008 Trial HSC Eamination MATHEMATICS General instructions Working time 3 hours. plus 5 minutes reading time) Write on the lined paper in the booklet provided. Each question is to commence on a new page. Write using blue or black pen. Class teacher please ) Mr Ireland Mr Lam Mr Lowe Mr Fletcher Mr Trenwith/Mr Taylor Board approved calculators may be used. All necessary working should be shown in every question. Attempt all questions. This paper must NOT be removed from the eamination room by the candidate. STUDENT NUMBER:... Marker s use only. QUESTION 3 5 6 7 8 9 0 Total % MARKS 0
008 Mathematics Trial HSC Eamination Question Marks) Commence a NEW page. Marks Evaluate ) e.5 correct to 3 significant figures. b) Epress + 5 with a rational denominator. c) Differentiate y = + ) 3 with respect to. d) Factorise y y fully. e) Solve the following for : i. 3 = 3. ii. =.
008 Mathematics Trial HSC Eamination 3 Question Marks) Commence a NEW page. Marks The line l has the equation + 3y + 6 = 0. It cuts the ais at A and the y ais at B and it intersects the line k at C. Line k is perpendicular to l and cuts the ais at D. y l k A + 3y + 6 = 0 B D C NOT TO SCALE Copy or trace the diagram on to your paper. i. Find the coordinates of A. ii. Find the coordinates of B. iii. If B is the midpoint of AC prove that the coordinates of C are 3, ). iv. Show that the equation of k is given by 3 y 7 = 0. v. Write the 3 inequalities required to define the interior region of ACD. 3 b) Find the equation of the tangent to the curve y = ln at the point P where = e. 3
008 Mathematics Trial HSC Eamination Question 3 Marks) Commence a NEW page. Marks Consider the parabola ) = 8y + ). i. Write down the focal length. ii. Write down the coordinates of the focus. iii. Find the equation of the directri. b) Differentiate with respect to : i. 3. ii. sin e. c) Evaluate e + ) d. d) Find an approimation for 3 g) d by using Simpson s Rule with the following function values in the table below, correct to decimal places. 3.5.5 3 g) 8 0 3 5
008 Mathematics Trial HSC Eamination 5 Question Marks) Commence a NEW page. Marks Michael is training for a local marathon. He has trained by completing practice runs over the marathon course. So far he has completed three practice runs with times shown below. Week Time hours) 3.7 3.3 i. Show that these times form a geometric series with common ratio r = 0.9. ii. iii. iv. If this series continues, what would be his epected time in Week 5, completed to the nearest minute? How many hours and minutes will he have run in total in his practice runs in these 5 weeks? If the previous winning time for the marathon was hour 5 min, how many weeks must he keep practising to be able to run the marathon in less that the previous winning time? b) A, B and C are markers in an orienteering course. AC = km and BC = 5 km. The bearing of C from B is 00 T. N C 5 km km B 0 A Copy or trace the diagram into your writing booklet. i. If the bearing of B from A is 60 T, show that CBA = 0, giving reasons. ii. Find CAB to the nearest degree. iii. Hence or otherwise, find the bearing of C from A.
cm 6 008 Mathematics Trial HSC Eamination Question 5 Marks) Commence a NEW page. Marks If f) =, i. Evaluate and epand f + h). ii. Hence or otherwise, differentiate f) = from first principles. 3 b) In ABC as shown in the diagram, ABC = 5, ADB = 55, AD = cm and DC = cm. A 5 55 D cm B C NOT TO SCALE i. Show that ABC BDC. 3 ii. Find, the length of BC. c) Let α and β be the solutions of 6 = 0. i. Find α + β. ii. Find αβ. iii. Hence, find 3α α.
008 Mathematics Trial HSC Eamination 7 Question 6 Marks) Commence a NEW page. Marks A particle is moving in a straight line. Its velocity for t 0 is given by v = t + t i. Find when the particle changes direction. ii. Find the eact distance travelled in the first seconds. 3 b) For the function y = 3 3 9 +, i. Find the coordinates of any stationary points and determine their nature. 3 ii. Find any points of infleion. iii. Neatly sketch the curve.
8 008 Mathematics Trial HSC Eamination Question 7 Marks) Commence a NEW page. Marks A farmer has a large tank full of water. The tank leaks water from a hole. The volume of water remaining in the tank, in litres, is given by V = 000 + 0 000e 0.0t where t is the time in hours after the leakage commenced. i. How many litres of water were in the tank when the leakage commenced? ii. At what rate is the water leaking after 5 hours? Answer correct to decimal place. iii. How many litres will eventually be in the tank after a long period of time? iv. If the farmer realises the tank is leaking when the volume of water remaining is 6 000 L, how long did it take him to realise there was a hole in the tank? Answer correct to the nearest minute. b) The diagram shows the graphs y = sin and y = cos, 0 π. The graphs intersect at A and B. y A y = sin y = cos π π B i. Show that A has coordinates ) π, and find the coordinates of B. 3 ii. Find the area enclosed by the two graphs. 3
008 Mathematics Trial HSC Eamination 9 Question 8 Marks) Commence a NEW page. Marks The diagram shows the region bounded by the curve y =, the line y = 6 and the and y aes. y 3 6 y = Find the volume of the solid of revolution formed when the region is rotated about the y ais. b) Kevin plays computer games competitively. From past eperience, Kevin has a 0.8 chance of winning a game of Sawcraft and a 0.6 chance of winning CounterStrife. During a LAN party he plays two games of Sawcraft and one of CounterStrife. What is the probability that he will win: i. all 3 games? ii. No games? iii. At least game? c) For the quadratic equation + p 3) p + ) = 0, i. Show that the discriminant is = p + p + 3. ii. Hence or otherwise, show that the quadratic equation +p 3) p+) = 0 will always have real, distinct roots for real valued p.
0 008 Mathematics Trial HSC Eamination Question 9 Marks) Commence a NEW page. Marks Consider the geometric series + 3 sin + 6 9 sin + 6 7 sin6 +. i. When the limiting sum eists, find its value in simplest form. ii. For what values of in the interval 0 < < π series eist? does the limiting sum of this b) The diagram below represents in metres) the dimensions of a small garden. 5 + y i. Show that y = 0 ). ii. iii. Write an epression, in terms of, for the perimeter P in metres) of the garden, and find a value of for which dp d = 0 Establish whether this value of gives a minimum or maimum value of P and find that value of P.
008 Mathematics Trial HSC Eamination Question 0 Marks) Commence a NEW page. Marks A city has a growing population at a rate proportional to the current population, that is dp dt = kp i. Verify that Pt) = P 0 e kt, t > 0 is a solution of the equation. ii. If the population on January 006, which is t =, was 7 00 and on January 007 when t = ) was 5 800, find the initial population and the value of k. Round your answer down to the nearest whole number. iii. Find the population on January 009. iv. Find the time it will take for the population to double. b) A car dealership has a car for sale for the cash price of $0 000. It can also be purchased on terms over 3 years. The first 6 months are interest free. Subsequently, interest is charged at % per annum, calculated monthly. Repayments are to be made in equal monthly instalments at the end of the first month. A customer purchases the car on these terms and agrees to monthly repayments of $M per month. Let $A n be the amount owing at the end of the n-th month. i. Find an epression for A 6. ii. Show that A 8 = 0 000 6M).0 M +.0). iii. Find an epression for A 36. iv. Find the value of M. End of paper.
STANDARD INTEGRALS n d = n + n+ + C, n ; 0 if n < 0 d = ln + C, > 0 e a d = a ea + C, a 0 cos a d = sina + C, a 0 a sin a d = cos a + C, a 0 a sec a d = tan a + C, a 0 a sec atan a d = seca + C, a 0 a a + d = a tan a + C, a 0 a d = sin + C, a > 0, a < < a a a d = ln + ) a + C, > a > 0 + a d = ln + ) + a + C NOTE: ln = log e, > 0
008 Mathematics Trial HSC Eamination SOLUTIONS 3 Suggested marking scheme Question marks) [] for correct value. [] for 3 significant figures. b) marks) ) e.5 = 0.83 3 s.f.) [] for multiplying by the fraction with appropriate conjugate surd. [] for final answer. + 5 5 0 0 = = 5 c) marks) [] for correct usage of chain rule. [] for final answer. dy d = dy du du d = 3 + ) d) marks) = + ) y = u 3 u = + y = 3u u = [] for correctly factorising y. [] for correctly factorising cubic. y y = y 3 y 3 ) = y y) + y + y ) ii. marks) [] for correctly factorising quadratic. [] for =,. Question i. mark) = = 0 ) + ) = 0 =, + 3y + 6 = 0 When y = 0, + 6 = 0 ii. mark) = 6 = 3 + 3y + 6 = 0 When = 0, 3y + 6 = 0 3y = 6 y = iii. marks) [] for using midpoint formula. [] for final answer C3, ). c + 3) 0, ) = c 3 = 0 c = 3 C3, ), y ) c + 0 y c = y c = iv. marks) [] for correct gradient of k. [] for correct y intercept of k. e) i. marks) [] for identifying 3 = 5. [] for correct solution. 3 = 3 = 5 3 = 5 = 8 = m l = 3 m k = 3 as l k y = 3 + b =3 y= = 3 3 + b b = 9 = 7 y = 3 7 3 y 7 = 0 LAST UPDATED SEPTEMBER, 008
008 Mathematics Trial HSC Eamination SOLUTIONS v. 3 marks) [] for each correct inequality. y 3 y 0 y 3 7 + 3y + 6 0 y 0 3 y 7 0 b) 3 marks) [] for application of product rule. [] for finding function value at = e. [] for final answer. y = ln u = v = ln u = v = dy d = uv + vu = + ln = + ln =e = e + eln e = 3e The function value at = e is y = ln = e =e Substituting e,e ) into equation of the tangent, Question 3 i. mark) y = 3e + b e = 3e + b b = e =e y=e y = 3e e ) = y + ) a = b) i. marks) [] for correct differentiation of each term. d 3 ) = 6 + d ii. mark) [] for correct application of product or quotient rule [] for correct final answer c) marks) dy d = vu uv v = e cos e sin e ) = e cos sin ) e e cos sin = e u = sin v = e u = cos v = e Alternatively, apply the product rule to y = e sin to obtain y = e cos sin ) [] for finding the primitive. [] for correct evaluation of limits. e + ) d) 3 marks) d = 3 3 + ln = 3 e e 3 ) + ln e ln ) = 3 e3 + 5 3 [] recollection of Simpson s Rule [] substitution of pronumerals. [] evaluation. ii. mark) V, ) a = F, + a) = F,) iii. mark) y = a = 3 A h 3 y + y even + y odd + y l ) = ) + 3 8 + 3) + 0 + 5 = 6 7 + ) = 6 6 LAST UPDATED SEPTEMBER, 008
008 Mathematics Trial HSC Eamination SOLUTIONS 5 Question i. mark) T =.7 T 3.0 = 0.9 T 3 =.3 T.70 = 0.9 ii. marks) [] for substitution of pronumerals. [] for correct answer to nearest minute. T n = ar n T 5 = 3 0.9 =.9683 = h 59 min iii. marks) [] for substitution of pronumerals. [] for correct answer to nearest minute. B 0 5 km N C W A A E B 60 W A AB = 0 since the bearing of B from A is 60. ABE B = 0 alt. on lines) CBA = 90 0 0 = 0. complementary ) ii. marks) [] for application of sine rule. [] for final answer. km b) S n = arn ) r S 5 = 3 0.95 ) 0.9 =.853 = h 7 min iv. marks) [] for correct evaluation resulting in n 9.3. [] for correct answer to n. T n =.5 3 h = 3 3 0.9n 0.9 n =.5 3 n )log 0.9 = log.5 3.5 log 3 n = log 0.9 + 9.3 Michael must run for 0 weeks to improve on the previous record of h 5 min. i. marks) [] for correct arithmetic. [] for correct reasoning. Any arithmetic/reasoning that is not acceptable will result in no marks awarded. 5 sin 0 sin CAB = 5 sin CAB = 5 5sin 0 0.803 CAB = 53 8 = 53 nearest ) iii. mark) The bearing of C from A is 53 8 + 60 = 33 8. Also accept 33. Question 5 i. mark) f + h) = + h) + h) = + h + h h ii. 3 marks) [] for recollection of limit. [] for substitution. [] for final answer. f + h) f) = + h + h h) ) = h + h h f f + h) f) ) = lim h 0 h h + h h = lim h 0 h h + h ) = lim = h 0 h LAST UPDATED SEPTEMBER, 008
6 008 Mathematics Trial HSC Eamination SOLUTIONS b) i. 3 marks) [] for each reason shown in summary. Question 6 B i. marks) A 5 cm [] for recollection of particle changing direction when v = 0. [] for correct arithmetic and reasoning to obtain t =. c) B 6 cm 5 55 D cm In ABC and BDC,. CDB = 80 55 = 5 = ABC supplementary ). BCD common to ABC and BDC. 3. DBC = BAC angle sum of ) since two other pairs of angles are equal. ABC BDC AAA) ii. marks) [] for relating corresponding sides in the same ratio. [] for correctly evaluating. Since ABC BDC, all corresponding sides are in the same ratio, i.e. i. mark) = 6 = 6 = 8 6 = 0 α + β = b a = 6 = 3 5.) ii. mark) iii. mark) αβ = c a = 3α α = α3 α) From 5.), β = 3 α C C Particle changes direction when v = 0 ii. 3 marks) t + = t = tt + ) t + t = 0 t + t = 0 t + )t ) = 0 t = since t 0 [] for applying absolute value to both terms of the distance. [] for d = ln + ln 3. [] for d = ln 3 + 3 m. [Note:] If d = 0 t+ t dt is used, a maimum of [] mark is awarded. d = t dt 0 t + + t dt t + = lnt + ) t 0 + lnt + ) t = ln ln ) 0 ) + ln 3 ln) ) = ln + ln 3 3 = ln ) + 3 ln 3 ) = ln ln 3 ) + = ln 3 + m α3 α) = αβ = LAST UPDATED SEPTEMBER, 008
008 Mathematics Trial HSC Eamination SOLUTIONS 7 b) i. 3 marks) [] correct identification of =, 3. [] testing nature of stationary points. [] finding coordinates & stating nature. y = 3 3 9 + dy d = 3 6 9 = 3 3) = 3 3) + ) iii. marks) [] shape of curve. [] coords of stationary pts, pt. of infleion. 0 y y = 3 3 9 + 6 3 Stationary pts. at y = 0. =,3 y = 3 3 9 + = = 6 y = 3 3 9 + = = 6 y = 3 3 9 + =0 = y = 3 3 9 + =3 y = 3 3 9 + = = 6 = 9 0 3 dy d + 0 0 + 6 y 9 6,6) is a local maimum and 3, 6) is a local minimum. Question 7 i. mark) V 0) = 000 + 0 000e 0 = 000 L ii. marks) [] for obtaining V t) = 00e 0.0t. [] for evaluating V 5) = 37.5 L/h. V t) = 000 + 0 000e 0.0t V t) = 0.0 0 000e 0.0t = 00e 0.0t V 5) = 37.5 L/h d.p.) ii. marks) [] obtaining y = 6 6. [] showing change in concavity when =. dy d = 3 6 9 d y d = 6 6 Pt. of infleion when y = 0 & concavity change occurs, i.e. 6 6 = 0 = y = 6 6 < 0 =0 y = 6 6 > 0 = iii. mark) lim t 000 + 0 000e 0.0t) = 000 L iv. marks) [] for obtaining e 0.0t = 5. [] for obtaining t = 0.3 = 0 h min. Let t be the time the farmer is aware of the leak. V t ) = 6 000 = 000 + 0 000e 0.0t 000 000 000 = 0 000 0 000 0 000 e 0.0t e 0.0t = 5 0.0t = ln 5 0 y 0 t = ln 5 0.0 = 0.3 h 0 h min have elapsed since the leak was discovered. LAST UPDATED SEPTEMBER, 008
8 008 Mathematics Trial HSC Eamination SOLUTIONS b) i. 3 marks) [] for solution of sin = cos. π [] for A, 5π ), B, ). sin = cos tan = ) π A, = π, 5π 5π B, ) ii. 3 marks) [] setting up integral. [] for successfully finding primitive. [] for A =. A = = 5π π 5π π sin d 5π π cos d sin cos ) d 5π = cos sin π = cos 5π cos π ) sin 5π sin π ) ) ) = = + = b) Tree diagram for this question: 0.8 0. i. mark) W s L s 0.8 0. 0.8 0. W s L s W s L s 0.6 0. 0.6 0. W c L c W c L c PW s W s W c ) = 0.8) 0.6) = 0.38 ii. marks) [] Find the complements of W s, W c : PL s ) = PW s ) = 0. PL c ) = PW c ) = 0. [] correct evaluation of PL s L s L c ) = 0.06 PL s L s L c ) = 0.) 0.) = 0.06 iii. marks) [] Find the complement: Pwin at least ) = PL s L s L c ) [] correct evaluation. Question 8 3 marks) [] for changing the subject to. [] for setting up integral. [] for solution V = 5π. y = y + = = y + V = π dy 6 = π = π 0 y + dy y + y 8 ) 6 = π ) 6 ) + 6) = 5π units 3 0 c) Pwin at least ) = Pwin none) i. marks) = PL s L s L c ) = 0.06 = 0.98 a = b = p 3) c = p + ) = b ac = p 3) p + ) = p 6p + 9 + 8p + = p + p + 3 ii. marks) [] correctly evaluating < 0. [] logical reasoning. = 3 = < 0 = p + p + 3 > 0 p Since > 0, therefore the quadratic always has real, distinct roots for real values of p. LAST UPDATED SEPTEMBER, 008
008 Mathematics Trial HSC Eamination SOLUTIONS 9 Question 9 i. marks) [] for correct identification a =, r = 3 sin. [] for correct substitution of a and r to the limiting sum S = 3 sin. ii. marks) [] for obtaining P = + 0 ) + 5 [] differentiating P) correctly. [] obtaining = 6. [] concluding = as is a length. S = = a = r = 3 sin a r = 3 sin 3 3 3 3 sin ii. marks) [] for equating 0 < 3 sin <. [] for finding 0 < < π 3. r < for limiting sum to eist. 3 sin < 0 < 3 sin < 0 < sin < 3 P = + + + y) + 5 = + 0 ) + 5 dp d = + ) 0 ) = = 0 0 0 = = 80 5 5 = 80 5 = 6 = since is a length & > 0 iii. marks) dp [] Checking d within a neighbourhood of =. [] Evaluating P) = 5 + 0. b) 0 < sin < 0 < < π 3 3 i. marks) [] for using Pythagoras Theorem [] for showing required y = 0 ) 5 + y = ) 5 + y = 5 = 0 y y = 0 y = 0 ) dp d eists when 0 and 0. dp d = 0 =.095 =3 dp d = 0 = 0.38 = 7 3 dy 7 d + 0 5 + 0 y 3.8.5 = makes P) = 5+0 a local maimum. LAST UPDATED SEPTEMBER, 008
0 008 Mathematics Trial HSC Eamination SOLUTIONS Question 0 b) i. mark) Differentiating Pt), =Pt) {}}{ P t) = P 0 ke kt = k P 0 e kt = kp ii. marks) [] for obtaining k = ln 387 368. [] for obtaining P 0 = 39 973. P) = 7 00 = P 0 e k 0.) P) = 5 800 = P 0 e k 0.) 0.) 0.): 5 800 7 00 = ek k = ln 387 368 0.3) Substitute 0.3) to 0.) 7 00 = P 0 ep ln 387 ) = P 0 387 368 368 7 00 368 P 0 = = 39 973 387 iii. mark) P) = 39 973ep ln 387 ) 368 = 7 97 iv. marks) [] for obtaining ln = t ln 387 368. [] for obtaining t years. Pt) = P 0 = P 0 ep ln 387 ) 368 t ln = t ln 387 368 t = ln ln 387 368 = 3.7688 = years i. mark) Since no interest is applied in the first 6 mths, then the amount owing will be A = 0 000 M A = 0 000 M. A 6 = 0 000 6M ii. marks) [] for correctly finding A 7. [] for correctly finding A 8. After the 6th month, % p.a. = 0.0 p.m. interest is applied to A 6. A 7 = A 6.0 M = 0 000 6M).0 M A 8 = A 7.0 M iii. mark) = 0 000 6M).0 M ).0 M = 0 000 6M).0.0M M = 0 000 6M).0 M +.0) A 9 = 0 000 6M).0 3 M +.0 +.0 ) A 36 = 0 000 6M).0 30 M } +.0 +.0 {{ + +.0 9 } ) 30 terms iv. marks) [] for finding S 30 = 00.0 30 ). [] for finding M = $68.78 S n = arn ) r S 30 =.030 ) = 00.0 30 ).0 A 36 = 0 as the loan is repaid and no amount is outstanding. Letting K =.0 30, 0 000 6M) K = 00M K ) 0 000K 6MK = 00M K ) M = 0 000K = 00M K ) + 6MK = M 00K 00 + 6K) = M 06K 00) 0 000.030 06.0 30 00 = 68.78 Their repayment is $68.78 per month. LAST UPDATED SEPTEMBER, 008