The curve C has parametric equations $x = n + 2$, $y = rac{1}{(t + 1)}$, $t > -1$. The finite region R between the curve C and the x-axis, bounded by the lines with equations $x = ext{ln} ext{2}$ and $x = ext{ln} ext{4}$, is shown shaded in Figure 3. (a) Show that the area of R is given by the integral $$egin{aligned} igg(rac{2}{0}^{1} rac{1}{(t + 1)(t + 2)} igg)dr. \\ \\ ext{(4)} \\ \\ \\ (b) Hence find an exact value for this area. \\ \\ ext{(6)} \\ \\ \ (c) Find a cartesian equation of the curve C, in the form $y = f(x)$. \\ \\ ext{(4)} \\ \\ \ (d) State the domain of values for x for this curve. \\ \\ ext{(1)}

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The curve C has parametric equations $x = n + 2$, $y = rac{1}{(t + 1)}$, $t > -1$ - Edexcel - A-Level Maths: Pure - Question 1 - 2008 - Paper 8

Question 1

The curve C has parametric equations $x = n + 2$, $y = rac{1}{(t + 1)}$, $t > -1$. The finite region R between the curve C and the x-axis, bounded by the li... show full transcript

Worked Solution & Example Answer:The curve C has parametric equations $x = n + 2$, $y = rac{1}{(t + 1)}$, $t > -1$ - Edexcel - A-Level Maths: Pure - Question 1 - 2008 - Paper 8

Step 1

Show that the area of R is given by the integral

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Answer

To find the area of region R, we use the formula for the area between a parametric curve and the x-axis. The area A can be expressed as:

ewline\int_{a}^{b} y rac{dx}{dt}igg|_t dt$$ In our case, we have: 1. We need to calculate $ rac{dx}{dt}$: $$\frac{dx}{dt} = \frac{1}{t + 2}$$ 2. We substitute this and our parametric equations into the integral: $$A = \int_{0}^{2} \left(\frac{1}{(t + 1)}\right) \left(\frac{1}{(t + 2)}\right) dt$$ 3. The limits of integration corresponding to $t+1=0$ (i.e., $t=-1$ or $x=ln(2)$) and $t= ext{ln}(2)$ (i.e., $x=ln(4)$). Thus, the area A of region R is given as: $$A = \int_{0}^{2} \frac{1}{(t + 1)(t + 2)} \, dt$$

Step 2

Hence find an exact value for this area.

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Answer

To evaluate the integral,

$\int_{0}^{2} \frac{1}{(t + 1)(t + 2)} \, dt$ ,

we can perform partial fraction decomposition:

$\frac{1}{(t + 1)(t + 2)} = \frac{A}{(t + 1)} + \frac{B}{(t + 2)}$ .

By equating, we find:

Multiply through by the denominator:

$1 = A(t + 2) + B(t + 1)$ .

Solving for A and B gives:

A = 1 and B = -1.

Consequently, we rewrite the integral:

$\int_{0}^{2} \left(\frac{1}{(t + 1)} - \frac{1}{(t + 2)}\right) dt$ .

Now, integrate:

$\left[\ln(t + 1) - \ln(t + 2)\right]_{0}^{2} = (\ln(3) - \ln(4)) - (\ln(1) - \ln(2))$

Therefore, simplifying the expression gives:

$A = \ln(\frac{3}{4}) - \ln(1) + \ln(2) = \ln(\frac{3}{2})$ .

Hence, the exact value of the area is:

$A = \ln(\frac{3}{2})$ .

Step 3

Find a cartesian equation of the curve C, in the form y = f(x).

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Answer

To find the cartesian equation from the parametric equations:

Given:
$x = \ln(t + 2)$
$y = \frac{1}{(t + 1)}$
Rearranging for t from the equation of x:
$t + 2 = e^{x} \\ t = e^{x} - 2$
Substitute t into the equation for y: $y = \frac{1}{(e^{x} - 1)}$

Thus, the cartesian equation of the curve in the form $y = f(x)$ is:

$y = \frac{1}{(e^{x} - 1)}$ .

Step 4

State the domain of values for x for this curve.

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Answer

The domain for x must ensure that the denominator is not zero:

From the equation $y = \frac{1}{(e^{x} - 1)}$ ,

We need $e^{x} - 1 > 0$ ⇔ $e^{x} > 1$ .
This implies $x > 0$ .

Thus, the domain of values for x is:

$x > 0$ .

The curve C has parametric equations $x = n + 2$, $y = rac{1}{(t + 1)}$, $t > -1$ - Edexcel - A-Level Maths: Pure - Question 1 - 2008 - Paper 8

Worked Solution & Example Answer:The curve C has parametric equations $x = n + 2$, $y = rac{1}{(t + 1)}$, $t > -1$ - Edexcel - A-Level Maths: Pure - Question 1 - 2008 - Paper 8

Show that the area of R is given by the integral

Hence find an exact value for this area.

Find a cartesian equation of the curve C, in the form y = f(x).

State the domain of values for x for this curve.

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