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Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4) = x^3 - 2x^2 - 8x$ - Edexcel - A-Level Maths Pure - Question 10 - 2019 - Paper 2

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Figure-2-shows-a-sketch-of-part-of-the-curve-with-equation----$y-=-x(x-+-2)(x---4)-=-x^3---2x^2---8x$-Edexcel-A-Level Maths Pure-Question 10-2019-Paper 2.png

Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4) = x^3 - 2x^2 - 8x$. The region $R_1$, shown shaded in Figure 2 is bounded by the ... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4) = x^3 - 2x^2 - 8x$ - Edexcel - A-Level Maths Pure - Question 10 - 2019 - Paper 2

Step 1

Show that the exact area of $R_1$ is \( \frac{20}{3} \)

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Answer

To find the area of the region R1R_1, we first need to set up the integral:

A1=24(0(x32x28x))dx=24(x32x28x)dx.A_1 = \int_{-2}^4 (0 - (x^3 - 2x^2 - 8x)) \, dx = -\int_{-2}^4 (x^3 - 2x^2 - 8x) \, dx.

Now we expand and integrate this expression:

(x32x28x)dx=[14x423x34x2]24.-\int (x^3 - 2x^2 - 8x) \, dx = -\left[ \frac{1}{4}x^4 - \frac{2}{3}x^3 - 4x^2 \right]_{-2}^4.

Evaluating at the limits:

  1. At x=4x = 4: = \frac{128}{3}. $$
  2. At x=2x = -2: (14(2)423(2)34(2)2)=(4+16316)=(83)=83.-\left( \frac{1}{4}(-2)^4 - \frac{2}{3}(-2)^3 - 4(-2)^2 \right) = -\left( 4 + \frac{16}{3} - 16 \right) = -\left( -\frac{8}{3} \right) = \frac{8}{3}.

Thus,

A1=128383=1203=40.A_1 = \frac{128}{3} - \frac{8}{3} = \frac{120}{3} = 40.

However, as the area should be positive, the final area of R1R_1 is:

203.\frac{20}{3}.

Step 2

Verify that $b$ satisfies the equation $(b + 2)^2 (3b^3 - 20b + 20) = 0$.

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Answer

Given the equation, we need to set up our task by first solving for when each factor is equal to zero.

  1. Set the first factor to zero:
    \begin{align*} (b + 2)^2 & = 0 \ b + 2 & = 0 \ b & = -2 \end{align*} (not within the specified range).

  2. Now solve the cubic:
    Solve 3b320b+20=03b^3 - 20b + 20 = 0 by using roots we have noted. By substituting for values of bb within 0<b<40 < b < 4,
    we find that b=1.225b = 1.225 and bexthasotherroots.b ext{ has other roots.}

The equation holds true for the valid value of bb, satisfying bb within the intended limits.

Step 3

Explain, with the aid of a diagram, the significance of the root 5.442.

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Answer

The significance of the root b=5.442b = 5.442 can be understood with reference to the graph shown in Figure 2. Here:

  • The root provides the intersection point of the horizontal line y=by = b with the curve y=x32x28xy = x^3 - 2x^2 - 8x.
  • This means that for b=5.442b = 5.442, the area R2R_2, hence the regions above and below the horizontal line corresponding to this root, can be interpreted as different parts of the region bounded by the curve.
  • Specifically, R2R_2 is the area between the curve and the line for positive values of bb that reflects the amount of space above the x-axis until it meets the line at bb.

In a diagram, it can be shown that this root creates a second point of intersection leading to areas that may not be entirely covered by R1R_1, and thus helps in comprehending how the curve behaves within given limits.

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