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Figure 2 shows part of the curve with equation y = (2x - 1) tan 2x, 0 < x < π/4 The curve has a minimum at the point P - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 4

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Figure-2-shows-part-of-the-curve-with-equation---y-=-(2x---1)-tan-2x,-0-<-x-<-π/4----The-curve-has-a-minimum-at-the-point-P-Edexcel-A-Level Maths Pure-Question 7-2006-Paper 4.png

Figure 2 shows part of the curve with equation y = (2x - 1) tan 2x, 0 < x < π/4 The curve has a minimum at the point P. The x-coordinate of P is k. (e) Show ... show full transcript

Worked Solution & Example Answer:Figure 2 shows part of the curve with equation y = (2x - 1) tan 2x, 0 < x < π/4 The curve has a minimum at the point P - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 4

Step 1

Show that k satisfies the equation

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Answer

To verify that k satisfies the equation, we start with the derivative of the function given by:

y=(2x1)an2xy = (2x - 1) an 2x

Using the product rule, we find:

dydx=(2tan2x+(2x1)2sec22x)\frac{dy}{dx} = (2 \tan 2x + (2x - 1) \cdot 2 \sec^2 2x)

Setting the derivative equal to zero for finding minimum points, we have:

2tan2x+2(2x1)sec22x=02 \tan 2x + 2(2x - 1) \sec^2 2x = 0

This simplifies to:

2tan2x+(2x1)2sec22x=02 \tan 2x + (2x - 1) \cdot 2 \sec^2 2x = 0

After solving and rearranging, we arrive at the equation:

4k+sin4k2=04k + \sin 4k - 2 = 0

Step 2

Calculate x_1, x_2, x_3, and x_4

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Answer

Using the iterative formula:

  1. Calculate x_1:
    x1=12(2sin4×0.3)x_1 = \frac{1}{2} (2 - \sin 4 \times 0.3)

Performing this calculation yields:
x1=0.2670x_1 = 0.2670
2. Now, calculate x_2:
x2=12(2sin4×0.2670)x_2 = \frac{1}{2} (2 - \sin 4 \times 0.2670)

This gives:
x2=0.2809x_2 = 0.2809
3. Next, for x_3:
x3=12(2sin4×0.2809)x_3 = \frac{1}{2} (2 - \sin 4 \times 0.2809)

Resulting in:
x3=0.2746x_3 = 0.2746
4. Finally, calculate x_4:
x4=12(2sin4×0.2746)x_4 = \frac{1}{2} (2 - \sin 4 \times 0.2746)

This yields:
x4=0.2774x_4 = 0.2774

Step 3

Show that k = 0.277, correct to 3 significant figures

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Answer

To finalize, we observe that through continuous iterations, we can approximate k as follows:
Given that x_4 = 0.2774, rounding to three significant figures gives k = 0.277.
Thus, we conclude that k approximates to 0.277.

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