CAIE P2 (Pure Mathematics 2) 2010 June

1 Solve the inequality \(| 2 x - 3 | > 5\).

2 Show that \(\int _ { 0 } ^ { 6 } \frac { 1 } { x + 2 } \mathrm {~d} x = 2 \ln 2\).

3

1. Show that the equation \(\tan \left( x + 45 ^ { \circ } \right) = 6 \tan x\) can be written in the form $$6 \tan ^ { 2 } x - 5 \tan x + 1 = 0$$
2. Hence solve the equation \(\tan \left( x + 45 ^ { \circ } \right) = 6 \tan x\), for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

4 The polynomial \(x ^ { 3 } + 3 x ^ { 2 } + 4 x + 2\) is denoted by \(\mathrm { f } ( x )\).

1. Find the quotient and remainder when \(\mathrm { f } ( x )\) is divided by \(x ^ { 2 } + x - 1\).
2. Use the factor theorem to show that \(( x + 1 )\) is a factor of \(\mathrm { f } ( x )\).

5

1. Given that \(y = 2 ^ { x }\), show that the equation $$2 ^ { x } + 3 \left( 2 ^ { - x } \right) = 4$$ can be written in the form $$y ^ { 2 } - 4 y + 3 = 0$$
2. Hence solve the equation $$2 ^ { x } + 3 \left( 2 ^ { - x } \right) = 4$$ giving the values of \(x\) correct to 3 significant figures where appropriate.

6 The equation of a curve is $$x ^ { 2 } y + y ^ { 2 } = 6 x$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 - 2 x y } { x ^ { 2 } + 2 y }\).
2. Find the equation of the tangent to the curve at the point with coordinates ( 1,2 ), giving your answer in the form \(a x + b y + c = 0\).

7

1. By sketching a suitable pair of graphs, show that the equation $$\mathrm { e } ^ { 2 x } = 2 - x$$ has only one root.
2. Verify by calculation that this root lies between \(x = 0\) and \(x = 0.5\).
3. Show that, if a sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( 2 - x _ { n } \right)$$ converges, then it converges to the root of the equation in part (i).
4. Use this iterative formula, with initial value \(x _ { 1 } = 0.25\), to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
5. By differentiating \(\frac { \cos x } { \sin x }\), show that if \(y = \cot x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
6. By expressing \(\cot ^ { 2 } x\) in terms of \(\operatorname { cosec } ^ { 2 } x\) and using the result of part (i), show that $$\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } \cot ^ { 2 } x \mathrm {~d} x = 1 - \frac { 1 } { 4 } \pi$$
7. Express \(\cos 2 x\) in terms of \(\sin ^ { 2 } x\) and hence show that \(\frac { 1 } { 1 - \cos 2 x }\) can be expressed as \(\frac { 1 } { 2 } \operatorname { cosec } ^ { 2 } x\). Hence, using the result of part (i), find $$\int \frac { 1 } { 1 - \cos 2 x } \mathrm {~d} x$$