Figure 2 shows a sketch of part of the curve C with equation y = x³ - 10x² + 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 Laws of Logarithms: Figure 2 shows a sketch of part

Figure 2 shows a sketch of part of the curve C with equation y = x³ - 10x² + kx, where k is a constant - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 3

Question 9

Figure 2 shows a sketch of part of the curve C with equation y = x³ - 10x² + kx, where k is a constant. The point P on C is the maximum turning point. Give... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation y = x³ - 10x² + kx, where k is a constant - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 3

Step 1

show that k = 28

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Answer

To find the value of k, we start by differentiating the equation of the curve.

First, we find the derivative:
$\frac{dy}{dx} = 3x^2 - 20x + k$

Since P is a maximum turning point, the derivative at this point must be equal to zero:
$\frac{dy}{dx}\bigg|_{x=2} = 0$

Substituting x = 2 into the derivative gives:
$3(2)^2 - 20(2) + k = 0$

Calculating this, we find:
$12 - 40 + k = 0$

Therefore, it follows that:
$k = 28$ .

Step 2

Use calculus to find the exact area of R

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Answer

To find the area of region R, we need to integrate the curve from the x-coordinate of P (which is 2) to the x-coordinate where the curve intersects the y-axis (x=0).

First, we rewrite the equation of the curve using k = 28:
$y = x^3 - 10x^2 + 28x$

Now we can set up our integral to find the area:
$\text{Area} = \int_0^2 (x^3 - 10x^2 + 28x) \, dx$

Calculating the integral:
$\int (x^3 - 10x^2 + 28x) \, dx = \frac{x^4}{4} - \frac{10x^3}{3} + 14x^2$

Evaluating from 0 to 2:
$\left[ \frac{2^4}{4} - \frac{10(2^3)}{3} + 14(2^2) \right] - \left[ 0 \right]$

Calculating the upper limit:
$= \left[ \frac{16}{4} - \frac{80}{3} + 56 \right] = [4 - \frac{80}{3} + 56]$

Putting everything over a common denominator gives:
$= \frac{12}{3} - \frac{80}{3} + \frac{168}{3} = \frac{100}{3}$

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 = x³ - 10x² + kx, where k is a constant - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 3

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation y = x³ - 10x² + kx, where k is a constant - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 3

show that k = 28

Use calculus to find the exact area of R

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