Grades 1-9

GCSE Further Maths Paper 1

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

13 questions - 60 marks - calculator allowed

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Questions

Question 1 [3 marks]

Algebraic Manipulation

Factorise fully 4x^2 - 20x + 24.

Question 2 [3 marks]

Coordinate Geometry

Find the equation of the straight line that passes through the point (-2, 5) and has gradient -4.

Give your answer in the form y = mx + c.

Question 3 [4 marks]

Matrices and Transformations

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

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

Find 3A + B, writing your answer as a matrix.

Question 4 [4 marks]

Functions and their Graphs

The function f is defined by f(x) = 2x^2 + x - 5.

Find f(3) and f(-2).

Question 5 [4 marks]

Geometric Proof

Triangle ABC has a point D on BC such that AD is the perpendicular bisector of BC.

Prove that triangle ABD is congruent to triangle ACD, and hence that triangle ABC is isosceles.

Question 6 [5 marks]

Surds and Indices

Rationalise the denominator of 12/sqrt(18), giving your answer in its simplest form.

Question 7 [5 marks]

Differentiation

y = x^4 - 8x^3.

Find d2y/dx2, and find the value(s) of x for which d2y/dx2 = 0.

Question 8 [5 marks]

Non-Right-Angled Triangle Trigonometry

Triangle XYZ has XY = 10 cm, XZ = 13 cm, and area 55 cm^2.

Find the possible size(s) of angle YXZ, giving your answer(s) to 1 decimal place.

Question 9 [5 marks]

Sequences

The nth term of a sequence is given by Un = an^3 + b, where a and b are constants.

Given that U1 = 7 and U2 = 21, find the values of a and b.

Question 10 [5 marks]

Trigonometric Identities and Equations

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

Question 11 [5 marks]

Functions and their Graphs

The function h is defined by h(x) = x^2 + 8x + 19 for all real x.

Express h(x) in the form (x + a)^2 + b, where a and b are integers.

Hence state the minimum value of h(x) and the value of x at which it occurs.

Question 12 [6 marks]

Simultaneous Equations and Inequalities

Solve the quadratic inequality 2x^2 + 3x - 20 > 0, giving your answer as two inequalities.

Question 13 [6 marks]

Applications of Differentiation

The number of bacteria, N, in a culture after t hours is modelled by N = 200 + 30t^2 - 2t^3, for 0 <= t <= 10.

Find the rate of growth dN/dt, and find the time t at which the rate of growth is at its maximum, together with this maximum rate.

Model solutions

Mark scheme for Question 1 [3 marks]
Question 1[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

Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or working	Marks
using y - 5 = -4(x + 2)	M1
expanding to y = -4x - 8 + 5	M1
y = -4x - 3	A1

Mark scheme for Question 3 [4 marks]
Question 3[4 marks]
Answer or working	Marks
tripling A to give the matrix with rows (6, 15) and (-9, 3)	M1
adding the corresponding entries of B	M1
the first row of the answer, (10, 13)	A1
the second row of the answer, (-8, 6)	A1
Final answer: the matrix with rows (10, 13) and (-8, 6)

Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or working	Marks
substituting x = 3 into f(x) = 2x^2 + x - 5	M1
f(3) = 16	A1
substituting x = -2 into f(x) = 2x^2 + x - 5	M1
f(-2) = 1	A1
Final answer: f(3) = 16, f(-2) = 1

Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or working	Marks
stating BD = DC, since AD is the perpendicular bisector of BC	B1
stating angle ADB = angle ADC = 90 degrees, and AD is common to both triangles	B1
concluding triangle ABD is congruent to triangle ACD by SAS	M1
concluding AB = AC, so triangle ABC is isosceles	A1
Final answer: Triangles ABD and ACD are congruent by SAS (BD = DC, angle ADB = angle ADC = 90 degrees, AD common), so AB = AC and triangle ABC is isosceles

Mark scheme for Question 6 [5 marks]
Question 6[5 marks]
Answer or working	Marks
writing sqrt(18) = 3sqrt(2)	M1
simplifying to 4/sqrt(2)	M1
multiplying numerator and denominator by sqrt(2)	M1
simplifying the numerator to 4sqrt(2)	M1
2sqrt(2)	A1

Mark scheme for Question 7 [5 marks]
Question 7[5 marks]
Answer or working	Marks
differentiating to get dy/dx = 4x^3 - 24x^2	M1
differentiating again to get d2y/dx2 = 12x^2 - 48x	M1
setting d2y/dx2 = 0	M1
factorising as 12x(x - 4) = 0	M1
x = 0 or x = 4	A1
Final answer: d2y/dx2 = 12x^2 - 48x; x = 0 or x = 4

Mark scheme for Question 8 [5 marks]
Question 8[5 marks]
Answer or working	Marks
using Area = (1/2)(XY)(XZ)sin(X)	M1
substituting to get 55 = (1/2)(10)(13)sin(X)	M1
rearranging to sin(X) = 55/65	M1
X = 57.8 degrees	A1
X = 122.2 degrees	A1
Final answer: angle YXZ = 57.8 degrees or 122.2 degrees (1 d.p.)

Mark scheme for Question 9 [5 marks]
Question 9[5 marks]
Answer or working	Marks
forming the equation a + b = 7 from U1	M1
forming the equation 8a + b = 21 from U2	M1
subtracting the equations to eliminate b, giving 7a = 14	M1
a = 2	A1
b = 5	A1
Final answer: a = 2, b = 5 (so Un = 2n^3 + 5)

Mark scheme for Question 10 [5 marks]
Question 10[5 marks]
Answer or working	Marks
using cos^2(x) = 1 - sin^2(x) to write the equation in terms of sin(x)	M1
rearranging to 2sin^2(x) - sin(x) - 1 = 0	M1
factorising as (2sin(x) + 1)(sin(x) - 1) = 0	M1
sin(x) = 1 giving x = 90 degrees	A1
sin(x) = -0.5 giving x = 210 degrees or x = 330 degrees	A1
Final answer: x = 90, 210 or 330 degrees

Mark scheme for Question 11 [5 marks]
Question 11[5 marks]
Answer or working	Marks
attempting to complete the square on x^2 + 8x	M1
(x + 4)^2 as the squared term	M1
(x + 4)^2 + 3	A1
identifying the minimum value as 3	B1
stating this occurs at x = -4	B1
Final answer: (x + 4)^2 + 3; minimum value 3 at x = -4

Mark scheme for Question 12 [6 marks]
Question 12[6 marks]
Answer or working	Marks
factorising 2x^2 + 3x - 20 as (2x - 5)(x + 4)	M1
identifying the critical values x = -4 and x = 2.5	M1
sketching or reasoning about the shape of the (positive) parabola	M1
reasoning that the expression is positive outside the roots	M1
x < -4	A1
x > 2.5	A1
Final answer: x < -4 or x > 2.5

Mark scheme for Question 13 [6 marks]
Question 13[6 marks]
Answer or working	Marks
differentiating N to get dN/dt = 60t - 6t^2	M1
differentiating again to get d2N/dt2 = 60 - 12t	M1
setting d2N/dt2 = 0 to find the stationary point of the rate	M1
t = 5 (hours)	A1
substituting t = 5 into dN/dt	M1
maximum rate = 150 bacteria per hour	A1
Final answer: dN/dt = 60t - 6t^2; rate of growth is greatest at t = 5 hours, when dN/dt = 150