A Level

A Level Maths Paper 1

Covers Proof and Algebraic Methods, Coordinate Geometry, Sequences, Series and the Binomial Expansion and 11 more.

15 questions - 60 marks - calculator allowed

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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

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or workingMarks
simplifying sqrt(45) = 3*sqrt(5) and sqrt(20) = 2*sqrt(5)M1
k = 5A1
Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or workingMarks
the antiderivative x^3 - x^2 + xM1
substituting the limits x = 2 and x = -1M1
the value 9A1
Final answer: 9
Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or workingMarks
identifying vertical equilibrium: R = weightM1
weight = mg = 4 x 9.8M1
R = 39.2 NA1
Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or workingMarks
standardising both limits to z = -0.833 (3 sf) and z = 1.25M1
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)
Mark scheme for Question 5 [3 marks]
Question 5[3 marks]
Answer or workingMarks
finding AB = b - a = 6i - 6jM1
finding AP = (1/3)AB = 2i - 2jM1
OP = a + AP = 5iA1
Final answer: OP = 5i (or 5i + 0j)
Mark scheme for Question 6 [4 marks]
Question 6[4 marks]
Answer or workingMarks
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)
Mark scheme for Question 7 [4 marks]
Question 7[4 marks]
Answer or workingMarks
taking moments about the pivot, a distance x from AM1
forming the equation 50x = 70(6 - x)M1
120x = 420A1
x = 3.5 mA1
Final answer: 3.5 m from A
Mark scheme for Question 8 [4 marks]
Question 8[4 marks]
Answer or workingMarks
using the substitution y = e^x to obtain the quadratic y^2 - 4y + 3 = 0M1
factorising to (y - 1)(y - 3) = 0A1
y = 1, giving x = 0A1
y = 3, giving x = ln(3)A1
Final answer: x = 0 or x = ln(3)
Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or workingMarks
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 proofA1
Final answer: Identity proved: the left-hand side simplifies to sec^2(x) cos^2(x) = 1
Mark scheme for Question 10 [5 marks]
Question 10[5 marks]
Answer or workingMarks
differentiating to find v = 6t^2 - 30t + 24M1
v = -12 m/s at t = 3A1
differentiating to find a = 12t - 30M1
a = 6 m/s^2 at t = 3A1
the conclusion: since v and a have opposite signs, P is decelerating at t = 3A1
Final answer: v = -12 m/s, a = 6 m/s^2 at t = 3; P is decelerating (v and a have opposite signs)
Mark scheme for Question 11 [5 marks]
Question 11[5 marks]
Answer or workingMarks
forming ar = 6 and ar^4 = 162M1
dividing to get r^3 = 27M1
r = 3 and a = 2A1
using S_n = a(r^n - 1)/(r - 1) with n = 6M1
S_6 = 728A1
Final answer: a = 2, r = 3; sum of first 6 terms = 728
Mark scheme for Question 12 [4 marks]
Question 12[4 marks]
Answer or workingMarks
using the coordinate (shoelace) formula for the area of a triangleM1
substituting the three coordinates correctly into the formulaM1
evaluating the expression inside the modulus to 32A1
the area = 16 square unitsA1
Final answer: Area = 16 square units
Mark scheme for Question 13 [4 marks]
Question 13[4 marks]
Answer or workingMarks
stating hypotheses H0: rho = 0, H1: rho > 0, where rho is the population correlation coefficientB1
comparing the sample value r = 0.62 with the critical value 0.497M1
noting 0.62 > 0.497A1
the conclusion: reject H0, there is evidence of positive correlation between the variables in the populationA1
Final answer: 0.62 > 0.497, so reject H0: evidence of positive correlation in the population
Mark scheme for Question 14 [6 marks]
Question 14[6 marks]
Answer or workingMarks
writing the ratio P(X = 3)/P(X = 2) using the binomial formulaM1
simplifying the combinatorial part of the ratio to C(n,3)/C(n,2) = (n - 2)/3M1
simplifying the probability part of the ratio to p/(1 - p) = 1/3A1
forming the equation (n - 2)/3 x 1/3 = 1, since P(X = 3) = P(X = 2)M1
simplifying to (n - 2)/9 = 1A1
n = 11A1
Mark scheme for Question 15 [6 marks]
Question 15[6 marks]
Answer or workingMarks
the sampling fraction = 45/900 = 0.05M1
multiplying the Science total by the fractionM1
the Science sample size = 17A1
the systematic sampling interval = population size / sample sizeM1
the interval = 340/17 = 20A1
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 thereafterB1
Final answer: 17 Science students needed; systematic sampling interval = 20