Year 11 Paper 1: Algebra and Graphs Recap
Revisits algebraic manipulation, surds and indices, simultaneous equations and inequalities, functions and their graphs, coordinate geometry, and sequences before the Year 11 topics are added.
Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [2 marks]
Surds and Indices
Simplify sqrt(50) - sqrt(18), giving your answer in the form k*sqrt(2).
Question 2 [2 marks]
Sequences
An arithmetic sequence has first term 5 and common difference 4.
Find the 12th term of the sequence.
Question 3 [2 marks]
Surds and Indices
Simplify sqrt(48) - sqrt(12), giving your answer in the form k*sqrt(3).
Question 4 [3 marks]
Functions and their Graphs
The function g is defined by g(x) = x^2 - 4 for all real x.
Find the values of x for which g(x) = 12.
Question 5 [4 marks]
Coordinate Geometry
Points C(-4, -1) and D(2, 7) are given.
Find the midpoint of CD and the length of CD.
Question 6 [4 marks]
Algebraic Manipulation
Expand and simplify (3x + 1)(x - 2) - (x - 1)^2.
Question 7 [4 marks]
Simultaneous Equations and Inequalities
Solve the quadratic inequality x^2 - 5x - 14 <= 0, giving your answer as a single inequality.
Question 8 [4 marks]
Algebraic Manipulation
Write as a single fraction in its simplest form: 3/(x + 1) + 2/(x - 2).
Question 9 [4 marks]
Coordinate Geometry
A circle has equation (x + 1)^2 + (y - 4)^2 = 20.
State the centre and radius of the circle, then determine whether the point (4, 5) lies inside, outside, or on the circle.
Question 10 [4 marks]
Algebraic Manipulation
Write as a single fraction in its simplest form: 4/(x - 3) - 1/(x + 2).
Question 11 [4 marks]
Functions and their Graphs
f(x) = 3x + 2 and g(x) = x^2 - 1.
Find fg(x), giving your answer in its simplest form.
Question 12 [5 marks]
Surds and Indices
Rationalise the denominator of 20/sqrt(32), giving your answer in its simplest form.
Question 13 [6 marks]
Algebraic Manipulation
Write as a single fraction in its simplest form: (x + 1)/(x^2 - 4) - 2/(x - 2).
Question 14 [6 marks]
Simultaneous Equations and Inequalities
A rectangular garden has length (x + 8) m and width (x - 3) m, where x > 3.
The area of the garden is at least 60 m^2.
Form and solve an inequality to find the range of possible values of x.
Question 15 [6 marks]
Algebraic Manipulation
Solve the equation (13 - x)/(x - 1) = x - 3.
Show each step of your working and give both solutions.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| writing sqrt(50) = 5sqrt(2) and sqrt(18) = 3sqrt(2) | M1 |
| 2sqrt(2) | A1 |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| using the formula a + (n - 1)d with n = 12 | M1 |
| T12 = 49 | A1 |
| Final answer: 49 | |
| Question 3[2 marks] | |
|---|---|
| Answer or working | Marks |
| writing sqrt(48) = 4sqrt(3) and sqrt(12) = 2sqrt(3) | M1 |
| 2sqrt(3) | A1 |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| forming the equation x^2 - 4 = 12 | M1 |
| rearranging to x^2 = 16 | M1 |
| x = 4 or x = -4 | A1 |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the midpoint formula ((x1 + x2)/2, (y1 + y2)/2) | M1 |
| midpoint (-1, 3) | A1 |
| using the distance formula sqrt((x2 - x1)^2 + (y2 - y1)^2) | M1 |
| length CD = 10 | A1 |
| Final answer: Midpoint (-1, 3); length CD = 10 | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| expanding (3x + 1)(x - 2) to give 3x^2 - 5x - 2 | M1 |
| expanding (x - 1)^2 to give x^2 - 2x + 1 | M1 |
| combining like terms | M1 |
| 2x^2 - 3x - 3 | A1 |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| factorising x^2 - 5x - 14 as (x - 7)(x + 2) | M1 |
| identifying the critical values x = -2 and x = 7 | M1 |
| reasoning about the sign of the quadratic between the roots | M1 |
| -2 <= x <= 7 | A1 |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the common denominator (x + 1)(x - 2) | M1 |
| writing 3(x - 2) + 2(x + 1) as the numerator | M1 |
| simplifying the numerator to 5x - 4 | A1 |
| (5x - 4)/((x + 1)(x - 2)) | A1 |
| Question 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| centre (-1, 4) | B1 |
| radius sqrt(20), i.e. 2sqrt(5) | B1 |
| calculating the distance from (4, 5) to the centre as sqrt(5^2 + 1^2) | M1 |
| concluding the point lies outside the circle, since sqrt(26) > sqrt(20) | A1 |
| Final answer: Centre (-1, 4), radius 2sqrt(5); the point (4, 5) lies outside the circle | |
| Question 10[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the common denominator (x - 3)(x + 2) | M1 |
| writing 4(x + 2) - (x - 3) as the numerator | M1 |
| simplifying the numerator to 3x + 11 | A1 |
| (3x + 11)/((x - 3)(x + 2)) | A1 |
| Question 11[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting g(x) into f | M1 |
| writing 3(x^2 - 1) + 2 | M1 |
| expanding to 3x^2 - 3 + 2 | M1 |
| fg(x) = 3x^2 - 1 | A1 |
| Question 12[5 marks] | |
|---|---|
| Answer or working | Marks |
| writing sqrt(32) = 4sqrt(2) | M1 |
| simplifying to 5/sqrt(2) | M1 |
| multiplying numerator and denominator by sqrt(2) | M1 |
| simplifying the numerator to 5sqrt(2) | M1 |
| (5/2)sqrt(2) | A1 |
| Final answer: 5sqrt(2)/2 (equivalently 2.5sqrt(2)) | |
| Question 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| factorising x^2 - 4 as (x - 2)(x + 2) | M1 |
| identifying the common denominator (x - 2)(x + 2) | M1 |
| writing 2/(x - 2) as 2(x + 2)/((x - 2)(x + 2)) | M1 |
| combining the numerators to give x + 1 - 2(x + 2) | M1 |
| simplifying the numerator to -x - 3 | A1 |
| -(x + 3)/((x - 2)(x + 2)) | A1 |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| forming the area expression (x + 8)(x - 3) | M1 |
| expanding to x^2 + 5x - 24 | M1 |
| forming the inequality x^2 + 5x - 84 >= 0 | M1 |
| factorising as (x - 7)(x + 12) >= 0 and identifying the critical values -12 and 7 | M1 |
| rejecting x <= -12 since x > 3 | A1 |
| x >= 7 | A1 |
| Question 15[6 marks] | |
|---|---|
| Answer or working | Marks |
| cross-multiplying to give 13 - x = (x - 3)(x - 1) | M1 |
| expanding the right side to x^2 - 4x + 3 | M1 |
| rearranging to x^2 - 3x - 10 = 0 | M1 |
| factorising as (x - 5)(x + 2) = 0 | M1 |
| x = 5 | A1 |
| x = -2 | A1 |
| Final answer: x = 5 or x = -2 | |