CAIE P2 (Pure Mathematics 2) 2019 June

1 Show that \(\ln \left( x ^ { 3 } - 4 x \right) - \ln \left( x ^ { 2 } - 2 x \right) \equiv \ln ( x + 2 )\).

2

1. Solve the inequality \(| 3 x - 5 | < | x + 3 |\).
2. Hence find the greatest integer \(n\) satisfying the inequality \(\left| 3 ^ { 0.1 n + 1 } - 5 \right| < \left| 3 ^ { 0.1 n } + 3 \right|\).

3 Find the equation of the normal to the curve $$x ^ { 2 } \ln y + 2 x + 5 y = 11$$ at the point \(( 3,1 )\).

4

1. Find \(\int \tan ^ { 2 } 3 x \mathrm {~d} x\).
2. Find the exact value of \(\int _ { 0 } ^ { 1 } \frac { \mathrm { e } ^ { 3 x } + 4 } { \mathrm { e } ^ { x } } \mathrm {~d} x\). Show all necessary working.

5 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = 5 x ^ { 3 } + a x ^ { 2 } + b x - 16$$ where \(a\) and \(b\) are constants. It is given that \(( x - 2 )\) is a factor of \(\mathrm { p } ( x )\) and that the remainder is 27 when \(\mathrm { p } ( x )\) is divided by \(( x + 1 )\).

1. Find the values of \(a\) and \(b\).
2. Hence factorise \(\mathrm { p } ( x )\) completely.

6
\includegraphics[max width=\textwidth, alt={}, center]{6694ccc1-c8b1-42a7-8b21-829a89af74c9-08_732_807_258_667} The diagram shows the curve with equation \(y = \frac { 8 + x ^ { 3 } } { 2 - 5 x }\). The maximum point is denoted by \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and determine the gradient of the curve at the point where the curve crosses the \(x\)-axis.
2. Show that the \(x\)-coordinate of the point \(M\) satisfies the equation \(x = \sqrt { } \left( 0.6 x + 4 x ^ { - 1 } \right)\).
3. Use an iterative formula, based on the equation in part (ii), to find the \(x\)-coordinate of \(M\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

7

1. Show that \(2 \operatorname { cosec } 2 \theta \cot \theta \equiv \operatorname { cosec } ^ { 2 } \theta\).
2. Hence show that \(\operatorname { cosec } ^ { 2 } 15 ^ { \circ } \tan 15 ^ { \circ } = 4\).
3. Solve the equation \(2 \operatorname { cosec } \phi \cot \frac { 1 } { 2 } \phi + \operatorname { cosec } \frac { 1 } { 2 } \phi = 12\) for \(- 360 ^ { \circ } < \phi < 360 ^ { \circ }\). Show all necessary working.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.