CAIE P1 2018 November — Question 4 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFunction Transformations
TypeTransformation of specific function type
DifficultyModerate -0.8 This is a straightforward P1 question requiring basic function composition (substituting g(x) into f) and solving a simple cosine equation, plus sketching a standard transformed cosine graph. Both parts are routine textbook exercises with no problem-solving insight needed, making it easier than average.
Spec1.02v Inverse and composite functions: graphs and conditions for existence1.05f Trigonometric function graphs: symmetries and periodicities
Functions f and g are defined by $$f : x \mapsto 2 - 3\cos x \text{ for } 0 \leqslant x \leqslant 2\pi,$$ $$g : x \mapsto \frac{1}{2}x \text{ for } 0 \leqslant x \leqslant 2\pi.$$
  1. Solve the equation \(\text{fg}(x) = 1\). [3]
  2. Sketch the graph of \(y = \text{f}(x)\). [3]
Question 4:
AnswerMarks
4(i)1
fg(x) = 2 – 3cos( x)
AnswerMarks Guidance
2B1 Correct fg
1 1 1 1   1
2 – 3cos( x) = 1 → cos( x) = →  x = cos−1 their 
AnswerMarks Guidance
2 2 3 2   3M1 M1 for correct order of operations to solve their fg(x) = 1 as
far as using inverse cos expect 1.23, ( or 70.5◦ ) condone x =.
4.7π( )
x = 2.46 awrt or 0.784π awrt
AnswerMarks Guidance
6A1 One solution only in the given range, ignore answers outside
the range.
Answer in degrees A0.
Alternative:
1
Solve f(y) = 1 → y = 1.23→ x=1.23 B1M1
2
→ x = 2.46 A1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
4(ii)B1 One cycle of ± cos curve, evidence of turning at the ends not
required at this stage. Can be a poor curve but not an inverted
“V”. If horizontal axis is not labelled mark everything to the
right of the vertical axis. If axis is clearly labelled mark 0 →
2π.
AnswerMarks
B1Start and finish at roughly the same negative y value.
Significantly more above the x axis than below or correct
range implied by labels .
AnswerMarks
B1Fully correct. Curves not lines.
Must be a reasonable curve clearly turning at both ends.
Labels not required but must be appropriate if present.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
--- 4(i) ---
4(i) | 1
fg(x) = 2 – 3cos( x)
2 | B1 | Correct fg
1 1 1 1   1
2 – 3cos( x) = 1 → cos( x) = →  x = cos−1 their 
2 2 3 2   3 | M1 | M1 for correct order of operations to solve their fg(x) = 1 as
far as using inverse cos expect 1.23, ( or 70.5◦ ) condone x =.
4.7π( )
x = 2.46 awrt or 0.784π awrt
6 | A1 | One solution only in the given range, ignore answers outside
the range.
Answer in degrees A0.
Alternative:
1
Solve f(y) = 1 → y = 1.23→ x=1.23 B1M1
2
→ x = 2.46 A1
3
Question | Answer | Marks | Guidance
--- 4(ii) ---
4(ii) | B1 | One cycle of ± cos curve, evidence of turning at the ends not
required at this stage. Can be a poor curve but not an inverted
“V”. If horizontal axis is not labelled mark everything to the
right of the vertical axis. If axis is clearly labelled mark 0 →
2π.
B1 | Start and finish at roughly the same negative y value.
Significantly more above the x axis than below or correct
range implied by labels .
B1 | Fully correct. Curves not lines.
Must be a reasonable curve clearly turning at both ends.
Labels not required but must be appropriate if present.
3
Question | Answer | Marks | Guidance
Functions f and g are defined by
$$f : x \mapsto 2 - 3\cos x \text{ for } 0 \leqslant x \leqslant 2\pi,$$
$$g : x \mapsto \frac{1}{2}x \text{ for } 0 \leqslant x \leqslant 2\pi.$$

\begin{enumerate}[label=(\roman*)]
\item Solve the equation $\text{fg}(x) = 1$. [3]
\item Sketch the graph of $y = \text{f}(x)$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2018 Q4 [6]}}
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