CAIE P2 (Pure Mathematics 2) 2011 November

1 Find the gradient of the curve \(y = \ln ( 5 x + 1 )\) at the point where \(x = 4\).

2 Solve the inequality \(| 2 x - 3 | \leqslant | 3 x |\).

3 Solve the equation \(2 \ln ( x + 3 ) - \ln x = \ln ( 2 x - 2 )\).

4

1. Express \(\cos ^ { 2 } x\) in terms of \(\cos 2 x\).
2. Hence show that $$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \left( \cos ^ { 2 } x + \sin 2 x \right) \mathrm { d } x = \frac { 1 } { 8 } \sqrt { } 3 + \frac { 1 } { 12 } \pi + \frac { 1 } { 4 }$$

5 Solve the equation \(5 \sec ^ { 2 } 2 \theta = \tan 2 \theta + 9\), giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

6

1. The polynomial \(x ^ { 4 } + a x ^ { 3 } - x ^ { 2 } + b x + 2\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { p } ( x )\). It is given that \(( x - 1 )\) and \(( x + 2 )\) are factors of \(\mathrm { p } ( x )\). Find the values of \(a\) and \(b\).
2. When \(a\) and \(b\) have these values, find the quotient when \(\mathrm { p } ( x )\) is divided by \(x ^ { 2 } + x - 2\).

7
\includegraphics[max width=\textwidth, alt={}, center]{e82fee05-0c55-4fe2-b781-e5e82186c153-2_608_999_1430_571} The diagram shows the curve \(y = ( x - 4 ) \mathrm { e } ^ { \frac { 1 } { 2 } x }\). The curve has a gradient of 3 at the point \(P\).

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = 2 + 6 \mathrm { e } ^ { - \frac { 1 } { 2 } x }$$
2. Verify that the equation in part (i) has a root between \(x = 3.1\) and \(x = 3.3\).
3. Use the iterative formula \(x _ { n + 1 } = 2 + 6 \mathrm { e } ^ { - \frac { 1 } { 2 } x _ { n } }\) to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

8 The equation of a curve is \(2 x ^ { 2 } - 3 x - 3 y + y ^ { 2 } = 6\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x - 3 } { 3 - 2 y }\).
2. Find the coordinates of the two points on the curve at which the gradient is - 1 .