OCR MEI C3 — Question 7 3 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| < constant (pure inequality)
DifficultyEasy -1.2 This is a straightforward application of the basic modulus inequality definition requiring only the standard technique of rewriting |x - 1| < 3 as -3 < x - 1 < 3, then solving to get -2 < x < 4. It's a single-step routine exercise testing recall of a fundamental concept with no problem-solving or insight required, making it easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities
7 Solve the inequality \(| x - 1 | < 3\).
Question 7:
AnswerMarks Guidance
\(x-1 < 3 \Rightarrow -3 < x-1 < 3\)
\(\Rightarrow -2 < x < 4\)A1 \(-2 <\)
B1\(< 4\) (penalise \(\leq\) once only)
[3] 
## Question 7:

| $|x-1| < 3 \Rightarrow -3 < x-1 < 3$ | M1 | or $x-1 = \pm 3$, or squaring $\Rightarrow$ correct quadratic $\Rightarrow (x+2)(x-4)$ (condone factorising errors); or correct sketch showing $y=3$ to scale |
|---|---|---|
| $\Rightarrow -2 < x < 4$ | A1 | $-2 <$ |
| | B1 | $< 4$ (penalise $\leq$ once only) |
| **[3]** | | |

---
7 Solve the inequality $| x - 1 | < 3$.

\hfill \mbox{\textit{OCR MEI C3  Q7 [3]}}
This paper (12 questions)
View full paper
Q1 4 Q2 2 Q3 6 Q4 4 Q5 4 Q6 4 Q7 3 Q8 3 Q9 4 Q10 5 Q11 3 Q12 3