CAIE P3 2021 November — Question 2 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2021
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |f(x)| compared to |g(x)| with parameters: equation or inequality only
DifficultyChallenging +1.2 This requires systematic case analysis of modulus inequalities with a parameter, testing critical points x = a/3 and x = -2a, then solving resulting linear inequalities in each region. More demanding than routine single-modulus problems due to the parameter and comparison of two modulus expressions, but follows standard techniques taught in P3.
Spec1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities

2 Solve the inequality \(| 3 x - a | > 2 | x + 2 a |\), where \(a\) is a positive constant.

Question 2:
AnswerMarks Guidance
AnswerMark Guidance
State or imply non-modular inequality \((3x-a)^2 > 2^2(x+2a)^2\), or corresponding quadratic equation, or pair of linear equations or linear inequalitiesB1 Need \(2^2\) seen or implied
Make reasonable attempt to solve a 3-term quadratic, or solve two linear equations for \(x\) in terms of \(a\)M1 \((5x^2 - 22ax - 15a^2 = 0)\)
Obtain critical values \(x = 5a\) and \(x = -\frac{3}{5}a\) and no othersA1 OE. Accept incorrect inequalities with correct critical values. Must state 2 values i.e. \(\frac{a \pm b}{c}\) is not sufficient
State final answer \(x > 5a\), \(x < -\frac{3}{5}a\)A1 Do not condone \(\geqslant\) for \(>\) or \(\leqslant\) for \(<\) in the final answer. \(5a < x < -\frac{3}{5}a\) is A0, 'and' is A0
Alternative method for Question 2:
AnswerMarks Guidance
AnswerMark Guidance
Obtain critical value \(x = 5a\) from graphical method, or by solving a linear equation or linear inequalityB1
Obtain critical value \(x = -\frac{3}{5}a\) similarlyB2 Maximum 2 marks if more than 2 critical values
State final answer \(x > 5a\), \(x < -\frac{3}{5}a\)B1 Do not condone \(\geq\) for \(>\) or \(\leq\) for \(<\) in the final answer. \(5a < x < -\frac{3}{5}a\) is B0, 'and' is B0
Total: 4 marks
## Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modular inequality $(3x-a)^2 > 2^2(x+2a)^2$, or corresponding quadratic equation, or pair of linear equations or linear inequalities | **B1** | Need $2^2$ seen or implied |
| Make reasonable attempt to solve a 3-term quadratic, or solve two linear equations for $x$ in terms of $a$ | **M1** | $(5x^2 - 22ax - 15a^2 = 0)$ |
| Obtain critical values $x = 5a$ and $x = -\frac{3}{5}a$ and no others | **A1** | OE. Accept incorrect inequalities with correct critical values. Must state 2 values i.e. $\frac{a \pm b}{c}$ is not sufficient |
| State final answer $x > 5a$, $x < -\frac{3}{5}a$ | **A1** | Do not condone $\geqslant$ for $>$ or $\leqslant$ for $<$ in the final answer. $5a < x < -\frac{3}{5}a$ is **A0**, 'and' is **A0** |

**Alternative method for Question 2:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain critical value $x = 5a$ from graphical method, or by solving a linear equation or linear inequality | **B1** | |
| Obtain critical value $x = -\frac{3}{5}a$ similarly | **B2** | Maximum 2 marks if more than 2 critical values |
| State final answer $x > 5a$, $x < -\frac{3}{5}a$ | **B1** | Do not condone $\geq$ for $>$ or $\leq$ for $<$ in the final answer. $5a < x < -\frac{3}{5}a$ is **B0**, 'and' is **B0** |

**Total: 4 marks**

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2 Solve the inequality $| 3 x - a | > 2 | x + 2 a |$, where $a$ is a positive constant.\\

\hfill \mbox{\textit{CAIE P3 2021 Q2 [4]}}