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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 4 - 2018 - Paper 5

Question 4

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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 a... 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 4 - 2018 - Paper 5

Step 1

Find the equation of l, writing your answer in the form y = mx + c, where m and c are constants.

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Answer

To find the normal line l at point A where the curve crosses the y-axis, we first determine the coordinates of A by setting x = 0:

$y = e^{2(0)} + (0)^{2} - 3 = 1 - 3 = -2$

Thus, point A is (0, -2). Next, we need to differentiate the equation of the curve to find the gradient:

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

At x = 0, the gradient is:

$\frac{dy}{dx} \bigg|_{x=0} = 2e^{0} + 0 = 2$

Since line l is normal to curve C at A, its slope is the negative reciprocal:

$m = -\frac{1}{2}$

Now we can use the point-slope form to write the equation of the normal line:

$y - y_1 = m(x - x_1)$

Substituting in the values, we have:

$y - (-2) = -\frac{1}{2}(x - 0)$

Simplifying this yields:

$y + 2 = -\frac{1}{2}x\nRightarrow y = -\frac{1}{2}x - 2$

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 find where line l meets curve C again at point B, we equate the standard equation of line l with the equation of curve C:

$-\frac{1}{2}x - 2 = e^{2x} + x^{2} - 3$

Rearranging gives us:

$e^{2x} + x^{2} + \frac{1}{2}x + 1 = 0$

While this doesn't directly give the x-coordinate of B, we note that this should be approximated to verify:

$x = \sqrt{1 + \frac{1}{2}x} - e^{-x}$ can be used to numerically verify the x-coordinate at point B.

Step 3

Find x_2 and x_3 to 3 decimal places.

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Answer

Given the iterative formula:

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

Starting from the initial value $x_1 = 1$ :

Calculate $x_2$ :
- Substitute $x_1 = 1$ : $x_2 = \sqrt{1 + \frac{1}{2}(1)} - e^{-1}$
- Calculate the values: $x_2 \approx 1.168$
Next, calculate $x_3$ using $x_2$ :
- Substitute $x_2 \approx 1.168$ : $x_3 = \sqrt{1 + \frac{1}{2}(1.168)} - e^{-1.168}$
- Continuing from prior calculations gives: $x_3 \approx 1.220$

Thus, the solutions are:

$x_2 \approx 1.168$
$x_3 \approx 1.220$

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 4 - 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 4 - 2018 - Paper 5

Find the equation of l, writing your answer in the form y = mx + c, where m and c are constants.

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

Find x_2 and x_3 to 3 decimal places.

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