GCSE Further Maths Short Paper A
Covers Algebraic Manipulation, Surds and Indices, Simultaneous Equations and Inequalities and 3 more.
Questions
Question 1 [3 marks]
Algebraic Manipulation
Expand and simplify (x - 5)(x + 5) - (x - 3)^2.
Question 2 [4 marks]
Simultaneous Equations and Inequalities
Solve the simultaneous equations 4x + y = 16 and 2x - 3y = -6.
Question 3 [4 marks]
Functions and their Graphs
The function f is defined by f(x) = x^2 - 3x + 1.
Find f(2) and f(-1).
Question 4 [4 marks]
Surds and Indices
Solve 3^(2x - 1) = 27^(x - 2), giving the exact value of x.
Question 5 [4 marks]
Coordinate Geometry
Line L1 has equation y = 2x - 1.
Line L2 is perpendicular to L1 and passes through the point (4, 5).
Find the equation of L2, giving your answer in the form y = mx + c.
Question 6 [5 marks]
Matrices and Transformations
A is the matrix with rows (1, 2) and (3, -1).
B is the matrix with rows (0, 4) and (2, 1).
Find BA, the product of B and A, writing your answer as a matrix, and state whether BA = AB for these two matrices.
Question 7 [5 marks]
Coordinate Geometry
A circle has equation x^2 + y^2 + 8x - 2y + 8 = 0.
Find the centre and radius of the circle by completing the square.
Question 8 [5 marks]
Algebraic Manipulation
Solve the equation 3/(x - 1) + 2/(x + 1) = 2.
Show that your equation reduces to a quadratic before solving it, and give both solutions.
Question 9 [6 marks]
Functions and their Graphs
The function k is defined by k(x) = -3x^2 + 24x - 41 for all real x.
Express k(x) in the form a - b(x - c)^2, where a, b and c are integers to be found.
Hence state the maximum value of k(x) and the value of x at which it occurs.
Model solutions
| Question 1[3 marks] | |
|---|---|
| Answer or working | Marks |
| expanding (x - 5)(x + 5) to x^2 - 25 | M1 |
| expanding (x - 3)^2 to x^2 - 6x + 9 | M1 |
| 6x - 34 | A1 |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging 4x + y = 16 to make y the subject, y = 16 - 4x | M1 |
| substituting into 2x - 3y = -6 | M1 |
| simplifying to 14x = 42 | M1 |
| x = 3 and y = 4 | A1 |
| Final answer: x = 3, y = 4 | |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting x = 2 into f(x) = x^2 - 3x + 1 | M1 |
| f(2) = -1 | A1 |
| substituting x = -1 into f(x) = x^2 - 3x + 1 | M1 |
| f(-1) = 5 | A1 |
| Final answer: f(2) = -1, f(-1) = 5 | |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| writing 27 as 3^3 | M1 |
| writing 27^(x - 2) as 3^(3x - 6) | M1 |
| equating exponents 2x - 1 = 3x - 6 | M1 |
| x = 5 | A1 |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the gradient of L1 as 2 | M1 |
| using the perpendicular gradient rule to find gradient -1/2 | M1 |
| using y - 5 = -0.5(x - 4) | M1 |
| y = -0.5x + 7 | A1 |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| attempting row-by-column multiplication for BA | M1 |
| the first row of BA, (12, -4) | M1 |
| the matrix BA with rows (12, -4) and (5, 3) | A1 |
| calculating AB, the matrix with rows (4, 6) and (-2, 11), for comparison | M1 |
| stating BA does not equal AB, since matrix multiplication is not commutative | B1 |
| Final answer: BA = the matrix with rows (12, -4) and (5, 3); BA does not equal AB | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| completing the square on the x terms: (x + 4)^2 - 16 | M1 |
| completing the square on the y terms: (y - 1)^2 - 1 | M1 |
| rearranging to (x + 4)^2 + (y - 1)^2 = 9 | M1 |
| centre (-4, 1) | A1 |
| radius 3 | A1 |
| Final answer: Centre (-4, 1), radius 3 | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying both sides by (x - 1)(x + 1) | M1 |
| expanding to 3(x + 1) + 2(x - 1) = 2(x - 1)(x + 1) | M1 |
| rearranging to 2x^2 - 5x - 3 = 0 | M1 |
| solving the quadratic, e.g. by factorising as (2x + 1)(x - 3) = 0 | M1 |
| x = 3 or x = -0.5 | A1 |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| taking out the factor -3 from the x terms: -3(x^2 - 8x) - 41 | M1 |
| completing the square inside the bracket: (x - 4)^2 - 16 | M1 |
| substituting back to get -3[(x - 4)^2 - 16] - 41 | M1 |
| 7 - 3(x - 4)^2 (a = 7, b = 3, c = 4) | A1 |
| identifying the maximum value as 7 | B1 |
| stating this occurs at x = 4 | B1 |
| Final answer: 7 - 3(x - 4)^2; maximum value 7 at x = 4 | |