Question 2 - A-Level Maths

CAIE P2 2022 November — Question 2 5 marks

Exam Board	CAIE
Module	P2 (Pure Mathematics 2)
Year	2022
Session	November
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Modulus function
Type	Evaluate modulus expression given equation
Difficulty	Moderate -0.3 This is a straightforward modulus equation requiring systematic case analysis (4 cases reduce to 2 meaningful ones), followed by simple substitution into another modulus expression. The technique is standard for P2 level with no conceptual surprises, making it slightly easier than average but not trivial due to the multi-step nature and need for careful algebraic manipulation.
Spec	1.02l Modulus function: notation, relations, equations and inequalities

2 The solutions of the equation \(| 4 x - 1 | = | x + 3 |\) are \(x = p\) and \(x = q\), where \(p < q\).
Find the exact values of \(p\) and \(q\), and hence determine the exact value of \(| p - 2 | - | q - 1 |\).

Show mark scheme Show mark scheme source

Question 2:

Answer	Marks	Guidance
Answer	Marks	Guidance
Solve \(4x - 1 = x + 3\) to obtain \(x = \frac{4}{3}\)	B1
Attempt solution of linear equation where signs of \(4x\) and \(x\) are different	M1
Obtain final value \(x = -\frac{2}{5}\)	A1
Substitute numerical values and apply modulus signs correctly to obtain \(\left	-\frac{12}{5}\right	- \left
Obtain \(\frac{31}{15}\)	A1	or exact equivalent

Alternative method for Question 2:

Answer	Marks	Guidance
Answer	Marks	Guidance
State or imply non-modulus equation \((4x-1)^2 = (x+3)^2\)	B1
Attempt solution of 3-term quadratic equation	M1
Obtain final values \(-\frac{2}{5}\) and \(\frac{4}{3}\)	A1

Question 2:

Answer	Marks	Guidance
Answer	Mark	Guidance
Substitute numerical values and apply modulus signs correctly to obtain \(\left	-\frac{12}{5}\right	- \left
Obtain \(\frac{31}{15}\)	A1	or exact equivalent

**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve $4x - 1 = x + 3$ to obtain $x = \frac{4}{3}$ | **B1** | |
| Attempt solution of linear equation where signs of $4x$ and $x$ are different | **M1** | |
| Obtain final value $x = -\frac{2}{5}$ | **A1** | |
| Substitute numerical values and apply modulus signs correctly to obtain $\left|-\frac{12}{5}\right| - \left|\frac{1}{3}\right|$ or equivalent, retaining exactness and with no subsequent squaring | **M1** | Allow *their* $p$ and $q$, $p < q$ |
| Obtain $\frac{31}{15}$ | **A1** | or exact equivalent |

**Alternative method for Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply non-modulus equation $(4x-1)^2 = (x+3)^2$ | **B1** | |
| Attempt solution of 3-term quadratic equation | **M1** | |
| Obtain final values $-\frac{2}{5}$ and $\frac{4}{3}$ | **A1** | |

## Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute numerical values and apply modulus signs correctly to obtain $\left|-\frac{12}{5}\right| - \left|\frac{1}{3}\right|$ or equivalent, retaining exactness and with no subsequent squaring | M1 | Allow *their* $p < q$ |
| Obtain $\frac{31}{15}$ | A1 | or exact equivalent |

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Show LaTeX source

2 The solutions of the equation $| 4 x - 1 | = | x + 3 |$ are $x = p$ and $x = q$, where $p < q$.\\
Find the exact values of $p$ and $q$, and hence determine the exact value of $| p - 2 | - | q - 1 |$.\\

\hfill \mbox{\textit{CAIE P2 2022 Q2 [5]}}

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