f(x) = 4 \, ext{cosec} \, x - 4x + 1, \text{ where } x \text{ is in radians.} (a) Show that there is a root $\alpha$ of $f(x) = 0$ in the interval [1.2, 1.3] - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 5

To rewrite $f(x) = 0$, we start from the original equation:

$4 \, \text{cosec} \, x - 4x + 1 = 0.$
Replacing $\text{cosec} \, x$ gives us:

$4 \cdot \frac{1}{\sin x} - 4x + 1 = 0.$
Rearranging this:

$4 \cdot \frac{1}{\sin x} = 4x - 1.$
Now divide both sides by 4:

$\frac{1}{\sin x} = x - \frac{1}{4}.$
Taking the reciprocal gives us the desired form:

$$x = \frac{1}{\sin x} + \frac{1}{4}.$

Using the initial value $x_0 = 1.25$, we can apply the iterative formula.

1. Calculate $x_1$:

   $x_1 = \frac{1}{\sin(1.25)} + \frac{1}{4}.$
   Performing the calculation gives:

   $x_1 \approx 1.2657.$

2. Calculate $x_2$:

   $x_2 = \frac{1}{\sin(1.2657)} + \frac{1}{4}.$
   Performing the calculation gives:

   $x_2 \approx 1.2838.$

3. Calculate $x_3$:

   $x_3 = \frac{1}{\sin(1.2838)} + \frac{1}{4}.$
   Performing the calculation gives:

   $x_3 \approx 1.2905.$

Thus, the results are:

• $x_1 \approx 1.2657$
• $x_2 \approx 1.2838$
• $x_3 \approx 1.2905$.

To verify that $\alpha = 1.291$ correct to three decimal places, we can evaluate the function $f(x)$ around this value:

1. Calculate $f(1.29)$:

   $f(1.29) = 4 \, \text{cosec}(1.29) - 4(1.29) + 1 \approx 0.0017 > 0.$

2. Calculate $f(1.29)$:

   $f(1.291) = 4 \, \text{cosec}(1.291) - 4(1.291) + 1 \approx -0.0001 < 0.$

Since $f(1.29) > 0$ and $f(1.291) < 0$, there is a change of sign between 1.29 and 1.291, confirming that $\alpha = 1.291$ is correct to three decimal places.