| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Evaluate composite at point |
| Difficulty | Moderate -0.5 This is a straightforward composite and inverse functions question with routine algebraic manipulation. Part (a) requires simple substitution, part (b) is standard inverse function technique, and part (c) involves solving a composite equation with basic algebra. All parts are textbook exercises requiring no novel insight, 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/Working | Mark | Guidance |
| \(ff(6) = f\!\left(\frac{9}{2}\right) = \frac{\frac{9}{2}+3}{\frac{9}{2}-4} = 15\) | M1; A1 | Substitutes \(x=6\) into f and substitutes result back into f |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f^{-1}(x) = \dfrac{4x+3}{x-1}\) | M1A1 | Changes subject of \(y = \dfrac{x+3}{x-4}\) to form \(y = \dfrac{...x\pm3}{...x\pm1}\) |
| \(x \in \mathbb{R},\ x \neq 1\) | B1 | Omission of \(x \in \mathbb{R}\) condoned. Accept alternative notations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. \(\left(\dfrac{x+3}{x-4}\right)^2 + 5 = 7\) or \(\dfrac{a+3}{a-4} = (\pm)\sqrt{7-5}\) | M1 | Correct attempt to set up equation in \(x\) or \(a\) |
| \(\Rightarrow x^2 - 22x + 23 = 0\) or \((a+3) = (\pm)\sqrt{2}(a-4) \Rightarrow a = ...\) | dM1 | Proceeds to solve for \(x\) or \(a\); forms 3TQ and solves, or cross multiplies and makes \(a\) subject |
| \(a = 11 + 7\sqrt{2}\) oe | A1 | Accept \(\dfrac{4\sqrt{2}+3}{\sqrt{2}-1}\) or \(\dfrac{-4\sqrt{2}-3}{1-\sqrt{2}}\). Ignore if \(11-7\sqrt{2}\) also given, but A0 if other extra solutions included |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $ff(6) = f\!\left(\frac{9}{2}\right) = \frac{\frac{9}{2}+3}{\frac{9}{2}-4} = 15$ | M1; A1 | Substitutes $x=6$ into f and substitutes result back into f |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f^{-1}(x) = \dfrac{4x+3}{x-1}$ | M1A1 | Changes subject of $y = \dfrac{x+3}{x-4}$ to form $y = \dfrac{...x\pm3}{...x\pm1}$ |
| $x \in \mathbb{R},\ x \neq 1$ | B1 | Omission of $x \in \mathbb{R}$ condoned. Accept alternative notations |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $\left(\dfrac{x+3}{x-4}\right)^2 + 5 = 7$ or $\dfrac{a+3}{a-4} = (\pm)\sqrt{7-5}$ | M1 | Correct attempt to set up equation in $x$ or $a$ |
| $\Rightarrow x^2 - 22x + 23 = 0$ or $(a+3) = (\pm)\sqrt{2}(a-4) \Rightarrow a = ...$ | dM1 | Proceeds to solve for $x$ or $a$; forms 3TQ and solves, or cross multiplies and makes $a$ subject |
| $a = 11 + 7\sqrt{2}$ oe | A1 | Accept $\dfrac{4\sqrt{2}+3}{\sqrt{2}-1}$ or $\dfrac{-4\sqrt{2}-3}{1-\sqrt{2}}$. Ignore if $11-7\sqrt{2}$ also given, but A0 if other extra solutions included |
---\begin{enumerate}
\item The function f is defined by
\end{enumerate}
$$\mathrm { f } ( x ) = \frac { x + 3 } { x - 4 } \quad x \in \mathbb { R } , x \neq 4$$
(a) Find ff(6)\\
(b) Find $f ^ { - 1 }$
The function $g$ is defined by
$$g ( x ) = x ^ { 2 } + 5 \quad x \in \mathbb { R } , x > 0$$
(c) Find the exact value of $a$ for which
$$\operatorname { gf } ( a ) = 7$$
\hfill \mbox{\textit{Edexcel P3 2023 Q2 [8]}}