OCR MEI C4 2011 June — Question 5 1 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2011
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeEvaluate claims about large data set
DifficultyEasy -2.5 This is a 1-mark question requiring only basic understanding of sampling concepts—identifying that qualitative observations or existence claims are less affected by sample size than quantitative estimates. It requires minimal statistical knowledge and no calculation, making it significantly easier than typical A-level maths questions.

The survey described in the article was based on a small sample. State one conclusion which is unlikely to be influenced by the size of the sample. [1]

Answer: \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\)
\(\Rightarrow 1 + \cot^2\theta = 1 + 2\cot\theta\)
\(\Rightarrow \cot^2\theta - 2\cot\theta = 0\)
\(\Rightarrow \cot\theta(\cot\theta - 2) = 0\)
\(\Rightarrow \cot\theta = 0,\) and \(\cot\theta = 2, \tan\theta = \frac{1}{2}\)
\(\theta = 26.6°, -153.4°, -90°, 90°\)
...................................................................
OR \(\frac{1}{\sin^2\theta} = 1 + \frac{2\cos\theta}{\sin\theta} + \frac{2\cos\theta}{\sin\theta}\)
\(\Rightarrow \sin^2\theta + 2\sin\theta\cos\theta - 1 = 0\)
\(\Rightarrow 2\sin\theta\cos\theta - \cos^2\theta = 0\)
\(\Rightarrow \cos\theta(\sin\theta - \cos\theta) = 0\)
\(\Rightarrow \cos\theta = 0,\) and \(\tan\theta = \frac{1}{2}\)
\(\theta = 26.6°, -153.4°, -90°, 90°\)
AnswerMarks
M1correct trig identity used
M1factorising oe
M1both needed and \(\cot\theta = 1/\tan\theta\) soi
B3,2,1,0accept 27°, -153° as above
(omission of \(\cot\theta=0\) could gain M1, M1, M0, B1) ...........................................
M1correct trig equivalents and a one line equation (or common denominator) formed
M1use of Pythagoras and factorising
M1accept 27°, -153° as above
B3,2,1,0
in both cases, -1 if extra solutions in the range are given ( dependent on at least B1 being scored)-not their incorrect solutions eg 26.6°,-153.4°, 0°, 180°, -180° would obtain B1 to lose both of these, at least B2 would need to be scored.
Total: [6]
**Answer:** $\operatorname{cosec}^2\theta = 1 + \cot^2\theta$

$\Rightarrow 1 + \cot^2\theta = 1 + 2\cot\theta$

$\Rightarrow \cot^2\theta - 2\cot\theta = 0$

$\Rightarrow \cot\theta(\cot\theta - 2) = 0$

$\Rightarrow \cot\theta = 0,$ **and** $\cot\theta = 2, \tan\theta = \frac{1}{2}$

$\theta = 26.6°, -153.4°, -90°, 90°$ 
...................................................................

**OR** $\frac{1}{\sin^2\theta} = 1 + \frac{2\cos\theta}{\sin\theta} + \frac{2\cos\theta}{\sin\theta}$

$\Rightarrow \sin^2\theta + 2\sin\theta\cos\theta - 1 = 0$

$\Rightarrow 2\sin\theta\cos\theta - \cos^2\theta = 0$

$\Rightarrow \cos\theta(\sin\theta - \cos\theta) = 0$

$\Rightarrow \cos\theta = 0,$ **and** $\tan\theta = \frac{1}{2}$

$\theta = 26.6°, -153.4°, -90°, 90°$

| M1 | correct trig identity used |
| M1 | factorising oe |
| M1 | both needed and $\cot\theta = 1/\tan\theta$ soi |
| B3,2,1,0 | accept 27°, -153° as above |
| | (omission of $\cot\theta=0$ could gain M1, M1, M0, B1) ........................................... |
| M1 | correct trig equivalents and a one line equation (or common denominator) formed |
| M1 | use of Pythagoras and factorising |
| M1 | accept 27°, -153° as above |
| B3,2,1,0 | |
| | **in both cases, -1 if extra solutions in the range are given ( dependent on at least B1 being scored)-not their incorrect solutions eg 26.6°,-153.4°, 0°, 180°, -180° would obtain B1 to lose both of these, at least B2 would need to be scored.** |

**Total: [6]**

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The survey described in the article was based on a small sample.

State one conclusion which is unlikely to be influenced by the size of the sample. [1]

\hfill \mbox{\textit{OCR MEI C4 2011 Q5 [1]}}