AS Previous Exams (2.6): Apply algebraic methods in solving problems 4 credits

Transcription

1 AS 9161 Previous Exams 9161 (.6): Apply algebraic methods in solving problems 4 credits

2 L MATHF 9903 Level Mathematics and Statistics, a.m. Thursday 4 November 016 FORMULAE SHEET for 9161, 916, 9167 Refer to this sheet to answer the questions in your Question and Answer Booklets. Check that this sheet is printed on the back. YOU MAY KEEP THIS SHEET AT THE END OF THE EXAMINATION. New Zealand Qualifications Authority, 016. All rights reserved. No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.

3 Quadratics If ax + bx + c = 0 then x = b ± and Δ = b 4ac Logarithms b 4ac a If y = b x then x = log b y log b ( x n ) = n log b x If y = e x then x = log e y = ln y Calculus ( ) = nx n 1 d dx xn ( ) If f (x) = x n, then f (x) = xn+1 n +1 + c Probability z = x µ σ L MATHF 0 z Standard Normal Distribution z = x µ σ Each entry gives the probability that the standardised normal random variable Z lies between 0 and z. Differences z

4 SUPERVISOR S Level Mathematics and Statistics, Apply algebraic methods in solving problems.00 pm Monday 19 November 01 Credits: Four Achievement Achievement with Merit Achievement with Excellence Apply algebraic methods in solving problems. Apply algebraic methods, using relational thinking, in solving problems. Apply algebraic methods, using extended abstract thinking, in solving problems. Check that the National Student Number (NSN) on your admission slip is the same as the number at the top of this page. You should attempt ALL the questions in this booklet. Show ALL working. If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly number the question. You are required to show algebraic working in this paper. Guess and check methods do not demonstrate relational thinking. Guess and check methods will limit grades to Achievement. Check that this booklet has pages 10 in the correct order and that none of these pages is blank. YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION. TOTAL New Zealand Qualifications Authority, 01. All rights reserved. No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.

5 You are advised to spend 60 minutes answering the questions in this booklet. QUESTION ONE (a) Solve (i) log x = 3log (ii) log 5 x = (b) Tara s aunt invests $000 for her when she is born. The interest rate is 3.5% per year. This rate does not change as long as the money stays invested. The interest is added to the amount she has invested on her birthday each year. The value of the investment after t years can be modelled by the equation A = 000 (1.035) t where the A is the value of the investment. (i) How long would it take for the value of the investment to be $50? Mathematics and Statistics 9161, 01

6 3 (ii) Tara reaches her 18th birthday. Calculate how much extra the investment will be worth if she leaves the money invested for another 3 years beyond her 18th birthday. (iii) Tara is calculating m (1.035 n 1) With reference to the investment, explain what Tara is calculating. (c) Solve 9 n (6 3 n ) 7 = 0 and explain why it has only one real solution. Hint: let 3 n = x Mathematics and Statistics 9161, 01

7 4 QUESTION TWO (a) (i) Factorise 5x 9x (ii) Solve 5x 9x = 0 (b) Solve x + 5 x + = 3 x + Show algebraic working. Mathematics and Statistics 9161, 01

8 5 (c) Mark solves the equation x x His working is shown below. x 5x + 6 = 4x + 4x 4 3x + 9x 30 = 0 3(x + 3x 10) = 0 3(x + 6)(x ) = 0 x = 6 or x = 5x+ 6 = 4 + x 6 Is Mark s answer correct? Fully justify your answer. (d) Find the value of c if x + x 6 x + x+ c = x (3x ++ 8) Mathematics and Statistics 9161, 01

9 (e) The width of a canal at ground level is 16 m. The sides of the canal can be modelled by a quadratic expression that would give a maximum depth of 16 m. However, the base of the canal is flat and has a width of 1 m. What is the actual depth of the canal? 6 Ground level Base of canal Mathematics and Statistics 9161, 01

10 7 QUESTION THREE (a) Simplify (i) (x 5 ) (x) 3 (ii) 8x 1 3 (iii) 8x x Mathematics and Statistics 9161, 01

11 8 (b) (i) Mark is solving (x 3)(x + 4) = 13 by using the quadratic formula = ± b b x 4ac a Give the values of a, b and c and hence solve the equation. (ii) The equation (x 3)(x + 4) = k has only one real solution. Find the value of k. Mathematics and Statistics 9161, 01

12 9 (c) Find the possible values of d if real solutions exist for x +5x 1 d(x +1) = 0. Mathematics and Statistics 9161, 01

13 NCEA Level Mathematics (9161) 01 page 1 of 3 Assessment Schedule 01 Mathematics with Statistics: Apply algebraic methods in solving problems (9161) Evidence Statement Q Expected Coverage Achievement (u) Merit (r) Excellence (t) ONE (a)(i) 8 Complete correct solution found. (ii) x = 5 = 5 Complete correct solution found. (b) (i) Log equivalent formed 50/000 = t t = log1.15/log1.035 = 3.4 years Establishing log equation. Problem solved using substitution (at least iterations). Accept 3.4 or 4 (years) or any other rounding. CRO of 3.4 allowed. Do not accept 3 unless accompanied by algebraic working. (ii) 000(1.035) 1 000(1.035) 18 = = = $ Value after 18 or 1 years found. Correct solution. CRO (iii) The additional amount in the account between Tara s m th and (m + n) th birthday. OR The difference in the amount from the m th year to the (m + n) th. Correct statement. (c) x 6x 7 = 0 (x + 3)(x 9) = 0 x = 3 or x = 9 3 n = 3 no solution Only solution is 3 n = 9 n = Quadratic equation in x formed and solved. Expression given for 3 n. Value of n found with algebraic evidence. NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence Attempt at one question 1 of u of u 3 of u 1 of r of r 1 of t of t

14 NCEA Level Mathematics (9161) 01 page of 3 Q Expected Coverage Achievement (u) Merit (r) Excellence (t) TWO (a)(i) (5x + 1)(x ) OR 5(x )(x + 0.) Factorise the expression. (ii) x = 1 5 or or equivalent. Equation solved giving TWO solutions. Accept in fractional form. Consistent with a(i) but not trivial. (b) x + 5x + = 3x + 6 x + x 4 = 0 x = 1.36, 3.36 Expanded and simplified to a quadratic equation = 0. CRO Truncate / rounding ok min 1dp. Equation solved giving TWO correct solutions. Truncate / rounding ok min 1dp. (c) (x 3)(x ) (x + 3)(x ) = 4 (x 3) (x + 3) = 4 3x = 15 x = 5 Two solutions x = 5, x =, with comments about incorrect factorisation (or the correct factorisation). OR An answer to the question with both values substituted showing these solutions do not =4. Correct solution of x=-5 only with one of the two aspects of the incorrect solution discussed. Correct solution and i) A comment about incorrect factorisation. ii) x = gives an invalid solution as it results in dividing by 0 or back substitution shows x = does not satisfy the equation. BOTH required. (d) (x + 3)(x ) 6x + 4x + c = x + 3 (3x + 8) Multiply numerator and denominator of RHS by (x ) (x )(3x + 8) Factorising, and recognising the need to multiply by (x ) to equate denominators. Or cross multiplication and expanding and simplifying correctly. Solving. = 6x + 4x 3 Therefore c = 3 (e) Equation d = a(x+ 8)(x 8) d = a(x 64) d = 0 64a = 16 General equation formed in any correct format. a calculated and equation formed. Depth = 7m a = ¼ d = ¼ (x + 8)(x 8) Width of 1, x = ±6 ¼ 14 = 7 m Problem solved. NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence Attempt at one question 1 of u of u 3 of u 1 of r of r 1 of t of t

15 NCEA Level Mathematics (9161) 01 page 3 of 3 Q Expected Coverage Achievement (u) Merit (r) Excellence (t) THREE (a)(i) 8x 13 Correct. (ii) 1 3 4x Correct (accept 0.3 as power). (iii) 1 3 4x x Consistent with 3a(ii). = x Correct. Or equivalent. (b)(i) x 3x + 8x 1 = 13 x + 5x 5 = 0 a =, b = 5 and c = 5 x =.5 or x = 5 Expanding and simplifying to = 0. Incorrect simplification, then correct use of quadratic formula giving two solutions. CRO. Solution including values for a, b, c. (ii) x + 5x 1 k = 0 For one solution b 4ac = (1+k) = 0 k = Knowledge of statement b 4ac = 0. Incorrect substitution into b 4ac. Correct substitution into b 4ac. Value of k calculated. (c) x + 5x 1 dx d = 0 x (1 d) + 5x (1 + d) = 0 To have solutions 5 + 4(1 d)(1 + d) > d > 0 4d < 9.69 < d <.69 Expansion and simplified equation collecting coefficients (line ). Correct substitution into the discriminant of b 4ac > 0. Including > or. Range for d calculated. Do not penalise for using 0. NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence Attempt at one question 1 of u of u 3 of u 1 of r of r 1 of t of t Judgement Statement Not Achieved Achievement Achievement with Merit Achievement with Excellence Score range

16 SUPERVISOR S Level Mathematics and Statistics, Apply algebraic methods in solving problems.00 pm Monday 18 November 013 Credits: Four Achievement Achievement with Merit Achievement with Excellence Apply algebraic methods in solving problems. Apply algebraic methods, using relational thinking, in solving problems. Apply algebraic methods, using extended abstract thinking, in solving problems. Check that the National Student Number (NSN) on your admission slip is the same as the number at the top of this page. You should attempt ALL the questions in this booklet. Show ALL working. If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly number the question. You are required to show algebraic working in this paper. Guess and check methods do not demonstrate relational thinking. Guess and check methods will limit grades to Achievement. Check that this booklet has pages 10 in the correct order and that none of these pages is blank. YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION. TOTAL New Zealand Qualifications Authority, 013. All rights reserved. No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.

17 You are advised to spend 60 minutes answering the questions in this booklet. QUESTION ONE (a) (i) Factorise 6x 11x 10 (ii) Solve 6x 11x 10 = 0 (b) Find the value of m so that only one value of x satisfies the equation: 4x 8x + m = 0 Mathematics and Statistics 9161, 013

18 3 (c) Simplify fully x 8 x x 8 (d) The equation ( x+ ) 3 ( x+ ) 4= 0 has only one real solution. Find the value of x. (Hint: Let a = ( x+ ) ) Mathematics and Statistics 9161, 013

19 4 (e) (i) Find expressions, in terms of m and n, for the roots of the equation: x m ( x+ m) = x n x+ n (ii) Give an inequality, in terms of m and n, so that the equation has two distinct roots. Mathematics and Statistics 9161, 013

20 6 QUESTION TWO (a) Simplify (4 a ) (8 a ) 3 5 (b) Simplify: (i) (16 x ) (ii) (16 x ) (9 x ) 1 (c) Lara says that she is thinking of a number. She: squares the number, multiplies the answer by 6, adds 1 times the number she was first thinking of, subtracts 48. Her answer is 0. What numbers could she be thinking of? Mathematics and Statistics 9161, 013

21 (d) Rearrange the formula a x = 5 (x 1) to make x the subject. 7 (e) The equation 3x + 4x k = 0 has two distinct real roots. If is a root of this equation, find the value of k and the second root. Mathematics and Statistics 9161, 013

22 8 QUESTION THREE (a) Solve the equations: (i) log x 64 = 3 (ii) 3 x 8 x+ 1 = 96 (b) At the beginning of his first year of study, Danny borrows $1 800 from his parents. His parents reduce the amount he owes them by 40% at the end of that year, and each subsequent year he continues his studies. Danny studies for several years and he does not make any repayments of the initial amount while he is studying. (i) Write an expression for the amount $A Danny owes his parents if he studies for n years. (ii) Use your expression to find the minimum number of years for which Danny studies if he owes his parents less than $100 when he finishes studying. Mathematics and Statistics 9161, 013

23 (c) Explain why the equation (3x + 1) = 7 does not have any real solutions, and explain what this means graphically. 9 (d) Solve the equation log x = log (mx) for x in terms of m. Mathematics and Statistics 9161, 013

24 NCEA Level Mathematics (9161) 013 page 1 of 4 Assessment Schedule 013 Mathematics with Statistics: Apply algebraic methods in solving problems (9161) Evidence Statement ONE Expected Coverage Achievement Merit Excellence (a)(i) (x 5) (3x + ) 6(x + 3 )(x 5 ) Correctly factorised. Correct decimal / rounding. (a)(ii) x = ( 0.67) or x = 5 (.5) Complete correct solutions 3 found. Any rounding / truncation. Consistency with a(i) but not trivial (coefficients of x > 1) (b) b 4ac = 0 16m = 64 m = 4 Perfect square. 4(x 1)(x 1) = 0 or equivalent x = 1 m = 4 Recognising the discriminant must = 0. OR CRO. Calculating the value of m. (c) (x + )(x ) (x + )(x 4) Factorised. OR Correctly simplified. = (x ) (x 4) accept (x 4) (x 4) Factorised and simplified with one error in factorising. (d) a 3 a 4 = 0 (a 4)( a + 1) = 0 a = 4 or a = 1 (x + ) = 4 or ( x + ) = 1 (not a solution) x = 14 ((x+) 4) = (3 (x + )) x 4x + 4 = 9(x + ) x 13x 14 = 0 (x + 1)(x 14) = 0 x = 1 and x = 14 Equation rearranged and factorised OR Solved using either method. (RANW= n) Solved for x. x = 14 and 1 but not disregarding x = 0. Recognition that x = 1 is not a solution. (e)(i) (ii) x mx + nx mn = (x nx + mx mn) x + 3mx 3nx mn = 0 x + (3m 3n)x mn = 0 3(m n) ± x = 9(m n) + 4mn = 3(m n) ± 9m 14mn + 9n Hence 9m 14mn + 9n > 0 OR 9(m n) + 4mn > 0 Cross multiplication and collection of like terms. Mei for one incorrect simplification step. Correct substitution into the quadratic formula, not necessarily simplified. OR No roots given but correct inequality in (ii). Roots and inequality found. NØ no response; no relevant evidence N1 attempt at one question N 1 of u A3 of u A4 3 of u M5 1 of r M6 of r E7 1 of t E8 of t

25 NCEA Level Mathematics (9161) 013 page of 4 TWO Expected Coverage Achievement Merit Excellence (a) 64a 6 /64a 10 = 1/ a 4 (or a 4 ) Correct simplification. (b)(i) x 0.5 ( or x, x 1 Correct simplification. (ii) x 0.5 3x 1.5 = 6x Correct simplification. Consistent with (b) (i). (c) 6x + 1x 48 = 0 x + x 8 = 0 (x + 4) (x ) = 0 x = 4 and x = Correct equation = 0 OR CAO or guess and check. OR If answer correct & x = 4 is eliminated. Equation solved showing equation. (d) x log a = (x 1)log5 = x log5 log5 x(log a log5) = log5 Expression written in log form and expanded. Expression for x correct. OR equivalent. x = log5 log a log5 (e) (3x + n)(x ) = 0 3x + (n 6)x n =0 n 6 = 4 n =10 (3x + n) =0 root is n / 3 = 10 / 3 k = n = 0 Establishing the relationship. OR CRO of k and other root. One value for n or k or the other root found with algebraic working. Solutions found for k and the other root with algebraic working. NØ no response; no relevant evidence N1 attempt at one question N 1 of u A3 of u A4 3 of u M5 1 of r M6 of r E7 1 of t E8 of t

26 NCEA Level Mathematics (9161) 013 page 3 of 4 THREE Expected Coverage Achievement Merit Excellence (a)(i) x 3 = 64 x = 4 (ii) x+1 = 3 3x = 5 x+1 3x = 5 1 x = 5 x = Correctly solved. CRO OR Whole equation in powers of two. OR Use of log, with exponents eliminated. eg: (x + 1)log = log3 +x log8 Correctly solved. (b)(i) n Expression correct. (ii) 100 > n 0.6 n < 1 18 nlog0.6 < log 1 18 n > years OR equivalent. In/Equation rearranged in index form. OR CRO. OR Solved by guess and check. OR 5.7 years with working. OR Consistent use of 0.4 n give n = 3.15 Number of years found as a whole number (n = 6) Consistent use of 0.4 n using logs give n = 4 years (whole number). (c) 9x + 6x + 8 = 0 b 4ac = 5 therefore no real roots. Graph of the parabola does not cut the x- axis. A squared term can never be negative hence there is no solution therefore the graphs do not intersect each other. Quadratic expression rearranged=0 OR Explanation of no x intercepts because the discriminant is less than zero, without 5. OR A squared term can never be negative hence there is no solution. Discriminant found and therefore no real roots but no x axis analysis. OR Quadratic expression rearranged = 0. AND Explanation of no x intercepts because the discriminant is less than zero, without 5. Discriminant calculated and explanation of no x intercepts given. OR Full explanation of the two graphs not intersecting. (d) x = (mx) x(m x 1) = 0 therefore either m x = 1 Equation given in index form. One solution found. x = 1 m Correctly solved with x = 0 disregarded. x = 1 if x 0. m OR x = 0 But log 0 is undefined Therefore x = 1 m OR Both solutions and x = 0 not disregarded. OR Use of log properties to solve completely. X = 0 still needs to be disregarded.

27 NCEA Level Mathematics (9161) 013 page 4 of 4 NØ no response; no relevant evidence N1 attempt at one question N 1 of u A3 of u A4 3 of u M5 1 of r M6 of r E7 1 of t E8 of t Judgement Statement Not Achieved Achievement Achievement with Merit Achievement with Excellence Score range

28 SUPERVISOR S Level Mathematics and Statistics, Apply algebraic methods in solving problems.00 pm Wednesday 19 November 014 Credits: Four Achievement Achievement with Merit Achievement with Excellence Apply algebraic methods in solving problems. Apply algebraic methods, using relational thinking, in solving problems. Apply algebraic methods, using extended abstract thinking, in solving problems. Check that the National Student Number (NSN) on your admission slip is the same as the number at the top of this page. You should attempt ALL the questions in this booklet. Make sure that you have Resource Sheet L MATHF. Show ALL working. If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly number the question. You are required to show algebraic working in this paper. Guess and check methods and correct answer only will generally limit grades to Achievement. Check that this booklet has pages 10 in the correct order and that none of these pages is blank. YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION. TOTAL New Zealand Qualifications Authority, 014. All rights reserved. No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.

29 QUESTION ONE (a) Simplify: (i) 5 a ( ) (ii) 0.5x 3 (iii) ( 8x ) 6 3 x 1 3 ( ) 4 (b) One root of the equation x + mx + 1 = 0 is three times the other. Find the values of m. Mathematics and Statistics 9161, 014

30 (c) The equation 3x nx + 5 = 0 has two distinct roots. Find the values of n. 3 (d) Solve 10x 4 13x + 4 = 0 You must show algebraic working. Mathematics and Statistics 9161, 014

31 5 QUESTION TWO (a) Factorise and solve 1a 11a 15 = 0 (b) (i) Write as a single fraction 3 x 4x x + 1 (ii) Solve the equation x + x 8 = 3 x x You must show algebraic working. Mathematics and Statistics 9161, 014

32 6 (c) (i) The height h metres of a tunnel is modelled by a function of the form h = r x t x where r and t are constants. Make x, the distance in metres from the left side of the tunnel, the subject of the equation. (ii) The shape of the tunnel can be modelled by a parabola. The maximum height of the tunnel is 6 m, and at ground level its width is 1 m. Find the equation of the parabola. Mathematics and Statistics 9161, 014

33 (iii) There are two lanes of equal width through the tunnel. The outside edge of each lane is marked by a line so that a car of height 1.8 m would have a minimum clearance of 0.1 m vertically from the top of the car to the tunnel roof. (Ignore the width of the line.) Find the width of each lane. 7 left side of lane Mathematics and Statistics 9161, 014

34 8 QUESTION THREE (a) (i) Find the value of x if x = log 3 81 (ii) Solve the equation log x 343 = 3 (b) Solve for x: 5 x x = 15 Mathematics and Statistics 9161, 014

35 (c) Thirty minutes after a patient is administered his first dose of a medication, the amount of medication in his blood stream reaches 4 mg. The amount of the medication in the blood stream decreases continuously by 0% each hour. The amount of the medication M mg in the patient s blood stream after it is administered can be modelled by the function M = t 0.5 where t is the time in hours since the drug was administered. 9 (i) Explain what the 0.8 represents in this function. (ii) Find the amount of medication administered initially. (iii) A second dose of the medication can be administered some time later, and again the amount of the medication in the patient s bloodstream from the second dose can be modelled by the same function as that for the first. The total amount of the drug in the blood stream must never exceed 300 mg. How long after administering the first dose can the second dose be administered? Mathematics and Statistics 9161, 014

36 NCEA Level Mathematics and Statistics (9161) 014 page 1 of 3 Assessment Schedule 014 Mathematics and Statistics: Apply algebraic methods in solving problems (9161) Evidence Statement One Expected Coverage Achievement (u) Merit (r) Excellence (t) (a)(i) a 1 15 Correct. (ii) 0.5x 1.5 or 0.5x 3 Correct equivalent. (iii) x ( ) 3x 8 = 3 x10 Numerator or denominator correct. Correct expression. (b) Let the roots be n and 3n. 3n = 1 n = 4 n = ± 4n = ±m m = ±8 Relationship shown. Values of n found. Values of m found. (c) 3x nx + 5 = 0 For solutions n > 0 n > 60 n > 7.7 or n < 7.7 Discriminant given. One value of n found. Both parts of inequality given. (d) (5x 4)(x 1) x = 0.8 or 0.5 x = ±0.894 or ±0.707 Expression factorised. Solved for x. All solutions given. NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence. Attempt at one question 1 of u of u 3 of u 1 of r of r 1 of t of t

37 NCEA Level Mathematics and Statistics (9161) 014 page of 3 Two Expected Coverage Achievement (u) Merit (r) Excellence (t) (a)(i) (3a 5)(4a + 3) a = 5 3 or 3 4 Correct factorising or solution. Correct solution showing factorisation. (b)(i) 3x + 3 4x + 8x (x )(x +1) = 4x +11x + 3 x x Correct single fraction Correctly simplified. Accept with denominator in factorised form. (ii) (x + 4)(x ) (x +1)(x ) = 3 x + 4 = 3x + 3 if x x = 1 x = 0.5 (c)(i) rx tx h = 0 Factorised and simplified CAO. Solved for x. Condition x. Answer with ± before surd. x = t ± t + 4rh r (ii) h = ax(x 1) when x = 6, h = 6 6 = 6a 6 General form of equation and recognition of point (6,6). Correct equation. a = 1 6 h = 1 x(x 1) 6 Or y = 1 6 x + x (iii) h = 1 x(x 1) = x +1x = 0 Solved for height of 1.9 m. Correct width of lane found. x = 1.04, width of lane m NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence. Attempt at one question 1 of u of u 3 of u 1 of r of r 1 of t of t

38 NCEA Level Mathematics and Statistics (9161) 014 page 3 of 3 Three Expected Coverage Achievement (u) Merit (r) Excellence (t) (a)(i) 3 x = 81 x = 4 Correctly solved. (ii) 343 = x 3 x = 7 Correctly solved. (b) 5 4 x = 15 x log1.5 = log15 x = log15 log1.5 x = 1.14 Expression simplified. Written in log form. x found. (c)(i) 0.8 is the fraction of medication remaining after an hour. Correct explanation. (ii) M = t 0.5 = = 50.4 mg Statement with t = 0 and attempt to solve. Correctly solved. (iii) 49.6 = t 0.8 t = t log0.8 = log t = 7.5 hours M = 49.6 recognised and attempt to solve. Correctly used logs in attempt to solve Correctly solved. NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence. Attempt at one question 1 of u of u 3 of u 1 of r of r 1 of t of t Cut Scores Not Achieved Achievement Achievement with Merit Achievement with Excellence Score range

39 SUPERVISOR S Level Mathematics and Statistics, Apply algebraic methods in solving problems.00 p.m. Tuesday 10 November 015 Credits: Four Achievement Achievement with Merit Achievement with Excellence Apply algebraic methods in solving problems. Apply algebraic methods, using relational thinking, in solving problems. Apply algebraic methods, using extended abstract thinking, in solving problems. Check that the National Student Number (NSN) on your admission slip is the same as the number at the top of this page. You should attempt ALL the questions in this booklet. Make sure that you have Resource Sheet L MATHF. Show ALL working. If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly number the question. You are required to show algebraic working in this paper. Guess-and-check methods and correct answer(s) only will generally limit grades to Achievement. Check that this booklet has pages 11 in the correct order and that none of these pages is blank. YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION. TOTAL New Zealand Qualifications Authority, 015. All rights reserved. No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.

40 QUESTION ONE (a) (i) Find the value of log 104. (ii) Solve the equation log 4 (3w + 1) =. (iii) Luka says that the equation log x (4x + 1) = has only one solution. Is he correct? Find the solution(s), justifying your answer. (b) Make x the subject of the equation a x = b x+1. Mathematics and Statistics 9161, 015

41 (c) The market value of Sue s house has been increasing at a constant exponential rate of 3% per annum since she bought it sixteen years ago at the start of At the start of 015 it was worth $ (i) 3 Assuming the exponential growth is of the form y = A r t, what was the value of the house at the start of 1999 when she bought it? (ii) A friend also bought a house at the start of 1999 that cost $ Its market value also has been steadily increasing, but at a slightly higher exponential rate of 3.5%. Its value, $y, t years after the start of 1999, is given by the function y = (1.035) t If the houses continue to keep increasing in value at the original rates, in which year will the two houses be worth the same amount? Mathematics and Statistics 9161, 015

42 4 QUESTION TWO (a) Simplify x + 7x 4 x 3 (b) 3 If a = y 4, find an expression for a 7 in terms of y. (c) 1 Solve the equation u 3 + 7u 3 = 4 Mathematics and Statistics 9161, 015

43 (d) Talia used timber to form the exterior sides of her rectangular garden. The length of the garden is x metres, and its area is 50 m. (i) Show that the perimeter of the garden is given by x x 5 (ii) If she uses 33 m of timber to build the sides, find the dimensions of the garden. Mathematics and Statistics 9161, 015

44 (e) David and Sione are competing in a cycle race of 150 km. Sione cycles on average 4 km per hour faster than David, and finishes half an hour earlier than David. Find David s average speed. You MUST use algebra to solve this problem. (Hint: average speed = distance ) time 6 Mathematics and Statistics 9161, 015

45 7 QUESTION THREE (a) Simplify, giving your answer with positive exponents: (i) a 10 4a 5 (ii) 5 3 x 5 3 (b) Solve the following equation for t: 1 t(t 1) 1 t = 3 t 1 Question Three continues on the following page. Mathematics and Statistics 9161, 015

46 (c) 8 For what value(s) of k does the graph of the quadratic function y = x + (3k 1)x + (k + 10) never touch the x-axis? Mathematics and Statistics 9161, 015

47 9 (d) The quadratic equation mx (m + )x + = 0 has two positive real roots. Find the possible value(s) of m, and the roots of the equation. Mathematics and Statistics 9161, 015

48 NCEA Level Mathematics and Statistics (9161) 015 page 1 of 5 Assessment Schedule 015 Mathematics and Statistics: Apply algebraic methods in solving problems (9161) Evidence One Expected Coverage Understanding (u) Relational thinking (r) Abstract thinking (t) (a)(i) x = 104 x = 10 Equation solved. (a)(ii) 3w + 1 = 4 3w = 15 and w = 5 Equation solved. (a)(iii) x = 4x + 1 x 4x 1 = 0 (x 6)(x + ) = 0 x = 6 or But base must be positive x = 6 is the only solution Sets up a quadratic equation. Solved problem using quadratic, but gives both values. Gives only valid solution with justification. (b) x log a = (x +1)logb x(log a logb) = logb x = logb log a logb Takes logs of both sides and multiplies by indices. Takes logs of both sides and rearranges. Correctly solved. (c)(i) P = A (1.03) t Beginning of 1999, t = 0, t = 16, P = = A (1.03) 16 A = So price was $ initially. Sets up model correctly. Answers question in context correctly. (c)(ii) (1.03) t = (1.035) t = t Set up correct equation. Solved for t. Correct year identified = t t = log log t = In 016. NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence. Attempt at one question 1 of u of u 3 of u 1 of r of r 1 of t of t

49 NCEA Level Mathematics and Statistics (9161) 015 page of 5 Two Expected Coverage Understanding (u) Relational thinking (r) Abstract thinking (t) (a) (x + 4)(x 1) (x 16) = = (x + 4)(x 1) (x 4)(x + 4) (x 1) (x 4) Provided x ±4 Factorised and simplified with one error. Correctly simplified. (b) 3 a 7 4 = y 7 Correct expression. 1 4 = y (c) 1 3 Let x = u x + 7x 4 = 0 (x 1)(x + 4) = 0 CAO or rewrites as quadratic. x = 1 or x = 4 Solves for x. 1 u 3 = 1 so u = 1 3 = 1 8 Solves completely with both solutions. 1 OR u 3 = 4 so u = ( 4) 3 = 64 (d)(i) Let x be the length and w the width. Then the perimeter is x + w. Area xw = 50 So w = 50 x So perimeter = x x or w = 100 x Shows relationship. (d)(ii) x x = 33 x 33x +100 = 0 (x 5)(x 4) = 0 x = 1.5 or x = 4 m Forms a quadratic equation Solved for x, or consistently solved from incorrect quadratic. Correctly solved and dimensions given. So the dimensions of the garden are 4 m and 1.5 m.

50 NCEA Level Mathematics and Statistics (9161) 015 page 3 of 5 (e) David s speed is x km / h Sione s speed is (x + 4) km / h Difference in time is half an hour. Sets up equation correctly and solves with an error. Correctly sets up equation and solves correctly. 0.5 = 150 x 150 x (x + 4) 150x 0.5 = x(x + 4) 0.5x(x + 4) = 600 x + 4x 100 = 0 x = 3.70 km / hr NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence. Attempt at one question 1 of u of u 3 of u 1 of r of r 1 of t of t

51 NCEA Level Mathematics and Statistics (9161) 015 page 4 of 5 Three Expected Coverage Understanding u Relational thinking r Abstract Thinking t (a)(i) a 10 4a 5 = 4 a 5 = 4a5 a 10 = 16 a 10 Evidence of correctly simplifying the negative or index or square or numerator or denominator correct Algebraic expression simplified. (ii) ( x ) x 5 = 3 x 5 3 = ( 3) 3 = = 8 x 3 x 3 Numerator or denominator correct. Algebraic expression simplified. (b) 1 t(t 1) t 1 t(t 1) 3t t(t 1) = 0 1 t +1 3t t(t 1) 4t t(t 1) = 0 (1 t) t(t 1) = 0 = 0 Partially solved by rewriting over correct common denominator. t = 1 Correctly solved. (c) Never touch the x-axis means < 0 (3k 1) 4(k +10) < 0 9k 6k +1 8k 40 < 0 9k 14k 39 < 0 If 9k 14k 39 = 0 Then (9k +13)(k 3) = 0 and k = 3 or 13 9 ( 1.44) < 0 Correct solutions for k. So if the graph does not cut the x-axis, then 13 9 < k < 3 Problem solved with correct inequality.

52 NCEA Level Mathematics and Statistics (9161) 015 page 5 of 5 (d) If both roots real, so > 0 ie [ (m + )] 4m > 0 So m 4m + 4 > 0 i.e. (m ) > o So m can be any real number but m, as any number squared except zero is always positive. Using the quadratic formula or otherwise, the roots are > 0 m + ± (m ) m m + ± (m ) = m x 1 = m m = 1 provided m 0 or x = 4 m = m provided m 0 So to fill all conditions including both roots are positive real, we have m > 0 m with roots 1 and m. (m ) > 0 and m OR BOTH roots found. Problem solved correctly. NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence. Attempt at one question 1 of u of u 3 of u 1 of r of r 1 of t of t Cut Scores Not Achieved Achievement Achievement with Merit Achievement with Excellence

53 SUPERVISOR S Level Mathematics and Statistics, Apply algebraic methods in solving problems 9.30 a.m. Thursday 4 November 016 Credits: Four Achievement Achievement with Merit Achievement with Excellence Apply algebraic methods in solving problems. Apply algebraic methods, using relational thinking, in solving problems. Apply algebraic methods, using extended abstract thinking, in solving problems. Check that the National Student Number (NSN) on your admission slip is the same as the number at the top of this page. You should attempt ALL the questions in this booklet. Make sure that you have Formulae Sheet L MATHF. Show ALL working. If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly number the question. You are required to show algebraic working in this paper. Guess-and-check methods and correct answer(s) only will generally limit grades to Achievement. Check that this booklet has pages 11 in the correct order and that none of these pages is blank. YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION. TOTAL New Zealand Qualifications Authority, 016. All rights reserved. No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.

54 QUESTION ONE (a) 3b Simplify c 4 leaving your answer with positive indices. (b) Write x 8x + 10 in the form (x p) + q. (c) (i) Show that the solutions of the equation x + x 56 = 0 are four times the solutions of the equation 4x + x 14 = 0. (ii) Find the relationship between the solutions of the equation dx + ex + f = 0 and the solutions of the equation x + ex + df = 0, where d, e, and f are real numbers. Mathematics and Statistics 9161, 016

55 3 (d) A quadratic equation of the form ax + bx + c = 0 has solutions 1 and 3. Find a possible set of values for a, b, and c. (e) Find positive integer value(s) for k so that the quadratic equation x + 4kx + (k + 3k 11) = 0 has real rational solutions. Justify your answer. Mathematics and Statistics 9161, 016

56 4 QUESTION TWO (a) Find the discriminant of the quadratic equation x = 10x + 3. ( ) (b) Simplify 4 log u 3 logu. (c) Marie buys a new car for $ The car s value decreases continuously by 1% each year. The value of the car, $P, t years after she first bought it, can be modelled by a function of the form P = A(r) t. How long will it take for the value of the car to halve? Mathematics and Statistics 9161, 016

57 5 (d) (i) Solve the equation log x = 3 8. (ii) Solve the equation 6(log 8 x) + log 8 x 4 = 0. Mathematics and Statistics 9161, 016

58 (e) The diagram below shows a triangular garden with a path around it. 6 The triangular garden has sides with lengths in the ratio 3:4:5. The path is 1 m wide. At each corner of the garden, the path is a sector (part) of a circle with a radius of 1 m. The difference between twice the total area of the path and the area of the garden is π m. Find the length of the longest side of the garden. (Area of circle = πr ) Mathematics and Statistics 9161, 016

59 7 QUESTION THREE (a) Where would the graph of y = 1x x 6 cut the x-axis? (b) For what value(s) of x does log x (16) = 3? 4x y( x + 3) (c) Rearrange the following formula to make x the subject: =. 5 Question Three continues on the following page. Mathematics and Statistics 9161, 016

60 8 (d) Solve the equation 9 = 7 3 8n+ 6 n 1 1 3n. (e) A symmetrical bridge has its central cable in the shape of a parabola, as shown in the diagram below. The towers supporting the cable are each 15 m high and 40 m apart. At the point midway between the towers, the height of the cable above the road is 3 m. A vertical post (shown dotted in the diagram) is placed 10 m from the centre of the bridge and just touches the cable. Diagram is NOT to scale 15 m road 40 m (i) Use algebra to show that the post is 6 m high. Mathematics and Statistics 9161, 016

61 9 (ii) The length of the bridge AB is 60 m. The outside cables are also parabolic and symmetrical in shape, and touch the road at their vertices A and B. Diagram is NOT to scale C 15 m D road A 40 m 60 m B Find the distance, CD, between the two parabolas at a height of 6 m above the road (the distance CD is shown in the diagram). Mathematics and Statistics 9161, 016

62 NCEA Level Mathematics and Statistics (9161) 016 page 1 of 6 Assessment Schedule 016 Mathematics and Statistics: Apply algebraic methods in solving problems (9161) Evidence Statement Q Expected Coverage Achievement (u) Merit (r) Excellence (t) ONE (a) 4 c 3b = c8 81b 4 Negative or fourth power correctly used. Correct answer. (b) x 8x + 10 = (x 4) 6 Correct arrangement or p and q given. (p = 4 and q = 6 not required) (c)(i) When x + x 56 = 0 (x + 8)(x 7) = 0 x = 8 or 7. When 4x + x 14 = 0 (4x 7)(x + ) = 0 Both equations factorised correctly if solution is incorrect. Both quadratics solved and relationship stated. x = 7 4 or So the solutions of the first quadratic are four times those of the second. (c)(ii) Solutions of dx + ex + f = 0 are e ± e 4df d and those of x + ex + df = 0 are e ± e 4df So the solutions of the second quadratic are d times those of the first. One set of solutions found. All solutions found. Devised a strategy and developed a chain of logical reasoning to solve the problem. (d) x + 1 x 3 = 0 (x + 1)(3x ) = 0 6x x = 0 So a = 6, b = 1 and c = or any other correct values of a, b, and c. Quadratic found with correct values of a, b, and c. Correct values of a, b and c stated.

63 NCEA Level Mathematics and Statistics (9161) 016 page of 6 (e) To have rational roots, the discriminant is 0 (accept > 0 at achieved and merit) a perfect square Hence 16k 4 (k + 3k 11) 0 4k , k 88 4 or 11 3 Cases, k integer Values substituted into discriminant. k 11 3 or unsimplified. OR One value for k found with reason. OR Discriminant used correctly with one restriction on k given. Both values of k found with logical chain of reasoning. k = 0 not possible as told k positive k = 1, = 64 which is a square k =, = 40 which is not a square k = 3, = 16 which is a square k 4 not possible as negative So only possible values are k = 1 or 3. NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence. Attempt at ONE question. 1 of u OR partial solution in TWO questions. of u 3 of u 1 of r of r 1 of t of t

64 NCEA Level Mathematics and Statistics (9161) 016 page 3 of 6 Q Expected Coverage Achievement (u) Merit (r) Excellence (t) TWO (a) x 10x 3 = 0 discriminant = = 11 Correct discriminant found. (b) 4 log( u 3 ) = 1logu logu logu = 1 Power rule for logs in numerator used. Correct answer. (c) P = 4 990(0.88) t = 4 990(0.88) t 0.5 = 0.88 t t = log0.5 log 0.88 CAO or equation set up and error made in solving. Correct equation solved to find value of t. Accept t = 6 if working shown. = 5.4 So it takes 5.4 years to halve in value. (d)(i) 3 x = 8 Correct value found. = = 4 (d)(ii) If u = log 8 x Then 6u + u 4 = 0 u = or 1 3 Either log 8 x = 3 or log 8 x = 1 x = 8 3 or 8 1 so x = 4 or 1 8 CAO. OR Quadratic formed.. Both values for u found. Devised a strategy and developed a chain of logical reasoning to solve the problem. Both values of x found.

65 NCEA Level Mathematics and Statistics (9161) 016 page 4 of 6 (e) Let the sides of the triangle be 3y, 4y, and 5y for some real positive number y. Area of triangle is ½ 3y 4y = 6y Path has width 1. So path area is 1y + π (1y + π) 6y = π 4y + π 6y π = 0 4y 6y = 0 Quadratic established. Quadratic solved for y, or consistently solved from incorrect quadratic. Correctly solved and dimensions given. 6y(4 y) = 0 So y = 4 ( as can t be 0) and length of longest side of triangle is 5 4 = 0 m. (Longest side.35m accepted as an alternative interpretation.) NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence. Attempt at ONE question. 1 of u OR partial solution in TWO questions. of u 3 of u 1 of r of r 1 of t of t

66 NCEA Level Mathematics and Statistics (9161) 016 page 5 of 6 Expected Coverage Achievement (u) Merit (r) Excellence (t) THREE (a) 1x x 6 = (4x 3)(3x + ) x = 3 4 or 3 Correct solutions. (b) log x 16 = 3 x 3 = 16 3 so x = 6 or 16 ( c) 4x y(x + 3) = 5 8x = 5y(x + 3) x(8 5y) = 15y x = 15y 8 5y Correct answer. Terms involving x collected to one side. Correctly solved. (d) 9 8n+6 = 7 n n ( 3 ) 8n+6 = 3 3(n 1) 3 1 3n 3 16n+1 = 3 3n 3+1 3n Base changed to 3 in all terms. Quadratic established. 16n +1 = 3n 3+1 3n 3n 19n 14 = 0 (3n + )(n 7) = 0 n = 3 or 7 Devised a strategy and developed a chain of logical reasoning to solve the problem. Correct values for n found. (e)(i) Assuming origin is at the centre of the bridge, vertex of middle parabola is (0,3), so form of parabola is y = ax + 3 x = 0, y = 15 gives 15 = a(400) + 3 So a = y = x When x = 10, y = = 6 as required. Equation formed and y = 6 when x = 10 shown.

67 NCEA Level Mathematics and Statistics (9161) 016 page 6 of 6 (e)(ii) For the second parabola on right using the same origin. y = a(x 30) When x = 0, y = 15 so 15 = a( 10) a = = 15 (x 30) 100 (x 30) = 40 x 30 = ±6.3 x = 30 ± 6.3 x = , as other value beyond end of bridge. So x = metres and the horizontal distance is or metres. Model for second parabola given and used to find value of x. (Accept other forms of parabolas.) Devised a strategy and developed a chain of logical reasoning to solve the problem. Correct length found. NØ N1 N A3 A4 M5 M6 E7 E8 No response; no relevant evidence. Attempt at ONE question. 1 of u OR partial solution in TWO questions. of u 3 of u 1 of r of r 1 of t of t Cut Scores Not Achieved Achievement Achievement with Merit Achievement with Excellence