CAIE P1 2023 June — Question 8 12 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeSketch function and inverse graphs
DifficultyModerate -0.3 This is a multi-part question on inverse functions and transformations of trigonometric graphs. Parts (a)-(c) involve standard techniques: reflecting to sketch an inverse, finding the inverse algebraically using arcsin, and determining whether a function has an inverse by checking one-to-one property. Part (d) requires describing transformations systematically but follows a routine procedure (stretch factors and translation). While it requires careful attention to order and detail for full marks, all components are standard P1 curriculum material with no novel problem-solving required, making it slightly easier than average.
Spec1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x)1.05f Trigonometric function graphs: symmetries and periodicities
\includegraphics{figure_8} The diagram shows the graph of \(y = f(x)\) where the function \(f\) is defined by $$f(x) = 3 + 2\sin \frac{1}{4}x \text{ for } 0 \leqslant x \leqslant 2\pi.$$
  1. On the diagram above, sketch the graph of \(y = f^{-1}(x)\). [2]
  2. Find an expression for \(f^{-1}(x)\). [2]
  3. \includegraphics{figure_8c} The diagram above shows part of the graph of the function \(g(x) = 3 + 2\sin \frac{1}{4}x\) for \(-2\pi \leqslant x \leqslant 2\pi\). Complete the sketch of the graph of \(g(x)\) on the diagram above and hence explain whether the function \(g\) has an inverse. [2]
  4. Describe fully a sequence of three transformations which can be combined to transform the graph of \(y = \sin x\) for \(0 \leqslant x \leqslant \frac{1}{2}\pi\) to the graph of \(y = f(x)\), making clear the order in which the transformations are applied. [6]
Question 8:
AnswerMarks Guidance
8(a)*B1 The line y = x correctly drawn. Can be implied by
reasonably correct graph of f1x.
AnswerMarks
DB1Fully correct (needs to reach y = 2π and x-axis and cross
the line y = x in the correct squares).
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
8(b)1 1 y3
y32sin x leading to sin x
AnswerMarks Guidance
4 4 2M1 1
Attempting to arrive at an expression for sin x; condone
4
 sign errors. Variables may be interchanged initially.
y3
M1 not implied by x .
1
2sin
4
 y3 x3
x4sin1 leading to [ f1xor y] 4sin1
   
AnswerMarks Guidance
 2   2 A1 x3 sin1x3
ISW Must clearly be sin1 NOT .
 
 2  2
3x 1
Allow but not  .
 
 2  4
2
AnswerMarks Guidance
8(c)B1 Continuing given graph from y intercept to –2π.
The correct shape needed between 0 and -2π, including
starting to level off (gradient in the final two squares needs
to be reducing) as -2π is approached.
The y co-ordinate at-2π must be in the correct square.
Yes it does have an inverse, because the graph is always increasing
OR because it is one-one OR because it passes the horizontal line test OR
AnswerMarks Guidance
it is not a many to one [function].B1 FT If there is no graph to the left of the y axis, no mark is
available.
FT an incorrect graph and if the answer is now ‘No’
provide an appropriate reason.
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
8(d){ } indicates different elements throughout.
{Stretch} {factor 4} {in x-direction}B2, 1,
0B2 for fully correct, B1 with two elements correct.
Condone use of ‘sf’ instead of factor and ‘co-ordinates’
stretched instead of graph stretched.
Allow any mention of x-axis, horizontally or y-axis
invariant.
Wavelength or period increased by a factor of 4 for B2 or
by 4 for B1.
AnswerMarks
{Stretch} {factor 2} {in y-direction}B2, 1,
0B2 for fully correct, B1 with two elements correct.
Condone use of ‘sf’ instead of factor and ‘co-ordinates’
stretched instead of graph stretched.
Allow any mention of y-axis, vertically or x-axis invariant.
Allow y ‘co-ordinates’ doubled or amplitude doubled for
B2.
0
{Translation}  
3
AnswerMarks
 B2, 1,
0B2 for fully correct, B1 with two elements correct.
Allow shift. Any mention of y axis, y-direction or
vertically implies 0, so shift by 3 vertically is B2, but
shift by a factor of 3 vertically or a translation of 3 ‘up’ is
B1.
AnswerMarks
6After scoring B2, B2 the final transformation can only be
awarded B2 if the order is fully correct i.e. the translation
must not be applied before the y stretch. If all correct
except the order award B2B2B1.
AnswerMarks Guidance
QuestionAnswer Marks 
Question 8:
--- 8(a) ---
8(a) | *B1 | The line y = x correctly drawn. Can be implied by
reasonably correct graph of f1x.
DB1 | Fully correct (needs to reach y = 2π and x-axis and cross
the line y = x in the correct squares).
2
Question | Answer | Marks | Guidance
--- 8(b) ---
8(b) | 1 1 y3
y32sin x leading to sin x
4 4 2 | M1 | 1
Attempting to arrive at an expression for sin x; condone
4
 sign errors. Variables may be interchanged initially.
y3
M1 not implied by x .
1
2sin
4
 y3 x3
x4sin1 leading to [ f1xor y] 4sin1
   
 2   2  | A1 | x3 sin1x3
ISW Must clearly be sin1 NOT .
 
 2  2
3x 1
Allow but not  .
 
 2  4
2
--- 8(c) ---
8(c) | B1 | Continuing given graph from y intercept to –2π.
The correct shape needed between 0 and -2π, including
starting to level off (gradient in the final two squares needs
to be reducing) as -2π is approached.
The y co-ordinate at-2π must be in the correct square.
Yes it does have an inverse, because the graph is always increasing
OR because it is one-one OR because it passes the horizontal line test OR
it is not a many to one [function]. | B1 FT | If there is no graph to the left of the y axis, no mark is
available.
FT an incorrect graph and if the answer is now ‘No’
provide an appropriate reason.
2
Question | Answer | Marks | Guidance
--- 8(d) ---
8(d) | { } indicates different elements throughout.
{Stretch} {factor 4} {in x-direction} | B2, 1,
0 | B2 for fully correct, B1 with two elements correct.
Condone use of ‘sf’ instead of factor and ‘co-ordinates’
stretched instead of graph stretched.
Allow any mention of x-axis, horizontally or y-axis
invariant.
Wavelength or period increased by a factor of 4 for B2 or
by 4 for B1.
{Stretch} {factor 2} {in y-direction} | B2, 1,
0 | B2 for fully correct, B1 with two elements correct.
Condone use of ‘sf’ instead of factor and ‘co-ordinates’
stretched instead of graph stretched.
Allow any mention of y-axis, vertically or x-axis invariant.
Allow y ‘co-ordinates’ doubled or amplitude doubled for
B2.
0
{Translation}  
3
  | B2, 1,
0 | B2 for fully correct, B1 with two elements correct.
Allow shift. Any mention of y axis, y-direction or
vertically implies 0, so shift by 3 vertically is B2, but
shift by a factor of 3 vertically or a translation of 3 ‘up’ is
B1.
6 | After scoring B2, B2 the final transformation can only be
awarded B2 if the order is fully correct i.e. the translation
must not be applied before the y stretch. If all correct
except the order award B2B2B1.
Question | Answer | Marks | Guidance
\includegraphics{figure_8}

The diagram shows the graph of $y = f(x)$ where the function $f$ is defined by
$$f(x) = 3 + 2\sin \frac{1}{4}x \text{ for } 0 \leqslant x \leqslant 2\pi.$$

\begin{enumerate}[label=(\alph*)]
\item On the diagram above, sketch the graph of $y = f^{-1}(x)$. [2]

\item Find an expression for $f^{-1}(x)$. [2]

\item \includegraphics{figure_8c}

The diagram above shows part of the graph of the function $g(x) = 3 + 2\sin \frac{1}{4}x$ for $-2\pi \leqslant x \leqslant 2\pi$.

Complete the sketch of the graph of $g(x)$ on the diagram above and hence explain whether the function $g$ has an inverse. [2]

\item Describe fully a sequence of three transformations which can be combined to transform the graph of $y = \sin x$ for $0 \leqslant x \leqslant \frac{1}{2}\pi$ to the graph of $y = f(x)$, making clear the order in which the transformations are applied. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2023 Q8 [12]}}
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