OCR MEI C3 — Question 12 3 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
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 a routine textbook exercise testing basic understanding of the modulus definition with minimal steps, making it easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities
12 Solve the equation \(| 3 x + 2 | = 1\).
Question 12:
AnswerMarks Guidance
\(3x+2 = 1 \Rightarrow x = -\frac{1}{3}\)B1 \(x = -\frac{1}{3}\) from a correct method – must be exact
\(3x+2 = -1\)M1
\(\Rightarrow x = -1\)A1
*or* \((3x+2)^2 = 1\)M1 Squaring and expanding correctly
\(\Rightarrow 9x^2 + 12x + 3 = 0\)
\(\Rightarrow 3x^2 + 4x + 1 = 0\)
\(\Rightarrow (3x+1)(x+1) = 0\)B1 \(x = -\frac{1}{3}\)
\(\Rightarrow x = -\frac{1}{3}\) or \(x = -1\)A1 \(x = -1\)
[3] 
## Question 12:

| $3x+2 = 1 \Rightarrow x = -\frac{1}{3}$ | B1 | $x = -\frac{1}{3}$ from a correct method – must be exact |
|---|---|---|
| $3x+2 = -1$ | M1 | |
| $\Rightarrow x = -1$ | A1 | |
| *or* $(3x+2)^2 = 1$ | M1 | Squaring and expanding correctly |
| $\Rightarrow 9x^2 + 12x + 3 = 0$ | | |
| $\Rightarrow 3x^2 + 4x + 1 = 0$ | | |
| $\Rightarrow (3x+1)(x+1) = 0$ | B1 | $x = -\frac{1}{3}$ |
| $\Rightarrow x = -\frac{1}{3}$ or $x = -1$ | A1 | $x = -1$ |
| **[3]** | | |
12 Solve the equation $| 3 x + 2 | = 1$.

\hfill \mbox{\textit{OCR MEI C3  Q12 [3]}}
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