Year 11

Year 11 Paper 2: Matrices and Coordinate Methods

Covers algebraic manipulation, simultaneous equations and inequalities, functions and their graphs, coordinate geometry, and matrices and transformations.

13 questions - 60 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]

Algebraic Manipulation

Expand and simplify (3x - 2)(x + 4).

Question 2 [3 marks]

Functions and their Graphs

The function g is defined by g(x) = 2x^2 - 5 for all real x.

Find the values of x for which g(x) = 45.

Question 3 [4 marks]

Matrices and Transformations

A is the matrix with rows (4, -2) and (3, 5).

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

Find 2A - B, writing your answer as a matrix.

Question 4 [4 marks]

Coordinate Geometry

Points A(-3, 4) and B(5, -2) are given.

Find the midpoint of AB and the length of AB.

Question 5 [4 marks]

Algebraic Manipulation

Factorise fully 6x^3 - 24x.

Question 6 [5 marks]

Simultaneous Equations and Inequalities

Solve algebraically the simultaneous equations y = 3x - 4 and y = x^2 + x - 7.

Question 7 [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 8 [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 9 [5 marks]

Algebraic Manipulation

Simplify fully (3x^2 + 5x - 2)/(9x^2 - 1).

Question 10 [5 marks]

Coordinate Geometry

A circle has centre (1, 2) and passes through the point (4, 6).

Find the equation of the circle.

Then find the equation of the tangent to the circle at the point (4, 6), giving your answer in the form y = mx + c.

Question 11 [6 marks]

Functions and their Graphs

The function k is defined by k(x) = -3x^2 + 24x - 41 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.

Question 12 [6 marks]

Matrices and Transformations

A single transformation is represented by the matrix N, the matrix with rows (a, b) and (c, d).

Under this transformation, the point (1, 0) maps to (3, -1) and the point (0, 1) maps to (2, 4).

Write down the matrix N, and hence find the image of the point (-1, 3) under this transformation.

Question 13 [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 working	Marks
expanding to give four terms with correct signs	M1
3x^2 + 10x - 8	A1

Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or working	Marks
forming the equation 2x^2 - 5 = 45	M1
rearranging to x^2 = 25	M1
x = 5 or x = -5	A1

Mark scheme for Question 3 [4 marks]
Question 3[4 marks]
Answer or working	Marks
doubling A to give the matrix with rows (8, -4) and (6, 10)	M1
subtracting the corresponding entries of B	M1
the first row of the answer, (7, -7)	A1
the second row of the answer, (8, 6)	A1
Final answer: the matrix with rows (7, -7) and (8, 6)

Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or working	Marks
using the midpoint formula ((x1 + x2)/2, (y1 + y2)/2)	M1
midpoint (1, 1)	A1
using the distance formula sqrt((x2 - x1)^2 + (y2 - y1)^2)	M1
length AB = 10	A1
Final answer: Midpoint (1, 1); length AB = 10

Mark scheme for Question 5 [4 marks]
Question 5[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 6 [5 marks]
Question 6[5 marks]
Answer or working	Marks
setting the two expressions for y equal: 3x - 4 = x^2 + x - 7	M1
rearranging to x^2 - 2x - 3 = 0	M1
factorising to (x - 3)(x + 1) = 0	M1
x = 3 or x = -1	A1
(x, y) = (3, 5) and (-1, -7)	A1
Final answer: (x, y) = (3, 5) or (-1, -7)

Mark scheme for Question 7 [5 marks]
Question 7[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

Mark scheme for Question 8 [5 marks]
Question 8[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 9 [5 marks]
Question 9[5 marks]
Answer or working	Marks
factorising the numerator as (3x - 1)(x + 2)	M1
the correct factorisation (3x - 1)(x + 2)	A1
factorising the denominator as the difference of two squares (3x - 1)(3x + 1)	M1
cancelling the common factor (3x - 1)	M1
(x + 2)/(3x + 1)	A1

Mark scheme for Question 10 [5 marks]
Question 10[5 marks]
Answer or working	Marks
finding the radius as sqrt(3^2 + 4^2) = 5	M1
the circle equation (x - 1)^2 + (y - 2)^2 = 25	A1
finding the gradient of the radius to (4, 6) as 4/3	M1
using the perpendicular gradient -3/4 with the point (4, 6)	M1
the tangent y = -0.75x + 9	A1
Final answer: Circle: (x - 1)^2 + (y - 2)^2 = 25; tangent: y = -0.75x + 9

Mark scheme for Question 11 [6 marks]
Question 11[6 marks]
Answer or working	Marks
taking out the factor -3 from the x terms: -3(x^2 - 8x) - 41	M1
completing the square inside the bracket: (x - 4)^2 - 16	M1
substituting back to get -3[(x - 4)^2 - 16] - 41	M1
7 - 3(x - 4)^2 (a = 7, b = 3, c = 4)	A1
identifying the maximum value as 7	B1
stating this occurs at x = 4	B1
Final answer: 7 - 3(x - 4)^2; maximum value 7 at x = 4

Mark scheme for Question 12 [6 marks]
Question 12[6 marks]
Answer or working	Marks
recognising the image of (1, 0) gives the first column of N, so a = 3 and c = -1	B1
recognising the image of (0, 1) gives the second column of N, so b = 2 and d = 4	B1
N = the matrix with rows (3, 2) and (-1, 4)	A1
multiplying N by the column vector (-1, 3)	M1
the image x-coordinate 3	A1
the image y-coordinate 13	A1
Final answer: N = the matrix with rows (3, 2) and (-1, 4); image of (-1, 3) is (3, 13)

Mark scheme for Question 13 [6 marks]
Question 13[6 marks]
Answer or working	Marks
taking out the factor -2 from the x terms: -2(x^2 - 6x) - 7	M1
completing the square inside the bracket: (x - 3)^2 - 9	M1
substituting back to get -2[(x - 3)^2 - 9] - 7	M1
11 - 2(x - 3)^2 (a = 11, b = 2, c = 3)	A1
identifying the maximum value as 11	B1
stating this occurs at x = 3	B1
Final answer: 11 - 2(x - 3)^2; maximum value 11 at x = 3