Edexcel C12 2016 June — Question 14 8 marks

Exam BoardEdexcel
ModuleC12 (Core Mathematics 1 & 2)
Year2016
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeGraphical equation solving with auxiliary line
DifficultyModerate -0.8 This question requires finding equations of line segments from coordinates (basic gradient formula), solving simultaneous linear equations, and applying a vertical stretch transformation. All techniques are routine C1/C2 skills with straightforward arithmetic. The multi-part structure and need for careful working elevate it slightly above trivial, but it remains easier than average A-level questions.
Spec1.02c Simultaneous equations: two variables by elimination and substitution

14. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{aa75f1c1-ee97-4fee-af98-957e6a3fbba1-21_831_919_127_509} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} Figure 4 shows a sketch of the graph of \(y = g ( x ) , - 3 \leqslant x \leqslant 4\) and part of the line \(l\) with equation \(y = \frac { 1 } { 2 } x\) The graph of \(y = \mathrm { g } ( x )\) consists of three line segments, from \(P ( - 3,4 )\) to \(Q ( 0,4 )\), from \(Q ( 0,4 )\) to \(R ( 2,0 )\) and from \(R ( 2,0 )\) to \(S ( 4,10 )\). The line \(l\) intersects \(y = \mathrm { g } ( x )\) at the points \(A\) and \(B\) as shown in Figure 4.
  1. Use algebra to find the \(x\) coordinate of the point \(A\) and the \(x\) coordinate of the point \(B\). Show each step of your working and give your answers as exact fractions.
  2. Sketch the graph with equation $$y = \frac { 3 } { 2 } g ( x ) , \quad - 3 \leqslant x \leqslant 4$$ On your sketch show the coordinates of the points to which \(P , Q , R\) and \(S\) are transformed.

Question 14:
Part (a):
AnswerMarks Guidance
WorkingMark Guidance
Sets \(\frac{1}{2}x = ax + 4\) where \(a < 0\)M1 Attempts the smaller solution; accept setting \(\frac{1}{2}x = ax + 4\) where \(a < 0\)
Solves \(\frac{1}{2}x = -2x + 4 \Rightarrow \frac{5}{2}x = 4 \Rightarrow x = \frac{8}{5}\)dM1, A1 Sets \(\frac{1}{2}x = -2x + 4\) and proceeds to \(x = ..\) by collecting terms; \(x = \frac{8}{5}\) oe, accept 1.6
Sets \(\frac{1}{2}x = 5x + b\) where \(b < 0\)M1 Attempts to find the larger solution; accept setting \(\frac{1}{2}x = 5x + b\) where \(b < 0\)
Solves \(\frac{1}{2}x = 5x - 10 \Rightarrow \frac{9}{2}x = 10 \Rightarrow x = \frac{20}{9}\)dM1, A1 Sets \(\frac{1}{2}x = 5x - 10\) and proceeds to \(x = ..\) by collecting terms; \(x = \frac{20}{9}\), accept exact equivalents such as \(2\frac{2}{9}\) but not 2.2 or \(2.\dot{2}\)
Part (b):
AnswerMarks Guidance
WorkingMark Guidance
Sketch with points \(P(-3, 6)\), \(Q(0, 6)\), \(R(2, 0)\), \(S(4, 15)\) — any two points correctB1 Any two points correct either in text or on sketch; accept 6 and 2 written on correct axes
Same shape with all four points correctB1 Shape + all four points correct. Candidates may adapt the given diagram. If coordinates given on diagram and in body of script, diagram takes precedence
## Question 14:

### Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| Sets $\frac{1}{2}x = ax + 4$ where $a < 0$ | M1 | Attempts the smaller solution; accept setting $\frac{1}{2}x = ax + 4$ where $a < 0$ |
| Solves $\frac{1}{2}x = -2x + 4 \Rightarrow \frac{5}{2}x = 4 \Rightarrow x = \frac{8}{5}$ | dM1, A1 | Sets $\frac{1}{2}x = -2x + 4$ and proceeds to $x = ..$ by collecting terms; $x = \frac{8}{5}$ oe, accept 1.6 |
| Sets $\frac{1}{2}x = 5x + b$ where $b < 0$ | M1 | Attempts to find the larger solution; accept setting $\frac{1}{2}x = 5x + b$ where $b < 0$ |
| Solves $\frac{1}{2}x = 5x - 10 \Rightarrow \frac{9}{2}x = 10 \Rightarrow x = \frac{20}{9}$ | dM1, A1 | Sets $\frac{1}{2}x = 5x - 10$ and proceeds to $x = ..$ by collecting terms; $x = \frac{20}{9}$, accept exact equivalents such as $2\frac{2}{9}$ but not 2.2 or $2.\dot{2}$ |

### Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| Sketch with points $P(-3, 6)$, $Q(0, 6)$, $R(2, 0)$, $S(4, 15)$ — any two points correct | B1 | Any two points correct either in text or on sketch; accept 6 and 2 written on correct axes |
| Same shape with all four points correct | B1 | Shape + all four points correct. Candidates may adapt the given diagram. If coordinates given on diagram and in body of script, diagram takes precedence |

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14.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{aa75f1c1-ee97-4fee-af98-957e6a3fbba1-21_831_919_127_509}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

Figure 4 shows a sketch of the graph of $y = g ( x ) , - 3 \leqslant x \leqslant 4$ and part of the line $l$ with equation $y = \frac { 1 } { 2 } x$

The graph of $y = \mathrm { g } ( x )$ consists of three line segments, from $P ( - 3,4 )$ to $Q ( 0,4 )$, from $Q ( 0,4 )$ to $R ( 2,0 )$ and from $R ( 2,0 )$ to $S ( 4,10 )$.

The line $l$ intersects $y = \mathrm { g } ( x )$ at the points $A$ and $B$ as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Use algebra to find the $x$ coordinate of the point $A$ and the $x$ coordinate of the point $B$.

Show each step of your working and give your answers as exact fractions.
\item Sketch the graph with equation

$$y = \frac { 3 } { 2 } g ( x ) , \quad - 3 \leqslant x \leqslant 4$$

On your sketch show the coordinates of the points to which $P , Q , R$ and $S$ are transformed.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C12 2016 Q14 [8]}}