Figure 2 shows a sketch of part of the curve C with equation $y = 2 \ln(2x + 5) - \frac{3x}{2}$, $x > -2.5$ 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) Taking $x_1 = 2$, find the values of $x_1$ and $x_2$, giving each answer to 4 decimal places.

A-Level Edexcel Maths Pure Differentiation: Figure 2 shows a sketch of part

Figure 2 shows a sketch of part of the curve C with equation $y = 2 \ln(2x + 5) - \frac{3x}{2}$, $x > -2.5$ The point P with x coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 4

Question 7

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

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation $y = 2 \ln(2x + 5) - \frac{3x}{2}$, $x > -2.5$ The point P with x coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 4

Step 1

Find an equation of the normal to C at P.

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Answer

To find the equation of the normal at point P where $x = -2$ :

First, calculate the $y$ coordinate at $x = -2$ :

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

Thus, the coordinates of P are $(-2, 3)$ .
Next, differentiate the function to find the slope of the tangent:

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

At $x = -2$ :

$\frac{dy}{dx} = \frac{4}{1} - \frac{3}{2} = 4 - 1.5 = 2.5$
The slope of the normal line is the negative reciprocal:

$m = -\frac{1}{2.5} = -\frac{2}{5}$
Using point-slope form, the equation of the normal line is:

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

Rearranging gives:

$2x + 5y = 20$ . Hence, the normal equation is $2x + 5y = 20$ .

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 is a solution of:

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

Substitute the normal line equation:

$5y + 2 = 11$

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

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

Simplifying leads to:

$\frac{20}{11} \ln(2x + 5) - 2 = x$ .
Therefore, the x coordinate of point Q can be represented as:

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

Step 3

Taking $x_1 = 2$, find the values of $x_1$ and $x_2$.

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Answer

Using the iteration formula:

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

Substitute $x_1 = 2$ :

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

Calculate:

$= \frac{20}{11} \ln(9) - 2 \approx 1.9950$
Continuing for $x_2$ :

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

Continue iterating:

$x_3 \approx 1.9920$

Thus, the approximations yield:

$x_1 = 2$
$x_2 = 1.9950$
$x_3 = 1.9920$ , all rounded to 4 decimal places.

Figure 2 shows a sketch of part of the curve C with equation $y = 2 \ln(2x + 5) - \frac{3x}{2}$, $x > -2.5$ The point P with x coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 4

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation $y = 2 \ln(2x + 5) - \frac{3x}{2}$, $x > -2.5$ The point P with x coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 7 - 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 $x_1 = 2$, find the values of $x_1$ and $x_2$.

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