OCR MEI C3 2005 June — Question 1 3 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2005
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| = constant
DifficultyEasy -1.2 This is a straightforward modulus equation requiring only the standard technique of splitting into two cases (3x+2=1 and 3x+2=-1), then solving two simple linear equations. It's below average difficulty as it involves minimal steps, no problem-solving insight, and is a textbook exercise testing basic recall of the modulus definition.
Spec1.02l Modulus function: notation, relations, equations and inequalities
1 Solve the equation \(| 3 x + 2 | = 1\).
AnswerMarks Guidance
\(3x + 2 = 1 \Rightarrow x = -1/3\) or \(3x + 2 = -1 \Rightarrow x = -1\)B1, M1, A1 \(x = -1/3\) from a correct method – must be exact
OR \((3x + 2)^2 = 1 \Rightarrow 9x^2 + 12x + 3 = 0 \Rightarrow 3x^2 + 4x + 1 = 0 \Rightarrow (3x + 1)(x + 1) = 0 \Rightarrow x = -1/3\) or \(x = -1\)M1, B1, A1 [3] Squaring and expanding correctly; \(x = -1/3\); \(x = -1\) 
$3x + 2 = 1 \Rightarrow x = -1/3$ or $3x + 2 = -1 \Rightarrow x = -1$ | B1, M1, A1 | $x = -1/3$ from a correct method – must be exact

OR $(3x + 2)^2 = 1 \Rightarrow 9x^2 + 12x + 3 = 0 \Rightarrow 3x^2 + 4x + 1 = 0 \Rightarrow (3x + 1)(x + 1) = 0 \Rightarrow x = -1/3$ or $x = -1$ | M1, B1, A1 [3] | Squaring and expanding correctly; $x = -1/3$; $x = -1$
1 Solve the equation $| 3 x + 2 | = 1$.

\hfill \mbox{\textit{OCR MEI C3 2005 Q1 [3]}}
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