OCR C4 — Question 5 6 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeFind value where max/min occurs
DifficultyStandard +0.3 This is a standard harmonic form question requiring routine application of R sin(x-α) = R sin x cos α - R cos x sin α, comparing coefficients to find R and α, then using knowledge that sin has maximum value 1. The working is methodical with no novel insight required, making it slightly easier than average.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc
5 Express \(\sqrt { 3 } \sin x - \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\). Express \(\alpha\) in the form \(k \pi\). Find the exact coordinates of the maximum point of the curve \(y = \sqrt { 3 } \sin x - \cos x\) for which \(0 < x < 2 \pi\).
Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sqrt{3}\sin x - \cos x = R\sin(x-\alpha) = R(\sin x\cos\alpha - \cos x\sin\alpha)\)
\(\Rightarrow \sqrt{3} = R\cos\alpha,\ 1 = R\sin\alpha\)M1 correct pairs soi
\(\Rightarrow R^2 = 3+1 = 4 \Rightarrow R = 2\)B1 \(R = 2\)
\(\tan\alpha = \frac{1}{\sqrt{3}} \Rightarrow \alpha = \frac{\pi}{6}\)M1 ft
\(\Rightarrow y = 2\sin(x - \frac{\pi}{6})\)A1 cao www
Max when \(x - \frac{\pi}{6} = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}\)B1 cao; ft their \(R\)
max value \(y = 2\)B1
So maximum is \(\left(\frac{2\pi}{3},\ 2\right)\)[6] SC B1 \((2,\ \frac{2\pi}{3})\) no working 
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{3}\sin x - \cos x = R\sin(x-\alpha) = R(\sin x\cos\alpha - \cos x\sin\alpha)$ | | |
| $\Rightarrow \sqrt{3} = R\cos\alpha,\ 1 = R\sin\alpha$ | M1 | correct pairs soi |
| $\Rightarrow R^2 = 3+1 = 4 \Rightarrow R = 2$ | B1 | $R = 2$ |
| $\tan\alpha = \frac{1}{\sqrt{3}} \Rightarrow \alpha = \frac{\pi}{6}$ | M1 | ft |
| $\Rightarrow y = 2\sin(x - \frac{\pi}{6})$ | A1 | cao www |
| Max when $x - \frac{\pi}{6} = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}$ | B1 | cao; ft their $R$ |
| max value $y = 2$ | B1 | |
| So maximum is $\left(\frac{2\pi}{3},\ 2\right)$ | [6] | SC B1 $(2,\ \frac{2\pi}{3})$ no working |

---
5 Express $\sqrt { 3 } \sin x - \cos x$ in the form $R \sin ( x - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$. Express $\alpha$ in the form $k \pi$.

Find the exact coordinates of the maximum point of the curve $y = \sqrt { 3 } \sin x - \cos x$ for which $0 < x < 2 \pi$.

\hfill \mbox{\textit{OCR C4  Q5 [6]}}
This paper (7 questions)
View full paper
Q1 6 Q2 7 Q3 19 Q4 7 Q5 6 Q6 7 Q7 6