OCR C1 2008 June — Question 7 7 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Year2008
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeSolve quadratic inequality
DifficultyModerate -0.8 This is a straightforward two-part question testing basic inequality manipulation and factorising a simple quadratic. Part (i) requires routine algebraic manipulation, and part (ii) involves factorising y(y+2) and determining sign regions—both are standard C1 exercises requiring recall of techniques rather than problem-solving, making this easier than average.
Spec1.02g Inequalities: linear and quadratic in single variable

7 Solve the inequalities
  1. \(8 < 3 x - 2 < 11\),
  2. \(y ^ { 2 } + 2 y \geqslant 0\).

Question 7(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(8 < 3x - 2 < 11\)M1 2 equations or inequalities both dealing with all 3 terms resulting in \(a < kx < b\)
\(10 < 3x < 13\)A1 10 and 13 seen
\(\frac{10}{3} < x < \frac{13}{3}\)A1
Question 7(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x(x+2) \geq 0\)M1 Correct method to solve a quadratic
\(0, -2\)A1
Correct method to solve inequalityM1
\(x \geq 0,\ x \leq -2\)A1
## Question 7(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $8 < 3x - 2 < 11$ | M1 | 2 equations or inequalities both dealing with all 3 terms resulting in $a < kx < b$ |
| $10 < 3x < 13$ | A1 | 10 and 13 seen |
| $\frac{10}{3} < x < \frac{13}{3}$ | A1 | |

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## Question 7(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x(x+2) \geq 0$ | M1 | Correct method to solve a quadratic |
| $0, -2$ | A1 | |
| Correct method to solve inequality | M1 | |
| $x \geq 0,\ x \leq -2$ | A1 | |

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7 Solve the inequalities\\
(i) $8 < 3 x - 2 < 11$,\\
(ii) $y ^ { 2 } + 2 y \geqslant 0$.

\hfill \mbox{\textit{OCR C1 2008 Q7 [7]}}