OCR MEI C3 — Question 9 4 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| < constant (pure inequality)
DifficultyEasy -1.2 This is a straightforward application of the standard method for solving modulus inequalities with a linear expression. It requires only the routine technique of splitting into -3 ≤ 2x - 1 ≤ 3 and solving for x, which is a basic skill tested early in C3 with minimal steps.
Spec1.02l Modulus function: notation, relations, equations and inequalities

9 Solve the inequality \(| 2 x - 1 | \leqslant 3\).

Question 9:
AnswerMarks Guidance
\(2x-1 \leq 3\)
\(\Rightarrow -3 \leq 2x-1 \leq 3\)
\(\Rightarrow -2 \leq 2x \leq 4\)
\(\Rightarrow -1 \leq x \leq 2\)
\(2x-1 \leq 3\) (or \(=\))M1
\(x \leq 2\)A1
\(2x-1 \geq -3\) (or \(=\))M1
\(x \geq -1\)A1
*or* \((2x-1)^2 \leq 9\)M1 squaring and forming quadratic \(= 0\) (or \(\leq\))
\(\Rightarrow 4x^2 - 4x - 8 \leq 0\)
\(\Rightarrow (4)(x+1)(x-2) \leq 0\)A1 factorising or solving to get \(x = -1, 2\)
\(\Rightarrow -1 \leq x \leq 2\)A1 \(x \geq -1\)
A1\(x \leq 2\) (www)
[4]
## Question 9:

| $|2x-1| \leq 3$ | | |
|---|---|---|
| $\Rightarrow -3 \leq 2x-1 \leq 3$ | | |
| $\Rightarrow -2 \leq 2x \leq 4$ | | |
| $\Rightarrow -1 \leq x \leq 2$ | | |
| $2x-1 \leq 3$ (or $=$) | M1 | |
| $x \leq 2$ | A1 | |
| $2x-1 \geq -3$ (or $=$) | M1 | |
| $x \geq -1$ | A1 | |
| *or* $(2x-1)^2 \leq 9$ | M1 | squaring and forming quadratic $= 0$ (or $\leq$) |
| $\Rightarrow 4x^2 - 4x - 8 \leq 0$ | | |
| $\Rightarrow (4)(x+1)(x-2) \leq 0$ | A1 | factorising or solving to get $x = -1, 2$ |
| $\Rightarrow -1 \leq x \leq 2$ | A1 | $x \geq -1$ |
| | A1 | $x \leq 2$ (www) |
| **[4]** | | |

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9 Solve the inequality $| 2 x - 1 | \leqslant 3$.

\hfill \mbox{\textit{OCR MEI C3  Q9 [4]}}