Edexcel C3 2016 June — Question 5 10 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2016
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeFind stationary points - mixed transcendental products
DifficultyStandard +0.3 Part (i) requires product rule with exponential and trig functions, then solving dy/dx=0 using a calculator - standard C3 technique. Part (ii) involves implicit differentiation with chain rule and trig identities to simplify to the required form - routine but requires careful algebraic manipulation. Both parts are typical C3 exercises with no novel problem-solving required.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation
5. (i) Find, using calculus, the \(x\) coordinate of the turning point of the curve with equation $$y = \mathrm { e } ^ { 3 x } \cos 4 x , \quad \frac { \pi } { 4 } \leqslant x < \frac { \pi } { 2 }$$ Give your answer to 4 decimal places.
(ii) Given \(x = \sin ^ { 2 } 2 y , \quad 0 < y < \frac { \pi } { 4 }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) as a function of \(y\). Write your answer in the form $$\frac { \mathrm { d } y } { \mathrm {~d} x } = p \operatorname { cosec } ( q y ) , \quad 0 < y < \frac { \pi } { 4 }$$ where \(p\) and \(q\) are constants to be determined. \includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-09_2258_47_315_37}
Question 5(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(y = e^{3x}\cos 4x \Rightarrow \frac{dy}{dx} = \cos 4x \times 3e^{3x} + e^{3x} \times -4\sin 4x\)M1A1 Use product rule \(uv' + vu'\); look for \(\frac{dy}{dx} = Ae^{3x}\cos 4x \pm Be^{3x}\sin 4x\), \(A,B \neq 0\)
Sets \(\cos 4x \times 3e^{3x} + e^{3x} \times -4\sin 4x = 0 \Rightarrow 3\cos 4x - 4\sin 4x = 0\)M1 Factorises/cancels by \(e^{3x}\) to form trig equation in \(\sin 4x\) and \(\cos 4x\) only
\(x = \frac{1}{4}\arctan\frac{3}{4}\)M1 Uses \(\frac{\sin 4x}{\cos 4x} = \tan 4x\), moves from \(\tan 4x = C\), \(C\neq 0\) using correct order of operations
\(x =\) awrt \(0.9463\) (4dp)A1 Ignore answers outside domain; withhold for additional answers inside domain
Question 5(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x = \sin^2 2y \Rightarrow \frac{dx}{dy} = 2\sin 2y \times 2\cos 2y\)M1A1 Use chain rule (or product rule) to achieve \(\pm P\sin 2y\cos 2y\) as derivative
Uses \(\sin 4y = 2\sin 2y\cos 2y\) in their expressionM1 May see \(4\sin 2y\cos 2y = 2\sin 4y\) stated
\(\frac{dx}{dy} = 2\sin 4y \Rightarrow \frac{dy}{dx} = \frac{1}{2\sin 4y} = \frac{1}{2}\text{cosec}\, 4y\)M1A1 Uses \(\frac{dy}{dx} = 1\big/\frac{dx}{dy}\); if \(\frac{dx}{dy} = a+b\) do not allow \(\frac{dy}{dx} = \frac{1}{a}+\frac{1}{b}\)
Alternative I:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x = \sin^2 2y \Rightarrow x = \frac{1}{2} - \frac{1}{2}\cos 4y\)2nd M1 From \(\cos 4y = \pm 1 \pm 2\sin^2 2y\)
\(\frac{dx}{dy} = 2\sin 4y\)1st M1 A1
\(\frac{dy}{dx} = \frac{1}{2\sin 4y} = \frac{1}{2}\text{cosec}\,4y\)M1A1
Alternative II:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^{\frac{1}{2}} = \sin 2y \Rightarrow \frac{1}{2}x^{-\frac{1}{2}} = 2\cos 2y\frac{dy}{dx}\)M1A1
Uses \(x^{\frac{1}{2}} = \sin 2y\) AND \(\sin 4y = 2\sin 2y\cos 2y\)M1
\(\frac{dy}{dx} = \frac{1}{2\sin 4y} = \frac{1}{2}\text{cosec}\,4y\)M1A1
Alternative III:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^{\frac{1}{2}} = \sin 2y \Rightarrow 2y = \text{invsin}\, x^{\frac{1}{2}} \Rightarrow 2\frac{dy}{dx} = \frac{1}{\sqrt{1-x}}\times\frac{1}{2}x^{-\frac{1}{2}}\)M1A1
Uses \(x^{\frac{1}{2}} = \sin 2y\), \(\sqrt{1-x} = \cos 2y\) and \(\sin 4y = 2\sin 2y\cos 2y\)M1
\(\frac{dy}{dx} = \frac{1}{2\sin 4y} = \frac{1}{2}\text{cosec}\,4y\)M1A1 
# Question 5(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = e^{3x}\cos 4x \Rightarrow \frac{dy}{dx} = \cos 4x \times 3e^{3x} + e^{3x} \times -4\sin 4x$ | M1A1 | Use product rule $uv' + vu'$; look for $\frac{dy}{dx} = Ae^{3x}\cos 4x \pm Be^{3x}\sin 4x$, $A,B \neq 0$ |
| Sets $\cos 4x \times 3e^{3x} + e^{3x} \times -4\sin 4x = 0 \Rightarrow 3\cos 4x - 4\sin 4x = 0$ | M1 | Factorises/cancels by $e^{3x}$ to form trig equation in $\sin 4x$ and $\cos 4x$ only |
| $x = \frac{1}{4}\arctan\frac{3}{4}$ | M1 | Uses $\frac{\sin 4x}{\cos 4x} = \tan 4x$, moves from $\tan 4x = C$, $C\neq 0$ using correct order of operations |
| $x =$ awrt $0.9463$ (4dp) | A1 | Ignore answers outside domain; withhold for additional answers inside domain |

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# Question 5(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \sin^2 2y \Rightarrow \frac{dx}{dy} = 2\sin 2y \times 2\cos 2y$ | M1A1 | Use chain rule (or product rule) to achieve $\pm P\sin 2y\cos 2y$ as derivative |
| Uses $\sin 4y = 2\sin 2y\cos 2y$ in their expression | M1 | May see $4\sin 2y\cos 2y = 2\sin 4y$ stated |
| $\frac{dx}{dy} = 2\sin 4y \Rightarrow \frac{dy}{dx} = \frac{1}{2\sin 4y} = \frac{1}{2}\text{cosec}\, 4y$ | M1A1 | Uses $\frac{dy}{dx} = 1\big/\frac{dx}{dy}$; if $\frac{dx}{dy} = a+b$ do not allow $\frac{dy}{dx} = \frac{1}{a}+\frac{1}{b}$ |

### Alternative I:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \sin^2 2y \Rightarrow x = \frac{1}{2} - \frac{1}{2}\cos 4y$ | 2nd M1 | From $\cos 4y = \pm 1 \pm 2\sin^2 2y$ |
| $\frac{dx}{dy} = 2\sin 4y$ | 1st M1 A1 | |
| $\frac{dy}{dx} = \frac{1}{2\sin 4y} = \frac{1}{2}\text{cosec}\,4y$ | M1A1 | |

### Alternative II:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^{\frac{1}{2}} = \sin 2y \Rightarrow \frac{1}{2}x^{-\frac{1}{2}} = 2\cos 2y\frac{dy}{dx}$ | M1A1 | |
| Uses $x^{\frac{1}{2}} = \sin 2y$ AND $\sin 4y = 2\sin 2y\cos 2y$ | M1 | |
| $\frac{dy}{dx} = \frac{1}{2\sin 4y} = \frac{1}{2}\text{cosec}\,4y$ | M1A1 | |

### Alternative III:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^{\frac{1}{2}} = \sin 2y \Rightarrow 2y = \text{invsin}\, x^{\frac{1}{2}} \Rightarrow 2\frac{dy}{dx} = \frac{1}{\sqrt{1-x}}\times\frac{1}{2}x^{-\frac{1}{2}}$ | M1A1 | |
| Uses $x^{\frac{1}{2}} = \sin 2y$, $\sqrt{1-x} = \cos 2y$ and $\sin 4y = 2\sin 2y\cos 2y$ | M1 | |
| $\frac{dy}{dx} = \frac{1}{2\sin 4y} = \frac{1}{2}\text{cosec}\,4y$ | M1A1 | |

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5. (i) Find, using calculus, the $x$ coordinate of the turning point of the curve with equation

$$y = \mathrm { e } ^ { 3 x } \cos 4 x , \quad \frac { \pi } { 4 } \leqslant x < \frac { \pi } { 2 }$$

Give your answer to 4 decimal places.\\
(ii) Given $x = \sin ^ { 2 } 2 y , \quad 0 < y < \frac { \pi } { 4 }$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ as a function of $y$.

Write your answer in the form

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = p \operatorname { cosec } ( q y ) , \quad 0 < y < \frac { \pi } { 4 }$$

where $p$ and $q$ are constants to be determined.

\includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-09_2258_47_315_37}\\

\hfill \mbox{\textit{Edexcel C3 2016 Q5 [10]}}
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