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.

A-Level Edexcel Maths Pure Trigonometric Equations: 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 9 - 2008 - Paper 2

Question 9

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-coord... 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 9 - 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 need to calculate the first derivative of the function:

$y' = 8 + 2x - 3x^2.$

Setting the derivative to zero to find the critical points:

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

Rearranging gives:

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

Using the quadratic formula, 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:

$(-2)^2 - 4 \cdot 3 \cdot (-8) = 4 + 96 = 100$

Now solving for x:

$x = \frac{2 \pm 10}{6}$

This gives:

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

Since A is a maximum, we confirm $x=2$ by checking the second derivative:

$y'' = 2 - 6x.$

At $x=2$ :

$y''(2) = 2 - 6(2) = -10 < 0,$

indicating a maximum point at A.

Step 2

Using calculus, find the exact area of R.

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Answer

To find the area of region R, we set up the following integral:

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

Now we calculate the integral:

Find the antiderivative: $\int (10 + 8x + x^2 - x^3) \, dx = 10x + 4x^2 + \frac{x^3}{3} - \frac{x^4}{4} + C.$
Evaluate it from 0 to 2: $= \left[ 10(2) + 4(2^2) + \frac{(2)^3}{3} - \frac{(2)^4}{4} \right] - \left[ 10(0) + 4(0^2) + 0 - 0 \right].$

Calculating each term: $= \left[ 20 + 16 + \frac{8}{3} - 4 \right]$ $= 32 + \frac{8}{3} - 4 = 28 + \frac{8}{3} = \frac{84 + 8}{3} = \frac{92}{3}.$

Thus, the exact area of region R is:

$\frac{92}{3}.$

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 9 - 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 9 - 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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