A Level Maths Paper 4
Covers Proof and Algebraic Methods, Coordinate Geometry, Sequences, Series and the Binomial Expansion and 11 more.
Questions
Question 1 [2 marks]
Differentiation
Without using a calculator, differentiate y = 4x^3 - 7x^2 + 2 with respect to x.
Question 2 [2 marks]
Proof and Algebraic Methods
Without using a calculator, simplify fully (3x^2 - 12) / (x^2 + x - 6).
Question 3 [3 marks]
Forces and Newton's Laws
A box of mass 4 kg rests on a horizontal surface.
Find the normal reaction force on the box. Use g = 9.8 m/s^2.
Question 4 [3 marks]
Sampling and Data Presentation
A researcher wants to select a sample of 50 diners from a large restaurant chain with 40 branches.
Describe how cluster sampling could be used to select the sample, and state one advantage of cluster sampling over stratified sampling in this context.
Question 5 [4 marks]
Sequences, Series and the Binomial Expansion
A geometric sequence has first term 16 and common ratio -1/2.
Find the 5th term, and find the sum to infinity of the sequence.
Question 6 [4 marks]
Exponentials and Logarithms
Without using a calculator, solve the equation e^(2x) - 4e^x + 3 = 0, giving both values of x as exact values.
Question 7 [4 marks]
Probability and the Binomial Distribution
A box contains 5 red and 3 blue balls. Three balls are selected at random without replacement.
Using combinations, find the probability that exactly 2 of the balls selected are red.
Question 8 [4 marks]
Kinematics
A particle travels in a straight line. Its velocity v m/s at time t seconds is given by v = 3t^2 - 12t + 9 for 0 <= t <= 5.
Find the values of t at which the particle is instantaneously at rest, and find the acceleration of the particle when t = 4.
Question 9 [5 marks]
Moments
A uniform beam AB has length 10 m and weight 200 N. It is supported horizontally by two vertical supports, one at A and one at B. A load of 60 N is placed 2 m from A, and a load of 90 N is placed 7 m from A.
Find the reactions at A and B, and verify your answer by taking moments about B.
Question 10 [6 marks]
Trigonometry
Solve 5 sin(x) + 2 = 5 for 0 <= x <= 360 deg, and hence solve 5 sin(2y) + 2 = 5 for 0 <= y <= 180 deg, giving all answers to 1 decimal place.
Question 11 [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 12 [6 marks]
Integration
The region R is bounded by the curve y = x^2 - 2x and the line y = 3x, between their two points of intersection.
Find the area of R.
Question 13 [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 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 |
| differentiating each term | M1 |
| dy/dx = 12x^2 - 14x | A1 |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| factorising the numerator as 3(x - 2)(x + 2) and the denominator as (x + 3)(x - 2) | M1 |
| cancelling the common factor (x - 2) to give 3(x + 2)/(x + 3) | A1 |
| Final answer: 3(x + 2)/(x + 3) | |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| identifying vertical equilibrium: R = weight | M1 |
| weight = mg = 4 x 9.8 | M1 |
| R = 39.2 N | A1 |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| a correct description of cluster sampling in context, e.g. randomly select a small number of branches (clusters) and sample all, or a proportion of, diners within each selected branch | B1 |
| a second valid detail of the method, e.g. the branches selected should themselves be chosen using simple random sampling | B1 |
| a valid advantage over stratified sampling, e.g. it does not require prior knowledge of every diner across all 40 branches, and is more practical to carry out | B1 |
| Final answer: Any valid description of cluster sampling (randomly select branches, then sample diners within them) and a valid advantage (e.g. no need for a full list of every diner). | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the nth term formula a*r^(n - 1) with n = 5 | M1 |
| the 5th term = 1 | A1 |
| using the sum to infinity formula a/(1 - r) | M1 |
| the sum to infinity = 32/3 | A1 |
| Final answer: 5th term = 1; sum to infinity = 32/3 | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the substitution y = e^x to obtain the quadratic y^2 - 4y + 3 = 0 | M1 |
| factorising to (y - 1)(y - 3) = 0 | A1 |
| y = 1, giving x = 0 | A1 |
| y = 3, giving x = ln(3) | A1 |
| Final answer: x = 0 or x = ln(3) | |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| using P = [C(5,2) x C(3,1)] / C(8,3) | M1 |
| C(5,2) = 10 and C(3,1) = 3 | A1 |
| C(8,3) = 56 | A1 |
| P = 30/56 = 15/28 | A1 |
| Final answer: 15/28 | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| setting v = 0 and simplifying to t^2 - 4t + 3 = 0 | M1 |
| t = 1 and t = 3 | A1 |
| differentiating to find a = 6t - 12 | M1 |
| a = 12 m/s^2 at t = 4 | A1 |
| Final answer: t = 1 s and t = 3 s; acceleration at t = 4 is 12 m/s^2 | |
| Question 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| taking moments about A: R_B x 10 = 200 x 5 + 60 x 2 + 90 x 7 | M1 |
| R_B = 175 N | A1 |
| using vertical equilibrium: R_A + R_B = 200 + 60 + 90 | M1 |
| R_A = 175 N | A1 |
| verifying by taking moments about B: R_A x 10 = 200 x 5 + 60 x 8 + 90 x 3 = 1750, confirming R_A = 175 N | M1 |
| Final answer: R_A = 175 N, R_B = 175 N | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging to sin(x) = 0.6 | M1 |
| x = 36.9 deg (1 dp) | A1 |
| x = 143.1 deg (1 dp), using the symmetry x = 180 - x | A1 |
| substituting 2y for x and solving sin(2y) = 0.6 for 0 <= 2y <= 360 deg to give 2y = 36.9 deg or 2y = 143.1 deg | M1 |
| y = 18.4 deg (1 dp) | A1 |
| y = 71.6 deg (1 dp) | A1 |
| Final answer: x = 36.9 deg or 143.1 deg; y = 18.4 deg or 71.6 deg (1 dp) | |
| Question 11[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 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| setting x^2 - 2x = 3x to find the intersections | M1 |
| x = 0 and x = 5 | A1 |
| recognising the line lies above the curve on (0, 5) and forming the integral of (3x - (x^2 - 2x)) dx | M1 |
| finding the antiderivative (5/2)x^2 - (1/3)x^3 | M1 |
| substituting the limits to obtain 125/6 | A1 |
| area = 125/6 square units, stated as an exact answer | A1 |
| Final answer: Area = 125/6 square units | |
| Question 13[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 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% | |