Edexcel C3 2012 June — Question 3 9 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2012
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind equation of tangent
DifficultyStandard +0.3 This is a straightforward application of the product rule to find dy/dx, then solving dy/dx=0 for part (a) and using point-normal form for part (b). The exponential-trig combination requires careful differentiation but follows standard techniques with no novel insight needed. Slightly above average due to the algebraic manipulation involved.
Spec1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3fbdfb55-5dd5-44ab-b031-d39e64bdfc3b-04_538_953_251_532} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of the curve \(C\) which has equation $$y = \mathrm { e } ^ { x \sqrt { 3 } } \sin 3 x , \quad - \frac { \pi } { 3 } \leqslant x \leqslant \frac { \pi } { 3 }$$
  1. Find the \(x\) coordinate of the turning point \(P\) on \(C\), for which \(x > 0\) Give your answer as a multiple of \(\pi\).
  2. Find an equation of the normal to \(C\) at the point where \(x = 0\)
AnswerMarks Guidance
(a) \(\frac{dy}{dx} = \sqrt{3}e^{\sqrt{3}x} \sin 3x + 3e^{\sqrt{3}x} \cos 3x\)M1A1 Applies the product rule \(vu' + uv'\) to \(e^{\sqrt{3}x} \sin 3x\). If the rule is quoted it must be correct and there must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by terms written out as \(u = ..., u' = ..., v = ..., v' = ...\)) followed by their \(vu' + uv'\)), only accept answers of the form \(\frac{dy}{dx} = Ae^{\sqrt{3}x} \sin 3x + e^{\sqrt{3}x} \times \pm B \cos 3x\) where \(A, B > 0\)
\(\frac{dy}{dx} = 0 \Rightarrow e^{\sqrt{3}x}(\sqrt{3} \sin 3x + 3 \cos 3x) = 0\)M1 Sets their \(\frac{dy}{dx} = 0\), factorises out or divides by \(e^{\sqrt{3}x}\) producing an equation in \(\sin 3x\) and \(\cos 3x\)
\(\tan 3x = -\sqrt{3}\)A1 Achieves either \(\tan 3x = -\sqrt{3}\) or \(\tan 3x = -\frac{3}{\sqrt{3}}\)
\(3x = \frac{2\pi}{3} \Rightarrow x = \frac{2\pi}{9}\)M1A1 Correct order of arctan, followed by ÷3. Accept \(3x = \frac{5\pi}{3} \Rightarrow x = \frac{5\pi}{9}\) or \(3x = \frac{-\pi}{3} \Rightarrow x = \frac{-\pi}{9}\) but not \(x = \arctan(-\sqrt{3})\). CS0 \(x = \frac{2\pi}{9}\). Ignore extra solutions outside the range. Withhold mark for extra inside the range.
(6)(6)
(b) At \(x=0\): \(\frac{dy}{dx} = 3\)B1 Sight of 3 for the gradient
Equation of normal is \(-\frac{1}{3} = \frac{y-0}{x-0}\) or any equivalent: \(y = -\frac{1}{3}x\)M1A1 A full method for finding an equation of the normal. Their tangent gradient \(m\) must be modified to \(-\frac{1}{m}\) and used together with \((0,0)\). Eg \(-\frac{1}{their\ m}\), \(\frac{y-0}{x-0}\) or equivalent is acceptable.
(3)(9 marks)
Notes for (a): M1 applies the product rule; A1 correct expression for \(\frac{dy}{dx}\); M1 sets their \(\frac{dy}{dx} = 0\), factorises out or divides by \(e^{\sqrt{3}x}\); A1 achieves either \(\tan 3x = -\sqrt{3}\) or tan \(3x = -\frac{3}{\sqrt{3}}\); M1 correct order of arctan or arccos etc to produce a value of \(x\); A1 cao \(x = \frac{2\pi}{9}\).
Alternative in part (a) using the form \(R\sin(3x + \alpha)\)
AnswerMarks Guidance
(a) \(\frac{dy}{dx} = \sqrt{3}e^{\sqrt{3}x} \sin 3x + 3e^{\sqrt{3}x} \cos 3x\)M1A1
\(\frac{dy}{dx} = 0 \Rightarrow e^{\sqrt{3}x}(\sqrt{3} \sin 3x + 3 \cos 3x) = 0\)M1
\((\sqrt{12}) \sin(3x + \frac{\pi}{3}) = 0\)A1 Achieves either \((\sqrt{12})\sin(3x + \frac{\pi}{3}) = 0\) or \((\sqrt{12})\cos(3x - \frac{\pi}{6}) = 0\)
\(3x = \frac{2\pi}{3} \Rightarrow x = \frac{2\pi}{9}\)M1A1 Correct order of arcsin or arccos, etc to produce a value of \(x\). Eg accept \(3x + \frac{\pi}{3} = 0\) or \(\pi\) or \(2\pi \Rightarrow x = ...\).
(6)(6)
A1Achieves either \((\sqrt{12})\sin(3x + \frac{\pi}{3}) = 0\) or \((\sqrt{12})\cos(3x - \frac{\pi}{6}) = 0\)
M1Correct order of arcsin or arccos, etc to produce a value of \(x\)
Eg accept \(3x + \frac{\pi}{3} = 0\) or \(\pi\) or \(2\pi \Rightarrow x = ....\)
A1Cao \(x = \frac{2\pi}{9}\) Ignore extra solutions outside the range. Withhold mark for extra inside the range.
Alternative to part (a) squaring both sides
AnswerMarks
(a) \(\frac{dy}{dx} = \sqrt{3}e^{\sqrt{3}x} \sin 3x + 3e^{\sqrt{3}x} \cos 3x\)M1A1
\(\frac{dy}{dx} = 0 \Rightarrow e^{\sqrt{3}x}(\sqrt{3} \sin 3x + 3 \cos 3x) = 0\)M1
\(\sqrt{3} \sin 3x = -3\cos 3x \Rightarrow \cos^2(3x) = \frac{1}{4}\) or \(\sin^2(3x) = \frac{3}{4}\)A1
\(x = \frac{1}{3}\arccos(\pm \sqrt{\frac{1}{4}})\) oeM1
\(x = \frac{2\pi}{9}\)A1
(6)(6) 
| (a) $\frac{dy}{dx} = \sqrt{3}e^{\sqrt{3}x} \sin 3x + 3e^{\sqrt{3}x} \cos 3x$ | M1A1 | Applies the product rule $vu' + uv'$ to $e^{\sqrt{3}x} \sin 3x$. If the rule is quoted it must be correct and there must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by terms written out as $u = ..., u' = ..., v = ..., v' = ...$) followed by their $vu' + uv'$), only accept answers of the form $\frac{dy}{dx} = Ae^{\sqrt{3}x} \sin 3x + e^{\sqrt{3}x} \times \pm B \cos 3x$ where $A, B > 0$ |
|---|---|---|
| $\frac{dy}{dx} = 0 \Rightarrow e^{\sqrt{3}x}(\sqrt{3} \sin 3x + 3 \cos 3x) = 0$ | M1 | Sets their $\frac{dy}{dx} = 0$, factorises out or divides by $e^{\sqrt{3}x}$ producing an equation in $\sin 3x$ and $\cos 3x$ |
| $\tan 3x = -\sqrt{3}$ | A1 | Achieves either $\tan 3x = -\sqrt{3}$ or $\tan 3x = -\frac{3}{\sqrt{3}}$ |
| $3x = \frac{2\pi}{3} \Rightarrow x = \frac{2\pi}{9}$ | M1A1 | Correct order of arctan, followed by ÷3. Accept $3x = \frac{5\pi}{3} \Rightarrow x = \frac{5\pi}{9}$ or $3x = \frac{-\pi}{3} \Rightarrow x = \frac{-\pi}{9}$ but not $x = \arctan(-\sqrt{3})$. CS0 $x = \frac{2\pi}{9}$. Ignore extra solutions outside the range. Withhold mark for extra inside the range. |
| **(6)** | **(6)** | |
| (b) At $x=0$: $\frac{dy}{dx} = 3$ | B1 | Sight of 3 for the gradient |
| Equation of normal is $-\frac{1}{3} = \frac{y-0}{x-0}$ or any equivalent: $y = -\frac{1}{3}x$ | M1A1 | A full method for finding an equation of the normal. Their tangent gradient $m$ must be modified to $-\frac{1}{m}$ and used together with $(0,0)$. Eg $-\frac{1}{their\ m}$, $\frac{y-0}{x-0}$ or equivalent is acceptable. |
| **(3)** | **(9 marks)** | |

**Notes for (a):** M1 applies the product rule; A1 correct expression for $\frac{dy}{dx}$; M1 sets their $\frac{dy}{dx} = 0$, factorises out or divides by $e^{\sqrt{3}x}$; A1 achieves either $\tan 3x = -\sqrt{3}$ or tan $3x = -\frac{3}{\sqrt{3}}$; M1 correct order of arctan or arccos etc to produce a value of $x$; A1 cao $x = \frac{2\pi}{9}$.

---

# Alternative in part (a) using the form $R\sin(3x + \alpha)$

| (a) $\frac{dy}{dx} = \sqrt{3}e^{\sqrt{3}x} \sin 3x + 3e^{\sqrt{3}x} \cos 3x$ | M1A1 | |
|---|---|---|
| $\frac{dy}{dx} = 0 \Rightarrow e^{\sqrt{3}x}(\sqrt{3} \sin 3x + 3 \cos 3x) = 0$ | M1 | |
| $(\sqrt{12}) \sin(3x + \frac{\pi}{3}) = 0$ | A1 | Achieves either $(\sqrt{12})\sin(3x + \frac{\pi}{3}) = 0$ or $(\sqrt{12})\cos(3x - \frac{\pi}{6}) = 0$ |
| $3x = \frac{2\pi}{3} \Rightarrow x = \frac{2\pi}{9}$ | M1A1 | Correct order of arcsin or arccos, etc to produce a value of $x$. Eg accept $3x + \frac{\pi}{3} = 0$ or $\pi$ or $2\pi \Rightarrow x = ...$. |
| **(6)** | **(6)** | |

| A1 | Achieves either $(\sqrt{12})\sin(3x + \frac{\pi}{3}) = 0$ or $(\sqrt{12})\cos(3x - \frac{\pi}{6}) = 0$ |
|---|---|
| M1 | Correct order of arcsin or arccos, etc to produce a value of $x$ |
| Eg accept $3x + \frac{\pi}{3} = 0$ or $\pi$ or $2\pi \Rightarrow x = ....$ |
| A1 | Cao $x = \frac{2\pi}{9}$ Ignore extra solutions outside the range. Withhold mark for extra inside the range. |

---

# Alternative to part (a) squaring both sides

| (a) $\frac{dy}{dx} = \sqrt{3}e^{\sqrt{3}x} \sin 3x + 3e^{\sqrt{3}x} \cos 3x$ | M1A1 | |
|---|---|---|
| $\frac{dy}{dx} = 0 \Rightarrow e^{\sqrt{3}x}(\sqrt{3} \sin 3x + 3 \cos 3x) = 0$ | M1 | |
| $\sqrt{3} \sin 3x = -3\cos 3x \Rightarrow \cos^2(3x) = \frac{1}{4}$ or $\sin^2(3x) = \frac{3}{4}$ | A1 | |
| $x = \frac{1}{3}\arccos(\pm \sqrt{\frac{1}{4}})$ oe | M1 | |
| $x = \frac{2\pi}{9}$ | A1 | |
| **(6)** | **(6)** | |

---
3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3fbdfb55-5dd5-44ab-b031-d39e64bdfc3b-04_538_953_251_532}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a sketch of the curve $C$ which has equation

$$y = \mathrm { e } ^ { x \sqrt { 3 } } \sin 3 x , \quad - \frac { \pi } { 3 } \leqslant x \leqslant \frac { \pi } { 3 }$$
\begin{enumerate}[label=(\alph*)]
\item Find the $x$ coordinate of the turning point $P$ on $C$, for which $x > 0$ Give your answer as a multiple of $\pi$.
\item Find an equation of the normal to $C$ at the point where $x = 0$
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3 2012 Q3 [9]}}
This paper (8 questions)
View full paper
Q1 4 Q2 9 Q3 9 Q4 7 Q5 9 Q6 14 Q7 11 Q8 12