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12. f(x) = 10e^{-0.25x} \, ext{sin} \, x, \, x \, > \, 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^{-0.25t} \, ext{sin} \, t| \, |t| > 0 showing the long-term behaviour of this curve - Edexcel - A-Level Maths: Pure - Question 13 - 2019 - Paper 1

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Question 13

12.-f(x)-=-10e^{-0.25x}-\,--ext{sin}-\,-x,-\,-x-\,->-\,-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^{-0.25t}-\,--ext{sin}-\,-t|-\,-|t|->-0---showing-the-long-term-behaviour-of-this-curve-Edexcel-A-Level Maths: Pure-Question 13-2019-Paper 1.png

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

Worked Solution & Example Answer:12. f(x) = 10e^{-0.25x} \, ext{sin} \, x, \, x \, > \, 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^{-0.25t} \, ext{sin} \, t| \, |t| > 0 showing the long-term behaviour of this curve - Edexcel - A-Level Maths: Pure - Question 13 - 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 f(x), we need to compute the first derivative and set it to zero. We have:

f(x)=2.5e0.25xextsinx+10e0.25xextcosxf'(x) = -2.5e^{-0.25x} \, ext{sin} \, x + 10e^{-0.25x} \, ext{cos} \, x

Setting this equal to zero gives:

2.5e0.25xextsinx+10e0.25xextcosx=0-2.5e^{-0.25x} \, ext{sin} \, x + 10e^{-0.25x} \, ext{cos} \, x = 0

Dividing by e^{-0.25x}, we obtain:

2.5extsinx+10extcosx=0-2.5 \, ext{sin} \, x + 10 \, ext{cos} \, x = 0

Rearranging this yields:

tanx=102.5=4\tan x = \frac{10}{2.5} = 4

This shows that the x coordinates of the turning points indeed satisfy the equation \tan x = 4.

Step 2

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

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Answer

To sketch the graph of H against t, we first note that H(t) oscillates due to the sine function while being damped by the exponential function.

  1. As t increases, the term e0.25te^{-0.25t} decreases, which means the amplitude of the oscillation reduces over time.
  2. The behavior of H(t) will exhibit peaks that gradually decrease in height, effectively showing a decaying sinusoidal wave.
  3. For the graph, ensure there is a smooth decay in the wave's height, illustrating the long-term behavior accurately.

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 above the ground, we need to maximize H(t). This occurs when sint=1\sin t = 1, since H(t) is proportional to sint\sin t:

Using:

H(t)=10e0.25tsintH(t) = |10e^{-0.25t} \sin t|

Maximizing gives:

H(t)=10e0.25tH(t) = 10e^{-0.25t}

We can find this maximum conditionally for the first bounce, considering the first maximum occurs approximately at t = 1.33 (from the turning points, solving \tan x = 4 yields a related t).

Thus, substituting this value:

H(1.33)=10e0.251.3310e0.33256.96extmH(1.33) = 10e^{-0.25 \cdot 1.33} \approx 10e^{-0.3325} \approx 6.96 ext{m}

Therefore, the maximum height is approximately 6.96 meters.

Step 4

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

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Answer

The model only accounts for the oscillation of height based on sine functions and exponential decay but does not incorporate real-world factors such as energy loss due to air resistance, impact duration, and ground friction. These can cause significant deviations from the predicted model, which assumes ideal bouncing without such losses. Additionally, each bounce duration may vary if the height significantly decreases over time, leading to unpredictability in bounce timings.

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