Figure 1 shows part of a curve C with equation $y = 2x + \frac{8}{x^2} - 5$, $x > 0$. The points P and Q lie on C and have x-coordinates 1 and 4 respectively. The region R, shaded in Figure 1, is bounded by C and the straight line joining P and Q. (a) Find the exact area of R. (b) Use calculus to show that $y$ is increasing for $x > 2$.

A-Level Edexcel Maths Pure Laws of Logarithms: Figure 1 shows part of a curve C

Figure 1 shows part of a curve C with equation $y = 2x + \frac{8}{x^2} - 5$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 2

Question 2

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Figure 1 shows part of a curve C with equation $y = 2x + \frac{8}{x^2} - 5$, $x > 0$. The points P and Q lie on C and have x-coordinates 1 and 4 respectively. The ... show full transcript

Worked Solution & Example Answer:Figure 1 shows part of a curve C with equation $y = 2x + \frac{8}{x^2} - 5$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 2

Step 1

Find the exact area of R.

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Answer

To find the area of the region R, we first need to determine the points P and Q on the curve.

For point P at $x = 1$ : $y = 2(1) + \frac{8}{1^2} - 5 = 2 + 8 - 5 = 5\rightarrow (1, 5)$
For point Q at $x = 4$ : $y = 2(4) + \frac{8}{4^2} - 5 = 8 + 0.5 - 5 = 3.5\rightarrow (4, 3.5)$
The line segment connecting points P (1,5) and Q (4,3.5) has the equation: $y - 5 = \frac{3.5 - 5}{4 - 1}(x - 1)\rightarrow y = -0.5x + 6.5$

Next, we calculate the area between the curve and the line between x = 1 and x = 4: $\text{Area} = \int_1^4 \left((2x + \frac{8}{x^2} - 5) - (-0.5x + 6.5)\right) dx$

Simplifying the integrand gives: $2.5x - \frac{8}{x^2} - 11.5$

Now compute the integral: $\int_1^4 \left(2.5x - \frac{8}{x^2} - 11.5\right) dx$

Evaluating this from 1 to 4 yields the exact area of region R.

Step 2

Use calculus to show that $y$ is increasing for $x > 2$.

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Answer

To determine where the function is increasing, we first need to find the derivative:
$\frac{dy}{dx} = 2 - \frac{16}{x^3}$

Next, we set the derivative greater than zero for $x > 2$ :
$2 - \frac{16}{x^3} > 0 \Rightarrow \frac{16}{x^3} < 2\Rightarrow 16 < 2x^3 \Rightarrow 8 < x^3\Rightarrow x > 2$

Thus, since the derivative is positive for $x > 2$ , it indicates that the function $y$ is indeed increasing for this range.

Figure 1 shows part of a curve C with equation $y = 2x + \frac{8}{x^2} - 5$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 2

Worked Solution & Example Answer:Figure 1 shows part of a curve C with equation $y = 2x + \frac{8}{x^2} - 5$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 2

Find the exact area of R.

Use calculus to show that $y$ is increasing for $x > 2$.

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