Edexcel C3 2013 January — Question 3 9 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2013
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFunction Transformations
TypeMultiple separate transformations (sketch-based, modulus involved)
DifficultyModerate -0.3 This is a standard C3 transformations question requiring recall of transformation rules (inverse, modulus, stretch/translation) and careful plotting of key points. Part (a) is trivial reading from the graph. Parts (b)-(d) are routine applications of well-practiced transformation techniques with no novel problem-solving required, though students must track multiple coordinate changes accurately.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x)

3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c78b0245-5c5a-407f-ad8a-602949a76e05-04_620_1095_223_420} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows part of the curve with equation \(y = \mathrm { f } ( x ) , x \in \mathbb { R }\).
The curve passes through the points \(Q ( 0,2 )\) and \(P ( - 3,0 )\) as shown.
  1. Find the value of ff(-3). On separate diagrams, sketch the curve with equation
  2. \(y = \mathrm { f } ^ { - 1 } ( x )\),
  3. \(y = \mathrm { f } ( | x | ) - 2\),
  4. \(y = 2 \mathrm { f } \left( \frac { 1 } { 2 } x \right)\). Indicate clearly on each sketch the coordinates of the points at which the curve crosses or meets the axes.

(a)
AnswerMarks Guidance
\(ff(-3) = f(0) = 2\)M1, A1 A full method of finding \(ff(-3)\). \(f(0)\) is acceptable but \(f(-3)=0\) is not. Accept a solution obtained from two substitutions into the equation \(y = \frac{2}{3}x + 2\) as the line passes through both points. Do not allow for \(y = \ln(x+4)\), which only passes through one of the points.
(b)
AnswerMarks Guidance
ShapeB1 For the correct shape. Award this mark for an increasing function in quadrants 3, 4 and 1 only. Do not award if the curve bends back on itself or has a clear minimum.
(0,-3) and (2,0)B1 This is independent to the first mark and for the graph passing through \((0,-3)\) and \((2, 0)\). Accept \(-3\) and \(2\) marked on the correct axes. Accept \(P'=(0,-3)\), \(Q'=(2,0)\) stated elsewhere as long as P' and Q' are marked in the correct place on the graph. There must be a graph for this to be awarded.
(c)
AnswerMarks Guidance
ShapeB1 Award for a correct shape 'roughly' symmetrical about the \(y\)- axis. It must have a cusp and a gradient that 'decreases' either side of the cusp. Do not award if the graph has a clear maximum \((0,0)\) lies on their graph. Accept the graph passing through the origin without seeing \((0,0)\) marked.
(0,0)B1
(d)
AnswerMarks Guidance
ShapeB1 Shape. The position is not important. The gradient should be always positive but decreasing. There should not be a clear maximum point.
(-6,0) or (0,4)B1 The graph passes through \((0,4)\) or \((-6,0)\). See part (b) for allowed variations.
(-6,0) and (0,4)B1 The graph passes through \((0,4)\) and \((-6,0)\). See part (b) for allowed variations.
(9 marks)
**(a)**

$ff(-3) = f(0) = 2$ | M1, A1 | A full method of finding $ff(-3)$. $f(0)$ is acceptable but $f(-3)=0$ is not. Accept a solution obtained from two substitutions into the equation $y = \frac{2}{3}x + 2$ as the line passes through both points. Do not allow for $y = \ln(x+4)$, which only passes through one of the points. | (2 marks)

**(b)**

| | Shape | B1 | For the correct shape. Award this mark for an increasing function in quadrants 3, 4 and 1 only. Do not award if the curve bends back on itself or has a clear minimum. |
| | (0,-3) and (2,0) | B1 | This is independent to the first mark and for the graph passing through $(0,-3)$ and $(2, 0)$. Accept $-3$ and $2$ marked on the correct axes. Accept $P'=(0,-3)$, $Q'=(2,0)$ stated elsewhere as long as **P' and Q' are marked in the correct place on the graph**. **There must be a graph for this to be awarded.** | (2 marks)

**(c)**

| | Shape | B1 | Award for a correct shape 'roughly' symmetrical about the $y$- axis. It must have a cusp and a gradient that 'decreases' either side of the cusp. Do not award if the graph has a clear maximum $(0,0)$ lies on their graph. Accept the graph passing through the origin without seeing $(0,0)$ marked. |
| | (0,0) | B1 | | (2 marks)

**(d)**

| | Shape | B1 | Shape. The position is not important. The gradient should be always positive but decreasing. There should not be a clear maximum point. |
| | (-6,0) or (0,4) | B1 | The graph passes through $(0,4)$ **or** $(-6,0)$. See part (b) for allowed variations. |
| | (-6,0) and (0,4) | B1 | The graph passes through $(0,4)$ **and** $(-6,0)$. See part (b) for allowed variations. | (3 marks)

| (9 marks) |

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3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{c78b0245-5c5a-407f-ad8a-602949a76e05-04_620_1095_223_420}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows part of the curve with equation $y = \mathrm { f } ( x ) , x \in \mathbb { R }$.\\
The curve passes through the points $Q ( 0,2 )$ and $P ( - 3,0 )$ as shown.
\begin{enumerate}[label=(\alph*)]
\item Find the value of ff(-3).

On separate diagrams, sketch the curve with equation
\item $y = \mathrm { f } ^ { - 1 } ( x )$,
\item $y = \mathrm { f } ( | x | ) - 2$,
\item $y = 2 \mathrm { f } \left( \frac { 1 } { 2 } x \right)$.

Indicate clearly on each sketch the coordinates of the points at which the curve crosses or meets the axes.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3 2013 Q3 [9]}}