OCR MEI C3 — Question 1 7 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyStandard +0.3 This is a straightforward multi-part question requiring standard techniques: using stationary point conditions and cosine properties to find constants, then applying the standard inverse function method (swap x and y, solve for y). The domain/range swap is routine. Slightly easier than average due to the guided structure and standard methods.
Spec1.02v Inverse and composite functions: graphs and conditions for existence1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae
1 Fig. 4 shows the curve \(y = \mathrm { f } ( x )\), where $$f ( x ) = a + \cos b x , 0 \leqslant x \leqslant 2 \pi ,$$ and \(a\) and \(b\) are positive constants. The curve has stationary points at \(( 0,3 )\) and \(( 2 \pi , 1 )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7825ba53-67eb-4050-a671-85e37a30150a-1_420_616_538_759} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
  1. Find \(a\) and \(b\).
  2. Find \(\mathrm { f } ^ { - 1 } ( x )\), and state its domain and range.
Question 1:
AnswerMarks
(i) \(a = 2\), \(b = \frac{1}{2}\)B1 B1
[2]
AnswerMarks
(ii) \(y = 2 + \cos \frac{1}{2} x\)
\(x = 2 + \cos \frac{1}{2} y\)M1
\(x - 2 = \cos \frac{1}{2} y\)M1
\(\arccos(x - 2) = \frac{1}{2} y\)A1
\(y = f^{-1}(x) = 2\arccos(x - 2)\)M1
Domain \(1 \leq x \leq 3\)A1
Range \(0 \leq y \leq 2\)
[5]
Guidance notes:
- may be seen later
- subtracting [their] \(a\) from both sides (first)
- arccos\((x -\)[their] \(a) = \)[their] \(b \times y\)
- cao or \(2 \cos^{-1}(x - 2)\)
- domain 1 to 3, range 0 to 2 correctly specified: must be \(\leq\), \(x\) for domain, \(y\) or \(f^{-1}\) or \(f^{-1}(x)\) for range
- need not substitute for \(a\), \(b\) or with \(x \leftrightarrow y\), need not substitute for \(a\), \(b\)
- may be implied by flow diagram
- if not stated, assume first is domain
- allow \([1, 3]\), \([0, 2]\) not \(360°\)
- (not \(f\))
Question 1:

(i) $a = 2$, $b = \frac{1}{2}$ | B1 B1
[2]

(ii) $y = 2 + \cos \frac{1}{2} x$ | 

$x = 2 + \cos \frac{1}{2} y$ | M1

$x - 2 = \cos \frac{1}{2} y$ | M1

$\arccos(x - 2) = \frac{1}{2} y$ | A1

$y = f^{-1}(x) = 2\arccos(x - 2)$ | M1

Domain $1 \leq x \leq 3$ | A1

Range $0 \leq y \leq 2$ | 
[5]

Guidance notes:
- may be seen later
- subtracting [their] $a$ from both sides (first)
- arccos$(x -$[their] $a) = $[their] $b \times y$
- cao or $2 \cos^{-1}(x - 2)$
- domain 1 to 3, range 0 to 2 correctly specified: must be $\leq$, $x$ for domain, $y$ or $f^{-1}$ or $f^{-1}(x)$ for range
- need not substitute for $a$, $b$ or with $x \leftrightarrow y$, need not substitute for $a$, $b$
- may be implied by flow diagram
- if not stated, assume first is domain
- allow $[1, 3]$, $[0, 2]$ not $360°$
- (not $f$)
1 Fig. 4 shows the curve $y = \mathrm { f } ( x )$, where

$$f ( x ) = a + \cos b x , 0 \leqslant x \leqslant 2 \pi ,$$

and $a$ and $b$ are positive constants. The curve has stationary points at $( 0,3 )$ and $( 2 \pi , 1 )$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{7825ba53-67eb-4050-a671-85e37a30150a-1_420_616_538_759}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}

(i) Find $a$ and $b$.\\
(ii) Find $\mathrm { f } ^ { - 1 } ( x )$, and state its domain and range.

\hfill \mbox{\textit{OCR MEI C3  Q1 [7]}}
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