OCR C3 2012 June — Question 1 5 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2012
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| > |linear|
DifficultyStandard +0.3 This requires squaring both sides to eliminate moduli, expanding and simplifying a quadratic inequality, then factorising to find critical values. While it involves multiple steps and careful algebraic manipulation, it's a standard C3 technique with no conceptual surprises—slightly above average due to the need to handle two moduli systematically.
Spec1.02l Modulus function: notation, relations, equations and inequalities

1 Solve the inequality \(| 2 x - 5 | > | x + 1 |\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
Attempt process for finding critical valuesM1 Squaring both sides, 2 linear eqns/ineqs. If using quadratic, need to go as far as factorising or substituting in formula for M1; if using two linear eqns or ineqs, signs of \(2x\) and \(x\) must be same in one, different in other
Obtain \(\frac{4}{3}\)A1
Obtain \(6\)A1
Attempt process for inequality involving two critical valuesM1 Sketch, table, …; implied by plausible solution
Obtain \(x < \frac{4}{3}\), \(x > 6\)A1 A0 for use of \(\leq\) and/or \(\geq\)
# Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt process for finding critical values | M1 | Squaring both sides, 2 linear eqns/ineqs. If using quadratic, need to go as far as factorising or substituting in formula for M1; if using two linear eqns or ineqs, signs of $2x$ and $x$ must be same in one, different in other |
| Obtain $\frac{4}{3}$ | A1 | |
| Obtain $6$ | A1 | |
| Attempt process for inequality involving two critical values | M1 | Sketch, table, …; implied by plausible solution |
| Obtain $x < \frac{4}{3}$, $x > 6$ | A1 | A0 for use of $\leq$ and/or $\geq$ |

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1 Solve the inequality $| 2 x - 5 | > | x + 1 |$.

\hfill \mbox{\textit{OCR C3 2012 Q1 [5]}}