Year 11

Year 11 Paper 5: Sequences, Trigonometry and Proof

Covers algebraic manipulation, sequences, matrices and transformations, non-right-angled triangle trigonometry, and geometric proof.

18 questions - 80 marks - calculator allowed

Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.

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Questions

Question 1 [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 2 [3 marks]

Geometric Proof

ABCD is a kite with AB = AD and CB = CD.

Prove that triangle ABC is congruent to triangle ADC.

Question 3 [3 marks]

Algebraic Manipulation

Expand and simplify (x - 5)(x + 5) - (x - 3)^2.

Question 4 [3 marks]

Non-Right-Angled Triangle Trigonometry

In triangle DEF, DE = 8 cm, DF = 10 cm, and angle EDF = 55 degrees.

Find the length of EF, giving your answer to 3 significant figures.

Question 5 [4 marks]

Geometric Proof

Triangle LMN and triangle LPN share the side LN.

Angle MLN = angle PLN, and LM = LP.

Prove that triangle LMN is congruent to triangle LPN.

Question 6 [3 marks]

Sequences

A geometric sequence has first term 5 and common ratio 3.

Find the 6th term of the sequence, and find the sum of the first 6 terms.

Question 7 [4 marks]

Geometric Proof

Triangle ABC is right-angled at B.

The point D lies on AC such that BD is perpendicular to AC.

Prove that triangle ABD is similar to triangle ABC.

Question 8 [4 marks]

Matrices and Transformations

The point Q(4, -1) is transformed by the matrix with rows (-1, 0) and (0, 1).

Find the coordinates of the image of Q, and describe fully the single geometric transformation that this matrix represents.

Question 9 [4 marks]

Algebraic Manipulation

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

Question 10 [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 11 [5 marks]

Geometric Proof

AB and CD are two chords of a circle with centre O that are equidistant from O.

Prove that AB = CD.

Question 12 [5 marks]

Sequences

A geometric sequence has third term 20 and sixth term 160.

Find the common ratio and the first term of the sequence.

Question 13 [5 marks]

Matrices and Transformations

M is the matrix with rows (k, 3) and (4, k - 1), where k is a constant.

Given that M is singular, find the possible values of k.

Question 14 [6 marks]

Sequences

The first three terms of a geometric sequence are (k + 1), (k + 4), and (3k + 2), where k is a positive constant.

Find the value of k, and find the common ratio.

Question 15 [6 marks]

Geometric Proof

In triangle ABC, D is the midpoint of AB and E is the midpoint of AC.

Prove that DE is parallel to BC and that DE = (1/2)BC.

Question 16 [6 marks]

Non-Right-Angled Triangle Trigonometry

Triangle ABC has AB = 9 cm, BC = 14 cm, and AC = 19 cm.

Find the largest angle of the triangle, giving your answer to 1 decimal place, and state which vertex it is at.

Question 17 [6 marks]

Geometric Proof

Triangle ABC is right-angled at B.

The point D lies on AC such that BD is perpendicular to AC.

Prove that BD^2 = AD x DC.

Question 18 [6 marks]

Algebraic Manipulation

Solve the equation (2x - 1)/(x + 3) = x - 3.

Show each step of your working and give both solutions.

Model solutions

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

Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or working	Marks
stating AB = AD (given)	B1
stating CB = CD (given), and AC is common to both triangles	B1
concluding triangle ABC is congruent to triangle ADC by SSS	B1
Final answer: Congruent by SSS: AB = AD, CB = CD (given), AC common

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

Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or working	Marks
using the cosine rule EF^2 = DE^2 + DF^2 - 2(DE)(DF)cos(55)	M1
substituting to get EF^2 = 164 - 160cos(55)	M1
EF = 8.50 cm	A1
Final answer: EF = 8.50 cm (3 s.f.)

Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or working	Marks
stating angle MLN = angle PLN (given)	B1
stating LM = LP (given)	B1
stating LN is common to both triangles	B1
concluding triangle LMN is congruent to triangle LPN by SAS	B1
Final answer: Congruent by SAS: angle MLN = angle PLN, LM = LP (given), LN common

Mark scheme for Question 6 [3 marks]
Question 6[3 marks]
Answer or working	Marks
using T6 = ar^5 = 5 x 3^5	M1
T6 = 1215	A1
the sum of the first 6 terms = 1820	A1
Final answer: T6 = 1215; sum of first 6 terms = 1820

Mark scheme for Question 7 [4 marks]
Question 7[4 marks]
Answer or working	Marks
stating angle ADB = 90 degrees, since BD is perpendicular to AC	B1
stating angle ABC = 90 degrees, given the triangle is right-angled at B	B1
stating angle BAD (equal to angle BAC) is common to both triangles	B1
concluding triangle ABD is similar to triangle ABC by the AA condition	B1
Final answer: Similar by AA: angle ADB = angle ABC = 90 degrees, and angle A is common to both triangles

Mark scheme for Question 8 [4 marks]
Question 8[4 marks]
Answer or working	Marks
multiplying the matrix by the column vector (4, -1)	M1
the image (-4, -1)	A1
identifying the transformation as a reflection	B1
stating the mirror line is the y-axis	B1
Final answer: Image (-4, -1); reflection in the y-axis

Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or working	Marks
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 10 [5 marks]
Question 10[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

Mark scheme for Question 11 [5 marks]
Question 11[5 marks]
Answer or working	Marks
stating OA = OC, since both are radii of the circle	B1
stating OM = ON (given the chords are equidistant from O), where M and N are the feet of the perpendiculars from O to AB and CD, and noting angle OMA = angle ONC = 90 degrees	B1
concluding triangle OMA is congruent to triangle ONC by RHS	M1
concluding AM = CN, and that M and N are the midpoints of AB and CD respectively (since the perpendicular from the centre bisects a chord)	A1
concluding AB = 2AM = 2CN = CD	A1
Final answer: Triangles OMA and ONC are congruent by RHS (OA = OC, OM = ON, right angles at M and N), so AM = CN, and since M and N bisect the chords, AB = 2AM = 2CN = CD

Mark scheme for Question 12 [5 marks]
Question 12[5 marks]
Answer or working	Marks
writing T3 = ar^2 = 20 and T6 = ar^5 = 160	M1
dividing to eliminate a, giving r^3 = 8	M1
r = 2	A1
substituting back into ar^2 = 20 to find a	M1
a = 5	A1
Final answer: common ratio r = 2; first term a = 5

Mark scheme for Question 13 [5 marks]
Question 13[5 marks]
Answer or working	Marks
writing the determinant k(k - 1) - 3(4)	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

Mark scheme for Question 14 [6 marks]
Question 14[6 marks]
Answer or working	Marks
using the geometric sequence property (k + 4)^2 = (k + 1)(3k + 2)	M1
expanding the left side to k^2 + 8k + 16	M1
expanding the right side to 3k^2 + 5k + 2	M1
rearranging to 2k^2 - 3k - 14 = 0 and solving, rejecting k = -2 since k is positive	M1
k = 3.5	A1
common ratio = 5/3	A1
Final answer: k = 3.5; common ratio = 5/3

Mark scheme for Question 15 [6 marks]
Question 15[6 marks]
Answer or working	Marks
stating AD/AB = 1/2, since D is the midpoint of AB	B1
stating AE/AC = 1/2, since E is the midpoint of AC	B1
stating angle DAE (equal to angle BAC) is common to both triangles	B1
concluding triangle ADE is similar to triangle ABC by the SAS condition (two sides in the same ratio with the included angle equal)	M1
concluding angle ADE = angle ABC (corresponding angles in similar triangles), so DE is parallel to BC	A1
concluding DE = (1/2)BC, from the scale factor of the similarity	A1
Final answer: Triangle ADE is similar to triangle ABC (SAS, ratio 1/2), so DE is parallel to BC and DE = (1/2)BC

Mark scheme for Question 16 [6 marks]
Question 16[6 marks]
Answer or working	Marks
identifying that the largest angle is opposite the longest side AC, so at vertex B	B1
using the cosine rule AC^2 = AB^2 + BC^2 - 2(AB)(BC)cos(B)	M1
substituting to get 361 = 277 - 252cos(B)	M1
rearranging to cos(B) = -84/252	M1
cos(B) = -0.333 (3 s.f.)	A1
angle B = 109.5 degrees (1 d.p.)	A1
Final answer: largest angle = 109.5 degrees (1 d.p.), at vertex B

Mark scheme for Question 17 [6 marks]
Question 17[6 marks]
Answer or working	Marks
stating angle ADB = angle BDC = 90 degrees, since BD is perpendicular to AC	B1
using angle ABC = 90 degrees to write angle DBA + angle DBC = 90 degrees	M1
using the angle sum of triangle ADB to write angle DAB + angle DBA = 90 degrees	M1
concluding angle DAB = angle DBC	A1
concluding triangle ADB is similar to triangle BDC by the AA condition	M1
writing the ratio AD/BD = BD/DC from the similarity, hence BD^2 = AD x DC	A1
Final answer: Triangle ADB is similar to triangle BDC (AA), giving AD/BD = BD/DC, so BD^2 = AD x DC

Mark scheme for Question 18 [6 marks]
Question 18[6 marks]
Answer or working	Marks
cross-multiplying to give 2x - 1 = (x - 3)(x + 3)	M1
expanding the right side to x^2 - 9	M1
rearranging to x^2 - 2x - 8 = 0	M1
factorising as (x - 4)(x + 2) = 0	M1
x = 4	A1
x = -2	A1
Final answer: x = 4 or x = -2