OCR MEI C3 2006 June — Question 1 3 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2006
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| = linear (non-modulus)
DifficultyModerate -0.8 This is a straightforward modulus equation requiring students to consider two cases (3x-2 ≥ 0 and 3x-2 < 0), then solve the resulting linear equations and check validity. It's a standard textbook exercise testing basic understanding of modulus definition with minimal algebraic complexity, making it easier than average but not trivial since it requires the case-splitting technique.
Spec1.02l Modulus function: notation, relations, equations and inequalities
1 Solve the equation \(| 3 x - 2 | = x\).
AnswerMarks Guidance
\(3x - 2 = x\)B1 M1 A1 \(x = 1\)
\(\Rightarrow 3x - 2 = x \Rightarrow 2x = 2 \Rightarrow x = 1\) or \(2 - 3x = x \Rightarrow 2 = 4x \Rightarrow x = \frac{1}{2}\) or \((3x - 2)^2 = x^2\)M1 A1 A1 solving correct quadratic
\(\Rightarrow 8x^2 - 12x + 4 = 0 \Rightarrow 2x^2 - 3x + 1 = 0\)M1
\(\Rightarrow (x - 1)(2x - 1) = 0\)A1
\(\Rightarrow x = 1, \frac{1}{2}\)[3] 
$3x - 2 = x$ | B1 M1 A1 | $x = 1$
$\Rightarrow 3x - 2 = x \Rightarrow 2x = 2 \Rightarrow x = 1$ or $2 - 3x = x \Rightarrow 2 = 4x \Rightarrow x = \frac{1}{2}$ or $(3x - 2)^2 = x^2$ | M1 A1 A1 | solving correct quadratic
$\Rightarrow 8x^2 - 12x + 4 = 0 \Rightarrow 2x^2 - 3x + 1 = 0$ | M1 | 
$\Rightarrow (x - 1)(2x - 1) = 0$ | A1 | 
$\Rightarrow x = 1, \frac{1}{2}$ | [3] |
1 Solve the equation $| 3 x - 2 | = x$.

\hfill \mbox{\textit{OCR MEI C3 2006 Q1 [3]}}
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Q1 3 Q2 6 Q3 7 Q4 6 Q5 6 Q6 8