f(x) = 2sin(x^2) + x - 2, 0 ≤ x < 2π (a) Show that f(x) = 0 has a root α between x = 0.75 and x = 0.85. The equation f(x) = 0 can be written as x = [arcsin(1 - 0.5x)]^{1/2}. (b) Use the iterative formula x_{n+1} = [arcsin(1 - 0.5x_n)]^{1/2}, x_0 = 0.8 to find the values of x_1, x_2, and x_3, giving your answers to 5 decimal places. (c) Show that α = 0.80157 is correct to 5 decimal places.

A-Level Edexcel Maths Pure Graphs of Functions: f(x) = 2sin(x^2) + x

f(x) = 2sin(x^2) + x - 2, 0 ≤ x < 2π (a) Show that f(x) = 0 has a root α between x = 0.75 and x = 0.85 - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 3

Question 3

f(x) = 2sin(x^2) + x - 2, 0 ≤ x < 2π (a) Show that f(x) = 0 has a root α between x = 0.75 and x = 0.85. The equation f(x) = 0 can be written as x = [arcsin(1 - 0.... show full transcript

Worked Solution & Example Answer:f(x) = 2sin(x^2) + x - 2, 0 ≤ x < 2π (a) Show that f(x) = 0 has a root α between x = 0.75 and x = 0.85 - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 3

Step 1

Show that f(x) = 0 has a root α between x = 0.75 and x = 0.85

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Answer

To show that there is a root α in the interval (0.75, 0.85), we evaluate the function at these two points:

$f(0.75) = 2sin(0.75^2) + 0.75 - 2$ Calculating, we find:

= 2sin(0.5625) + 0.75 - 2 \\ ≈ -0.184 $$. Now for x = 0.85: $$f(0.85) = 2sin(0.85^2) + 0.85 - 2$$ Evaluating: $$f(0.85) \\ = 2sin(0.7225) + 0.85 - 2 \\ ≈ -0.171 $$. Because f(0.75) < 0 and f(0.85) < 0, we notice that the function changes signs (a change of sign) as we approach these values. Hence, there is at least one root α in the interval (0.75, 0.85).

Step 2

Use the iterative formula to find x_1

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Answer

We will use the iterative formula:

$x_{n+1} = [arcsin(1 - 0.5 x_n)]^{1/2}$ Substituting the initial value x_0 = 0.8:

= [arcsin(1 - 0.4)]^{1/2} \\ = [arcsin(0.6)]^{1/2} \\ ≈ [0.6435]^{1/2} \\ ≈ 0.80219$$ Therefore, x_1 ≈ 0.80219.

Step 3

Use the iterative formula to find x_2

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Answer

We continue using the iterative formula with our new value:

= [arcsin(1 - 0.401095)]^{1/2} \\ = [arcsin(0.598905)]^{1/2} \\ ≈ [0.6337]^{1/2} \\ ≈ 0.80133$$ Thus, x_2 ≈ 0.80133.

Step 4

Use the iterative formula to find x_3

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Answer

For the third iteration, we use:

= [arcsin(1 - 0.400665)]^{1/2} \\ = [arcsin(0.599335)]^{1/2} \\ ≈ [0.6347]^{1/2} \\ ≈ 0.80167$$ Hence, x_3 ≈ 0.80167.

Step 5

Show that α = 0.80157 is correct to 5 decimal places

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Answer

To show that α = 0.80157 is correct to 5 decimal places, we evaluate:

= 2sin(0.801565^2) + 0.801565 - 2$$ Calculating gives: $$f(0.801565) ≈ -2.7 imes 10^{-6}$$ And for f(0.801575): $$f(0.801575) = 2sin(0.801575^2) + 0.801575 - 2 \\ ≈ 8.6 imes 10^{-6}$$ The change of sign between f(0.801565) and f(0.801575) confirms that there is a root near 0.80157, validating its correctness to 5 decimal places.

f(x) = 2sin(x^2) + x - 2, 0 ≤ x < 2π (a) Show that f(x) = 0 has a root α between x = 0.75 and x = 0.85 - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 3

Worked Solution & Example Answer:f(x) = 2sin(x^2) + x - 2, 0 ≤ x < 2π (a) Show that f(x) = 0 has a root α between x = 0.75 and x = 0.85 - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 3

Show that f(x) = 0 has a root α between x = 0.75 and x = 0.85

Use the iterative formula to find x_1

Use the iterative formula to find x_2

Use the iterative formula to find x_3

Show that α = 0.80157 is correct to 5 decimal places

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