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 usual. Questio 3 he argumet is doubled ad the modulus is less sice z. Questio 4 Stadard u u P., the Questio 5 Questio 6 Questio 7 Questio 8 he regio is clearly a aulus, so (B). Complete the square the use table of stadard itegrals. Use the fact that circumferece is r. Stadard slicig problem, where the base legth is y, so the area is 4y. Questio 9 We ow the limit as x is, ad the fuctio is eve, so (B). Questio First fid the total, igorig restrictios, the subtract the excess cases where the rooms have more tha 4 people. For the total, each perso has 4 choices, so 6 4. Case #: 5 i oe room, i aother. Choose perso to be left out, so we have 5 people have 4 rooms to choose from, the last perso has 3 rooms, so 4 3. 6. Case #: All 6 i oe room. Four rooms to choose from, so 4 ways. Hece the total is 6 4 643 4 4.
Writte Respose Questio (a) (i) z w i 3 i 3 3 i 3 Questio (a) (ii) w cis 3 Questio (a) (iii) w 4 4 4 cis 3 4 cis 8 4... sice cis, for all... by De Moivre's heorem Questio (b) x x x 8 4 3 x 3 x x 3 x Questio (c) By ispectio, z i is a solutio. We ca see that the sum of roots is 4i, so let the other root be. i 4i 5i Hece, P z z 5i z i.
Questio (d) We use the substitutio u x du x dx. x u x u Hece we have I u u u du u u u 3 3 5 3 5 3 5 u u du u du u u du du u u 3 Questio (e) Lettig z x iy, we have x ixy y x ixy y 8x y 4 his is a rectagular hyperbola with x itercepts at x. 3
Questio (a) sec x dt dx x ta dx t dx dt dx t dt dx 4 5cos x t t 4 5 t t t 3t3t t t 4 5 4 4t 5 5t 6 6 dt 3t 3t dt 3 3 t 3 t l 3 l 3 3 3 t l 3 3 t l 3 dt dt dt 4
Questio (b) Differetiate both sides with respect to x. y y y y 5 y y y 5 y y y y y 5 y y y 5 y y y 5 y y 5 Questio (c) We use the method of cylidrical shells. We ca see that the radius of the shell will be 4 x ad that its height will be y. Ufoldig the shell, we roughly have a rectagular prism of height y, legth 4 x ad thicess x. Hece, the volume of the shell is 5
V 4x y x x 4x e x V lim x x3 3 3 3 e 3e e e 3 x 3 x e 4x e dx 3 e 4e 4 e e 4 x e 4 x e x dx x x Questio (d) (i) We first use parametric differetiatio. dy dx dy dp dx dp c p c p Now we proceed with the poit/gradiet formula as usual, usig P. x c y p p p y cp x cp p y cp x cp 6
Questio (d) (ii) We ow that OB maes a right agle with OA, sice they are the coordiate axes. We must ow prove that P is the midpoit of AB, so we first fid A ad B. For A, let y, so x cp ad hece, A cp. For B, let x, so c p y cp y ad hece p c B, p. Clearly, P is the midpoit of AB. Questio (d) (iii) Sice C ad D are x ad y itercepts of the taget at Q, we ca apply the result of (ii) to deduce that C, Q ad D lie o a circle cetred at Q. Sice P ad Q are the cetres of their respective circles, we have the ratios AP DQ AB DC However, otice that the itervals PQ ad BC form itercepts o trasversals, so sice the ratios are the same, we ca deduce that BC is parallel to PQ. 7
Questio 3 (a) (i) We use Itegratio by Parts, with the followig choices. u x dv dx du x x dx v x I I I I x x x x dx x x dx I x x x x dx I I dx Questio 3 (a) (ii) 5 5 3 5 3 5 I5 I3 I, usig the area of a quarter circle with radius. 6 6 4 6 4 4 3 Questio 3 (b) (i) his is essetially the square root curve, except reflected about the x axis. 8
Questio 3 (b) (i) We first draw y f x, the we draw its reciprocal curve. Questio 3 (c) (i) We eed to somehow get, ad the oly way is by somehow cosiderig the exterior agle of ADC, costructed by extedig CD to say F. But otice if we do this, that is also the opposite exterior agle of the cyclic quadrilateral ABCD, so ADF ABC. ACB 9 sice AB is a diameter ad C is a poit o the circle. Now, we use trig ratios. si ABC si AC r AC r si 9
Questio 3 (c) (ii) AE AD cos, so we must prove that AD r si. However, if we costruct BD, we see that ADB 9 sice AB is a diameter. Furthermore, AD ABD sice it is subteded by chord AD, which subteds, so si ad hece r we have the result. Questio 3 (c) (iii) We ow that AE r cossi. Similarly, EC r sicos. But AC AE EC, so we have rsi rcos si r si cos si cos si si cos Questio 4 (a) We ca see that the area uder the curve A C is greater tha the area of the right agled triagle A. A C t l x dx x l x x t l t t t... usig Itegratio by Parts A t l t But we ow that AC A for all t, so t l t t t l t t l t t t l t l t t l t l t t t l t t
Questio 4 (b) Base Case: z, which is ideed true sice i. Iductio Hypothesis: z. Iductive Step: i z z z i z z z i z z i i Ad thus prove by Iductio. Questio 4 (c) (i) sec ta sec ta
Questio 4 (c) (ii) 8 6 sec sec sec 3 ta sec 4 6 ta ta sec 3 ta 8 3 3 3 3 4 6 sec d ta ta ta sec 3 3 3 5 7 ta ta ta ta C 5 7 Questio 4 (d) (i) AD AE AC AB BAC is commo. Hece, ABC AED, sice two sides are i ratio ad the icluded agle is equal. Questio 4 (d) (ii) Sice ABC AED, we ca deduce that DEA DBC. But oe is the opposite exterior agle of the other, so BCED must be a cyclic quadrilateral, by the coverse of the opposite iterior agle theorem.
Questio 4 (d) (iii) We use the Cosie Rule o DCE i both DCE ad DCA. I DCE : CD 3 CD 5 cos CD 4CD I DCA : 6 CD 3 7 CD cos 6CD CD Equate the two CD 5 7 CD 4CD CD 7 CD CD 5 3 3CD 5 7 CD CD 4 CD CD 3
Questio 4 (d) (iv) First let O be the cetre of the circle ad costruct chord CD ad perpedicular bisector OP. O P By similar triagle ratios, we ca deduce that BC 6 ad so we ca wor out that cos But 8 6 6 ABC 86 3 ABC COD sice agles at the cetre are twice agles at circumferece. cos COD cos ABC 3 9 Now, we use the Cosie Rule o r coscod r r r 9 COD. Ad solvig for r yields r 3 5 4
Questio 5 (a) Let z Acis ad w Bcis.. Sice the area of the triagle is A, we ca deduce that A ABsi AB cis zw wz Acis B cis Acis B cis AB cis AB... sice AB cis cis cis cis ABi Im cis ABi s i 4iA Questio 5 (b) (i) Let the roots of the polyomial be,,,. From the sum of roots i triples, we have We ow use the sum of roots. b b From the product of roots, we have e, so b e ad thus be. We ow that P 3, P ad P sice it has a double root at x. 3 ad P a b c e P a b c e Subtractig the two equatios, we have 3 b 3 b. Differetiate the polyomial so we ca let P, by the Multiple Root heorem. 3 4 3 ad thus P x ax bx cx Hece we have 9 4a c 3b P 4a 3b c 5
Questio 5 (b) (ii) he slope of the taget whe x is P 4a 3b c. But sice 9 4a c, we have 9 3 P 3 9 9 9 Questio 5 (c) (i) Sice it completes all four days, we have Questio 5 (c) (i) 4.7.7.7.7.7. he required sceario is the coverse of the followig cases. Let 4 p.7. Case : No cars complete. 8 P X p p 8 Case : car completes. 8 P X p p 7 Case 3: cars complete. 8 P X p p 6 Hece, the probability is P X 6
Questio 5 (d) (i) For termial velocity, let mv. v v v mg mg mg Questio 5 (d) (ii) We use the formula dv vv, but the required equatio of motio is dx mv mg v dv m v mg v dx H v dx m dv mg v m H l H l m mg u l mg v u mg v m mg mg u u l g v u 7
Questio 5 (d) (iii) We derive similar equatios for w. dv m v mg v dx H w v dx m dv mg v m H l mg v m mg w H l mg v w l g v v v w l g v v v l g v w Equate the two heights. v v v u l l g v w g v v v u w v u w v v v 4 w v v w u v u w 4 u v v w u w w u v 8
Questio 6 (a) (i) o fid the miimum value, let the first derivative be equal to. 6x 3x 4 x 5x 4 x x 4 x, 4 Sice it is a positive cubic, we ow that the miimum value must come after the maximum value, so 4 x must be where the miimum occurs. Hece, it is 4,, by fidig 4 P. Questio 6 (a) (ii) From (i), we foud that x 4 is a double root, so we ow that our curve loos lie this ow. 4 Hece, we see that for all x, the curve is positive, so P x. We ow maipulate our polyomial. 3 x 5x 4x6 x x 4x6 5x 3 x 5x x 8 x x 4x 8 x x 8x 6 3 x x x 5x 5x 5x 4 5x 9
Questio 6 (a) (iii) From our expressio i (ii), we have x x4 Settig x m, we have 5x x, sice x. m m 4 5 m m 5 m m m 5m m m m m... sice m m Questio 6 (b) (i) Sice the legth of the strig is a, so PS PS a, ad SS ae, where e, we ca deduce that P lies o a ellipse with foci S ad S. Questio 6 (b) (ii) PS epm a e acos e a ecos Questio 6 (b) (iii) QS si PS a cos ae epm a cos e a ecos e cos e cos
Questio 6 (b) (iv) Resolvig forces, we have si si Now, we must fid si QPS si QPS QS PS ae a cos epm ae a cos a ecos e cos e cos Substitutig this bac i, we have ecos ecos mg ecos ecos QPS mg., which is very similar to (iii). e cos ecos e cos ecos e cos e e cos cos e cos e e cos cos e cos mg e cos e cos e cos mg mg
Questio 6 (b) (v) Resolvig forces horizotally, we have cos cos cos cos cos cos QPS mr a PQ a e a e PQ a PQ PQ mr PS PS PS PS PQ mr PS PS e cos QPS mr mr mr bsi mr a e cos a e si mr a e cos.... where b is the legth of the mior axis... sice b a e Questio 6 (b) (vi) Divide the expressios from (iv) ad (v). r g e si e c os e ta e ta e r ta e g