| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Find range of k for number of roots |
| Difficulty | Standard +0.3 This is a straightforward modulus function question requiring interpretation of a given graph. Part (a) is direct reading from the graph, part (b) involves solving a linear equation with modulus (split into two cases), and part (c) requires identifying where a horizontal line intersects the V-shaped graph twice. All techniques are standard and the graph provides visual guidance, making this slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b| |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) \geqslant 5\) | B1 | Also allow \(f(x)\in[5,\infty)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(-2(3-x)+5 = \frac{1}{2}x+30\) | M1 | Deduces solution to \(f(x)=\frac{1}{2}x+30\) found by solving \(-2(3-x)+5=\frac{1}{2}x+30\) |
| \(\frac{3}{2}x = 31\) | M1 | Correct method; multiplies out bracket, collects terms |
| \(x = \frac{62}{3}\) only | A1 | Do not allow \(20.6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Makes connection that there must be two intersections; implied by either end point \(k>5\) or \(k\leqslant 11\) | M1 | Deduces two distinct roots occur when \(y=k\) intersects \(y=f(x)\) in two places |
| \(\{k : k\in\mathbb{R},\, 5 < k \leqslant 11\}\) | A1 | Correct solution only |
# Question 11(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) \geqslant 5$ | B1 | Also allow $f(x)\in[5,\infty)$ |
# Question 11(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $-2(3-x)+5 = \frac{1}{2}x+30$ | M1 | Deduces solution to $f(x)=\frac{1}{2}x+30$ found by solving $-2(3-x)+5=\frac{1}{2}x+30$ |
| $\frac{3}{2}x = 31$ | M1 | Correct method; multiplies out bracket, collects terms |
| $x = \frac{62}{3}$ only | A1 | Do not allow $20.6$ |
# Question 11(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Makes connection that there must be two intersections; implied by either end point $k>5$ or $k\leqslant 11$ | M1 | Deduces two distinct roots occur when $y=k$ intersects $y=f(x)$ in two places |
| $\{k : k\in\mathbb{R},\, 5 < k \leqslant 11\}$ | A1 | Correct solution only |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a75c9ef7-b648-47be-bad1-fc8b315be3df-14_570_556_205_758}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the graph $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = 2 | 3 - x | + 5 , \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item State the range of f
\item Solve the equation
$$f ( x ) = \frac { 1 } { 2 } x + 30$$
Given that the equation $\mathrm { f } ( x ) = k$, where $k$ is a constant, has two distinct roots, (c) state the set of possible values for $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 Q11 [6]}}