MATHEMATICS Learner’s Study and Revision Guide for Grade 12
EXAMINATION TYPE QUESTIONS IN ALGEBRA
Solution Hints by Roseinnes Phahle Examination Questions by the Department of Basic Education 1
Preparation for the Mathematics examination brought to you by Kagiso Trust
Contents Unit 2 Examination questions with solution hints and answers
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More questions from past examination papers
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Answers
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How to use this revision and study guide 1. This guide assumes that you have worked through the basic techniques of algebra covered in Unit 1. 2. The first set of examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes. 3. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty. 4. What follows next are more questions taken from past examination papers. 5. Answers to the extra past examination questions appear at the end. 6. Finally, don’t be a loner. Work through this guide in a team with your classmates.
Examination questions in algebra
REVISION UNIT 2: EXAMINATION TYPE QUESTIONS IN ALGEBRA We have no need to introduce the techniques of manipulating algebraic expressions, equations and inequalities as this has been done thoroughly in UNIT 1. Now you can try your hand at the examination type of questions asked in algebra. PAPER 1 QUESTIONS 1
DoE/ADDITIONAL EXEMPLAR 2008
PAPER 1 QUESTIONS 1
DoE/NOVEMBER 2008
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PAPER 1 QUESTION 1 Number Hints and answers 1.1.1 Find the LCM of the LHS. Express the LHS as a single fraction. Cross multiply. Put all terms on LHS so LHS = zero. Simplify by collecting like terms. Now solve the equation.
1.1.2
Answer: x =1,91 or x =0,09 Put all terms on LHS so the RHS becomes zero. Factorise LHS. Solve using a number line or from a sketch of the quadratic function. Answer: x ≤ −4 or x ≥ 7 You can also write the answer as
x ∈ (− ∞;−4] [7; ∞ )
1.2
Use the first equation to substitute for x or y in the second equation. Always choose the linear equation for this purpose and simpler still if you choose to substitute for the unknown that has one as a coefficient. In this case, you are better off choosing y . Do this and get a solution for x and y. Answer:
1 or x =2 5 17 or y =1 y =− 5
x=−
DoE/ADDITIONAL EXEMPLAR 2008 Work out the solutions in the boxes below
Examination questions in algebra
PAPER 1 QUESTION 1
DoE/NOVEMBER 2008
Number Hints and answers 1.1.1 As always when solving equations put all the terms to one side so that LHS=0 and then solve.
1.1.2
Work out the solutions in the boxes below
Answer: x = 4 or x = 1 Make LHS=0 and solve. Answer:
x = 3,79 or x = −0,79
1.1.3
Again put all the terms to one side and make the coefficient of x 2 positive. What you will then have is LHS>0. Hence solve the inequality. Answer:
3 or x > 1 ; or write as 2 3 x ∈ − ∞;− (1; ∞ ) 2
x<−
1.2
Its your choice whether you substitute for x or y in the second equation using the first one. Always best to choose to make the substitution that will make the working out easier. Which one is that? Answer:
x= 1.3
1.4
2 5 or 3; and y = or -3 3 3
Simplify by factorizing the numerator and something will cancel out. Answer: 1 000 000 000 001 or 1012 + 1 Again put all the terms to one side so that LHS=0. See whether what you get has a solution. If cannot have a solution, explain why: Answer: you write the explanation. 5
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PAPER 1 QUESTIONS 8
PAPER 1 QUESTION 8 Hints and answers This is an example of a problem which you cannot be taught how to solve. All your need are your wits to see what is needed. Replace any of the numbers by p or by any letter of the alphabet. Repeat: any of the numbers. Suppose we make p = 2007 Then 2008 = p +1, 2000 = p -7 and so on. Replace all thenumbers you seem by what you have worked them out to be in terms of p . Simplify by multiplying, adding & subtracting. As an exercise, let p be any of the other numbers 2008 or 2009 or 2006 or 2010 or 2016 or 2000. See whether you get the same value for an answer.
Answer: 67
DoE/ADDITIONAL EXEMPLAR 2008
DoE/ADDITIONAL EXEMPLAR 2008 Work out the solutions in the box below
Examination questions in algebra
MORE QUESTIONS FROM PAST EXAMINATION PAPERS Exemplar 2008
Additional Exemplar 2008
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Preparatory Examination 2008
November 2008
Examination questions in algebra
Feb â&#x20AC;&#x201C; March 2009
November 2009
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November 2009(1)
Examination questions in algebra
Feb â&#x20AC;&#x201C; March 2010
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ANSWERS Exemplar 2008 1.1.1 x = 12 or x = −2 1.1.2 x = 0,41 or x = −24,41
1.1.3 − 1 ≤ x ≤ 4 or x ∈ [− 1;4]
1 x = 2 1.2 y = 3 2
or
x = 2 y = 1
1.3 1 000 000 000 001 or 1012 + 1 1.4 Requires you to do the proof. Feb/March 2009
1 or x = 1 3 1.1.2 x = 3,13 or x = −0,13 1.1.3 x < −1 and x > 3 or x ∈ (− ∞;−1) (3; ∞ ) 1.1.1 x =
Additional Exemplar 2008 1.1.1 x = 1,91 or x = 0,09
x = −3 x = 4 1.2 4 or y = 1 y = − 3
1 x = − 5 x = 2 1.2 or y = 1 y = − 17 5
1.3 2
Preparatory Examination 2008 1.1 x = 5 or x = −4 1.2 x = 0 or x = 3
1.1.3
1.1.2 x ≤ −4 and x ≥ 7 or x ∈ (− ∞;−4] [7; ∞ )
x = 1,7 x = −2,28 or y = 1,35 y = −0,64 1.4 x ≤ −5 and x ≥ 2 or x ∈ (− ∞;−5] [2;−∞ )
November 2009 1.1.1 x = 5 or x = −1 1.1.2 x = 4,95 or x = 0,05
x = 3 x = −1 or y = y = 0 y = −4
1.2 25
1.3
November 2009(1) 1.1.1 x = 6 or x = −5 1.1.2 x = 1,4 or x = 0,2
November 2008 1.1.1 x = 4 or x = 1 1.1.2 x = 3,79 or x = −0,79
1.1.3 x <
3 3 1.1.3 x < − and x > 1 or x ∈ − ∞;− (1; ∞ ) 2 2
2 = x x = 3 3 1.2 or y = −3 y = 5 3
4 1 1 4 and x > or x ∈ − ∞; ; ∞ 3 3 3 3
1 x = 5 x = 2 1.2 2 or y = 2 y = −2 1.3
10 99
1.4 1 + 2 x 2
Examination questions in algebra
Feb/March 2010 1.1.1 x = 4 or x = −6 1.1.2 −
1 1 ≤ x ≤ 2 or x ∈ − ;2 2 2
11 x = 9 x = −2 1.2 or y = 2 y = − 40 9 1.3 18 1.4 a = 2 and b = 7 Also a = 1 and b = 28
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