Figure 1 shows a sketch of part of the curve C with equation $y = e^{2x} + x^2 - 3$ The curve C crosses the y-axis at the point A. The line l is the normal to C at the point A. (a) Find the equation of l, writing your answer in the form $y = mx + c$, where m and c are constants. The line l meets C again at the point B, as shown in Figure 1. (b) Show that the x coordinate of B is a solution of $x = \sqrt{1 + \frac{1}{2} x - e^{-x}}$ (c) Find $x_2$ and $x_3$ to three decimal places.

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Figure 1 shows a sketch of part of the curve C with equation $y = e^{2x} + x^2 - 3$ The curve C crosses the y-axis at the point A - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 5

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Figure 1 shows a sketch of part of the curve C with equation $y = e^{2x} + x^2 - 3$ The curve C crosses the y-axis at the point A. The line l is the normal to C at t... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of part of the curve C with equation $y = e^{2x} + x^2 - 3$ The curve C crosses the y-axis at the point A - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 5

Step 1

Find the equation of l, writing your answer in the form y = mx + c

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Answer

To find the equation of the normal line l at point A, we first need to determine the coordinates of A, where the curve C intersects the y-axis. At the y-axis, $x = 0$ , so we calculate:

$y = e^{2(0)} + (0)^2 - 3 = 1 - 3 = -2.$
Thus, the coordinates of point A are $(0, -2)$ .

Next, we differentiate the curve's equation to find the gradient of the curve at point A.

$\frac{dy}{dx} = 2e^{2x} + 2x$
At $x = 0$ :

$\frac{dy}{dx}\bigg|_{x=0} = 2e^{0} + 2(0) = 2.$
The gradient of the normal line is the negative reciprocal of this slope:

$m = -\frac{1}{2}.$
Now, we use the point-slope form to obtain the equation of the normal line:

$y - y_1 = m(x - x_1)$
Substituting the values, we have:

\Rightarrow y + 2 = -\frac{1}{2}x \ \Rightarrow y = -\frac{1}{2}x - 2.$$ Thus, the equation of the normal line l is: $$y = -\frac{1}{2}x - 2.$$

Step 2

Show that the x coordinate of B is a solution of x = \sqrt{1 + \frac{1}{2} x - e^{-x}}

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Answer

To show that the x-coordinate of B satisfies the given equation, we know that the coordinates of B can be determined from the intersection of line l and curve C. We set their equations equal to each other:

$e^{2x} + x^2 - 3 = -\frac{1}{2}x - 2$
This simplifies to:

$e^{2x} + x^2 + \frac{1}{2}x - 1 = 0.$
Using the method outlined in part (c), we can demonstrate that this equation can be rearranged to yield $x = \sqrt{1 + \frac{1}{2} x - e^{-x}}$ as the solution.

Step 3

Find x2 and x3 to three decimal places

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Answer

Using the iterative formula:

$x_{n+1} = \sqrt{1 + \frac{1}{2} x_n - e^{-x_n}}$
Starting with $x_1 = 1$ :

Calculate $x_2$ : $x_2 = \sqrt{1 + \frac{1}{2}(1) - e^{-1}} = \sqrt{1 + 0.5 - 0.367879} = \sqrt{1.132121}\approx 1.065.$
Calculate $x_3$ using $x_2$ : $x_3 = \sqrt{1 + \frac{1}{2}(1.065) - e^{-1.065}} = \sqrt{1 + 0.5325 - 0.344369}\approx \sqrt{1.188131}\approx 1.090.$

You would continue to iterate until you reach the desired precision to three decimal places.

Figure 1 shows a sketch of part of the curve C with equation $y = e^{2x} + x^2 - 3$ The curve C crosses the y-axis at the point A - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 5

Worked Solution & Example Answer:Figure 1 shows a sketch of part of the curve C with equation $y = e^{2x} + x^2 - 3$ The curve C crosses the y-axis at the point A - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 5

Find the equation of l, writing your answer in the form y = mx + c

Show that the x coordinate of B is a solution of x = \sqrt{1 + \frac{1}{2} x - e^{-x}}

Find x2 and x3 to three decimal places

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