| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Determine if inverse exists |
| Difficulty | Moderate -0.8 This is a straightforward question on function composition, domains/ranges, and inverse functions. Parts (a)-(c)(i) require only basic recall and substitution. Part (c)(ii) tests understanding that a function must be one-to-one to have an inverse, which students can verify by noting h(x) = ln(|x| + 1) fails the horizontal line test or that h(-x) = h(x). All parts are standard textbook exercises with no novel problem-solving required. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks |
|---|---|
| 17(a) | Deduces correct region |
| Answer | Marks | Guidance |
|---|---|---|
| Condonef or y 1 | 2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1,) | 2.5 | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 17(b) | Obtains x : x 0 or ( 0 , ) | |
| Accept y:y 0 but not y0 | 1.1b | B1 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 17(c)(i) | ( ) |
| Answer | Marks | Guidance |
|---|---|---|
| ISW | 1.1b | B1 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 17(c)(ii) | States that h does not have an |
| Answer | Marks | Guidance |
|---|---|---|
| or that h is many-to-one | 2.2a | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| that h is a function of f. | 2.4 | E1 |
| Subtotal | 2 | |
| Question 17 Total | 6 | |
| Q | Marking instructions | AO |
Question 17:
--- 17(a) ---
17(a) | Deduces correct region
(x)1
eg f or y1
(x)1
Condonef or y 1 | 2.2a | M1 | y : y 1
Obtains correct answer in set
notation
eg
x : x 1
(x) (x)1
f :f
1,) | 2.5 | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 17(b) ---
17(b) | Obtains x : x 0 or ( 0 , )
Accept y:y 0 but not y0 | 1.1b | B1 | x 0
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 17(c)(i) ---
17(c)(i) | ( )
Obtains l n x + 1
Accept l n x + 1
ISW | 1.1b | B1 | h ( x ) = l n ( x + 1 )
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 17(c)(ii) ---
17(c)(ii) | States that h does not have an
inverse
And
States that h is not one-to-one
or that h is many-to-one | 2.2a | E1 | The function h does not have an
inverse as h is not one-to-one.
For example
h ( 1 ) = l n 2
and
h ( − 1 ) = l n 2
Explains why h is not one-to-one
or why h is many-to-one
For example
Gives two x values such that
h ( x ) = h ( x )
1 2
Or
Explains that using the positive
and negative of the same x-
value will result in the same y-
value
Or
Sketches a graph of y=h(x) with
a horizontal line meeting the
curve in two places.
eg
Or
States that f is many-to-one and
that h is a function of f. | 2.4 | E1
Subtotal | 2
Question 17 Total | 6
Q | Marking instructions | AO | Marks | Typical solutionThe function f is defined by
$$f(x) = |x| + 1 \quad \text{for } x \in \mathbb{R}$$
The function g is defined by
$$g(x) = \ln x$$
where g has its greatest possible domain.
\begin{enumerate}[label=(\alph*)]
\item Using set notation, state the range of f
[2 marks]
\item State the domain of g
[1 mark]
\item The composite function h is given by
$$h(x) = g f(x) \quad \text{for } x \in \mathbb{R}$$
\begin{enumerate}[label=(\roman*)]
\item Write down an expression for $h(x)$ in terms of $x$
[1 mark]
\item Determine if h has an inverse.
Fully justify your answer.
[2 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2024 Q17 [6]}}