AQA C4 2011 January — Question 1 6 marks

Exam BoardAQA
ModuleC4 (Core Mathematics 4)
Year2011
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeFind value where max/min occurs
DifficultyModerate -0.3 This is a standard harmonic form question requiring routine application of the R sin(x+α) formula. Part (a) involves straightforward use of R=√(a²+b²) and tan α=b/a, while parts (b)(i) and (b)(ii) follow directly from understanding that the maximum of R sin(x+α) is R, occurring when sin(x+α)=1. This is a textbook exercise with no novel problem-solving required, making it slightly easier than average.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc

1
  1. Express \(2 \sin x + 5 \cos x\) in the form \(R \sin ( x + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
    1. Write down the maximum value of \(2 \sin x + 5 \cos x\).
    2. Find the value of \(x\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\) at which this maximum occurs, giving the value of \(x\) to the nearest \(0.1 ^ { \circ }\).

Question 1:
Part (a)
AnswerMarks Guidance
\(R\sin(x+\alpha) = R\sin x\cos\alpha + R\cos x\sin\alpha\)M1 Expanding and comparing coefficients
\(R\cos\alpha = 2\), \(R\sin\alpha = 5\)
\(R = \sqrt{4+25} = \sqrt{29}\)A1
\(\alpha = \arctan\left(\frac{5}{2}\right) = 68.2°\)A1 Accept 68.2°
Part (b)(i)
AnswerMarks Guidance
Maximum value \(= \sqrt{29}\)B1 ft their \(R\)
Part (b)(ii)
AnswerMarks Guidance
Maximum when \(\sin(x + 68.2°) = 1\), so \(x + 68.2° = 90°\)M1
\(x = 21.8°\)A1 ft their \(\alpha\)
# Question 1:

## Part (a)
| $R\sin(x+\alpha) = R\sin x\cos\alpha + R\cos x\sin\alpha$ | M1 | Expanding and comparing coefficients |
|---|---|---|
| $R\cos\alpha = 2$, $R\sin\alpha = 5$ | | |
| $R = \sqrt{4+25} = \sqrt{29}$ | A1 | |
| $\alpha = \arctan\left(\frac{5}{2}\right) = 68.2°$ | A1 | Accept 68.2° |

## Part (b)(i)
| Maximum value $= \sqrt{29}$ | B1 | ft their $R$ |
|---|---|---|

## Part (b)(ii)
| Maximum when $\sin(x + 68.2°) = 1$, so $x + 68.2° = 90°$ | M1 | |
|---|---|---|
| $x = 21.8°$ | A1 | ft their $\alpha$ |

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1
\begin{enumerate}[label=(\alph*)]
\item Express $2 \sin x + 5 \cos x$ in the form $R \sin ( x + \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$. Give your value of $\alpha$ to the nearest $0.1 ^ { \circ }$.
\item \begin{enumerate}[label=(\roman*)]
\item Write down the maximum value of $2 \sin x + 5 \cos x$.
\item Find the value of $x$ in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$ at which this maximum occurs, giving the value of $x$ to the nearest $0.1 ^ { \circ }$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C4 2011 Q1 [6]}}