Figure 1 shows a sketch of part of the curve with equation $y = g(x)$, where $g(x) = |4e^{2x} - 25|, \, x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 3

To find the $x$-coordinate at point $B$, we set $g(x) = 0$:

$|4e^{2x} - 25| = 0 \Rightarrow 4e^{2x} - 25 = 0 \Rightarrow 4e^{2x} = 25 \Rightarrow e^{2x} = \frac{25}{4} \Rightarrow 2x = \ln\left(\frac{25}{4}\right) \Rightarrow x = \frac{1}{2}\ln\left(\frac{25}{4}\right).$

Calculating further gives:

$x = \frac{1}{2}(\ln(25) - \ln(4)) = \frac{1}{2}(\ln(25) - 2\ln(2)) = \frac{1}{2}(2 \ln(5) - 2 \ln(2)) = \ln\left(\frac{5}{2}\right).$

So, the exact $x$-coordinate of point $B$ is $\ln\left(\frac{5}{2}\right)$.

The constant $k$ is the horizontal asymptote of the function. As $x \to \infty$, $g(x)$ approaches:

$g(x) = |4e^{2x} - 25| \sim 4e^{2x} \to \infty.$ Thus, as $x$ becomes very large, $g(x)$ increases without bound. Therefore, at large negative $x$, we have:

$g(x) \to -25 \Rightarrow k = 25.$

Thus, the value of the constant $k$ is $25$.

Starting from the equation:

$g(x) = 2x + 43,$ we set:

$4e^{2x} - 25 = 2x + 43.$

Rearranging gives:

$4e^{2x}= 2x + 43 + 25 = 2x + 68.$ Dividing by 4 gives:

$e^{2x} = \frac{1}{2}(2x + 68) = \frac{1}{2}(2x + 17 + 34) = \frac{1}{2}(2x + 17) + 17.$

By logarithmic manipulation:

$2x = \ln \left(\frac{1}{2(2x + 17)}\right).$

Thus, we have shown that:\n$\alpha$ is indeed a solution.

Using the iteration formula:

$x_{n+1} = \frac{1}{2} \ln \left( \frac{1}{2x_n + 17} \right),$ with $x_1 = 1.4$:

1. For $x_1 = 1.4$: $x_2 = \frac{1}{2} \ln \left( \frac{1}{2 \cdot 1.4 + 17} \right) = \frac{1}{2} \ln \left( \frac{1}{19.8} \right) = \frac{1}{2} \cdot - ext{ln}(19.8)$. Approximating gives $\approx 1.4368$ (to 4 decimal places).

2. For $x_2$, calculate $x_3$ using the same formula: $x_3 = \frac{1}{2} \ln \left( \frac{1}{2 \cdot 1.4368 + 17} \right) = \frac{1}{2} \ln \left( \frac{1}{19.8736} \right) \approx 1.4373$ (to 4 decimal places).

To show that $1.437$ is a solution to 3 decimal places, we can perform evaluations near $x=1.437$:

$f(1.436) = g(1.436) - (2 \times 1.436 + 43)$ gives a negative result,

$f(1.438) = g(1.438) - (2 \times 1.438 + 43)$ gives a positive result.

Since $f(x)$ changes sign over the interval $[1.436, 1.438]$, we can conclude by the Intermediate Value Theorem that a root exists and the solution is approximately $1.437$ to 3 decimal places.