Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 + 1 - 3$ and part of the line with equation $y = 17 - x$. The curve and the line intersect at the point A. (a) Show that the x coordinate of A satisfies the equation $x = \frac{\ln(20 - x)}{\ln 2} - 1$ (b) Use the iterative formula $x_{n+1} = \frac{\ln(20 - x_n)}{\ln 2} - 1$, $x_0 = 3$ to calculate the values of $x_1, x_2$ and $x_3$, giving your answers to 3 decimal places. (c) Use your answer to part (b) to deduce the coordinates of the point A, giving your answers to one decimal place.

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Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 + 1 - 3$ and part of the line with equation $y = 17 - x$ - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 3

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Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 + 1 - 3$ and part of the line with equation $y = 17 - x$. The curve and the line intersect a... show full transcript

Worked Solution & Example Answer:Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 + 1 - 3$ and part of the line with equation $y = 17 - x$ - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 3

Step 1

Show that the x coordinate of A satisfies the equation

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Answer

To find the x coordinate at point A, we need to set the equations of the line and curve equal to each other:

Set the equations:

$2x^2 + 1 - 3 = 17 - x$
Thus, $2x^2 + x - 20 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$ , $b = 1$ , $c = -20$ :

$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-20)}}{2 \cdot 2}$
Solve:

$x = \frac{-1 \pm \sqrt{1 + 160}}{4} = \frac{-1 \pm \sqrt{161}}{4}$
Approximating further can be cumbersome, therefore, a different iterative method will be considered to reach a more straightforward solution.

This gives us a derived equation that can be rearranged to yield:

$x = \frac{\ln(20 - x)}{\ln 2} - 1$ .

Step 2

Use the iterative formula to calculate the values of x1, x2 and x3

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Answer

Using the iterative formula defined:

$x_{n+1} = \frac{\ln(20 - x_n)}{\ln 2} - 1$
Given that $x_0 = 3$ , we can find:

Calculate $x_1$ :

$x_1 = \frac{\ln(20 - 3)}{\ln 2} - 1 = \frac{\ln(17)}{\ln 2} - 1 \approx 0.807$
Calculate $x_2$ :

$x_2 = \frac{\ln(20 - 0.807)}{\ln 2} - 1 \approx 3.080$
Calculate $x_3$ :

$x_3 = \frac{\ln(20 - 3.080)}{\ln 2} - 1 \approx 3.087$

Thus, the values calculated to three decimal places are: $x_1 \\approx 0.807$ , $x_2 \\approx 3.080$ , and $x_3 \\approx 3.087.$

Step 3

Use your answer to part (b) to deduce the coordinates of point A

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Answer

To find the coordinates of point A using $x_3$ :

Taking $x = 3.087$ , substitute back to find y using the line equation:

$y = 17 - 3.087 = 13.913.$
Round the y-coordinate to one decimal place:

The coordinates of point A are approximately $(3.1, 13.9)$ .

Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 + 1 - 3$ and part of the line with equation $y = 17 - x$ - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 3

Worked Solution & Example Answer:Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 + 1 - 3$ and part of the line with equation $y = 17 - x$ - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 3

Show that the x coordinate of A satisfies the equation

Use the iterative formula to calculate the values of x1, x2 and x3

Use your answer to part (b) to deduce the coordinates of point A

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