| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y=a|bx+c|+d given: solve equation or inequality |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on modulus functions and rational functions requiring standard techniques: function composition (routine substitution), reading range from a graph, finding inverse of a rational function (standard algebraic manipulation), and interpreting the graph to find when a horizontal line intersects twice. All parts are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02l 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 existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(fg(1) = f(2) = 7\) | M1A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Either \(g(0)=3\) or \(g(x\to\infty)\to 0.5\) | M1 | — |
| \(0.5 < g(x) \leq 3\) | A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempt change of subject: \(y = \frac{x+9}{2x+3} \Rightarrow y(2x+3)=x+9\) | M1 | — |
| \(\Rightarrow 2xy - x = 9 - 3y \Rightarrow x(2y-1) = 9-3y \Rightarrow x = \frac{9-3y}{2y-1}\) | dM1 | Dependent on first M1 |
| \(g^{-1}(x) = \frac{9-3x}{2x-1},\quad 0.5 < x \leq 3\) | A1, B1ft | A1: Correct expression. B1ft: Correct domain (follow through from part (b)) (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(0) = 2(3)+5=11 \Rightarrow k \leq 11\) or \(f(3)=2(0)+5=5 \Rightarrow k>5\) | M1A1 | — |
| \(5 < k \leq 11\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Full method for \(g(1)\) substituted into \(f\) | M1 | Can be scored for seeing 1 substituted into \(2\ |
| \(= 7\) | A1 | cso; do not accept multiple answers; just '7' scores both marks if no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Calculates value of \(g\) at either end; sight of 3 or 0.5 | M1 | |
| \(0.5 < g(x) \leqslant 3\) | A1 | Accept \(0.5 < y \leqslant 3\); also accept \((0.5, 3]\); do not accept in terms of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to make \(x\) the subject; must cross multiply to get both \(x\) terms on same side | M1 | Allow slips |
| Collect like terms, factorise and divide | dM1 | Condone one numerical/sign slip; accept \(x\) as function of \(y\) |
| \(g^{-1}(x) = \frac{9-3x}{2x-1}\) or \(g^{-1}(x) = \frac{3x-9}{1-2x}\) | A1 | Accept \(y =\) or \(g^{-1} =\) but must be in terms of \(x\) |
| Domain: \(0.5 < x \leqslant 3\) | B1ft | Accept \((0.5, 3]\); follow through on part (b); domain cannot be in terms of \(y\); do not follow through on \(y \in \mathbb{R} \Rightarrow x \in \mathbb{R}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt \(f(0)\) or \(f(3)\); sight of \(2\ | 3-0\ | +5\) or \(2\ |
| Range with both ends, e.g. \(5 < k < 11\), \(5 \leqslant k \leqslant 11\), \(5 \leqslant k < 11\) | A1 | Or one end completely correct: \(k \leqslant 11\) or \(k > 5\) |
| \(5 < k \leqslant 11\) | A1 | Accept \((5,11]\), \(k \leqslant 11\) and \(k > 5\); do not accept \(k \leqslant 11\) or \(k > 5\) or \(5 < y \leqslant 11\) |
## Question 4:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $fg(1) = f(2) = 7$ | M1A1 | **(2 marks)** |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Either $g(0)=3$ or $g(x\to\infty)\to 0.5$ | M1 | — |
| $0.5 < g(x) \leq 3$ | A1 | **(2 marks)** |
**Part (c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt change of subject: $y = \frac{x+9}{2x+3} \Rightarrow y(2x+3)=x+9$ | M1 | — |
| $\Rightarrow 2xy - x = 9 - 3y \Rightarrow x(2y-1) = 9-3y \Rightarrow x = \frac{9-3y}{2y-1}$ | dM1 | Dependent on first M1 |
| $g^{-1}(x) = \frac{9-3x}{2x-1},\quad 0.5 < x \leq 3$ | A1, B1ft | A1: Correct expression. B1ft: Correct domain (follow through from part (b)) **(4 marks)** |
**Part (d):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0) = 2(3)+5=11 \Rightarrow k \leq 11$ or $f(3)=2(0)+5=5 \Rightarrow k>5$ | M1A1 | — |
| $5 < k \leq 11$ | A1 | **(3 marks)** |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Full method for $g(1)$ substituted into $f$ | M1 | Can be scored for seeing 1 substituted into $2\|3 - \frac{x+9}{2x+3}\| + 5$ |
| $= 7$ | A1 | cso; do not accept multiple answers; just '7' scores both marks if no incorrect working |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Calculates value of $g$ at either end; sight of 3 or 0.5 | M1 | |
| $0.5 < g(x) \leqslant 3$ | A1 | Accept $0.5 < y \leqslant 3$; also accept $(0.5, 3]$; do not accept in terms of $x$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to make $x$ the subject; must cross multiply to get both $x$ terms on same side | M1 | Allow slips |
| Collect like terms, factorise and divide | dM1 | Condone one numerical/sign slip; accept $x$ as function of $y$ |
| $g^{-1}(x) = \frac{9-3x}{2x-1}$ or $g^{-1}(x) = \frac{3x-9}{1-2x}$ | A1 | Accept $y =$ or $g^{-1} =$ but must be in terms of $x$ |
| Domain: $0.5 < x \leqslant 3$ | B1ft | Accept $(0.5, 3]$; follow through on part (b); domain cannot be in terms of $y$; do not follow through on $y \in \mathbb{R} \Rightarrow x \in \mathbb{R}$ |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt $f(0)$ or $f(3)$; sight of $2\|3-0\|+5$ or $2\|3-3\|+5$ | M1 | Evidence of 5 or 11 |
| Range with both ends, e.g. $5 < k < 11$, $5 \leqslant k \leqslant 11$, $5 \leqslant k < 11$ | A1 | Or one end completely correct: $k \leqslant 11$ or $k > 5$ |
| $5 < k \leqslant 11$ | A1 | Accept $(5,11]$, $k \leqslant 11$ **and** $k > 5$; do not accept $k \leqslant 11$ **or** $k > 5$ **or** $5 < y \leqslant 11$ |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5b698944-41ac-4072-b5e1-c580b7752c39-10_606_613_285_278}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5b698944-41ac-4072-b5e1-c580b7752c39-10_602_608_287_1062}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the graph $y = \mathrm { f } ( x )$, where
$$f ( x ) = 2 | 3 - x | + 5 , \quad x \geqslant 0$$
Figure 2 shows a sketch of part of the graph $y = \mathrm { g } ( x )$, where
$$\operatorname { g } ( x ) = \frac { x + 9 } { 2 x + 3 } , \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\mathrm { fg } ( 1 )$
\item State the range of g
\item Find $\mathrm { g } ^ { - 1 } ( x )$ and state its domain.
Given that the equation $\mathrm { f } ( x ) = k$, where $k$ is a constant, has exactly two roots,
\item state the range of possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q4 [11]}}