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by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the...

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Page 1 HAEF IB – FURTHER MATH HL TEST 2 SETS, RELATIONS AND GROUPS by Christos Nikolaidis SOLUTIONS Date: 8/12/2017 Questions Solution
Transcript
Page 1: by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the equation aAb = 4Ab , for a #4. (a) ... ¥42 for all correct, Al for 2 correct. .41A1

Page 1

HAEF IB – FURTHER MATH HL

TEST 2

SETS, RELATIONS AND GROUPS

by Christos Nikolaidis

SOLUTIONS Date: 8/12/2017

Questions

Solution

Page 2: by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the equation aAb = 4Ab , for a #4. (a) ... ¥42 for all correct, Al for 2 correct. .41A1

Page 2

Solution

Page 3: by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the equation aAb = 4Ab , for a #4. (a) ... ¥42 for all correct, Al for 2 correct. .41A1

Page 3

Solution

(a) (i)

A1A1

[8 marks]

Page 4: by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the equation aAb = 4Ab , for a #4. (a) ... ¥42 for all correct, Al for 2 correct. .41A1

Page 4

Solution

Page 5: by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the equation aAb = 4Ab , for a #4. (a) ... ¥42 for all correct, Al for 2 correct. .41A1

Page 5

5. [Maximum mark: 10]

(d) Using the set Z+ as both domain and codomain, give an example of a surjective

function that is not injective. [3 marks]

Solution

(d) ���� = �1, = 1 − 1, > 1

Surjective as range = codomain

Non-injective as ��1� = 1 = ��2�

Page 6: by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the equation aAb = 4Ab , for a #4. (a) ... ¥42 for all correct, Al for 2 correct. .41A1

Page 6

6. [Maximum mark: 12]

Solution

Page 7: by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the equation aAb = 4Ab , for a #4. (a) ... ¥42 for all correct, Al for 2 correct. .41A1

Page 7

7. [Maximum mark: 19]

Solution

(ii)

OR the sum of two half-integers is a half integer A1R1

Page 8: by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the equation aAb = 4Ab , for a #4. (a) ... ¥42 for all correct, Al for 2 correct. .41A1

Page 8

8. [Maximum mark: 15]

Solution

Page 9: by Christos Nikolaidis SOLUTIONS · a2+a (a — b) (a + b + l) O ... Find the solutions of the equation aAb = 4Ab , for a #4. (a) ... ¥42 for all correct, Al for 2 correct. .41A1

Page 9


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