Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) − 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) Taking x₁ = 2, find the values of x₁ and x₂, giving each answer to 4 decimal places.

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Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) − 3x/2 , x > −2.5 - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 4

Question 5

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

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) − 3x/2 , x > −2.5 - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 4

Step 1

Find an equation of the normal to C at P.

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Answer

At the point P where x = -2, substitute into the curve's equation:

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

y = 2ln(1) + 3 = 3.$$

Now, calculate the derivative ( \frac{dy}{dx} ) at P:

Use the quotient rule and simplify: $\frac{dy}{dx} = \frac{4}{2x + 5} - \frac{3}{2}$ at x = -2: $\frac{dy}{dx}_{x=-2} = \frac{4}{1} - \frac{3}{2} = 4 - 1.5 = 2.5.$
The slope of the normal is the negative reciprocal: $m_{normal} = -\frac{1}{2.5} = -\frac{2}{5}.$
Use the point-slope form to find the equation of the normal at P: $y - 3 = -\frac{2}{5}(x + 2).$ Rearranging to the form ( ax + by = c ): $2x + 5y = 21.$

Step 2

Show that the x coordinate of Q is a solution of the equation.

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Answer

To show that the x-coordinate of Q satisfies the equation:

Start from the equation of the normal: $5y + 2x = 11$ and the equation of the curve: $y = 2ln(2x + 5) - \frac{3}{2}x.$
Substitute for y and rearrange: $5(2ln(2x + 5) - \frac{3}{2}x) + 2x = 11$ Simplifying gives: $10ln(2x + 5) - \frac{15}{2}x + 2x = 11$ which can be further written as: $10ln(2x + 5) - \frac{11}{2}x = 11.$
This leads to the final expression: $x = \frac{20}{11}ln(2x + 5) - 2.$

Step 3

Taking x₁ = 2, find the values of x₁ and x₂.

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Answer

Using the iteration formula:

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

For x₁ = 2: $x_1 = 2, \quad x_2 = \frac{20}{11}ln(2(2) + 5) - 2$ Substitute: $x_2 = \frac{20}{11}ln(9) - 2.$
Calculate: $ln(9) \approx 2.1972 \Rightarrow x_2 \approx \frac{20}{11}(2.1972) - 2 \approx 0.9952.$
For the next iteration (using x₂): $x_3 = \frac{20}{11}ln(2(0.9952) + 5) - 2 \Rightarrow x_3 \approx 1.995 :)$ Thus, provide answers to four decimal places:
x₁ = 2.0000, x₂ = 0.9952 (further iterations as needed).

Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) − 3x/2 , x > −2.5 - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 4

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) − 3x/2 , x > −2.5 - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 4

Find an equation of the normal to C at P.

Show that the x coordinate of Q is a solution of the equation.

Taking x₁ = 2, find the values of x₁ and x₂.

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