Edexcel P3 2021 October — Question 2 10 marks

Exam BoardEdexcel
ModuleP3 (Pure Mathematics 3)
Year2021
SessionOctober
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeGraph y=a|bx+c|+d given: solve equation or inequality
DifficultyModerate -0.3 This is a straightforward modulus function question testing standard techniques: finding vertex coordinates, range, function composition, solving modulus inequalities, and transformations. All parts follow routine procedures taught in P3 with no novel problem-solving required. Part (c) requires splitting into cases but this is a standard textbook exercise. Part (d) on transformations is mechanical application of rules. Slightly easier than average due to the step-by-step scaffolding across multiple parts.
Spec1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02w Graph transformations: simple transformations of f(x)

2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9b0b8db0-79fd-4ad5-88c9-737447d9f894-06_570_604_255_673} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of part of the graph with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = | 3 x - 13 | + 5 \quad x \in \mathbb { R }$$ The vertex of the graph is at point \(P\), as shown in Figure 1.
  1. State the coordinates of \(P\).
    1. State the range of f .
    2. Find the value of ff(4)
  2. Solve, using algebra and showing your working, $$16 - 2 x > | 3 x - 13 | + 5$$ The graph with equation \(y = \mathrm { f } ( x )\) is transformed onto the graph with equation \(y = a \mathrm { f } ( x + b )\) The vertex of the graph with equation \(y = a \mathrm { f } ( x + b )\) is \(( 4,20 )\) Given that \(a\) and \(b\) are constants,
  3. find the value of \(a\) and the value of \(b\).

Question 2:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\left(\frac{13}{3}, 5\right)\)B1 B1 One correct coordinate; both coordinates correct; condone missing brackets
Part (b)(i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(f(x) \geqslant 5\)B1ft Follow through on \(y\)-coordinate from (a); allow \(y\geqslant 5\), \(y\in[5,\infty)\), \(f\geqslant 5\); do not allow "range \(\geqslant 5\)"
Part (b)(ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(10\)B1
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Attempts to solve \(16-2x \ldots 3x-13+5\) or \(16-2x \ldots -3x+13+5\)M1 Ignore direction of inequality; allow slips in rearrangement
Both critical values \(2,\ \frac{24}{5}\)A1 Correct critical values; may be part of incorrect inequality
Selects inside region for critical valuesdM1 Selects inside region or both correct inequalities seen
\(2 < x < \frac{24}{5}\)A1 Allow \(x\in\left(2,\frac{24}{5}\right)\) or \(\frac{24}{5}>x>2\); must be on one line; accept \(2
Part (d)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(a=4,\ b=\frac{1}{3}\)B1ft B1 One correct value ft on part (a); both correct; may be embedded within \(y=af(x+b)\)
# Question 2:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{13}{3}, 5\right)$ | B1 B1 | One correct coordinate; both coordinates correct; condone missing brackets |

## Part (b)(i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) \geqslant 5$ | B1ft | Follow through on $y$-coordinate from (a); allow $y\geqslant 5$, $y\in[5,\infty)$, $f\geqslant 5$; do not allow "range $\geqslant 5$" |

## Part (b)(ii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $10$ | B1 | |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to solve $16-2x \ldots 3x-13+5$ or $16-2x \ldots -3x+13+5$ | M1 | Ignore direction of inequality; allow slips in rearrangement |
| Both critical values $2,\ \frac{24}{5}$ | A1 | Correct critical values; may be part of incorrect inequality |
| Selects inside region for critical values | dM1 | Selects inside region or both correct inequalities seen |
| $2 < x < \frac{24}{5}$ | A1 | Allow $x\in\left(2,\frac{24}{5}\right)$ or $\frac{24}{5}>x>2$; must be on one line; accept $2<x<\frac{48}{10}$ |

## Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $a=4,\ b=\frac{1}{3}$ | B1ft B1 | One correct value ft on part (a); both correct; may be embedded within $y=af(x+b)$ |
2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{9b0b8db0-79fd-4ad5-88c9-737447d9f894-06_570_604_255_673}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a sketch of part of the graph with equation $y = \mathrm { f } ( x )$, where

$$f ( x ) = | 3 x - 13 | + 5 \quad x \in \mathbb { R }$$

The vertex of the graph is at point $P$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of $P$.
\item \begin{enumerate}[label=(\roman*)]
\item State the range of f .
\item Find the value of ff(4)
\end{enumerate}\item Solve, using algebra and showing your working,

$$16 - 2 x > | 3 x - 13 | + 5$$

The graph with equation $y = \mathrm { f } ( x )$ is transformed onto the graph with equation $y = a \mathrm { f } ( x + b )$ The vertex of the graph with equation $y = a \mathrm { f } ( x + b )$ is $( 4,20 )$

Given that $a$ and $b$ are constants,
\item find the value of $a$ and the value of $b$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel P3 2021 Q2 [10]}}