A Level Maths Paper 2
Covers Proof and Algebraic Methods, Coordinate Geometry, Sequences, Series and the Binomial Expansion and 11 more.
Questions
Question 1 [2 marks]
Exponentials and Logarithms
Without using a calculator, find the exact value of log_2(32).
Question 2 [3 marks]
Sequences, Series and the Binomial Expansion
A sequence is defined by u_1 = 3 and u_(n+1) = 2u_n - 1 for n >= 1.
Find the values of u_2, u_3 and u_4.
Question 3 [3 marks]
Moments
A see-saw pivots at its centre. A child of weight 300 N sits 2 m from the pivot on one side.
Find the distance from the pivot at which a child of weight 250 N must sit on the other side for the see-saw to balance.
Question 4 [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 5 [4 marks]
Sampling and Data Presentation
A company has 200 employees, numbered 1 to 200 on a staff list.
Describe how a simple random sample of 15 employees could be selected using this list and a random number generator, and state one advantage and one disadvantage of simple random sampling compared with stratified sampling.
Question 6 [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 7 [4 marks]
Integration
Evaluate the definite integral of sin(2x) with respect to x between x = 0 and x = pi/3.
Question 8 [4 marks]
Differentiation
A curve has equation y = ln(5x^2 + 1).
Find dy/dx using the chain rule, and find the gradient of the curve at the point where x = 2, giving your answer to 3 significant figures.
Question 9 [5 marks]
Forces and Newton's Laws
Forces of (3i + 4j) N and (-7i + 2j) N act on a particle of mass 2 kg.
Find the acceleration of the particle as a vector, and find the magnitude of the acceleration.
Question 10 [5 marks]
Trigonometry
Without using a calculator, prove the identity sin(x)/(1 - cos(x)) + sin(x)/(1 + cos(x)) = 2/sin(x).
Question 11 [5 marks]
Probability and the Binomial Distribution
A random variable X ~ B(25, 0.16).
Find the mean and variance of X, and find P(X = 3), giving your answer to 3 significant figures.
Question 12 [6 marks]
Vectors
Vectors p = 2i - 3j + k and q = i + 2j - 2k.
Find the value of the scalar t for which p + tq is perpendicular to p, and find p + tq in this case.
Question 13 [6 marks]
Coordinate Geometry
A circle passes through the points A(2, 8), B(7, 3) and C(-3, 3).
Using the general equation x^2 + y^2 + 2gx + 2fy + c = 0, find the values of g, f and c, and hence state the centre and radius of the circle.
Question 14 [6 marks]
The Normal Distribution and Hypothesis Testing
A seed supplier claims that 40% of a particular type of seed germinate within one week of planting. A gardener believes the true proportion is lower and plants a random sample of 25 of the seeds.
Using the binomial distribution X ~ B(25, 0.4), find the critical region for a test of H0: p = 0.4 against H1: p < 0.4 at the 5% significance level, stating the actual significance level. Given that 6 of the 25 seeds germinate within a week, state the conclusion of the test.
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 |
| substituting n = 1 to find u_2 = 2(3) - 1 = 5 | M1 |
| u_3 = 2(5) - 1 = 9 | A1 |
| u_4 = 2(9) - 1 = 17 | A1 |
| Final answer: u_2 = 5, u_3 = 9, u_4 = 17 | |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| taking moments about the pivot | M1 |
| forming the equation 300 x 2 = 250 x d | M1 |
| d = 2.4 m | A1 |
| Final answer: 2.4 m | |
| Question 4[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 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| numbering the 200 employees 1 to 200 and using a random number generator to generate numbers in this range | B1 |
| continuing until 15 distinct numbers are generated (discarding repeats or out-of-range numbers) and selecting the matching employees | B1 |
| a valid advantage, e.g. every possible sample of size 15 is equally likely, so the method is free from selection bias | B1 |
| a valid disadvantage, e.g. it does not guarantee proportional representation of subgroups, unlike stratified sampling | B1 |
| Final answer: Number the employees 1-200 and use a random number generator to select 15 distinct numbers; simple random sampling is unbiased, but (unlike stratified sampling) does not guarantee subgroups are proportionally represented. | |
| Question 6[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 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| the antiderivative -1/2 cos(2x) | M1 |
| substituting the limits x = pi/3 and x = 0 | M1 |
| -1/2 cos(2pi/3) - (-1/2 cos(0)) | A1 |
| the value 3/4 | A1 |
| Final answer: 3/4 | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the chain rule d/dx[ln(u)] = (1/u)(du/dx) | M1 |
| dy/dx = 10x/(5x^2 + 1) | A1 |
| substituting x = 2 | M1 |
| the gradient = 20/21 (or 0.952 to 3 sf) | A1 |
| Final answer: dy/dx = 10x/(5x^2 + 1); gradient at x = 2 is 20/21 (0.952 to 3 sf) | |
| Question 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| summing the forces: (3 - 7)i + (4 + 2)j | M1 |
| the resultant force = -4i + 6j (N) | A1 |
| using a = F/m | M1 |
| a = -2i + 3j (m/s^2) | A1 |
| the magnitude of a = sqrt(2^2 + 3^2) = 3.61 m/s^2 (3 sf) | A1 |
| Final answer: a = -2i + 3j (m/s^2); magnitude = 3.61 m/s^2 (3 sf) | |
| Question 10[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 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| the mean = np = 25 x 0.16 | M1 |
| the mean = 4 and the variance = np(1 - p) = 25 x 0.16 x 0.84 = 3.36 | A1 |
| using the binomial formula C(25,3)(0.16)^3(0.84)^22 for P(X = 3) | M1 |
| evaluating to an unrounded value of approximately 0.2033 | A1 |
| P(X = 3) = 0.203 (3 sf) | A1 |
| Final answer: mean = 4, variance = 3.36; P(X = 3) = 0.203 (3 sf) | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| forming the condition (p + tq).p = 0 | M1 |
| expanding to p.p + t(q.p) = 0 | M1 |
| p.p = 14 | A1 |
| q.p = -6 | A1 |
| solving to t = 14/6 = 7/3 | M1 |
| p + tq = (13/3)i + (5/3)j - (11/3)k | A1 |
| Final answer: t = 7/3; p + tq = (13/3)i + (5/3)j - (11/3)k | |
| Question 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| substituting each point into the general equation to form three equations | M1 |
| the three equations 4g + 16f + c = -68, 14g + 6f + c = -58 and -6g + 6f + c = -18 (or equivalent) | A1 |
| eliminating c to obtain two equations in g and f | M1 |
| g = -2 and f = -3 | A1 |
| substituting back to find c = -12 | M1 |
| the centre (2, 3) and radius 5 | A1 |
| Final answer: g = -2, f = -3, c = -12; centre (2, 3), radius 5 | |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| identifying X ~ B(25, 0.4) and attempting cumulative probabilities under H0 | M1 |
| P(X <= 5) = 0.0294 (accept | A1awrt |
| P(X <= 6) = 0.0736 (accept awrt), confirming this exceeds 0.05 | A1 |
| the critical region X <= 5, with actual significance level 0.0294 (2.94%) | A1 |
| noting that 6 does not lie in the critical region, since 6 > 5 | A1 |
| the conclusion: insufficient evidence at the 5% level that the true proportion germinating is lower than 40% | A1 |
| Final answer: Critical region X <= 5 (significance level 0.0294); since 6 is not in the critical region, there is insufficient evidence that the true proportion is lower than 40% | |