Figure 1 is a sketch showing part of the curve with equation $y = 2^{x+1} - 3$ 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.

A-Level Edexcel Maths Pure Modelling with Functions: Figure 1 is a sketch showing par

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

Question 7

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

Worked Solution & Example Answer:Figure 1 is a sketch showing part of the curve with equation $y = 2^{x+1} - 3$ 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 of point A, we set the equations of the curve and the line equal to each other. This results in:

2^{x+1} - 3 = 17 - x

Rearranging gives:

2^{x+1} + x - 20 = 0

Taking the natural logarithm of both sides leads us to:

\ln(2^{x+1}) = \ln(20 - x)

Using properties of logarithms, we can rewrite this as:

(x + 1)\ln 2 = \ln(20 - x)

Dividing both sides by \ln 2 yields:

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

Finally, solving for x gives:

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

Step 2

Use the iterative formula

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Answer

To find the values of $x_1$ , $x_2$ , and $x_3$ using the iterative formula:

Start with $x_0 = 3$ .
Calculate $x_1$ :

x_1 = \frac{\ln(20 - 3)}{\ln 2} - 1 \approx 3.087

3. Now, use $x_1$ to calculate $x_2$:

x_2 = \frac{\ln(20 - 3.087)}{\ln 2} - 1 \approx 3.081

4. Next, use $x_2$ to calculate $x_3$:

x_3 = \frac{\ln(20 - 3.081)}{\ln 2} - 1 \approx 3.078

Step 3

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

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Answer

From part (b), we found that $x \approx 3.081$ for the x-coordinate of point A. To find the y-coordinate, substitute this value into either of the original equations. Using $y = 17 - x$ :

y = 17 - 3.081 \approx 13.919 \approx 13.9 (to \, one \, decimal \, place)

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

Figure 1 is a sketch showing part of the curve with equation $y = 2^{x+1} - 3$ 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 $y = 2^{x+1} - 3$ 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

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

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