and g are defined as follows, where $a$ and $b$ are constants.
$$\begin{aligned}
& \mathrm { f } ( x ) = 1 + \frac { 2 a } { x - a } \text { for } x > a \\
& \mathrm {~g} ( x ) = b x - 2 \text { for } x \in \mathbb { R }
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { f } ( 7 ) = \frac { 5 } { 2 }$ and $\mathrm { gf } ( 5 ) = 4$, find the values of $a$ and $b$.\\
For the rest of this question, you should use the value of $a$ which you found in (a).
\item Find the domain of $\mathrm { f } ^ { - 1 }$.
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q8 [8]}}