Figure 1 shows a sketch of part of the curve C with equation $$y = e^{2x} + x^2 - 3$$ The curve C crosses the y-axis at the point A - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 5

To show that the x-coordinate of B satisfies the equation:

$x = \sqrt{1 + \frac{1}{2} x - e^{-2x}}$

We need to find where the line l intersects the curve C again. We substitute the equation of line l into the equation of the curve:

Substituting: $-\frac{1}{2}x - 2 = e^{2x} + x^2 - 3$

Rearranging this, we get: $e^{2x} + x^2 + \frac{1}{2}x - 1 = 0$

Analyzing this equation, we can use numerical or iterative methods to find its root, which will correspond to the x-coordinate of point B.

We are given the iterative formula: $x_{n+1} = \sqrt{1 + \frac{1}{2} x_n - e^{-x_n}}$

Starting with $x_1 = 1$:

1. Calculate $x_2$: $x_2 = \sqrt{1 + \frac{1}{2} \cdot 1 - e^{-1}}$ Approximating: $e^{-1} \approx 0.367879$ gives: $x_2 = \sqrt{1 + 0.5 - 0.367879} \approx \sqrt{1.132121} \approx 1.065$(approximately).

2. Calculate $x_3$: Now using $x_2$: $x_3 = \sqrt{1 + \frac{1}{2} \cdot 1.065 - e^{-1.065}}$ Approximating: $e^{-1.065} \approx 0.344202$ gives: $x_3 = \sqrt{1 + 0.5325 - 0.344202} \approx \sqrt{1.188298} \approx 1.091$

To three decimal places: $x_2 \approx 1.065$ and $x_3 \approx 1.091$.