CAIE P3 (Pure Mathematics 3) 2020 June

1 Find the quotient and remainder when \(6 x ^ { 4 } + x ^ { 3 } - x ^ { 2 } + 5 x - 6\) is divided by \(2 x ^ { 2 } - x + 1\).

2 \includegraphics[max width=\textwidth, alt={}, center]{3149080d-ad1a-4d2e-8e20-eb9977ced619-03_515_901_260_623} The variables \(x\) and \(y\) satisfy the equation \(y ^ { 2 } = A \mathrm { e } ^ { k x }\), where \(A\) and \(k\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points (1.5, 1.2) and (5.24, 2.7) as shown in the diagram. Find the values of \(A\) and \(k\) correct to 2 decimal places.

3 Find the exact value of $$\int _ { 1 } ^ { 4 } x ^ { \frac { 3 } { 2 } } \ln x \mathrm {~d} x$$

4 A curve has equation \(y = \cos x \sin 2 x\).
Find the \(x\)-coordinate of the stationary point in the interval \(0 < x < \frac { 1 } { 2 } \pi\), giving your answer correct to 3 significant figures.

5

1. Express \(\sqrt { 2 } \cos x - \sqrt { 5 } \sin x\) in the form \(R \cos ( x + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the exact value of \(R\) and the value of \(\alpha\) correct to 3 decimal places.
2. Hence solve the equation \(\sqrt { 2 } \cos 2 \theta - \sqrt { 5 } \sin 2 \theta = 1\), for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

6 \includegraphics[max width=\textwidth, alt={}, center]{3149080d-ad1a-4d2e-8e20-eb9977ced619-08_318_750_260_699} The diagram shows the curve \(y = \frac { x } { 1 + 3 x ^ { 4 } }\), for \(x \geqslant 0\), and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.
2. Using the substitution \(u = \sqrt { 3 } x ^ { 2 }\), find by integration the exact area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 1\).

7 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 1 } { ( x + 1 ) ( x + 3 ) }$$ It is given that \(y = 2\) when \(x = 0\).
Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

8

1. Solve the equation \(( 1 + 2 \mathrm { i } ) w + \mathrm { i } w ^ { * } = 3 + 5 \mathrm { i }\). Give your answer in the form \(x + \mathrm { i } y\), where \(x\) and \(y\) are real.
2. 1. On a sketch of an Argand diagram, shade the region whose points represent complex numbers \(z\) satisfying the inequalities \(| z - 2 - 2 \mathrm { i } | \leqslant 1\) and \(\arg ( z - 4 \mathrm { i } ) \geqslant - \frac { 1 } { 4 } \pi\).
   2. Find the least value of \(\operatorname { Im } z\) for points in this region, giving your answer in an exact form.

9 \includegraphics[max width=\textwidth, alt={}, center]{3149080d-ad1a-4d2e-8e20-eb9977ced619-14_558_686_260_726} The diagram shows the curves \(y = \cos x\) and \(y = \frac { k } { 1 + x }\), where \(k\) is a constant, for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The curves touch at the point where \(x = p\).

1. Show that \(p\) satisfies the equation \(\tan p = \frac { 1 } { 1 + p }\).
2. Use the iterative formula \(p _ { n + 1 } = \tan ^ { - 1 } \left( \frac { 1 } { 1 + p _ { n } } \right)\) to determine the value of \(p\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
3. Hence find the value of \(k\) correct to 2 decimal places.

10 With respect to the origin \(O\), the points \(A\) and \(B\) have position vectors given by \(\overrightarrow { O A } = 6 \mathbf { i } + 2 \mathbf { j }\) and \(\overrightarrow { O B } = 2 \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k }\). The midpoint of \(O A\) is \(M\). The point \(N\) lying on \(A B\), between \(A\) and \(B\), is such that \(A N = 2 N B\).

1. Find a vector equation for the line through \(M\) and \(N\).
   The line through \(M\) and \(N\) intersects the line through \(O\) and \(B\) at the point \(P\).
2. Find the position vector of \(P\).
3. Calculate angle \(O P M\), giving your answer in degrees.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.