CAIE P2 (Pure Mathematics 2) 2022 November

1 Solve the equation \(\sec \theta = 5 \operatorname { cosec } \theta\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

2 The solutions of the equation \(| 4 x - 1 | = | x + 3 |\) are \(x = p\) and \(x = q\), where \(p < q\).
Find the exact values of \(p\) and \(q\), and hence determine the exact value of \(| p - 2 | - | q - 1 |\).

3
\includegraphics[max width=\textwidth, alt={}, center]{68f4b2dc-a05d-4061-aaf0-de15cfe186a9-04_714_515_262_804} The variables \(x\) and \(y\) satisfy the equation \(y = A x ^ { k }\), where \(A\) and \(k\) are constants. The graph of \(\ln y\) against \(\ln x\) is a straight line passing through the points ( \(0.56,2.87\) ) and ( \(0.81,3.47\) ), as shown in the diagram. Find the value of \(k\), and the value of \(A\) correct to 2 significant figures.

4 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = a x ^ { 3 } + 23 x ^ { 2 } - a x - 8$$ where \(a\) is a constant. It is given that \(( 2 x + 1 )\) is a factor of \(\mathrm { p } ( x )\).

1. Find the value of \(a\) and hence factorise \(\mathrm { p } ( x )\) completely.
2. Hence solve the equation \(\mathrm { p } \left( \mathrm { e } ^ { 4 y } \right) = 0\), giving your answer correct to 3 significant figures.

5 The curve with equation \(y = x \ln ( 4 x + 1 ) - 3 x\) has one stationary point \(P\).

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = \frac { 2 x + 0.75 } { \ln ( 4 x + 1 ) } - 0.25$$
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 1.8 and 1.9.
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

6
\includegraphics[max width=\textwidth, alt={}, center]{68f4b2dc-a05d-4061-aaf0-de15cfe186a9-08_616_531_269_799} The diagram shows the curves \(y = \frac { 6 } { 3 x + 2 }\) and \(y = 3 \mathrm { e } ^ { - x } - 3\) for values of \(x\) between 0 and 4. The shaded region is bounded by the two curves and the lines \(x = 0\) and \(x = 4\). Find the exact area of the shaded region, giving your answer in the form \(\ln a + b + c \mathrm { e } ^ { d }\).

7
\includegraphics[max width=\textwidth, alt={}, center]{68f4b2dc-a05d-4061-aaf0-de15cfe186a9-10_657_792_269_664} The diagram shows the curve with parametric equations $$x = 3 \cos 2 \theta , \quad y = 4 \sin \theta ,$$ for \(\pi \leqslant \theta \leqslant \frac { 3 } { 2 } \pi\). Points \(P\) and \(Q\) lie on the curve. The gradient of the curve at \(P\) is 2 . The straight line \(3 x + y = 0\) meets the curve at \(Q\).

1. Find the value of \(\theta\) at \(P\), giving your answer correct to 3 significant figures.
2. Find the gradient of the curve at \(Q\), giving your answer correct to 3 significant figures.
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