A Level Maths Paper 1
Covers Proof and Algebraic Methods, Coordinate Geometry, Sequences, Series and the Binomial Expansion and 11 more.
Questions
Question 1 [2 marks]
Proof and Algebraic Methods
Without using a calculator, show that sqrt(45) + sqrt(20) can be written as k*sqrt(5), stating the value of k.
Question 2 [3 marks]
Integration
Evaluate the definite integral of (3x^2 - 2x + 1) with respect to x between x = -1 and x = 2.
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]
The Normal Distribution and Hypothesis Testing
The random variable X ~ N(75, 12^2).
Find P(65 < X < 90).
Question 5 [3 marks]
Vectors
Points A and B have position vectors a = 3i + 2j and b = 9i - 4j relative to a fixed origin O.
Find the position vector of the point P that divides AB such that AP:PB = 1:2.
Question 6 [4 marks]
Differentiation
Without using a calculator, find dy/dx for y = sin(3x) + cos(2x), and find the exact gradient of the curve at x = pi/6.
Question 7 [4 marks]
Moments
A light rod AB has length 6 m. Weights of 50 N and 70 N are attached at A and B respectively.
Find the distance from A at which the rod must be pivoted for it to balance horizontally.
Question 8 [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 9 [4 marks]
Trigonometry
Without using a calculator, prove the identity sec^2(x) - sin^2(x) sec^2(x) = 1.
Question 10 [5 marks]
Kinematics
A particle P moves along the x-axis such that its displacement from a fixed point O at time t seconds is given by x = 2t^3 - 15t^2 + 24t + 5 for t >= 0.
Find the velocity and the acceleration of P when t = 3, and determine whether P is accelerating or decelerating at this instant.
Question 11 [5 marks]
Sequences, Series and the Binomial Expansion
The second term of a geometric sequence is 6 and the fifth term is 162.
Find the first term and the common ratio, and hence find the sum of the first 6 terms.
Question 12 [4 marks]
Coordinate Geometry
Using coordinates, find the area of the triangle with vertices A(1, 2), B(7, 4) and C(3, 8).
Question 13 [4 marks]
Sampling and Data Presentation
A sample of 12 pairs of data gives a product moment correlation coefficient of r = 0.62. The critical value for a sample of size 12 at the 5% significance level (one-tail) is 0.497.
Test, at the 5% significance level, whether this provides evidence of positive correlation between the variables in the population.
Question 14 [6 marks]
Probability and the Binomial Distribution
A random variable X ~ B(n, 0.25).
Given that P(X = 2) = P(X = 3), find the value of n.
Question 15 [6 marks]
Sampling and Data Presentation
A college has 900 students: 340 study Sciences, 260 study Humanities and 300 study Arts. A stratified sample of 45 students is to be selected by subject, and within Sciences, systematic sampling is then used to choose the required number from a numbered list of all 340 Science students.
Find the number of Science students needed in the sample, and find the sampling interval that should be used for the systematic sampling.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| simplifying sqrt(45) = 3*sqrt(5) and sqrt(20) = 2*sqrt(5) | M1 |
| k = 5 | A1 |
| Question 2[3 marks] | |
|---|---|
| Answer or working | Marks |
| the antiderivative x^3 - x^2 + x | M1 |
| substituting the limits x = 2 and x = -1 | M1 |
| the value 9 | A1 |
| Final answer: 9 | |
| 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 |
| standardising both limits to z = -0.833 (3 sf) and z = 1.25 | M1 |
| using P(65 < X < 90) = Phi(1.25) - (1 - Phi(0.833)) | M1 |
| P(65 < X < 90) = 0.692 (3 sf) | A1 |
| Final answer: 0.692 (3 sf) | |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| finding AB = b - a = 6i - 6j | M1 |
| finding AP = (1/3)AB = 2i - 2j | M1 |
| OP = a + AP = 5i | A1 |
| Final answer: OP = 5i (or 5i + 0j) | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating sin(3x) using the chain rule to give 3cos(3x) | M1 |
| differentiating cos(2x) using the chain rule to give -2sin(2x) | M1 |
| dy/dx = 3cos(3x) - 2sin(2x) | A1 |
| substituting x = pi/6 to give the gradient = 3cos(pi/2) - 2sin(pi/3) = 0 - sqrt(3) = -sqrt(3) | A1 |
| Final answer: dy/dx = 3cos(3x) - 2sin(2x); gradient at x = pi/6 is -sqrt(3) | |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| taking moments about the pivot, a distance x from A | M1 |
| forming the equation 50x = 70(6 - x) | M1 |
| 120x = 420 | A1 |
| x = 3.5 m | A1 |
| Final answer: 3.5 m from A | |
| Question 8[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 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| factorising the left-hand side as sec^2(x)(1 - sin^2(x)) | M1 |
| using 1 - sin^2(x) = cos^2(x) | A1 |
| simplifying to sec^2(x) cos^2(x) | M1 |
| using sec(x) = 1/cos(x) to give cos^2(x)/cos^2(x) = 1, completing the proof | A1 |
| Final answer: Identity proved: the left-hand side simplifies to sec^2(x) cos^2(x) = 1 | |
| Question 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating to find v = 6t^2 - 30t + 24 | M1 |
| v = -12 m/s at t = 3 | A1 |
| differentiating to find a = 12t - 30 | M1 |
| a = 6 m/s^2 at t = 3 | A1 |
| the conclusion: since v and a have opposite signs, P is decelerating at t = 3 | A1 |
| Final answer: v = -12 m/s, a = 6 m/s^2 at t = 3; P is decelerating (v and a have opposite signs) | |
| Question 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| forming ar = 6 and ar^4 = 162 | M1 |
| dividing to get r^3 = 27 | M1 |
| r = 3 and a = 2 | A1 |
| using S_n = a(r^n - 1)/(r - 1) with n = 6 | M1 |
| S_6 = 728 | A1 |
| Final answer: a = 2, r = 3; sum of first 6 terms = 728 | |
| Question 12[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the coordinate (shoelace) formula for the area of a triangle | M1 |
| substituting the three coordinates correctly into the formula | M1 |
| evaluating the expression inside the modulus to 32 | A1 |
| the area = 16 square units | A1 |
| Final answer: Area = 16 square units | |
| Question 13[4 marks] | |
|---|---|
| Answer or working | Marks |
| stating hypotheses H0: rho = 0, H1: rho > 0, where rho is the population correlation coefficient | B1 |
| comparing the sample value r = 0.62 with the critical value 0.497 | M1 |
| noting 0.62 > 0.497 | A1 |
| the conclusion: reject H0, there is evidence of positive correlation between the variables in the population | A1 |
| Final answer: 0.62 > 0.497, so reject H0: evidence of positive correlation in the population | |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| writing the ratio P(X = 3)/P(X = 2) using the binomial formula | M1 |
| simplifying the combinatorial part of the ratio to C(n,3)/C(n,2) = (n - 2)/3 | M1 |
| simplifying the probability part of the ratio to p/(1 - p) = 1/3 | A1 |
| forming the equation (n - 2)/3 x 1/3 = 1, since P(X = 3) = P(X = 2) | M1 |
| simplifying to (n - 2)/9 = 1 | A1 |
| n = 11 | A1 |
| Question 15[6 marks] | |
|---|---|
| Answer or working | Marks |
| the sampling fraction = 45/900 = 0.05 | M1 |
| multiplying the Science total by the fraction | M1 |
| the Science sample size = 17 | A1 |
| the systematic sampling interval = population size / sample size | M1 |
| the interval = 340/17 = 20 | A1 |
| a correct statement of how the systematic sample would then be chosen, e.g. select a random start between 1 and 20 and then every 20th student thereafter | B1 |
| Final answer: 17 Science students needed; systematic sampling interval = 20 | |