Edexcel P3 2023 January — Question 2 6 marks

Exam BoardEdexcel
ModuleP3 (Pure Mathematics 3)
Year2023
SessionJanuary
Marks6
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 R cos(x - α) = R cos α cos x + R sin α sin x, followed by straightforward substitution to find maximum values. The multi-part structure and need for exact/decimal answers adds slight complexity, but the techniques are textbook exercises with no novel problem-solving required.
Spec1.02w Graph transformations: simple transformations of f(x)1.05f Trigonometric function graphs: symmetries and periodicities1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc

2. $$f ( x ) = \cos x + 2 \sin x$$
  1. Express \(\mathrm { f } ( x )\) in the form \(R \cos ( x - \alpha )\), where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\) Give the exact value of \(R\) and give the value of \(\alpha\), in radians, to 3 decimal places. $$g ( x ) = 3 - 7 f ( 2 x )$$
  2. Using the answer to part (a),
    1. write down the exact maximum value of \(\mathrm { g } ( x )\),
    2. find the smallest positive value of \(x\) for which this maximum value occurs, giving your answer to 2 decimal places.

Question 2:
Part (a)
AnswerMarks Guidance
AnswerMark Guidance
\(R = \sqrt{5}\)B1 Correct exact value. Condone \(R = \pm\sqrt{5}\). isw after correct answer e.g. \(R = \sqrt{5} = 2.24\)
\(\tan\alpha = \frac{2}{1} \Rightarrow \alpha = \ldots\)M1 Allow: \(\tan\alpha = \pm\frac{2}{1}\), \(\tan\alpha = \pm\frac{1}{2}\), \(\cos\alpha = \pm\frac{1}{"R"}\), \(\sin\alpha = \pm\frac{2}{"R"}\) leading to a value for \(\alpha\). If no method shown, imply by sight of awrt 1.1 rads or awrt \(63°\)
\(\alpha = 1.107\)A1 awrt 1.107
Part (b)(i)
AnswerMarks Guidance
AnswerMark Guidance
\(\text{Max} = 3 + 7\sqrt{5}\)B1ft Award for \(3 + 7\times\) their \(R\) where \(R > 0\). Follow through on decimal answers from (a). Condone solutions such as \(3 - 7\times-\sqrt{5} = 18.65\)
Part (b)(ii)
AnswerMarks Guidance
AnswerMark Guidance
\((2x - \text{"1.107"}) = \pi \Rightarrow x = \ldots\)M1 Attempt to solve \((2x \pm\text{"1.107"}) = \pi\) or \((2x \pm\text{"1.107"}) = -\pi\). May be implied by awrt 2.12. Condone bracketing slip \(\cos(2(x\pm\text{"1.107"}))=-1\). Also condone attempt to solve \((2x\pm\text{"63°"})=180°\) but not mixed units
\(x = \frac{\pi + \text{"1.107"}}{2} = 2.12\)A1 awrt 2.12. Cannot be given in a list; 2.12 must be selected
# Question 2:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = \sqrt{5}$ | B1 | Correct exact value. Condone $R = \pm\sqrt{5}$. isw after correct answer e.g. $R = \sqrt{5} = 2.24$ |
| $\tan\alpha = \frac{2}{1} \Rightarrow \alpha = \ldots$ | M1 | Allow: $\tan\alpha = \pm\frac{2}{1}$, $\tan\alpha = \pm\frac{1}{2}$, $\cos\alpha = \pm\frac{1}{"R"}$, $\sin\alpha = \pm\frac{2}{"R"}$ leading to a value for $\alpha$. If no method shown, imply by sight of awrt 1.1 rads or awrt $63°$ |
| $\alpha = 1.107$ | A1 | awrt 1.107 |

## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Max} = 3 + 7\sqrt{5}$ | B1ft | Award for $3 + 7\times$ their $R$ where $R > 0$. Follow through on decimal answers from (a). Condone solutions such as $3 - 7\times-\sqrt{5} = 18.65$ |

## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(2x - \text{"1.107"}) = \pi \Rightarrow x = \ldots$ | M1 | Attempt to solve $(2x \pm\text{"1.107"}) = \pi$ or $(2x \pm\text{"1.107"}) = -\pi$. May be implied by awrt 2.12. Condone bracketing slip $\cos(2(x\pm\text{"1.107"}))=-1$. Also condone attempt to solve $(2x\pm\text{"63°"})=180°$ but not mixed units |
| $x = \frac{\pi + \text{"1.107"}}{2} = 2.12$ | A1 | awrt 2.12. Cannot be given in a list; 2.12 must be selected |

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2.

$$f ( x ) = \cos x + 2 \sin x$$
\begin{enumerate}[label=(\alph*)]
\item Express $\mathrm { f } ( x )$ in the form $R \cos ( x - \alpha )$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$\\
Give the exact value of $R$ and give the value of $\alpha$, in radians, to 3 decimal places.

$$g ( x ) = 3 - 7 f ( 2 x )$$
\item Using the answer to part (a),
\begin{enumerate}[label=(\roman*)]
\item write down the exact maximum value of $\mathrm { g } ( x )$,
\item find the smallest positive value of $x$ for which this maximum value occurs, giving your answer to 2 decimal places.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Edexcel P3 2023 Q2 [6]}}