Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2-x^3$. The curve has a maximum turning point $A$. (a) Using calculus, show that the x-coordinate of $A$ is 2. (b) The region $R$, shown shaded in Figure 2, is bounded by the curve, the y-axis and the line from $O$ to $A$, where $O$ is the origin. Using calculus, find the exact area of $R$.

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

Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2-x^3$ - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 2

Question 1

Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2-x^3$. The curve has a maximum turning point $A$. (a) Using calculus, show that the x-coordi... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2-x^3$ - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 2

Step 1

Using calculus, show that the x-coordinate of A is 2.

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Answer

To find the x-coordinate of the maximum turning point $A$ , we first need to differentiate the given equation with respect to $x$ :

$\frac{dy}{dx} = 8 + 2x - 3x^2$

Setting the derivative equal to zero to find critical points:

$0 = 8 + 2x - 3x^2$

Rearranging the equation gives:

$3x^2 - 2x - 8 = 0$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 3$ , $b = -2$ , and $c = -8$ :

$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3}$

Calculating the discriminant:

$= \frac{2 \pm \sqrt{4 + 96}}{6} = \frac{2 \pm 10}{6}$

This gives two values:

$x = 2\quad \text{and} \quad x = -\frac{4}{3}$

Since we are looking for the maximum in our context, we evaluate the second derivative:

$\frac{d^2y}{dx^2} = 2 - 6x$

At $x = 2$ :

$\frac{d^2y}{dx^2} = 2 - 12 = -10 < 0$

Thus, $A$ is a maximum at $x = 2$ .

Step 2

Using calculus, find the exact area of R.

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Answer

To find the area of the region $R$ , we set up the integral from $x = 0$ to $x = 2$ :

$\text{Area} = \int_0^2 (10 + 8x + x^2 - x^3) \, dx$

First, we integrate:

$= \left[10x + 4x^2 + \frac{x^3}{3} - \frac{x^4}{4}\right]_0^2$

Evaluating at the bounds:

$= \left(10(2) + 4(2)^2 + \frac{(2)^3}{3} - \frac{(2)^4}{4}\right) - \left(10(0) + 4(0) + \frac{(0)^3}{3} - \frac{(0)^4}{4}\right)$

Calculating the terms:

$= (20 + 16 + \frac{8}{3} - 4)$

Combining all terms gives:

$= 32 + \frac{8}{3} = \frac{96}{3} + \frac{8}{3} = \frac{104}{3}$

Thus, the exact area of $R$ is:

$\text{Area of } R = \frac{104}{3} \text{ square units}$

Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2-x^3$ - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 2

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2-x^3$ - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 2

Using calculus, show that the x-coordinate of A is 2.

Using calculus, find the exact area of R.

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