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12. f(x) = 10e^{-0.25x} ext{ sin }x, ext{ } x ext{ } extgreater 0 (a) Show that the x coordinates of the turning points of the curve with equation y = f(x) satisfy the equation \tan x = 4 (4) (b) Sketch the graph of H against t where H(t) = |10e^{-0.25x} ext{ sin }x| ext{ } | t ext{ } > 0 showing the long-term behaviour of this curve - Edexcel - A-Level Maths Pure - Question 12 - 2019 - Paper 1

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12.-f(x)-=-10e^{-0.25x}--ext{-sin-}x,--ext{-}-x--ext{-}--extgreater-0---(a)-Show-that-the-x-coordinates-of-the-turning-points-of-the-curve-with-equation-y-=-f(x)-satisfy-the-equation-\tan-x-=-4---(4)--(b)-Sketch-the-graph-of-H-against-t-where---H(t)-=-|10e^{-0.25x}--ext{-sin-}x|--ext{-}-|-t--ext{-}->-0---showing-the-long-term-behaviour-of-this-curve-Edexcel-A-Level Maths Pure-Question 12-2019-Paper 1.png

12. f(x) = 10e^{-0.25x} ext{ sin }x, ext{ } x ext{ } extgreater 0 (a) Show that the x coordinates of the turning points of the curve with equation y = f(x) s... show full transcript

Worked Solution & Example Answer:12. f(x) = 10e^{-0.25x} ext{ sin }x, ext{ } x ext{ } extgreater 0 (a) Show that the x coordinates of the turning points of the curve with equation y = f(x) satisfy the equation \tan x = 4 (4) (b) Sketch the graph of H against t where H(t) = |10e^{-0.25x} ext{ sin }x| ext{ } | t ext{ } > 0 showing the long-term behaviour of this curve - Edexcel - A-Level Maths Pure - Question 12 - 2019 - Paper 1

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 find the derivative of f(x):

f(x)=10e0.25x(extcosx0.25extsinx)f'(x) = 10e^{-0.25x} ( ext{cos } x - 0.25 ext{ sin } x)

Setting the derivative to zero for turning points gives us:

10e0.25x(extcosx0.25extsinx)=010e^{-0.25x} ( ext{cos } x - 0.25 ext{ sin } x) = 0

Since the exponential term is never zero, we simplify to:

extcosx0.25extsinx=0 ext{cos } x - 0.25 ext{ sin } x = 0

Rearranging this equation, we have:

extcosx=0.25extsinx ext{cos } x = 0.25 ext{ sin } x

Taking the ratio gives

tanx=4\tan x = 4

Hence, the x-coordinates of the turning points satisfy \tan x = 4.

Step 2

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

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Answer

To sketch the graph of H(t), we consider the behavior of the function over large values of t. Since the exponential term e0.25xe^{-0.25x} will lead to H approaching zero as t increases, the long-term behavior is characterized by oscillations of decreasing amplitude.

The graph should have loops that gradually decrease in height, illustrating the damping effect due to the exponential decay.

Step 3

Find 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, we can evaluate H(t) at the x-coordinates found from the turning points.

Calculating H(4.47):

H(4.47)=10e0.25imes4.47extsin(4.47)H(4.47) = |10e^{-0.25 imes 4.47} ext{ sin }(4.47)|

Calculating this gives:

H(4.47)3.18extmetresH(4.47) \approx 3.18 ext{ metres}

Therefore, the maximum height is approximately 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 does not take into account various real-life factors that affect a ball's bounce, such as air resistance or energy loss due to impacts.

The heights and timing may vary significantly due to these external factors, making the model unreliable for precise predictions of bounces.

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