| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|linear| then solve equation or inequality (numeric coefficients) |
| Difficulty | Moderate -0.8 This is a straightforward modulus function question requiring standard techniques: sketching a simple linear modulus (V-shape), solving by cases, function composition, and finding range of a quadratic on a restricted domain. All parts are routine C3 exercises with no novel problem-solving required, making it easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02m Graphs of functions: difference between plotting and sketching1.02n Sketch curves: simple equations including polynomials1.02u 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 |
| V-shape graph with vertex on the \(x\)-axis | M1 | V or \(\diagup\) or \(\diagdown\) graph with vertex on \(x\)-axis |
| \(\left(\frac{5}{2}, 0\right)\) and \(\left(0, 5\right)\) seen, graph in first and second quadrants | A1 | Both intercepts correct and graph in Q1 and Q2 |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 20\) | B1 | |
| \(2x - 5 = -(15 + x) \Rightarrow x = -\frac{10}{3}\) | M1; A1 oe | Either \(2x-5=-(15+x)\) or \(-(2x-5)=15+x\) |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(fg(2) = f(-3) = \ | 2(-3)-5\ | = \ |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(g(x) = x^2 - 4x + 1 = (x-2)^2 - 4 + 1 = (x-2)^2 - 3\), hence \(g_{\min} = -3\) | M1 | Full method to find minimum: \((x \pm \alpha)^2 + \beta\) leading to \(g_{\min} = \beta\), or differentiate, set to zero, substitute back. |
| Either \(g_{\min} = -3\) or \(g(x) \geqslant -3\), or \(g(5) = 25 - 20 + 1 = 6\) | B1 | For finding correct minimum value of \(g\) (can be implied by \(g(x) \geqslant -3\) or \(g(x) > -3\)) or stating \(g(5)=6\) |
| \(-3 \leqslant g(x) \leqslant 6\) or \(-3 \leqslant y \leqslant 6\) | A1 | Note: \(-3 \leqslant x \leqslant 6\) is A0. \(-3 \leqslant f(x) \leqslant 6\) is A0. If candidate writes \(-3 < g(x) < 6\) or \(-3 < y < 6\): award M1B1A0. |
| (3) [10] |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| V-shape graph with vertex on the $x$-axis | M1 | V or $\diagup$ or $\diagdown$ graph with vertex on $x$-axis |
| $\left(\frac{5}{2}, 0\right)$ and $\left(0, 5\right)$ seen, graph in first and second quadrants | A1 | Both intercepts correct and graph in Q1 and Q2 |
| | **(2)** | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 20$ | B1 | |
| $2x - 5 = -(15 + x) \Rightarrow x = -\frac{10}{3}$ | M1; A1 oe | Either $2x-5=-(15+x)$ or $-(2x-5)=15+x$ |
| | **(3)** | |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $fg(2) = f(-3) = \|2(-3)-5\| = \|-11\| = 11$ | M1; A1 | M1: Full method of inserting $g(2)$ into $f(x) = \|2x-5\|$ or inserting $x=2$ into $\|2(x^2-4x+1)-5\|$. Must show modulus applied. |
| | **(2)** | |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $g(x) = x^2 - 4x + 1 = (x-2)^2 - 4 + 1 = (x-2)^2 - 3$, hence $g_{\min} = -3$ | M1 | Full method to find minimum: $(x \pm \alpha)^2 + \beta$ leading to $g_{\min} = \beta$, or differentiate, set to zero, substitute back. |
| Either $g_{\min} = -3$ or $g(x) \geqslant -3$, or $g(5) = 25 - 20 + 1 = 6$ | B1 | For finding correct minimum value of $g$ (can be implied by $g(x) \geqslant -3$ or $g(x) > -3$) or stating $g(5)=6$ |
| $-3 \leqslant g(x) \leqslant 6$ or $-3 \leqslant y \leqslant 6$ | A1 | **Note:** $-3 \leqslant x \leqslant 6$ is A0. $-3 \leqslant f(x) \leqslant 6$ is A0. If candidate writes $-3 < g(x) < 6$ or $-3 < y < 6$: award M1B1A0. |
| | **(3) [10]** | |4. The function $f$ is defined by
$$f : x \mapsto | 2 x - 5 | , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph with equation $y = \mathrm { f } ( x )$, showing the coordinates of the points where the graph cuts or meets the axes.
\item Solve $\mathrm { f } ( x ) = 15 + x$.
The function $g$ is defined by
$$g : x \mapsto x ^ { 2 } - 4 x + 1 , \quad x \in \mathbb { R } , \quad 0 \leqslant x \leqslant 5$$
\item Find fg(2).
\item Find the range of g.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2010 Q4 [10]}}