CAIE P2 2012 June — Question 1 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2012
SessionJune
Marks4
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Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |f(x)| = |g(x)| with non-linear or exponential expressions
DifficultyModerate -0.3 This is a straightforward modulus equation requiring students to split into two cases (x³ - 14 = 13 or x³ - 14 = -13), then solve simple cubic equations (x³ = 27 or x³ = 1) to get x = 3 or x = 1. The working is routine with no conceptual challenges beyond basic modulus definition, making it slightly easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities
1 Solve the equation \(\left| x ^ { 3 } - 14 \right| = 13\), showing all your working.
AnswerMarks Guidance
Either: Obtain value \(x^2 = 27\) from inspection, equation, …B1
Obtain value \(x^2 = 1\) similarlyB2
Obtain \(x = 1\) and \(x = 3\)B1
Or: Attempt to square both sides obtaining 3 terms on LHSM1
Attempt solution for \(x^3\) of 3-term quadraticDM1
Obtain \(x^3 = 1\) and \(x^3 = 27\)A1
Obtain \(x = 1\) and \(x = 3\)A1 [4] 
Either: Obtain value $x^2 = 27$ from inspection, equation, … | B1 |
Obtain value $x^2 = 1$ similarly | B2 |
Obtain $x = 1$ and $x = 3$ | B1 |
Or: Attempt to square both sides obtaining 3 terms on LHS | M1 |
Attempt solution for $x^3$ of 3-term quadratic | DM1 |
Obtain $x^3 = 1$ and $x^3 = 27$ | A1 |
Obtain $x = 1$ and $x = 3$ | A1 | [4]
1 Solve the equation $\left| x ^ { 3 } - 14 \right| = 13$, showing all your working.

\hfill \mbox{\textit{CAIE P2 2012 Q1 [4]}}
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