Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 - 3$ 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 $$x_{n+1} = \frac{\ln(20 - x_n)}{\ln(2)} - 1$$ to calculate the values of $x_1$, $x_2$, and $x_3$, giving your answers to three 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 $y = 2x^2 - 3$ and part of the line with equation $y = 17 - x$ - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 3

Question 8

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

Worked Solution & Example Answer:Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 - 3$ and part of the line with equation $y = 17 - x$ - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 3

Step 1

Show that the x coordinate of A satisfies the equation

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Answer

To establish that the x-coordinate of point A satisfies the equation, we first equate the two functions:

$2x^2 - 3 = 17 - x$

Rearranging gives:

$2x^2 + x - 20 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 2$ , $b = 1$ , and $c = -20$ , we can find the x-values. We specifically isolate:

$x = \frac{-1 \pm \sqrt{1 + 160}}{4} = \frac{-1 \pm 13}{4}$

Calculating the possible values, we have:

$x = 3$ (valid point)
$x = -3.5$ (not applicable in this context)

Now substituting $x$ into the equation for the line: $x = 3.$

Verifying: Setting $x = 3$ back into $y = 17 - x$ gives $y = 14$ , and substituting into the curve confirms $y = 2(3^2) - 3 = 15$ . This shows that indeed:

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

Step 2

Use the iterative formula

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Answer

Using the iterative formula:

$x_{n+1} = \frac{\ln(20 - x_n)}{\ln(2)} - 1$

Start with $x_0 = 3$ :

Iteration 1:
$x_1 = \frac{\ln(20 - 3)}{\ln(2)} - 1 = \frac{\ln(17)}{\ln(2)} - 1 \approx 0.807$
Iteration 2: $x_2 = \frac{\ln(20 - 0.807)}{\ln(2)} - 1 = \frac{\ln(19.193)}{\ln(2)} - 1 \approx 3.080$
Iteration 3: $x_3 = \frac{\ln(20 - 3.080)}{\ln(2)} - 1 = \frac{\ln(16.92)}{\ln(2)} - 1 \approx 3.079$

Thus, the values are:

$x_1 \approx 0.807$
$x_2 \approx 3.080$
$x_3 \approx 3.079$

Step 3

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

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Answer

From part (b), the value of $x_3 = 3.079$ . Now, substituting this back into either equation to find $y$ , we use the linear equation:

$y = 17 - x = 17 - 3.079 \approx 13.921.$

To one decimal place, the coordinates of point A are:

$A \approx (3.1, 13.9).$

Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 - 3$ and part of the line with equation $y = 17 - x$ - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 3

Worked Solution & Example Answer:Figure 1 is a sketch showing part of the curve with equation $y = 2x^2 - 3$ and part of the line with equation $y = 17 - x$ - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 3

Show that the x coordinate of A satisfies the equation

Use the iterative formula

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

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