Grades 1-9

GCSE Further Maths Paper 3

Covers Algebraic Manipulation, Surds and Indices, Simultaneous Equations and Inequalities and 9 more.

16 questions - 60 marks - calculator allowed

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Questions

Question 1 [2 marks]

Surds and Indices

Simplify sqrt(48) - sqrt(12), giving your answer in the form k*sqrt(3).

Question 2 [2 marks]

Functions and their Graphs

The function f is defined by f(x) = 3x^2 - 2x.

Find f(-2).

Question 3 [2 marks]

Matrices and Transformations

A is the matrix with rows (5, -2) and (1, 3).

B is the matrix with rows (-3, 4) and (2, -1).

Find A - B, writing your answer as a matrix.

Question 4 [2 marks]

Trigonometric Identities and Equations

Show that sin(x)/cos(x) + cos(x)/sin(x) can be simplified to 1/(sin(x)cos(x)).

Question 5 [3 marks]

Applications of Differentiation

A curve has equation y = x^2 - 6x + 10.

Find the equation of the tangent to the curve at the point (4, 2), giving your answer in the form y = mx + c.

Question 6 [3 marks]

Trigonometric Identities and Equations

Solve the equation 2sin(x) = 1 for 0 <= x <= 360 degrees, giving all solutions.

Question 7 [3 marks]

Simultaneous Equations and Inequalities

Solve the inequality 4x + 3 > 7x - 12, giving your answer in the form x < k.

Question 8 [4 marks]

Non-Right-Angled Triangle Trigonometry

In triangle DEF, DE = 11 cm, EF = 8 cm, and DF = 6 cm.

Find the size of angle DEF, giving your answer to 1 decimal place.

Question 9 [4 marks]

Differentiation

A curve has equation y = x^3 - 6x^2 + 9x + 1.

Find the coordinates of the stationary points and use the second derivative to determine the nature of each one.

Question 10 [4 marks]

Geometric Proof

Two tangents are drawn from an external point T to a circle with centre O, touching the circle at points A and B.

Prove that TA = TB.

Question 11 [4 marks]

Algebraic Manipulation

Factorise fully 6x^3 - 24x.

Question 12 [4 marks]

Sequences

The nth term of a sequence is given by Un = 3n^2 - 2n.

Find the 5th term of the sequence, and determine whether 780 is a term of the sequence, showing your working.

Question 13 [5 marks]

Coordinate Geometry

Find the equation of the straight line passing through the points E(3, -4) and F(-1, 8), giving your answer in the form y = mx + c.

State the coordinates of the point where the line crosses the x-axis.

Question 14 [6 marks]

Algebraic Manipulation

Solve the equation (13 - x)/(x - 1) = x - 3.

Show each step of your working and give both solutions.

Question 15 [6 marks]

Surds and Indices

Show that (4 + sqrt(3))/(4 - sqrt(3)) can be written in the form a + b*sqrt(3), where a and b are rational numbers to be found.

Question 16 [6 marks]

Functions and their Graphs

f(x) = 3/(x + 1) - 2 for x not equal to -1.

Find f^-1(x), the inverse function of f, stating the value excluded from its domain.

Model solutions

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or working	Marks
writing sqrt(48) = 4sqrt(3) and sqrt(12) = 2sqrt(3)	M1
2sqrt(3)	A1

Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or working	Marks
substituting x = -2 into f(x) = 3x^2 - 2x	M1
f(-2) = 16	A1

Mark scheme for Question 3 [2 marks]
Question 3[2 marks]
Answer or working	Marks
subtracting corresponding entries	M1
the matrix with rows (8, -6) and (-1, 4)	A1

Mark scheme for Question 4 [2 marks]
Question 4[2 marks]
Answer or working	Marks
combining the two fractions over a common denominator sin(x)cos(x)	M1
using sin^2(x) + cos^2(x) = 1 to obtain 1/(sin(x)cos(x))	A1
Final answer: 1/(sin(x)cos(x)), shown using sin^2(x) + cos^2(x) = 1

Mark scheme for Question 5 [3 marks]
Question 5[3 marks]
Answer or working	Marks
finding the gradient at x = 4 as dy/dx = 2	M1
using y - 2 = 2(x - 4)	M1
y = 2x - 6	A1

Mark scheme for Question 6 [3 marks]
Question 6[3 marks]
Answer or working	Marks
rearranging to sin(x) = 0.5	M1
finding the principal value x = 30 degrees	M1
x = 30 degrees or x = 150 degrees	A1

Mark scheme for Question 7 [3 marks]
Question 7[3 marks]
Answer or working	Marks
collecting x terms on one side	M1
collecting number terms on the other side	M1
x < 5	A1

Mark scheme for Question 8 [4 marks]
Question 8[4 marks]
Answer or working	Marks
using the cosine rule DF^2 = DE^2 + EF^2 - 2(DE)(EF)cos(E)	M1
substituting to get 36 = 185 - 176cos(E)	M1
rearranging to cos(E) = 149/176	M1
angle DEF = 32.2 degrees	A1
Final answer: angle DEF = 32.2 degrees (1 d.p.)

Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or working	Marks
differentiating and setting dy/dx = 3x^2 - 12x + 9 = 0 to find x = 1 and x = 3	M1
the points (1, 5) and (3, 1)	A1
finding the second derivative d2y/dx2 = 6x - 12 and evaluating it at each x value	M1
identifying (1, 5) as a maximum and (3, 1) as a minimum	A1
Final answer: (1, 5) is a maximum; (3, 1) is a minimum

Mark scheme for Question 10 [4 marks]
Question 10[4 marks]
Answer or working	Marks
stating angle OAT = angle OBT = 90 degrees, since a tangent is perpendicular to the radius at the point of contact	B1
stating OA = OB, since both are radii of the circle, and OT is common to both triangles	B1
concluding triangle OAT is congruent to triangle OBT by RHS	M1
concluding TA = TB	A1
Final answer: Triangles OAT and OBT are congruent by RHS (right angles at A and B, OA = OB, OT common), so TA = TB

Mark scheme for Question 11 [4 marks]
Question 11[4 marks]
Answer or working	Marks
taking out the common factor 6x	M1
6x(x^2 - 4)	A1
recognising x^2 - 4 as the difference of two squares	M1
6x(x - 2)(x + 2)	A1

Mark scheme for Question 12 [4 marks]
Question 12[4 marks]
Answer or working	Marks
substituting n = 5 into Un = 3n^2 - 2n	M1
the 5th term = 65	A1
forming the equation 3n^2 - 2n - 780 = 0 and using the discriminant	M1
concluding 780 is not a term since n is not a positive integer, as the discriminant 9364 is not a perfect square	A1
Final answer: 5th term = 65; 780 is not a term of the sequence

Mark scheme for Question 13 [5 marks]
Question 13[5 marks]
Answer or working	Marks
using gradient = (y2 - y1)/(x2 - x1)	M1
gradient = -3	A1
using y + 4 = -3(x - 3)	M1
y = -3x + 5	A1
the x-intercept (5/3, 0)	B1
Final answer: y = -3x + 5; crosses the x-axis at (5/3, 0)

Mark scheme for Question 14 [6 marks]
Question 14[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

Mark scheme for Question 15 [6 marks]
Question 15[6 marks]
Answer or working	Marks
multiplying numerator and denominator by (4 + sqrt(3))	M1
the denominator simplifying to 16 - 3 = 13	M1
expanding the numerator (4 + sqrt(3))^2	M1
the numerator simplifying to 19 + 8sqrt(3)	A1
a = 19/13	A1
b = 8/13	A1
Final answer: a = 19/13, b = 8/13, so 19/13 + (8/13)sqrt(3)

Mark scheme for Question 16 [6 marks]
Question 16[6 marks]
Answer or working	Marks
adding 2 to both sides to give y + 2 = 3/(x + 1)	M1
taking reciprocals of both sides to give 1/(y + 2) = (x + 1)/3	M1
multiplying both sides by 3 to give 3/(y + 2) = x + 1	M1
subtracting 1 to give x = 3/(y + 2) - 1	M1
swapping x and y to give f^-1(x) = 3/(x + 2) - 1	A1
stating x = -2 is excluded from the domain of f^-1	B1
Final answer: f^-1(x) = 3/(x + 2) - 1, x not equal to -2