OCR C3 2010 June — Question 5 7 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2010
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| < |linear|
DifficultyStandard +0.8 This question requires systematic case analysis of absolute value inequalities (splitting into regions based on sign changes at x=-1/2 and x=3), then solving multiple linear inequalities and combining solution sets. Part (ii) adds an optimization element requiring understanding of the solution set's geometry. While methodical, it demands careful organization and goes beyond routine absolute value exercises.
Spec1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities

  1. Solve the inequality \(|2x + 1| \leqslant |x - 3|\). [5]
  2. Given that \(x\) satisfies the inequality \(|2x + 1| \leqslant |x - 3|\), find the greatest possible value of \(|x + 2|\). [2]

(i)
AnswerMarks Guidance
Attempt process for finding both critical valuesM1 squaring both sides to obtain 3 terms on each side or considering 2 different linear eqns/inequalities
Obtain \(-4\)A1
Obtain \(\frac{2}{3}\)A1
Attempt process for solving inequalityM1 table, sketch, ...; needs two critical values; implied by plausible answer
Obtain \(-4 \le x \le \frac{2}{3}\)A1 5 with \(\le\) and not \(<\)
(ii)
AnswerMarks Guidance
Use correct process to find value of \(\leftx + 2 \right \) using any value
Obtain \(2\frac{2}{3}\) or \(\frac{8}{3}\)A1 2 dependent on 5 marks awarded in part (i)
## (i)

Attempt process for finding both critical values | M1 | squaring both sides to obtain 3 terms on each side or considering 2 different linear eqns/inequalities

Obtain $-4$ | A1 | —
Obtain $\frac{2}{3}$ | A1 | —

Attempt process for solving inequality | M1 | table, sketch, ...; needs two critical values; implied by plausible answer

Obtain $-4 \le x \le \frac{2}{3}$ | A1 5 | with $\le$ and not $<$

## (ii)

Use correct process to find value of $\left| x + 2 \right|$ using any value | M1 | ... whether part of answer to (i) or not

Obtain $2\frac{2}{3}$ or $\frac{8}{3}$ | A1 2 | dependent on 5 marks awarded in part (i)

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\begin{enumerate}[label=(\roman*)]
\item Solve the inequality $|2x + 1| \leqslant |x - 3|$. [5]
\item Given that $x$ satisfies the inequality $|2x + 1| \leqslant |x - 3|$, find the greatest possible value of $|x + 2|$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR C3 2010 Q5 [7]}}