OCR MEI C4 — Question 1 8 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeFind maximum or minimum value
DifficultyStandard +0.3 This is a standard harmonic form question requiring routine application of the R cos(x - α) transformation formula and solving for parameters given a maximum value. The steps are well-practiced in C4: use R² = 1 + λ² and tan α = λ, then apply the condition that max = R = 2. No novel insight required, just methodical application of textbook techniques.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc
1 You are given that \(\mathrm { f } ( x ) = \cos x + \lambda \sin x\) where \(\lambda\) is a positive constant.
  1. Express \(\mathrm { f } ( x )\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving \(R\) and \(\alpha\) in terms of \(\lambda\).
  2. Given that the maximum value (as \(x\) varies) of \(\mathrm { f } ( x )\) is 2 , find \(R , \lambda\) and \(\alpha\), giving your answers in exact form.
Question 1
(i)
\(\cos x + \sin x = R \cos(x - \alpha)\)
\(= R \cos x \cos \alpha + R \sin x \sin \alpha\)
M1 — Correct pairs. Condone sign error (so accept \(R\sin\alpha\))
\(R \cos \alpha = 1\), \(R \sin \alpha = 1\)
B1
\(R^2 = 1 + 1\), \(R = \sqrt{1 + 1}\)
M1 — Positive square root only - isw. Accept \(R = 1/\cos(\arctan(1))\) or \(R = 1/\sin(\arctan(1))\)
\(\tan \alpha = 1\) (or equivalent)
A1 — Follow through their pairs. \(\tan \alpha = 1\) with no working implies both M marks. However, \(\cos \alpha = 1\), \(\sin \alpha = 1\), \(\tan \alpha = 1\) scores M0M1. First two M marks may be implied by combining one of the pairs with \(R\), e.g., \(\frac{1}{\cos \alpha}\) or \(\frac{1}{\sin \alpha}\)
\(\alpha = \arccos(1)\), \(\arcsin(1)\) — Accept embedded answers, e.g., \(\sqrt{1 + 1}\) \(\cos(x - \arctan(1))\) for full marks
[4]
(ii)
max is \(R\) so \(R = 2\)
B1
\(1 + 2 = 4\)
\(2 = \sqrt{3}\)
M1 — using their \(R\) from (i), A0 for \(3\) as final answer
\(\alpha = \arctan \sqrt{3} = \pi/3\)
B1 — M1 for using their \((1/2)R_{\text{max}}\), www (e.g., \(2 = 1\) and \(\cos(1) = 1/3\) is B0). Exact answers only for final A and B marks
[4]
# Question 1

## (i)

$\cos x + \sin x = R \cos(x - \alpha)$

$= R \cos x \cos \alpha + R \sin x \sin \alpha$

M1 — Correct pairs. Condone sign error (so accept $R\sin\alpha$)

$R \cos \alpha = 1$, $R \sin \alpha = 1$

B1

$R^2 = 1 + 1$, $R = \sqrt{1 + 1}$

M1 — Positive square root only - isw. Accept $R = 1/\cos(\arctan(1))$ or $R = 1/\sin(\arctan(1))$

$\tan \alpha = 1$ (or equivalent)

A1 — Follow through their pairs. $\tan \alpha = 1$ with no working implies both M marks. However, $\cos \alpha = 1$, $\sin \alpha = 1$, $\tan \alpha = 1$ scores M0M1. First two M marks may be implied by combining one of the pairs with $R$, e.g., $\frac{1}{\cos \alpha}$ or $\frac{1}{\sin \alpha}$

$\alpha = \arccos(1)$, $\arcsin(1)$ — Accept embedded answers, e.g., $\sqrt{1 + 1}$ $\cos(x - \arctan(1))$ for full marks

[4]

## (ii)

max is $R$ so $R = 2$

B1

$1 + 2 = 4$

$2 = \sqrt{3}$

M1 — using their $R$ from (i), A0 for $3$ as final answer

$\alpha = \arctan \sqrt{3} = \pi/3$

B1 — M1 for using their $(1/2)R_{\text{max}}$, www (e.g., $2 = 1$ and $\cos(1) = 1/3$ is B0). Exact answers only for final A and B marks

[4]
1 You are given that $\mathrm { f } ( x ) = \cos x + \lambda \sin x$ where $\lambda$ is a positive constant.\\
(i) Express $\mathrm { f } ( x )$ in the form $R \cos ( x - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$, giving $R$ and $\alpha$ in terms of $\lambda$.\\
(ii) Given that the maximum value (as $x$ varies) of $\mathrm { f } ( x )$ is 2 , find $R , \lambda$ and $\alpha$, giving your answers in exact form.

\hfill \mbox{\textit{OCR MEI C4  Q1 [8]}}
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