Let $$f(x) = x^2 - 3x + 2 ext{cos} \left( \frac{x}{2} \right), \quad 0 \leq x \leq \pi.$$ (a) Show that the equation $f(x) = 0$ has a solution in the interval $0.8 < x < 0.9.$ (b) The curve with equation $y = f(x)$ has a minimum point $P.$ Show that the $x$-coordinate of $P$ is the solution of the equation $$x = \frac{3 + \sin \left( \frac{x}{2} \right)}{2}.$$ (c) Using the iteration formula $$x_{n+1} = \frac{3 + \sin \left( \frac{x_n}{2} \right)}{2}, \quad x_0 = 2$$ fine the values of $x_1, x_2,$ and $x_3,$ giving your answers to 3 decimal places - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 6

To find the minimum point $P,$ we calculate the derivative of $f(x)$ and set it to zero:

1. Find ( f'(x) ): $f'(x) = 2x - 3 - \frac{1}{2} \sin \left( \frac{x}{2} \right)$

2. Set ( f'(x) = 0 ): $2x - 3 - \frac{1}{2} \sin \left( \frac{x}{2} \right) = 0$ Rearranging, we get: $2x = 3 + \frac{1}{2} \sin \left( \frac{x}{2} \right)$ Therefore, this can be expressed as: $x = \frac{3 + \sin \left( \frac{x}{2} \right)}{2},$ which confirms that the $x$-coordinate of $P$ is indeed the solution of this equation.

Using the iteration formula

$x_{n+1} = \frac{3 + \sin \left( \frac{x_n}{2} \right)}{2} \quad x_0 = 2,$ we find the iteration values:

1. Calculate ( x_1 ): $x_1 = \frac{3 + \sin \left( \frac{2}{2} \right)}{2} = \frac{3 + \sin(1)}{2} \approx \frac{3 + 0.8415}{2} \approx 1.921$

2. Calculate ( x_2 ): $x_2 = \frac{3 + \sin \left( \frac{1.921}{2} \right)}{2} \approx \frac{3 + \sin(0.9605)}{2} \approx \frac{3 + 0.8252}{2} \approx 1.910$

3. Calculate ( x_3 ): $x_3 = \frac{3 + \sin \left( \frac{1.910}{2} \right)}{2} \approx \frac{3 + \sin(0.955)}{2} \approx \frac{3 + 0.8176}{2} \approx 1.908$

Thus, the values are:

• $x_1 \approx 1.921$
• $x_2 \approx 1.910$
• $x_3 \approx 1.908$.

We can refine our search for the root by examining the intervals around our solutions. Noticing that:

• From previous calculations, we found that $f(1.905) > 0$ and $f(1.910) < 0$, we can assess the interval $(1.905, 1.910)$:

1. Calculate ( f(1.907) ): $f(1.907) = (1.907)^2 - 3(1.907) + 2 \cos \left( \frac{1.907}{2} \right) \approx 0.0001$

2. Calculate ( f(1.908) ): $f(1.908) = (1.908)^2 - 3(1.908) + 2 \cos \left( \frac{1.908}{2} \right) \approx -0.00001$

Since we have verified a sign change from $1.907$ to $1.908$, we can conclude that the root lies within $(1.9078, 1.908)$, thus confirming that the $x$-coordinate of $P$ is approximately 1.9078 correct to 4 decimal places.