GCSE Further Maths Paper 4
Covers Algebraic Manipulation, Surds and Indices, Simultaneous Equations and Inequalities and 9 more.
Questions
Question 1 [2 marks]
Differentiation
Find dy/dx when y = 5x^3 + 2x^2 - 9.
Question 2 [2 marks]
Geometric Proof
Triangle ABC is isosceles with AB = AC.
D is the midpoint of BC.
Prove that triangle ABD is congruent to triangle ACD.
Question 3 [3 marks]
Algebraic Manipulation
Factorise fully 4x^2 - 20x + 24.
Question 4 [3 marks]
Non-Right-Angled Triangle Trigonometry
In triangle PQR, PQ = 9 cm, QR = 7 cm, and angle PQR = 65 degrees.
Find the length of PR, giving your answer to 3 significant figures.
Question 5 [4 marks]
Surds and Indices
Simplify sqrt(75) + sqrt(12) - sqrt(27), giving your answer in the form k*sqrt(3).
Question 6 [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 7 [4 marks]
Applications of Differentiation
A curve has equation y = x^3 - 3x + 2.
Find the equation of the normal to the curve at the point where x = 0, giving your answer in the form y = mx + c.
Question 8 [5 marks]
Simultaneous Equations and Inequalities
Solve the inequality 3(2x - 1) < 5(x + 2), giving your answer in the form x < k.
Question 9 [5 marks]
Sequences
A geometric sequence has second term 6 and fifth term 162.
Find the common ratio and the first term of the sequence.
Question 10 [5 marks]
Matrices and Transformations
M is the matrix with rows (k, 2) and (6, k + 1), where k is a constant.
Given that M is singular, find the possible values of k.
Question 11 [5 marks]
Coordinate Geometry
A circle has centre (-2, 3) and passes through the point (2, 6).
Find the equation of the circle.
Then find the equation of the tangent to the circle at the point (2, 6), giving your answer in the form y = mx + c.
Question 12 [6 marks]
Trigonometric Identities and Equations
Prove the identity cos(x)/(1 + sin(x)) + cos(x)/(1 - sin(x)) = 2/cos(x).
Question 13 [6 marks]
Functions and their Graphs
f(x) = 5/(x - 2) + 1 for x not equal to 2.
Find f^-1(x), the inverse function of f, stating the value excluded from its domain.
Question 14 [6 marks]
Non-Right-Angled Triangle Trigonometry
A ship sails from port P on a bearing of 040 degrees for 32 km to reach point Q.
It then sails from Q on a bearing of 130 degrees for 21 km to reach point R.
Find angle PQR, and hence find the distance PR using the cosine rule, giving your answer to 3 significant figures.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating each term using the power rule | M1 |
| dy/dx = 15x^2 + 4x | A1 |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| stating BD = CD (D is the midpoint of BC) and AD is common to both triangles | B1 |
| concluding triangle ABD is congruent to triangle ACD by SSS, using AB = AC given | B1 |
| Final answer: Congruent by SSS: AB = AC (given), BD = CD (D midpoint), AD common | |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| taking out the common factor of 4 | M1 |
| factorising x^2 - 5x + 6 into two brackets | M1 |
| 4(x - 2)(x - 3) | A1 |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| using the cosine rule PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)cos(65) | M1 |
| substituting to get PR^2 = 130 - 126cos(65) | M1 |
| PR = 8.76 cm | A1 |
| Final answer: PR = 8.76 cm (3 s.f.) | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| writing sqrt(75) = 5sqrt(3) | M1 |
| writing sqrt(12) = 2sqrt(3) | M1 |
| writing sqrt(27) = 3sqrt(3) | M1 |
| 4sqrt(3) | A1 |
| Question 6[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 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding y = 2 and dy/dx = -3 at x = 0 | M1 |
| using the tangent gradient to find the normal gradient as 1/3 | M1 |
| using y - 2 = (1/3)(x - 0) | M1 |
| y = (1/3)x + 2 | A1 |
| Final answer: y = x/3 + 2 | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| expanding the left side to 6x - 3 | M1 |
| expanding the right side to 5x + 10 | M1 |
| collecting x terms to give x on the left | M1 |
| collecting number terms to give 13 on the right | M1 |
| x < 13 | A1 |
| Question 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| writing T2 = ar = 6 and T5 = ar^4 = 162 | M1 |
| dividing to eliminate a, giving r^3 = 27 | M1 |
| r = 3 | A1 |
| substituting back into ar = 6 to find a | M1 |
| a = 2 | A1 |
| Final answer: common ratio r = 3; first term a = 2 | |
| Question 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| writing the determinant k(k + 1) - 2(6) | M1 |
| expanding to k^2 + k - 12 | M1 |
| setting the determinant equal to 0, since M is singular | M1 |
| factorising as (k + 4)(k - 3) = 0 | M1 |
| k = -4 or k = 3 | A1 |
| Question 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding the radius as sqrt(4^2 + 3^2) = 5 | M1 |
| the circle equation (x + 2)^2 + (y - 3)^2 = 25 | A1 |
| finding the gradient of the radius to (2, 6) as 3/4 | M1 |
| using the perpendicular gradient -4/3 with the point (2, 6) | M1 |
| the tangent y = -(4/3)x + 26/3 | A1 |
| Final answer: Circle: (x + 2)^2 + (y - 3)^2 = 25; tangent: y = -(4/3)x + 26/3 | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| combining the two fractions over the common denominator (1 + sin(x))(1 - sin(x)) | M1 |
| expanding the denominator to 1 - sin^2(x) | M1 |
| using sin^2(x) + cos^2(x) = 1 to write the denominator as cos^2(x) | B1 |
| combining the numerators to give cos(x)(1 - sin(x)) + cos(x)(1 + sin(x)) | M1 |
| simplifying the numerator to 2cos(x) | A1 |
| concluding the left side equals 2cos(x)/cos^2(x) = 2/cos(x), completing the proof | A1 |
| Final answer: Shown: cos(x)/(1 + sin(x)) + cos(x)/(1 - sin(x)) = 2cos(x)/cos^2(x) = 2/cos(x) | |
| Question 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| subtracting 1 from both sides to give y - 1 = 5/(x - 2) | M1 |
| taking reciprocals of both sides to give 1/(y - 1) = (x - 2)/5 | M1 |
| multiplying both sides by 5 to give 5/(y - 1) = x - 2 | M1 |
| adding 2 to both sides to give x = 2 + 5/(y - 1) | M1 |
| swapping x and y to give f^-1(x) = 2 + 5/(x - 1) | A1 |
| stating x = 1 is excluded from the domain of f^-1 | B1 |
| Final answer: f^-1(x) = 2 + 5/(x - 1), x not equal to 1 | |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| finding the bearing of P from Q as 040 + 180 = 220 degrees | M1 |
| finding angle PQR as 220 - 130 = 90 degrees | M1 |
| angle PQR = 90 degrees | A1 |
| using the cosine rule PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)cos(PQR) | M1 |
| substituting to get PR^2 = 1024 + 441 - 0, since cos(90) = 0 | M1 |
| PR = 38.3 km (3 s.f.) | A1 |
| Final answer: angle PQR = 90 degrees; PR = 38.3 km (3 s.f.) | |