| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|linear| then solve equation or inequality (numeric coefficients) |
| Difficulty | Moderate -0.3 This is a straightforward modulus function question requiring standard techniques: sketching a V-shaped graph by finding the critical point (x=4), solving a modulus equation by cases, function composition, and finding range via completing the square. All parts are routine C3 exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02m Graphs of functions: difference between plotting and sketching1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| V shape just in Quadrant 1, correct position | B1 | V shape just in quadrant one; left end meets \(y\)-axis, minimum on \(x\)-axis, right section at least as high as left; no curved base |
| Meets/cuts \(y\)-axis at \((0, 8)\) | B1 | Graph must be present; condone \((8,0)\) on correct axis |
| Meets \(x\)-axis at \((4, 0)\) | B1 | Graph must be present; condone \((0,4)\) on correct axis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 1\) | B1 | |
| \(x + 5 = -(8-2x) \Rightarrow x = 13\) | M1A1 | M1 for attempt at second solution; accept \((x+5)^2=(8-2x)^2\); \(x=13\) and no other solutions; do not condone invisible brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(fg(5) = f(2) = -1\) | M1A1 | M1 for full method: sub \(x=5\) into \( |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 2x-3 \Rightarrow\) min at \(x=\frac{3}{2} \Rightarrow\) min \(= -\frac{5}{4}\) | M1A1 | M1 for full method finding turning point (calculus or completing the square); A1 for \(y=-\frac{5}{4}\) (award for \(y > -1.25\)) |
| Maximum value \(= 5\) | B1 | May be scored from inequality |
| Range: \(-\frac{5}{4} \leqslant f(x) \leqslant 5\) | A1 | CSO; allow \([-1.25, 5]\); do not allow "or", \(f(x)<5\), or open interval at \(-\frac{5}{4}\) |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| V shape just in Quadrant 1, correct position | B1 | V shape just in quadrant one; left end meets $y$-axis, minimum on $x$-axis, right section at least as high as left; no curved base |
| Meets/cuts $y$-axis at $(0, 8)$ | B1 | Graph must be present; condone $(8,0)$ on correct axis |
| Meets $x$-axis at $(4, 0)$ | B1 | Graph must be present; condone $(0,4)$ on correct axis |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 1$ | B1 | |
| $x + 5 = -(8-2x) \Rightarrow x = 13$ | M1A1 | M1 for attempt at second solution; accept $(x+5)^2=(8-2x)^2$; $x=13$ and no other solutions; do not condone invisible brackets |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $fg(5) = f(2) = -1$ | M1A1 | M1 for full method: sub $x=5$ into $|8-2x|$, then substitute result into $x^2-3x+1$; A1 for $-1$ only |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 2x-3 \Rightarrow$ min at $x=\frac{3}{2} \Rightarrow$ min $= -\frac{5}{4}$ | M1A1 | M1 for full method finding turning point (calculus or completing the square); A1 for $y=-\frac{5}{4}$ (award for $y > -1.25$) |
| Maximum value $= 5$ | B1 | May be scored from inequality |
| Range: $-\frac{5}{4} \leqslant f(x) \leqslant 5$ | A1 | CSO; allow $[-1.25, 5]$; do not allow "or", $f(x)<5$, or open interval at $-\frac{5}{4}$ |
---3. The function $g$ is defined by
$$\mathrm { g } : x \mapsto | 8 - 2 x | , \quad x \in \mathbb { R } , \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph with equation $y = \mathrm { g } ( x )$, showing the coordinates of the points where the graph cuts or meets the axes.
\item Solve the equation
$$| 8 - 2 x | = x + 5$$
The function f is defined by
$$\mathrm { f } : x \mapsto x ^ { 2 } - 3 x + 1 , \quad x \in \mathbb { R } , \quad 0 \leqslant x \leqslant 4$$
\item Find fg(5).
\item Find the range of f. You must make your method clear.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q3 [12]}}