A Level

A Level Paper 1: Pure Foundations

Covers proof and algebraic methods, coordinate geometry, sequences and series with the binomial expansion, trigonometry, exponentials and logarithms, and differentiation.

15 questions - 60 marks - calculator allowed

Download printable PDF

Questions

Question 1 [2 marks]

Exponentials and Logarithms

Without using a calculator, solve the equation 2^x = 1/8, giving the exact value of x.

Question 2 [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 3 [3 marks]

Coordinate Geometry

A line passes through the point (2, -5) and is parallel to the line 3x - y + 4 = 0.

Find the equation of the line in the form y = mx + c.

Question 4 [3 marks]

Trigonometry

Without using a calculator, given that sin(theta) = 3/5 where theta is acute, find the exact values of cos(theta) and tan(theta).

Question 5 [3 marks]

Differentiation

Without using a calculator, find dy/dx for y = e^(3x) - 4x.

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]

Exponentials and Logarithms

The number of bacteria in a culture is modelled by N = 200 e^(0.15t), where t is the time in hours after the culture is started.

Find the number of bacteria after 10 hours, and find the time taken for the number of bacteria to reach 1000, giving your answer to 1 decimal place.

Question 8 [4 marks]

Differentiation

A curve has equation y = (4x - 1)/(x + 2).

Find dy/dx using the quotient rule, and find the gradient of the curve at the point where x = 1.

Question 9 [4 marks]

Exponentials and Logarithms

Given that log_a(5) = p and log_a(2) = q, express log_a(20) in terms of p and q.

Question 10 [5 marks]

Differentiation

Without using a calculator, differentiate y = (5x^2 - 3x)^4 with respect to x, using the chain rule.

Question 11 [5 marks]

Differentiation

A curve has parametric equations x = t^2, y = t^3 - 3t.

Find dy/dx in terms of t, and find the gradient of the curve at the point where t = 2.

Question 12 [5 marks]

Exponentials and Logarithms

Solve the equation 5^x = 3^(x + 1), giving your answer to 3 significant figures.

Question 13 [5 marks]

Sequences, Series and the Binomial Expansion

In the binomial expansion of (2 + kx)^5, the coefficient of the x^2 term is 720.

Find the possible value(s) of k.

Question 14 [5 marks]

Exponentials and Logarithms

Without using a calculator, solve the equation (log_2(x))^2 - 5 log_2(x) + 6 = 0, giving both values of x as exact values.

Question 15 [6 marks]

Differentiation

A curve has equation y = 2x^3 - 3x^2 - 12x + 5.

Find the coordinates of the two stationary points on the curve, and determine the nature of each.

Model solutions

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or workingMarks
writing 1/8 as 2^-3M1
x = -3A1
Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or workingMarks
common ratio r = 12/8 = 1.5M1
using the 6th term = a*r^5M1
6th term = 60.75A1
Final answer: r = 1.5, 6th term = 60.75
Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or workingMarks
finding the gradient of the given line as 3 (from y = 3x + 4)M1
forming the equation y - (-5) = 3(x - 2)M1
y = 3x - 11A1
Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or workingMarks
using Pythagoras' theorem (a 3-4-5 triangle) to find the third side = 4M1
cos(theta) = 4/5A1
tan(theta) = 3/4A1
Final answer: cos(theta) = 4/5, tan(theta) = 3/4
Mark scheme for Question 5 [3 marks]
Question 5[3 marks]
Answer or workingMarks
using the chain rule to differentiate e^(3x)M1
the derivative of e^(3x) is 3e^(3x)A1
dy/dx = 3e^(3x) - 4A1
Mark scheme for Question 6 [3 marks]
Question 6[3 marks]
Answer or workingMarks
testing successive values of n (e.g. n = 1, 2, 3) and finding each gives a primeM1
testing n = 4 to obtain n^2 + n + 1 = 21M1
identifying 21 = 3 x 7 is not prime, so the statement is falseA1
Final answer: False: n = 4 gives n^2 + n + 1 = 21 = 3 x 7, which is not prime.
Mark scheme for Question 7 [4 marks]
Question 7[4 marks]
Answer or workingMarks
substituting t = 10 into the modelM1
896 bacteria (nearest whole number)A1
setting 200 e^(0.15t) = 1000 and taking logarithmsM1
t = 10.7 hours (1 dp)A1
Final answer: 896 bacteria after 10 hours; t = 10.7 hours
Mark scheme for Question 8 [4 marks]
Question 8[4 marks]
Answer or workingMarks
applying the quotient rule with u' = 4 and v' = 1M1
dy/dx = 9/(x + 2)^2 after simplificationA1
substituting x = 1M1
gradient = 1A1
Final answer: dy/dx = 9/(x + 2)^2; gradient at x = 1 is 1
Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or workingMarks
writing 20 as 4 x 5 = 2^2 x 5M1
using the multiplication law log_a(2^2 x 5) = log_a(2^2) + log_a(5)M1
using the power law log_a(2^2) = 2 log_a(2) = 2qA1
combining to give log_a(20) = p + 2qA1
Final answer: log_a(20) = p + 2q
Mark scheme for Question 10 [5 marks]
Question 10[5 marks]
Answer or workingMarks
identifying the inner function u = 5x^2 - 3x and outer function u^4M1
differentiating the outer function: 4u^3M1
differentiating the inner function: du/dx = 10x - 3A1
applying the chain rule dy/dx = 4u^3 x du/dxM1
dy/dx = 4(5x^2 - 3x)^3(10x - 3)A1
Final answer: dy/dx = 4(5x^2 - 3x)^3 (10x - 3)
Mark scheme for Question 11 [5 marks]
Question 11[5 marks]
Answer or workingMarks
finding dx/dt = 2tM1
finding dy/dt = 3t^2 - 3M1
dy/dx = (3t^2 - 3)/(2t)A1
substituting t = 2M1
the gradient = 9/4A1
Final answer: dy/dx = (3t^2 - 3)/(2t); gradient at t = 2 is 9/4
Mark scheme for Question 12 [5 marks]
Question 12[5 marks]
Answer or workingMarks
taking logarithms of both sidesM1
using the power law to give x ln(5) = (x + 1) ln(3)M1
rearranging to x(ln(5) - ln(3)) = ln(3)A1
x = ln(3) / (ln(5) - ln(3)) (unrounded)A1
x = 2.15 (3 sf)A1
Mark scheme for Question 13 [5 marks]
Question 13[5 marks]
Answer or workingMarks
identifying the general term C(5,2)(2)^3(kx)^2M1
simplifying this term to 80k^2 x^2M1
forming the equation 80k^2 = 720A1
solving to give k^2 = 9M1
k = 3 or k = -3A1
Mark scheme for Question 14 [5 marks]
Question 14[5 marks]
Answer or workingMarks
substituting y = log_2(x) to obtain the quadratic y^2 - 5y + 6 = 0M1
factorising to (y - 2)(y - 3) = 0A1
y = 2 or y = 3A1
converting back using x = 2^yM1
x = 4 or x = 8A1
Mark scheme for Question 15 [6 marks]
Question 15[6 marks]
Answer or workingMarks
differentiating to dy/dx = 6x^2 - 6x - 12M1
factorising/solving to x = 2 and x = -1M1
the y-coordinates (2, -15) and (-1, 12)A1
finding the second derivative d^2y/dx^2 = 12x - 6M1
at x = 2, d^2y/dx^2 = 18 > 0, so a minimumA1
at x = -1, d^2y/dx^2 = -18 < 0, so a maximumA1
Final answer: (2, -15) is a minimum; (-1, 12) is a maximum