OCR MEI Paper 3 Specimen — Question 9 7 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
SessionSpecimen
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeRange of rational function with harmonic denominator
DifficultyStandard +0.3 This is a standard harmonic form question with a straightforward application to find a constant. Part (a) is routine bookwork (expressing in R cos(θ-α) form), and part (b) requires recognizing that max of 1/(k+R cos(θ-α)) occurs when the denominator is minimized, leading to a simple algebraic equation. The algebra with surds is slightly involved but follows a clear method with no novel insight required.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc

9
  1. Express \(\cos \theta + 2 \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(0 < \alpha < \frac { 1 } { 2 } \pi\) and \(R\) is positive and given in exact form. The function \(\mathrm { f } ( \theta )\) is defined by \(\mathrm { f } ( \theta ) = \frac { 1 } { ( k + \cos \theta + 2 \sin \theta ) } , 0 \leq \theta \leq 2 \pi , k\) is a constant.
  2. The maximum value of \(\mathrm { f } ( \theta )\) is \(\frac { ( 3 + \sqrt { 5 } ) } { 4 }\). Find the value of \(k\).

Question 9(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(\cos\theta + 2\sin\theta \equiv R\cos(\theta - \alpha)\)M1 1.1a
\(\Rightarrow R\cos\alpha = 1,\ R\sin\alpha = 2\)
\(\Rightarrow R^2 = 5,\ R = \sqrt{5}\)B1 1.1
\(\tan\alpha = 2,\ \alpha = 1.107\)M1, A1 1.1, 1.1
Question 9(b):
AnswerMarks Guidance
AnswerMarks Guidance
max value is \(\frac{1}{(k - \sqrt{5})}\)M1 3.1a
\(\frac{1}{(k-\sqrt{5})} = \frac{(3+\sqrt{5})}{4}\)M1 1.1
\(4 = 3k - 5 + k\sqrt{5} - 3\sqrt{5}\)
[This is independent of \(\sqrt{5}\) so] \(k = 3\)A1 1.1
## Question 9(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta + 2\sin\theta \equiv R\cos(\theta - \alpha)$ | M1 | 1.1a |
| $\Rightarrow R\cos\alpha = 1,\ R\sin\alpha = 2$ | | |
| $\Rightarrow R^2 = 5,\ R = \sqrt{5}$ | B1 | 1.1 |
| $\tan\alpha = 2,\ \alpha = 1.107$ | M1, A1 | 1.1, 1.1 | **[4]** |

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## Question 9(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| max value is $\frac{1}{(k - \sqrt{5})}$ | M1 | 3.1a |
| $\frac{1}{(k-\sqrt{5})} = \frac{(3+\sqrt{5})}{4}$ | M1 | 1.1 |
| $4 = 3k - 5 + k\sqrt{5} - 3\sqrt{5}$ | | |
| [This is independent of $\sqrt{5}$ so] $k = 3$ | A1 | 1.1 | **[3]** |

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9
\begin{enumerate}[label=(\alph*)]
\item Express $\cos \theta + 2 \sin \theta$ in the form $R \cos ( \theta - \alpha )$, where $0 < \alpha < \frac { 1 } { 2 } \pi$ and $R$ is positive and given in exact form.

The function $\mathrm { f } ( \theta )$ is defined by $\mathrm { f } ( \theta ) = \frac { 1 } { ( k + \cos \theta + 2 \sin \theta ) } , 0 \leq \theta \leq 2 \pi , k$ is a constant.
\item The maximum value of $\mathrm { f } ( \theta )$ is $\frac { ( 3 + \sqrt { 5 } ) } { 4 }$.

Find the value of $k$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Paper 3  Q9 [7]}}