OCR MEI C3 — Question 11 3 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
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), solve two linear equations, and check validity. It's a standard textbook exercise testing basic understanding of modulus 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
11 Solve the equation \(| 3 x - 2 | = x\).
Question 11:
AnswerMarks Guidance
\(3x-2 = x\)
\(3x-2 = x \Rightarrow 2x = 2 \Rightarrow x = 1\)B1 \(x=1\)
or \(-3x = x \Rightarrow 2 = 4x \Rightarrow x = \frac{1}{2}\)M1 A1
*or* \((3x-2)^2 = x^2\)
\(\Rightarrow 8x^2 - 12x + 4 = 0 \Rightarrow 2x^2 - 3x + 1 = 0\)M1 solving correct quadratic
\(\Rightarrow (x-1)(2x-1) = 0\)
\(\Rightarrow x = 1, \frac{1}{2}\)A1 A1
[3] 
## Question 11:

| $|3x-2| = x$ | | |
|---|---|---|
| $3x-2 = x \Rightarrow 2x = 2 \Rightarrow x = 1$ | B1 | $x=1$ |
| or $-3x = x \Rightarrow 2 = 4x \Rightarrow x = \frac{1}{2}$ | M1 A1 | |
| *or* $(3x-2)^2 = x^2$ | | |
| $\Rightarrow 8x^2 - 12x + 4 = 0 \Rightarrow 2x^2 - 3x + 1 = 0$ | M1 | solving correct quadratic |
| $\Rightarrow (x-1)(2x-1) = 0$ | | |
| $\Rightarrow x = 1, \frac{1}{2}$ | A1 A1 | |
| **[3]** | | |

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11 Solve the equation $| 3 x - 2 | = x$.

\hfill \mbox{\textit{OCR MEI C3  Q11 [3]}}
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