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. (3) 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. (b) Using calculus, find the exact area of $R$. (8)

A-Level Edexcel Maths Pure Proof: 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 ... 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 maximum turning point, we first need to compute the derivative of the curve:

$f(x) = 10 + 8x + x^2 - x^3$

The derivative is:

$f'(x) = 8 + 2x - 3x^2$

Setting the derivative to zero to find critical points:

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

Rearranging gives:

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

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 3$ , $b = -2$ , and $c = -8$ . Thus,

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-8)}}{2(3)} = \frac{2 \pm \sqrt{4 + 96}}{6} = \frac{2 \pm 10}{6}$

This simplifies to two possible solutions:

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

Since the problem states it is a maximum turning point, we can verify by substituting back to find that $x = 2$ gives us a maximum as anticipated.

Step 2

Using calculus, find the exact area of R.

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Answer

To find the area of region $R$ , we can use integration from the origin $O(0,0)$ to point $A(2, f(2))$ :

The area $A$ is given by:

$A = \int_0^2 (10 + 8x + x^2 - x^3) \, dx$

Calculating this integral, we get:

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

Evaluating the integral at the bounds gives:

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

Calculating each term yields:

$A = 20 + 16 + \frac{8}{3} - 4 = 32 + \frac{8}{3} = \frac{96}{3} + \frac{8}{3} = \frac{104}{3}$

Thus, the exact area of region $R$ is:

$\frac{104}{3}$

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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