CAIE P2 (Pure Mathematics 2) 2005 November

1 Solve the inequality \(( 0.8 ) ^ { x } < 0.5\).

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

1. Find the remainder when \(\mathrm { p } ( x )\) is divided by \(x - 1\).
2. Find the quotient and remainder when \(\mathrm { p } ( x )\) is divided by \(x ^ { 2 } + x - 1\).

3

1. Express \(12 \cos \theta - 5 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation $$12 \cos \theta - 5 \sin \theta = 10$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

4 The equation of a curve is \(x ^ { 3 } + y ^ { 3 } = 9 x y\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 y - x ^ { 2 } } { y ^ { 2 } - 3 x }\).
2. Find the equation of the tangent to the curve at the point ( 2,4 ), giving your answer in the form \(a x + b y = c\).

5

1. By sketching a suitable pair of graphs, show that there is only one value of \(x\) that is a root of the equation $$\frac { 1 } { x } = \ln x$$
2. Verify by calculation that this root lies between 1 and 2 .
3. Show that this root also satisfies the equation $$x = \mathrm { e } ^ { \frac { 1 } { x } }$$
4. Use the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { \frac { 1 } { x _ { n } } }$$ with initial value \(x _ { 1 } = 1.8\), to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 x } - 2 \mathrm { e } ^ { - x }\). The point \(( 0,1 )\) lies on the curve.

1. Find the equation of the curve.
2. The curve has one stationary point. Find the \(x\)-coordinate of this point and determine whether it is a maximum or a minimum point.

7
\includegraphics[max width=\textwidth, alt={}, center]{d527d21f-0ab5-40fa-8cfd-ebfb4aba0a87-3_493_863_264_641} The diagram shows the part of the curve \(y = \sin ^ { 2 } x\) for \(0 \leqslant x \leqslant \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin 2 x\).
2. Hence find the \(x\)-coordinates of the points on the curve at which the gradient of the curve is 0.5 . [3]
3. By expressing \(\sin ^ { 2 } x\) in terms of \(\cos 2 x\), find the area of the region bounded by the curve and the \(x\)-axis between 0 and \(\pi\).