f(x) = 2 sin(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 4 - 2011 - Paper 3

We will start with x_0 = 0.8 and use the iterative formula:

$x_{n+1} = [ ext{arcsin}(1 - 0.5x_n)]^{1/2}$

1. For x_1: $x_1 = [ ext{arcsin}(1 - 0.5 imes 0.8)]^{1/2} \\ = [ ext{arcsin}(0.6)]^{1/2} \\ ≈ 0.80219$

2. For x_2: $x_2 = [ ext{arcsin}(1 - 0.5 imes 0.80219)]^{1/2} \\ = [ ext{arcsin}(0.598905)]^{1/2} \\ ≈ 0.80133$

3. For x_3: $x_3 = [ ext{arcsin}(1 - 0.5 imes 0.80133)]^{1/2} \\ = [ ext{arcsin}(0.598335)]^{1/2} \\ ≈ 0.80167$

To show that α = 0.80157 is correct to 5 decimal places, we will evaluate the function at x = 0.80157:

• Calculating f(0.80157): $f(0.80157) = 2 ext{sin}(0.80157^2) + 0.80157 - 2$

Next, we check the values for f(0.801565) and f(0.801575):

• $f(0.801565) = -2.7...×10^{-5}$
• $f(0.801575) = +8.6...×10^{-6}$

Since f(0.801565) < 0 and f(0.801575) > 0, it confirms that there is a sign change around α, hence verifying that α = 0.80157 is indeed correct to 5 decimal places.