Question 10 - A-Level Maths

Edexcel C34 2018 January — Question 10 12 marks

Exam Board	Edexcel
Module	C34 (Core Mathematics 3 & 4)
Year	2018
Session	January
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Composite & Inverse Functions
Type	Find inverse function
Difficulty	Standard +0.2 This is a straightforward multi-part question testing standard C3/C4 techniques: sketching an exponential function, finding range of a rational function, finding an inverse (using standard algebraic manipulation), and solving a composite function equation. All parts are routine textbook exercises requiring no novel insight, making it easier than average.
Spec	1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence 1.06a Exponential function: a^x and e^x graphs and properties

It is given that

$$\begin{gathered} \mathrm { f } ( x ) = \mathrm { e } ^ { - 2 x } \quad x \in \mathbb { R } \\ \mathrm {~g} ( x ) = \frac { x } { x - 3 } \quad x > 3 \end{gathered}$$

Sketch the graph of $y = \mathrm { f } ( x )$, showing the coordinates of any points where the graph crosses the axes.
Find the range of g
Find $\mathrm { g } ^ { - 1 } ( x )$, stating the domain of $\mathrm { g } ^ { - 1 }$
Using algebra, find the exact value of $x$ for which $\operatorname { fg } ( x ) = 3$

Show mark scheme Show mark scheme source

Question 10(a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Correct shape or intercept at 1	M1	Curve passing through 1 on +ve $y$-axis, gradient always negative and increasing
Fully correct graph	A1	Correct shape with $y$-intercept at 1, asymptotic to $x$-axis

Question 10(b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$g(x) > 1$	M1 A1

Question 10(c):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$y = \frac{x}{x-3} \Rightarrow (x-3)y = x \Rightarrow xy - x = 3y$	M1	Attempt to rearrange
$\Rightarrow x = \frac{3y}{y-1}$	dM1	Collects $x$ terms and factorises
$g^{-1}(x) = \frac{3x}{x-1}$, with $x > 1$	A1, B1ft	A1 for expression; B1ft for domain

Question 10(d):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Way 1: $e^{-2\left(\frac{x}{x-3}\right)} = 3$	M1	Sets up correct equation
$-2\left(\frac{x}{x-3}\right) = \ln 3$	A1
$-2x = \ln 3(x-3)$, solves for $x$	dM1	Rearranges and solves
$x = \frac{3\ln 3}{2 + \ln 3}$	A1	(outside range of $g(x)$, so no solutions)
Way 2: $g(x) = f^{-1}(3)$; $g(x) = -\frac{1}{2}\ln 3$; $x = g^{-1}\!\left(-\frac{1}{2}\ln 3\right)$	M1, A1, dM1, A1

Question (from first page - appears to be part of a functions question):

Part (b):

Answer	Marks	Guidance
$g(x) > 1$, $g > 1$, $y > 1$, $(1,\infty)$	M1	Finds value 1 but incorrect inequality possible
$g(x) > 1$, $g > 1$, $y > 1$, $(1,\infty)$ or $(1;\infty)$	A1	Needs $g(x)>1$, $g>1$, $y>1$; do not accept $f(x)>1$

Part (c):

Answer	Marks	Guidance
Setting $y =$ multiplying across, collecting $x$ terms; award for $\pm xy \pm x = \pm 3y$	M1	Condone numerical slips; alternatively starting with $x = \frac{y}{y-3}$
Attempt making $x$ or replaced $y$ the subject; look for $x = \frac{\pm 3y}{\pm y \pm 1}$	dM1	Dependent on previous M mark; condone numerical slips
$g^{-1}(x) = \frac{3x}{x-1}$ or exact equivalent e.g. $g^{-1}(x) = -\frac{3x}{1-x}$ or $g^{-1}(x) = 3+\frac{3}{x-1}$ or $g^{-1}(x) = 3-\frac{3}{1-x}$	A1	Do not allow $y = \frac{3x}{x-1}$ or $f^{-1}(x) = \frac{3x}{x-1}$
Domain $x > 1$ or ft range from (b); set form $(1,\infty)$	B1ft	Condone $(1;\infty)$; don't follow through on $y \in \mathbb{R}$ following $x \in \mathbb{R}$

Part (d):

Answer	Marks	Guidance
Setting $fg(x) = 3$ (correct order) or using $g(x) = f^{-1}(3)$	M1	Condone slips but order of operations must be correct
Undoes exponentials to reach correct equation in $x$: $-2\left(\frac{x}{x-3}\right) = \ln 3$ or $\frac{x}{x-3} = -\frac{1}{2}\ln 3$ or $g(x) = -\frac{1}{2}\ln 3$	A1
Full attempt to make $x$ the subject from two $x$ terms; or attempts $g^{-1}f^{-1}(3)$	dM1	Apply same rules for change of subject as M1 dM1 in (c)
$x = \frac{3\ln 3}{2 + \ln 3}$ or exact equivalent; $\ln 3$ may appear as $-\ln(1/3)$ or $-\frac{1}{2}(\ln 9)$	A1	Condone lack of final conclusion

# Question 10(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct shape or intercept at 1 | M1 | Curve passing through 1 on +ve $y$-axis, gradient always negative and increasing |
| Fully correct graph | A1 | Correct shape with $y$-intercept at 1, asymptotic to $x$-axis |

---

# Question 10(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $g(x) > 1$ | M1 A1 | |

---

# Question 10(c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{x}{x-3} \Rightarrow (x-3)y = x \Rightarrow xy - x = 3y$ | M1 | Attempt to rearrange |
| $\Rightarrow x = \frac{3y}{y-1}$ | dM1 | Collects $x$ terms and factorises |
| $g^{-1}(x) = \frac{3x}{x-1}$, with $x > 1$ | A1, B1ft | A1 for expression; B1ft for domain |

---

# Question 10(d):

| Answer/Working | Marks | Guidance |
|---|---|---|
| **Way 1:** $e^{-2\left(\frac{x}{x-3}\right)} = 3$ | M1 | Sets up correct equation |
| $-2\left(\frac{x}{x-3}\right) = \ln 3$ | A1 | |
| $-2x = \ln 3(x-3)$, solves for $x$ | dM1 | Rearranges and solves |
| $x = \frac{3\ln 3}{2 + \ln 3}$ | A1 | (outside range of $g(x)$, so no solutions) |
| **Way 2:** $g(x) = f^{-1}(3)$; $g(x) = -\frac{1}{2}\ln 3$; $x = g^{-1}\!\left(-\frac{1}{2}\ln 3\right)$ | M1, A1, dM1, A1 | |

# Question (from first page - appears to be part of a functions question):

## Part (b):
| $g(x) > 1$, $g > 1$, $y > 1$, $(1,\infty)$ | M1 | Finds value 1 but incorrect inequality possible |
| $g(x) > 1$, $g > 1$, $y > 1$, $(1,\infty)$ or $(1;\infty)$ | A1 | Needs $g(x)>1$, $g>1$, $y>1$; do not accept $f(x)>1$ |

## Part (c):
| Setting $y =$ multiplying across, collecting $x$ terms; award for $\pm xy \pm x = \pm 3y$ | M1 | Condone numerical slips; alternatively starting with $x = \frac{y}{y-3}$ |
| Attempt making $x$ or replaced $y$ the subject; look for $x = \frac{\pm 3y}{\pm y \pm 1}$ | dM1 | Dependent on previous M mark; condone numerical slips |
| $g^{-1}(x) = \frac{3x}{x-1}$ or exact equivalent e.g. $g^{-1}(x) = -\frac{3x}{1-x}$ or $g^{-1}(x) = 3+\frac{3}{x-1}$ or $g^{-1}(x) = 3-\frac{3}{1-x}$ | A1 | Do not allow $y = \frac{3x}{x-1}$ or $f^{-1}(x) = \frac{3x}{x-1}$ |
| Domain $x > 1$ or ft range from (b); set form $(1,\infty)$ | B1ft | Condone $(1;\infty)$; don't follow through on $y \in \mathbb{R}$ following $x \in \mathbb{R}$ |

## Part (d):
| Setting $fg(x) = 3$ (correct order) or using $g(x) = f^{-1}(3)$ | M1 | Condone slips but order of operations must be correct |
| Undoes exponentials to reach correct equation in $x$: $-2\left(\frac{x}{x-3}\right) = \ln 3$ or $\frac{x}{x-3} = -\frac{1}{2}\ln 3$ or $g(x) = -\frac{1}{2}\ln 3$ | A1 | |
| Full attempt to make $x$ the subject from two $x$ terms; or attempts $g^{-1}f^{-1}(3)$ | dM1 | Apply same rules for change of subject as M1 dM1 in (c) |
| $x = \frac{3\ln 3}{2 + \ln 3}$ or exact equivalent; $\ln 3$ may appear as $-\ln(1/3)$ or $-\frac{1}{2}(\ln 9)$ | A1 | Condone lack of final conclusion |

---

Show LaTeX source

\begin{enumerate}
  \item It is given that
\end{enumerate}

$$\begin{gathered}
\mathrm { f } ( x ) = \mathrm { e } ^ { - 2 x } \quad x \in \mathbb { R } \\
\mathrm {~g} ( x ) = \frac { x } { x - 3 } \quad x > 3
\end{gathered}$$

(a) Sketch the graph of $y = \mathrm { f } ( x )$, showing the coordinates of any points where the graph crosses the axes.\\
(b) Find the range of g\\
(c) Find $\mathrm { g } ^ { - 1 } ( x )$, stating the domain of $\mathrm { g } ^ { - 1 }$\\
(d) Using algebra, find the exact value of $x$ for which $\operatorname { fg } ( x ) = 3$

\begin{center}

\end{center}

\hfill \mbox{\textit{Edexcel C34 2018 Q10 [12]}}

This paper (14 questions)

View full paper

Q1 6 Q2 7 Q3 7 Q4 6 Q5 7 Q6 5 Q7 13 Q8 4 Q9 9 Q10 12 Q11 12 Q12 12 Q13 13 Q14 12

Answer	Marks	Guidance
\(g(x) > 1\), \(g > 1\), \(y > 1\), \((1,\infty)\)	M1	Finds value 1 but incorrect inequality possible
\(g(x) > 1\), \(g > 1\), \(y > 1\), \((1,\infty)\) or \((1;\infty)\)	A1	Needs \(g(x)>1\), \(g>1\), \(y>1\); do not accept \(f(x)>1\)

Answer	Marks	Guidance
Setting \(y =\) multiplying across, collecting \(x\) terms; award for \(\pm xy \pm x = \pm 3y\)	M1	Condone numerical slips; alternatively starting with \(x = \frac{y}{y-3}\)
Attempt making \(x\) or replaced \(y\) the subject; look for \(x = \frac{\pm 3y}{\pm y \pm 1}\)	dM1	Dependent on previous M mark; condone numerical slips
\(g^{-1}(x) = \frac{3x}{x-1}\) or exact equivalent e.g. \(g^{-1}(x) = -\frac{3x}{1-x}\) or \(g^{-1}(x) = 3+\frac{3}{x-1}\) or \(g^{-1}(x) = 3-\frac{3}{1-x}\)	A1	Do not allow \(y = \frac{3x}{x-1}\) or \(f^{-1}(x) = \frac{3x}{x-1}\)
Domain \(x > 1\) or ft range from (b); set form \((1,\infty)\)	B1ft	Condone \((1;\infty)\); don't follow through on \(y \in \mathbb{R}\) following \(x \in \mathbb{R}\)

Answer	Marks	Guidance
Setting \(fg(x) = 3\) (correct order) or using \(g(x) = f^{-1}(3)\)	M1	Condone slips but order of operations must be correct
Undoes exponentials to reach correct equation in \(x\): \(-2\left(\frac{x}{x-3}\right) = \ln 3\) or \(\frac{x}{x-3} = -\frac{1}{2}\ln 3\) or \(g(x) = -\frac{1}{2}\ln 3\)	A1
Full attempt to make \(x\) the subject from two \(x\) terms; or attempts \(g^{-1}f^{-1}(3)\)	dM1	Apply same rules for change of subject as M1 dM1 in (c)
\(x = \frac{3\ln 3}{2 + \ln 3}\) or exact equivalent; \(\ln 3\) may appear as \(-\ln(1/3)\) or \(-\frac{1}{2}(\ln 9)\)	A1	Condone lack of final conclusion