CAIE P3 (Pure Mathematics 3) 2013 November

1 The equation of a curve is \(y = \frac { 1 + x } { 1 + 2 x }\) for \(x > - \frac { 1 } { 2 }\). Show that the gradient of the curve is always negative.

2 Solve the equation \(2 \left| 3 ^ { x } - 1 \right| = 3 ^ { x }\), giving your answers correct to 3 significant figures.

3 Find the exact value of \(\int _ { 1 } ^ { 4 } \frac { \ln x } { \sqrt { } x } \mathrm {~d} x\).

4 The parametric equations of a curve are $$x = \mathrm { e } ^ { - t } \cos t , \quad y = \mathrm { e } ^ { - t } \sin t$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \tan \left( t - \frac { 1 } { 4 } \pi \right)\).

5

1. Prove that \(\cot \theta + \tan \theta \equiv 2 \operatorname { cosec } 2 \theta\).
2. Hence show that \(\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 3 } \pi } \operatorname { cosec } 2 \theta \mathrm {~d} \theta = \frac { 1 } { 2 } \ln 3\).

6
\includegraphics[max width=\textwidth, alt={}, center]{dd7b2aee-4318-48e8-97c0-541e47f2e83a-2_551_567_1416_788} In the diagram, \(A\) is a point on the circumference of a circle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the circumference at \(B\) and \(C\). The angle \(O A B\) is \(\theta\) radians. The shaded region is bounded by the circumference of the circle and the arc with centre \(A\) joining \(B\) and \(C\). The area of the shaded region is equal to half the area of the circle.

1. Show that \(\cos 2 \theta = \frac { 2 \sin 2 \theta - \pi } { 4 \theta }\).
2. Use the iterative formula $$\theta _ { n + 1 } = \frac { 1 } { 2 } \cos ^ { - 1 } \left( \frac { 2 \sin 2 \theta _ { n } - \pi } { 4 \theta _ { n } } \right)$$ with initial value \(\theta _ { 1 } = 1\), to determine \(\theta\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.
   \(7 \quad\) Let \(\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } - 7 x - 1 } { ( x - 2 ) \left( x ^ { 2 } + 3 \right) }\).
3. Express \(\mathrm { f } ( x )\) in partial fractions.
4. Hence obtain the expansion of \(\mathrm { f } ( x )\) in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\).

1. The complex numbers \(u\) and \(v\) satisfy the equations $$u + 2 v = 2 \mathrm { i } \quad \text { and } \quad \mathrm { i } u + v = 3$$ Solve the equations for \(u\) and \(v\), giving both answers in the form \(x + \mathrm { i } y\), where \(x\) and \(y\) are real.
2. On an Argand diagram, sketch the locus representing complex numbers \(z\) satisfying \(| z + \mathrm { i } | = 1\) and the locus representing complex numbers \(w\) satisfying \(\arg ( w - 2 ) = \frac { 3 } { 4 } \pi\). Find the least value of \(| z - w |\) for points on these loci.

8 Throughout this question the use of a calculator is not permitted.

1. The complex numbers \(u\) and \(v\) satisfy the equations $$u + 2 v = 2 \mathrm { i } \quad \text { and } \quad \mathrm { i } u + v = 3$$ Solve the equations for \(u\) and \(v\), giving both answers in the form \(x + \mathrm { i } y\), where \(x\) and \(y\) are real.
2. On an Argand diagram, sketch the locus representing complex numbers \(z\) satisfying \(| z + \mathrm { i } | = 1\) and the locus representing complex numbers \(w\) satisfying \(\arg ( w - 2 ) = \frac { 3 } { 4 } \pi\). Find the least value of \(| z - w |\) for points on these loci.

9
\includegraphics[max width=\textwidth, alt={}, center]{dd7b2aee-4318-48e8-97c0-541e47f2e83a-3_704_714_1272_717} The diagram shows three points \(A , B\) and \(C\) whose position vectors with respect to the origin \(O\) are given by \(\overrightarrow { O A } = \left( \begin{array} { r } 2
- 1
2 \end{array} \right) , \overrightarrow { O B } = \left( \begin{array} { l } 0
3
1 \end{array} \right)\) and \(\overrightarrow { O C } = \left( \begin{array} { l } 3
0
4 \end{array} \right)\). The point \(D\) lies on \(B C\), between \(B\) and \(C\), and is such that \(C D = 2 D B\).

1. Find the equation of the plane \(A B C\), giving your answer in the form \(a x + b y + c z = d\).
2. Find the position vector of \(D\).
3. Show that the length of the perpendicular from \(A\) to \(O D\) is \(\frac { 1 } { 3 } \sqrt { } ( 65 )\).

10
\includegraphics[max width=\textwidth, alt={}, center]{dd7b2aee-4318-48e8-97c0-541e47f2e83a-4_335_875_262_635} A tank containing water is in the form of a cone with vertex \(C\). The axis is vertical and the semivertical angle is \(60 ^ { \circ }\), as shown in the diagram. At time \(t = 0\), the tank is full and the depth of water is \(H\). At this instant, a tap at \(C\) is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt { } h\), where \(h\) is the depth of water at time \(t\). The tank becomes empty when \(t = 60\).

1. Show that \(h\) and \(t\) satisfy a differential equation of the form $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A h ^ { - \frac { 3 } { 2 } } ,$$ where \(A\) is a positive constant.
2. Solve the differential equation given in part (i) and obtain an expression for \(t\) in terms of \(h\) and \(H\).
3. Find the time at which the depth reaches \(\frac { 1 } { 2 } H\).
   [0pt] [The volume \(V\) of a cone of vertical height \(h\) and base radius \(r\) is given by \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]