CAIE P3 (Pure Mathematics 3) 2019 June

1 Find the coefficient of \(x ^ { 3 }\) in the expansion of \(( 3 - x ) ( 1 + 3 x ) ^ { \frac { 1 } { 3 } }\) in ascending powers of \(x\).

2 Showing all necessary working, solve the equation \(9 ^ { x } = 3 ^ { x } + 12\). Give your answer correct to 2 decimal places.

3 Showing all necessary working, solve the equation \(\cot 2 \theta = 2 \tan \theta\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

4 Find the exact coordinates of the point on the curve \(y = \frac { x } { 1 + \ln x }\) at which the gradient of the tangent is equal to \(\frac { 1 } { 4 }\).

5 Throughout this question the use of a calculator is not permitted. It is given that the complex number \(- 1 + ( \sqrt { } 3 ) \mathrm { i }\) is a root of the equation $$k x ^ { 3 } + 5 x ^ { 2 } + 10 x + 4 = 0$$ where \(k\) is a real constant.

1. Write down another root of the equation.
2. Find the value of \(k\) and the third root of the equation.

6
\includegraphics[max width=\textwidth, alt={}, center]{772393d7-6e81-4b99-913a-63c9f87d1af2-08_492_812_260_664} In the diagram, \(A\) is the mid-point of the semicircle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the semicircle at \(B\) and \(C\). The angle \(O A B\) is equal to \(x\) radians. The area of the shaded region bounded by \(A B , A C\) and the arc with centre \(A\) is equal to half the area of the semicircle.

1. Use triangle \(O A B\) to show that \(A B = 2 r \cos x\).
2. Hence show that \(x = \cos ^ { - 1 } \sqrt { } \left( \frac { \pi } { 16 x } \right)\).
3. Verify by calculation that \(x\) lies between 1 and 1.5.
4. Use an iterative formula based on the equation in part (ii) to determine \(x\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

7 The variables \(x\) and \(y\) satisfy the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = x \mathrm { e } ^ { x + y }\). It is given that \(y = 0\) when \(x = 0\).

1. Solve the differential equation, obtaining \(y\) in terms of \(x\).
2. Explain why \(x\) can only take values that are less than 1 .

8 Let \(\mathrm { f } ( x ) = \frac { 10 x + 9 } { ( 2 x + 1 ) ( 2 x + 3 ) ^ { 2 } }\).

1. Express \(\mathrm { f } ( x )\) in partial fractions.
2. Hence show that \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x = \frac { 1 } { 2 } \ln \frac { 9 } { 5 } + \frac { 1 } { 5 }\).

9 The points \(A\) and \(B\) have position vectors \(\mathbf { i } + 2 \mathbf { j } - \mathbf { k }\) and \(3 \mathbf { i } + \mathbf { j } + \mathbf { k }\) respectively. The line \(l\) has equation \(\mathbf { r } = 2 \mathbf { i } + \mathbf { j } + \mathbf { k } + \mu ( \mathbf { i } + \mathbf { j } + 2 \mathbf { k } )\).

1. Show that \(l\) does not intersect the line passing through \(A\) and \(B\).
2. The plane \(m\) is perpendicular to \(A B\) and passes through the mid-point of \(A B\). The plane \(m\) intersects the line \(l\) at the point \(P\). Find the equation of \(m\) and the position vector of \(P\).

10
\includegraphics[max width=\textwidth, alt={}, center]{772393d7-6e81-4b99-913a-63c9f87d1af2-16_524_689_260_726} The diagram shows the curve \(y = \sin 3 x \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\) and its minimum point \(M\). The shaded region \(R\) is bounded by the curve and the \(x\)-axis.

1. By expanding \(\sin ( 3 x + x )\) and \(\sin ( 3 x - x )\) show that $$\sin 3 x \cos x = \frac { 1 } { 2 } ( \sin 4 x + \sin 2 x ) .$$
2. Using the result of part (i) and showing all necessary working, find the exact area of the region \(R\).
3. Using the result of part (i), express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\cos 2 x\) and hence find the \(x\)-coordinate of \(M\), giving your answer correct to 2 decimal places.
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