Year 12. Mathematics EAS. Workbook. Robert Lakeland & Carl Nugent. ulake Ltd. Innovative Publisher of Mathematics Texts

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1 Year 12 Mathematics EAS Workbook Robert Lakeland & Carl Nugent Innovative Publisher of Mathematics Texts

2 Inside front cover iv v vi Contents Explanation of icons used in the book Introduction Customising the Workbook for our own learning stle Stud Skills Topic Section Page Algebraic Methods 2.6 Calculus 2.7 Achievement Standard Expanding Factorising Exponents Changing the Subject Logarithms Rational Expressions Linear Equations Linear Inequations Quadratic Equations Quadratic Formula The Nature of the Roots of a Quadratic Excellence Questions for Algebra Practice External Assessment Achievement Standard The Concept of Gradient x 2 + 5x + 6 = 3 The Gradient Function ? h = V πr 2 5 log 3 The Gradient at a Point The Derivative from First Principles Calculus Notation The Tangent to a Curve Maximum and Minimum Points Increasing and Decreasing Functions Applications of Maximum and Minimum Kinematics Calculating Rates of Change The Reverse of Differentiation Anti-differentiation Kinematics Merit and Excellence Questions Practice External Assessment

3 Topic Section Page Probabilit Methods 2.12 Achievement Standard Theoretical Probabilit Relative Frequenc Informal Conditional Probabilit Multiplicative Principle and Expected Number Tree Diagrams Risk, Absolute and Relative The Standard Normal Distribution Inverse Normal Calculating X Values Inverse Normal Calculating Parameters Experimental Distributions Practice External Assessment Answers Algebraic Methods Calculus Probabilit Methods Normal Distribution Table

4 Algebraic Methods Algebraic Methods 2.6 This achievement standard involves appling algebraic methods in solving problems. Achievement Achievement with Merit Achievement with Excellence Appl algebraic methods in solving problems. Appl algebraic methods, using relational thinking, in solving problems. Appl algebraic methods, using extended abstract thinking, in solving problems. This achievement standard is derived from Level 7 of The New Zealand Curriculum and is related to the achievement objectives manipulate rational, exponential, and logarithmic algebraic expressions form and use linear and quadratic equations. Appl algebraic methods in solving problems involves: selecting and using methods demonstrating knowledge of algebraic concepts and terms communicating using appropriate representations. Relational thinking involves one or more of: selecting and carring out a logical sequence of steps connecting different concepts or representations demonstrating understanding of concepts forming and using a model; and also relating findings to a context, or communicating thinking using appropriate mathematical statements. Extended abstract thinking involves one or more of: devising a strateg to investigate or solve a problem identifing relevant concepts in context developing a chain of logical reasoning, or proof forming a generalisation; and also using correct mathematical statements, or communicating mathematical insight. Problems are situations that provide opportunities to appl knowledge or understanding of mathematical concepts and methods. Situations will be set in real-life or mathematical contexts. Methods include a selection from those related to: manipulating algebraic expressions, including rational expressions manipulating expressions with exponents, including fractional and negative exponents determining the nature of the roots of a quadratic equation solving exponential equations (which ma include manipulating logarithms) forming and solving linear and quadratic equations. EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

5 Algebraic Methods Factorising Complex Quadratics When we are asked to factorise a quadratic in the form ax 2 + bx + c where the a term is not 1 then we need to approach the problem differentl. To factorise 5x 2 14x 3 1. Multipl the first and last terms together. (5x 2 x 3) = 15x 2 2. Find two numbers that multipl to give First check if there is a common factor 15x 2 and add to give the middle term of 14x. 3. The numbers are 15x and x. If a 1 first check to see whether there is a 4. Rewrite the quadratic, using these two values common factor that can be removed. as the middle term. Consider 3x x x 2 14x 3 = 5x 2 15x + x 3 Taking out the common factor of 3 Factorise each pair of terms using the grouping technique. 3x x + 18 = 3(x 2 + 5x + 6) 5x 2 14x 3 = 5x 2 15x + x 3 Then factorising bracketed terms gives = 5x(x 3) + 1(x 3) 3x x + 18 = 3(x + 2)(x + 3) 6. Taking out (x 3) as the common factor. If there is no common factor use the technique on = (x 3)(5x + 1) the left. Example Example Factorise the quadratic 2x 2 4x 126 Factorise the quadratic 3x 2 11x + 6 We recognise that there is a common There is no common factor so we begin factor for 2x 2, 4x and 126. So we start b finding two terms that multipl to give b taking out this common factor of 2. 3x 2 x 6 = 18x 2 and add to give 11x. 2x 2 4x 126 = 2(x 2 2x 63) The terms are 9x and 2x. We then factorise the bracketed term x 2 2x 63 to Rewriting the quadratic with these new terms get 3x 2 11x + 6 = 3x 2 Note: The 9x 2x + 6 sign changes 2x 2 4x 126 = 2(x 2 2x 63) Grouping = (3x 2 9x) (2x 6) as there is a = 2(x 9)(x + 7) Factorising = 3x(x 3) 2(x 3) = (x 3)(3x 2) minus outside the brackets. Achievement Factorise the following x 2 6x x 2 12x x x b b x 2 7x n 2 23n x 2 9x x x n n + 24 EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

6 Algebraic Methods Merit Solve the following A compan pas 214. Auckland house prices $ for a new forklift have been inflating at which the compan can 14.5% pa. A house is depreciate at the rate of originall purchased for 13% per annum. $ The formula The formula n r Vn = Ab1 - l can be 100 V = U 1+ r T used to represent depreciation, where A is 100 can the initial cost, r the rate of depreciation, n be used to represent inflation, where U is the the number of ears of depreciation and V n initial cost, r the rate of inflation, T the number the value of the article after n ears. of ears of inflation and V the value of the house after T ears. a) Find the forklift s value after 5 ears. a) What is the expected price in ten ears? b) Form an equation and solve it to find out how b) How long would ou have to own the long it takes (in ears) before the forklift is $ house for it to be worth $ ? worth onl $3000. Excellence Form an equation and solve it to answer the following problems Penn has been left $ from her Aunt and instead of spending it she hopes to invest it at 4.5% pa and grow it until she has a million dollars. The formula V = U 1+ r T can be used to 100 represent the compound interest, where U is the initial amount, r the rate of interest, T the number of ears and V the value after T ears. The interest is alwas credited at the end of the investment ear. Penn made her investment during In what ear will Penn s investment be worth one million dollars? 216. Auckland house prices have been inflating at 14.5% pa while Wellington prices have onl been inflating at 8.4% pa. In 2016, a house is purchased in Auckland for $ and one in Wellington for $1.2 million. The formula V = U 1+ r T 100 can be used to represent inflation, where U is the initial cost, r the rate of inflation, T the number of ears and V the value of the house after T ears. In what ear would the house in Auckland be worth more than the house in Wellington? EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

7 Algebraic Methods Quadratic Equations Quadratic Equations Be on the look out for quadratic equations disguised as something else. If we are able to factorise a quadratic equation then finding the solutions is straightforward. A quadratic equation is a polnomial where the If we have a product of two factors that equal 0 highest power of the unknown is two. A example is then one (or both) of the factors must equal 0. 3x 2 + x 2 = 0 A.B = 0 (3x 2)(x + 1) = 0 then either A = 0 or B = 0 x = 1, 2 Therefore for a quadratic equation 3 but if we divide all terms b x and manipulate it (x + 3)(x + 2) = 0 the equation at first glance does not appear as a then either (x + 2) = 0 quadratic x = 2 3x +1 = 2 x or (x + 3) = 0 Yet this is an identical equation. Two other equations that can be solved as quadratic equations x = 3 are The solution is written as 3x 4 + x 2 2 = 0 x = 3, 2 and 3x + x = 2 If the quadratic is not alread factorised, then we In the first we substitute z = x 2 to get our must factorise it for this method to work. quadratic and in the second we substitute z = x. Example Example Solve the equation Solve the equation 2x 2 x = 36 3x + x = 2 2x 2 x = 36 Rearrange so that the equation equals zero first Factorising 2x 2 x 36 = 0 (2x 9)(x + 4) = 0 Setting each factor equal to 0 gives (2x 9) = 0 x = 4.5 and (x + 4) = 0 x = 4 x = 4, 4.5 Substitute z = x 3z 2 + z = 2 Rearrange so that the equation equals zero first 3z 2 + z 2 = 0 Factorising (3z 2)(z + 1) = 0 Setting each factor equal to 0 gives z = 1, 2 3 x = z 2 This would appear to give answers x = 1, 4 9 but substitution into the original equation we find the onl answer is x = 4 9 This is because the square root sign represents the positive root onl (i.e. 1) which does not solve the original equation. We should alwas confirm these disguised questions b substituting back into the original equation. EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

8 76 Calculus Methods 2.7 The Derivative from First Principles The Derivative If we continue to reduce the value of h then the gradient of the chord PQ gets closer and closer to the gradient of a tangent at the point P. Q f(b) f(a) P Q f(x + h) f(x) P f(x+h) - f(x) h a b x x x + h x fb ( )- fa ( ) If the interval over which we are finding the average m = b- a rate of change is made smaller and smaller, then the Q P h Q f(x + h) f(x + h) Q P f(x) f(x+h) - f(x) P f(x) x x + h x h Ltd ulake We have alread established that the average rate of change of a graph is given b finding the gradient of a chord. To calculate the gradient or average rate of the chord PQ we use the formula If we let the gap between a and b be h and we change our notation so P is now (x, f(x)) then Q becomes (x + h, f(x + h)) x h x + h The gradient of the chord is now fx ( + h) -f( x) m = ( x+ h) -x Now if we reduce the value of h, then Q gets closer to P. x limiting value of the chord s gradient as h approaches zero [ h 0] is the gradient of the tangent and the gradient of the curve at this point. This limiting value of the gradient at a point becomes lim fx ( + h) -f( x) f (x) = h " 0 ( x+ h) -x f (x) = lim fx ( + h) -f( x). h " 0 h f(x + h) f(x) Q Q P f(x+h) - f(x) h x x + h x This derived function f (x), is known as the gradient function and the process is called differentiation. When we use the formula above to calculate the derived function we are differentiating b first principles. The notation f (x) is used to denote the gradient function. EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

9 82 Calculus Methods 2.7 Example Differentiate the following using the correct notation a) = 4x 3 + 5x 2 x + 2 b) f(x) = 2x 7 5x c) = 2x 2 7x 15 find d. find f (x). find the derivative of. dx If the problem starts f(x) etc. we use function notation. If it starts with = we use Leibniz notation. d a) dx = b) f (x) = 14x 6 25x 4 c) Using Leibniz notation 12x2 + 10x 1 differentiating once d f (x) = 84x 5 100x 3 = 4x 7 differentiating once dx differentiating once differentiating again Note: We acknowledge that onl first derivatives are in the sllabus, but the authors expect a number of teachers will teach the use of second derivatives in solving maximum and minimum problems. Achievement / Merit Differentiate using the appropriate notation. 61. f(x) = 9x 3 + 2x + 3 find f (x) 62. f(x) = 3x 7 5x 3 find f (x) 63. = 5x 3x 5 find d 64. = x 2 4 find d dx dx 65. f(x) = 3x 5 15x find f (x) 66. = 4x 3 6x 2 + 7x + 2 find d2 dx 2 67 = 1 2 x2 3 4 x3 find d dx 68. = 1.2x x find d dx Ltd ulake 69. f(x) = 3 4 x3 2 3 x2 + 4x 1 find f (x) 70. = 16x 4 24x find d2 dx f(x) = 4 x x x2 2x + 5 find f (x) 72. = 3x 2 4x3 2 9 x 2 7 find d2 dx 2 EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

10 94 Calculus Methods 2.7 Example A wire frame is to be constructed out of 1200 mm of wire. w Find the height and width of the frame so that the h total area is as large as possible. Using DELS DEBT Draw a diagram. w h h h h w Express the problem that needs to be maximised or minimised (often in terms of two variables). Area = height x whole width = h x w = hw Link the two variables above with an equation = 4h + 2w Substitute for one variable using the link above w h = b l 4 = w Ltd ulake Area = w( w) = 300w 0.5w 2 Differentiate. Area = 300 w Equate to zero so that ou can find the turning points. 300 w = 0 For Max / Min w = 300 Back substitute into the original problem to find the maximum or minimum value. Area = 300 x (300) 2 = mm 2 Test our answer b re-reading the question. No. The original question called for the height h and width w, so the correct answer is w = 300 mm and h = x 300 = 150 mm EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

11 Calculus Methods Find the function whose derivative is d = 3 x dx and has roots at 2 and Find the function whose derivative is f (x) = 1 3x 2 and crosses the axis at = 1. x x 158. Find the function whose derivative is 159. Find the function whose derivative is f (x) = 3x 2 2x+ 1 f (x) = 1 x 2 + x and crosses the axis at 2. and passes through the origin. Ltd ulake x x EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

12 130 Probabilit Methods 2.12 Example A fair die and a coin are thrown. a) List all the possible outcomes (sample space). Using the list of possible outcomes from a) find the probabilit that b) a head occurs. c) a head and an even number occur. d) the number on the die is greater than or equal to three and a tail occurs. a) Sample space Die H 1,H 2,H 3,H 4,H 5,H 6,H Coin T 1,T 2,T 3,T 4,T 5,T 6,T 6 b) P(head) = (1,H), (2,H), (3,H), 12 (4,H), (5,H), (6,H) out = 1 of 12 possibilities. 2 c) P(H and even) = = d) P(die 3 and T) = 3 (2,H), (4,H), (6,H) out 12 of 12 possibilities (3,T), (4, T), (5,T), (6,T) 12 out of 12 possibilities. = 1 3 Relative Frequenc Results that we obtain from repeated experiments or observations can be used to predict outcomes of future events. We express these results as a proportion of the whole called relative frequenc. For example, we ma have observations on the different transport options used b students travelling to school and we use these to predict the probabilit of a random student using a particular method. Alternativel, we ma want to know the probabilit of a drawing pin landing pin up and we conduct an experiment to get some data. As we repeat an experiment a large number of times the probabilit approaches a consistent value. This is called long run relative frequenc. EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

13 Probabilit Methods The local netball team is ver good, but it has a poor awa record. On the basis of its record calculate the following probabilities. Win Lose Total Home games Awa games Total The breakdown of the sex and age of emploees working at a compan is given in the table below. Male Female Total Less than or over Total a) What is the probabilit of it winning a home game? b) If we know the team won a game, what is the probabilit it was an awa game? c) If we know the team lost a game, what is the probabilit it was a home game? a) What is the probabilit of an emploee selected at random being under 40 ears of age? b) Given an emploee is female what is the probabilit the are 40 ears of age or over? c) Given an emploee is under 40 ears of age, what is the probabilit the are male? 8. An analsis at ulake a local hospital of the cause of 9. Free blood pressure Ltd tests conducted at a death of people aged between 50 and 65 and pharmac ielded the following results. whether the smoked is given below. Male Female Total Smoking Normal Nil Light Heav Pre-hpertensive Other causes Hpertensive Cancer / heart disease Total Find the probabilit that a randoml selected death is a) of a non-smoker. a) What is the probabilit of a random person who was tested being hpertensive? b) from cancer or heart disease. b) Given a person is pre-hpertensive what is the probabilit the are male? c) from cancer or heart disease given the person smokes. c) Given a person is female what is the probabilit the have a normal blood pressure reading? d) from cancer or heart disease given the person is a heav smoker. e) a heav smoker who has died of cancer or heart disease. EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

14 148 Probabilit Methods 2.12 Risk exposed to cancer is 0.25 while the probabilit of a non-smoker being exposed to cancer is Risk Therefore the relative risk = 0.25 In statistics, risk is the probabilit or likelihood of an 0.02 = event occurring. Risk is often expressed as a decimal So smokers are 12.5 times more likel than nonsmokers to develop lung cancer. or percentage. Relative risk is a value that identifies how much When ou attempt something that could result something ou do, such as maintaining a health in harm or injur ou are putting ourself at risk. weight, can change our risk compared to the risk if Risk is generall considered to have this negative ou re ver overweight. connotation so in situations where this is not the case ou should look to use another term to avoid Values for relative risk (RR) can be an number confusing readers. For example, if ou were looking greater than zero. If our relative risk (RR) value at the data for two groups gaining emploment is 1 it means the outcome is not influenced b the it would be inappropriate even if mathematicall factor present, i.e. the likelihood of lung cancer is correct to use the term risk of getting a job as ou not affected b smoking. are likel to confuse our audience. You could just If our relative risk (RR) value is greater than 1 it use the term probabilit of getting a job. There are means the outcome is greater when the factor is man different kinds of risk: business risk, economic present, i.e. ou are more likel to get lung cancer risk, health risk etc. if ou smoke. In probabilit we can define two tpes of risk: If our relative risk (RR) is less than 1 it means the absolute risk and relative risk. outcome is less likel when the factor is present. An example ma be the relative risk of having a Absolute Risk heart attack when taking a low dose of aspirin Absolute risk is the probabilit or percentage of compared to having a heart attack and not taking subjects in an group that experience an outcome. an aspirin. For example, the absolute ulake risk of a person Heart Ltd No heart developing lung cancer over their lifetime could be Attack attack expressed as 0.11 or 11% or 1 in 9. Takes aspirin 0.94% 99.06% Relative Risk No medication 1.71% 98.29% Relative risk is used to compare the risk in two sub If ou now calculated the relative risk of having a groups of our population. For example, we could heart attack with or without a low dose of aspirin measure the risk of people getting lung cancer as a then consequence of smoking b comparing the risk of Risk of heart attack with aspirin getting lung cancer for people who smoke with the RR = Risk of heart attack with no aspirin risk of getting lung cancer for the group that do not smoke. RR = 0.94 We can Ltd calculate ulake the relative risk of getting lung 1.71 cancer as a consequence of smoking (or the risk RR = 0.55 ratio) b using the formula A relative risk less than 1 is correct but difficult Risk or Prob. with condition present to interpret so often the problem is restated so RR = Risk or Prob. with no condition present that the increased risk is the numerator and the relative risk is then greater than 1. Where RR is the relative risk. Consider the table below which shows that the risk of getting lung cancer among smokers is 25% while onl 2% among non-smokers. Lung No lung cancer cancer Smoker Non-smoker To calculate the relative risk of getting lung cancer as a consequence of smoking we use the formula given above. The probabilit of a smoker being In this case we would state that the relative risk of having a heart attack b taking no medication compared to taking a low dose of aspirin is Risk of heart attack with no aspirin RR = Risk of heart attack with aspirin RR = RR = 1.82 The risk of a heart attack b taking no medication is 1.8 times the risk compared to taking a low dose of aspirin. EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

15 162 Probabilit Methods 2.12 Example (Using graphics calculator) The mean mark in a school s Year 12 mathematics exam was 51% with a standard deviation of 16%. Assuming the marks are normall distributed find the probabilit that a student selected at random scored a) less than 55%. b) between 35% and 45%. Hence find how man out of 250 students ou expect to get between 35% and 45%. Using our TI-84 Plus Using our Casio 9750GII Even when using our calculator sketch at least The sketch for part a) is on the left. It is one normal curve so ou can visualise the area ou important that ou sketch at least one normal require. We can see we expect an answer over 0.5. curve so ou can visualize the area ou require. From the Main menu select STAT then DIST, NORM and Ncd. STAT DIST NORM Ncd 2 F5 F1 F2 For the Casio use the format Lower, Upper, σ and µ a) With P(X < 55) with σ = 16 and µ = 51. enter Lower : 0, Upper : 1.5, σ : 16 and Access the distribution menu ulake and either scroll µ : 51. Ltd down or select 2 to get normalcdf(. 0 EXE 5 5 EXE 1 6 2nd DISTR 2 EXE 5 1 EXE EXE a) With P(X < 55) and µ = 51, σ = 16, enter lower : 0, upper : 55, µ : 51 and σ : 16. Normal C.D normalcdf lower:0 upper:55 µ : 51 σ : 16 Paste Data :Variable Lower :0 Upper :55 σ : 16 µ : 51 getting the probabilit = (4 dp). b) For P(35 < X < 45) This gives or (4 dp). b) The sketch for part b) is on the right. For P(35 < X < 45) and µ = 51, σ = 16. Enter lower : 35, upper : 45, µ : 51 and σ : 16. This gives (4 dp). Number of students = x 250 = 48 (rounding down) P( 35 < X < 45) Enter Lower : 35, Upper : 45, σ : 16 and µ : EXE 4 5 EXE 1 6 EXE 5 1 EXE x EXE This gives (4 dp). Number of students = x 250 = 48 (rounding down) EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

16 194 Answers Algebraic Methods 2.6 Page 57 Ltd 418. a) 5000 = 50 (1.095) T a) A = A e) 25 litres a month at a cost of 100 = (1.095) T k = $26 A log 100 t = 13.7 ears M T = log Sept 2021 or Jan 2022 E Sufficienc T = 50.7 ears Depends upon whether t For Question 1 students require T = 51 ears round up rounded or not as to the two of A for Achievement or two of month it reaches 200. i.e. in M for Merit or one or two of E for b) 500 (1.095) N = 25 x 68 (1.030) N b) i) x = (4 sf) A Excellence. ii) x = 5, N = 20 ears rounding up 2 A For Question 2 students require Page 58 Excludes x = 2 M two of A for Achievement or two of as all log expressions M for Merit or one or two of E for 419. a) k = (8) must be positive. E Excellence. (d + 0.5) 1.2 = c) 2n 14 > 9n For Question 3 students require d = 5.6 km (13 t) 7n > 14 two of A for Achievement or two of b) 48 = 120 x M for Merit or one or two of E for n < t = 74.1 or 74 weeks 2 A Excellence. c) t 1 + t 2 = d) p(x + 2) 2 + q(x + 2) + r Overall students require t 1 = 0.45c = p(x 2 + 4x + 4) + qx + 2q + r Two or more Achievement = px t 2 = c 2 + (4p + q)x + 4p + 2q + r questions or better for overall Equating to 2x 2 7x c c 2 = Achievement. p = 2, q = 15, r = 18 M 0.015c c = 0 Two or more Merit questions plus c = 56.21, 26.2 Question Three one Achievement question or Answer 26 characters better for overall Merit. a) i) (18 2x)(15 x) = Two or more Excellence questions 2x 2 48x = 0 plus one Achievement question or Pages ulake 4x 2 96x = 0 A better for overall Ltd Excellence. Practice External Assessment Task ii) (2x 43)(2x 5) A Question One x = 2.5, 21.5 In the external examinations NZQA a) 2x 3 5x 2 13x + 30 A Ignore 21.5 uses a different approach to marking Length = 13 m based on understanding (u), b) log A3 B relational thinking (r) and abstract C A Width = 12.5 m M thinking (t). The then allocate b) 4x = (x + 1) 2 marks to these concepts and add 1 c) i) x =, 1 A x 2 2x + 1 = 0 them up to decide upon the overall 3 x = 1 M grade. This approach is not as eas ii) Discriminant c) x 2 + ( x k) 2 = 8 for students to self mark as the 0 = k 2 4 x 3 x 4 2x 2 + 2kx + k 2 8 = 0 NuLake approach, but the results k =! 4 3 (±6.928) E Discriminant > 0 should be broadl similar. 4k 2 4 x 2(k 2 8) > 0 d) x + 3x + (6x 12) = 128 4k > 0 Mar is 72 ears M 4 k 4 E A part eqn. or M eqn. & soln. A correct simplified e) i) Factorise numerator and denominator quadratic or M discriminant > 0. ] x- 4g] x+ 1g = 3] x -4g x = 4 M A some simplification or d) T 1 = a T n = a + (n 1)d M complete simplification first + last S and solution. n = n x 2 ii) Both x = 2 and x = 2 make the original equation undefined. E Question Two S n = n a + S n = n 2 S n = n 2 ( a + (n 1)d) 2 ( a + a + (n 1)d) ( 2a + (n 1)d) E Question Three cont... EAS Workbook Year 12 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent