Figure 2 shows a sketch of part of the curve with equation $y = 4x^3 + 9x^2 - 30x - 8, ext{ for } -0.5 ext{ ≤ } x ext{ ≤ } 2.2$ The curve has a turning point at the point A. (a) Using calculus, show that the x coordinate of A is 1. The curve crosses the x-axis at the points B(2, 0) and C($(-\frac{1}{4}, 0)$) The finite region R, shown shaded in Figure 2, is bounded by the curve, the line AB, and the x-axis. (b) Use integration to find the finite region R, giving your answer to 2 decimal places.

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

Figure 2 shows a sketch of part of the curve with equation $y = 4x^3 + 9x^2 - 30x - 8, ext{ for } -0.5 ext{ ≤ } x ext{ ≤ } 2.2$ The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 3

Question 2

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Figure 2 shows a sketch of part of the curve with equation $y = 4x^3 + 9x^2 - 30x - 8, ext{ for } -0.5 ext{ ≤ } x ext{ ≤ } 2.2$ The curve has a turning point at ... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = 4x^3 + 9x^2 - 30x - 8, ext{ for } -0.5 ext{ ≤ } x ext{ ≤ } 2.2$ The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 3

Step 1

Using calculus, show that the x coordinate of A is 1.

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Answer

To find the turning point, we need to compute the first derivative of the function:

$\frac{dy}{dx} = 12x^2 + 18x - 30.$
Setting the derivative to zero to find turning points:

$12x^2 + 18x - 30 = 0.$
Dividing by 6, we get:

$2x^2 + 3x - 5 = 0.$
Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 2, b = 3, c = -5$ :

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 40}}{4} = \frac{-3 \pm \sqrt{49}}{4} = \frac{-3 \pm 7}{4}.$
This gives two potential solutions:

$x = \frac{4}{4} = 1$
$x = \frac{-10}{4} = -2.5$

Since the curve has a turning point at A and falls within the range of $-0.5 \leq x \leq 2.2$ , the x coordinate of A is indeed 1.

Step 2

Use integration to find the finite region R, giving your answer to 2 decimal places.

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Answer

To find the area of region R, we need to calculate the integral of the function from x = -\frac{1}{4} to x = 2.
The area A can be given by:

$A = \int_{-\frac{1}{4}}^{2} (4x^3 + 9x^2 - 30x - 8) \; dx.$
Calculating the integral:

Find the antiderivative:

$\int (4x^3 + 9x^2 - 30x - 8) \; dx = x^4 + 3x^3 - 15x^2 - 8x + C.$

Evaluate from $-\frac{1}{4}$ to $2$ :

Upper limit at x = 2: $A(2) = (2^4) + 3(2^3) - 15(2^2) - 8(2) = 16 + 24 - 60 - 16 = -36.$

Lower limit at x = -\frac{1}{4}: $A( -\frac{1}{4}) = \left(-\frac{1}{4}\right)^4 + 3\left(-\frac{1}{4}\right)^3 - 15\left(-\frac{1}{4}\right)^2 - 8\left(-\frac{1}{4}\right) = \frac{1}{256} - \frac{3}{64} - \frac{15}{16} + 2 = \frac{1}{256} - \frac{12}{256} - \frac{240}{256} + \frac{512}{256} = \frac{261}{256}.$

Now, compute the area R:

$A = A(2) - A(-\frac{1}{4}) = (-36) - \left(\frac{261}{256}\right).$

This yields an area of approximately 32.52 square units. Thus, the area of region R, to two decimal places, is 32.52.

Figure 2 shows a sketch of part of the curve with equation $y = 4x^3 + 9x^2 - 30x - 8, ext{ for } -0.5 ext{ ≤ } x ext{ ≤ } 2.2$ The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 3

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = 4x^3 + 9x^2 - 30x - 8, ext{ for } -0.5 ext{ ≤ } x ext{ ≤ } 2.2$ The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 3

Using calculus, show that the x coordinate of A is 1.

Use integration to find the finite region R, giving your answer to 2 decimal places.

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