8. (a) Express $2 ext{cos}3x - 3 ext{sin}3x$ in the form $R ext{cos}(3x + u)$, where $R$ and $\nu$ are constants, $R > 0$ and $0 < \nu < \frac{\pi}{2}$. Give your answers to 3 significant figures. $f(x) = e^{2x} ext{cos}3x$ (b) Show that $f'(x)$ can be written in the form $f'(x) = R e^{2x} ext{cos}(3x + \alpha)$ where $R$ and $\alpha$ are the constants found in part (a). (c) Hence, or otherwise, find the smallest positive value of $x$ for which the curve with equation $y = f(x)$ has a turning point.

A-Level Edexcel Maths Pure Rational Expressions: 8. (a) Express $2 ext{cos}3x

8. (a) Express $2 ext{cos}3x - 3 ext{sin}3x$ in the form $R ext{cos}(3x + u)$, where $R$ and $\nu$ are constants, $R > 0$ and $0 < \nu < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 4

Question 2

$8.-(a)-Express-$2-ext{cos}3x---3-ext{sin}3x$-in-the-form-$R-ext{cos}(3x-+--u)$,-where-$R$-and-$\nu$-are-constants,-$R->-0$-and-$0-<-\nu-<-\frac{\pi}{2}$-Edexcel-A-Level Maths Pure-Question 2-2011-Paper 4.png$

8. (a) Express $2 ext{cos}3x - 3 ext{sin}3x$ in the form $R ext{cos}(3x + u)$, where $R$ and $\nu$ are constants, $R > 0$ and $0 < \nu < \frac{\pi}{2}$. Give your a... show full transcript

Worked Solution & Example Answer:8. (a) Express $2 ext{cos}3x - 3 ext{sin}3x$ in the form $R ext{cos}(3x + u)$, where $R$ and $\nu$ are constants, $R > 0$ and $0 < \nu < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 4

Step 1

Express $2\text{cos}3x - 3\text{sin}3x$ in the form $R\text{cos}(3x + \alpha)$

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Answer

To express $2\text{cos}3x - 3\text{sin}3x$ in the form $R\text{cos}(3x + \alpha)$ , we start by defining:

Calculate $R$ : $R^2 = 2^2 + (-3)^2 = 4 + 9 = 13$ Thus, $R = \sqrt{13} \approx 3.61$ .
Calculate $\alpha$ using:
$\alpha = \tan^{-1}\left(\frac{-3}{2}\right) \approx -0.983$$ Therefore, we convert $\alpha$ to the positive equivalent in the range $0 < \alpha < \frac{\pi}{2}$ so we get $\alpha \approx 2.158$ radians.$

Step 2

Show that $f'(x)$ can be written in the form $f'(x) = R e^{2x} \text{cos}(3x + \alpha)$

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Answer

To show that $f'(x)$ can be expressed in the specified form, we derive:

The derivative: $f'(x) = e^{2x}(-3\text{sin}3x) + 2e^{2x}\text{cos}3x = e^{2x}(2\text{cos}3x - 3\text{sin}3x)$
Substitute the expression from part (a): $f'(x) = e^{2x}(R\text{cos}(3x + \alpha))$ Thus: $f'(x) = Re^{2x}\text{cos}(3x + \alpha)$

Step 3

Find the smallest positive value of $x$ for which the curve has a turning point

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Answer

To find the turning points, set $f'(x) = 0$ :

From part (b): $R\text{cos}(3x + \alpha) = 0$
Solve for $3x + \alpha = \frac{\pi}{2}$ , which gives: $3x = \frac{\pi}{2} - \alpha$
Substitute $\alpha \approx 2.158$ : $3x = \frac{\pi}{2} - 2.158 \approx 1.413$
Therefore, solving for $x$ yields: $x = \frac{1.413}{3} \approx 0.196$ Hence the smallest positive value of $x$ is approximately $0.20$ .

8. (a) Express $2 ext{cos}3x - 3 ext{sin}3x$ in the form $R ext{cos}(3x + u)$, where $R$ and $\nu$ are constants, $R > 0$ and $0 < \nu < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 4

Worked Solution & Example Answer:8. (a) Express $2 ext{cos}3x - 3 ext{sin}3x$ in the form $R ext{cos}(3x + u)$, where $R$ and $\nu$ are constants, $R > 0$ and $0 < \nu < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 4

Express $2\text{cos}3x - 3\text{sin}3x$ in the form $R\text{cos}(3x + \alpha)$

Show that $f'(x)$ can be written in the form $f'(x) = R e^{2x} \text{cos}(3x + \alpha)$

Find the smallest positive value of $x$ for which the curve has a turning point

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