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 \(P\).