AS Paper 6: Full Course Review
Covers proof and algebraic methods, sequences and series with the binomial expansion, trigonometry, differentiation, vectors, sampling and data presentation, the binomial probability distribution, and forces and Newton's laws.
Questions
Question 1 [2 marks]
Trigonometry
Without using a calculator, find the exact value of cos(60 deg) + sin(30 deg) - tan(45 deg).
Question 2 [2 marks]
Differentiation
Without using a calculator, differentiate y = 4x^3 - 7x^2 + 2 with respect to x.
Question 3 [2 marks]
Forces and Newton's Laws
A resultant force of 36 N gives a body an acceleration of 4 m/s^2.
Find the mass of the body.
Question 4 [3 marks]
Sequences, Series and the Binomial Expansion
The first three terms of a geometric sequence are 8, 12, 18.
Find the common ratio and the 6th term.
Question 5 [3 marks]
Proof and Algebraic Methods
Without using a calculator, solve the inequality x^2 - 2x - 8 <= 0, giving your answer as a single inequality.
Question 6 [3 marks]
Sampling and Data Presentation
State one advantage and one disadvantage of using stratified sampling rather than simple random sampling to select a sample of students from a school, and state what is meant by a sampling frame.
Question 7 [4 marks]
Probability and the Binomial Distribution
A bag contains 5 red and 3 blue counters. A counter is drawn at random, its colour is noted, and it is then replaced. A second counter is then drawn.
Find the probability that both counters are the same colour.
Question 8 [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 9 [4 marks]
Trigonometry
Using the small angle approximation cos(x) = 1 - (x^2)/2, valid for small values of x in radians, show that (1 - cos(x))/x^2 is approximately 1/2 for small x.
Question 10 [3 marks]
Proof and Algebraic Methods
By finding a suitable counter-example, show that the statement "n^2 + n + 1 is a prime number for every positive integer n" is false.
Question 11 [4 marks]
Vectors
Find the angle that the vector v = 2i - j + 2k makes with the positive x-axis, giving your answer to 1 decimal place.
Question 12 [3 marks]
Proof and Algebraic Methods
The function f is defined by f(x) = (2x - 1)/(x + 3) for x != -3.
Find f^-1(x) and state its domain.
Question 13 [5 marks]
Differentiation
Without using a calculator, differentiate y = (5x^2 - 3x)^4 with respect to x, using the chain rule.
Question 14 [4 marks]
Sampling and Data Presentation
A scatter diagram for 10 pairs of data on hours revised (x) and test score (y) has product moment correlation coefficient r = 0.82.
Interpret this value in context, and state, giving a reason, whether it would be appropriate to use the regression line of y on x to predict the score of a student who revised for 15 hours, given that the data ranges from 2 to 12 hours of revision.
Question 15 [5 marks]
Forces and Newton's Laws
Two particles A and B, of mass 7 kg and 5 kg respectively, are connected by a light inextensible string passing over a smooth fixed pulley. The particles are released from rest, hanging freely on either side of the pulley.
Find the acceleration of the system, and find the tension in the string. Use g = 9.8 m/s^2.
Question 16 [5 marks]
Differentiation
A curve is defined implicitly by x^2 + y^2 - 4x + 6y = 12. The point (5, 1) lies on the curve.
Using implicit differentiation, find dy/dx in terms of x and y, and find the gradient of the curve at (5, 1).
Question 17 [5 marks]
Trigonometry
Without using a calculator, prove the identity sin(x)/(1 - cos(x)) + sin(x)/(1 + cos(x)) = 2/sin(x).
Question 18 [5 marks]
Differentiation
A curve has equation y = x^3 - 3x + 2.
Find the equation of the normal to the curve at the point where x = 2, giving your answer in the form y = mx + c.
Question 19 [6 marks]
Forces and Newton's Laws
Particles A (mass 4 kg) and B (mass 6 kg) are connected by a light inextensible string passing over a smooth pulley fixed at the top of two smooth inclined planes, back to back. A rests on a plane inclined at 30 degrees to the horizontal, and B rests on a plane inclined at 40 degrees to the horizontal, on the other side of the pulley.
The system is released from rest with B moving down its plane and A moving up its plane. Find the acceleration of the system and the tension in the string. Use g = 9.8 m/s^2.
Question 20 [5 marks]
Differentiation
Without using a calculator, find the set of values of x for which f(x) = 2x^3 - 3x^2 - 12x + 7 is decreasing.
Question 21 [6 marks]
Forces and Newton's Laws
A particle of mass 2 kg is held in equilibrium on a rough plane inclined at 30 degrees to the horizontal by a horizontal force of magnitude P newtons, acting in the vertical plane containing the line of greatest slope. The coefficient of friction between the particle and the plane is 0.2, and the particle is on the point of slipping down the plane.
Find the value of P. Use g = 9.8 m/s^2.
Question 22 [6 marks]
Probability and the Binomial Distribution
A multiple-choice test has 8 questions, each with 5 possible answers, only one of which is correct. A student guesses the answer to every question independently.
Find the probability that the student gets at least 2 questions correct, giving your answer to 3 significant figures.
Question 23 [5 marks]
Trigonometry
Without using a calculator, solve 2 cos^2(x) + 3 sin(x) = 3 for 0 <= x <= 360 deg.
Question 24 [6 marks]
Forces and Newton's Laws
A block of mass 4 kg is pushed up a rough plane inclined at 25 degrees to the horizontal by a force of 45 N acting parallel to the plane. After travelling 3 m up the plane from rest, the force is removed. The coefficient of friction between the block and the plane is 0.35.
Using g = 9.8 m/s^2, find the speed of the block when the force is removed, and find the additional distance the block travels up the plane before coming to rest.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| stating cos(60 deg) = 1/2, sin(30 deg) = 1/2 and tan(45 deg) = 1 | M1 |
| 0 | A1 |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating each term | M1 |
| dy/dx = 12x^2 - 14x | A1 |
| Question 3[2 marks] | |
|---|---|
| Answer or working | Marks |
| using F = ma | M1 |
| m = 9 kg | A1 |
| Final answer: 9 kg | |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| common ratio r = 12/8 = 1.5 | M1 |
| using the 6th term = a*r^5 | M1 |
| 6th term = 60.75 | A1 |
| Final answer: r = 1.5, 6th term = 60.75 | |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| factorising x^2 - 2x - 8 as (x - 4)(x + 2) | M1 |
| identifying the critical values x = -2 and x = 4 | M1 |
| the solution -2 <= x <= 4 | A1 |
| Final answer: -2 <= x <= 4 | |
| Question 6[3 marks] | |
|---|---|
| Answer or working | Marks |
| a valid advantage of stratified sampling, e.g. it ensures proportional representation from every subgroup, reducing the risk of a biased sample | B1 |
| a valid disadvantage, e.g. it requires the population to be divided into known, non-overlapping strata before sampling, which needs prior information | B1 |
| a correct description of a sampling frame, e.g. a list of all the members of the population from which the sample is drawn | B1 |
| Final answer: Any valid advantage (e.g. proportional representation), disadvantage (e.g. requires known strata) and correct definition of a sampling frame (a list of all population members). | |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| P(both red) = (5/8)(5/8) | M1 |
| P(both blue) = (3/8)(3/8) | M1 |
| the two probabilities 25/64 and 9/64 | A1 |
| the total 34/64 = 17/32 | A1 |
| Final answer: 17/32 | |
| Question 8[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 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting the approximation cos(x) = 1 - (x^2)/2 | M1 |
| 1 - cos(x) = (x^2)/2 | A1 |
| dividing by x^2 | M1 |
| (1 - cos(x))/x^2 = 1/2, as required | A1 |
| Final answer: (1 - cos(x))/x^2 is approximately 1/2 for small x, since 1 - cos(x) = (x^2)/2 using the small angle approximation | |
| Question 10[3 marks] | |
|---|---|
| Answer or working | Marks |
| testing successive values of n (e.g. n = 1, 2, 3) and finding each gives a prime | M1 |
| testing n = 4 to obtain n^2 + n + 1 = 21 | M1 |
| identifying 21 = 3 x 7 is not prime, so the statement is false | A1 |
| Final answer: False: n = 4 gives n^2 + n + 1 = 21 = 3 x 7, which is not prime. | |
| Question 11[4 marks] | |
|---|---|
| Answer or working | Marks |
| using cos(theta) = (v.i)/|v| | M1 |
| v.i = 2 and |v| = 3 | A1 |
| cos(theta) = 2/3 | M1 |
| theta = 48.2 deg (1 dp) | A1 |
| Question 12[3 marks] | |
|---|---|
| Answer or working | Marks |
| setting y = (2x - 1)/(x + 3) and rearranging to y(x + 3) = 2x - 1 | M1 |
| collecting x terms to give x(y - 2) = -1 - 3y | M1 |
| f^-1(x) = (1 + 3x)/(2 - x), with domain x != 2 | A1 |
| Final answer: f^-1(x) = (1 + 3x)/(2 - x), x != 2 | |
| Question 13[5 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the inner function u = 5x^2 - 3x and outer function u^4 | M1 |
| differentiating the outer function: 4u^3 | M1 |
| differentiating the inner function: du/dx = 10x - 3 | A1 |
| applying the chain rule dy/dx = 4u^3 x du/dx | M1 |
| dy/dx = 4(5x^2 - 3x)^3(10x - 3) | A1 |
| Final answer: dy/dx = 4(5x^2 - 3x)^3 (10x - 3) | |
| Question 14[4 marks] | |
|---|---|
| Answer or working | Marks |
| stating r = 0.82 indicates a strong positive correlation between hours revised and test score | B1 |
| a correct contextual interpretation, e.g. as hours revised increases, test score tends to increase | B1 |
| stating it would not be appropriate, because 15 hours is outside the range of the data (2 to 12 hours) | B1 |
| correctly naming this extrapolation, and noting the relationship may not hold outside the observed range | B1 |
| Final answer: r = 0.82 shows strong positive correlation; predicting for 15 hours would be extrapolation and unreliable, since it is outside the data range 2 to 12 hours. | |
| Question 15[5 marks] | |
|---|---|
| Answer or working | Marks |
| the equation of motion for A: 7g - T = 7a | M1 |
| the equation of motion for B: T - 5g = 5a | M1 |
| adding the two equations to eliminate T: 2g = 12a | M1 |
| a = 1.63 m/s^2 (3 sf) | A1 |
| T = 57.2 N (3 sf) | A1 |
| Final answer: a = 1.63 m/s^2 (3 sf), T = 57.2 N (3 sf) | |
| Question 16[5 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating each term implicitly: 2x + 2y(dy/dx) - 4 + 6(dy/dx) = 0 | M1 |
| collecting the dy/dx terms: dy/dx (2y + 6) = 4 - 2x | M1 |
| dy/dx = (2 - x)/(y + 3) | A1 |
| substituting the point (5, 1) | M1 |
| the gradient = -3/4 | A1 |
| Final answer: dy/dx = (2 - x)/(y + 3); gradient at (5, 1) is -3/4 | |
| Question 17[5 marks] | |
|---|---|
| Answer or working | Marks |
| writing the left-hand side as a single fraction over (1 - cos(x))(1 + cos(x)) | M1 |
| simplifying the denominator to 1 - cos^2(x) | M1 |
| using 1 - cos^2(x) = sin^2(x) | A1 |
| simplifying the numerator sin(x)(1 + cos(x)) + sin(x)(1 - cos(x)) to 2 sin(x) | M1 |
| combining to 2 sin(x)/sin^2(x) = 2/sin(x), completing the proof | A1 |
| Final answer: Identity proved: the left-hand side simplifies to 2/sin(x) | |
| Question 18[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding the y-coordinate at x = 2: y = 4 | M1 |
| differentiating to get dy/dx = 3x^2 - 3 | M1 |
| the gradient at x = 2 is 9 | A1 |
| the normal gradient = -1/9 and forming y - 4 = -1/9(x - 2) | M1 |
| y = -x/9 + 38/9 (or equivalent) | A1 |
| Final answer: y = -x/9 + 38/9 | |
| Question 19[6 marks] | |
|---|---|
| Answer or working | Marks |
| the equation of motion for A along its plane: T - mg sin(30) = 4a | M1 |
| the equation of motion for B along its plane: 6g sin(40) - T = 6a | M1 |
| adding the two equations to eliminate T | M1 |
| a = 1.82 m/s^2 (3 sf) | A1 |
| substituting back to find T, e.g. T = 4(a + g sin(30)) | M1 |
| T = 26.9 N (3 sf) | A1 |
| Final answer: a = 1.82 m/s^2 (3 sf); T = 26.9 N (3 sf) | |
| Question 20[5 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating to find f'(x) = 6x^2 - 6x - 12 | M1 |
| factorising f'(x) as 6(x - 2)(x + 1) | M1 |
| the critical values x = -1 and x = 2 | A1 |
| testing the sign of f'(x) between and outside the critical values | M1 |
| f is decreasing for -1 < x < 2 | A1 |
| Final answer: -1 < x < 2 | |
| Question 21[6 marks] | |
|---|---|
| Answer or working | Marks |
| resolving perpendicular to the plane: R = mg cos(30) + P sin(30) | M1 |
| resolving along the plane with friction F = mu R acting up the slope (point of slipping down): P cos(30) + F = mg sin(30) | M1 |
| substituting F = 0.2R to form P cos(30) + 0.2(mg cos(30) + P sin(30)) = mg sin(30) | A1 |
| collecting terms in P to give P(cos(30) + 0.2 sin(30)) = mg(sin(30) - 0.2 cos(30)) | M1 |
| P = mg(sin(30) - 0.2 cos(30)) / (cos(30) + 0.2 sin(30)) | A1 |
| P = 6.63 N (3 sf) | A1 |
| Question 22[6 marks] | |
|---|---|
| Answer or working | Marks |
| identifying X ~ B(8, 0.2) | M1 |
| using the complement P(X >= 2) = 1 - P(X = 0) - P(X = 1) | M1 |
| computing P(X = 0) = (0.8)^8 = 0.168 (3 sf) | M1 |
| computing P(X = 1) = C(8,1)(0.2)(0.8)^7 = 0.336 (3 sf) | A1 |
| the sum = 0.503 (3 sf) | A1 |
| P(X >= 2) = 1 - 0.503 = 0.497 (3 sf) | A1 |
| Final answer: P(X >= 2) = 0.497 (3 sf) | |
| Question 23[5 marks] | |
|---|---|
| Answer or working | Marks |
| substituting cos^2(x) = 1 - sin^2(x) | M1 |
| rearranging to the quadratic 2 sin^2(x) - 3 sin(x) + 1 = 0 | M1 |
| factorising to (2 sin(x) - 1)(sin(x) - 1) = 0 | A1 |
| sin(x) = 1/2 giving x = 30 deg and x = 150 deg | A1 |
| sin(x) = 1 giving x = 90 deg | A1 |
| Final answer: x = 30 deg, 90 deg or 150 deg | |
| Question 24[6 marks] | |
|---|---|
| Answer or working | Marks |
| resolving perpendicular to find R = mg cos(25) and friction = mu R | M1 |
| R = 35.5 N (3 sf) and friction = 12.4 N (3 sf) | A1 |
| the equation of motion while the force acts: 45 - mg sin(25) - friction = 4a | M1 |
| a = 4.00 m/s^2 (3 sf), giving v^2 = 2 x 4.00 x 3 = 24.0 and v = 4.90 m/s (3 sf) | A1 |
| the equation of motion after the force is removed: mg sin(25) + friction = 4a2 | M1 |
| a2 = 7.25 m/s^2 (3 sf), giving the additional distance = 24.0/(2 x 7.25) = 1.65 m (3 sf) | A1 |
| Final answer: v = 4.90 m/s (3 sf); additional distance = 1.65 m (3 sf) | |