OCR C4 (Core Mathematics 4) 2013 January

1 Find \(\int x \cos 3 x \mathrm {~d} x\).

2 Find the first three terms in the expansion of \(( 9 - 16 x ) ^ { \frac { 3 } { 2 } }\) in ascending powers of \(x\), and state the set of values for which this expansion is valid.

3 The equation of a curve is \(x y ^ { 2 } = x ^ { 2 } + 1\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\), and hence find the coordinates of the stationary points on the curve.

4 The equations of two lines are $$\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + \lambda ( 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k } ) \text { and } \mathbf { r } = 6 \mathbf { i } + 8 \mathbf { j } + \mathbf { k } + \mu ( \mathbf { i } + 4 \mathbf { j } - 5 \mathbf { k } ) .$$

1. Show that these lines meet, and find the coordinates of the point of intersection.
2. Find the acute angle between these lines.

5 The parametric equations of a curve are $$x = 2 + 3 \sin \theta \text { and } y = 1 - 2 \cos \theta \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$$

1. Find the coordinates of the point on the curve where the gradient is \(\frac { 1 } { 2 }\).
2. Find the cartesian equation of the curve.

6 Use the substitution \(u = 2 x + 1\) to evaluate \(\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 4 x - 1 } { ( 2 x + 1 ) ^ { 5 } } \mathrm {~d} x\).

7

1. Given that \(y = \ln ( 1 + \sin x ) - \ln ( \cos x )\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \cos x }\).
2. Using this result, evaluate \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \sec x \mathrm {~d} x\), giving your answer as a single logarithm.

8 The points \(A ( 3,2,1 ) , B ( 5,4 , - 3 ) , C ( 3,17 , - 4 )\) and \(D ( 1,6,3 )\) form a quadrilateral \(A B C D\).

1. Show that \(A B = A D\).
2. Find a vector equation of the line through \(A\) and the mid-point of \(B D\).
3. Show that \(C\) lies on the line found in part (ii).
4. What type of quadrilateral is \(A B C D\) ?

9 The temperature of a freezer is \(- 20 ^ { \circ } \mathrm { C }\). A container of a liquid is placed in the freezer. The rate at which the temperature, \(\theta ^ { \circ } \mathrm { C }\), of a liquid decreases is proportional to the difference in temperature between the liquid and its surroundings. The situation is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta + 20 ) ,$$ where time \(t\) is in minutes and \(k\) is a positive constant.

1. Express \(\theta\) in terms of \(t , k\) and an arbitrary constant. Initially the temperature of the liquid in the container is \(40 ^ { \circ } \mathrm { C }\) and, at this instant, the liquid is cooling at a rate of \(3 ^ { \circ } \mathrm { C }\) per minute. The liquid freezes at \(0 ^ { \circ } \mathrm { C }\).
2. Find the value of \(k\) and find also the time it takes (to the nearest minute) for the liquid to freeze. The procedure is repeated on another occasion with a different liquid. The initial temperature of this liquid is \(90 ^ { \circ } \mathrm { C }\). After 19 minutes its temperature is \(0 ^ { \circ } \mathrm { C }\).
3. Without any further calculation, explain what you can deduce about the value of \(k\) in this case.

10

1. Use algebraic division to express \(\frac { x ^ { 3 } - 2 x ^ { 2 } - 4 x + 13 } { x ^ { 2 } - x - 6 }\) in the form \(A x + B + \frac { C x + D } { x ^ { 2 } - x - 6 }\), where \(A , B , C\) and \(D\) are constants.
2. Hence find \(\int _ { 4 } ^ { 6 } \frac { x ^ { 3 } - 2 x ^ { 2 } - 4 x + 13 } { x ^ { 2 } - x - 6 } \mathrm {~d} x\), giving your answer in the form \(a + \ln b\).