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Figure 2 shows a sketch of part of the curve with equation y = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 1 - 2015 - Paper 2

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Figure-2-shows-a-sketch-of-part-of-the-curve-with-equation--y-=-4x³-+-9x²---30x---8,---0.5-≤-x-≤-2.2--The-curve-has-a-turning-point-at-the-point-A-Edexcel-A-Level Maths Pure-Question 1-2015-Paper 2.png

Figure 2 shows a sketch of part of the curve with equation y = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A. (a) Using calcul... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation y = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 1 - 2015 - Paper 2

Step 1

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

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Answer

To find the x-coordinate of the turning point A, we first need to calculate the derivative of the function.

Starting with the given function:

y=4x3+9x230x8y = 4x^3 + 9x^2 - 30x - 8

The first derivative is:

dydx=12x2+18x30\frac{dy}{dx} = 12x^2 + 18x - 30

Now, we set the derivative equal to zero to find the turning points:

12x2+18x30=012x^2 + 18x - 30 = 0

Dividing the entire equation by 6 gives:

2x2+3x5=02x^2 + 3x - 5 = 0

Using the quadratic formula, where (a = 2), (b = 3), and (c = -5):

x=b±b24ac2a=3±324(2)(5)2(2)x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)}

Calculating the discriminant:

b24ac=9+40=49b^2 - 4ac = 9 + 40 = 49

Thus, the solutions for x are:

x=3±74x = \frac{-3 \pm 7}{4}

This produces two potential solutions:

  1. x=44=1x = \frac{4}{4} = 1
  2. x=104=2.5x = \frac{-10}{4} = -2.5

Since we're considering the range (-0.5 ≤ x ≤ 2.2), we conclude that the x-coordinate of point A is indeed 1.

Step 2

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

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Answer

To find the area of region R enclosed by the curve, line AB, and the x-axis, we first need to establish the integration limits which are between C(−\frac{1}{4}, 0) and B(2, 0).

Thus, the area A can be defined as:

A=142(f(x)0)dxA = \int_{-\frac{1}{4}}^{2} (f(x) - 0) \, dx

where f(x) is the given curve equation. Substituting the curve's equation into the integral gives:

A=142(4x3+9x230x8)dxA = \int_{-\frac{1}{4}}^{2} (4x^3 + 9x^2 - 30x - 8) \, dx

Calculating the integral, we have:

  1. Find the antiderivative:
    • 4x3dx=x4\int 4x^3 dx = x^4
    • 9x2dx=3x3\int 9x^2 dx = 3x^3
    • 30xdx=15x2\int -30x dx = -15x^2
    • 8dx=8x\int -8 dx = -8x

So the antiderivative is:

F(x)=x4+3x315x28xF(x) = x^4 + 3x^3 - 15x^2 - 8x

  1. Evaluating from (-\frac{1}{4}) to 2:
    • F(2)=24+3(23)15(22)8(2)F(2) = 2^4 + 3(2^3) - 15(2^2) - 8(2)

    • =16+246016=36= 16 + 24 - 60 - 16 = -36

    • F(14)=(14)4+3(14)315(14)28(14)F(-\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)

    • This simplifies to:

    • =12563641516+2=112240+512256=261256= \frac{1}{256} - \frac{3}{64} - \frac{15}{16} + 2 = \frac{1 - 12 - 240 + 512}{256} = \frac{261}{256}

Finally, we can calculate the area:

A=F(2)F(14)=36261256A = F(2) - F(-\frac{1}{4}) = -36 - \frac{261}{256}

Calculating this gives us: A=361.019=37.019A = -36 - 1.019 = -37.019

Since we're interested in the absolute area, we can express:

A=37.019.|A| = 37.019.

Thus, giving the final answer to two decimal places results in:

A37.02.A \approx 37.02.

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