CAIE P2 2021 June — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2021
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeEvaluate modulus expression given equation
DifficultyModerate -0.3 This is a straightforward modulus equation requiring case analysis (x≥0 and x<0), solving two linear equations, then substituting into a modulus expression. The steps are routine and mechanical with no conceptual difficulty beyond basic modulus properties, making it slightly easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities

2 The solutions of the equation \(5 | x | = 5 - 2 x\) are \(x = a\) and \(x = b\), where \(a < b\).
Find the value of \(| 3 a - 1 | + | 7 b - 1 |\).

Question 2:
AnswerMarks Guidance
AnswerMark Guidance
Solve \(5x = 5 - 2x\) to obtain \(x = \frac{5}{7}\)B1 Allow AWRT 0.714
Attempt solution of linear equation where signs of \(5x\) and \(2x\) are the sameM1
Obtain \(x = -\frac{5}{3}\)A1 Allow AWRT −1.67
Substitute *their* values correctlyM1 Substitution must be seen unless implied by a correct answer. *Their* values must come from consideration of \(5
Obtain \(-6 +
Alternative: State or imply non-modulus equation \(25x^2 = (5-2x)^2\)B1
Attempt solution of 3-term quadratic equationM1
Obtain \(-\frac{5}{3}\) and \(\frac{5}{7}\)A1 Allow AWRT 0.714 and AWRT −1.67
Substitute *their* values correctlyM1 Substitution must be seen unless implied by a correct answer. *Their* values must come from consideration of \(5
Obtain \(-6 +
Total5
## Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| Solve $5x = 5 - 2x$ to obtain $x = \frac{5}{7}$ | B1 | Allow AWRT 0.714 |
| Attempt solution of linear equation where signs of $5x$ and $2x$ are the same | M1 | |
| Obtain $x = -\frac{5}{3}$ | A1 | Allow AWRT −1.67 |
| Substitute *their* values correctly | M1 | Substitution must be seen unless implied by a correct answer. *Their* values must come from consideration of $5|x| = 5 - 2x$ |
| Obtain $|-6| + |4|$ and hence 10 | A1 | |
| **Alternative:** State or imply non-modulus equation $25x^2 = (5-2x)^2$ | B1 | |
| Attempt solution of 3-term quadratic equation | M1 | |
| Obtain $-\frac{5}{3}$ and $\frac{5}{7}$ | A1 | Allow AWRT 0.714 and AWRT −1.67 |
| Substitute *their* values correctly | M1 | Substitution must be seen unless implied by a correct answer. *Their* values must come from consideration of $5|x| = 5 - 2x$ |
| Obtain $|-6| + |4|$ and hence 10 | A1 | |
| **Total** | **5** | |

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2 The solutions of the equation $5 | x | = 5 - 2 x$ are $x = a$ and $x = b$, where $a < b$.\\
Find the value of $| 3 a - 1 | + | 7 b - 1 |$.\\

\hfill \mbox{\textit{CAIE P2 2021 Q2 [5]}}