AQA C3 2011 January — Question 4 14 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2011
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyStandard +0.3 This is a standard C3 inverse function question with routine steps: finding range of a cosine function, deriving inverse using arccos, solving a simple equation, composing with absolute value, sketching a transformed trig graph, and describing transformations. All techniques are textbook exercises requiring careful execution but no novel insight. Slightly easier than average due to straightforward function forms.
Spec1.02l 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 existence1.02w Graph transformations: simple transformations of f(x)1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2

4 The functions f and g are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = 3 \cos \frac { 1 } { 2 } x , & \text { for } 0 \leqslant x \leqslant 2 \pi \\ \mathrm {~g} ( x ) = | x | , & \text { for all real values of } x \end{array}$$
  1. Find the range of f .
  2. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
    1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
    2. Solve the equation \(\mathrm { f } ^ { - 1 } ( x ) = 1\), giving your answer in an exact form.
    1. Write down an expression for \(\mathrm { gf } ( x )\).
    2. Sketch the graph of \(y = \operatorname { gf } ( x )\) for \(0 \leqslant x \leqslant 2 \pi\).
  3. Describe a sequence of two geometrical transformations that maps the graph of \(y = \cos x\) onto the graph of \(y = 3 \cos \frac { 1 } { 2 } x\).

Question 4:
Part (a)
AnswerMarks Guidance
AnswerMark Guidance
\(-3 \leq f(x) \leq 3\)B1 B1 B1 for each correct bound; accept \([-3, 3]\)
Part (b)(i)
AnswerMarks Guidance
AnswerMark Guidance
\(x = 3\cos\frac{1}{2}y \Rightarrow \cos\frac{1}{2}y = \frac{x}{3}\)M1 Attempt to rearrange
\(\frac{1}{2}y = \cos^{-1}\left(\frac{x}{3}\right)\)M1 Applying \(\cos^{-1}\)
\(f^{-1}(x) = 2\cos^{-1}\left(\frac{x}{3}\right)\)A1 Correct final answer
Part (b)(ii)
AnswerMarks Guidance
AnswerMark Guidance
\(2\cos^{-1}\left(\frac{1}{3}\right) = 1\) — solve \(\cos^{-1}\left(\frac{1}{3}\right) = \frac{1}{2}\)M1 Setting up equation
\(x = 3\cos\frac{1}{2}\)A1 Accept exact form
Part (c)(i)
AnswerMarks Guidance
AnswerMark Guidance
\(\text{gf}(x) = \left3\cos\frac{1}{2}x\right \)
Part (c)(ii)
AnswerMarks Guidance
AnswerMark Guidance
Correct shape of \(3\cos\frac{1}{2}x \) for \(0 \leq x \leq 2\pi\)
Touches \(x\)-axis at \(x = \pi\)B1
Ends at \((2\pi, 3)\) with correct W/V shapeB1
Part (d)
AnswerMarks Guidance
AnswerMark Guidance
Stretch parallel to \(x\)-axis, scale factor 2B1 M1
Stretch parallel to \(y\)-axis, scale factor 3A1 Both transformations required for full marks
# Question 4:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $-3 \leq f(x) \leq 3$ | B1 B1 | B1 for each correct bound; accept $[-3, 3]$ |

## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 3\cos\frac{1}{2}y \Rightarrow \cos\frac{1}{2}y = \frac{x}{3}$ | M1 | Attempt to rearrange |
| $\frac{1}{2}y = \cos^{-1}\left(\frac{x}{3}\right)$ | M1 | Applying $\cos^{-1}$ |
| $f^{-1}(x) = 2\cos^{-1}\left(\frac{x}{3}\right)$ | A1 | Correct final answer |

## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $2\cos^{-1}\left(\frac{1}{3}\right) = 1$ — solve $\cos^{-1}\left(\frac{1}{3}\right) = \frac{1}{2}$ | M1 | Setting up equation |
| $x = 3\cos\frac{1}{2}$ | A1 | Accept exact form |

## Part (c)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{gf}(x) = \left|3\cos\frac{1}{2}x\right|$ | B1 | Correct expression |

## Part (c)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct shape of $3|\cos\frac{1}{2}x|$ for $0 \leq x \leq 2\pi$ | B1 | Starts at $(0,3)$ |
| Touches $x$-axis at $x = \pi$ | B1 | |
| Ends at $(2\pi, 3)$ with correct W/V shape | B1 | |

## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Stretch parallel to $x$-axis, scale factor 2 | B1 M1 | |
| Stretch parallel to $y$-axis, scale factor 3 | A1 | Both transformations required for full marks |

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

$$\begin{array} { l l } 
\mathrm { f } ( x ) = 3 \cos \frac { 1 } { 2 } x , & \text { for } 0 \leqslant x \leqslant 2 \pi \\
\mathrm {~g} ( x ) = | x | , & \text { for all real values of } x
\end{array}$$
\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 $\mathrm { f } ^ { - 1 } ( x ) = 1$, giving your answer in an exact form.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Write down an expression for $\mathrm { gf } ( x )$.
\item Sketch the graph of $y = \operatorname { gf } ( x )$ for $0 \leqslant x \leqslant 2 \pi$.
\end{enumerate}\item Describe a sequence of two geometrical transformations that maps the graph of $y = \cos x$ onto the graph of $y = 3 \cos \frac { 1 } { 2 } x$.
\end{enumerate}

\hfill \mbox{\textit{AQA C3 2011 Q4 [14]}}