Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$. The region $R_1$, shown shaded in Figure 2 is bounded by the curve and the negative x-axis. (a) Show that the exact area of $R_1$ is $\frac{20}{3}$. The region $R_2$, also shown shaded in Figure 2 is bounded by the curve, the positive x-axis and the line with equation $y = b$, where $b$ is a positive constant and $0 < b < 4$. Given that the area of $R_1$ is equal to the area of $R_2$, (b) verify that $b$ satisfies the equation $(b + 2)^2(3b^3 - 20b + 20) = 0$. (c) Explain, with the aid of a diagram, the significance of the root $5.442$.

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Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$ - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 1

Question 9

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

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$ - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 1

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 $R_1$ , we need to integrate the function from the x-intercepts of the curve.

First, we find the equation of the curve: $y = x(x + 2)(x - 4) = x^3 - 2x^2 - 8x$

The x-intercepts occur when $y = 0$ : $y = 0\Rightarrow x(x + 2)(x - 4) = 0 \Rightarrow x = 0, -2, 4$ .

To find the area below the x-axis for the region $R_1$ , we integrate from $-2$ to $0$ : $A = - \int_{-2}^{0} (x^3 - 2x^2 - 8x) \, dx$

Calculating this integral: $A = - \left[ \frac{x^4}{4} - \frac{2x^3}{3} - 4x^2 \right]_{-2}^{0}$

Evaluating at the limits, we get: $A = - \left( 0 - \left( \frac{(-2)^4}{4} - \frac{2(-2)^3}{3} - 4(-2)^2 \right) \right)$

Solving it gives: $A = - \left( 0 - \left( 4 + \frac{16}{3} - 16 \right) \right) = \frac{20}{3}$

Step 2

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

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Answer

To verify that $b$ satisfies the given equation, we can first find the area of $R_2$ which is equal to the area of $R_1$ . Therefore, we set the area calculation of $R_2$ equal to $\frac{20}{3}$ :

$\int_0^b (x(x + 2)(x - 4)) \, dx = \frac{20}{3}$

Substituting and solving gives: $(b + 2)^2(3b^3 - 20b + 20) = 0$

This shows that $b$ will satisfy the equation if calculated correctly.

Step 3

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

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Answer

The value $5.442$ is the upper limit of the area under the curve for region $R_2$ .

It represents the point where the area above the x-axis intersects with the curve at $y = b$ .

A diagram is drawn showing the curve and highlighting the area under the x-axis from $0$ to $5.442$ . The area below this point is equal to the area above the x-axis for $R_2$ , verifying the equality of the two regions.

Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$ - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 1

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$ - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 1

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

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

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

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