AS Paper 1: Algebra and Trigonometry
Covers proof and algebraic methods, coordinate geometry, sequences and series with the binomial expansion, trigonometry and exponentials and logarithms.
Questions
Question 1 [3 marks]
Coordinate Geometry
Find the equation of the perpendicular bisector of the line segment joining the points A(2, 5) and B(-4, 1), giving your answer in the form y = mx + c.
Question 2 [3 marks]
Proof and Algebraic Methods
Without using a calculator, solve the inequality x^2 - 2x - 8 <= 0, giving your answer as a single inequality.
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 [4 marks]
Sequences, Series and the Binomial Expansion
An arithmetic series has first term 4 and common difference 5.
Find the 25th term, and find the sum of the first 25 terms.
Question 5 [4 marks]
Exponentials and Logarithms
Without using a calculator, solve the equation 3^(x + 2) = 27^(x - 1), giving the exact value of x.
Question 6 [4 marks]
Coordinate Geometry
The line l1 has equation x + 2y = 16.
The line l2 is perpendicular to l1 and passes through the point (4, 1).
Find the equation of l2 in the form y = mx + c, and find the coordinates of the point where l1 and l2 intersect.
Question 7 [3 marks]
Proof and Algebraic Methods
The function f is defined by f(x) = (2x - 1)/(x + 3) for x != -3.
Find f^-1(x) and state its domain.
Question 8 [4 marks]
Trigonometry
Using the small angle approximation cos(x) = 1 - (x^2)/2, valid for small values of x in radians, show that (1 - cos(x))/x^2 is approximately 1/2 for small x.
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 [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 11 [4 marks]
Sequences, Series and the Binomial Expansion
A geometric series has first term 10 and common ratio 0.8.
Find the sum to infinity of the series, and find the smallest number of terms needed for the sum of the series to exceed 49.
Question 12 [4 marks]
Trigonometry
Solve the equation 3 tan(x) - 2 = 4 for 0 <= x <= 360 deg.
Question 13 [5 marks]
Proof and Algebraic Methods
Use proof by contradiction to show that there is no largest even integer.
Question 14 [6 marks]
Sequences, Series and the Binomial Expansion
A sequence is defined by u_1 = 4 and u_(n+1) = (1/2)u_n + 3 for n >= 1.
Find the values of u_2, u_3 and u_4, and determine, with a reason, whether the sequence is increasing or decreasing.
Question 15 [6 marks]
Proof and Algebraic Methods
The cubic polynomial f(x) = x^3 + ax^2 - 7x + 10 has (x - 1) as a factor.
Find the value of a, and hence find all three roots of the equation f(x) = 0.
Model solutions
| Question 1[3 marks] | |
|---|---|
| Answer or working | Marks |
| finding the midpoint of AB as (-1, 3) | M1 |
| finding the gradient of AB = 2/3, so the perpendicular gradient = -3/2 | M1 |
| y = -3/2 x + 3/2 | A1 |
| Question 2[3 marks] | |
|---|---|
| Answer or working | Marks |
| factorising x^2 - 2x - 8 as (x - 4)(x + 2) | M1 |
| identifying the critical values x = -2 and x = 4 | M1 |
| the solution -2 <= x <= 4 | A1 |
| Final answer: -2 <= x <= 4 | |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| 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 - 11 | A1 |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| using term = a + (n - 1)d | M1 |
| 25th term = 124 | A1 |
| using S_n = n/2(2a + (n - 1)d) | M1 |
| S_25 = 1600 | A1 |
| Final answer: 25th term = 124; sum of first 25 terms = 1600 | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| writing 27 as 3^3 | M1 |
| equating exponents x + 2 = 3(x - 1) | M1 |
| simplifying to x + 2 = 3x - 3 | A1 |
| x = 2.5 | A1 |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| gradient of l1 = -1/2, so gradient of l2 = 2 | M1 |
| forming l2 as y - 1 = 2(x - 4) | M1 |
| l2: y = 2x - 7 | A1 |
| the intersection point (6, 5) | A1 |
| Final answer: l2: y = 2x - 7; intersection (6, 5) | |
| Question 7[3 marks] | |
|---|---|
| Answer or working | Marks |
| setting y = (2x - 1)/(x + 3) and rearranging to y(x + 3) = 2x - 1 | M1 |
| collecting x terms to give x(y - 2) = -1 - 3y | M1 |
| f^-1(x) = (1 + 3x)/(2 - x), with domain x != 2 | A1 |
| Final answer: f^-1(x) = (1 + 3x)/(2 - x), x != 2 | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting the approximation cos(x) = 1 - (x^2)/2 | M1 |
| 1 - cos(x) = (x^2)/2 | A1 |
| dividing by x^2 | M1 |
| (1 - cos(x))/x^2 = 1/2, as required | A1 |
| Final answer: (1 - cos(x))/x^2 is approximately 1/2 for small x, since 1 - cos(x) = (x^2)/2 using the small angle approximation | |
| Question 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| writing 20 as 4 x 5 = 2^2 x 5 | M1 |
| 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) = 2q | A1 |
| combining to give log_a(20) = p + 2q | A1 |
| Final answer: log_a(20) = p + 2q | |
| Question 10[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 11[4 marks] | |
|---|---|
| Answer or working | Marks |
| sum to infinity = a/(1 - r) | M1 |
| sum to infinity = 50 | A1 |
| setting up the inequality 50(1 - 0.8^n) > 49 and solving using logarithms | M1 |
| n = 18 | A1 |
| Final answer: Sum to infinity = 50; smallest n = 18 | |
| Question 12[4 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging to tan(x) = 2 | M1 |
| x = 63.4 deg (1 dp) | A1 |
| using the periodicity of tan to find a second solution x + 180 deg | M1 |
| x = 243.4 deg (1 dp) | A1 |
| Final answer: x = 63.4 deg or x = 243.4 deg (1 dp) | |
| Question 13[5 marks] | |
|---|---|
| Answer or working | Marks |
| assuming, for contradiction, that there is a largest even integer, N | B1 |
| writing N = 2k for some integer k, since N is even | M1 |
| considering N + 2 = 2k + 2 = 2(k + 1) | M1 |
| recognising 2(k + 1) is also an even integer | A1 |
| identifying that N + 2 > N, which contradicts N being the largest even integer, so no largest even integer exists | A1 |
| Final answer: Proof: assuming a largest even integer N = 2k exists, N + 2 = 2(k + 1) is also even and greater than N, a contradiction, so no largest even integer exists. | |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| substituting n = 1 to find u_2 = (1/2)(4) + 3 = 5 | M1 |
| u_3 = (1/2)(5) + 3 = 5.5 | A1 |
| u_4 = (1/2)(5.5) + 3 = 5.75 | A1 |
| comparing successive terms | M1 |
| noting each term is greater than the one before (4 < 5 < 5.5 < 5.75) | A1 |
| concluding the sequence is increasing | A1 |
| Final answer: u_2 = 5, u_3 = 5.5, u_4 = 5.75; the sequence is increasing | |
| Question 15[6 marks] | |
|---|---|
| Answer or working | Marks |
| substituting x = 1 into f(x) and setting f(1) = 0 | M1 |
| forming the equation a + 4 = 0, giving a = -4 | A1 |
| dividing f(x) = x^3 - 4x^2 - 7x + 10 by (x - 1) | M1 |
| the quotient x^2 - 3x - 10 | A1 |
| factorising the quotient as (x - 5)(x + 2) | M1 |
| the full solution set x = 1, x = 5, x = -2, stated together | A1 |
| Final answer: a = -4; x = 1, x = 5, x = -2 | |