CAIE P3 (Pure Mathematics 3) 2008 June

1 Solve the inequality \(| x - 2 | > 3 | 2 x + 1 |\).

2 Solve, correct to 3 significant figures, the equation $$\mathrm { e } ^ { x } + \mathrm { e } ^ { 2 x } = \mathrm { e } ^ { 3 x }$$

3
\includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-2_337_828_657_657} In the diagram, \(A B C D\) is a rectangle with \(A B = 3 a\) and \(A D = a\). A circular arc, with centre \(A\) and radius \(r\), joins points \(M\) and \(N\) on \(A B\) and \(C D\) respectively. The angle \(M A N\) is \(x\) radians. The perimeter of the sector \(A M N\) is equal to half the perimeter of the rectangle.

1. Show that \(x\) satisfies the equation $$\sin x = \frac { 1 } { 4 } ( 2 + x ) \text {. }$$
2. This equation has only one root in the interval \(0 < x < \frac { 1 } { 2 } \pi\). Use the iterative formula $$x _ { n + 1 } = \sin ^ { - 1 } \left( \frac { 2 + x _ { n } } { 4 } \right) ,$$ with initial value \(x _ { 1 } = 0.8\), to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

4

1. Show that the equation \(\tan \left( 30 ^ { \circ } + \theta \right) = 2 \tan \left( 60 ^ { \circ } - \theta \right)\) can be written in the form $$\tan ^ { 2 } \theta + ( 6 \sqrt { } 3 ) \tan \theta - 5 = 0$$
2. Hence, or otherwise, solve the equation $$\tan \left( 30 ^ { \circ } + \theta \right) = 2 \tan \left( 60 ^ { \circ } - \theta \right) ,$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

5 The variable complex number \(z\) is given by $$z = 2 \cos \theta + \mathrm { i } ( 1 - 2 \sin \theta ) ,$$ where \(\theta\) takes all values in the interval \(- \pi < \theta \leqslant \pi\).

1. Show that \(| z - \mathrm { i } | = 2\), for all values of \(\theta\). Hence sketch, in an Argand diagram, the locus of the point representing \(z\).
2. Prove that the real part of \(\frac { 1 } { z + 2 - \mathrm { i } }\) is constant for \(- \pi < \theta < \pi\).

6 The equation of a curve is \(x y ( x + y ) = 2 a ^ { 3 }\), where \(a\) is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the \(x\)-axis, and find the coordinates of this point.

7 Let \(\mathrm { f } ( x ) \equiv \frac { x ^ { 2 } + 3 x + 3 } { ( x + 1 ) ( x + 3 ) }\).

1. Express \(\mathrm { f } ( x )\) in partial fractions.
2. Hence show that \(\int _ { 0 } ^ { 3 } \mathrm { f } ( x ) \mathrm { d } x = 3 - \frac { 1 } { 2 } \ln 2\).

8
\includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-3_597_951_1471_598} In the diagram the tangent to a curve at a general point \(P\) with coordinates \(( x , y )\) meets the \(x\)-axis at \(T\). The point \(N\) on the \(x\)-axis is such that \(P N\) is perpendicular to the \(x\)-axis. The curve is such that, for all values of \(x\) in the interval \(0 < x < \frac { 1 } { 2 } \pi\), the area of triangle \(P T N\) is equal to \(\tan x\), where \(x\) is in radians.

1. Using the fact that the gradient of the curve at \(P\) is \(\frac { P N } { T N }\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 } y ^ { 2 } \cot x .$$
2. Given that \(y = 2\) when \(x = \frac { 1 } { 6 } \pi\), solve this differential equation to find the equation of the curve, expressing \(y\) in terms of \(x\).

9
\includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-4_547_1401_264_370} The diagram shows the curve \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } \sqrt { } ( 1 + 2 x )\) and its maximum point \(M\). The shaded region between the curve and the axes is denoted by \(R\).

1. Find the \(x\)-coordinate of \(M\).
2. Find by integration the volume of the solid obtained when \(R\) is rotated completely about the \(x\)-axis. Give your answer in terms of \(\pi\) and e.

10 The points \(A\) and \(B\) have position vectors, relative to the origin \(O\), given by $$\overrightarrow { O A } = \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k } .$$ The line \(l\) has vector equation $$\mathbf { r } = ( 1 - 2 t ) \mathbf { i } + ( 5 + t ) \mathbf { j } + ( 2 - t ) \mathbf { k }$$

1. Show that \(l\) does not intersect the line passing through \(A\) and \(B\).
2. The point \(P\) lies on \(l\) and is such that angle \(P A B\) is equal to \(60 ^ { \circ }\). Given that the position vector of \(P\) is \(( 1 - 2 t ) \mathbf { i } + ( 5 + t ) \mathbf { j } + ( 2 - t ) \mathbf { k }\), show that \(3 t ^ { 2 } + 7 t + 2 = 0\). Hence find the only possible position vector of \(P\). \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }