Question 6 - A-Level Maths

AQA C3 2010 January — Question 6 11 marks

Exam Board	AQA
Module	C3 (Core Mathematics 3)
Year	2010
Session	January
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Composite & Inverse Functions
Type	Find inverse function
Difficulty	Moderate -0.3 This is a standard C3 inverse function question with routine techniques: finding range of exponential function, finding inverse by swapping and rearranging, solving simple equations, and function composition. All parts follow textbook procedures with no novel insight required, 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

6 The functions $f$ and $g$ are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } - 3 , & \text { for all real values of } x \\ \mathrm {~g} ( x ) = \frac { 1 } { 3 x + 4 } , & \text { for real values of } x , x \neq - \frac { 4 } { 3 } \end{array}$$

Find the range of $f$.
The inverse of f is $\mathrm { f } ^ { - 1 }$.
1. Find $\mathrm { f } ^ { - 1 } ( x )$.
2. Solve the equation $\mathrm { f } ^ { - 1 } ( x ) = 0$.
1. Find an expression for $\operatorname { gf } ( x )$.
2. Solve the equation $\mathrm { gf } ( x ) = 1$, giving your answer in an exact form.

Show mark scheme Show mark scheme source

Question 6(a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$f(x) > -3$	M1
A1	Allow $y > -3$; also accept $> -3$, $x > -3$, $f(x) \geq -3$

Question 6(b)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$y = e^{2x} - 3$, so $y + 3 = e^{2x}$, so $\ln(y+3) = 2x$	M1	swap $x$ and $y$
M1	attempt to isolate: $\ln(y \pm A) = Bx$ or reverse
$f^{-1}(x) = \frac{1}{2}\ln(x+3)$	A1	OE with no further incorrect working; condone $y = \ldots$

Alternative:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x \to \times 2 \to e \to -3$; $\div 2 \leftarrow \ln \leftarrow +3 \leftarrow x$	(M1)(M1)
$y = \frac{\ln(x+3)}{2}$	(A1)

Question 6(b)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x + 3 = 1$	M1	for putting their $p(x) = 1$ from $k\ln(p(x))$ in their part (b)(i)
$x = -2$	A1	CSO. SC: B2 $x = -2$ with no working, if full marks gained in part (b)(i)

Question 6(c)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$gf(x) = \frac{1}{3(e^{2x}-3)+4}$ or $= \frac{1}{3e^{2x}-5}$	B1	substituting $f$ into $g$; ISW

Question 6(c)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\frac{1}{3e^{2x}-5} = 1$, so $1 = 3e^{2x} - 5$	M1	Correct removal of their fraction
$e^{2x} = 2$, so $2x = \ln 2$	m1	Correct use of logs leading to $kx = \ln\frac{a}{b}$
$x = \frac{1}{2}\ln 2$	A1	CSO. No ISW except for numerical evaluation

## Question 6(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) > -3$ | M1 | |
| | A1 | Allow $y > -3$; also accept $> -3$, $x > -3$, $f(x) \geq -3$ |

## Question 6(b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = e^{2x} - 3$, so $y + 3 = e^{2x}$, so $\ln(y+3) = 2x$ | M1 | swap $x$ and $y$ |
| | M1 | attempt to isolate: $\ln(y \pm A) = Bx$ or reverse |
| $f^{-1}(x) = \frac{1}{2}\ln(x+3)$ | A1 | OE with no further incorrect working; condone $y = \ldots$ |

**Alternative:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x \to \times 2 \to e \to -3$; $\div 2 \leftarrow \ln \leftarrow +3 \leftarrow x$ | (M1)(M1) | |
| $y = \frac{\ln(x+3)}{2}$ | (A1) | |

## Question 6(b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x + 3 = 1$ | M1 | for putting their $p(x) = 1$ from $k\ln(p(x))$ in their part (b)(i) |
| $x = -2$ | A1 | CSO. SC: B2 $x = -2$ with no working, if full marks gained in part (b)(i) |

## Question 6(c)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $gf(x) = \frac{1}{3(e^{2x}-3)+4}$ or $= \frac{1}{3e^{2x}-5}$ | B1 | substituting $f$ into $g$; ISW |

## Question 6(c)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{3e^{2x}-5} = 1$, so $1 = 3e^{2x} - 5$ | M1 | Correct removal of their fraction |
| $e^{2x} = 2$, so $2x = \ln 2$ | m1 | Correct use of logs leading to $kx = \ln\frac{a}{b}$ |
| $x = \frac{1}{2}\ln 2$ | A1 | CSO. No ISW except for numerical evaluation |

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Show LaTeX source

6 The functions $f$ and $g$ are defined with their respective domains by

$$\begin{array} { l l } 
\mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } - 3 , & \text { for all real values of } x \\
\mathrm {~g} ( x ) = \frac { 1 } { 3 x + 4 } , & \text { for real values of } x , x \neq - \frac { 4 } { 3 }
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find the range of $f$.
\item The inverse of f is $\mathrm { f } ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { f } ^ { - 1 } ( x )$.
\item Solve the equation $\mathrm { f } ^ { - 1 } ( x ) = 0$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find an expression for $\operatorname { gf } ( x )$.
\item Solve the equation $\mathrm { gf } ( x ) = 1$, giving your answer in an exact form.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C3 2010 Q6 [11]}}

This paper (8 questions)

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Q1 8 Q2 11 Q3 8 Q4 6 Q5 12 Q6 11 Q7 8 Q8 11

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(f(x) > -3\)	M1
A1	Allow \(y > -3\); also accept \(> -3\), \(x > -3\), \(f(x) \geq -3\)