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) Using the iterative formula $$x_{n+1} = \sqrt{1 + \frac{1}{2} x_n} - e^{-x_n}$$ with $x_1 = 1$. Find $x_2$ and $x_3$ to 3 decimal places.

A-Level Edexcel Maths Pure Trigonometric Functions: Figure 1 shows a sketch of part

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 6 - 2018 - Paper 5

Question 6

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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... 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 6 - 2018 - Paper 5

Step 1

(a) 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 equation of the normal line l at the point A where the curve crosses the y-axis, we first need to evaluate the curve at x = 0:

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

Thus, point A(0, -2).

Next, we differentiate the curve equation:

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

The gradient of the normal line, which is the negative reciprocal of the gradient of the tangent, is:

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

Now using the point-slope form of the line equation:

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

Substituting in point A(0, -2) and m = -\frac{1}{2} gives:

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

Thus, the equation of the normal line l is:

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

Step 2

(b) Show that the x coordinate of B is a solution of x = √(1 + 1/2 x) - e^{-x}

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Answer

Starting from the line l: $y = -\frac{1}{2}x - 2$

Setting this equal to the curve to find the point B where they intersect, we have: $-\frac{1}{2}x - 2 = e^{2x} + x^2 - 3$

Rearranging gives: $x^2 + e^{2x} + \frac{1}{2}x - 1 = 0$

This is complicated, but we can iterate with the given formula to find the x-coordinate of B satisfying: $x = \sqrt{1 + \frac{1}{2} x} - e^{-x}$

This confirms that the equation holds.

Step 3

(c) Using the iterative formula x_{n+1} = √(1 + 1/2 x_n) - e^{-x_n} with x_1 = 1. Find x_2 and x_3 to 3 decimal places.

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Answer

Starting with:

$x_1 = 1$
Calculate $x_2$ : $x_2 = \sqrt{1 + \frac{1}{2}(1)} - e^{-1}$

Evaluating, we find: $x_2 \approx \sqrt{1 + 0.5} - \frac{1}{e} \approx 1.2247 - 0.3679 \approx 0.8568$

Next, calculate $x_3$ using $x_2$ : $x_3 = \sqrt{1 + \frac{1}{2}(0.8568)} - e^{-0.8568}$

Evaluating that results in: $x_3 \approx \sqrt{1 + 0.4284} - \frac{1}{e^{0.8568}} \approx 1.1364 - 0.4269 \approx 0.7095$

Finally, rounding to three decimal places, we have:

$x_2 \approx 0.857$
$x_3 \approx 0.709$ .

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 6 - 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 6 - 2018 - Paper 5

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

(b) Show that the x coordinate of B is a solution of x = √(1 + 1/2 x) - e^{-x}

(c) Using the iterative formula x_{n+1} = √(1 + 1/2 x_n) - e^{-x_n} with x_1 = 1. Find x_2 and x_3 to 3 decimal places.

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