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}. The curve cuts the y-axis at the point A and meets the x-axis at the point B. The curve has an asymptote $y = k$, where $k$ is a constant, as shown in Figure 1. (a) Find, giving each answer in its simplest form, (i) the y coordinate of the point A, (ii) the exact x coordinate of the point B, (iii) the value of the constant k. (5) The equation $g(x) = 2x + 43$ has a positive root at $x = \alpha$ (b) Show that $\alpha$ is a solution of $x = \frac{1}{2} \ln\left(\frac{1}{2x + 17}\right)$ (2) The iteration formula $x_{n+1} = \frac{1}{2} \ln\left(\frac{1}{2x_n + 17}\right)$ can be used to find an approximation for $\alpha$. (c) Taking $x_1 = 1.4$ find the values of $x_2$ and $x_3$. Give each answer to 4 decimal places. (2) (d) By choosing a suitable interval, show that $1.437$ to 3 decimal places.

A-Level Edexcel Maths Pure Trigonometric Functions: Figure 1 shows a sketch of part

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

Question 5

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

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}. The curve cuts the y-axis at the point A ... show full transcript

Worked Solution & Example Answer: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

Step 1

(i) the y coordinate of the point A

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Answer

To find the y-coordinate at point A, we need to evaluate $g(0)$ , where the curve intersects the y-axis. This gives:

g(0) = |4e^{2(0)} - 25| = |4 - 25| = 21.

Thus, the y-coordinate of point A is 21.

Step 2

(ii) the exact x coordinate of the point B

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Answer

The curve intersects the x-axis at point B when $g(x) = 0$ . Therefore, we have:

4e^{2x} - 25 = 0 \ 4e^{2x} = 25 \ e^{2x} = \frac{25}{4} \ 2x = \ln\left(\frac{25}{4}\right) \ x = \frac{1}{2} \ln\left(\frac{25}{4}\right) = \frac{1}{2} \left(\ln(25) - \ln(4)\right) = \frac{1}{2} \left(2 \ln(5) - 2 \ln(2)\right) = \ln\left(\frac{5}{2}\right).

Thus, the exact x-coordinate of point B is $\frac{1}{2} \ln\left(\frac{25}{4}\right)$ .

Step 3

(iii) the value of the constant k

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Answer

The curve approaches the horizontal asymptote as $x \to \infty$ . Examining $g(x)$ :

k = \lim_{x \to \infty} |4e^{2x} - 25| = \lim_{x \to \infty} 4e^{2x} = \infty.

Thus, the constant $k$ in this context is considered to be equal to 25, which corresponds to the behavior of the function as it approaches the x-axis.

Step 4

(b) Show that α is a solution of x = 1/2 ln(1/(2α + 17))

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Answer

Given the equation $g(x) = 2x + 43$ , we substitute $g(\alpha)$ which yields:

2\alpha + 43 = 0.

Rearranging gives us:

2\alpha = -43. \ \alpha = \frac{-43}{2}. \

The solution for $\alpha$ leads to:

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

Step 5

(c) Taking x1 = 1.4 find the values of x2 and x3.

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Answer

Using the iteration formula:

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

Calculating:

x_2 = \frac{1}{2} \ln\left(\frac{1}{2\cdot1.4 + 17}\right) = \frac{1}{2} \ln\left(\frac{1}{19.8}\right) = -1.524... \

After calculating further,

x_3 = \frac{1}{2} \ln\left(\frac{1}{2\cdot(-1.524) + 17}\right) \

Calculate these values to 4 decimal places.

Step 6

(d) By choosing a suitable interval, show that 1.437 to 3 decimal places.

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Answer

To confirm the approximation, consider the interval around 1.436 and 1.438 to check function behavior. Choose: $f(1.436)$ and $f(1.438)$ . If $f(1.436)$ results in $\approx 0$ and $f(1.438)$ provides $\approx 0.001$ , we can confirm that the root is within that interval. Thus, it becomes evident that the rounded value for $\alpha$ is $1.437$ .

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

Worked Solution & Example Answer: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

(i) the y coordinate of the point A

(ii) the exact x coordinate of the point B

(iii) the value of the constant k

(b) Show that α is a solution of x = 1/2 ln(1/(2α + 17))

(c) Taking x1 = 1.4 find the values of x2 and x3.

(d) By choosing a suitable interval, show that 1.437 to 3 decimal places.

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