The straight line with equation $y = x + 4$ cuts the curve with equation $y = -x^3 + 2x + 24$ at the points A and B, as shown in Figure 3. (a) Use algebra to find the coordinates of the points A and B. (4) The finite region R is bounded by the straight line and the curve and is shown shaded in Figure 3. (b) Use calculus to find the exact area of R. (7)

A-Level Edexcel Maths Pure Polynomials: The straight line with equation

The straight line with equation $y = x + 4$ cuts the curve with equation $y = -x^3 + 2x + 24$ at the points A and B, as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 3

Question 2

The-straight-line-with-equation-$y-=-x-+-4$-cuts-the-curve-with-equation-$y-=--x^3-+-2x-+-24$-at-the-points-A-and-B,-as-shown-in-Figure-3-Edexcel-A-Level Maths Pure-Question 2-2010-Paper 3.png

The straight line with equation $y = x + 4$ cuts the curve with equation $y = -x^3 + 2x + 24$ at the points A and B, as shown in Figure 3. (a) Use algebra to find t... show full transcript

Worked Solution & Example Answer:The straight line with equation $y = x + 4$ cuts the curve with equation $y = -x^3 + 2x + 24$ at the points A and B, as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 3

Step 1

Use algebra to find the coordinates of the points A and B.

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Answer

To find the coordinates of points A and B where the line and curve intersect, we set their equations equal to each other:

$-x^3 + 2x + 24 = x + 4$

Rearranging this gives:

$-x^3 + x + 20 = 0$

Using trial and error or synthetic division, we can find the roots of this polynomial. By testing potential rational roots, we find that $x = -4$ is a root. Performing synthetic division to factor the cubic polynomial results in:

$(x + 4)(-x^2 + 4) = 0$

The quadratic can be factored further:

$(x + 4)(-1)(x - 2)(x + 2) = 0$

Thus, the solutions for x are $x = -4$ , $x = 2$ , and $x = -2$ . To find the corresponding y-values:

For $x = -4$ : $y = -4 + 4 = 0$ ; so point A is $(-4, 0)$ .
For $x = 2$ : $y = 2 + 4 = 6$ ; so point B is $(2, 6)$ .
For $x = -2$ : $y = -2 + 4 = 2$ ; but this doesn't match with the curve's output since we only need A and B.

The points are A $(-4, 0)$ and B $(2, 6)$ .

Step 2

Use calculus to find the exact area of R.

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Answer

To determine the area of region R, we need to integrate the difference of the functions representing the curve and the line between their intersection points:

The area R can be calculated as:

$A = \int_{-4}^{2} ((-x^3 + 2x + 24) - (x + 4)) \, dx$

Simplifying the integrand gives:

$A = \int_{-4}^{2} (-x^3 + x + 20) \, dx$
Now, we calculate the definite integral:

$A = \left[ -\frac{x^4}{4} + \frac{x^2}{2} + 20x \right]_{-4}^{2}$
Evaluating at the bounds:

For $x = 2$ :

$= -\frac{2^4}{4} + \frac{2^2}{2} + 20(2) = -4 + 2 + 40 = 38$

For $x = -4$ :

$= -\frac{(-4)^4}{4} + \frac{(-4)^2}{2} + 20(-4) = -64 + 8 - 80 = -136$
Thus,

$A = 38 - (-136) = 38 + 136 = 174$

The exact area of region R is thus $A = 174$ square units.

The straight line with equation $y = x + 4$ cuts the curve with equation $y = -x^3 + 2x + 24$ at the points A and B, as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 3

Worked Solution & Example Answer:The straight line with equation $y = x + 4$ cuts the curve with equation $y = -x^3 + 2x + 24$ at the points A and B, as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 3

Use algebra to find the coordinates of the points A and B.

Use calculus to find the exact area of R.

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