Figure 2 shows a sketch of part of the curve with equation y = f(x) where f(x) = 8 sin \left( \frac{1}{2} x \right) - 3x + 9 x > 0 and x is measured in radians - Edexcel - A-Level Maths Pure - Question 8 - 2022 - Paper 2

To explain why the root a must lie in the interval [4, 5], we analyze the values of f(x) at these boundaries:

• $f(4) = 4.274 > 0$
• $f(5) = -1.212 < 0$

Since f(4) is positive and f(5) is negative, and considering that f(x) is continuous, by the Intermediate Value Theorem, there must be at least one root in the interval [4, 5].

To apply the Newton-Raphson method, we start with an initial guess:

$x_0 = 5$

Using the Newton-Raphson formula:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

We first evaluate:

• $f(5) = -1.212$
• $f'(5) = 4 \cos \left( \frac{1}{2} \times 5 \right) - 3 \approx 4 \times 0.877582 - 3 \approx 3.510328$

Now substituting into the formula gives:

$x_1 = 5 - \frac{-1.212}{3.510328} \approx 5 + 0.345\approx 5.345$

For a more accurate second approximation, repeat the process: Evaluate:

• $f(5.345) ext{ and } f'(5.345)$ Calculate: $x_2 = 5.345 - \frac{f(5.345)}{f'(5.345)}$ Solve this to find the x-value that provides a clear approximation to a, rounded to three significant figures.