AS Paper 5: Mixed Pure and Applied
Covers coordinate geometry, sequences and series with the binomial expansion, exponentials and logarithms, integration, vectors, sampling and data presentation, and kinematics.
Questions
Question 1 [2 marks]
Exponentials and Logarithms
Without using a calculator, find the exact value of log_2(32).
Question 2 [3 marks]
Coordinate Geometry
A line passes through the point (2, -5) and is parallel to the line 3x - y + 4 = 0.
Find the equation of the line in the form y = mx + c.
Question 3 [4 marks]
Kinematics
A cyclist decelerates uniformly from 15 m/s to rest in 12 seconds.
Find the deceleration, and find the distance travelled while decelerating.
Question 4 [4 marks]
Vectors
Find the magnitude of the vector v = 3i - 4j, and find the vector of magnitude 15 in the same direction as v.
Question 5 [4 marks]
Sequences, Series and the Binomial Expansion
Evaluate the sum of (4r - 3) for r = 1 to r = 25.
Question 6 [4 marks]
Coordinate Geometry
A circle has equation (x - 2)^2 + (y + 3)^2 = 16.
Find the length of the tangent from the point P(10, -3) to the circle, giving your answer as an exact surd.
Question 7 [5 marks]
Integration
The curve y = 6x - x^2 meets the x-axis at the origin and at the point (6, 0).
Find the area enclosed between the curve and the x-axis.
Question 8 [5 marks]
Vectors
Vectors p = 2i - j + 2k and q = 3i + 2j + ck.
Given that p and q are perpendicular, find the value of c.
Question 9 [6 marks]
Sampling and Data Presentation
The number of pets owned by each of 8 families is: 0, 1, 1, 2, 2, 3, 4, 7.
Find the mean number of pets, and find the standard deviation, giving your answers to 3 significant figures. State, with a reason, whether the mean or the median would better represent a typical family in this data set.
Question 10 [5 marks]
Coordinate Geometry
The circle C has equation (x - 4)^2 + (y - 1)^2 = 20.
The line l has equation y = 2x + 3.
Show that l is a tangent to C, and find the coordinates of the point of contact.
Question 11 [6 marks]
Kinematics
A particle P moves in a straight line so that its displacement s metres from a fixed point O at time t seconds (t >= 0) is given by s = t^3 - 9t^2 + 24t.
Find the velocity and acceleration of P as functions of t, find the times at which P is instantaneously at rest, and find the total distance travelled by P in the first 5 seconds.
Question 12 [6 marks]
Sampling and Data Presentation
Two classes sat the same test. Class A has median 62, lower quartile 55, upper quartile 70. Class B has median 58, lower quartile 50, upper quartile 62.
Compare the two classes' test scores, referring to both location and spread, and comment on the skewness of each distribution.
Question 13 [6 marks]
Kinematics
Two particles P and Q are 200 m apart on a straight horizontal road. At time t = 0, P starts moving towards Q with constant velocity 6 m/s, and at the same instant Q starts moving towards P with constant acceleration 0.5 m/s^2 from rest.
Find the time at which the particles meet, and find the distance travelled by P before they meet.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| recognising 32 = 2^5 | M1 |
| log_2(32) = 5 | A1 |
| Final answer: 5 | |
| Question 2[3 marks] | |
|---|---|
| Answer or working | Marks |
| finding the gradient of the given line as 3 (from y = 3x + 4) | M1 |
| forming the equation y - (-5) = 3(x - 2) | M1 |
| y = 3x - 11 | A1 |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| using a = (v - u)/t | M1 |
| a = -1.25 m/s^2 (a deceleration of 1.25 m/s^2) | A1 |
| using s = (u + v)/2 x t | M1 |
| s = 90 m | A1 |
| Final answer: deceleration = 1.25 m/s^2; distance = 90 m | |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| |v| = sqrt(3^2 + 4^2) | M1 |
| |v| = 5 | A1 |
| the scale factor 15/5 = 3 | M1 |
| the vector 9i - 12j | A1 |
| Final answer: |v| = 5; the vector of magnitude 15 is 9i - 12j | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the sum as an arithmetic series with first term 1 and common difference 4 | M1 |
| using S_n = n/2(2a + (n - 1)d) with n = 25 | M1 |
| substituting to 25/2(2 + 96) | A1 |
| the sum = 1225 | A1 |
| Final answer: 1225 | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the centre (2, -3) and radius 4 | M1 |
| finding the distance from P to the centre using the distance formula | M1 |
| the distance = 8 | A1 |
| the tangent length = sqrt(8^2 - 4^2) = 4*sqrt(3) | A1 |
| Final answer: Tangent length = 4*sqrt(3) | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the limits x = 0 and x = 6, from solving 6x - x^2 = 0 | M1 |
| finding the antiderivative 3x^2 - x^3/3 | M1 |
| substituting x = 6 to get 108 - 72 = 36 | M1 |
| substituting x = 0 to get 0 | A1 |
| area = 36 square units | A1 |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| using the dot product condition p.q = 0 for perpendicular vectors | M1 |
| computing p.q = (2)(3) + (-1)(2) + (2)(c) | M1 |
| simplifying to 6 - 2 + 2c | A1 |
| forming the equation 4 + 2c = 0 | A1 |
| c = -2 | A1 |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| summing the data to give 20 | M1 |
| the mean = 2.5 | A1 |
| finding the sum of squared deviations from the mean = 34 | M1 |
| the variance = 34/8 = 4.25 and the standard deviation = 2.06 (3 sf) | A1 |
| the median = 2 (the average of the two middle values 2 and 2) | B1 |
| a reasoned conclusion: since the value 7 is an outlier that inflates the mean, the median is a better measure of a typical family | B1 |
| Final answer: mean = 2.5, standard deviation = 2.06 (3 sf); median = 2, which better represents a typical family since the mean is inflated by the outlier value 7 | |
| Question 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| substituting y = 2x + 3 into the equation of C | M1 |
| expanding and simplifying the resulting equation to 5x^2 = 0 | M1 |
| the repeated root x = 0, showing l meets C at exactly one point | A1 |
| substituting x = 0 into y = 2x + 3 | M1 |
| the point of contact (0, 3) | A1 |
| Final answer: Point of contact (0, 3) | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating to find v = 3t^2 - 18t + 24 | M1 |
| differentiating to find a = 6t - 18 | A1 |
| setting v = 0 and solving to get t^2 - 6t + 8 = 0 | M1 |
| t = 2 and t = 4 | A1 |
| evaluating s at t = 0, 2, 4, 5 (s = 0, 20, 16, 20) and using these to find the distance travelled in each interval | M1 |
| total distance = 28 m | A1 |
| Final answer: v = 3t^2 - 18t + 24, a = 6t - 18; at rest at t = 2 s and t = 4 s; total distance in first 5 s = 28 m | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| comparing location: Class A has a higher median (62) than Class B (58), so Class A generally scored higher | B1 |
| finding IQR_A = 70 - 55 = 15 and IQR_B = 62 - 50 = 12 | M1 |
| comparing spread: Class A has a larger interquartile range, so Class A's scores are more variable | B1 |
| comparing (median - lower quartile) and (upper quartile - median) for each class | M1 |
| correctly identifying Class A as roughly symmetric (8 vs 7, close in size) | B1 |
| correctly identifying Class B as negatively skewed (median - lower quartile = 8 is greater than upper quartile - median = 4) | B1 |
| Final answer: Class A has a higher median and larger spread; Class A is roughly symmetric, Class B is negatively skewed | |
| Question 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| setting up displacement equations for P (6t) and Q (0.25t^2) from their respective starting points | M1 |
| forming the equation 6t + 0.25t^2 = 200 (the sum of the displacements equals the initial gap) | M1 |
| rearranging to t^2 + 24t - 800 = 0 | A1 |
| solving using the quadratic formula, taking the positive root | M1 |
| t = 18.7 s (3 sf) | A1 |
| the distance travelled by P = 6 x 18.7 = 112 m (3 sf) | A1 |
| Final answer: t = 18.7 s (3 sf); distance travelled by P = 112 m (3 sf) | |