8. (a) Express 2 cos 3 x - 3 sin 3 x in the form R cos (3 x + α), where R and α are constants, R > 0 and 0 < α < π/2. Give your answers to 3 significant figures. f(x) = e^{2x} cos 3x (b) Show that f'(x) can be written in the form f'(x) = R e^{2x} cos (3 x + 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 Modelling with Functions: 8. (a) Express 2 cos 3 x

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

Question 2

8. (a) Express 2 cos 3 x - 3 sin 3 x in the form R cos (3 x + α), where R and α are constants, R > 0 and 0 < α < π/2. Give your answers to 3 significant figures. f(... show full transcript

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

Step 1

Express 2 cos 3 x - 3 sin 3 x in the form R cos (3 x + α)

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Answer

To express the function in the required form, we first calculate

$R^2 = 2^2 + (-3)^2 = 4 + 9 = 13$

thus,

oot{13}{3.61} $$ Next, we find the angle α using $$ an α = \frac{-3}{2} \implies α = \tan^{-1}(-1.5) $$ This gives us α ≈ 0.983 radians. Therefore, we have: $$ R cos(3x + α) = R \left( 2 cos(3x) - 3 sin(3x) \right) $$ The final expression is: $$ 2 cos 3x - 3 sin 3x = R cos(3x + α) $$ with R = 3.61 and α = 0.983.

Step 2

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

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Answer

To derive f'(x), we apply the product and chain rules:

$f'(x) = rac{d}{dx} (e^{2x} cos 3x) = e^{2x} (-3 sin 3x) + 2 e^{2x} (cos 3x)$

This simplifies to:

$f'(x) = e^{2x} (2 cos 3x - 3 sin 3x)$

Substituting the result from part (a), we find:

$f'(x) = R e^{2x} cos(3x + α)$

showing that R and α are constants found in part (a).

Step 3

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

Setting the derivative equal to zero:

$f'(x) = 0 \implies cos(3x + α) = 0$

From this, we identify:

$3x + α = \frac{π}{2} + kπ$

for k being any integer. Simplifying, we find:

$x = \frac{\frac{π}{2} - α + kπ}{3}$

Taking k = 0 gives:

$x = \frac{\frac{π}{2} - 0.983}{3} \approx 0.196$

Thus, for the smallest positive value:

$x \approx 0.196... awrt 0.20$ .

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

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

Express 2 cos 3 x - 3 sin 3 x in the form R cos (3 x + α)

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

Find the smallest positive value of x for which the curve with equation y = f(x) has a turning point

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