Mathematics Extension 1 SOLUTIONS

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1 Bored of Studies Trial Examiatios Mathematics Extesio SOLUTIONS Disclaimer: These solutios may cotai small errors. If ay are foud, please feel free to cotact either Carrotstics or Trebla o reardi them. Thas: To Trebla, for his may hours spet verifyi solutios ad suesti alterate methods.

2 Multiple Choice. D. C 3. D 4. A 5. C 6. B 7. D 8. D 9. B. D Brief Explaatios Questio Re-arrae ito stadard form v A x. Questio Let the expoet of x i the eeral term be zero to acquire 3,. Questio 3 Split umerator ito two terms ad draw a diaram. Questio 4 Observe limit as x, ad that x caot lie i x. Questio 5 Ale betwee two lies formula, ad let the expressio be. Questio 6 Stadard permutatios problem. Note that they are i a circle, so it s!. Questio 7 Questio 8 Stadard Newto s method of approximatio questio. Fid the coordiates of C, the substitute ito the lie. Questio 9 Neative quartic, with a triple root at the orii ad a sile root at x 4. Questio Biomial probability questio. Use uess/chec to acquire closest solutio.

3 Writte Respose Questio (a) We will use the t formula substitutios. Let t ta So our expressio is: t t t t t Re-arrae: t t t t 3 t t Form a cubic polyomial i t, the solve: 3 t t t t t t t t t t t, ta, Solve for : 3, So therefore we have,.

4 Questio (b) (i) There are a total of letters, ad we have H s ad 3 O s. So the umber of permutatios is!!3!. Questio (b) (ii) There are 4 vowels ad 7 cosoats, ad the vowels are to be rouped toether. Example: HCL(OIOO)PMHR Arrae all the cosoats ad the roup (OIOO) to et 8!. Note we have! i the! deomiator because we have H s. Arrae the vowels i the roup, oti that we have three O, to et 4! 3!. So therefore the aswer is 8! 4!.! 3! Questio (b) (iii) We will cout the umber of aps betwee cosoats, ad the isert the vowels ito these aps. We have 7 cosoats, so therefore 8 aps. We isert 4 vowels ito these 8 aps, ad thus we 8 have 4. We must ow permute the vowels to acquire 4! 3!. Permute the cosoats to acquire 7!!. 8 4! 7! So therefore the aswer is. 4 3!! 3

5 Questio (c) Let u dx u. x ta x such that d u x 4 u x cos ta x x dx 4 cos 4 u du cos u si u u du Questio (d) We ow that P p p ad Pq Px x px pqx Ax B Usi the above coditios: P p Ap B p P q Aq B q : A p q p q A Substitute ito : p B p B Hece the remaider is exactly x. q. 4

6 Questio (e) From AOC, h OA ad similarly i BOC, we have ta h OB. ta Usi Pythaoras Theorem, h h d ta ta h ta h ta h d ta ta ta ta h ta ta d d OA OB d. Ad therefore: h d ta ta ta ta Sice 9, 9, we have ta, ta. Also, we must have h ad hece: h d tata ta ta 5

7 Questio (a) Whe x, x 3, so we have 3 ad thus 3b. b Whe x, x, so we have b ad thus b. Solvi simultaeously yields b ad thus 6. So therefore yields: 6 x x ad thus d 6. Iterati both sides with respect to x dx x 6l x C We are ive that whe x, v, so: 5 6 l 3 C C 5 6 l 3 So our expressio is ow: x 6 l 5 6 l 3 x l l 3 x l 3 Let 7 : x l 7 3 x l 89 3 x l x x 4.9m So Ji JUST maes it out. 6

8 Alteratively From x l l 3 x l 3 Substitute x 5: Ad hece, Ji JUST maes it out. 7

9 Questio (b) (i) Base Case:. LHS p 3 p 6 8 RHS Therefore true for. Iductive Hypothesis: p 3 p 4 Iductive Step:. Required to prove: p p LHS p p. p p RHS Hece true by iductio for all. 8

10 Questio (b) (ii) lim S lim p p 3 lim 4 3 lim Questio (c) (i) There are a couple of ways to do this questio. Method #: The equatio of the ormal is ive to be x py ap p lies o it, so we will substitute i the poit T at, at. at apt ap p 3 at apt ap ap Re-arrae: 3 ap apt ap at ap p t a p t... Noti that t ap p t a p p t ap p t p t a p t p p pt. But we ow that the poit T 9

11 Method #: The equatio of the ormal itersects the parabola twice, but we ow oe of the roots is x ap. We could easily do it the other way aroud, by substituti x ito x would be quite tedious. Substitute the equatio of the ormal ito the parabola: x py ap p y a p Hece we have: x 4a a p 4 x p x p 4a x a p p 4ay, but that Re-arrai: 4a x x 4a p p 4a Sum of roots is x x. But we already ow that oe of the roots is x ap ad the p other is x at, so therefore we have: 4a ap at p p t p Ad hece the result p pt.

12 Method #3: The chord PT must be perpedicular to the taet at P. PT ap at ap at a p t a p t p t p t The radiet of the taet at P is dy dy dp dx dx dp ap a p Hece p t p p pt p pt Questio (c) (ii) Similarly to (i), we ca deduce the same expressio, except with q. So we have: p pt q qt Subtract the two equatios: p q t p q p q p q t p q... ote that p q p q t

13 Questio (c) (iii) So we ow have p q t ad p pt. Mae t the subject to acquire t p q, the substitute ito p pt : p p p q p p pq pq Questio (d) (i) First, we costruct PT ad TQ. P B A O T Q C Let PBT APT 9 (Ale subteded from diameter) Therefore a circle ca be costructed throuh poits B, P ad T such that BT is a diameter (coverse of Thale s Theorem). This implies that AT is taetial to the circle (also sice ATB 9 ). Hece, PBT PTA (Alterate Semet Theorem) But PTA PQA (Ale subteded by commo chord)

14 Therefore PBT PQA. Hece PBCQ is a cyclic quadrilateral (coverse of exterior ale from cyclic quadrilateral theorem). Alteratively Let PTA PQA (Ale subteded by a commo chord) APT 9 (Ale subteded from diameter) PAT 9 (Ale sum of PAT ) But BTA is also riht-aled, so PBT 9 9 (Ale sum of BTA) Hece, by the coverse of Exterior Ale = Opposite Iterior Ale Theorem: PBCQ is a cyclic quadrilateral Questio (d) (ii) A basic ale chase yields the result immediately. BCQ APQ (Exterior ale opposite iterior ale of a cyclic quadrilateral) APQ ATQ (Ale subteded by commo chord) Hece by the coverse of the Alterate Semet Theorem, we have the result. Alteratively We ca simply observe that AQT 9, sice it is a ale subteded from a diameter. It follows, by supplemetary ales, that TQC 9 ad hece the result by the coverse of Thale s Theorem (Ale subteded from diameter is 9 ). 3

15 Questio 3 (a) dt dt. We bei with the differetial equatio E T Separati the terms ad roupi them appropriately, we have: dt T E dt Note that we mae the arraemet from ETto T Esice E T. Iterate both sides with respect to the appropriate variable: T t T dt T E t dt l T E t T t T t Substituti ad re-arrai, we have: T E l T E t t l T l T Ad hece: E t E T E l T E t t t But recall that t, hece: T l E t T E 4

16 Questio 3 (b) (i) We are ive the domai x, from which we observe that x. Multiply both sides by a b x a b a b : This is allowed sice ab. Expad ad re-arrae: a x b x a b b a x a b x We carefully square root both sides, owi that the iequality is still preserved. b a x a b x Flip both sides, ad thus the iequality: a b x b a x Hece f x x. Ad so the other iequality follows. Alteratively Let f x x : a b x b a x a b x b a x Square both sides carefully, oti that the iequality is preserved. a b x x a b b a a x b Hece x ad thus x, sice x. The other directio of the iequality follows. 5

17 Questio 3 (b) (ii) This is a ormal volumes problem ow. Sice for x, we have f x x, we ca compute. b a x a b x ta ab ab ta ta ta ab ta ab ax bx ta b a a b b a a b b a a b b a a b ab dx Questio 3 (b) (iii) This is essetially the same thi, with differet limits ad f x x. ta ab a b x b a x bx ax ta a b b a b a ta ta ta ta ab a b a b dx Note that as, Hece: ta b a ad ta a b. ta ab ta ab b a ta a b a b ta b a Ad this is the same expressio as (i). 6

18 Questio 3 (c) (i) We will use the idetity si cos. Sice A B, we have: t t t x Acos si B Asi t A B Asi B A A cos t A B B A cos t Differetiate oce with respect to t : B A x si t Differetate aai with respect to t : B A x cos t A B x Hece, the particle moves i Simple Harmoic Motio, with cetre of motio bei A B x. 7

19 Alteratively Usi the results cos cos ad si cos, we have: t t x Acos Bsi A B A B A Bcos t cos t cos t Differetiate oce with respect to t : x A Bsi t Differetiate aai with respect to t : x A Bcos t A B x 8

20 Questio 3 (c) (ii) Observe that the cetre of motio is A B x ad amplitude is B A. AB B A Oe edpoit is x B. AB B A The other edpoit is x A. Hece A x B. Questio 3 (d) (i) Usi the Sie Rule i l si si APO But we also have: APO 8 AOP So: l si si 9 cos s i l cs o AOP, we have: 9

21 Questio 3 (d) (ii) () Usi the Chai Rule, we have d d dl d S. dt dl dt dl si cos dl cos cos si d cos cos d cos dl cos d dt d dl dl dt cos S cos Let : cos S cos cos S cos S cs o

22 Questio 3 (d) (ii) () d We will use the formula. d It may seem urecoisable ow, but it is actually more commoly ow as which is much more well-ow (as it is tauht that way). d d d cos S d cos cos si S cos cos S cs o Let : si si cos S cos cossi S cos S si dv av, dx But si l cos ad whe, si l cos sicos cos si Hece: si cos S cos cossi S cos Sl

23 Alteratively d dt d d d dt d S cos d cos But recall that S cos si d cos S d cos S si cos Hece : S si Substitute : S si S sicos But recall that S cos. Also, similarly to the alterative solutio above, l si. Hece S l

24 Questio 4 (a) (i) Cosider the expasio x x x Coefficiet of Coefficiet of x x x x x x... x x x from RHS: x from LHS: m m m m m m m m Hece coefficiet of m m. x from LHS is: m m m m... Ad hece the result. 3

25 Questio 4 (a) (ii) We mae the followi substitutios,, m,. The the idetity from (i) ow becomes:... Simplifyi this:... But recall the idetity : Hece,,,. Therefore:... 4

26 Questio 4 (a) (iii) We ow mae the substitutio r r i the summatio o the riht. This will mea that i a similar fashio to Iteratio by Substitutio, r r ad r. But also ote that if we sum from to, or from to, it maes o differece so we ca just write the bottom limit as ad the top limit as, to follow eeral covetio. Hece, our ew expressio is: r r r r r r Expadi the RHS, we et: r r r r r r r r But recall that. Applyi it, we have: r r r r r r r r r r r r r Re-arrai the terms, we have: r r r r r Ad this is possible, sice r r r r r r. Substitute the result i (ii): r r r r r Dividi both sides by yields the required result. 5

27 Alteratively Or i a more compact form 6

28 Questio 4 (b) (i) By symmetry of the parabola, we have. Similarly for other trajectories,, 3,. ad iductively, we have 3... Hece, for all. Questio 4 (b) (ii) We will first fid the time for the first trajectory. Let y : si t si... sice t is whe it is at the orii s i t t t si So therefore, we have t. Similarly: t si si, usi the recurrece. si si si t t si So summi up a ifiite umber of these, we have: 7

29 T lim r r si si lim r si si r... sice Questio 4 (b) (iii) We will first fid the distace of the first trajectory. x t cos si sicos Similarly, x si cos si si x 4 si si si x si Hece, the total distace is: 8

30 R lim r r r si lim r si s i si... sice Questio 4 (b) (iv) We ow that T R R si. So placi it as a ratio: T T R R si si R R si si So we must ow fid R. Equate R with the rae of ay ormal trajectory: 9

31 R si si R R R Hece, substituti it i: T T R R Questio 4 (b) (v) We ow that. Hece, ad ad therefore. Square rooti both sides yields the same bouds, so. But recall that T T R, so therefore TR, ad hece T TR T. Physically, this meas that it will always tae less time to et the ball to a locatio via ladi it there i a sile larer trajectory, as opposed to bouci it there with a smaller oe. 3

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