Question 10 - A-Level Maths

CAIE P1 2022 June — Question 10 13 marks

Exam Board	CAIE
Module	P1 (Pure Mathematics 1)
Year	2022
Session	June
Marks	13
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 covering standard inverse function techniques. Parts (a)-(d) involve routine procedures: reading domain from a graph, finding inverse of a rational function using standard algebraic manipulation, function composition, and explaining why a parabola isn't one-to-one. Part (e) requires algebraic manipulation and finding a tangent equation, but follows standard methods. While comprehensive, each component is a textbook exercise requiring no novel insight, making it slightly easier than average.
Spec	1.02k Simplify rational expressions: factorising, cancelling, algebraic division 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence 1.07m Tangents and normals: gradient and equations 1.08e Area between curve and x-axis: using definite integrals

10 Functions $f$ and $g$ are defined as follows: $$\begin{array} { l l } \mathrm { f } ( x ) = \frac { 2 x + 1 } { 2 x - 1 } & \text { for } x \neq \frac { 1 } { 2 } \\ \mathrm {~g} ( x ) = x ^ { 2 } + 4 & \text { for } x \in \mathbb { R } \end{array}$$

\includegraphics[max width=\textwidth, alt={}, center]{bb7595c9-93ae-49e8-9cc5-9ecc802e6060-16_773_1182_555_511} The diagram shows part of the graph of $y = \mathrm { f } ( x )$.
State the domain of $\mathrm { f } ^ { - 1 }$.
Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
Find $\mathrm { gf } ^ { - 1 } ( 3 )$.
Explain why $\mathrm { g } ^ { - 1 } ( x )$ cannot be found.
Show that $1 + \frac { 2 } { 2 x - 1 }$ can be expressed as $\frac { 2 x + 1 } { 2 x - 1 }$. Hence find the area of the triangle enclosed by the tangent to the curve $y = \mathrm { f } ( x )$ at the point where $x = 1$ and the $x$ - and $y$-axes.

Show mark scheme Show mark scheme source

Question 10(a):

Answer	Marks	Guidance
Answer	Marks	Guidance
$x \neq 1$ or $x < 1,\ x > 1$ or $(-\infty, 1),(1,\infty)$ $[x \in \mathbb{R}]$	B1	Must be $x$ not $f^{-1}(x)$ or $y$. Do not accept $1 < x < 1$.

Question 10(b):

Answer	Marks	Guidance
Answer	Marks	Guidance
$y = \frac{2x+1}{2x-1}$ leading to $(2x-1)y = 2x+1$ leading to $2xy - y = 2x+1$	*M1	Setting $y =\ $, removing fraction and expanding brackets
$2xy - 2x = y + 1$ leading to $2x(y-1) = y+1$ leading to $x = \frac{y+1}{2(y-1)}$	DM1	Reorganising to get $x =$. Condone $\pm$ sign errors only
$[f^{-1}(x)] = \frac{x+1}{2(x-1)},\ \frac{x+1}{x-1} \times \frac{1}{2}$ or $\frac{1}{x-1} + \frac{1}{2}$	A1	OE. Must be in terms of $x$. Do not allow $\frac{x+1}{x-1} \div 2$

Question 10(c):

Answer	Marks	Guidance
Answer	Marks	Guidance
$\left(\text{their } f^{-1}(3)\right)$ leading to $\left(\text{their } f^{-1}(3)\right)^2 + 4\ \ [f^{-1}(3) = 1, 1+4 =]$	M1	Correct order of operations and substitution of $x = 3$ needed
$5$	A1

Question 10(d):

Answer	Marks	Guidance
Answer	Marks	Guidance
Sight of 'not one to one' or 'many to one' or 'one to many'	B1	Any reason mentioning 2 values, or $+$ and $-$, such as: square root gives 2 values or horizontal line test crosses curve twice or 2 values because of turning point or 2 values because it is a quadratic

Question 10(e):

Answer	Marks	Guidance
Answer	Marks	Guidance
$f(x) = 1 + \frac{2}{2x-1} = \frac{2x-1}{2x-1} + \frac{2}{2x-1} = \frac{2x+1}{2x-1}$	B1	AG. Do not condone equating expressions and verification
$f'(x) = -4(2x-1)^{-2}$ or $2(2x-1)^{-1} + \{-(2x+1)2(2x-1)^{-2}\}$ or $\frac{(2x-1)2 - 2(2x+1)}{(2x-1)^2}$	*M1	For $k(2x-1)^{-2}$ and no other terms or correct use of product or quotient rule then ISW
Gradient $m = -4$	A1	Differentiation must have clearly taken place
Equation of tangent is $y - 3 = -4(x-1)\ [\Rightarrow y = -4x+7]$	DM1	Using $(1,3)$ in the equation of a line with their gradient
Crosses axes at $\left(\frac{7}{4}, 0\right)$ and $(0, 7)$	A1 FT	SOI from their straight line or by integration from 0 to 'their $7/4$'
[Area $=$] $\frac{49}{8}$	A1	OE e.g. 6.13 AWRT. If M0 A0 DM0, SC B2 available for correct answer

## Question 10(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x \neq 1$ or $x < 1,\ x > 1$ or $(-\infty, 1),(1,\infty)$ $[x \in \mathbb{R}]$ | B1 | Must be $x$ not $f^{-1}(x)$ or $y$. Do not accept $1 < x < 1$. |

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## Question 10(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{2x+1}{2x-1}$ leading to $(2x-1)y = 2x+1$ leading to $2xy - y = 2x+1$ | *M1 | Setting $y =\ $, removing fraction and expanding brackets |
| $2xy - 2x = y + 1$ leading to $2x(y-1) = y+1$ leading to $x = \frac{y+1}{2(y-1)}$ | DM1 | Reorganising to get $x =$. Condone $\pm$ sign errors only |
| $[f^{-1}(x)] = \frac{x+1}{2(x-1)},\ \frac{x+1}{x-1} \times \frac{1}{2}$ or $\frac{1}{x-1} + \frac{1}{2}$ | A1 | OE. Must be in terms of $x$. Do not allow $\frac{x+1}{x-1} \div 2$ |

---

## Question 10(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\text{their } f^{-1}(3)\right)$ leading to $\left(\text{their } f^{-1}(3)\right)^2 + 4\ \ [f^{-1}(3) = 1, 1+4 =]$ | M1 | Correct order of operations and substitution of $x = 3$ needed |
| $5$ | A1 | |

---

## Question 10(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Sight of 'not one to one' or 'many to one' or 'one to many' | B1 | Any reason mentioning 2 values, or $+$ and $-$, such as: square root gives 2 values or horizontal line test crosses curve twice or 2 values because of turning point or 2 values because it is a quadratic |

---

## Question 10(e):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = 1 + \frac{2}{2x-1} = \frac{2x-1}{2x-1} + \frac{2}{2x-1} = \frac{2x+1}{2x-1}$ | B1 | AG. Do not condone equating expressions and verification |
| $f'(x) = -4(2x-1)^{-2}$ or $2(2x-1)^{-1} + \{-(2x+1)2(2x-1)^{-2}\}$ or $\frac{(2x-1)2 - 2(2x+1)}{(2x-1)^2}$ | *M1 | For $k(2x-1)^{-2}$ and no other terms or correct use of product or quotient rule then ISW |
| Gradient $m = -4$ | A1 | Differentiation must have clearly taken place |
| Equation of tangent is $y - 3 = -4(x-1)\ [\Rightarrow y = -4x+7]$ | DM1 | Using $(1,3)$ in the equation of a line with *their* gradient |
| Crosses axes at $\left(\frac{7}{4}, 0\right)$ and $(0, 7)$ | A1 FT | SOI from *their* straight line or by integration from 0 to '*their* $7/4$' |
| [Area $=$] $\frac{49}{8}$ | A1 | OE e.g. 6.13 AWRT. If M0 A0 DM0, **SC B2** available for correct answer |

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Show LaTeX source

10 Functions $f$ and $g$ are defined as follows:

$$\begin{array} { l l } 
\mathrm { f } ( x ) = \frac { 2 x + 1 } { 2 x - 1 } & \text { for } x \neq \frac { 1 } { 2 } \\
\mathrm {~g} ( x ) = x ^ { 2 } + 4 & \text { for } x \in \mathbb { R }
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{bb7595c9-93ae-49e8-9cc5-9ecc802e6060-16_773_1182_555_511}

The diagram shows part of the graph of $y = \mathrm { f } ( x )$.\\
State the domain of $\mathrm { f } ^ { - 1 }$.
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
\item Find $\mathrm { gf } ^ { - 1 } ( 3 )$.
\item Explain why $\mathrm { g } ^ { - 1 } ( x )$ cannot be found.
\item Show that $1 + \frac { 2 } { 2 x - 1 }$ can be expressed as $\frac { 2 x + 1 } { 2 x - 1 }$. Hence find the area of the triangle enclosed by the tangent to the curve $y = \mathrm { f } ( x )$ at the point where $x = 1$ and the $x$ - and $y$-axes.
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2022 Q10 [13]}}

This paper (11 questions)

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Q1 4 Q2 4 Q3 4 Q4 5 Q5 9 Q6 5 Q7 6 Q8 8 Q9 7 Q10 13 Q11 10

Answer	Marks	Guidance
Answer	Marks	Guidance
\(x \neq 1\) or \(x < 1,\ x > 1\) or \((-\infty, 1),(1,\infty)\) \([x \in \mathbb{R}]\)	B1	Must be \(x\) not \(f^{-1}(x)\) or \(y\). Do not accept \(1 < x < 1\).