Question 6 - A-Level Maths

Edexcel P3 2018 Specimen — Question 6 14 marks

Exam Board	Edexcel
Module	P3 (Pure Mathematics 3)
Year	2018
Session	Specimen
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Composite & Inverse Functions
Type	Find inverse function
Difficulty	Moderate -0.3 This is a standard P3/C3 composite and inverse functions question covering routine techniques: finding range, composition, solving exponential equations, finding inverse of e^x + 2, and sketching. All parts are textbook exercises requiring only direct application of learned methods with no novel problem-solving, 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 1.06a Exponential function: a^x and e^x graphs and properties 1.06d Natural logarithm: ln(x) function and properties

The functions f and g are defined by

$$\mathrm { f } : x \mapsto \mathrm { e } ^ { x } + 2 \quad x \in \mathbb { R }$$ $$\mathrm { g } : x \mapsto \ln x \quad x > 0$$

State the range of f .
Find $\mathrm { fg } ( x )$, giving $y$ our answer in its simplest form.
Find the exact value of $x$ for which $\mathrm { f } ( 2 x + 3 ) = 6$
Find $\mathrm { f } ^ { - 1 }$ stating its domain.
On the same axes sketch the curves with equation $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$, giving the coordinates of all the points where the curves cross the axes.

Show mark scheme Show mark scheme source

Question 6:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$f(x) > 2$	B1	Accept $y>2$, $(2,\infty)$, $f>2$; do not accept $x>2$

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$fg(x) = e^{\ln x} + 2 = x + 2$	M1 A1	Correct order of operations; $\ln e^x + 2$ is M0

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$e^{2x+3} + 2 = 6 \Rightarrow e^{2x+3} = 4$	M1 A1
$2x + 3 = \ln 4$	M1	Takes ln both sides
$x = \frac{\ln 4 - 3}{2}$ or $\ln 2 - \frac{3}{2}$	M1 A1

Part (d):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$y = e^x + 2 \Rightarrow y - 2 = e^x \Rightarrow \ln(y-2) = x$	M1
$f^{-1}(x) = \ln(x-2)$, $x > 2$	A1, B1ft

Part (e):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Correct shape for $f(x)$	B1
$(0, 3)$ marked	B1
Correct shape for $f^{-1}(x)$	B1
$(3, 0)$ marked	B1

Question 6:

Part (d)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x = \frac{\ln 4 - 3}{2}$ oe. eg $\ln 2 - \frac{3}{2}$	A1	Remember to isw any incorrect working after a correct answer

Part (d) - Inverse function

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Starts with $y = e^x + 2$ or $x = e^y + 2$ and attempts to change subject	M1	All ln work must be correct. The 2 must be dealt with first. Eg $y = e^x + 2 \Rightarrow \ln y = x + \ln 2 \Rightarrow x = \ln y - \ln 2$ is M0
$f^{-1}(x) = \ln(x-2)$ or $y = \ln(x-2)$ or \(y = \ln	x-2	\)
Either $x > 2$, or follow through on part (a) provided it wasn't $y \in \Re$	B1ft	Do not accept $y > 2$ or $f^{-1}(x) > 2$

Part (e)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Shape for $y = e^x$	B1	Graph only in quadrants 1 and 2. Start with gradient approx. 0 above x-axis in Q2, increasing gradient into Q1. No minimum point
$(0, 3)$ lies on the curve	B1	Accept 3 written on y-axis as long as point lies on curve
Shape for $y = \ln x$	B1	Graph only in quadrants 4 and 1. Gradient approx. infinite to right of y-axis in Q4, decreasing in Q1. No maximum point. Must not intersect $y = e^x$
$(3, 0)$ lies on the curve	B1	Accept 3 written on x-axis as long as point lies on curve

## Question 6:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) > 2$ | B1 | Accept $y>2$, $(2,\infty)$, $f>2$; do **not** accept $x>2$ |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $fg(x) = e^{\ln x} + 2 = x + 2$ | M1 A1 | Correct order of operations; $\ln e^x + 2$ is M0 |

### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{2x+3} + 2 = 6 \Rightarrow e^{2x+3} = 4$ | M1 A1 | |
| $2x + 3 = \ln 4$ | M1 | Takes ln both sides |
| $x = \frac{\ln 4 - 3}{2}$ or $\ln 2 - \frac{3}{2}$ | M1 A1 | |

### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = e^x + 2 \Rightarrow y - 2 = e^x \Rightarrow \ln(y-2) = x$ | M1 | |
| $f^{-1}(x) = \ln(x-2)$, $x > 2$ | A1, B1ft | |

### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape for $f(x)$ | B1 | |
| $(0, 3)$ marked | B1 | |
| Correct shape for $f^{-1}(x)$ | B1 | |
| $(3, 0)$ marked | B1 | |

## Question 6:

### Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \frac{\ln 4 - 3}{2}$ oe. eg $\ln 2 - \frac{3}{2}$ | A1 | Remember to isw any incorrect working after a correct answer |

### Part (d) - Inverse function
| Answer/Working | Mark | Guidance |
|---|---|---|
| Starts with $y = e^x + 2$ or $x = e^y + 2$ and attempts to change subject | M1 | All ln work must be correct. The 2 must be dealt with first. Eg $y = e^x + 2 \Rightarrow \ln y = x + \ln 2 \Rightarrow x = \ln y - \ln 2$ is M0 |
| $f^{-1}(x) = \ln(x-2)$ or $y = \ln(x-2)$ or $y = \ln|x-2|$ | A1 | There must be some form of bracket |
| Either $x > 2$, or follow through on part (a) provided it wasn't $y \in \Re$ | B1ft | Do not accept $y > 2$ or $f^{-1}(x) > 2$ |

### Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Shape for $y = e^x$ | B1 | Graph only in quadrants 1 and 2. Start with gradient approx. 0 above x-axis in Q2, increasing gradient into Q1. No minimum point |
| $(0, 3)$ lies on the curve | B1 | Accept 3 written on y-axis as long as point lies on curve |
| Shape for $y = \ln x$ | B1 | Graph only in quadrants 4 and 1. Gradient approx. infinite to right of y-axis in Q4, decreasing in Q1. No maximum point. Must not intersect $y = e^x$ |
| $(3, 0)$ lies on the curve | B1 | Accept 3 written on x-axis as long as point lies on curve |

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Show LaTeX source

\begin{enumerate}
  \item The functions f and g are defined by
\end{enumerate}

$$\mathrm { f } : x \mapsto \mathrm { e } ^ { x } + 2 \quad x \in \mathbb { R }$$

$$\mathrm { g } : x \mapsto \ln x \quad x > 0$$

(a) State the range of f .\\
(b) Find $\mathrm { fg } ( x )$, giving $y$ our answer in its simplest form.\\
(c) Find the exact value of $x$ for which $\mathrm { f } ( 2 x + 3 ) = 6$\\
(d) Find $\mathrm { f } ^ { - 1 }$ stating its domain.\\
(e) On the same axes sketch the curves with equation $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$, giving the coordinates of all the points where the curves cross the axes.

\begin{center}

\end{center}

\hfill \mbox{\textit{Edexcel P3 2018 Q6 [14]}}

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Q1 4 Q2 8 Q3 5 Q4 7 Q5 6 Q6 14 Q7 7 Q8 7 Q9 9 Q10 8