CAIE P2 (Pure Mathematics 2) 2023 June

1 Use logarithms to solve the equation \(12 ^ { x } = 3 ^ { 2 x + 1 }\). Give your answer correct to 3 significant figures.

2 A curve has equation \(y = \frac { 2 + 3 \ln x } { 1 + 2 x }\).
Find the equation of the tangent to the curve at the point \(\left( 1 , \frac { 2 } { 3 } \right)\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

3 It is given that \(\int _ { 0 } ^ { a } \left( 3 \mathrm { e } ^ { 2 x } - 1 \right) \mathrm { d } x = 12\), where \(a\) is a positive constant.

1. Show that \(a = \frac { 1 } { 2 } \ln \left( 9 + \frac { 2 } { 3 } a \right)\).
2. Use an iterative formula, based on the equation in (a), to find the value of \(a\) correct to 4 significant figures. Use an initial value of 1 and give the result of each iteration to 6 significant figures. [3]

4 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = 2 x ^ { 3 } + 3 x ^ { 2 } + k x - 30$$ where \(k\) is a constant. It is given that \(( x - 3 )\) is a factor of \(\mathrm { p } ( x )\).

1. Find the value of \(k\).
2. Hence find the quotient when \(\mathrm { p } ( x )\) is divided by ( \(x - 3\) ) and factorise \(\mathrm { p } ( x )\) completely.
3. It is given that \(a\) is one of the roots of the equation \(\mathrm { p } ( x ) = 0\). Given also that the equation \(| 4 y - 5 | = a\) is satisfied by two real values of \(y\), find these two values of \(y\).

5
\includegraphics[max width=\textwidth, alt={}, center]{3966e088-0a2f-434a-94fc-40765cd157a7-06_376_848_269_644} The diagram shows the curve with parametric equations $$x = 4 \mathrm { e } ^ { 2 t } , \quad y = 5 \mathrm { e } ^ { - t } \cos 2 t$$ for \(- \frac { 1 } { 4 } \pi \leqslant t \leqslant \frac { 1 } { 4 } \pi\). The curve has a maximum point \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the coordinates of \(M\), giving each coordinate correct to 3 significant figures.

6 Show that \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } \left( 4 \cos ^ { 2 } 2 x + \frac { 1 } { \cos ^ { 2 } x } \right) \mathrm { d } x = \frac { 3 } { 4 } \sqrt { 3 } + \frac { 1 } { 6 } \pi - 1\).

7

1. Express \(7 \cos \theta + 24 \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the value of \(\alpha\) correct to 2 decimal places.
2. Solve the equation \(7 \cos \theta + 24 \sin \theta = 18\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
3. As \(\beta\) varies, the greatest possible value of $$\frac { 150 } { 7 \cos \frac { 1 } { 2 } \beta + 24 \sin \frac { 1 } { 2 } \beta + 50 }$$ is denoted by \(V\).
   Find the value of \(V\) and determine the smallest positive value of \(\beta\) (in degrees) for which the value of \(V\) occurs.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.