Figure 2 shows a sketch of part of the curve C with equation $y = x^3 - 10x^2 + kx$, where $k$ is a constant. The point P on C is the maximum turning point. Given that the x-coordinate of P is 2, (a) show that $k = 28$. The line through P parallel to the x-axis cuts the y-axis at the point N. The region R is bounded by C, the y-axis and PN, as shown shaded in Figure 2. (b) Use calculus to find the exact area of R.

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

Figure 2 shows a sketch of part of the curve C with equation $y = x^3 - 10x^2 + kx$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 10 - 2010 - Paper 3

Question 10

Figure 2 shows a sketch of part of the curve C with equation $y = x^3 - 10x^2 + kx$, where $k$ is a constant. The point P on C is the maximum turning point. Gi... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation $y = x^3 - 10x^2 + kx$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 10 - 2010 - Paper 3

Step 1

(a) show that k = 28

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Answer

To show that $k = 28$ , we first need to find the derivative of the function:

$\frac{dy}{dx} = 3x^2 - 20x + k$

Since point P is a maximum turning point, the derivative at this point must equal zero. Therefore, we substitute $x = 2$ into the derivative:

$\frac{dy}{dx} \bigg|_{x=2} = 3(2)^2 - 20(2) + k = 0$

Calculating this gives:

$\frac{dy}{dx} \bigg|_{x=2} = 3(4) - 40 + k = 0$

This simplifies to:

$12 - 40 + k = 0$

Thus, we can isolate $k$ :

$k = 40 - 12 = 28.$

Therefore, we have shown that $k = 28$ .

Step 2

(b) Use calculus to find the exact area of R.

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Answer

To find the area of region R, we first need the y-coordinate of point P, which we find using the function with $k = 28$ . The function becomes:

$y = x^3 - 10x^2 + 28x.$

Substituting $x = 2$ :

$y = 2^3 - 10(2^2) + 28(2) = 8 - 40 + 56 = 24.$

Now, we can set up the integral for the area of region R bounded by the curve and the y-axis:

$\text{Area} = \int_0^2 (x^3 - 10x^2 + 28x) \, dx.$

Calculating the integral:

$\int_0^2 (x^3 - 10x^2 + 28x) \, dx = \left[ \frac{x^4}{4} - \frac{10x^3}{3} + 14x^2 \right]_0^2.$

Now substituting the limits:

$= \left[ \frac{2^4}{4} - \frac{10(2^3)}{3} + 14(2^2) \right] - \left[ 0 \right]$ $= \left[ 4 - \frac{80}{3} + 56 \right]$

Calculating further:

$= 4 + 56 - \frac{80}{3} = \frac{12}{3} + \frac{168}{3} - \frac{80}{3} = \frac{100}{3}.$

Thus, the exact area of region R is:

$\text{Area} = \frac{100}{3}.$

Figure 2 shows a sketch of part of the curve C with equation $y = x^3 - 10x^2 + kx$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 10 - 2010 - Paper 3

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation $y = x^3 - 10x^2 + kx$, where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 10 - 2010 - Paper 3

(a) show that k = 28

(b) Use calculus to find the exact area of R.

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