| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|linear| and y=linear with unknown constants, then solve |
| Difficulty | Standard +0.3 This is a straightforward modulus function question requiring standard sketching techniques and solving a linear modulus inequality by considering cases. The sketches are routine (V-shapes with translations), and part (b) involves splitting into two cases based on the critical point x=a, then solving linear inequalities—all standard C3/C4 techniques with no novel insight required. Slightly above average difficulty due to the parameter manipulation and need for careful case analysis. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| V shape with vertex on \(x\)-axis but not at the origin | B1 | V shape with vertex on \(x\)-axis but not at origin |
| Correct V shape with \((0, a)\) or just \(a\) and \((a, 0)\) or just \(a\) marked in correct places. Left branch must cross or touch the \(y\)-axis | B1 | Allow coordinates the wrong way round if marked in the correct place |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Their part (i) translated down (by any amount) but clearly not left or right, or correct V with vertex in 4th quadrant | B1ft | Follow through from part (i) |
| A \(y\)-intercept of \(a-b\) on positive \(y\)-axis or intercepts of \(a-b\) and \(a+b\) on positive \(x\)-axis with \(a+b\) to right of \(a-b\) | B1 | |
| A fully correct diagram | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x - a - b = \frac{1}{2}x \Rightarrow x = \ldots\) or \(-x + a - b = \frac{1}{2}x \Rightarrow x = \ldots\) | M1 | Solves \(x - a - b = \frac{1}{2}x\) or solves \(-x + a - b = \frac{1}{2}x\) as far as \(x = \ldots\) Allow \(<\) or \(>\) for \(=\) |
| \(x - a - b = \frac{1}{2}x \Rightarrow x = \ldots\) and \(-x + a - b = \frac{1}{2}x \Rightarrow x = \ldots\) | M1 | Solves both equations as far as \(x = \ldots\) Allow \(<\) or \(>\) for \(=\) |
| \(\frac{2}{3}(a-b) < x < 2(a+b)\) | ddM1A1 | ddM1: Chooses inside region. A1: Allow alternatives e.g. \(x < 2(a+b)\) and \(x > \frac{2}{3}(a-b)\), or \(\left(\frac{2}{3}(a-b),\ 2(a+b)\right)\) but not \(x < 2(a+b)\), \(x > \frac{2}{3}(a-b)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((x-a)^2 = \left(\frac{1}{2}x + b\right)^2 \Rightarrow 3x^2 - 4x(2a+b) + 4(a^2 - b^2) = 0\) | M1 | Squares both sides and obtains 3TQ \(= 0\) |
| \(x = \frac{4(2a+b) \pm 4(a+2b)}{6} \left(= 2(a+b),\ \frac{2}{3}(a-b)\right)\) | M1 | Attempt to solve 3TQ applying usual rules |
| \(\frac{2}{3}(a-b) < x < 2(a+b)\) | ddM1A1 | Dependent on both previous M marks. Expressions must have just one term in \(a\) and one term in \(b\) |
| Total: 4 marks | Total: 9 marks |
## Question 6:
### Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| V shape with vertex on $x$-axis but **not** at the origin | B1 | V shape with vertex on $x$-axis but not at origin |
| Correct V shape with $(0, a)$ or just $a$ and $(a, 0)$ or just $a$ marked in correct places. Left branch must cross or touch the $y$-axis | B1 | Allow coordinates the wrong way round if marked in the correct place |
**Total: 2 marks**
### Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Their part (i) translated down (by any amount) but clearly not left or right, or correct V with vertex in 4th quadrant | B1ft | Follow through from part (i) |
| A $y$-intercept of $a-b$ on positive $y$-axis or intercepts of $a-b$ and $a+b$ on positive $x$-axis with $a+b$ to right of $a-b$ | B1 | |
| A fully correct diagram | B1 | |
**Total: 3 marks**
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x - a - b = \frac{1}{2}x \Rightarrow x = \ldots$ **or** $-x + a - b = \frac{1}{2}x \Rightarrow x = \ldots$ | M1 | Solves $x - a - b = \frac{1}{2}x$ **or** solves $-x + a - b = \frac{1}{2}x$ as far as $x = \ldots$ Allow $<$ or $>$ for $=$ |
| $x - a - b = \frac{1}{2}x \Rightarrow x = \ldots$ **and** $-x + a - b = \frac{1}{2}x \Rightarrow x = \ldots$ | M1 | Solves **both** equations as far as $x = \ldots$ Allow $<$ or $>$ for $=$ |
| $\frac{2}{3}(a-b) < x < 2(a+b)$ | ddM1A1 | ddM1: Chooses inside region. A1: Allow alternatives e.g. $x < 2(a+b)$ **and** $x > \frac{2}{3}(a-b)$, or $\left(\frac{2}{3}(a-b),\ 2(a+b)\right)$ but **not** $x < 2(a+b)$, $x > \frac{2}{3}(a-b)$ |
**Alternative (squaring method):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(x-a)^2 = \left(\frac{1}{2}x + b\right)^2 \Rightarrow 3x^2 - 4x(2a+b) + 4(a^2 - b^2) = 0$ | M1 | Squares both sides and obtains 3TQ $= 0$ |
| $x = \frac{4(2a+b) \pm 4(a+2b)}{6} \left(= 2(a+b),\ \frac{2}{3}(a-b)\right)$ | M1 | Attempt to solve 3TQ applying usual rules |
| $\frac{2}{3}(a-b) < x < 2(a+b)$ | ddM1A1 | Dependent on both previous M marks. Expressions must have just one term in $a$ and one term in $b$ |
**Total: 4 marks | Total: 9 marks**
---6. Given that $a$ and $b$ are constants and that $a > b > 0$
\begin{enumerate}[label=(\alph*)]
\item on separate diagrams, sketch the graph with equation
\begin{enumerate}[label=(\roman*)]
\item $y = | x - a |$
\item $y = | x - a | - b$
Show on each sketch the coordinates of each point at which the graph crosses or meets the $x$-axis and the $y$-axis.
\end{enumerate}\item Hence or otherwise find the complete set of values of $x$ for which
$$| x - a | - b < \frac { 1 } { 2 } x$$
giving your answer in terms of $a$ and $b$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2016 Q6 [9]}}