2013 Bored of Studies Trial Examinations. Mathematics Extension 2 SOLUTIONS

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1 Bored of Studies Trial Eamiatios Mathematics Etesio SOLUTIONS

2 Multiple Choice. D. D. C 4. D 5. C 6. A 7. A 8. D 9. C. B Brief Eplaatios Questio (I) is ot ecessarily true because the roots could be i, i, 4. If (II) is true, that meas is a real umber, sice the real umbers ca be ordered. However, just because is a real umber, it does t ecessarily mea that there are at least two o-real roots, sice the roots could be 4 i,,. The root 4i does t have a cojugate because the polyomial may ot be real, so the roots do t have to occur i cojugate pairs. For (III), agai the roots may ot be cojugate pairs. So a couter-eample is a polyomial with roots i, i, 5. Hece, the aswer is (D). Questio Although (C) yields the correct aswer if the formula were blidly used, it would be icorrect because it oly taes the solutio i the pricipal rage of which is withi that domai. Hece, the aswer is (D). y ta, ta. However, sice z is i the third quadrat, it clearly caot eist Questio Stadard use of a for directri ad ae for foci. e Questio 4 To mae a diagoal, we eed poits out of, so we have diagoals. However, of these diagoals are actually sides of the polygo, so the total umber of actual diagoals is. Hece, we solve 44, ad the solutio is (D). Questio 5 We ca start from (D) ad use various substitutios. To get (A), it loos lie the substitutio ta u. To get (B), it loos lie the substitutio u lie we mae the substitutio. For (C), it loos u e, but the limits are icorrect. Hece, the aswer is (C).

3 Questio 6 To get the origial curve, we multiply the give curve by y, otig that we actually do ot have a discotiuity at the origi i the origial curve, sice we have i a sese itroduced the discotiuity i the process of multiplyig the curve by y. Questio 7 Use the Multiple Root Theorem. Questio 8 Neither (A) or (C) are correct sice they have m s, whe the epressios should be idepedet of m. (B) is icorrect because it represets cetripetal acceleratio, ot tagetial acceleratio. Hece, the aswer is (D), which is tagetial velocity. Questio 9 Usig either the Washer or Cylidrical Shells method will ot yield ay of the optios immediately. The correct aswer (C) ca be foud by usig the Washer Method ad the the substitutio y si ad u. Questio We thi about the situatio geometrically. I (A), that is obviously ot always the other choices because the fuctios could be egative. (C) caot be the aswer ad to see this, cosider the sie curve. (D) caot be correct for similar reasos. However, (B) guaratees that the curve is above or o the ais, thereby maimisig the itegral which will also be the area uder the curve.

4 Writte Respose Questio (a) Divide top ad bottom by cos. a bsi a sec bsec ta d b asi bsec a ta bsec a ta cos b a si b d Questio (b) Differetiatig we have implicitly ad substitutig the poit of itersectio, y a P y, yy y y Similarly for y c, we have y y y y Ad we see that y y, hece the two curves itersect at right agles, regardless of the value of a ad c.

5 Questio (c) (i) The equatio ca be traslated as The set of all poits w that are equidistat from ad the origi. The locus is the the perpedicular bisector of the iterval from to the origi. Hece, it is a straight lie. Questio (c) (ii) Let, where i,,. Thus we have w zi z i. w Note that we have zi c c. This is because c is the cetre of the circle, so c is the radius of the circle. However, all of the poits z i, where i,,, must be equidistat from c, sice they are poits o a circle. We ow mae the substitutio z i. w c w cw w cw c w c w c c c w w w However, settig c, we have the epressio from (i). Hece, the locus of w is a straight lie, as we have foud earlier. However, we also have w zi, hece z, z ad z are colliear. 4

6 Questio (d) Usig polyomial log divisio, we have y ad that as, y., so we ca see that as, We ow that the curve has vertical asymptotes where the deomiator is zero, amely. We also ca easily fid ad y itercepts, beig ad y respectively. Puttig all the pieces together, we get the setch of the curve. y 5

7 Questio (e) (i) The volume V ca be foud by subtractig two idetical coes from a larger cylider. A r O B We first fid the volume of the coe. V coe R h, where cos R r ad h rsi. V coe r cos si. Now, to fid the volume of the cylider. V cylide r R H, where cos R r ad H rsi. Vcylid er r cos si We ow have eough to fid V. V V V cylider coe cos cos 4 r cos s i Questio (e) (ii) r si r si 6

8 We ca either use cylidrical shells or the washer method. Here, we will use the method of cylidrical shells, though the washer method will of course yield the same result. Costruct a thi vertical strip of thicess from a arbitrary poit P, y o the mior segmet AB. Rotate this thi strip about the y ais to roughly form a cylider. We ow ufold the cylider to acquire a rectagular prism with legth, height y ad width. A O B V 4 y V 4 r lim r cos r 4 4 r r cos r r 4 r d r r r cos 4 r r cos r r cos 4 r cos cos 4 r si Sice V V, we have 4 4 r cos si r si cos si ta ta 4 Questio (a) (i) 7

9 Base Case: LHS a RHS a Iductive Hypothesis:... a a a a a a a... a Iductive Step: Required to prove:.... a a a a a a a a... a a LHS... a a a a a a a a... a a a aaa... a a a a... a a, sice a aa... a a a a... a a RHS Questio (a) (ii) a a aa... a aa... a a a... a a a... a a a... a a aa... a aa... a aa... a aa... a a a... a a a Questio (a) (iii) 8

10 Sice a a, we ow that the sequece a is icreasig for all. Hece, sice a, we ca deduce that as, we have aa... a a. Therefore as,. a Questio (b) Resolvig forces vertically ad horizotally, we have Horizotally: T si T si mr Vertically: Tcos Tcos mg Applyig the iequality T T to both epressios, we have T si T si mr T si T si T cos T cos mg T cos T cos Tcos Tcos mg T cos T cos Sice cos cos, we ow that all the quatities are positive. Multiply both iequalities. mr Tsi Tsi mg T cos T cos si si cos cos g si si r cos cos Questio (c) (i) 9

11 We view the slice from two perspectives, ad costruct lies such that we have similar triagles. The followig is the frot view of the solid. a H h Usig similar triagles, we have the followig relatio. b b a H b a H h a a Re-arragig to mae the subject, we have Now, cosider the side-view of the solid. b ah h a. H b y H h Usig similar triagles, we have the followig relatio. a

12 b a H b a b ah y a h y a y a H But we also have V yh, so we put the pieces together. V b a H h b a h a a h H H ah b ah b a H ah b ah h H a H h b a b a hh a b a H h H ah a b a b a hh h h H abh b a hh h h H Questio (c) (ii) To fid the volume, we itegrate with respect to h. b a hh h V lim abh h h h HH H H H H H abh abh abhh b a hh h dh H b a h b a h dh H b a b a h H b a H b a abh H H b a abh 6 H a 4ab b 6 Questio (a) (i) h H

13 We first pic a umber to have eactly two of, so p. q We must ow pic the colour of the two cards. There are q possible colours, so. Now, there remai cards to pic out from a remaiig total of p sice oe of them ca be the same umber as those which we have already chose, so p. We have ow piced cards, two of which have their colours chose already. Each of the remaiig cards eeds a colour allocated to it. Hece, we must multiply by each of the cards have q possible colours. q, sice The total umber of ways of choosig cards from is Thus, the probability is p q p q.. Questio (a) (ii) Settig pq, we have p p p p p p p p p p p p p p Questio (b) (i) From Newto s d Law, we have Termial velocity occurs as a, so Questio (b) (ii) ma mg mv, so u g u a g v. g

14 dv Sice the required epressio has time, we use a. dt dv g v dt dv dt g v We ow itegrate both sides with respect to their variables. T dt V dv g v V dv g v V dv u v V dv u u v u v l u u v l u u v u v l u v uv T l u u V We ow re-arrage to mae v the subject. e ut uv u V ut u V e u V ut ut V e u e ut e V u e ut ut d e u ut dt e ut e d u dt ut e V V

15 X * T ut e d u dt ut e * T ut ut e e u dt ut ut e e * T l ut ut e e * * ut ut e e X l But u g g u, so we thus have u * * gt gt u u e e X l Questio (c) (i) BAC DAP... give CBA ADP... eterior of cyclic quad. BAC DAP... equiagular By similar triagle ratios, we have DP AD ad hece the result. BC AB Questio (c) (ii) ABD ACD... agles from same chord BAD CAD... where BAC DAP CAP ABD ACP... equiagular By similar triagle ratios, we have CP AC ad hece the result. BD AB 4

16 Questio (c) (iii) AC BD ABCP AB CD DP ABCD AB DP ABCD AD BC... from (ii)... from (i) Questio (c) (iv) If we use the epressio from (iii) for the quadrilateral ABPC, we have PA BC PBCA PC BA But the triagle is equilateral, so AB BC CA. Substitutig that i, we have the required result. PA BC PB BC PC BC PA PB PC Questio 4 (a) (i) P, sice P Questio 4 (a) (ii) Suppose all roots eist iside the uit circle. The all roots have modulus. However, we ow from (i) that if is a root of P z, is also a root of P z. But if, the, which cotradicts all roots havig modulus. Hece, there eists at least oe root such that. 5

17 Questio 4 (a) (iii) Substitute z ad re-arrage our origial polyomial. Questio 4 (a) (iv) We ow use the fact that Let cis, where. cis cis cos i si cos i si cos cos si si cos si cos cos cos si cos cos 6

18 Questio 4 (a) (v) We ow that.. Similarly,, so we must have We deduced (iv) from assumig that all roots have modulus. However, this cotradicts the fact that. Hece, all the roots must have modulus. Questio 4 (b) (i) Let P cis, so P. d P P cos isi cos si Questio 4 (b) (ii) cos si cos d cos cos si sice... Questio 4 (b) (iii) d But recall that..., so.... d. Hece, we have 7

19 Questio 4 (c) (i) Cosider the followig y y y y y Settig a ad A b y yields the result immediately. B Questio 4 (c) (ii) From (i), we ow that ab A A B A B a b a b B But ote that A a, so A a. Hece, we have a b ab. AB A B So our epressio simplifies to ab AB ab AB a b 8

20 Questio 4 (c) (iii) The difficult part ow is maig the correct choice of a ad b. We otice that oe of the terms has to be some form of S S i order to acquire the result. However, we must tae the square root because we otice that if we were to set a S S, S S the we will be stuc with, which is udesirable. So we set a. We S S must ow choose b. Notice that the required result s RHS is idepedet of S ad. Hece, we must choose somethig that will result i a costat, so the appropriate choice would be b S S. S Settig a S ad b S S S S S S S S S S S S S S S S S S S S Questio 5 (a) (i) si si si cos si cos si cos si cos cossi 9

21 Questio 5 (a) (ii) First, we re-arrage (i) ad establish a idetity. cos si si si To match S, we let. We must ow mae the appropriate choice of, so we loo at the required result. We see that i the deomiator, we have si, so it implies that we must set. Hece, we have cos si si si si si si Suppose. We sum the above epressio ad form a Telescopig Sum. S cos si si si si si si si si si si Suppose. We use the origial sum. S cos... sice all terms ecept the first ad last cacel

22 Questio 5 (b) We use Itegratio by Parts, usig u ad dv cos. cos d si si d si d cos cos Note that we have acquired the last epressio usig the fact that the cosie of iteger multiples of ca be or, depedig o, so cos. Questio 5 (c) (i) Notice that f is just S, where. I lim S d a cos d... sice S is defied at 4 4 a cos d cos d cos d 4

23 Questio 5 (c) (ii) I Questio 5 (c) (iii) I lim f d a a lim a a si 4 d si lim si 4 d a si a We use Itegratio by Parts with u ad si choices because the questio defied g the term that is differetiated. dv si 4. We mae these d, which implies that it has to be d si 4 4 I lim cos lim g cos d 4 a si 4 a a a 4 a a 4 lim cos lim g cos 4 a si a 4 a a d

24 Note that a large term cacelled i the substitutios because 4 cos, sice we are taig the cosie of a odd multiple of. We also otice the opportuity to use the familiar limit result. si lim as well as 4 a lim cos. Hece, we have the required a 4 I lim g cos d 4 a a Questio 5 (c) (iv) Sice we ow that g i the domai, we have 4 cos 4 g g cos g Itegrate both sides with respect to from to. Note that the itegral of because g d. d si 4 lim g cos lim si si a a a a 4 a lim g cos lim a a a a si si si si 4 lim g cos d a a a g is si

25 Questio 5 (c) (v) From (iv), we have 4 I I 4 4 Hece, as, I as, I. is squeezed betwee two terms, both of which approach. Hece, Questio 5 (c) (vi) We ow from (ii) that Hece, lim I. But from (v), I as. 8. Cosider the sum S lim. For simplicity, assume is 8 eve. This does ot chage aythig sice we are taig the limit as ayway. S lim lim... 4 lim lim S S 8 4 S 4 8 S 6 4

26 Questio 6 (a) (i) From the Reflectio Property, we ow that SPT QPS. But sice QPR is a reflectio of SP Q about the lie QT, QP R QPS, so therefore QP R TP S Sice QP T is a straight lie, we have vertically opposite agles beig equal, so therefore SPR must be a straight lie, so S, P ad R are colliear. T Q P R S Questio 6 (a) (ii) We ow that TR TS ad TR TS, sice they are reflectios of each other about TP ad TP respectively. Recall the ofte forgotte idetity PS PS a for some costat as we do ot ow aythig about the dimesios of this ellipse. Usig that, we have PS PS. But PS PR so PS PR. Similarly, we have PS PS PR PS. However, sice we have prove that S, P ad R are colliear from (i), we ca coclude that SR S R, which proves that the third side is equal. Hece, TRS TS R sice all three correspodig sides are equal. 5

27 Questio 6 (a) (iii) RTS RTS S TS RTS RTS S TS But from (ii), we ow that RTS RTS, sice they are correspodig agles i cogruet triagles. Hece, we ca coclude that RTS STS RTS STS RTS RTS But ote that by reflectio about TP ad TP, we have RTS STP ad that RTS STP. Thus, we have STP STP STP STP Questio 6 (b) (i) Applyig Az to all poits scales the triagle ad applyig B to all poits traslates the triagle. Hece, the origial triagle ad the triagle with T applied to all poits are similar. Alteratively: We will prove that ay two sides are i the ratio A. By doig so, we prove that all sides are i the same ratio A ad are therefore similar. Let ay two sides be z i ad z j. So the legth of that side is zi zj poit.. Now, apply T to each T z T z Az B Az B i j i i Az i A z Az i z i j Hece, we have prove similarity. 6

28 Questio 6 (b) (ii) T PT z z T z z T z z T z T z T z T z T z Az B Az B Az B Az B Az B Az B A z z A z z Az z A z z z z z z A P z P T z T z T z Questio 6 (b) (iii) Differetiate both sides of the epressio i (ii) with respect to z. T P z... usig the Chai Rule, where z PT T z A P z P T z A A T A If is a root of P z, the P. T A P P T Questio 6 (b) (iv) Say we have some real cubic polyomial P with eactly oe real root ad thus two o-real roots that are also cojugates. These roots form P. If we apply T to all the roots, we get aother similar triagle (from (i)) has some derivative P. T T, which are the roots of PT z. This ew polyomial Now, P has two roots (either of which deoted by ), sice it is a quadratic. If we apply T to both roots, we get T. What (iii) proved is that T is also the same as the root of P! T Geometrically, this meas that whe the roots of the cubic polyomial P are trasformed by T, the roots of its derivative are also trasformed by T. 7

29 Questio 6 (c) (i) Sice the roots of P z are i, i, r, where P is moic, we ca re-costruct the cubic. z z r P z z i z i z r z rz z r P z z rz Questio 6 (c) (ii) Solvig the quadratic eplicitly, we have z r 4r r r i r 6 r Sice the real compoet is r, we ow that the roots are betwee the y ais ad R. Now, we must prove that the positive imagiary compoet is below the lie AR whe ad by provig the positive case, we also prove the egative case due to symmetry. r By ispectig the gradiet ad y itercept of AR, we ca deduce its equatio y. r 8

30 Settig r y r 4 r, we have r Thus, the roots of the derivative lie i the iterior of the triagle. Questio 6 (d) I (c) (i), we proved a fairly specific case, for whe the roots are i, i, r, formig some isosceles triagle. However, ote that ay real cubic polyomial with distict roots will have its roots formig a isosceles triagle with its ais of symmetry o the real ais, due to the Cojugate Root Theorem. But ay isosceles triagle ca have a similar image of it costructed from the poits i, i, r, whe r is chose appropriately Sice we have prove that the roots of P lie withi the iterior of the triagle formed by the roots of P (for the case with i, i, r ), we have thus prove it for ay isosceles triagle with two purely imagiary roots. But from (b), we ow that if the roots of P are trasformed by T, so are the roots of P. Hece, they stay i the same positio relative to each other. Thus, we have prove the case for ANY isosceles triagle, ot just oes with two purely imagiary roots. 9

Mathematics Extension 2 SOLUTIONS

Mathematics Extension 2 SOLUTIONS 3 HSC Examiatio Mathematics Extesio SOLUIONS Writte by Carrotstics. Multiple Choice. B 6. D. A 7. C 3. D 8. C 4. A 9. B 5. B. A Brief Explaatios Questio Questio Basic itegral. Maipulate ad calculate as