1.02y Partial fractions: decompose rational functions

7 Fig. 7 illustrates the growth of a population with time. The proportion of the ultimate (long term) population is denoted by \(x\), and the time in years by \(t\). When \(t = 0 , x = 0.5\), and as \(t\) increases, \(x\) approaches 1 . \begin{figure}[h] \end{figure} One model for this situation is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ( 1 - x )$$

1. Verify that \(x = \frac { 1 } { 1 + \mathrm { e } ^ { - t } }\) satisfies this differential equation, including the initial condition.
2. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. An alternative model for this situation is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ^ { 2 } ( 1 - x ) ,$$ with \(x = 0.5\) when \(t = 0\) as before.
3. Find constants \(A , B\) and \(C\) such that \(\frac { 1 } { x ^ { 2 } ( 1 - x ) } = \frac { A } { x ^ { 2 } } + \frac { B } { x } + \frac { C } { 1 - x }\).
4. Hence show that \(t = 2 + \ln \left( \frac { x } { 1 - x } \right) - \frac { 1 } { x }\).
5. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value.

7 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).

1. Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
2. Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\). In a second model, \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when \(t = 0 , v = 0\).
3. Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
4. Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
5. Which of the two models fits the data better?

8 The growth of a tree is modelled by the differential equation $$10 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 20 - h ,$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).

1. Write down the value of \(h\) for which \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0\), and interpret this in terms of the growth of the tree.
2. Verify that \(h = 20 \left( 1 - \mathrm { e } ^ { - 0.1 t } \right)\) satisfies this differential equation and its initial condition. The alternative differential equation $$200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 400 - h ^ { 2 }$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
3. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } } .$$
4. What does this solution indicate about the long-term height of the tree?
5. After a year, the tree has grown to a height of 2 m . Which model fits this information better?

7 A drug is administered by an intravenous drip. The concentration, \(x\), of the drug in the blood is measured as a fraction of its maximum level. The drug concentration after \(t\) hours is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k \left( 1 + x - 2 x ^ { 2 } \right) ,$$ where \(0 \leqslant x < 1\), and \(k\) is a positive constant. Initially, \(x = 0\).

1. Express \(\frac { 1 } { ( 1 + 2 x ) ( 1 - x ) }\) in partial fractions.
2. Hence solve the differential equation to show that \(\frac { 1 + 2 x } { 1 - x } = \mathrm { e } ^ { 3 k t }\).
3. After 1 hour the drug concentration reaches \(75 \%\) of its maximum value and so \(x = 0.75\). Find the value of \(k\), and the time taken for the drug concentration to reach \(90 \%\) of its maximum value.
4. Rearrange the equation in part (ii) to show that \(x = \frac { 1 - \mathrm { e } ^ { - 3 k t } } { 1 + 2 \mathrm { e } ^ { - 3 k t } }\). Verify that in the long term the drug concentration approaches its maximum value. \section*{END OF QUESTION PAPER} \section*{Tuesday 16 J une 2015 - Afternoon} \section*{A2 GCE MATHEMATICS (MEI)} 4754/01B Applications of Advanced Mathematics (C4) Paper B: Comprehension \section*{QUESTION PAPER} \section*{Candidates answer on the Question Paper.} \section*{OCR supplied materials:}
   • Insert (inserted)
   • MEI Examination Formulae and Tables (MF2)
   \section*{Other materials required:}
   • Scientific or graphical calculator
   • Rough paper
   Duration: Up to 1 hour \includegraphics[max width=\textwidth, alt={}, center]{132ae754-bd4c-4819-80ef-4823ac2ead4f-05_117_495_1014_1308} PLEASE DO NOT WRITE IN THIS SPACE 2 In line 79 it says "For most journeys, more than half the journey time is composed of load time and transfer time". For what percentage of the journey time for the round trip made by car A in Table 4 is the car stationary?
   \includegraphics[max width=\textwidth, alt={}]{132ae754-bd4c-4819-80ef-4823ac2ead4f-07_645_1746_388_164}
   3 Using the expression on line 51, work out the answer to the question on lines 39 and 40 for the case where there are 10 upper floors and 7 people. Give your answer to 2 decimal places.
   \includegraphics[max width=\textwidth, alt={}]{132ae754-bd4c-4819-80ef-4823ac2ead4f-07_488_1746_1233_164}
   4 In lines 89 and 90 it says "... on average there will be approximately 8 stops per trip. A round trip with 8 stops would take between 188 and 200 seconds". Explain how the figure of 188 seconds has been derived. 5
5. Referring to Strategy 3 and lines 99 to 101, complete the table below for car C .
6. Calculate the time car C will take to transport all the people who work on floors 7 and 8 , and return to the ground floor.

   5	\includegraphics[max width=\textwidth, alt={}]{132ae754-bd4c-4819-80ef-4823ac2ead4f-08_1095_816_484_700}

   68 people make independent visits to any one of the upper floors of a building with 10 upper floors. What is the probability that at least one of the visitors goes to the top floor?

   6

   7 On lines 94 and 95 it says "Table 4 gives the timings for round trips in which the cars are required to stop at every floor they serve; Table 2 suggests this is a common occurrence in this case". Explain how Table 2 is used to make this claim. \includegraphics[max width=\textwidth, alt={}, center]{132ae754-bd4c-4819-80ef-4823ac2ead4f-09_1093_1740_1238_166} END OF QUESTION PAPER

5 You are given that \(\frac { 3 } { ( 5 + 3 x ) ( 2 + 3 x ) } \equiv \frac { 1 } { 2 + 3 x } - \frac { 1 } { 5 + 3 x }\).

1. Use this result to find \(\sum _ { r = 1 } ^ { 100 } \frac { 1 } { ( 5 + 3 r ) ( 2 + 3 r ) }\), giving your answer as an exact fraction.
2. Write down the limit to which \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 5 + 3 r ) ( 2 + 3 r ) }\) converges as \(n\) tends to infinity.

5

1. Show that \(\frac { 1 } { 5 r - 2 } - \frac { 1 } { 5 r + 3 } \equiv \frac { 5 } { ( 5 r - 2 ) ( 5 r + 3 ) }\) for all integers \(r\).
2. Hence use the method of differences to show that \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 5 r - 2 ) ( 5 r + 3 ) } = \frac { n } { 3 ( 5 n + 3 ) }\).

7 A curve has equation \(y = \frac { ( x + 2 ) ( 3 x - 5 ) } { ( 2 x + 1 ) ( x - 1 ) }\).

1. Write down the coordinates of the points where the curve crosses the axes.
2. Write down the equations of the three asymptotes.
3. Determine whether the curve approaches the horizontal asymptote from above or below for
   (A) large positive values of \(x\),
   (B) large negative values of \(x\).
4. Sketch the curve.

7 Fig. 7 shows an incomplete sketch of \(y = \frac { ( 2 x - 1 ) ( x + 3 ) } { ( x - 3 ) ( x - 2 ) }\). \begin{figure}[h] \end{figure}

1. Find the coordinates of the points where the curve cuts the axes.
2. Write down the equations of the three asymptotes.
3. Determine whether the curve approaches the horizontal asymptote from above or below for large positive values of \(x\), justifying your answer. Copy and complete the sketch.
4. Solve the inequality \(\frac { ( 2 x - 1 ) ( x + 3 ) } { ( x - 3 ) ( x - 2 ) } < 2\).

7 A curve has equation \(y = \frac { ( x + 9 ) ( 3 x - 8 ) } { x ^ { 2 } - 4 }\).

1. Write down the coordinates of the points where the curve crosses the axes.
2. Write down the equations of the three asymptotes.
3. Determine whether the curve approaches the horizontal asymptote from above or below for
   (A) large positive values of \(x\),
   (B) large negative values of \(x\).
4. Sketch the curve.

5

1. Show that \(\frac { 1 } { 2 r + 1 } - \frac { 1 } { 2 r + 3 } \equiv \frac { 2 } { ( 2 r + 1 ) ( 2 r + 3 ) }\).
2. Use the method of differences to find \(\sum _ { r = 1 } ^ { 30 } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }\), expressing your answer as a fraction.

7 A curve has equation \(y = \frac { x ^ { 2 } - 25 } { ( x - 3 ) ( x + 4 ) ( 3 x + 2 ) }\).

1. Write down the coordinates of the points where the curve crosses the axes.
2. Write down the equations of the asymptotes.
3. Determine how the curve approaches the horizontal asymptote for large positive values of \(x\), and for large negative values of \(x\).
4. Sketch the curve.

5 You are given that \(\frac { 4 } { ( 4 n - 3 ) ( 4 n + 1 ) } \equiv \frac { 1 } { 4 n - 3 } - \frac { 1 } { 4 n + 1 }\). Use the method of differences to show that $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 4 r - 3 ) ( 4 r + 1 ) } = \frac { n } { 4 n + 1 }$$

7 Fig. 7 shows an incomplete sketch of \(y = \frac { c x ^ { 2 } } { ( b x - 1 ) ( x + a ) }\) where \(a , b\) and \(c\) are integers. The asymptotes of the curve are also shown. \begin{figure}[h] \end{figure}

1. Determine the values of \(a , b\) and \(c\). Use these values of \(a , b\) and \(c\) throughout the rest of the question.
2. Determine how the curve approaches the horizontal asymptote for large positive values of \(x\), and for large negative values of \(x\), justifying your answer. On the copy of Fig. 7, sketch the rest of the curve.
3. Find the \(x\) coordinates of the points on the curve where \(y = 1\). Write down the solution to the inequality \(\frac { c x ^ { 2 } } { ( b x - 1 ) ( x + a ) } < 1\).

4 Use the identity \(\frac { 1 } { 2 r + 3 } - \frac { 1 } { 2 r + 5 } \equiv \frac { 2 } { ( 2 r + 3 ) ( 2 r + 5 ) }\) and the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r + 3 ) ( 2 r + 5 ) }\), expressing your answer as a single fraction.

6 Prove by induction that \(\frac { 1 } { 1 \times 3 } + \frac { 1 } { 3 \times 5 } + \frac { 1 } { 5 \times 7 } + \ldots + \frac { 1 } { ( 2 n - 1 ) ( 2 n + 1 ) } = \frac { n } { 2 n + 1 }\).

7 A curve has equation \(y = \frac { x ^ { 2 } - 5 } { ( x + 3 ) ( x - 2 ) ( a x - 1 ) }\), where \(a\) is a constant.

1. Find the coordinates of the points where the curve crosses the \(x\)-axis and the \(y\)-axis.
2. You are given that the curve has a vertical asymptote at \(x = \frac { 1 } { 2 }\). Write down the value of \(a\) and the equations of the other asymptotes.
3. Sketch the curve.
4. Find the set of values of \(x\) for which \(y > 0\).

7 A curve has equation \(\mathrm { y } = \frac { ( 3 \mathrm { x } + 2 ) ( \mathrm { x } - 3 ) } { ( \mathrm { x } - 2 ) ( \mathrm { x } + 1 ) }\).

1. Write down the equations of the three asymptotes and the coordinates of the points where the curve crosses the axes.
2. Sketch the curve, justifying how it approaches the horizontal asymptote.
3. Find the set of values of \(x\) for which \(y \geqslant 3\).

1 Express \(\frac { 5 x } { ( x - 1 ) \left( x ^ { 2 } + 4 \right) }\) in partial fractions.

4 Express \(\frac { x ^ { 3 } } { ( x - 2 ) \left( x ^ { 2 } + 4 \right) }\) in partial fractions.

8 The curve \(C _ { 1 }\) has equation \(y = \frac { \mathrm { p } ( x ) } { \mathrm { q } ( x ) }\), where \(\mathrm { p } ( x )\) and \(\mathrm { q } ( x )\) are polynomials of degree 2 and 1 respectively. The asymptotes of the curve are \(x = - 2\) and \(y = \frac { 1 } { 2 } x + 1\), and the curve passes through the point \(\left( - 1 , \frac { 17 } { 2 } \right)\).

1. Express the equation of \(C _ { 1 }\) in the form \(y = \frac { \mathrm { p } ( x ) } { \mathrm { q } ( x ) }\).
2. For the curve \(C _ { 1 }\), find the range of values that \(y\) can take.
3. For the curve \(C _ { 1 }\), find the range of values that \(y\) can take.
   Another curve, \(C _ { 2 }\), has equation \(y ^ { 2 } = \frac { \mathrm { p } ( x ) } { \mathrm { q } ( x ) }\), where \(\mathrm { p } ( x )\) and \(\mathrm { q } ( x )\) are the polynomials found in part (i).
4. It is given that \(C _ { 2 }\) intersects the line \(y = \frac { 1 } { 2 } x + 1\) exactly once. Find the coordinates of the point of intersection. Another curve, \(C _ { 2 }\), has equation \(y ^ { 2 } = \frac { \mathrm { p } ( x ) } { \mathrm { q } ( x ) }\), where \(\mathrm { p } ( x )\) and \(\mathrm { q } ( x )\) are the polynomials found in part (i).
5. It is given that \(C _ { 2 }\) intersects the line \(y = \frac { 1 } { 2 } x + 1\) exactly once. Find the coordinates of the point of
   intersection. \section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE.}

7 The equation of a curve is \(y = \frac { x ^ { 2 } + 1 } { ( x + 1 ) ( x - 7 ) }\).

1. Write down the equations of the asymptotes.
2. Find the coordinates of the stationary points on the curve.
3. Find the coordinates of the point where the curve meets one of its asymptotes.
4. Sketch the curve.

5 A curve has equation \(y = \frac { x ^ { 2 } - 8 } { x - 3 }\).

1. Find the equations of the asymptotes of the curve.
2. Prove that there are no points on the curve for which \(4 < y < 8\).
3. Sketch the curve. Indicate the asymptotes in your sketch.

7 It is given that \(\mathrm { f } ( x ) = \frac { x ^ { 2 } - 25 } { ( x - 1 ) ( x + 2 ) }\).

1. Express \(\mathrm { f } ( x )\) in partial fractions.
2. Write down the equations of the asymptotes of the curve \(y = \mathrm { f } ( x )\).
3. Find the value of \(x\) where the graph of \(y = \mathrm { f } ( x )\) cuts the horizontal asymptote.
4. Sketch the graph of \(y ^ { 2 } = \mathrm { f } ( x )\).

9 The curve \(C\) has equation $$y = \frac { x ^ { 2 } - 2 x + \lambda } { x + 1 }$$ where \(\lambda\) is a constant. Show that the equations of the asymptotes of \(C\) are independent of \(\lambda\). Find the value of \(\lambda\) for which the \(x\)-axis is a tangent to \(C\), and sketch \(C\) in this case. Sketch \(C\) in the case \(\lambda = - 4\), giving the exact coordinates of the points of intersection of \(C\) with the \(x\)-axis.

1 Express \(\frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }\) in partial fractions and hence use the method of differences to find $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }$$ Deduce the value of $$\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }$$