THGS Mthemtics etension Tril 00 Yer Tril Emintion Mthemtics Etension Question One mrks (Strt on new pge) Mrks ) If P is the point (-, 5) nd Q is the point (, -), find the co-ordintes of the point R which divides the intervl PQ eternlly in the rtio :. b) When ( + ) ( ) + is divided by k, the reminder is k. Find the vlue of k. c) Solve d) Find the generl solution of sin cos in rdins e) Find the ect vlue of sin d 0 Question Two mrks (Strt on new pge) dy d y 0 nd 0 d d ) Drw grph of function y f for such tht for b) Differentite: i) 4 ii) e loge c) i) Show tht 4 9 d ii) Hence find 4 d) Evlute d using the substitution tht 0 u Pge of 6
THGS Mthemtics etension Tril 00 Question Three mrks (Strt on new pge) Mrks ) Given f i) Write n epression for the inverse function f ii) Write down the domin nd rnge of f b) B A P PT is tngent to circle ABT. PAB is secnt intersecting the circle in A nd B. PA = 8 cm nd AB = 0 cm. Find the length of PT giving resons. T c) Find the grdients of the lines which mke ngles of 45 with the line whose eqution is y 6 0. d) i) Epress cos sin in the form r cos where r 0nd 0 ii) Hence solve cos sin for Question Four mrks (Strt on new pge) ) R Q km 7 h K The ngle of elevtion of hill top from plce P due south of it is 7. The ngle of elevtion of this sme hill top from plce Q, due west of P, is. The distnce of Q from P is km. If the height of the hill is h km. P South Est i) Prove tht PK = h cot 7 ii) Find similr epression for QK iii) Hence, or otherwise clculte the height of the hill to two deciml plces. Pge of 6
THGS Mthemtics etension Tril 00 Question Four continued Mrks b) 0 0 Wter is running out of filled conicl funnel t the rte of 5cm s. The rdius of the funnel is 0 cm nd the height is 0 cm: i) How fst is the wter level dropping when the wter is 0 cm deep? (nswer in ect form) 4 ii) How long does it tke for the wter to drop to 0 cm deep? (nswer to deciml plces) Question Five mrks (Strt on new pge) ) T P B C A PT is tngent nd PAB is secnt. TC = TA. Prove BTC TPA b) Given θ is cute: i) Write sin in terms of cos ii) Prove tht sin tn cos iii) If sin 4, find the vlue of tn 5 d d c) Find cos sin Pge of 6
THGS Mthemtics etension Tril 00 Question Si mrks (Strt on new pge) Mrks ) i) Show tht the sum of the cubes of three consecutive integers n, n, n is n 6n. ii) Using prt i), prove by mthemticl induction, for ll positive integers n, n tht the sum of the cubes of the three consecutive integers is divisible by 9. b) Consider the vrible point P, y on the prbol is given by = t: y. The vlue of P i) write its y vlue in terms of t ii) write n epression, in terms of t, for the squre of the distnce, m from P to the point (6, 0) iii) hence find the co-ordintes of P such tht P is closest to the point (6, 0). 5 Question Seven mrks (Strt on new pge) ) Use the binomil epnsion of n to show tht: i) n n n n 4... 4 n ii) n n n n n 4... n4 n n Pge 4 of 6
THGS Mthemtics etension Tril 00 Question Seven continued Mrks b) Mr Mc hits golf bll from point O towrds flt, elevted green s shown in the digrm below. The hole t the bse of the flg is situted in the centre of the green: y Flight pth of golf bll Flg with hole t its bse Elevted green Digrm not to scle 5 m Point O 58 5 m The golf bll is projected from the point O with initil velocity of v ms - t n ngle of α to the horizontl. You my ssume the only force cting on the golf bll in flight is grvity which is pproimtely 0ms - : i) Tking the point O s the origin, show the prmetric equtions of the flight pth of the golf bll re given by vt cos nd y 5t vt sin ii) If the ngle of projection is given by tn find v which will enble Mr Mc to hit the bll directly into the hole on the green. The hole is situted 58.5 m horizontlly nd 5 m verticlly from point O. Answer to the nerest integer. iii) Using your vlue of v from prt (ii), t wht ngle does the bll strike the green? ( nswer to the nerest degree) END OF EXAMINATION Pge 5 of 6
THGS Mthemtics etension Tril 00 STANDARD INTEGRALS n d n n, n ; 0, if n 0 d ln, 0 e d e, 0 cos d sin d sec d sin, 0 cos, 0 tn, 0 sec tn d sec, 0 d tn, 0 d sin, 0, d ln( ), 0 d ln( ) NOTE: ln = log e, > 0 Pge 6 of 6
THGS Mthemtics etension tril solutions 00 QUESTION ) P,5 G, 6 9 0 6 y 6 R 9, 6 b) P k k k k k 4 0 k 4 c), 0 0 d) sin = cos tn = = tn = n 4 n e) sin d 0 0 cos = sin sin 0 d 0
THGS Mthemtics etension tril solutions 00 QUESTION bi) y 4 y 8 4 bii e ) y = loge y e ln ci) RHS = 9 = 4 4 9 = 4 = LHS cii) d 4 d 9 = tn c ), d d u du d 0 = 4 u du 4, u 4 = u 0, u 7 = Question i) y
THGS Mthemtics etension tril solutions 00 Question i) y y y y y y y f ii) Domin ll rel, rnge ll rel, y b PT BP PA p ) rod of secnt nd tngent 88 PT m m c) tn mm m m m m 5 m m 5 since 45 45 90, m 5
THGS Mthemtics etension tril solutions 00 di) cos sin r cos rcos cos rsin sin r cos r sin tn cos sin cos ii) cos sin cos cos 5 5 7,,,, 4,,0,, QUESTION 4 PK i) tn7 h PK = h cot 7 h ii) tn = QK QK h cot 4 iii) h cot 7 h cot h cot cot 7 9 h h.54km 9 cot cot 7
THGS Mthemtics etension tril solutions 00 4 bi) v r h h h h, r h dv h dt 4 dh dh dv dt dv dt 4 5 h 0 h dh when h 0, cm / s dt 5 dv ii) 5 dt v 5t c 50 when t 0, v 0 50 000 5t 750 5t 50 t t 67s
THGS Mthemtics etension tril solutions 00 Question 5 5) PTA BTA ngle between tn nd chord TPA PTA TAC et ngle of TCA TPA PTA equl ngles of isos BTC TCB BTA equl ngles of isos BTC TPA bi ) cos sin sin c os cos sin sin bii) RHS cos sin cos cos sin cos cos sin cos tn RHS
THGS Mthemtics etension tril solutions 00 sin biii) tn cos 4 5 5 c d d ) cos sin d sin d sin cos cos for ist nd 4th qud = for nd nd rd qud, cos 0
THGS Mthemtics etension tril solutions 00 Question 6 i) S n n n n n n n n n n n n 6 RHS n ii) S 9, S is divisible by 9 k Assume tht S is divisible by 9 when n k ie S n 9 m, then 6 k S k k k 6k 9 k k 9m 9 k k S k is divisible by 9. Hence by the pple of MI, the sttement is true for ll n b) y i) t, t y t y ii m t ) t 6 t t 6 dm iii t t dt ) 0 t t t, 6 0 4 t 4 6 0 t t t P, 4 4 0 no soln
THGS Mthemtics etension tril solutions 00 QUESTION 7 n n 0 n n n n 0 n i) C C C... C let n n n n C C... C n n n n n n n C C Cn 4... ie. C 4 C... 4 C s reqd n n n n n ii) Differentite wrt n n n n n n n 0 C C C... n C n now let n n n n n n n C 4 C C... n Cn. n n n n n n C 4 C C... n C 4. n n n n n n n C 4 C C... n C 4 s reqd n
THGS Mthemtics etension tril solutions 00 Question 7b continued 7 bi) horizontl 0 verticl y 0 c y 0t c v cos y vsin v cosdt t 0, vt cos c c v sin t 0, 0c 0 y 0t v sin vt cos y 0t vsindt y t vt 5 sin ii) tn cos sin vt cos t v cos sub into y t vt 5 sin v 5 sin v cos vcos 5 sec tn v 5 tn tn v using 58.5, y 5, tn 558.5 then 5 58.5 v v v 558.5 4 58.5 5 95.98 v 4.999 m / s v 44 m / s
THGS Mthemtics etension tril solutions 00 5 7 biii) y tn v cos 5 44 5 45 5 y 76 m tn 7 5 76 cute tn 7 bll strikes t 5 End of ssessment