Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) - \frac{3x}{2}, \, x > -2.5 The point P with x-coordinate -2 lies on C. (a) Find an equation of the normal to C at P. Write your answer in the form ax + by = c, where a, b and c are integers. The normal to C at P cuts the curve again at the point Q, as shown in Figure 2. (b) Show that the x coordinate of Q is a solution of the equation x = \frac{20}{11}ln(2x + 5) - 2 (c) The iteration formula x_{n+1} = \frac{20}{11}ln(2x_n + 5) - 2 can be used to find an approximation for the x coordinate of Q. (c) Taking x_1 = 2, find the values of x_n and x_{n+1}, giving each answer to 4 decimal places.

A-Level Edexcel Maths Pure Modelling with Functions: Figure 2 shows a sketch of part

Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) - \frac{3x}{2}, \, x > -2.5 The point P with x-coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 4

Question 6

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Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) - \frac{3x}{2}, \, x > -2.5 The point P with x-coordinate -2 lies on C. (a) Find an e... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) - \frac{3x}{2}, \, x > -2.5 The point P with x-coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 4

Step 1

Find an equation of the normal to C at P. Write your answer in the form ax + by = c, where a, b and c are integers.

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Answer

First, we determine the y-coordinate for point P when x = -2:

$y = 2ln(2(-2) + 5) - \frac{3(-2)}{2}$

Calculating the value gives: $y = 2ln(1) + 3 = 3.$

Since the coordinates of P are (-2, 3), the next step is to calculate the derivative of the curve at this point:

$\frac{dy}{dx} = \frac{4}{2x + 5} - \frac{3}{2}.$

Substituting x = -2:

$\frac{dy}{dx} = \frac{4}{2(-2) + 5} - \frac{3}{2} = \frac{4}{1} - \frac{3}{2} = 4 - 1.5 = 2.5.$

The slope of the normal is then the negative reciprocal of 2.5, giving us:

$m_{normal} = -\frac{1}{2.5} = -\frac{2}{5}.$

Using point-slope form of the line equation: $y - 3 = -\frac{2}{5}(x + 2).$

Rearranging this yields: $2x + 5y = 11.$

Thus, the equation of the normal in the required form is:

$2x + 5y = 11.$

Step 2

Show that the x coordinate of Q is a solution of the equation x = \frac{20}{11}ln(2x + 5) - 2.

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Answer

To find the x-coordinate of Q, we need to combine the equations of the curve and the normal:

Substituting the normal's equation into the curve’s equation:

$5y + 2 = 11$ and $y = 2ln(2x + 5) - \frac{3x}{2}.$

This gives us: $5(\frac{11 - 2}{5}) + 2ln(2x + 5) - \frac{3x}{2} = 11,$

which simplifies to:

$10ln(2x + 5) = 20,$

leading to:

$x = \frac{20}{11}ln(2x + 5) - 2.$

Step 3

Taking x_1 = 2, find the values of x_n and x_{n+1}, giving each answer to 4 decimal places.

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Answer

Using the iteration formula:

$x_{n+1} = \frac{20}{11}ln(2x_n + 5) - 2,$

we start with: $x_1 = 2.$

Calculating:

$x_2 = \frac{20}{11}ln(2(2) + 5) - 2$

Calculates to approximately: $x_2 \approx 1.9950.$

Next calculate:

$x_3 = \frac{20}{11}ln(2(1.9950) + 5) - 2$

Calculates to approximately: $x_3 \approx 1.9929.$

Rounding each result to four decimal places: $x_1 = 2.0000, x_2 = 1.9950, x_3 = 1.9929.$

Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) - \frac{3x}{2}, \, x > -2.5 The point P with x-coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 4

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) - \frac{3x}{2}, \, x > -2.5 The point P with x-coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 4

Find an equation of the normal to C at P. Write your answer in the form ax + by = c, where a, b and c are integers.

Show that the x coordinate of Q is a solution of the equation x = \frac{20}{11}ln(2x + 5) - 2.

Taking x_1 = 2, find the values of x_n and x_{n+1}, giving each answer to 4 decimal places.

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