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Show that the x coordinates of the turning points of the curve with equation y = f(x) satisfy the equation tan x = 4 Figure 3 shows a sketch of part of the curve with equation y = f(x) - Edexcel - A-Level Maths Pure - Question 13 - 2019 - Paper 2

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Question 13

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Show that the x coordinates of the turning points of the curve with equation y = f(x) satisfy the equation tan x = 4 Figure 3 shows a sketch of part of the curve wi... show full transcript

Worked Solution & Example Answer:Show that the x coordinates of the turning points of the curve with equation y = f(x) satisfy the equation tan x = 4 Figure 3 shows a sketch of part of the curve with equation y = f(x) - Edexcel - A-Level Maths Pure - Question 13 - 2019 - Paper 2

Step 1

Show that the x coordinates of the turning points of the curve with equation y = f(x) satisfy the equation tan x = 4

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Answer

To find the turning points of the function, we first need to differentiate it:

f(x)=2.5e0.25xsin(x)+10e0.25xcos(x)f'(x) = -2.5e^{-0.25x}sin(x) + 10e^{-0.25x}cos(x)

Setting the derivative to zero gives us:

2.5e0.25xsin(x)+10e0.25xcos(x)=0-2.5e^{-0.25x}sin(x) + 10e^{-0.25x}cos(x) = 0

Dividing through by e0.25xe^{-0.25x} (which is never zero), we simplify this to:

10cos(x)2.5sin(x)=010cos(x) - 2.5sin(x) = 0

Rearranging, we get:

rac{sin(x)}{cos(x)} = rac{10}{2.5}

Thus:

tan(x)=4tan(x) = 4

This shows that the x coordinates of the turning points of the curve must satisfy the equation tan(x)=4tan(x) = 4.

Step 2

Sketch the graph of H against τ where H(τ) = |10e^{-0.25τ} sin(τ)| showing the long-term behaviour of this curve.

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Answer

To sketch the graph of H(τ)=10e0.25τsin(τ)H(τ) = |10e^{-0.25τ} sin(τ)|, we note that:

  • As ττ increases, e0.25τe^{-0.25τ} decreases, resulting in a decay of the amplitude of the sine function.
  • The sine function oscillates between -1 and 1, thus sin(τ)|sin(τ)| will oscillate between 0 and 1.

The graph should show oscillations that decrease in height over time, giving a shape with peaks that gradually become lower, resembling a decaying sinusoidal wave.

Step 3

the maximum height of the ball above the ground between the first and second bounce.

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Answer

To find the maximum height of the ball, we evaluate:

H(τ)=10e0.25τsin(τ)H(τ) = |10e^{-0.25τ}sin(τ)|

At the maximum point where H(τ)H(τ) is maximized, we determine: dH/dτ=0dH/dτ = 0 and evaluate it at the turning points found earlier. Substituting τ=1.33extsecondsτ = 1.33 ext{ seconds} gives:

H(1.33)=10e0.251.33sin(1.33)H(1.33) = |10e^{-0.25 * 1.33}sin(1.33)|

This yields an approximate maximum height of about 3.18 metres.

Step 4

Explain why this model should not be used to predict the time of each bounce.

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Answer

The model H(τ)H(τ) provides an ideal estimation of the height of the ball, assuming no energy loss due to factors like air resistance or ground absorption. In reality, the height will decrease with each bounce due to dissipative forces that are not accounted for in this function. Therefore, while it can give a rough estimate of maxima, it cannot reliably predict actual bounce timings or heights after the first bounce.

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