Grades 1-9

GCSE Further Maths Paper 2

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

14 questions - 60 marks - calculator allowed

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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]

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 [4 marks]

Trigonometric Identities and Equations

Show that (cos(x) - sin(x))^2 can be simplified to 1 - 2sin(x)cos(x).

Question 4 [4 marks]

Geometric Proof

Triangle PQR and triangle PSR share the side PR.

Angle PQR = angle PSR = 90 degrees, and QR = SR.

Prove that triangle PQR is congruent to triangle PSR.

Question 5 [4 marks]

Simultaneous Equations and Inequalities

Solve the simultaneous equations 4x + y = 16 and 2x - 3y = -6.

Question 6 [4 marks]

Algebraic Manipulation

Simplify fully (x^2 - 9)/(x^2 + x - 6).

Question 7 [4 marks]

Sequences

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

Find the 6th term of the sequence, and determine whether 200 is a term of the sequence, showing your working.

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]

Coordinate Geometry

A circle has equation x^2 + y^2 - 6x + 4y - 12 = 0.

Find the centre and radius of the circle by completing the square.

Question 10 [5 marks]

Applications of Differentiation

A curve has equation y = x^2 + 5x - 2.

Find the equation of the normal to the curve at the point where x = -2, giving your answer in the form y = mx + c.

Question 11 [5 marks]

Matrices and Transformations

The matrix N represents an enlargement, scale factor 2, centre the origin, followed by a reflection in the x-axis.

Write down the matrix representing the enlargement and the matrix representing the reflection, and hence find N as a single 2x2 matrix.

State which of the two matrices is applied first when N is used to transform a point.

Question 12 [5 marks]

Differentiation

A curve has equation y = x^3 - 3x^2 - 9x + 4.

Find the range of values of x for which the curve is decreasing.

Question 13 [5 marks]

Sequences

The sum of the first n terms of an arithmetic series is given by Sn = n(2n - 3).

Find the first term and the common difference of the series.

Hence find the 15th term of the series.

Question 14 [6 marks]

Functions and their Graphs

The function k is defined by k(x) = -2x^2 + 12x - 7 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

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or workingMarks
writing sqrt(50) = 5sqrt(2) and sqrt(18) = 3sqrt(2)M1
2sqrt(2)A1
Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or workingMarks
stating BD = CD (D is the midpoint of BC) and AD is common to both trianglesB1
concluding triangle ABD is congruent to triangle ACD by SSS, using AB = AC givenB1
Final answer: Congruent by SSS: AB = AC (given), BD = CD (D midpoint), AD common
Mark scheme for Question 3 [4 marks]
Question 3[4 marks]
Answer or workingMarks
expanding the brackets to give cos^2(x) - 2sin(x)cos(x) + sin^2(x)M1
stating the identity sin^2(x) + cos^2(x) = 1B1
substituting sin^2(x) + cos^2(x) = 1M1
1 - 2sin(x)cos(x)A1
Final answer: 1 - 2sin(x)cos(x), shown using sin^2(x) + cos^2(x) = 1
Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or workingMarks
stating angle PQR = angle PSR = 90 degrees (given)B1
stating PR is common to both triangles, and is the hypotenuse of eachB1
stating QR = SR (given)B1
concluding triangle PQR is congruent to triangle PSR by RHSB1
Final answer: Congruent by RHS: angle PQR = angle PSR = 90 degrees, PR common (hypotenuse), QR = SR (given)
Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or workingMarks
rearranging 4x + y = 16 to make y the subject, y = 16 - 4xM1
substituting into 2x - 3y = -6M1
simplifying to 14x = 42M1
x = 3 and y = 4A1
Final answer: x = 3, y = 4
Mark scheme for Question 6 [4 marks]
Question 6[4 marks]
Answer or workingMarks
factorising the numerator as (x - 3)(x + 3)M1
factorising the denominator as (x + 3)(x - 2)M1
cancelling the common factor (x + 3)M1
(x - 3)/(x - 2)A1
Mark scheme for Question 7 [4 marks]
Question 7[4 marks]
Answer or workingMarks
substituting n = 6 into Un = 2n^2 + 3nM1
the 6th term = 90A1
forming the equation 2n^2 + 3n - 200 = 0 and using the discriminantM1
concluding 200 is not a term since n is not a positive integer, as the discriminant 1609 is not a perfect squareA1
Final answer: 6th term = 90; 200 is not a term of the sequence
Mark scheme for Question 8 [5 marks]
Question 8[5 marks]
Answer or workingMarks
using Area = (1/2)(XY)(XZ)sin(X)M1
substituting to get 55 = (1/2)(10)(13)sin(X)M1
rearranging to sin(X) = 55/65M1
X = 57.8 degreesA1
X = 122.2 degreesA1
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 workingMarks
completing the square on the x terms: (x - 3)^2 - 9M1
completing the square on the y terms: (y + 2)^2 - 4M1
rearranging to (x - 3)^2 + (y + 2)^2 = 25M1
centre (3, -2)A1
radius 5A1
Final answer: Centre (3, -2), radius 5
Mark scheme for Question 10 [5 marks]
Question 10[5 marks]
Answer or workingMarks
finding y = -8 at x = -2M1
differentiating to get dy/dx = 2x + 5M1
finding the tangent gradient = 1 at x = -2M1
using the perpendicular gradient rule to get normal gradient -1M1
y = -x - 10A1
Mark scheme for Question 11 [5 marks]
Question 11[5 marks]
Answer or workingMarks
the enlargement matrix, the matrix with rows (2, 0) and (0, 2)B1
the reflection matrix, the matrix with rows (1, 0) and (0, -1)B1
multiplying the reflection matrix by the enlargement matrix in that orderM1
N = the matrix with rows (2, 0) and (0, -2)A1
stating the enlargement is applied first, since it is written on the right of the productB1
Final answer: N = the matrix with rows (2, 0) and (0, -2), found as (reflection matrix) x (enlargement matrix)
Mark scheme for Question 12 [5 marks]
Question 12[5 marks]
Answer or workingMarks
differentiating to get dy/dx = 3x^2 - 6x - 9M1
setting dy/dx = 0M1
solving to find x = -1 and x = 3, e.g. factorising as 3(x - 3)(x + 1) = 0M1
reasoning that dy/dx < 0 between the two rootsM1
-1 < x < 3A1
Mark scheme for Question 13 [5 marks]
Question 13[5 marks]
Answer or workingMarks
using S1 = a to find the first term a = -1M1
finding S2 and using T2 = S2 - S1M1
common difference d = 4A1
using T15 = a + 14dM1
T15 = 55A1
Final answer: first term = -1; common difference = 4; 15th term = 55
Mark scheme for Question 14 [6 marks]
Question 14[6 marks]
Answer or workingMarks
taking out the factor -2 from the x terms: -2(x^2 - 6x) - 7M1
completing the square inside the bracket: (x - 3)^2 - 9M1
substituting back to get -2[(x - 3)^2 - 9] - 7M1
11 - 2(x - 3)^2 (a = 11, b = 2, c = 3)A1
identifying the maximum value as 11B1
stating this occurs at x = 3B1
Final answer: 11 - 2(x - 3)^2; maximum value 11 at x = 3