Year 12 A-Level Bridging Project Mathematics Teacher Contact

Year 12

A-Level Bridging Project

Mathematics

Teacher Contact:

Mrs Smith [email protected]

Name ………………………………

2

Year 12 A level Mathematics Bridging Project

Topic: A-Level Foundation Skills Task 1: Create revisions guides, cards, posters, pod casts, PowerPoints, puzzles, games etc.

on each of the topics in the following checklist. CGP have a FREE Headstart to A-Level textbook available on the kindle. Youtube - haberdashers adams maths videos are also useful Task 2: Complete the included questions provided on the topics in the checklist and self-mark.

Task 3: Complete the Algebraic Expressions assessment in assessment conditions (1 hour) and the Quadratic Inequalities and Simultaneous Equations assessment (45 minutes) – to be handed in on the first day back to Mrs Smith. If you are able to scan it you can also email it to

Mrs Smith on [email protected] Task 4: In the second week on the A-Level course you will be given a test on the topics listed above. Use your revision cards, topic questions, Mathswatch and any other resources and

useful websites to revise and prepare for this assessment. The Task 3 assessment will allow us to provide extra help where needed prior to this.

Task 5: Complete the included Discovery worksheet on the trapezium rule and compare

answers. Learn the trapezium rule. We will look at answer upon return to school Task 6: Complete the included Algebra Award paper and self mark. There will be some topics you have not covered yet but you have covered the majority of them. 60/90 is a pass. You will

be sitting the algebra award in May, this gives us a good basis.

Task 7: Become familiar with the Excel large data set and fill in the included worksheet. Please do not worry if you do not understand this, we have plenty of time to go through it next year. This is for familiarisation only.

OPTIONAL Task 8: Research Project – we will look at your answers upon return.

All your work must be handed in in the first lesson back, this and the assessment will identify if you have a good understanding of the core skills needed to start A-Level Maths. If you fail to complete the tasks to a high standard this will imply that you are not committed to

the course and we will be considering your suitability for the course. This will result in meetings with the Head of Sixth form to discuss.

Enjoy the summer break, Maths Department

3

Checklist

Topic Skill Revised and

created revision

material

Assessed and

completed the

questions.

Task 2

Algebra Expanding brackets

Solving equations

Changing the subject

Laws of Indices Multiplying, dividing, power

Fractional

Negative

Surds Simplifying

Rationalising

Surds and indices Challenge

Quadratics Solving through factorising

Solving through formula

Solving through completing

the square

Difference of two squares

Simultaneous

equations

Solving through substitution

Solving through eliminations

Linear and quadratic

Inequalities Solving inequalities

EXTRA WORKSHEETS - OPTIONAL

Coordinate Geometry Gradients and midpoints

Transformation of

Graphs

Graphs of function

Graph transformations

Task 3

Assessment

Task 5

Discovery

Worksheet

Task 6

Algebra Award Paper

Task 7

Large Data set

familiarisation

Large Data Set

Worksheet

Task 8

Research Project

4

Section 1: ALGEBRA

EXPANDING BRACKETS EXERCISE A Multiply out the following brackets and simplify.

1. 7(4x + 5)

2. -3(5x - 7)

3. 5a – 4(3a - 1)

4. 4y + y(2 + 3y)

5. 5(2x - 1) – (3x - 4)

6. (x + 2)(x + 3)

7. (t - 5)(t - 2)

8. (2x + 3y)(3x – 4y)

9. (2y - 1)(2y + 1)

10. (3 + 5x)(4 – x)

EXERCISE B Multiply out

1. (x - 1)2

2. (3x + 5)2

3. (7x - 2)2

4. (x + 2)(x - 2)

5. (3x + 1)(3x - 1)

6. (5y - 3)(5y + 3)

FACTORISING EXERCISE C: Factorise each of the following

1) 3x + xy 2) 4x2 – 2xy

3) pq2 – p2q 4) 3pq - 9q2

5) 2x3 – 6x2

6) 8a5b2 – 12a3b4 7) 5y(y – 1) + 3(y – 1)

EXERCISE D Factorise

1) 2 6x x− −

2) 2 6 16x x+ −

3) 22 5 2x x+ +

4) 22 3x x−

5) 210 5 30x x+ −

6) 24 25x −

7) 24 12 8x x− +

8) 2 216 81m n−

9) 3 24 9y a y−

10) 28( 1) 2( 1) 10x x+ − + −

5

SOLVING EXERCISE E Solving

1) 2x + 5 = 19

2) 5x – 2 = 13

3) 11 – 4x = 5

4) 5 – 7x = -9

5) 11 + 3x = 8 – 2x

EXERCISE F Solving

6) 7x + 2 = 4x – 5

7) 5(2x – 4) = 4

8) 4(2 – x) = 3(x – 9)

9) 8 – (x + 3) = 4

10) 14 – 3(2x + 3) = 2

CHANGING THE SUBJECT EXERCISE H

Solomon Press

ALGEBRA C1 Worksheet B 1 Evaluate

a 82 b 63 c 70 d (−5)4 e (−3)5 f 412( )

g 323( ) h 31

4( )− i 213(1 ) j 41

2(1 ) k (0.1)5 l (−0.2)3 2 Write in the form 2n

a 25 × 23 b 2 × 26 c 1 d 26 ÷ 22 e 215 ÷ 26 f (27)2

3 Simplify

a 2p2 × 4p5 b x2 × x3 × x5 c 12n7 ÷ 2n2 d (y3)4

e (2b)3 ÷ 4b2 f p3q × pq2 g x4y3 ÷ xy2 h 2r2s × 3s2

i 6x5y8 ÷ 3x2y j 6a4b5 × 23 ab3 k (5rs2)3 ÷ (10rs)2 l 3p4q3 ÷ 1

5 pq2 4 Evaluate

a 3−2 b 025( ) c (−2)−6 d 21

6( )− e 312(1 )− f

129

g 1416 h

13( 27)− i

121

49( ) j 13125 k

124

9( ) l 1236−

m 1481− n

13( 64)−− o

151

32( )− p 138

125( )− q 121

4(2 ) r 133

8(3 )− 5 Evaluate

a 324 b

2327 c

3416 d

23( 125)− e

529 f

238−

g 3236− h

431

8( ) i 324

9( ) j 231

216( )− k 329

16( )− l 4327

64( )−

m 12(0.04) n

32(2.25)− o

23(0.064) p

329

16(1 )− q 341

16(5 ) r 4310

27(2 )− 6 Work out

a 124 ×

1327 b

1416 +

1225 c

138− ÷

1236 d

13( 64)− ×

329

e 213( )− −

13( 8)− f

121

25( ) × 214( )− g

3481 −

121

49( )− h 131

27( )− × 324

9( )−

i 121

9( )− × 35( 32)− j (121)0.5 + (32)0.2 k (100)0.5 ÷ (0.25)1.5 l (16)−0.25 × (243)0.4

7 Simplify

a x8 × x−6 b y−2 × y−4 c 6p3 ÷ 2p7 d (2x−4)3

e y3 × 12y− f

232b ×

144b g

35x ÷

13x h

12a ÷

43a

i 14p ÷

15p− j

25 2(3 )x k y ×

56y ×

32y− l

324t ÷

1212t

m 14

12

2b b

b

× n 1132y y

y× o

2 13 6

34

4 3

6

x x

x

−× p 34

12

2

8

a a

a−

×

PMT

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8 Solve each equation.

a 12x = 6 b

13x = 5 c

12x− = 2 d

14x− = 1

3

e 32x = 8 f

23x = 16 g

43x = 81 h

32x− = 27

9 Express in the form xk

a x b 31x

c x2 × x d 4 xx

e 3x f x × 3 x g 5( )x h 3 2x × 3( )x 10 Express each of the following in the form axb, where a and b are rational constants.

a 4x

b 12x

c 33

4x d 2

1(3 )x

e 32

5 x f

3

1

9x

11 Express in the form 2k

a 82 b 214( )− c

131

2( ) d 1616− e

258 f 31

32( )− 12 Express each of the following in the form 3y, where y is a function of x.

a 9x b 81x + 1 c 427x d 1

3( )x e 92x − 1 f 2127( )x +

13 Given that y = 2x, express each of the following in terms of y.

a 2x + 1 b 2x − 2 c 22x d 8x e 24x + 3 f 312( )x −

14 Find the value of x such that

a 2x = 64 b 5x − 1 = 125 c 3x + 4 − 27 = 0 d 8x − 2 = 0

e 32x − 1 = 9 f 16 − 43x − 2 = 0 g 9x − 2 = 27 h 82x + 1 = 16

i 49x + 1 = 7 j 33x − 2 = 3 9 k 316( )x + = 36 l 3 11

2( ) x − = 8 15 Solve each equation.

a 2x + 3 = 4x b 53x = 25x + 1 c 92x = 3x − 3 d 16x = 41 − x

e 4x + 2 = 8x f 272x = 93 − x g 63x − 1 = 36x + 2 h 8x = 162x − 1

i 125x = 5x − 3 j 13( )x = 3x − 4 k 11

2( ) x− = 218( ) x l 11

4( )x + = 8x 16 Expand and simplify

a x(x2 − x−1) b 2x3(x−1 + 3) c x−1(3x − x3) d 4x−2(3x5 + 2x3)

e 12 x2(6x + 4x−1) f

31 12 2 23 ( )x x x− − g

3 72 22(5 )x x x− + h

51 43 3 3(3 )x x x−−

i (x2 + 1)(x4 − 3) j (2x5 + x)(x4 + 3) k (x2 − 2x−1)(x − x−2) l 3 12 22( )( )x x x x− −

17 Simplify

a 3 2x x

x+ b

5 3

24 6

2t t

t− c

32

12

3x x

x

− d 2 3( 6)

3y y

y−

e 32

34

p p

p

+ f 12

12

8 2

4

w w

w−

− g 1 12 2

1x

x x−

+

+ h 3 1

2 2

32 4

2

t t

t t−

−

−

C1 ALGEBRA Worksheet B continued

PMT

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ALGEBRA C1 Worksheet A 1 Evaluate

a 49 b 121 c 19 d 4

25 e 0.01 f 0.09

g 3 8 h 3 1000 i 4 81 j 9161 k 3 0.125 l 53

815 2 Simplify

a 7 × 7 b 4 5 × 5 c (3 3 )2 d ( 6 )4

e ( 2 )5 f (2 3 )3 g 2 × 8 h 2 3 × 27

i 322

j 312

k ( 3 6 )3 l (3 3 2 )3

3 Express in the form k 2

a 18 b 50 c 8 d 98 e 200 f 162 4 Simplify

a 12 b 28 c 80 d 27 e 24 f 128

g 45 h 40 i 75 j 112 k 99 l 147

m 216 n 800 o 180 p 60 q 363 r 208 5 Simplify

a 18 + 50 b 48 − 27 c 2 8 + 72

d 360 − 2 40 e 2 5 − 45 + 3 20 f 24 + 150 − 2 96 6 Express in the form a + b 3

a 3 (2 + 3 ) b 4 − 3 − 2(1 − 3 ) c (1 + 3 )(2 + 3 )

d (4 + 3 )(1 + 2 3 ) e (3 3 − 4)2 f (3 3 + 1)(2 − 5 3 ) 7 Simplify

a ( 5 + 1)(2 5 + 3) b (1 − 2 )(4 2 − 3) c (2 7 + 3)2

d (3 2 − 1)(2 2 + 5) e ( 5 − 2 )( 5 + 2 2 ) f (3 − 8 )(4 + 2 ) 8 Express each of the following as simply as possible with a rational denominator.

a 15

b 23

c 18

d 147

e 3 23

f 515

g 13 7

h 1272

i 180

j 32 54

k 4 203 18

l 3 1752 27

PMT

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9 Simplify

a 8 + 62

b 48 − 103

c 6 82

−

d 45 520

− e 118

+ 132

f 23

− 672

10 Solve each equation, giving your answers as simply as possible in terms of surds.

a x(x + 4) = 4(x + 8) b x − 48 = 2 3 − 2x

c x 18 − 4 = 8 d x 5 + 2 = 20 (x − 1) 11 a Simplify (2 − 3 )(2 + 3 ).

b Express 22 3−

in the form a + b 3 .

12 Express each of the following as simply as possible with a rational denominator.

a 12 1+

b 43 1−

c 16 2−

d 32 3+

e 12 5+

f 22 1−

g 67 3+

h 13 2 2+

i 14 2 3−

j 33 2 4+

k 2 37 4 3−

l 65 3−

13 Solve the equation

3x = 5 (x + 2),

giving your answer in the form a + b 5 , where a and b are rational. 14

(3 2 − 3) cm

l cm

The diagram shows a rectangle measuring (3 2 − 3) cm by l cm.

Given that the area of the rectangle is 6 cm2, find the exact value of l in its simplest form. 15 Express each of the following as simply as possible with a rational denominator.

a 22 6+

b 1 32 3

++

c 1 1010 3+

− d 3 2

4 3 2−+

e 1 23 8

−−

f 3 52 3 4

−−

g 12 33 3

+−

h 3 7 22 7 5

−−

C1 ALGEBRA Worksheet A continued

PMT

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ALGEBRA C1 Worksheet C 1 Express each of the following in the form a 2 + b 3 , where a and b are integers.

a 27 + 2 50

b 6 ( 3 − 8 )

2 Given that x > 0, find in the form k 3 the value of x such that

x(x − 2) = 2(6 − x). 3 Solve the equation

25x = 54x + 1. 4 a Express 3 24 in the form 3 3k .

b Find the integer n such that

3 24 + 3 81 = 3 n . 5 Show that

10 315

+ 45 7−

can be written in the form k 7 , where k is an integer to be found. 6 Showing your method clearly,

a express 37.5 in the form a 6 ,

b express 359 − 2

36 in the form b 15 . 7 Given that x = 2t − 1 and y = 23t,

a find expressions in terms of t for

i xy ii 2y2

b Hence, or otherwise, find the value of t for which

2y2 − xy = 0. 8 Solve the equation

2 (3x − 1) = 2(2x + 3),

giving your answer in the form a + b 2 , where a and b are integers. 9 Given that 6 y + 1 = 36x − 2,

a express y in the form ax + b,

b find the value of 1 24x y− .

10 Express each of the following in the form a + b 2 , where a and b are integers.

a (3 − 2 )(1 + 2 )

b 22 1−

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11 Solve the equation

16x + 1 = 82x + 1. 12 Given that (a − 2 3 )2 = b − 20 3 ,

find the values of the integers a and b. 13 a Find the value of t such that

314( )t − = 8.

b Solve the equation

13( ) y = 27y + 1.

14 Express each of the following in the form a + b 5 , where a and b are integers.

a 20 ( 5 − 3)

b (1 − 5 )(3 + 2 5 )

c 1 55 2+

−

15 Given that 13a =

34b , and that a > 0 and b > 0,

a find an expression for 12a in terms of b,

b find an expression for 12b in terms of a.

16 A 2 3 − 1

3 + 2

In triangle ABC, AB = 2 3 − 1, BC = 3 + 2 and ∠ABC = 90°.

a Find the exact area of triangle ABC in its simplest form.

b Show that AC = 2 5 .

c Show that tan (∠ACB) = 5 3 − 8. 17 a Given that y = 2x, express each of the following in terms of y.

i 2x + 2 ii 4x

b Hence, or otherwise, find the value of x for which

4x − 2x + 2 = 0.

18 Given that the point with coordinates (1 + 3 , 5 3 ) lies on the curve with the equation

y = 2x2 + px + q,

find the values of the rational constants p and q.

C1 ALGEBRA Worksheet C continued

B C

PMT

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ALGEBRA C1 Worksheet E 1 Factorise

a x2 + 4x + 3 b x2 + 7x + 10 c y2 − 3y + 2 d x2 − 6x + 9

e y2 − y − 2 f a2 + 2a − 8 g x2 − 1 h p2 + 9p + 14

i x2 − 2x − 15 j 16 − 10m + m2 k t 2 + 3t − 18 l y2 − 13y + 40

m r2 − 16 n y2 − 2y − 63 o 121 + 22a + a2 p x2 + 6x − 72

q 26 − 15x + x2 r s2 + 23s + 120 s p2 + 14p − 51 t m2 − m − 90 2 Factorise

a 2x2 + 3x + 1 b 2 + 7p + 3p2 c 2y2 − 5y + 3 d 2 − m − m2

e 3r2 − 2r − 1 f 5 − 19y − 4y2 g 4 − 13a + 3a2 h 5x2 − 8x − 4

i 4x2 + 8x + 3 j 9s2 − 6s + 1 k 4m2 − 25 l 2 − y − 6y2

m 4u2 + 17u + 4 n 6p2 + 5p − 4 o 8x2 + 19x + 6 p 12r2 + 8r − 15 3 Using factorisation, solve each equation.

a x2 − 4x + 3 = 0 b x2 + 6x + 8 = 0 c x2 + 4x − 5 = 0 d x2 − 7x = 8

e x2 − 25 = 0 f x(x − 1) = 42 g x2 = 3x h 27 + 12x + x2 = 0

i 60 − 4x − x2 = 0 j 5x + 14 = x2 k 2x2 − 3x + 1 = 0 l x(x − 1) = 6(x − 2)

m 3x2 + 11x = 4 n x(2x − 3) = 5 o 6 + 23x − 4x2 = 0 p 6x2 + 10 = 19x

q 4x2 + 4x + 1 = 0 r 3(x2 + 4) = 13x s (2x + 5)2 = 5 − x t 3x(2x − 7) = 2(7x + 3) 4 Factorise fully

a 2y2 − 10y + 12 b x3 + x2 − 2x c p3 − 4p d 3m3 + 21m2 + 18m

e a4 + 4a2 + 3 f t 4 + 3t 2 − 10 g 12 + 20x − 8x2 h 6r2 − 9r − 42

i 6x3 − 26x2 + 8x j y4 + 3y3 − 18y2 k m4 − 1 l p5 − 4p3 + 4p 5 Sketch each curve showing the coordinates of any points of intersection with the coordinate axes.

a y = x2 − 3x + 2 b y = x2 + 5x + 6 c y = x2 − 9

d y = x2 − 2x e y = x2 − 10x + 25 f y = 2x2 − 14x + 20

g y = −x2 + 5x − 4 h y = 2 + x − x2 i y = 2x2 − 3x + 1

j y = 2x2 + 13x + 6 k y = 3 − 8x + 4x2 l y = 2 + 7x − 4x2

m y = 5x2 − 17x + 6 n y = −6x2 + 7x − 2 o y = 6x2 + x − 5 6 Solve each of the following equations.

a x − 5 + 4x

= 0 b x − 10x

= 3 c 2x3 − x2 − 3x = 0 d x2(10 − x2) = 9

e 25x

+ 4x

− 1 = 0 f 64

xx

−−

= x g x + 5 = 33x +

h x2 − 24x

= 3

i 4x4 + 7x2 = 2 j 23

xx−

= 12x +

k 2 13

xx

++

= 2x

l 72x +

− 3x = 2

PMT

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ALGEBRA C1 Worksheet G 1 By completing the square, show that the roots of the equation ax2 + bx + c = 0 are given by

x = 2 4

2b b ac

a− ± − .

2 Use the quadratic formula to solve each equation, giving your answers as simply as possible in

terms of surds where appropriate.

a x2 + 4x + 1 = 0 b 4 + 8t − t 2 = 0 c y2 − 20y + 91 = 0 d r2 + 2r − 7 = 0

e 6 + 18a + a2 = 0 f m(m − 5) = 5 g x2 + 11x + 27 = 0 h 2u2 + 6u + 3 = 0

i 5 − y − y2 = 0 j 2x2 − 3x = 2 k 3p2 + 7p + 1 = 0 l t 2 − 14t = 14

m 0.1r2 + 1.4r = 0.9 n 6u2 + 4u = 1 o 12 y2 − 3y = 2

3 p 4x(x − 3) = 11 − 4x 3 y

y = 2x2 − 8x + 3 O x The diagram shows the curve with equation y = 2x2 − 8x + 3.

Find and simplify the exact coordinates of the points where the curve crosses the x-axis. 4 State the condition for which the roots of the equation ax2 + bx + c = 0 are

a real and distinct b real and equal c not real 5 Sketch the curve y = ax2 + bx + c and the x-axis in the cases where

a a > 0 and b2 − 4ac > 0 b a < 0 and b2 − 4ac < 0

c a > 0 and b2 − 4ac = 0 d a < 0 and b2 − 4ac > 0 6 By evaluating the discriminant, determine whether the roots of each equation are real and

distinct, real and equal or not real.

a x2 + 2x − 7 = 0 b x2 + x + 3 = 0 c x2 − 4x + 5 = 0 d x2 − 6x + 3 = 0

e x2 + 14x + 49 = 0 f x2 − 9x + 17 = 0 g x2 + 3x = 11 h 2 + 3x + 2x2 = 0

i 5x2 + 8x + 3 = 0 j 3x2 − 7x + 5 = 0 k 9x2 − 12x + 4 = 0 l 13x2 + 19x + 7 = 0

m 4 − 11x + 8x2 = 0 n x2 + 23 x = 1

4 o x2 − 34 x + 1

8 = 0 p 25 x2 + 3

5 x + 13 = 0

7 Find the value of the constant p such that the equation x2 + x + p = 0 has equal roots. 8 Given that q ≠ 0, find the value of the constant q such that the equation x2 + 2qx − q = 0

has a repeated root. 9 Given that the x-axis is a tangent to the curve with the equation

y = x2 + rx − 2x + 4,

find the two possible values of the constant r.

PMT

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ALGEBRA C1 Worksheet F 1 Express in the form (x + a)2 + b

a x2 + 2x + 4 b x2 − 2x + 4 c x2 − 4x + 1 d x2 + 6x

e x2 + 4x + 8 f x2 − 8x − 5 g x2 + 12x + 30 h x2 − 10x + 25

i x2 + 6x − 9 j 18 − 4x + x2 k x2 + 3x + 3 l x2 + x − 1

m x2 − 18x + 100 n x2 − x − 12 o 20 + 9x + x2 p x2 − 7x − 2

q 5 − 3x + x2 r x2 − 11x + 37 s x2 + 23 x + 1 t x2 − 1

2 x − 14

2 Express in the form a(x + b)2 + c

a 2x2 + 4x + 3 b 2x2 − 8x − 7 c 3 − 6x + 3x2 d 4x2 + 24x + 11

e −x2 − 2x − 5 f 1 + 10x − x2 g 2x2 + 2x − 1 h 3x2 − 9x + 5

i 3x2 − 24x + 48 j 3x2 − 15x k 70 + 40x + 5x2 l 2x2 + 5x + 2

m 4x2 + 6x − 7 n −2x2 + 4x − 1 o 4 − 2x − 3x2 p 13 x2 + 1

2 x − 14

3 Solve each equation by completing the square, giving your answers as simply as possible in terms

of surds where appropriate.

a y2 − 4y + 2 = 0 b p2 + 2p − 2 = 0 c x2 − 6x + 4 = 0 d 7 + 10r + r2 = 0

e x2 − 2x = 11 f a2 − 12a − 18 = 0 g m2 − 3m + 1 = 0 h 9 − 7t + t 2 = 0

i u2 + 7u = 44 j 2y2 − 4y + 1 = 0 k 3p2 + 18p = −23 l 2x2 + 12x = 9

m −m2 + m + 1 = 0 n 4x2 + 49 = 28x o 1 − t − 3t 2 = 0 p 2a2 − 7a + 4 = 0 4 By completing the square, find the maximum or minimum value of y and the value of x for which

this occurs. State whether your value of y is a maximum or a minimum in each case.

a y = x2 − 2x + 7 b y = x2 + 2x − 3 c y = 1 − 6x + x2

d y = x2 + 10x + 35 e y = −x2 + 4x + 4 f y = x2 + 3x − 2

g y = 2x2 + 8x + 5 h y = −3x2 + 6x i y = 7 − 5x − x2

j y = 4x2 − 12x + 9 k y = 4x2 + 20x − 8 l y = 17 − 2x − 2x2 5 Sketch each curve showing the exact coordinates of its turning point and the point where it

crosses the y-axis.

a y = x2 − 4x + 3 b y = x2 + 2x − 24 c y = x2 − 2x + 5

d y = 30 + 8x + x2 e y = x2 + 2x + 1 f y = 8 + 2x − x2

g y = −x2 + 8x − 7 h y = −x2 − 4x − 7 i y = x2 − 5x + 4

j y = x2 + 3x + 3 k y = 3 + 8x + 4x2 l y = −2x2 + 8x − 15

m y = 1 − x − 2x2 n y = 25 − 20x + 4x2 o y = 3x2 − 4x + 2 6 a Express x2 − 4 2 x + 5 in the form a(x + b)2 + c.

b Write down an equation of the line of symmetry of the curve y = x2 + 4 2 x + 5. 7 f(x) ≡ x2 + 2kx − 3.

By completing the square, find the roots of the equation f(x) = 0 in terms of the constant k.

PMT

Difference of Squares

Factor fully

1. X2 – 1 2. 64x2 – 4 3. 36 – 100x2

4. 1 – 36x2 5. 49 – 4x2 6. 81x2 – 9

7. 9x2 – 4 8. 16x2 – 100 9. 169 – 25x2

10. 49x2 – 25

Simplify and then factor, showing all your steps.

1. 2x2 – 6 + 4x2 + 3x2 – 10

2. X2 – 24 – 3x2 + 28 – 7x2

3. 4 – 2x2 + 8 – 7x2 + 13

4. 7 + 10x2 – 9 + 90x2 – 2

5. – 32 + 60x2 – 4 + 4x2

6. 3x2 + 6 +83x2 – 22 – x2

7. -105 + 4x2 + 5

8. 7x2 + 3 – 32x2 + 33

9. 2 – 6x2 + 2 – 58x2 + 32

10.10x2 – 5 – x2 + 1 + 7x2

Solomon Press

ALGEBRA C1 Worksheet I 1 Solve each pair of simultaneous equations.

a y = 3x b y = x − 6 c y = 2x + 6

y = 2x + 1 y = 12 x − 4 y = 3 − 4x

d x + y − 3 = 0 e x + 2y + 11 = 0 f 3x + 3y + 4 = 0

x + 2y + 1 = 0 2x − 3y + 1 = 0 5x − 2y − 5 = 0 2 Find the coordinates of the points of intersection of the given straight line and curve in each case.

a y = x + 2 b y = 4x + 11 c y = 2x − 1

y = x2 − 4 y = x2 + 3x − 1 y = 2x2 + 3x − 7 3 Solve each pair of simultaneous equations.

a x2 − y + 3 = 0 b 2x2 − y − 8x = 0 c x2 + y2 = 25

x − y + 5 = 0 x + y + 3 = 0 2x − y = 5 d x2 + 2xy + 15 = 0 e x2 − 2xy − y2 = 7 f 3x2 − x − y2 = 0

2x − y + 10 = 0 x + y = 1 x + y − 1 = 0 g 2x2 + xy + y2 = 22 h x2 − 4y − y2 = 0 i x2 + xy = 4

x + y = 4 x − 2y = 0 3x + 2y = 6 j 2x2 + y − y2 = 8 k x2 − xy + y2 = 13 l x2 − 5x + y2 = 0

2x − y = 3 2x − y = 7 3x + y = 5 m 3x2 − xy + y2 = 36 n 2x2 + x − 4y = 6 o x2 + x + 2y2 − 52 = 0

x − 2y = 10 3x − 2y = 4 x − 3y + 17 = 0 4 Solve each pair of simultaneous equations.

a x − 1y

− 4y = 0 b xy = 6 c 3x

− 2y + 4 = 0

x − 6y − 1 = 0 x − y = 5 4x + y − 7 = 0 5 The line y = 5 − x intersects the curve y = x2 − 3x + 2 at the points P and Q.

Find the length PQ in the form k 2 . 6 Solve the simultaneous equations

3x − 1 = 92y

8x − 2 = 41 + y 7 Given that

(A + 2 3 )(B − 3 ) ≡ 9 3 − 1,

find the values of the integers A and B.

PMT

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ALGEBRA C1 Worksheet J 1 Find the set of values of x for which

a 2x + 1 < 7 b 3x − 1 ≥ 20 c 2x − 5 > 3 d 6 + 3x ≤ 42

e 5x + 17 ≥ 2 f 13 x + 7 < 8 g 9x − 4 ≥ 50 h 3x + 11 < 7

i 18 − x > 4 j 10 + 4x ≤ 0 k 12 − 3x < 10 l 9 − 12 x ≥ 4

2 Solve each inequality.

a 2y − 3 > y + 4 b 5p + 1 ≤ p + 3 c x − 2 < 3x − 8

d a + 11 ≥ 15 − a e 17 − 2u < 2 + u f 5 − b ≥ 14 − 3b

g 4x + 23 < x + 5 h 12 + 3y ≥ 2y − 1 i 16 − 3p ≤ 36 + p

j 5(r − 2) > 30 k 3(1 − 2t) ≤ t − 4 l 2(3 + x) ≥ 4(6 − x)

m 7(y + 3) − 2(3y − 1) < 0 n 4(5 − 2x) > 3(7 − 2x) o 3(4u − 1) − 5(u − 3) < 9 3 Find the set of values of x for which

a x2 − 4x + 3 < 0 b x2 − 4 ≤ 0 c 15 + 8x + x2 < 0 d x2 + 2x ≤ 8

e x2 − 6x + 5 > 0 f x2 + 4x > 12 g x2 + 10x + 21 ≥ 0 h 22 + 9x − x2 > 0

i 63 − 2x − x2 ≤ 0 j x2 + 11x + 30 > 0 k 30 + 7x − x2 > 0 l x2 + 91 ≥ 20x 4 Solve each inequality.

a 2x2 − 9x + 4 ≤ 0 b 2r2 − 5r − 3 < 0 c 2 − p − 3p2 ≥ 0

d 2y2 + 9y − 5 > 0 e 4m2 + 13m + 3 < 0 f 9x − 2x2 ≤ 10

g a2 + 6 < 8a − 9 h x(x + 4) ≤ 7 − 2x i y(y + 9) > 2(y − 5)

j x(2x + 1) > x2 + 6 k u(5 − 6u) < 3 − 4u l 2t + 3 ≥ 3t(t − 2)

m (y − 2)2 ≤ 2y − 1 n (p + 2)(p + 3) ≥ 20 o 2(13 + 2x) < (6 + x)(1 − x) 5 Giving your answers in terms of surds, find the set of values of x for which

a x2 + 2x − 1 < 0 b x2 − 6x + 4 > 0 c 11 − 6x − x2 > 0 d x2 + 4x + 1 ≥ 0 6 Find the value or set of values of k such that

a the equation x2 − 6x + k = 0 has equal roots,

b the equation x2 + 2x + k = 0 has real and distinct roots,

c the equation x2 − 3x + k = 0 has no real roots,

d the equation x2 + kx + 4 = 0 has real roots,

e the equation kx2 + x − 1 = 0 has equal roots,

f the equation x2 + kx − 3k = 0 has no real roots,

g the equation x2 + 2x + k − 2 = 0 has real and distinct roots,

h the equation 2x2 − kx + k = 0 has equal roots,

i the equation x2 + kx + 2k − 3 = 0 has no real roots,

j the equation 3x2 + kx − x + 3 = 0 has real roots.

PMT

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COORDINATE GEOMETRY C1 Worksheet A 1 Find the gradient of the line segment joining each pair of points.

a (3, 1) and (5, 5) b (4, 7) and (10, 9) c (6, 1) and (2, 5) d (−2, 2) and (2, 8)

e (1, 3) and (7, −1) f (4, 5) and (−5, −7) g (−2, 0) and (0, −8) h (8, 6) and (−7, −2) 2 Write down the gradient and y-intercept of each line.

a y = 4x − 1 b y = 13 x + 3 c y = 6 − x d y = −2x − 3

5 3 Find the gradient and y-intercept of each line.

a x + y + 3 = 0 b x − 2y − 6 = 0 c 3x + 3y − 2 = 0 d 4x − 5y + 1 = 0 4 Write down, in the form y − y1 = m(x − x1), the equation of the straight line with the given

gradient which passes through the given point.

a gradient 2, point (4, 1) b gradient 5, point (2, −5)

c gradient −3, point (−1, 1) d gradient 12 , point (1, 6)

e gradient −2, point ( 34 ,

14− ) f gradient

15− , point (−3, −7)

5 Find, in the form y = mx + c, the equation of the straight line with the given gradient which

passes through the given point.

a gradient 3, point (1, 2) b gradient −1, point (5, 3)

c gradient 4, point (−2, −3) d gradient −2, point (−4, 1)

e gradient 13 , point (−3, 1) f gradient

56− , point (9, −2)

6 Find, in each case, the equation of the straight line with gradient m which passes through the

point P. Give your answers in the form ax + by + c = 0, where a, b and c are integers.

a m = 1, P (2, −4) b m = 12 , P (6, 1) c m = −4, P (−1, 8)

d m = 25 , P (−3, 5) e m = −3, P ( 3

2 , 18− ) f m = 3

4− , P ( 23 , −7)

7 Find, in the form y = mx + c, the equation of the straight line passing through each pair of points.

a (0, 1) and (4, 13) b (2, 9) and (7, −1) c (−4, 3) and (2, 7)

d ( 12− , −2) and (2, 8) e (3, −2) and (18, −5) f (−3.2, 4) and (−2, 0.4)

8 Find, in the form ax + by + c = 0, where a, b and c are integers, the equation of the straight line

which passes through each pair of points.

a (3, 0) and (5, 2) b (−1, 8) and (5, −4) c (−5, 3) and (7, 5)

d (−4, −1) and (8, −17) e (2, −1.5) and (7, 0) f ( 35− , 1

10 ) and (3, 1) 9 The straight line l passes through the points A (−6, 8) and B (3, 2).

a Find an equation of the line l.

b Show that the point C (9, −2) lies on l. 10 The point M (k, 2k) lies on the line with equation x − 3y + 15 = 0.

Find the value of the constant k.

PMT

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GRAPHS OF FUNCTIONS C1 Worksheet A 1 Sketch and label each pair of graphs on the same set of axes showing the coordinates of any

points where the graphs intersect. Write down the equations of any asymptotes.

a y = x2 and y = x3 b y = x2 and y = x4

c y = 1x

and y = 21x

d y = x and y = x

e y = x2 and y = 3x2 f y = 1x

and y = 2x

2 f(x) = (x − 1)(x − 3)(x − 4).

a Find f(0).

b Write down the solutions of the equation f(x) = 0.

c Sketch the curve y = f(x). 3 Sketch each graph showing the coordinates of any points of intersection with the coordinate axes.

a y = (x + 1)(x − 1)(x − 3) b y = 2x(x − 1)(x − 5)

c y = −(x + 2)(x + 1)(x − 2) d y = x2(x − 4)

e y = 3x(2 + x)(1 − x) f y = (x + 2)(x − 1)2

4 a Factorise fully x3 + 6x2 + 9x.

b Hence, sketch the curve y = x3 + 6x2 + 9x, showing the coordinates of any points where the curve meets the coordinate axes.

5 Given that the constants p and q are such that p > q > 0, sketch each of the following graphs

showing the coordinates of any points of intersection with the coordinate axes.

a y = (x − p)(x − q)2 b y = (x − p)(x2 − q2) 6 y O x −5 y = f(x)

The diagram shows the curve with equation y = f(x) which has a turning point at (1, −2) and crosses the y-axis at the point (0, −5).

Given that f(x) is a quadratic function, find an expression for f(x). 7 y

1 2 −2 O x

−8 y = ax3 + bx2 + cx + d The diagram shows the curve with equation y = ax3 + bx2 + cx + d.

Given that the curve crosses the y-axis at the point (0, −8) and crosses the x-axis at the points (−2, 0), (1, 0) and (2, 0), find the values of the constants a, b, c and d.

(1, −2)

PMT

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GRAPHS OF FUNCTIONS C1 Worksheet B 1 Describe how the graph of y = f(x) is transformed to give the graph of

a y = f(x − 1) b y = f(x) − 3 c y = 2f(x) d y = f(4x)

e y = −f(x) f y = 15 f(x) g y = f(−x) h y = f( 2

3 x) 2 y (0, 3) y = f(x)

O (4, 0) x The diagram shows the curve with equation y = f(x) which crosses the coordinate axes at the

points (0, 3) and (4, 0).

Showing the coordinates of any points of intersection with the axes, sketch on separate diagrams the graphs of

a y = 3f(x) b y = f(x + 4) c y = −f(x) d y = f( 12 x)

3 Find and simplify an equation of the graph obtained when

a the graph of y = 2x + 5 is translated by 1 unit in the positive y-direction,

b the graph of y = 1 − 4x is stretched by a factor of 3 in the y-direction, about the x-axis,

c the graph of y = 3x + 1 is translated by 4 units in the negative x-direction,

d the graph of y = 4x − 7 is reflected in the x-axis. 4 y

y = f(x) (0, 6)

(2, 4) O x

The diagram shows the curve with equation y = f(x) which has a turning point at (2, 4) and crosses the y-axis at the point (0, 6).

Showing the coordinates of the turning point and of any points of intersection with the axes, sketch on separate diagrams the graphs of

a y = f(x) − 3 b y = f(x + 2) c y = f(2x) d y = 12 f(x)

5 Describe a single transformation that would map the graph of y = x3 onto the graph of

a y = 4x3 b y = (x − 2)3 c y = −x3 d y = x3 + 5

6 Describe a single transformation that would map the graph of y = x2 + 2 onto the graph of

a y = 2x2 + 4 b y = x2 − 5 c y = 19 x2 + 2 d y = x2 + 4x + 6

PMT

mathsgenie.co.uk

Write your name hereSurname Other Names

AS/A Level Mathematics

Quadratic Inequalities and Simultaneous Equations

Instructions

• Use black ink or ball-point pen.• If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).• Fill in the boxes at the top of this page with your name.• Answer all questions and ensure that your answers to parts of questions are clearly labelled..• Answer the questions in the spaces provided– there may be more space than you need.• You should show sufficient working to make your methods clear.Answers without working may not gain full credit.• Answers should be given to three significant figures unless otherwise stated.

Information

• The marks for each question are shown in brackets– use this as a guide as to how much time to spend on each question.

Advice

• Read each question carefully before you start to answer it.• Try to answer every question.• Check your answers if you have time at the end.

Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.

1 (a) Solve the inequality x2 + 8x > 20

(b) Find the set of values for x which satisfy both of the inequalities

x2 + 8x > 20

18 + 3x < 23 + x

(Total for question 1 is 4 marks)

2 Find the set of values of x for which (x + 5)(x + 1) < 32


3 Solve the simultaneous equations

x + y = 3 x2 + 2y2 – 8x = 6


4 Solve the inequality x(x + 1) ≤ 12


5 Find the coordinates of the points where where the the circle C with equation x2 + y2 – 2x = 19 meets the line L with equation y = 3x – 1


6 The curve C has the equation y = x2 – 2x + 7 The line L has the equation x + y = 7 Find the coordinates of the points where L and C intersect.


7 Solve the simultaneous equations

x + 2y = 3 x2 + y2 – 2xy = 6


8 (a) Solve the inequality x2 + 3x – 10 < 0

(b) Find the set of values for x which satisfy both of the inequalities

x2 + 3x – 10 < 0

9 + 3x ≤ 12 + x


(3)

(1)

(3)

(2)

mathsgenie.co.uk

Write your name hereSurname Other Names

AS/A Level Mathematics

Algebraic Expressions

Instructions

• Use black ink or ball-point pen.• If pencil is used for diagrams/sketches/graphs it must be dark (HB or B).• Fill in the boxes at the top of this page with your name.• Answer all questions and ensure that your answers to parts of questions are clearly labelled..• Answer the questions in the spaces provided– there may be more space than you need.• You should show sufficient working to make your methods clear.Answers without working may not gain full credit.• Answers should be given to three significant figures unless otherwise stated.

Information

• The marks for each question are shown in brackets– use this as a guide as to how much time to spend on each question.

Advice

• Read each question carefully before you start to answer it.• Try to answer every question.• Check your answers if you have time at the end.

Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation and integration, or have retrievable mathematical formulae stored in them.

1 Simplify (Total for question 1 is 2 marks)

( 125 x6

64 )13

(2 x12 )

3

4 x2

2 Simplify


4 Express 93x + 2 in the form 3y, giving y in the form ax + b, where a and b are constants. (Total for question 4 is 2 marks)

5 Express 82x – 5 in the form 2y, giving y in the form ax + b, where a and b are constants.


6 Given y = 2x

(a) Express 4x in terms of y.

(b) Hence, or otherwise, solve 4x – 6(2x) – 16 = 0 (Total for question 6 is 5 marks)

7 Solve: 22x + 1 – 5(2x) – 12 = 0


( 216 x6

27 y3 )−

23

3 Simplify



8 Solve the equation 82x – 5 = 2x + 1

9 (a) Solve the equation: x2 – 9x + 8 = 0

(b) Hence solve the equation: y3 – 9 y32 + 8=0


(2)

(2)

(2)

(3)

15 Find the value of x such that1 + x

= 5x

Giving your answer in the form a + b 5 where a and b are rational numbers.



16 Show that can be written as

5 + 3

2 – 3 13 + 7 3



5 + 2 32 + 3

4 – 3



1 – 2

3 + 2 27 – 5 2

11 Factorise completely: x – 16x3


12 Factorise completely: 75x – 12x3


10 Find the value of x such that 8x + 1 = 43x – 1


13 Expand and Simplify: (2x – 1)(x + 2)(x – 3)


14 Expand and Simplify: (3x – 2)(x – 5)2


2 cm 2 cm 2 cm 2 cm

5 cm

5.5

cm

6 c

m

7 cm

9 cm

Discovering the Trapezium Rule

1) Find the total areas of the shaded shapes using the formula for the area of a trapezium. Simplify your

answers and leave them as fractions. a) b)

Challenge: Can you think of a more efficient way of working out the areas rather than adding up all the individual areas of the trapeziums?

Hint: Try factorising out ℎ

2 from the first line of working in c).

2) The curve shown has equation y = x2 + 4. We can estimate the area under the curve using trapeziums.

Find the area of the shaded shape.

Ask for a hint if you need it.

Challenge: Can you find a formula for working out an estimation of the area using trapeziums for a general

curve?

h cm h cm h cm h cm 5

cm

5.5

cm

6 c

m

7 cm

9 cm

h cm h cm h cm h cm

a c

m

b c

m

c cm

d c

m

e cm

c)

×

×

×

×

×

1 2 3 4 0

y = x2 + 4

Centre Number Candidate Number

Write your name hereSurname Other names

Total Marks

Paper Reference

Turn over

P46442A©2016 Pearson Education Ltd.

6/6/6/2/

*P46442A0124*

AlgebraLevel 3Calculator NOT allowed

Tuesday 10 May 2016 – MorningTime: 2 hours AAL30/01

You must have: Ruler graduated in centimetres and millimetres, pair of compasses, pen, HB pencil, eraser.

Instructions

• Use black ink or ball-point pen.• Fill in the boxes at the top of this page with your name, centre number and candidate number.• Answer all questions.• Answer the questions in the spaces provided

– there may be more space than you need.• Calculators are not allowed.

Information

• The total mark for this paper is 90• The marks for each question are shown in brackets

– use this as a guide as to how much time to spend on each question.

Advice

• Read each question carefully before you start to answer it.• Keep an eye on the time.• Try to answer every question.• Check your answers if you have time at the end.

Pearson

Edexcel Award

2

*P46442A0224*

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Answer ALL questions.

Write your answers in the spaces provided.

You must write down all stages in your working.

You must NOT use a calculator.

1 (a) On the grid, construct the graph of x2 + y2 – 25 = 0

6

4

2

–6 –4 –2 O 2 4 6

–2

–4

–6

x

y

(2)

3

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A is the point (3, 4)

(b) (i) Draw the tangent to the graph at the point A.

(ii) Write down the size of the angle between the tangent to the graph at A and the normal to the graph at A.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(Total for Question 1 is 4 marks)

2 w = 42

2

2t

t +

Make t the subject of the formula.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


4

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3 On the grid, shade the region that satisfies all these inequalities.

x + y < 5 y > 2x + 1 y < –3x

Label the region R.


6

4

2

O

–2

–4

8

10

–4 –2 2 4 6

y

x

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4 (a) Expand and simplify (2x – 4)(x + 3)

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(2)

(b) Factorise 10d 2e2 + 15de3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(c) Factorise 3p2 – 12q2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


6

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5 The straight line L passes through the points A and B with coordinates (1, 3) and (–1, –1) respectively.

(a) Find an equation for L in the form ax + by + c = 0 where a, b and c are integers.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)

(b) Find an equation of a straight line which is perpendicular to the line 4y = x + 8

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(2)


7


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6 T is inversely proportional to the cube of x.

When x = 3, T = 154

(a) Find a formula for T in terms of x.

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(3)

(b) Calculate the value of x when T = 4

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(2)


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7 (a) Simplify e2 × e–3

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(1)

(b) Simplify 23

2

n⎛⎝⎜

⎞⎠⎟

−

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(c) Express 23

xx +

+ 73x −

as a single fraction.

Give your answer in its simplest form.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)


9


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8 Solve, algebraically, the simultaneous equations

y = x2 – x – 6 x – y = 3

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


10

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9 Use the quadratic formula to solve the equation 3x2 + 4x – 5 = 0

Give your solutions in the form p qr

± where p, q and r are integers.

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10 (a) Solve 8 – 3y < 11

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(b) Solve x2 + 3x – 4 < 0

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)


11


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11 Here is a graph for values of x from 1 to 8

3

2

1

O 1 2 3 4 5 6 7 8

y

x

4

Use the trapezium rule to find an estimate of the area of the region under the curve, between x = 2 and x = 6, and above y = 0

Use 4 strips of equal width.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


12

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12 (a) Find the value of (2 7 )2

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(1)

(b) Express 432 27− in the form a 3 where a is an integer.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(c) Rationalise the denominator of 27 3+

Give your answer in the form a bc

− where a, b and c are integers.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)


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13 The sum of the first three terms of an arithmetic series is 39 The ninth term of this series is 41

(a) Find the first term of the series and the common difference of the series.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

common difference .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)

The nth term of a different arithmetic series is 13n – 6

(b) Find an expression, in terms of n, for the sum of the first n terms of this series. Give your answer in its simplest form.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)


14

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14 Here is the graph of y = f(x)

8

7

6

5

4

3

2

1

–1

–2

–6 –5 –4 –3 –2 –1 O 1 2 3 4 5 6 x

y

(a) On the grid below, draw the graph of y = f(x) + 3

8

7

6

5

4

3

2

1

–1

–2

–6 –5 –4 –3 –2 –1 O 1 2 3 4 5 6 x

y

(2)

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(b) On the grid below, draw the graph of y = f(x – 2)

x

y8

7

6

5

4

3

2

1

–1

–2

–6 –5 –4 –3 –2 –1 O 1 2 3 4 5 6

(2)


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15 (a) On the grid below, draw the graph of y = 2x + 1 for values of x from –4 to 3

(4)

17


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(b) Use your graph to find an estimate for the solution of 2x + 1 = 12 Give your answer to 1 decimal place.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)


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16 The equation x2 + 4x + c = 0 has real roots.

(a) Find the range of possible values of c.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(b) Sketch the graph of y = (x – 3)2

You must label the coordinates of any points at which the graph meets the coordinate axes.

(3)


19

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Turn over

17 x2 + 4x + 1 = (x + p)2 + q for all values of x.

(a) Find the value of p and the value of q.

p = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

q = .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(b) Solve the equation 2x2 – 5x – 3 = 0

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


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18 7x2 + 6x – 3 = 0 is a quadratic equation.

For this quadratic equation, write down the sum of its roots and the product of its roots.

sum of roots.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

product of roots.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


21

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19 On the axes below, sketch the graph of y = 12x +

Show clearly any asymptotes and the coordinates of any points of intersection of the graph with the axes.

O x

y


Turn over

22

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20 Here is a speed-time graph for part of a journey.

120

90

60

30

1030 1045 1100 1115 1130 1145

Time of day

Speed (km/h)

150

0

(a) Between which two times is the acceleration greatest?

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

(b) Calculate the acceleration, in km/h², in the first 15 minutes of the journey.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . km/h²(2)

23

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(c) Work out the total distance travelled in the first 45 minutes of the journey.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . km(2)


TOTAL FOR PAPER IS 90 MARKS

24

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BLANK PAGE

Large Data Set

Data recorded

Daily Mean

Temperature

Units and accuracy

n/a means Range of times covered Range of temperatures

Daily Total

Rainfall

Units and accuracy

What does tr mean?

Max rainfall UK/worldwide Range of times covered

Daily Total Sunshine

Units and accuracy (how is it measured)

Max/min

Daily Maximum

Relative Humidity

What is it?

How is it measured?

Wind What is measured and how (4 types)?

How are they measured? Max/min UK/worldwide

Visibility What is it?

How is it measured?

Pressure How is it measured?

Max/min

Daily mean total

cloud

What is it?

What units are used to measure it?

Cities included 3. Heathrow 6.Leeming

1. Beijing 4. Hurn 7. Leuchars

2. Camborne 5. Jacksonville 8. Perth

Months included:

____________ to ______________

Years included:

__________ & __________

Months included

with 30 days:

Daily Mean Temperature

Which months have the highest average temperature in the UK?

Which locations are warmest in the UK? Which locations are coldest in the UK?

Which locations are warmer worldwide?

Which locations are colder worldwide? Are there any cities that have very different weather to the others?

What range of temperatures do you get in the UK? Worldwide?

Is there a difference between 1987 average temperatures and 2015?

Daily Total Rainfall

Which months are wettest for UK cities? Is there a difference in rainfall between 1987 and 2015

Do all cities follow a similar pattern for rainfall? Are there any with very different patterns?

Which of the cities is the wettest in the UK? What

about worldwide?

Which of the cities is the driest in the UK? What

about worldwide?

Windspeed

Typical average windspeed Windiest months UK? Any difference with worldwide cities?

Any differences between 1987 and 2015?

10

Research: Research the binomial expansion and its relationship with Pascals triangle. Try the questions below

Binomial:

(1) A company owns 400 laptops. Each laptop has an 8% probability of not working. You

randomly select 20 laptops for your salespeople.

(a) What is the likelihood that 5 will be broken?

(b) What is the likelihood that they will all work?

(c) What is the likelihood that they will all be broken?

Research the factor and remainder theorem. Try the questions below Factor and Remainder Theorem:

11

SOLUTIONS TO THE EXERCISES

Ex A Multiplying Out Brackets

1) 28x + 35

2) -15x + 21

3) -7a + 4

4) 6y + 3y2

6) 7x – 1

7) x2 + 5x + 6

8) t2 – 3t + 10

9) 6x2 + xy – 12y2

10) 4x2 + 4x – 24

11) 4y2 – 1

12) 12 + 17x – 5x2

Ex B

1) x2 – 2x + 1

2) 9x2 + 30x + 25

3) 49x2 – 28x + 4

4) x2 – 4

5) 9x2 -1

6) 25y2 – 9

Ex A Factorising

1) x(3 + y)

2) 2x(2x – y)

3) pq(q – p)

4) 3q(p – 3q)

5) 2x2(x - 3)

6) 4a3b2(2a2 – 3b2)

7) (y – 1)(5y

+ 3) Ex B

1) (x – 3)(x + 2)

2) (x + 8)(x – 2)

3) (2x + 1)(x + 2)

4) x(2x – 3)

8) 5(2x – 3)(x + 2)

9) (2x + 5)(2x – 5)

11) 4(x – 2)(x – 1)

12) (4m – 9n)(4m + 9n) 13) y(2y – 3a)(2y + 3a) 14) 2(4x + 5)(x – 4)

Ex A Solving

1) 7 2) 3 3) 1½ 4) 2 5) -3/5 6) -7/3

Ex B

1) 2.4 2) 5 3) 1

4) ½

Ex C

1) 7 2) 15 3) 24/7 4) 35/3 5) 3 6) 2 7) 9/5 8) 5

Ex A – rearranging

These will be

marked in

school

Solomon Press

ALGEBRA C1 Answers - Worksheet B

1 a = 64 b = 216 c = 1

d = 625 e = −243 f = 116

g = 827 h = 1

64− i = 243( ) = 16

9 or 791

j = 432( ) = 81

16 or 1165 k = 0.000 01 l = −0.008

2 a 28 b 27 c 20 d 24 e 29 f 214

3 a = 8p7 b = x10 c = 6n5 d = y12

e = 2b f = p4q3 g = x3y h = 6r2s3

i = 2x3y7 j = 4a5b8 k = 125r3s6 ÷ 100r2s2 l = 15p3q = 5

4 rs4

4 a 2

1 193

= = b = 1 c 61 1

64( 2)−= =

d 26 36= = e 3 33 822 3 27( ) ( )−= = = f 9 3= =

g 4 16 2= = h 3 27 3= − = − i 1 149 7= =

j 3 125 5= = k 4 29 3= = l 1 1

636= =

m 41 1

381= = n 3

1 1464−

= = − o 5 32 2= =

p 8 23125 5= − = − q 9 3 1

4 2 2 or 1= = r 1327 8 23

8 27 3( )−= = =

5 a 3 3( 4) 2 8= = = b 2 23( 27) 3 9= = =

c 3 34( 16) 2 8= = = d 2 23( 125) ( 5) 25= − = − =

e 5 5( 9) 3 243= = = f 2 231 1 1

4( 8) 2= = =

g 3 31 1 1

216( 36 ) 6= = = h 4 41 1 13

8 2 16( ) ( )= = =

i 3 3 84 29 3 27( ) ( )= = = j 2 23( 216) 6 36= = =

k 3 316 64 1049 3 27 27( ) ( ) or 2= = = l 4 427 3 813

64 4 256( ) ( )= − = − =

m 4 2 1100 10 5 or 0.2= = = n

32 3 39 84 2

4 9 3 27( ) ( ) ( )−= = = =

o 2 264 4 431000 10 25( ) ( ) or 0.16= = = p

32 3 325 16 644

16 25 5 125( ) ( ) ( )−= = = =

q 34 3 381 81 3 27 34

16 16 2 8 8( ) ( ) ( ) or 3= = = = r 43 4 464 27 3 813

27 64 4 256( ) ( ) ( )−= = = =

PMT

C1 ALGEBRA Answers - Worksheet B page 2

Solomon Press

6 a = 34 27× b = 4 16 25+ c = 3

18

36÷ d = 33 64 ( 9)− ×

= 2 × 3 = 6 = 2 + 5 = 7 = 12 ÷ 6 = 1

12 = −4 × 27 = −108 e = 2 33 8− − f = 21

25 4× g = 34( 81) 49− h = 33 9427 ( )×

= 9 − (−2) = 11 = 15 × 16 = 16 1

5 5 or 3 = 27 − 7 = 20 = 3 × 278 = 81 1

8 8 or 10 i = 359 ( 32)× − j = 5121 32+ k = 31

4100 ( )÷ l = 4251

16( 243)×

= 3 × (−8) = −24 = 11 + 2 = 13 = 10 ÷ 18 = 80 = 1

2 × 9 = 9 12 2 or 4

7 a = x2 b = y−6 c = 3p−4 d = 8x−12

e = 52y f

2 1 113 4 128 8b b+= = g

3 1 45 3 15x x−= = h

51 42 3 6a a− −= =

i 91 1

4 5 20( )p p− −= = j = 459x k

5 3 16 2 31y y+ −= = l = 1

3 t

m 71 1

4 2 42b b+ −= = n 1 1 12 3 61y y+ − −= = o

32 1 13 6 4 4( )2 2x x+ − − −= = p

3 914 2 41 ( )1 1

4 4a a+ − −= =

8 a x = 62 = 36 b x = 53 = 125 c

12x = 1

2 d 14x = 3

x = ( 12 )2 = 1

4 x = 34 = 81

e 12x = 3 8 = 2 f

13x = ± 16 = ± 4 g

13x = ± 4 81 = ± 3 h

32x = 1

27

x = 22 = 4 x = (± 4)3 = ± 64 x = (± 3)3 = ± 27 12x = 13

27 = 13

x = ( 13 )2 = 1

9

9 a =

12x b =

13x− c

512 22x x x= × = d

14x

x=

34x−=

e 31

2 23( )x x= = f 511

3 62x x x= × = g 51

2 25( )x x= = h 132 3

3 62x x x= × =

10 a

124x− b 11

2 x− c 334 x− d 21

9 x− e 132

5 x− f 321

3 x−

11 a = (23)2 = 26 b = (2−2)−2 = 24 c =

1 13 31(2 ) 2−− =

d = 1 26 34(2 ) 2− −= e =

625 53(2 ) 2= f = (2−5)−3 = 215

12 a = (32)x = 32x b = (34)x + 1 = 34x + 4 c =

34 43(3 ) 3x x=

d = (3−1)x = 3−x e = (32)2x − 1 = 34x − 2 f = (3−3)x + 2 = 3−3x − 6

13 a = 2 × 2x = 2y b = 2−2 × 2x = 1

4 y c = (2x)2 = y2

d = (23)x = 23x = (2x)3 = y3 e = 23 × 24x = 8y4 f = (2−1)x − 3 = 23 × 2−x = 8y

PMT

C1 ALGEBRA Answers - Worksheet B page 3

Solomon Press

14 a 2x = 26 b 5x − 1 = 53 c 3x + 4 = 27 = 33 d (23)x = 23x = 2 x = 6 x − 1 = 3 x + 4 = 3 3x = 1 x = 4 x = −1 x = 1

3

e 32x − 1 = 32 f 16 = 42 = 43x − 2 g (32)x − 2 = 32x − 4 = 33 h (23)2x + 1 = 26x + 3 = 24 2x − 1 = 2 2 = 3x − 2 2x − 4 = 3 6x + 3 = 4 x = 3

2 x = 43 x = 7

2 x = 16

i (72)x+1 = 72x+2 = 127 j 33x − 2 =

1 23 32(3 ) 3= k (6−1)x+3 = 6−x−3 = 62 l (2−1)3x − 1 = 21 − 3x = 23

2x + 2 = 12 3x − 2 = 2

3 −x − 3 = 2 1 − 3x = 3 x = 3

4− x = 89 x = −5 x = 2

3−

15 a 2x + 3 = (22)x = 22x b 53x = (52)x+1 = 52x+2 c (32)2x = 34x = 3x − 3 d (42)x = 42x = 41 − x x + 3 = 2x 3x = 2x + 2 4x = x − 3 2x = 1 − x x = 3 x = 2 x = −1 x = 1

3 e (22)x + 2 = (23)x f (33)2x = (32)3 − x g 63x − 1 = (62)x + 2 h (23)x = (24)2x − 1 22x + 4 = 23x 36x = 36 − 2x 63x − 1 = 62x + 4 23x = 28x − 4 2x + 4 = 3x 6x = 6 − 2x 3x − 1 = 2x + 4 3x = 8x − 4 x = 4 x = 3

4 x = 5 x = 45

i (53)x = 5x − 3 j (3−1)x = 3x − 4 k (2−1)1 − x = (2−3)2x l (2−2)x + 1 = (23)x 53x = 5x − 3 3−x = 3x − 4 2x − 1 = 2−6x 2−2x − 2 = 23x 3x = x − 3 −x = x − 4 x − 1 = −6x −2x − 2 = 3x x = 3

2− x = 2 x = 17 x = 2

5−

16 a = x3 − 1 b = 2x2 + 6x3 c = 3 − x2 d = 12x3 + 8x

e = 3x3 + 2x f = 3 − 3x2 g = 12 25x x+ h = 3x2 − x−1

i = x6 + x4 − 3x2 − 3 j = 2x9 + 6x5

+ x5 + 3x k = x3 − 1 − 2 + 2x−3 l =

5 52 23 2x x x x− − +

= 2x9 + 7x5 + 3x = x3 − 3 + 2x−3 =523 22x x x− +

17 a = x2 + 2 b = 2t3 − 3t c =

123x x− d =

5 263

y yy

−

= 413 2y y−

e = 31

4 4p p+ f = 32 1

22w w− g = 12 ( 1)

1x x

x+

+ h =

12 2

22 ( 2)

2t t t

t× −

−

= 12x =

322t

PMT

Solomon Press

ALGEBRA C1 Answers - Worksheet A

1 a = 7 b = 11 c = 1

3

d = 25 e = 0.1 f = 0.3

g = 2 h = 10 i = 3

j = 2516 = 5

4 or 141 k = 13

8 = 12 or 0.5 l = 1253

8 = 52 or 1

22

2 a = 7 b = 20 c = 27 d = 36

e = 4 2 f = 24 3 g = 16 = 4 h = 2 81 = 18

i = 16 = 4 j = 14 = 1

2 k = 6 l = 54

3 a 9 2 3 2= × = b 25 2 5 2= × = c 4 2 2 2= × =

d 49 2 7 2= × = e 100 2 10 2= × = f 81 2 9 2= × =

4 a 4 3 2 3= × = b 4 7 2 7= × = c 16 5 4 5= × =

d 9 3 3 3= × = e 4 6 2 6= × = f 64 2 8 2= × =

g 9 5 3 5= × = h 4 10 2 10= × = i 25 3 5 3= × =

j 16 7 4 7= × = k 9 11 3 11= × = l 49 3 7 3= × =

m 36 6 6 6= × = n 400 2 20 2= × = o 36 5 6 5= × =

p 4 15 2 15= × = q 121 3 11 3= × = r 16 13 4 13= × =

5 a = 3 2 5 2 8 2+ = b = 4 3 3 3 3− = c = 4 2 6 2 10 2+ =

d = 6 10 4 10 2 10− = e = 2 5 3 5 6 5 5 5− + = f = 2 6 5 6 8 6 6+ − = −

6 a = 3 + 2 3 b = 4 − 3 − 2 + 2 3 c = 2 + 3 + 2 3 + 3 = 2 + 3 = 5 + 3 3

d = 4 + 8 3 + 3 + 6 e = 27 − 24 3 + 16 f = 6 3 − 45 + 2 − 5 3 = 10 + 9 3 = 43 − 24 3 = −43 + 3

7 a = 10 + 3 5 + 2 5 + 3 b = 4 2 − 3 − 8 + 3 2 c = 28 + 12 7 + 9 = 13 + 5 5 = 7 2 − 11 = 37 + 12 7

d = 12 + 15 2 − 2 2 − 5 e = 5 + 2 10 − 10 − 4 f = (3 − 2 2 )(4 + 2 ) = 7 + 13 2 = 1 + 10 = 12 + 3 2 − 8 2 − 4 = 8 − 5 2

PMT

C1 ALGEBRA Answers - Worksheet A page 2

Solomon Press

8 a = 1 55 5

× = 15 5 b = 2 3

3 3× = 2

3 3 c = 1 22 2 2

× = 14 2

d = 14 77 7

× = 2 7 e = 3 2 33 3

× = 6 f = 5 1 33 5 3 3

×= = 13 3

g = 1 73 7 7

× = 121 7 h = 12 2

6 2 2× = 2 i = 1 5

4 5 5× = 1

20 5

j = 3 66 6 6

× = 112 6 k = 8 5 2

9 2 2× = 4

9 10 l = 15 7 36 3 3

× = 56 21

9 a = 2 2 + 6 22 2

× b = 4 3 − 10 33 3

× c = 6 2 2 22 2

− ×

= 2 2 + 3 2 = 4 3 − 103 3 = 6 2 4

2−

= 5 2 = 23 3 = 3 2 − 2

d = 3 5 5 52 5 5

− × e = 1 23 2 2

× + 1 24 2 2

× f = 2 33 3

× − 2 36 2

= 15 5 510− = 1

6 2 + 18 2 = 2

3 3 − 16 3

= 12 (3 − 5 ) = 7

24 2 = 12 3

10 a x2 + 4x = 4x + 32 b x − 4 3 = 2 3 − 2x x2 = 32 3x = 6 3 x = ± 32 x = 2 3 x = ± 4 2

c 3 2 x − 4 = 2 2 d 5 x + 2 = 2 5 (x − 1) 6x − 4 2 = 4 5x + 2 5 = 10(x − 1) 6x = 4 + 4 2 5x = 10 + 2 5 x = 2

3 (1 + 2 ) x = 2 + 25 5

11 a = 4 − ( 3 )2 = 4 − 3 = 1

b = 2 2 32 3 2 3

+×− +

= 2(2 3)1+ = 4 + 2 3

12 a 1 2 1 2 12 12 1 2 1

− −= × =−+ −

= 2 1−

b 4 3 1 4( 3 1)3 13 1 3 1

+ += × =−− +

= 2( 3 1)+

c 1 6 2 6 26 46 2 6 2

+ += × =−− +

= 1 12 2( 6 2) or 6 1+ +

d 3 2 3 3(2 3)4 32 3 2 3

− −= × =−+ −

= 3(2 3)−

e 1 2 5 2 54 52 5 2 5

− −= × =−+ −

= 5 2−

PMT

C1 ALGEBRA Answers - Worksheet A page 3

Solomon Press

f 2 2 1 2( 2 1)2 12 1 2 1

+ += × =−− +

= 2 2+

g 6 7 3 6( 7 3)7 97 3 7 3

− −= × =−+ −

= 3(3 7)−

h 1 3 2 2 3 2 29 83 2 2 3 2 2

− −= × =−+ −

= 3 2 2−

i 1 4 2 3 4 2 316 124 2 3 4 2 3

+ += × =−− +

= 1 12 2(2 3) or 1 3+ +

j 3 3 2 4 3(3 2 4)18 163 2 4 3 2 4

− −= × =−+ −

= 3 92 2(3 2 4) or 2 6− −

k 2 3 7 4 3 2 3(7 4 3)49 487 4 3 7 4 3

+ += × =−− +

= 2(7 3 12)+

l 6 5 3 6( 5 3)5 35 3 5 3

+ += × =−− +

= 3( 5 3)+

13 3x = 5 x + 2 5 x(3 − 5 ) = 2 5

x = 2 53 5−

= 2 53 5−

× 3 53 5

++

= 2 5(3 5)9 5

+−

x = 6 5 104+ = 5

2 + 32 5

14 l = 63 2 3−

= 63 2 3−

× 3 2 33 2 3

++

= 6(3 2 3)18 9

+−

l = 18( 2 1)9

+ = 2 2 + 2

15 a 2 2 6 2( 2 6)2 62 6 2 6

− −= × =−+ −

= 1 14 2(2 2 3) ( 3 1)− − = −

b 1 3 2 3 (1 3)(2 3)4 32 3 2 3

+ − + −= × =−+ −

= 2 3 2 3 3 3 1− + − = −

c 1 10 10 3 (1 10)( 10 3)10 910 3 10 3

+ + + += × =−− +

= 10 3 10 3 10 13 4 10+ + + = +

d 3 2 4 3 2 (3 2)(4 3 2) 12 9 2 4 2 616 18 24 3 2 4 3 2

− − − − − − += × = =− −+ −

= 1312 2(13 2 18) or 2 9− −

e 1 2 3 2 2 (1 2)(3 2 2)9 83 2 2 3 2 2

− + − += × =−− +

= 3 2 2 3 2 4 1 2+ − − = − −

f 3 5 2 3 4 ( 3 5)(2 3 4) 6 4 3 10 3 2012 16 42 3 4 2 3 4

− + − + + − −= × = =− −− +

= 12 (7 3 3)+

g 2 3 3 3 3 (2 3 3)(3 3) 6 3 6 9 3 39 3 63 3 3 3

+ + + + + + += × = =−− +

= 12 (3 3 5)+

h 3 7 2 2 7 5 (3 7 2)(2 7 5) 42 15 7 4 7 1028 25 32 7 5 2 7 5

− + − + + − −= × = =−− +

= 13 (32 11 7)+

PMT

Solomon Press

ALGEBRA C1 Answers - Worksheet C

1 a = 9 3 2 25 2+ 2 x2 − 2x = 12 − 2x = 10 2 3 3+ x2 = 12 b = 18 48− x = ± 12 = ± 2 3 = 9 2 16 3− x > 0 ∴ x = 2 3

= 3 2 4 3−

3 25x = (52)x = 54x + 1 4 a = 3 8 × 3 3 = 2 3 3 52x = 54x + 1 b 3 81 = 3 27 × 3 3 = 3 3 3 2x = 4x + 1 ∴ 3 24 + 3 81 = 2 3 3 + 3 3 3 = 5 3 3 x = 1

2− = 3 125 3× = 3 375 ∴ n = 375

5 10 3

15 = 10 3 10 10 5

5 3 5 5 5= = × = 2 5 6 a = 75 5 3 5 3 2

2 2 2 2= = × = 5

2 6

4 5 7 4( 5 7)5 75 7 5 7

+ +× =−− +

= 2 5 2 7− − b = 48 20 4 3 2 55 3 5 3

− = −

∴ 10 315

+ 45 7−

= 2 5 2 5 2 7− − = 4 3 5 2 5 35 5 3 3

× − ×

= 2 7− [ k = −2 ] = 4 25 315 15−

= 215 15

7 a i xy = 2t − 1 × 23t = 24t − 1 8 3x 2 − 2 = 4x + 6 ii 2y2 = 2 × (23t)2 = 2 × 26t = 26t + 1 x(3 2 − 4) = 6 + 2

b 26t + 1 − 24t − 1 = 0 x = 6 2 6 2 3 2 43 2 4 3 2 4 3 2 4

+ + += ×− − +

= (6 2)(3 2 4)18 16

+ +−

26t + 1 = 24t − 1 = 12 (18 2 + 24 + 6 + 4 2 )

6t + 1 = 4t − 1 = 12 (30 + 22 2 )

t = −1 = 15 + 11 2

9 a 6 y + 1 = 36x − 2 = (62)x − 2 10 a = 3 + 3 2 − 2 − 2 6 y + 1 = 62x − 4 = 1 + 2 2 y + 1 = 2x − 4 b = 2 2 1 2( 2 1)

2 12 1 2 1+ +× =

−− +

y = 2x − 5 = 2 ( 2 + 1) b x − 1

2 y = x − 12 (2x − 5) = x − x + 5

2 = 52 = 2 + 2

∴ 1 24x y− =

52 54 ( 4)= = 25 = 32

PMT

C1 ALGEBRA Answers - Worksheet C page 2

Solomon Press

11 (24)x + 1 = (23)2x + 1 12 a2 − 4a 3 + 12 = b − 20 3 24x + 4 = 26x + 3 a and b integers ∴ −4a = −20 4x + 4 = 6x + 3 a = 5 x = 1

2 also a2 + 12 = b b = 37

13 a (2−2)t − 3 = 23 14 a = 2 5 ( 5 − 3) 26 − 2t = 23 = 10 − 6 5 6 − 2t = 3 b = 3 + 2 5 − 3 5 − 10 t = 3

2 = −7 − 5

b (3−1)y = (33)y + 1 c = 1 5 5 2 (1 5)( 5 2)5 45 2 5 2

+ + + +× =−− +

3−y = 33y + 3 = (1 + 5 )( 5 + 2) −y = 3y + 3 = 5 + 2 + 5 + 2 5 y = 3

4− = 7 + 3 5

15 a a =

3 94 43( )b b= 16 a area = 1

2 (2 3 − 1)( 3 + 2) 991 182 4 2( )a b b= = = 1

2 (6 + 4 3 − 3 − 2)

b b = 1 4 43 3 9( )a a= = 1

2 (4 + 3 3 ) or 2 + 32 3

4 21 19 92 2( )b a a= = b AC2 = (2 3 − 1)2 + ( 3 + 2)2

= 12 − 4 3 + 1 + 3 + 4 3 + 4 = 20 ∴ AC = 20 = 4 5 = 2 5

c tan (∠ACB) = 2 3 1 3 2 (2 3 1)( 3 2)3 43 2 3 2

− − − −× =−+ −

= −(2 3 − 1)( 3 − 2) = −(6 − 4 3 − 3 + 2)

= −(8 − 5 3 ) = 5 3 − 8

17 a i 2x + 2 = 22 × 2x = 4y 18 5 3 = 2(1 + 3 )2 + p(1 + 3 ) + q ii 4x = (22)x = 22x = (2x)2 = y2 5 3 = 2 + 4 3 + 6 + p + p 3 + q

b y2 − 4y = 0 p, q rational ∴ 5 3 = 4 3 + p 3 y(y − 4) = 0 p = 1 y = 0 or 4 and 0 = 2 + 6 + p + q 2x = 0 (no solutions) or 2x = 4 q = −9 x = 2

PMT

Solomon Press

ALGEBRA C1 Answers - Worksheet E

1 a (x + 1)(x + 3) b (x + 2)(x + 5) c (y − 1)(y − 2) d (x − 3)2

e (y + 1)(y − 2) f (a + 4)(a − 2) g (x + 1)(x − 1) h (p + 2)(p + 7) i (x + 3)(x − 5) j (m − 2)(m − 8) k (t + 6)(t − 3) l (y − 5)(y − 8) m (r + 4)(r − 4) n (y + 7)(y − 9) o (a + 11)2 p (x + 12)(x − 6) q (x − 2)(x − 13) r (s + 8)(s + 15) s (p + 17)(p − 3) t (m − 10)(m + 9)

2 a (2x + 1)(x + 1) b (3p + 1)(p + 2) c (2y − 3)(y − 1) d (2 + m)(1 − m) e (3r + 1)(r − 1) f (5 + y)(1 − 4y) g (3a − 1)(a − 4) h (5x + 2)(x − 2) i (2x + 1)(2x + 3) j (3s − 1)2 k (2m + 5)(2m − 5) l (2 + 3y)(1 − 2y) m (4u + 1)(u + 4) n (3p + 4)(2p − 1) o (8x + 3)(x + 2) p (6r − 5)(2r + 3)

3 a (x − 1)(x − 3) = 0 b (x + 4)(x + 2) = 0 c (x + 5)(x − 1) = 0 d x2 − 7x − 8 = 0 x = 1 or 3 x = −4 or −2 x = −5 or 1 (x + 1)(x − 8) = 0 x = −1 or 8 e (x + 5)(x − 5) = 0 f x2 − x − 42 = 0 g x2 − 3x = 0 h (x + 9)(x + 3) = 0 x = −5 or 5 (x + 6)(x − 7) = 0 x(x − 3) = 0 x = −9 or −3 x = −6 or 7 x = 0 or 3 i x2 + 4x − 60 = 0 j x2 − 5x − 14 = 0 k (2x − 1)(x − 1) = 0 l x2 − x = 6x − 12 (x + 10)(x − 6) = 0 (x + 2)(x − 7) = 0 x = 1

2 or 1 x2 − 7x + 12 = 0 x = −10 or 6 x = −2 or 7 (x − 3)(x − 4) = 0

x = 3 or 4 m 3x2 + 11x − 4 = 0 n 2x2 − 3x − 5 = 0 o 4x2 − 23x − 6 = 0 p 6x2 − 19x + 10 = 0 (3x − 1)(x + 4) = 0 (2x − 5)(x + 1) = 0 (4x + 1)(x − 6) = 0 (3x − 2)(2x − 5) = 0 x = −4 or 1

3 x = −1 or 52 x = 1

4− or 6 x = 23 or 5

2 q (2x + 1)2 = 0 r 3x2 − 13x + 12 = 0 s 4x2

+20x+25 = 5−x t 6x2 − 21x = 14x + 6 x = 1

2− (3x − 4)(x − 3) = 0 4x2 + 21x + 20 = 0 6x2 − 35x − 6 = 0 x = 4

3 or 3 (4x + 5)(x + 4) = 0 (6x + 1)(x − 6) = 0 x = −4 or 5

4− x = 16− or 6

4 a = 2(y2 − 5y + 6) b = x(x2 + x − 2) c = p(p2 − 4) d = 3m(m2 + 7m + 6) = 2(y − 3)(y − 2) = x(x − 1)(x + 2) = p(p + 2)(p − 2) = 3m(m + 1)(m + 6) e = (a2 + 1)(a2 + 3) f = (t 2 + 5)(t 2 − 2) g = 4(3 + 5x − 2x2) h = 3(2r2 − 3r − 14) = 4(3 − x)(1 + 2x) = 3(2r − 7)(r + 2) i = 2x(3x2 − 13x + 4) j = y2(y2 + 3y − 18) k = (m2 + 1)(m2 − 1) l = p(p4 − 4p2 + 4) = 2x(3x − 1)(x − 4) = y2(y + 6)(y − 3) =(m2

+1)(m +1)(m −1) = p(p2 − 2)2

PMT

C1 ALGEBRA Answers - Worksheet E page 2

Solomon Press

5 a x2 − 3x + 2 = 0 b x2 + 5x + 6 = 0 c x2 − 9 = 0 (x − 1)(x − 2) = 0 (x + 3)(x + 2) = 0 (x + 3)(x − 3) = 0 x = 1 or 2 x = −3 or −2 x = −3 or 3 y y y (0, 6)

(0, 2) (−3, 0) (3, 0)

O x

O (1, 0) (2, 0) x (−3, 0) (−2, 0) O x (0, −9)

d x2 − 2x = 0 e x2 − 10x + 25 = 0 f 2x2 − 14x + 20 = 0 x(x − 2) = 0 (x − 5)2 = 0 2(x − 2)(x − 5) = 0 x = 0 or 2 x = 5 x = 2 or 5 y y y (0, 25) (0, 20) (0, 0)

O (2, 0) x O (5, 0) x O (5, 0) x (2, 0)

g −x2 + 5x − 4 = 0 h 2 + x − x2 = 0 i 2x2 − 3x + 1 = 0 x2 − 5x + 4 = 0 x2 − x − 2 = 0 (2x − 1)(x − 1) = 0 (x − 1)(x − 4) = 0 (x + 1)(x − 2) = 0 x = 1

2 or 1 x = 1 or 4 x = −1 or 2 y y y

(0, 2)

O (1, 0) (4, 0) x (0, 1) (−1, 0) (2, 0)

(0, −4) O x

O (1, 0) x ( 1

2 , 0)

j 2x2 + 13x + 6 = 0 k 3 − 8x + 4x2 = 0 l 2 + 7x − 4x2 = 0 (2x + 1)(x + 6) = 0 (2x − 1)(2x − 3) = 0 4x2 − 7x − 2 = 0 x = −6 or 1

2− x = 12 or 3

2 (4x + 1)(x − 2) = 0 x = 1

4− or 2 y y y (0, 6) (0, 3) (0, 2)

(−6, 0) O x O ( 32 , 0) x ( 1

4− , 0) O (2, 0) x

( 12− , 0) ( 1

2 , 0)

PMT

C1 ALGEBRA Answers - Worksheet E page 3

Solomon Press

m 5x2 − 17x + 6 = 0 n −6x2 + 7x − 2 = 0 o 6x2 + x − 5 = 0 (5x − 2)(x − 3) = 0 6x2 − 7x + 2 = 0 (6x − 5)(x + 1) = 0 x = 2

5 or 3 (2x − 1)(3x − 2) = 0 x = −1 or 56

x = 12 or 2

3 y y y

( 12 , 0)

O ( 23 , 0) x (−1, 0) O ( 5

6 , 0) x

O (3, 0) x (0, −2) (0, −5) ( 2

5 , 0)

6 a x2 − 5x + 4 = 0 b x2 − 10 = 3x c x(2x2 − x − 3) = 0 d 10x2 − x4 = 9 (x − 1)(x − 4) = 0 x2 − 3x − 10 = 0 x(2x − 3)(x + 1) = 0 x4 − 10x2 + 9 = 0 x = 1 or 4 (x + 2)(x − 5) = 0 x = −1, 0 or 3

2 (x2 − 1)(x2 − 9) = 0 x = −2 or 5 x2 = 1 or 9 x = ± 1 or ± 3 e 5 + 4x − x2 = 0 f x − 6 = x(x − 4) g (x + 5)(x + 3) = 3 h x4 − 4 = 3x2 x2 − 4x − 5 = 0 x − 6 = x2 − 4x x2 + 8x + 15 = 3 x4 − 3x2 − 4 = 0 (x + 1)(x − 5) = 0 x2 − 5x + 6 = 0 x2 + 8x + 12 = 0 (x2 + 1)(x2 − 4) = 0 x = −1 or 5 (x − 2)(x − 3) = 0 (x + 6)(x + 2) = 0 x2 = −1 (no sol’s) or 4 x = 2 or 3 x = −6 or −2 x = ± 2 i 4x4 + 7x2 − 2 = 0 j 2x(x + 2) = 3 − x k x(2x + 1) = 2(x + 3) l 7 − 3x(x + 2) = 2(x + 2) (4x2 − 1)(x2 + 2) = 0 2x2 + 4x = 3 − x 2x2 + x = 2x + 6 7 − 3x2 − 6x = 2x + 4 x2 = −2 (no sol’s) or 1

4 2x2 + 5x − 3 = 0 2x2 − x − 6 = 0 3x2 + 8x − 3 = 0

x = ± 12 (2x − 1)(x + 3) = 0 (2x + 3)(x − 2) = 0 (3x − 1)(x + 3) = 0

x = −3 or 12 x = 3

2− or 2 x = −3 or 13

(0, 6)

PMT

Solomon Press

ALGEBRA C1 Answers - Worksheet G

1 ax2 + bx + c = 0 x2 + b

ax + c

a = 0

(x + 2ba

)2 − 2

24ba

+ ca

= 0

(x + 2ba

)2 = 2

24ba

− ca

= 2

24

4b ac

a−

x + 2ba

= ±2

24

4b ac

a− = ±

2 42

b aca−

x = −2ba

±2 42

b aca− =

2 42

b b aca

− ± −

2 a x = 4 16 4

2− ± − b t = 8 64 16

2− ± +

− c y = 20 400 364

2± − d r = 2 4 28

2− ± +

x = 4 2 32

− ± t = 8 4 52

− ±−

y = 20 62± r = 2 4 2

2− ±

x = −2 ± 3 t = 4 ± 2 5 y = 7 or 13 r = −1 ± 2 2

e a = 18 324 242

− ± − f m2 − 5m − 5 = 0 g x = 11 121 1082

− ± − h u = 6 36 244

− ± −

a = 18 10 32

− ± m = 5 25 202

± + x = 12 (−11 ± 13 ) u = 6 2 3

4− ±

a = −9 ± 5 3 m = 12 (5 ± 3 5 ) u = 1

2 (−3 ± 3 )

i y = 1 1 202

± +−

j 2x2 − 3x − 2 = 0 k p = 7 49 126

− ± − l t2 − 14t − 14 = 0

y = 12− (1 ± 21 ) x = 3 9 16

4± + p = 1

6 (−7 ± 37 ) t = 14 196 562

± +

x = 3 54± t = 14 6 7

2±

x = 12− or 2 t = 7 ± 3 7

m r2 + 14r − 9 = 0 n 6u2 + 4u − 1 = 0 o 3y2 − 18y − 4 = 0 p 4x2 − 8x − 11 = 0

r = 14 196 362

− ± + u = 4 16 2412

− ± + y = 18 324 486

± + x = 8 64 1768

± +

r = 14 2 582

− ± u = 4 2 1012

− ± y = 18 2 936

± x = 8 4 158

±

r = −7 ± 58 u = 16 (−2 ± 10 ) y = 3 ± 1

3 93 x = 1 ± 12 15

3 2x2 − 8x + 3 = 0

x = 8 64 244

± − = 8 2 104

± = 2 ± 12 10

∴ (2 − 12 10 , 0) and (2 + 1

2 10 , 0)

PMT

C1 ALGEBRA Answers - Worksheet G page 2

Solomon Press

4 a b2 − 4ac > 0 b b2 − 4ac = 0 c b2 − 4ac < 0

5 a b c d x x x x

6 a b2 − 4ac = 32 b b2 − 4ac = −11 c b2 − 4ac = −4 d b2 − 4ac = 24 ∴ real and distinct ∴ not real ∴ not real ∴ real and distinct e b2 − 4ac = 0 f b2 − 4ac = 13 g b2 − 4ac = 53 h b2 − 4ac = −7 ∴ real and equal ∴ real and distinct ∴ real and distinct ∴ not real i b2 − 4ac = 4 j b2 − 4ac = −11 k b2 − 4ac = 0 l b2 − 4ac = −3 ∴ real and distinct ∴ not real ∴ real and equal ∴ not real m b2 − 4ac = −7 n b2 − 4ac = 13

9 o b2 − 4ac = 116 p b2 − 4ac = 13

75− ∴ not real ∴ real and distinct ∴ real and distinct ∴ not real

7 equal roots

∴ b2 − 4ac = 0 1 − 4p = 0 p = 1

4

8 repeated root

∴ b2 − 4ac = 0 4q2 + 4q = 0 4q(q + 1) = 0 q ≠ 0 ∴ q = −1

9 x2 + rx − 2x + 4 = 0 has equal roots

∴ b2 − 4ac = 0 (r − 2)2 − 16 = 0 r2 − 4r − 12 = 0 (r + 2)(r − 6) = 0 r = −2 or 6

PMT

Solomon Press

ALGEBRA C1 Answers - Worksheet F

1 a = (x + 1)2 − 1 + 4 b = (x − 1)2 − 1 + 4 c = (x − 2)2 − 4 + 1 d = (x + 3)2 − 9 = (x + 1)2 + 3 = (x − 1)2 + 3 = (x − 2)2 − 3 e = (x + 2)2 − 4 + 8 f = (x − 4)2 − 16 − 5 g = (x + 6)2 − 36 + 30 h = (x − 5)2 − 25 + 25 = (x + 2)2 + 4 = (x − 4)2 − 21 = (x + 6)2 − 6 = (x − 5)2 i = (x + 3)2 − 9 − 9 j = (x − 2)2 − 4 + 18 k = (x + 3

2 )2 − 94 + 3 l = (x + 1

2 )2 − 14 − 1

= (x + 3)2 − 18 = (x − 2)2 + 14 = (x + 32 )2 + 3

4 = (x + 12 )2 − 5

4 m = (x − 9)2

− 81 + 100 n = (x − 12 )2 − 1

4 − 12 o = (x +

92 )2 − 81

4 + 20 p = (x − 72 )2 − 49

4 − 2 = (x − 9)2 + 19 = (x − 1

2 )2 − 34 = (x + 9

2 )2 − 14 = (x − 7

2 )2 − 574

q = (x − 3

2 )2 − 94 + 5 r = (x −

112 )2

− 1214 + 37 s = (x + 1

3 )2 − 19 + 1 t = (x − 1

4 )2 − 116 − 1

4 = (x − 3

2 )2 + 114 = (x − 11

2 )2 + 274 = (x + 1

3 )2 + 89 = (x − 1

4 )2 − 516

2 a = 2[x2 + 2x] + 3 b = 2[x2 − 4x] − 7 c = 3[x2 − 2x] + 3 d = 4[x2 + 6x] + 11 = 2[(x + 1)2 − 1] + 3 = 2[(x − 2)2 − 4] − 7 = 3[(x − 1)2 − 1] + 3 = 4[(x + 3)2 − 9] + 11 = 2(x + 1)2 + 1 = 2(x − 2)2 − 15 = 3(x − 1)2 = 4(x + 3)2 − 25 e = −[x2 + 2x] − 5 f = −[x2 − 10x] + 1 g = 2[x2 + x] − 1 h = 3[x2 − 3x] + 5 = −[(x + 1)2 − 1] − 5 = −[(x − 5)2 − 25] +1 = 2[(x + 1

2 )2 − 14 ] −1 = 3[(x − 3

2 )2 − 94 ] + 5

= −(x + 1)2 − 4 = −(x − 5)2 + 26 = 2(x + 12 )2 − 3

2 = 3(x − 32 )2 − 7

4 i = 3[x2 − 8x] + 48 j = 3[x2 − 5x] k = 5[x2 + 8x] + 70 l = 2[x2 + 5

2 x] + 2 = 3[(x − 4)2

−16] + 48 = 3[(x − 52 )2 − 25

4 ] = 5[(x + 4)2 −16] + 70 = 2[(x +

54 )2 −

2516 ] + 2

= 3(x − 4)2 = 3(x − 52 )2 − 75

4 = 5(x + 4)2 − 10 = 2(x + 54 )2 − 9

8 m = 4[x2 + 3

2 x] − 7 n = −2[x2 − 2x] − 1 o = −3[x2 + 23 x] + 4 p = 1

3 [x2 + 32 x] − 1

4 = 4[(x +

34 )2 − 9

16 ] −7 = −2[(x − 1)2 −1] − 1 = −3[(x + 13 )2

− 19 ] +4 = 1

3 [(x + 34 )2 − 9

16 ] − 14

= 4(x + 34 )2 − 37

4 = −2(x − 1)2 + 1 = −3(x + 13 )2 + 13

3 = 13 (x + 3

4 )2 − 716

3 a (y − 2)2 − 4 + 2 = 0 b (p + 1)2 − 1 − 2 = 0 c (x − 3)2 − 9 + 4 = 0 d (r + 5)2 − 25 + 7 = 0 (y − 2)2 = 2 (p + 1)2 = 3 (x − 3)2 = 5 (r + 5)2 = 18 y − 2 = ± 2 p + 1 = ± 3 x − 3 = ± 5 r + 5 = 18± = 3 2± y = 2 ± 2 p = −1 ± 3 x = 3 ± 5 r = −5 ± 3 2 e (x − 1)2 − 1 = 11 f (a − 6)2 − 36 − 18 = 0 g (m − 3

2 )2 − 94 +1 = 0 h (t − 7

2 )2 − 494 + 9 = 0

(x − 1)2 = 12 (a − 6)2 = 54 (m − 32 )2 = 5

4 (t − 72 )2 = 13

4

x − 1= 12± = 2 3± a − 6 = 54± = 3 6± m − 32 = ± 5

2 t − 72 = ± 13

2

x = 1 ± 2 3 a = 6 ± 3 6 m = 12 (3 ± 5 ) t = 1

2 (7 ± 13 )

PMT

C1 ALGEBRA Answers - Worksheet F page 2

Solomon Press

i (u + 7

2 )2 − 494 = 44 j y2 − 2y + 1

2 = 0 k p2 + 6p = 233− l x2 + 6x = 9

2 (u + 7

2 )2 = 2254 (y − 1)2 − 1 + 1

2 = 0 (p + 3)2 − 9 = 233− (x + 3)2 − 9 = 9

2 u + 7

2 = ± 152 (y − 1)2 = 1

2 (p + 3)2 = 43 (x + 3)2 = 27

2

u = − 72 ± 15

2 y − 1 = ± 12

= ± 12 2 p + 3 = ± 2

3 = ± 2

3 3 x + 3 = ± 272 = ± 3

2 6

u = −11 or 4 y = 1 ± 12 2 p = −3 ± 2

3 3 x = −3 ± 32 6

m m2 − m = 1 n 4x2 − 28x + 49 = 0 o t2 + 1

3 t = 13 p a2 − 7

2 a + 2 = 0 (m − 1

2 )2 − 14 = 1 x2 − 7x + 49

4 = 0 (t + 16 )2 − 1

36 = 13 (a − 7

4 )2 − 4916 + 2 = 0

(m − 12 )2 = 5

4 (x −

72 )2

− 494 +

494 = 0 (t + 1

6 )2 = 1336 (a − 7

4 )2 = 1716

m − 12 = ± 5

2 (x − 72 )2 = 0 t + 1

6 = ± 136 a − 7

4 = 174

m = 12 (1 ± 5 ) x = 7

2 t = 16 (−1 ± 13 ) a = 1

4 (7 ± 17 )

4 a y = (x − 1)2 − 1 + 7 b y = (x + 1)2 − 1 − 3 c y = (x − 3)2 − 9 + 1 y = (x − 1)2 + 6 y = (x + 1)2 − 4 y = (x − 3)2 − 8 y = 6 at x = 1, minimum y = −4 at x = −1, minimum y = −8 at x = 3, minimum d y = (x + 5)2 − 25 + 35 e y = −[x2 − 4x] + 4 f y = (x + 3

2 )2 − 94 − 2

y = (x + 5)2 + 10 y = −[(x − 2)2 − 4] + 4 y = (x + 32 )2 − 17

4 y = 10 at x = −5, minimum y = −(x − 2)2 + 8 y = − 17

4 at x = − 32 , minimum

y = 8 at x = 2, maximum g y = 2[x2 + 4x] + 5 h y = −3[x2 − 2x] i y = −[x2 + 5x] + 7 y = 2[(x + 2)2 − 4] + 5 y = −3[(x − 1)2 − 1] y = −[(x + 5

2 )2 − 254 ] + 7

y = 2(x + 2)2 − 3 y = −3(x − 1)2 + 3 y = −(x + 52 )2 + 53

4 y = −3 at x = −2, minimum y = 3 at x = 1, maximum y = 53

4 at x = − 52 , maximum

j y = 4[x2 − 3x] + 9 k y = 4[x2 + 5x] − 8 l y = −2[x2 + x] + 17 y = 4[(x − 3

2 )2 − 94 ] + 9 y = 4[(x + 5

2 )2 − 254 ] − 8 y = −2[(x + 1

2 )2 − 14 ] + 17

y = 4(x − 32 )2 y = 4(x + 5

2 )2 − 33 y = −2(x + 12 )2 + 35

2 y = 0 at x = 3

2 , minimum y = −33 at x = − 52 , minimum y = 35

2 at x = − 12 , maximum

5 a y = (x − 2)2 − 4 + 3 b y = (x + 1)2 − 1 − 24 c y = (x − 1)2 − 1 + 5 y = (x − 2)2 − 1 y = (x + 1)2 − 25 y = (x − 1)2 + 4 minimum (2, −1) minimum (−1, −25) minimum (1, 4)

y y y

(0, 3) O x (0, 5) O x (0, −24) O x

(2, −1) (−1, −25)

(1, 4)

PMT

C1 ALGEBRA Answers - Worksheet F page 3

Solomon Press

d y = (x + 4)2 − 16 + 30 e y = (x + 1)2 − 1 + 1 f y = −[x2 − 2x] + 8 y = (x + 4)2 + 14 y = (x + 1)2 y = −[(x − 1)2 − 1] + 8 minimum (−4, 14) minimum (−1, 0) y = −(x − 1)2 + 9 maximum (1, 9)

y y y

(0, 30) (0, 8)

(−4, 14) (0, 1)

O x (−1, 0) O x O x g y = −[x2 − 8x] − 7 h y = −[x2 + 4x] − 7 i y = (x − 5

2 )2 − 254 + 4

y = −[(x − 4)2 − 16] − 7 y = −[(x + 2)2 − 4] − 7 y = (x − 52 )2 − 9

4 y = −(x − 4)2 + 9 y = −(x + 2)2 − 3 minimum ( 5

2 , − 94 )

maximum (4, 9) maximum (−2, −3)

y y y (4, 9)

(−2, −3) O x (0, 4) O x (0, −7) (0, −7) O x

j y = (x + 32 )2 − 9

4 + 3 k y = 4[x2 + 2x] + 3 l y = −2[x2 − 4x] − 15 y = (x + 3

2 )2 + 34 y = 4[(x + 1)2 − 1] + 3 y = −2[(x − 2)2 − 4] − 15

minimum ( 32− , 3

4 ) y = 4(x + 1)2 − 1 y = −2(x − 2)2 − 7 minimum (−1, −1) maximum (2, −7)

y y y (0, 3) (0, 3) O (2, −7) x ( 3

2− , 34 )

O x O x

m y = −2[x2 + 1

2 x] + 1 n y = 4[x2 − 5x] + 25 o y = 3[x2 − 43 x] + 2

y = −2[(x + 14 )2 − 1

16 ] + 1 y = 4[(x − 52 )2 − 25

4 ] + 25 y = 3[(x − 23 )2 − 4

9 ] + 2 y = −2(x + 1

4 )2 + 98 y = 4(x − 5

2 )2 y = 3(x − 23 )2 + 2

3 maximum ( 1

4− , 98 ) minimum ( 5

2 , 0) minimum ( 23 , 2

3 )

y y y

(0, 1) (0, 25) (0, 2) ( 2

3 , 23 )

O x O ( 52 , 0) x O x

6 a = (x − 2 2 )2 − 8 + 5 7 x2 + 2kx − 3 = 0 = (x − 2 2 )2 − 3 (x + k)2 − k2 − 3 = 0 b x = 2 2 (x + k)2 = k2 + 3 x + k = ± 2 3k +

x = −k ± 2 3k +

(1, 9)

(−1, −1)

(0, −15)

( 14− , 9

8 )

( 52 , 9

4− )

PMT

Difference of squares Answers

1. (x+1)(x-1)

2. (8x+2)(8x-2)

3. (6-10x)(6+10x)

4. (1+6x)(1-6x)

5. (7+2x)(7-2x)

6. (9x+3)(9x-3)

7. (3x+2)(3x-2)

8. (4x+10)(4x-10)

9. (13-5x)(13+5x)

10.(7x+5)(7x-5)

Simplify and then factor, showing all your steps.

1. 2x2 – 6 + 4x2 + 3x2 – 10 = 9x2 – 16 = (3x – 4)(3x + 4)

2. X2 – 24 – 3x2 + 28 – 7x2 = - 9x2 + 4 = 4 – 9x2 = (3x – 2)(3x + 2)

3. 4 – 2x2 + 8 – 7x2 + 13 = - 9x2 + 25 = 25 – 9x2 = (5 – 3x)(5 +3x)

4. 7 + 10x2 – 9 + 90x2 – 2 = 100x2 – 4 = (10x – 2)(10x + 2)

5. – 32 + 60x2 – 4 + 4x2 = 64x2 – 36 = (8x – 6)(8x + 6)

6. 3x2 + 6 +83x2 – 22 –5 x2 = 81x2 – 16 = (9x – 4)(9x + 4)

7. -105 + 4x2 + 5 = 4x2 – 100 = (2x – 10)(2x + 10)

8. 7x2 + 3 – 32x2 + 33 = - 25x2 + 36 =36 - 25x2 = (6 – 5x)(6 + 5x)

9. 2 – 6x2 + 2 – 58x2 + 32 = - 64x2 + 36 = 36 -64x2 = (6 – 8x)(6 + 8x)

10.10x2 – 5 – x2 + 1 + 7x2 = 16x2 – 4 = (4x – 2)(4x + 2)

Solomon Press

ALGEBRA C1 Answers - Worksheet I

1 a 3x = 2x + 1 b x − 6 = 1

2 x − 4 c 2x + 6 = 3 − 4x x = 1 x = 4 x = 1

2− ∴ x = 1, y = 3 ∴ x = 4, y = −2 ∴ x = 1

2− , y = 5 d subtracting e 2x + 4y + 22 = 0 f 6x + 6y + 8 = 0 y + 4 = 0 2x − 3y + 1 = 0 15x − 6y − 15 = 0 y = −4 subtracting adding ∴ x = 7, y = −4 7y + 21 = 0 21x − 7 = 0 y = −3 x = 1

3 ∴ x = −5, y = −3 ∴ x = 1

3 , y = 53−

2 a x + 2 = x2 − 4 b 4x + 11 = x2 + 3x − 1 c 2x − 1 = 2x2 + 3x − 7 x2 − x − 6 = 0 x2 − x − 12 = 0 2x2 + x − 6 = 0 (x + 2)(x − 3) = 0 (x + 3)(x − 4) = 0 (2x − 3)(x + 2) = 0 x = −2 or 3 x = −3 or 4 x = −2 or 3

2 ∴ (−2, 0) and (3, 5) ∴ (−3, −1) and (4, 27) ∴ (−2, −5) and ( 3

2 , 2)

3 a subtracting b adding c y = 2x − 5 x2 − x − 2 = 0 2x2 − 7x + 3 = 0 sub (x + 1)(x − 2) = 0 (2x − 1)(x − 3) = 0 x2 + (2x − 5)2 = 25 x = −1 or 2 x = 1

2 or 3 x2 − 4x = 0 ∴ x = −1, y = 4 ∴ x = 1

2 , y = 72− x(x − 4) = 0

or x = 2, y = 7 or x = 3, y = −6 x = 0 or 4 ∴ x = 0, y = −5 or x = 4, y = 3 d y = 2x + 10 e y = 1 − x f y = 1 − x sub. sub. sub. x2 + 2x(2x + 10) + 15 = 0 x2 − 2x(1 − x) − (1 − x)2 = 7 3x2 − x − (1 − x)2 = 0 x2 + 4x + 3 = 0 x2 = 4 2x2 + x − 1 = 0 (x + 3)(x + 1) = 0 x = ± 2 (2x − 1)(x + 1) = 0 x = −3 or −1 ∴ x = −2, y = 3 x = −1 or 1

2 ∴ x = −3, y = 4 or x = 2, y = −1 ∴ x = −1, y = 2 or x = −1, y = 8 or x = 1

2 , y = 12

g y = 4 − x h x = 2y i y = 3 − 3

2 x sub. sub. sub. 2x2 + x(4 − x) + (4 − x)2 = 22 (2y)2 − 4y − y2 = 0 x2 + x(3 − 3

2 x) = 4 x2 − 2x − 3 = 0 3y2 − 4y = 0 x2 − 6x + 8 = 0 (x + 1)(x − 3) = 0 y(3y − 4) = 0 (x − 2)(x − 4) = 0 x = −1 or 3 y = 0 or 4

3 x = 2 or 4 ∴ x = −1, y = 5 ∴ x = 0, y = 0 ∴ x = 2, y = 0 or x = 3, y = 1 or x = 8

3 , y = 43 or x = 4, y = −3

PMT

C1 ALGEBRA Answers - Worksheet I page 2

Solomon Press

j y = 2x − 3 k y = 2x − 7 l y = 5 − 3x sub. sub. sub. 2x2 + (2x − 3) − (2x − 3)2 = 8 x2 − x(2x − 7) + (2x − 7)2 = 13 x2 − 5x + (5 − 3x)2 = 0 x2 − 7x + 10 = 0 x2 − 7x + 12 = 0 2x2 − 7x + 5 = 0 (x − 2)(x − 5) = 0 (x − 3)(x − 4) = 0 (2x − 5)(x − 1) = 0 x = 2 or 5 x = 3 or 4 x = 1 or 5

2 ∴ x = 2, y = 1 ∴ x = 3, y = −1 ∴ x = 1, y = 2 or x = 5, y = 7 or x = 4, y = 1 or x = 5

2 , y = 52−

m x = 2y + 10 n y = 3

2 x − 2 o x = 3y − 17 sub. sub. sub. 3(2y+10)2−y(2y+10)+y2=36 2x2 + x − 4( 3

2 x − 2) = 6 (3y−17)2+(3y−17)+2y2−52=0 y2 + 10y + 24 = 0 2x2 − 5x + 2 = 0 y2 − 9y + 20 = 0 (y + 6)(y + 4) = 0 (2x − 1)(x − 2) = 0 (y − 4)(y − 5) = 0 y = −6 or −4 x = 1

2 or 2 y = 4 or 5 ∴ x = −2, y = −6 ∴ x = 1

2 , y = 54− ∴ x = −5, y = 4

or x = 2, y = −4 or x = 2, y = 1 or x = −2, y = 5 4 a subtracting b y = x − 5 c y = 7 − 4x − 1

y + 2y + 1 = 0 sub. sub.

−1 + 2y2 + y = 0 x(x − 5) = 6 3x

− 2(7 − 4x) + 4 = 0

2y2 + y − 1 = 0 x2 − 5x − 6 = 0 3 − 2x(7 − 4x) + 4x = 0 (2y − 1)(y + 1) = 0 (x + 1)(x − 6) = 0 8x2 − 10x + 3 = 0 y = −1 or 1

2 x = −1 or 6 (4x − 3)(2x − 1) = 0 ∴ x = −5, y = −1 ∴ x = −1, y = −6 x = 1

2 or 34

or x = 4, y = 12 or x = 6, y = 1 ∴ x = 1

2 , y = 5 or x = 3

4 , y = 4

5 5 − x = x2 − 3x + 2 6 3x − 1 = (32)2y ∴ x − 1 = 4y x2 − 2x − 3 = 0 (23)x − 2 = (22)1 + y ∴ 3x − 6 = 2 + 2y (x + 1)(x − 3) = 0 6x − 16 = 4y x = −1 or 3 ⇒ 6x − 16 = x − 1 P and Q are the points (−1, 6) and (3, 2) x = 3 PQ2 = (3 + 1)2 + (2 − 6)2 ∴ x = 3, y = 1

2

PQ = 32 = 4 2

7 AB − A 3 + 2B 3 − 6 ≡ 9 3 − 1 A and B integers ∴ AB − 6 = −1 (1) and −A + 2B = 9 (2) (2) ⇒ A = 2B − 9 sub. (1) B(2B − 9) − 6 = −1 ⇒ 2B2 − 9B − 5 = 0 (2B + 1)(B − 5) = 0 B = 1

2− or 5 B integer ∴ B = 5 ⇒ A = 1, B = 5

PMT

Solomon Press

ALGEBRA C1 Answers - Worksheet J

1 a 2x < 6 b 3x ≥ 21 c 2x > 8 d 3x ≤ 36 x < 3 x ≥ 7 x > 4 x ≤ 12

e 5x ≥ −15 f 13 x < 1 g 9x ≥ 54 h 3x < −4

x ≥ −3 x < 3 x ≥ 6 x < 43−

i x < 14 j 4x ≤ −10 k 2 < 3x l 5 ≥ 12 x

x ≤ 52− x > 2

3 x ≤ 10

2 a y > 7 b 4p ≤ 2 c 6 < 2x p ≤ 1

2 x > 3

d 2a ≥ 4 e 15 < 3u f 2b ≥ 9 a ≥ 2 u > 5 b ≥ 9

2

g 3x < −18 h y ≥ −13 i −20 ≤ 4p x < −6 p ≥ −5

j r − 2 > 6 k 3 − 6t ≤ t − 4 l 6 + 2x ≥ 24 − 4x r > 8 7 ≤ 7t 6x ≥ 18 t ≥ 1 x ≥ 3

m 7y + 21 − 6y + 2 < 0 n 20 − 8x > 21 − 6x o 12u − 3 − 5u + 15 < 9 y < −23 −1 > 2x 7u < −3 x < 1

2− u < 37−

3 a (x − 1)(x − 3) < 0 b (x + 2)(x − 2) ≤ 0 c (x + 5)(x + 3) < 0 d x2 + 2x − 8 ≤ 0 (x + 4)(x − 2) ≤ 0 1 3 −2 2 −5 −3 −4 2 ∴ 1 < x < 3 ∴ −2 ≤ x ≤ 2 ∴ −5 < x < −3 ∴ −4 ≤ x ≤ 2 e (x − 1)(x − 5) > 0 f x2 + 4x − 12 > 0 g (x + 7)(x + 3) ≥ 0 h x2 − 9x − 22 < 0 (x + 6)(x − 2) > 0 (x + 2)(x − 11) < 0 1 5 −6 2 −7 −3 −2 11 ∴ x < 1 or x > 5 ∴ x < −6 or x > 2 ∴ x ≤ −7 or x ≥ −3 ∴ −2 < x < 11 i x2 + 2x − 63 ≥ 0 j (x + 6)(x + 5) > 0 k x2 − 7x − 30 < 0 l x2 − 20x + 91 ≥ 0 (x + 9)(x − 7) ≥ 0 (x + 3)(x − 10) < 0 (x − 7)(x − 13) ≥ 0 −9 7 −6 −5 −3 10 7 13 ∴ x ≤ −9 or x ≥ 7 ∴ x < −6 or x > −5 ∴ −3 < x < 10 ∴ x ≤ 7 or x ≥ 13

PMT

C1 ALGEBRA Answers - Worksheet J page 2

Solomon Press

4 a (2x − 1)(x − 4) ≤ 0 b (2r + 1)(r − 3) < 0 c 3p2 + p − 2 ≤ 0 (3p − 2)(p + 1) ≤ 0

12 4 1

2− 3 −1 23

∴ 1

2 ≤ x ≤ 4 ∴ 12− < r < 3 ∴ −1 ≤ p ≤ 2

3 d (2y − 1)(y + 5) > 0 e (4m + 1)(m + 3) < 0 f 2x2 − 9x + 10 ≥ 0 (2x − 5)(x − 2) ≥ 0

−5 12 −3 1

4− 2 52

∴ y < −5 or y > 1

2 ∴ −3 < m < 14− ∴ x ≤ 2 or x ≥ 5

2 g a2 − 8a + 15 < 0 h x2 + 4x ≤ 7 − 2x i y2 + 9y > 2y − 10 (a − 3)(a − 5) < 0 x2 + 6x − 7 ≤ 0 y2 + 7y + 10 > 0 (x + 7)(x − 1) ≤ 0 (y + 5)(y + 2) > 0 3 5 −7 1 −5 −2 ∴ 3 < a < 5 ∴ −7 ≤ x ≤ 1 ∴ y < −5 or y > −2 j 2x2 + x > x2 + 6 k 5u − 6u2 < 3 − 4u l 2t + 3 ≥ 3t2 − 6t x2 + x − 6 > 0 2u2 − 3u + 1 > 0 3t2 − 8t − 3 ≤ 0 (x + 3)(x − 2) < 0 (2u − 1)(u − 1) > 0 (3t + 1)(t − 3) ≤ 0

−3 2 12 1 1

3− 3 ∴ −3 < x < −2 ∴ u < 1

2 or u > 1 ∴ 13− ≤ t ≤ 3

m y2 − 4y + 4 ≤ 2y − 1 n p2 + 5p + 6 ≥ 20 o 26 + 4x < 6 − 5x − x2 y2 − 6y + 5 ≤ 0 p2 + 5p − 14 ≥ 0 x2 + 9x + 20 < 0 (y − 1)(y − 5) ≤ 0 (p + 7)(p − 2) ≥ 0 (x + 5)(x + 4) < 0 1 5 −7 2 −5 −4 ∴ 1 ≤ y ≤ 5 ∴ p ≤ −7 or p ≥ 2 ∴ −5 < x < −4

PMT

C1 ALGEBRA Answers - Worksheet J page 3

Solomon Press

5 a for critical values b for critical values c for critical values d for critical values

x = 2 4 42

− ± + x = 6 36 162

± − x = 6 36 442

± +−

x = 4 16 42

− ± −

x = 2 2 22

− ± x = 6 2 52

± x = 6 4 52

±−

x = 4 2 32

− ±

x = −1 ± 2 x = 3 ± 5 x = −3 ± 2 5 x = −2 ± 3

∴ ∴ ∴ ∴ −1− 2 < x < −1+ 2 x < 3− 5 or x > 3+ 5 −3−2 5 < x <−3+2 5 x ≤−2− 3 or x ≥−2+ 3 6 a equal roots b real and distinct roots ∴ b2 − 4ac = 0 ∴ b2 − 4ac > 0 36 − 4k = 0 4 − 4k > 0 k = 9 4 > 4k

k < 1

c no real roots d real roots ∴ b2 − 4ac < 0 ∴ b2 − 4ac ≥ 0 9 − 4k < 0 k2 − 16 ≥ 0 9 < 4k (k + 4)(k − 4) ≥ 0 −4 4 k > 9

4 k ≤ −4 or k ≥ 4

e equal roots f no real roots ∴ b2 − 4ac = 0 ∴ b2 − 4ac < 0

1 + 4k = 0 k2 + 12k < 0 k = 1

4− k(k + 12) < 0 −12 0 −12 < k < 0

g real and distinct roots h equal roots

∴ b2 − 4ac > 0 ∴ b2 − 4ac = 0 4 − 4(k − 2) > 0 k2 − 8k = 0 12 > 4k k(k − 8) = 0 k < 3 k = 0 or 8

i no real roots j real roots

∴ b2 − 4ac < 0 ∴ b2 − 4ac ≥ 0 k2 − 4(2k − 3) < 0 (k − 1)2 − 36 ≥ 0 k2 − 8k + 12 < 0 k2 − 2k − 35 ≥ 0 (k − 2)(k − 6) < 0 2 6 (k + 5)(k − 7) ≥ 0 −5 7 2 < k < 6 k ≤ −5 or k ≥ 7

PMT

Solomon Press

COORDINATE GEOMETRY C1 Answers - Worksheet A

1 a = 5 1

5 3−−

= 2 b = 9 710 4

−−

= 13 c = 5 1

2 6−−

= −1 d = 8 22 2

−+

= 32

e = 1 37 1

− −−

= 23− f = 7 5

5 4− −− −

= 43 g = 8 0

0 2− −

+ = −4 h = 2 6

7 8− −− −

= 815

2 a grad = 4 b grad = 1

3 c grad = −1 d grad = −2 y-int = −1 y-int = 3 y-int = 6 y-int = 3

5−

3 a y = −x − 3 b 2y = x − 6 c 3y = −3x + 2 d 5y = 4x + 1 grad = −1 y = 1

2 x − 3 y = −x + 23 y = 4

5 x + 15

y-int = −3 grad = 12 grad = −1 grad = 4

5 y-int = −3 y-int = 2

3 y-int = 15

4 a y − 1 = 2(x − 4) b y + 5 = 5(x − 2)

c y − 1 = −3(x + 1) d y − 6 = 12 (x − 1)

e y + 14 = −2(x − 3

4 ) f y + 7 = 15− (x + 3)

5 a y − 2 = 3(x − 1) b y − 3 = −(x − 5) y = 3x − 1 y = −x + 8

c y + 3 = 4(x + 2) d y − 1 = −2(x + 4) y = 4x + 5 y = −2x − 7

e y − 1 = 13 (x + 3) f y + 2 = 5

6− (x − 9) y = 1

3 x + 2 y = 56− x + 11

2

6 a y + 4 = x − 2 b y − 1 = 1

2 (x − 6) c y − 8 = −4(x + 1) x − y − 6 = 0 2y − 2 = x − 6 y − 8 = −4x − 4 x − 2y − 4 = 0 4x + y − 4 = 0

d y − 5 = 25 (x + 3) e y + 1

8 = −3(x − 32 ) f y + 7 = 3

4− (x − 23 )

5y − 25 = 2x + 6 8y + 1 = −24x + 36 4y + 28 = −3x + 2 2x − 5y + 31 = 0 24x + 8y − 35 = 0 3x + 4y + 26 = 0 7 a grad = 13 1

4 0−

− = 3 b grad = 1 9

7 2− −

− = −2 c grad = 7 3

2 4−+

= 23

y = 3x + 1 y − 9 = −2(x − 2) y − 3 = 23 (x + 4)

y = −2x + 13 y = 23 x + 17

3

d grad = 12

8 22

++

= 4 e grad = 5 218 3− +

− = 1

5− f grad = 0.4 42 3.2

−− +

= −3

y − 8 = 4(x − 2) y + 2 = 15− (x − 3) y − 4 = −3(x + 3.2)

y = 4x y = 15− x − 7

5 y = −3x − 5.6

PMT

C1 COORDINATE GEOMETRY Answers - Worksheet A page 2

Solomon Press

8 a grad = 2 0

5 3−−

= 1 b grad = 4 85 1

− −+

= −2 c grad = 5 37 5

−+

= 16

y = x − 3 y − 8 = −2(x + 1) y − 3 = 16 (x + 5)

x − y − 3 = 0 y − 8 = −2x − 2 6y − 18 = x + 5 2x + y − 6 = 0 x − 6y + 23 = 0

d grad = 17 18 4

− ++

= 43− e grad = 0 1.5

7 2+−

= 0.3 f grad = 1

1035

1

3

−

+ = 1

4

y + 1 = 43− (x + 4) y = 0.3(x − 7) y − 1 = 1

4 (x − 3) 3y + 3 = −4x − 16 10y = 3x − 21 4y − 4 = x − 3 4x + 3y + 19 = 0 3x − 10y − 21 = 0 x − 4y + 1 = 0 9 a grad = 2 8

3 6−+

= 23− 10 k − 3(2k) + 15 = 0

∴ y − 8 = 23− (x + 6) 15 = 5k

[ 2x + 3y − 12 = 0 ] k = 3 b sub. 2(9) + 3(−2) − 12 = 18 − 6 − 12 = 0 ∴ C lies on l 11 2(4p) − 4(p2) + 5 = 0 4p2 − 8p − 5 = 0 (2p + 1)(2p − 5) = 0 p = 1

2− or 52

12 a x = 0: y = 5 b x = 0: y = 2 c x = 0: y = 3

4 d x = 0: y = 103−

y = 0: x = 52− y = 0: x = −6 y = 0: x = 3

2 y = 0: x = 2 ( 5

2− , 0) and (0, 5) (−6, 0) and (0, 2) (0, 34 ) and ( 3

2 , 0) (0, 103− ) and (2, 0)

13 a x = 0 ⇒ y = 5

3− y = 0 ⇒ x = 6 ∴ (0, 5

3− ) and (6, 0)

b area = 12 × 6 × 5

3 = 5 14 a = 2 23 4+ b = 2 23 1+ c = 2 28 15+ = 25 = 5 = 10 = 289 = 17

d = 2 216 12+ e = 2 22 5+ f = 2 28 4+ = 400 = 20 = 29 = 80 = 4 5 15 let centre be C ∴ radius = CP = 2 220 15+ = 625 = 25 ∴ CQ2 = 152 + c2 = 252 c2 = 625 − 225 = 400 c = ± 20 CR2 = (k − 2)2 + 242 = 252 (k − 2)2 = 625 − 576 = 49 k − 2 = ± 7 k = −5 or 9

PMT

Solomon Press

GRAPHS OF FUNCTIONS C1 Answers - Worksheet A

1 a y b y y = x2

y = x2 (1, 1) (−1, 1) (1, 1) O (0, 0) x y = x4

y = x3 O (0, 0) x

c y d y y = 2

1x

y = x

(1, 1) y = x

O x (1, 1)

y = 1x

O (0, 0) x

asymptotes: y = 0 and x = 0

e y y = 3x2 f y

y = x2 y = 1x

y = 2x

O (0, 0) O x asymptotes: y = 0 and x = 0 2 a = (−1) × (−3) × (−4) = −12

b x = 1, 3, 4

c y O 1 3 4 x −12 3 a y b y (0, 3) (0, 0) (1, 0) (5, 0)

(−1, 0) (1, 0) (3, 0) O x

O x

PMT

C1 GRAPHS OF FUNCTIONS Answers - Worksheet A page 2

Solomon Press

c y d y

(0, 0) (4, 0)

(0, 4) O x (−2, 0) (−1, 0) (2, 0)

O x e y f y (−2, 0) (0, 0) (1, 0)

O x (0, 2) (−2, 0) O (1, 0) x 4 a = x(x2 + 6x + 9) = x(x + 3)2

b y

(−3, 0) (0, 0)

O x 5 a y b y = (x − p)(x + q)(x − q) y (q, 0) (0, pq2)

O (p, 0) x

(−q, 0) O (q, 0) (p, 0) x

(0, −pq2)

6 TP at (1, −2) 7 crosses x-axis at (−2, 0), (1, 0) and (2, 0) ∴ f(x) = k(x − 1)2 − 2 ∴ y = k(x + 2)(x − 1)(x − 2) crosses y-axis at (0, −5) crosses y-axis at (0, −8) ∴ −5 = k − 2 ∴ −8 = 4k k = −3 k = −2 ∴ f(x) = −3(x − 1)2 − 2 ∴ y = −2(x + 2)(x − 1)(x − 2) [ f(x) = −3x2 + 6x − 5 ] = −2(x + 2)(x2 − 3x + 2) = −2(x3 − 3x2 + 2x + 2x2 − 6x + 4) = −2x3 + 2x2 + 8x − 8 ∴ a = −2, b = 2, c = 8, d = −8 8 a 4 b 0 c 2 d 3

PMT

Solomon Press

GRAPHS OF FUNCTIONS C1 Answers - Worksheet B

1 a translated 1 unit in positive x-direction b translated 3 units in negative y-direction

c stretched by a factor of 2 in y-direction d stretched by a factor of 14 in x-direction

e reflected in the x-axis f stretched by a factor of 15 in y-direction

g reflected in the y-axis h stretched by a factor of 32 in x-direction

2 a b c d (0, 9) O (4, 0) x (0, 3) (0, −3) (0, 0) O x O (8, 0) x O (4, 0) x 3 a y = 2x + 5 + 1 ⇒ y = 2x + 6 b y = 3(1 − 4x) ⇒ y = 3 − 12x

c y = 3(x + 4) + 1 ⇒ y = 3x + 13 d y = −(4x − 7) ⇒ y = 7 − 4x 4 a b c d (0, 3) (0, 6) (0, 3) (0, 4) (1, 4) O x O x O x O x

5 a stretch by a factor of 4 in y-direction b translation by 2 units in positive x-direction

c reflection in the x-axis d translation by 5 units in positive y-direction 6 a y = 2(x2 + 2) b y = (x2 + 2) − 7 stretch by a factor of 2 in y-direction translation by 7 units in negative y-direction

c y = ( 13 x)2 + 2 d y = (x + 2)2 + 2

stretch by a factor of 3 in x-direction translation by 2 units in negative x-direction 7 a y = (x − 1)2 + 2(x − 1) ⇒ y = x2 − 1

b y = (3x)2 − 4(3x) + 5 ⇒ y = 9x2 − 12x + 5

c y = (−x)2 + (−x) − 6 ⇒ y = x2 − x − 6

d y = 2( 12 x)2 − 3( 1

2 x) ⇒ y = 12 x2 − 3

2 x

8 a f(x) = (x − 2)2 − 4 ∴ turning point (2, −4)

b i y ii y iii y

y = 3+f(x) y = f(2x) y = f(x) y = f(x) y = f(x −−−− 2) y = f(x) (2,−1)

O x O x O x (2, −4) (2, −4) (4, −4) (1, −4) (2, −4)

y y y y

y y y y

(2, 1) (2, 2)

PMT

Mark Scheme (Results) Summer 2016 Pearson Edexcel Level 3 Award in Algebra (AAL30)

Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. Pearson: helping people progress, everywhere Pearson aspires to be the world’s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: www.pearson.com/uk Summer 2016 Publications Code AAL30_01_1606_MS All the material in this publication is copyright © Pearson Education Ltd 2016

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NOTES ON MARKING PRINCIPLES 1 Types of mark

M marks: method marks A marks: accuracy marks

B marks: unconditional accuracy marks (independent of M marks) 2 Abbreviations cao – correct answer only ft – follow through isw – ignore subsequent working SC: special case oe – or equivalent (and appropriate) dep – dependent indep - independent 3 No working If no working is shown then correct answers normally score full marks

If no working is shown then incorrect (even though nearly correct) answers score no marks. 4 With working

If there is a wrong answer indicated on the answer line always check the working in the body of the script (and on any diagrams), and award any marks appropriate from the mark scheme. If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative work. If it is clear from the working that the “correct” answer has been obtained from incorrect working, award 0 marks. Send the response to review, and discuss each of these situations with your Team Leader. If there is no answer on the answer line then check the working for an obvious answer. Any case of suspected misread loses A (and B) marks on that part, but can gain the M marks. Discuss each of these situations with your Team Leader. If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makes clear the method that has been used.

5 Follow through marks Follow through marks which involve a single stage calculation can be awarded without working since you can check the answer yourself, but if ambiguous do not award. Follow through marks which involve more than one stage of calculation can only be awarded on sight of the relevant working, even if it appears obvious that there is only one way you could get the answer given.

6 Ignoring subsequent work

It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question: e.g. incorrect cancelling of a fraction that would otherwise be correct It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect e.g. algebra. Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer line; mark the correct answer.

7 Parts of questions

Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another. 8 Use of ranges for answers If an answer is within a range this is inclusive, unless otherwise stated.

PAPER: AAL30_01 Question Working Answer Mark Notes 1 (a)

Circle centre (0,0) radius 5 drawn

2 M1 for using (0,0) as the centre of a circle or a

circle of radius 5 drawn A1 cao

(b)(i) Tangent drawn 2 B1

(ii) 90˚ or 𝜋2 radians B1

2

t = ±�−2𝑤𝑤−4

3 M1 for multiplying by t² +2

M1 for 𝑡² = −2𝑤𝑤−4

oe

A1 for t = ±�−2𝑤𝑤−4

oe

3 Shaded region

5 M3 for drawing all 3 lines correctly

(M2 for drawing 2 lines correctly) (M1 for drawing 1 line correctly) A2 for fully correct shading of region (A1 for correct shading for 2 inequalities)


2𝑥² + 6𝑥 − 4𝑥 − 12 2𝑥² + 2𝑥 − 12

2 M1 for expanding bracket to obtain 4 terms

with all 4 correct without considering signs or for 3 terms out of 4 correct with correct signs A1 cao

(b) 5𝑑𝑒²(2𝑑 + 3𝑒)

2 B2 cao (B1 for any correct partial factorisation with at least two of the factors 5, d, e.)

(c) 3(𝑝 − 2𝑞)(𝑝 + 2𝑞)

2 M1 for a correct partial factorisation with 2 linear factors A1 cao

5 (a)

2𝑥 − 𝑦 + 1 = 0

3 M1 for a −1−3−1−1

or 3−−11−−1

or gradient = 2 or setting up a pair of simultaneous equations . M1 for complete method. A1 oe

(b) 𝑦 = −4𝑥 + 𝑐

2 M1 for use of product of gradient equals −1 A1 oe

6 (a)

T = 12𝑥³

3 M1 for T =𝑘𝒙³

or T ∝ 1𝑥³

oe M1 for substitution to find k A1 oe

(b) 0.5

2 M1 for substitution of 4 into their formula A1 for 0.5 oe


𝑒−1

1 B1 oe

(b) 𝑛6

4

2 M1 for inverting the fraction, or squaring the fraction A1 cao

(c) 2𝑥(𝑥 − 3) + 7(𝑥 + 3)(𝑥 + 3)(𝑥 − 3)

2𝑥² + 𝑥 + 21(𝑥 + 3)(𝑥 − 3)

3 M1 for using a correct common denominator, eg (𝑥 + 3)(𝑥 − 3) M1 for 2𝑥(𝑥−3)+7(𝑥+3)

(𝑥+3)(𝑥−3) oe

A1 for 2𝑥²+𝑥+21(𝑥+3)(𝑥−3)

or 2𝑥²+𝑥+21𝑥²−9

8 𝑦 = 𝑥 − 3 𝑥2 − 2𝑥 − 3 = 0 (𝑥 − 3)(𝑥 + 1) = 0 𝑥 = 3 or − 1 𝑦 = 0 or − 4 OR x = y + 3 y = (y + 3)² − (y + 3) – 6 y² + 4y = 0 𝑦 = 0 or − 4 𝑥 = 3 or − 1

x = 3, y = 0 and x = −1, y = −4

5 M1 for method to eliminate one variable M1(dep M1) for simplifying to get a quadratic (= 0) in one variable M1(dep M2) for correct method to solve their quadratic A1 x = 3, x = −1 or y = 0, y = −4 A1 x = 3, y = 0 and x = −1, y = −4

9 −4 ± √766

2 M1 Stating the quadratic formula or substitution into formula A1 −4±√76

6 oe


y > −1

2 M1 for isolating the term in y, eg −3y < 3,

3y ˃ −3 A1 cao

(b) −4 < x < 1

3 M1 (x−1)(x+4) M1 for critical values 1 and −4 A1 cao

11 0.5(2 + 2(1.3 + 1 + 0.8) + 0.6) to 0.5(2 + 2(1.4 + 1 + 0.8 )+ 0.7)

4.4 - 4.6

3 M1 for reading off values from the graph 2, 1.3 to 1.4, 1, 0.8, 0.6 to 0.7 oe M1 for substituting values into trapezium rule A1 4.4 - 4.6

12 (a)

28

1 B1 cao

(b) 9√3

2 M1 for correct first step, eg 12√3, 3√3 A1 cao

(c) 7 − √323

3 M1 for multiplying by 7−√37−√3

M1 for 14 − 2√3 and 49 − 3 A1 cao


9

4

3 M1 for 39 = 32

{2𝑎 + (3 − 1)𝑑} oe or a + 8d = 41 oe A1 for a = 9 A1 for d = 4

(b)

𝑛2(1 + 13n)

3 B1 for a = 7

M1 for 𝑛2

{2 × "7" + (𝑛 − 1) × "13"} oe or 𝑛2

{ "7" + 13𝑛 − 6} oe

A1 for 𝑛2(1 + 13n) or 13𝑛²+𝑛

2

14 (a)

Graph drawn

2 B2 fully correct graph drawn (B1 for a translation parallel to the y-axis)

(b) Graph drawn

2 B2 fully correct graph drawn (B1 for a translation parallel to the x- axis)

15 (a)

(−4, 18) (−3, 1

4) (−2, 1

2) (−1, 1)

(0, 2) (1, 4) (2, 8) (3, 16)

Correct curve

4 B1 for drawing suitable axes on grid M1 for calculating at least 4 points for values of x from x = −4 to 3 A1 for all correct points calculated A1 for correct curve drawn

(b) 2.5 - 2.7

1 B1


c ≤ 4

2 M1 for use of b² − 4ac ≥ 0 or b² − 4ac ˃ 0

A1 cao

(b) Graph drawn

3 B3 fully correct graph drawn with labels at (3, 0) and (0, 9) (B2 correct shape of graph drawn with one label) (B1 correct shape of graph or for (3, 0) and (0, 9) indicated)

17 (a)

2 −3

2 M1 for (x + 2)² or p = 2 A1 for p = 2 and 𝑞 = −3

(b) −12 and 3

2 M1 for a complete method A1 cao

18 −𝑏𝑎

= −67

𝑐𝑎

= −37

−67 −3

7

2 B1 for sum of roots = −67

B1 for product of roots = −37

19 Graph drawn

4 B1 for asymptote of 𝑥 = −2 B1 for intercept at (0, 1

2)

M1 for correct shape A1 for fully correct graph.


10 30 to 10 45

1 B1 cao

(b) 360

2 M1 for a method to find the gradient, eg 900.25

, 9015

A1 cao

(c) 12 × 90 × 1

4 + 1

2(90 + 120) × 1

2

= 11.25 + 52.5 or 12 × 90 × 1

4 + 1

2 × 90 + 1

2 × 30 × 1

2

=11.25 + 45 + 7.5

63.75

2 M1 a complete method to find the correct area A1 cao

Question 3

R 2

4

6

8

10

-2

-4

2 4 6 -2 -4 0 x

y

x + y = 5

y = 2x + 1

y = −3x

Question 14 (a) (b)

1

2

3

4

5

6

7

8

-1

-2

1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 0 x

y

y = f(x) + 3

1

2

3

4

5

6

7

8

-1

-2

1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 0 x

y

y = f(x) − 2

Question 15(a)

2

4

6

8

10

12

14

16

1 2 3 4 -1 -2 -3 -4 -5 0 x

y

y = 2x + 1

Question 16(b)

x

y

y = (x − 3)2

O 3

9

Question 19

O x

y

y = 1𝑥+2

−2 12

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