| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find range of composite |
| Difficulty | Standard +0.3 This is a multi-part question on composite and inverse functions with standard techniques. Parts (a)-(c) involve routine procedures: solving f(x)=3/2, algebraic manipulation to partial fractions form, and stating the domain of g as the range of g^(-1). Part (d) requires composing functions and finding the range, which needs careful tracking of domains/ranges but uses standard A-level methods. Slightly above average due to the composition in part (d), but all techniques are well-practiced. |
| 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 |
| Attempts to solve \(\frac{3}{2} = \frac{8x+5}{2x+3} \Rightarrow x = \ldots\) or substitutes \(x = \frac{3}{2}\) into \(\frac{5-3x}{2x-8}\) | M1 | Condone poor algebra as long as they reach a value for \(x\). Note attempts to find e.g. \(f'(x)\) or \(\frac{1}{f(x)}\) which may be implied by values such as \(\frac{6}{17}, \frac{17}{6}, \frac{7}{18}, \frac{18}{7}\) score M0 |
| \(f^{-1}\!\left(\frac{3}{2}\right) = -\frac{1}{10}\) | A1 | Do not be concerned what they call it, just look for the value. \(x = -\frac{1}{10}\) or just \(-\frac{1}{10}\) is fine. Correct answer with no (or minimal) working scores both marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{8x+5}{2x+3} = 4 \pm \frac{\ldots}{2x+3}\) | M1 | Look for \(4 \pm \frac{\ldots}{2x+3}\) where … is a constant, or \(8x+5 = A(2x+3)+B\) with \(A\) or \(B\) correct (which may be in a fraction), or a long division attempt obtaining a quotient of 4, or attempts to express the numerator in terms of the denominator e.g. \(\frac{8x+5}{2x+3} = \frac{4(2x+3)+\ldots}{2x+3}\) |
| \(\frac{8x+5}{2x+3} = 4 - \frac{7}{2x+3}\) | A1 | Also allow correct values e.g. \(A=4, B=-7\). Do not isw here e.g. if they obtain \(A=4, B=-7\) and then write \(-7+\frac{4}{2x+3}\) score A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 \leqslant g^{-1}(x) \leqslant 4\) | B1 | E.g. \(0 \leqslant y \leqslant 4\), \(0 \leqslant \text{range} \leqslant 4\), \(g^{-1}(x) \leqslant 4\) and \(g^{-1}(x) \geqslant 0\), \(0 \leqslant g^{-1} \leqslant 4\), \([0,4]\) but not e.g. \(0 \leqslant x \leqslant 4\), \(0 \leqslant g(x) \leqslant 4\), \((0,4)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts either boundary: \(f(0) = \frac{8\times0+5}{2\times0+3}\) or \(f(4) = \frac{8\times4+5}{2\times4+3}\) | M1 | Or uses (b) e.g. \(f(0) = 4 - \frac{7}{2\times0+3}\) or \(f(4) = 4 - \frac{7}{2\times4+3}\) |
| Attempts both boundaries: \(f(0) = \frac{8\times0+5}{2\times0+3}\) and \(f(4) = \frac{8\times4+5}{2\times4+3}\) | dM1 | Or uses (b) e.g. \(f(0) = 4 - \frac{7}{2\times0+3}\) and \(f(4) = 4 - \frac{7}{2\times4+3}\) |
| Range: \(\frac{5}{3} \leqslant \text{fg}^{-1}(x) \leqslant \frac{37}{11}\) | A1 | Correct answer written in the correct form. E.g. \(\frac{5}{3} \leqslant \text{fg}^{-1}(x) \leqslant \frac{37}{11}\), \(\frac{5}{3} \leqslant \text{range} \leqslant \frac{37}{11}\), \(\frac{5}{3} \leqslant y \leqslant \frac{37}{11}\), \(\left[\frac{5}{3}, \frac{37}{11}\right]\) but not e.g. \(\frac{5}{3} \leqslant x \leqslant \frac{37}{11}\). Note that \(\frac{37}{11}\) is sometimes obtained fortuitously from incorrect working so check working carefully. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(g^{-1}(x) = \sqrt{16-x} \Rightarrow \text{fg}^{-1}(x) = \frac{8\sqrt{16-x}+5}{2\sqrt{16-x}+3}\); attempts either boundary using \(x=0\) or \(x=16\) | M1 | Attempt to substitute \(\sqrt{16-x}\) (condone \(\pm\sqrt{16-x}\)) into \(f\) |
| Attempts both boundaries | dM1 | |
| Range \(\frac{5}{3} \leqslant \text{fg}^{-1}(x) \leqslant \frac{37}{11}\) | A1 | Correct answer in correct form with exact values |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to solve $\frac{3}{2} = \frac{8x+5}{2x+3} \Rightarrow x = \ldots$ **or** substitutes $x = \frac{3}{2}$ into $\frac{5-3x}{2x-8}$ | M1 | Condone poor algebra as long as they reach a value for $x$. Note attempts to find e.g. $f'(x)$ or $\frac{1}{f(x)}$ which may be implied by values such as $\frac{6}{17}, \frac{17}{6}, \frac{7}{18}, \frac{18}{7}$ score M0 |
| $f^{-1}\!\left(\frac{3}{2}\right) = -\frac{1}{10}$ | A1 | Do not be concerned what they call it, just look for the value. $x = -\frac{1}{10}$ or just $-\frac{1}{10}$ is fine. Correct answer with no (or minimal) working scores both marks. |
**(2 marks)**
---
## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{8x+5}{2x+3} = 4 \pm \frac{\ldots}{2x+3}$ | M1 | Look for $4 \pm \frac{\ldots}{2x+3}$ where … is a constant, or $8x+5 = A(2x+3)+B$ with $A$ or $B$ correct (which may be in a fraction), or a long division attempt obtaining a quotient of 4, or attempts to express the numerator in terms of the denominator e.g. $\frac{8x+5}{2x+3} = \frac{4(2x+3)+\ldots}{2x+3}$ |
| $\frac{8x+5}{2x+3} = 4 - \frac{7}{2x+3}$ | A1 | Also allow correct values e.g. $A=4, B=-7$. Do not isw here e.g. if they obtain $A=4, B=-7$ and then write $-7+\frac{4}{2x+3}$ score A0 |
**(2 marks)**
---
## Question 10(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 \leqslant g^{-1}(x) \leqslant 4$ | B1 | E.g. $0 \leqslant y \leqslant 4$, $0 \leqslant \text{range} \leqslant 4$, $g^{-1}(x) \leqslant 4$ and $g^{-1}(x) \geqslant 0$, $0 \leqslant g^{-1} \leqslant 4$, $[0,4]$ but not e.g. $0 \leqslant x \leqslant 4$, $0 \leqslant g(x) \leqslant 4$, $(0,4)$ |
**(1 mark)**
---
## Question 10(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts either boundary: $f(0) = \frac{8\times0+5}{2\times0+3}$ **or** $f(4) = \frac{8\times4+5}{2\times4+3}$ | M1 | Or uses (b) e.g. $f(0) = 4 - \frac{7}{2\times0+3}$ **or** $f(4) = 4 - \frac{7}{2\times4+3}$ |
| Attempts both boundaries: $f(0) = \frac{8\times0+5}{2\times0+3}$ **and** $f(4) = \frac{8\times4+5}{2\times4+3}$ | dM1 | Or uses (b) e.g. $f(0) = 4 - \frac{7}{2\times0+3}$ **and** $f(4) = 4 - \frac{7}{2\times4+3}$ |
| Range: $\frac{5}{3} \leqslant \text{fg}^{-1}(x) \leqslant \frac{37}{11}$ | A1 | Correct answer written in the correct form. E.g. $\frac{5}{3} \leqslant \text{fg}^{-1}(x) \leqslant \frac{37}{11}$, $\frac{5}{3} \leqslant \text{range} \leqslant \frac{37}{11}$, $\frac{5}{3} \leqslant y \leqslant \frac{37}{11}$, $\left[\frac{5}{3}, \frac{37}{11}\right]$ but not e.g. $\frac{5}{3} \leqslant x \leqslant \frac{37}{11}$. Note that $\frac{37}{11}$ is sometimes obtained fortuitously from incorrect working so check working carefully. |
**Alternative via $\text{fg}^{-1}(x)$:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g^{-1}(x) = \sqrt{16-x} \Rightarrow \text{fg}^{-1}(x) = \frac{8\sqrt{16-x}+5}{2\sqrt{16-x}+3}$; attempts either boundary using $x=0$ or $x=16$ | M1 | Attempt to substitute $\sqrt{16-x}$ (condone $\pm\sqrt{16-x}$) into $f$ |
| Attempts both boundaries | dM1 | |
| Range $\frac{5}{3} \leqslant \text{fg}^{-1}(x) \leqslant \frac{37}{11}$ | A1 | Correct answer in correct form with exact values |
**(3 marks) — Total: 8 marks**
---\begin{enumerate}
\item The function f is defined by
\end{enumerate}
$$f ( x ) = \frac { 8 x + 5 } { 2 x + 3 } \quad x > - \frac { 3 } { 2 }$$
(a) Find $\mathrm { f } ^ { - 1 } \left( \frac { 3 } { 2 } \right)$\\
(b) Show that
$$\mathrm { f } ( x ) = A + \frac { B } { 2 x + 3 }$$
where $A$ and $B$ are constants to be found.
The function $g$ is defined by
$$g ( x ) = 16 - x ^ { 2 } \quad 0 \leqslant x \leqslant 4$$
(c) State the range of $\mathrm { g } ^ { - 1 }$\\
(d) Find the range of $\mathrm { fg } ^ { - 1 }$
\hfill \mbox{\textit{Edexcel Paper 2 2022 Q10 [8]}}