8. (a) Express 2cos3x − 3sin3x in the form Rcos(3x + α), where R and α are constants, R > 0 and 0 < α < π/2. Give your answers to 3 significant figures. (b) Show that f′(x) can be written in the form f′(x) = R2e2x cos(3x + α) where R and α 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 Applications of Differentiation: 8. (a) Express 2cos3x − 3sin3x i

8. (a) Express 2cos3x − 3sin3x in the form Rcos(3x + α), where R and α are constants, R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 3

Question 8

8. (a) Express 2cos3x − 3sin3x in the form Rcos(3x + α), where R and α are constants, R > 0 and 0 < α < π/2. Give your answers to 3 significant figures. (b) Show th... show full transcript

Worked Solution & Example Answer:8. (a) Express 2cos3x − 3sin3x in the form Rcos(3x + α), where R and α are constants, R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 3

Step 1

Express 2cos3x − 3sin3x in the form Rcos(3x + α)

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Answer

To express the function in the desired form, we use the identity: $R = \\sqrt{a^2 + b^2}$ where a = 2 and b = -3. Hence:

$R = \\sqrt{2^2 + (-3)^2} = \\sqrt{4 + 9} = \\sqrt{13}.$

Next, we find α using: $\tan(α) = \\frac{-3}{2}.$

This gives:

parg{\frac{-3}{2}} = 0.983...$$ Thus, the expression becomes: $$2cos3x - 3sin3x = Rcos(3x + α) = \\sqrt{13}cos(3x + 0.983).$$

Step 2

Show that f′(x) can be written in the form f′(x) = R2e2x cos(3x + α)

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Answer

We start with the given function:
$f(x) = e^{2x} (2cos3x - 3sin3x).$

Differentiating, we apply the product rule:

This can be rewritten using the value of R from part (a):
$f′(x) = R2e^{2x} cos(3x + α).$

Step 3

Hence, or otherwise, find the smallest positive value of x for which the curve with equation y = f(x) has a turning point.

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Answer

A turning point occurs when: $f′(x) = 0 \\Rightarrow cos(3x + α) = 0.$

Thus, we find: $3x + α = \\frac{π}{2} + nπ,$
where n is an integer. For the smallest positive value, use n = 0: $3x + 0.983 = \\frac{π}{2} \\Rightarrow 3x = \\frac{π}{2} - 0.983.$

Calculating this gives: $x = \\frac{\frac{π}{2} - 0.983}{3} ≈ 0.196....$

Thus, the smallest positive value of x is approximately: $x ≈ 0.20.$

8. (a) Express 2cos3x − 3sin3x in the form Rcos(3x + α), where R and α are constants, R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 3

Worked Solution & Example Answer:8. (a) Express 2cos3x − 3sin3x in the form Rcos(3x + α), where R and α are constants, R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 3

Express 2cos3x − 3sin3x in the form Rcos(3x + α)

Show that f′(x) can be written in the form f′(x) = R2e2x cos(3x + α)

Hence, or otherwise, find the smallest positive value of x for which the curve with equation y = f(x) has a turning point.

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