# MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS

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2 QUESTION QUESTION Answer is A Answer is C QUESTION The correct nswer is option B. Formul: b y ± ( x h) + k WORKINGS Substitute 4, b, h, k : y ± ( x + ) y 4 ± ( x + ) 4 y 4 ( x + ) or 4 y 4 ( x + ) 4 y x or 4y + x 5 QUESTION The correct nswer is option B. Require x px q Δ ( p ) 4()( q) < p + 4q < p < 4q QUESTION The correct nswer is option B. sec( x) cos( x) sec( x) Pythgoren Identity: sin ( x ) + cos ( x) sin ( x) cos ( x) sin( x) ± π sin( x) since sin( x ) < for π x < cosec( x ) sin( x) The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge

3 QUESTION 4 The correct nswer is option E. Function Domin Rnge y cos ( x),] [ [, π ] Diltion by fctor from y xis cos ( x) Diltion by fctor b from x xis bcos ( x) Reflection in x xis bcos ( x) [, π ], [, bπ ], [ bπ, ], π Trnsltion by long y xis π bcos ( x) +, π π b π +, π ( b) π, QUESTION 5 The correct nswer is option D. / 8π 4π ( z ) 6 / cis + mπ 4cis + m z π, m Z 4π 4π 4π m : z 4cis 4cos + i sin 4 i i π π π m : z 4 cis 4cos + i sin 4 i + i + ( ( i), s expected) The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge

4 QUESTION 6 The correct nswer is option C. / 7 Since is rel, ll solutions lie on circle of rdius nd centre t the origin. π 4π The spcing between the rguments of ech solution is. 7 4 Another solution is therefore The Crtesin form of this solution is / 7 π 4π / 7 π cis + cis. 4 4 / 7 i, purely imginry number. QUESTION 7 The correct nswer is option C. Since the coefficients of p (z) re ll rel, z i is liner fctor of p (z) by the conjugte root theorem. + Therefore ( z + i)( z i) z 9 is qudrtic fctor. z z + 7z 8 ( z + 9)(z ) z is obviously rel root of p (z) nd so option C is flse. QUESTION 8 The correct nswer is option E. A rhombus is four-sided figure with opposite sides prllel nd ll sides of the sme length. Option A: The condition is insufficient becuse it does not ensure tht opposite sides AD nd BC re prllel. Option B: The condition is insufficient becuse it does not ensure tht ll sides hve the sme length. Option C: The condition is necessry but not sufficient. For exmple, the digonls of kite stisfy this condition but kite is not lwys rhombus. Option D: The condition is insufficient becuse it does not ensure tht ll sides hve the sme length. The condition in option E is therefore sufficient by the process of elimintion. The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 4

5 QUESTION 9 The correct nswer is option D. Option A is flse becuse α is NOT the ngle between NOT til-to-til. Option B is flse becuse b + c c. nd Option C is flse becuse c + b c + b + b. b - these vectors re Option E is flse becuse the vectors re linerly dependent: Option D is therefore true by the process of elimintion. c + b. Proof tht option D is true: A c c cos β nd b c bc cosγ. α Option D: c b c. b Therefore: c cos β bc cosγ cosγ () b cos β O β c γ C γ Eqution () is true for b AND β γ, tht is, for n isosceles tringle. It now needs to be shown tht eqution () is true ONLY for b AND β γ, otherwise there would exist NON-isosceles tringles stisfying option D. Sine rule (true for ll tringles): sin γ b sin β () b sin γ sin β Substitute eqution () into eqution (): tn γ tn β γ β sin γ cosγ sin β cos β Therefore the tringle OAC will be isosceles if c b c (option D). The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 5

6 QUESTION The correct nswer is option C. I π / 4 π / 4 π / 4 8 sec x dx sec x(sec x) dx sec x( + tn Substitute u tn x : Therefore: du dx x u π x u 4 du sec x dx sec x du I sec x( + u ) ( + u sec x ) du x) dx QUESTION The correct nswer is option D. When x, y Sin 4 sin y y sin ( x) x V V V where: π π π V Volume of cylinder of rdius nd height V volume of solid of revolution formed when the re enclosed by curve π y sin ( x) nd the line y is rotted bout the y-xis 6 y π / 6 sin y π π x dy π dy sin y dy. 4 y π / 6 π / 6.5 π π 6 π 6 y.5.5 x Therefore V π / 6 π / 6 π π π π sin y dy sin y dy The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 6

7 QUESTION The correct nswer is option B. Amount of tblet undissolved fter time t seconds is ( 9 x) mg. dx 5 Therefore: (9 x) 5 + x QUESTION The correct nswer is option A. dx t x, t, h., sin tn xn+ xn + h sin nd tn+ tn +. t x x + h sin + sin nd t t +.. ( ( )) t. x x+ h sin +. sin +. sin.5 nd t t+.. ( ( )) t. x x + hsin + sin.5 +. sin ( ( )) ( ) ( ) + sin.5 +. sin. nd t t +.. QUESTION 4 The correct nswer is option A. The direction (slope) field shows tht the line y x is member of the fmily of solution curves. Furthermore, the direction (slope) field suggests tht ll solution curves pproch the line y x s x +. This elimintes ll options except option A. The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 7

8 QUESTION 5 The correct nswer is option A. Chin rule: dh dh dv dv Given rte: 4 At time t : From similr tringles: h V π 6 π V r h dv 6 h h r 5 r 4 dv h π dh 6 dh dv 6 π h Rte of decrese of h : dh 6 (4) π h 64 π h QUESTION 6 The correct nswer is option D. d Option A is true: v dx + x v constnt 6 6 d x Option B is true: 6 constnt Option C is true: Option D is flse: dv + t v constnt d x x ± t constnt dv Option E is true: constnt QUESTION 7 The correct nswer is option C. dv v cos( x) ( sin(x)) cos(x) sin(x) sin(6x) dx The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 8

9 QUESTION 8 The correct nswer is option D. The equtions for stright line motion under constnt ccelertion cn be used: Tke the upwrds direction s positive: Dt: u 9 m/s v? 9.8 m/s t 8 s s h m Substitute the dt into s ut + t : h (9)(8) + ( 9.8)(8) 6 6 h 6.6 metres QUESTION 9 The correct nswer is option E. Forces cting on the smller mss: R Rection force of lrge mss on smll mss (Newtons rd Lw) m R Rection force of ground on smll mss mg weight force The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 9

10 QUESTION The correct nswer is option B. Let the lift hve mss m nd ccelertion. The mgnitude of the ccelertion is. Tke the downwrds direction s positive. Note tht the direction of the tension force in the cble is lwys upwrds. Eqution of motion of lift: m mg T T m( g ) Constnt ccelertion, tht is, : T T m( g ) Constnt velocity, tht is, : T T mg Constnt retrdtion, tht is, : T T m( g ( )) m( g+ ) T < < therefore option B. T T QUESTION The correct nswer is option A. Equtions of motion of kg object: T Forces (mesured in newton) in verticl direction: R + g 5 T R g. () 5 4T R Forces (mesured in newton) in horizontl direction: () 5 5 Substitute eqution () into eqution (): 4T 5 T g 5 5 T 5g 5 T 5g 5 Substitute T 6 : mx 68 5g 68 5g 5 5 R T T sinθ T 5 θ T cosθ 4T 5 μ R R 5 g The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge

11 QUESTION The correct nswer is option C. F μ R ( 5)(6)(9 8) 94 Since μ R 94 N is greter thn the force exerted by the worker ( N), the crte remins t rest becuse the friction is ble to djust to this vlue ( N) to prevent motion. In other words, the crte remins t rest since F. Note: A simple force digrm shows tht if friction is greter thn N, then there is net force in the direction of the friction. The crte would therefore move in the direction of the frictionl force. But by definition the frictionl force must oppose motion nd so there is contrdiction. Options B nd A re therefore clerly wrong. The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge

12 SECTION EXTENDED ANSWER QUESTIONS QUESTION () (i) z ( + i) i i (ii) zw ( 8+ 8 i)( + i) 6 6i+ 6 i 6 ( 6 6 ) ( 6 6) + (b) (i) π z + i 4cis π w + i cis 4 i (c) (ii) π π zw 4cis cis 4 π π 6 cis cis 4 π π 6 cis + 4 π cis From (ii) nd b (ii) π cos 6 6 π 6 6 cos ( ) 6 + ( ) + ( ) + 4 The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge

13 QUESTION () (i) () ( ) ( ) rt t i t j ( ) ( ) Distnce rt ( ) t + t+ 4 4 t 4t t 4t+ t 4t+ 5 (ii) Let D t 4t+ 5 4 D t 4t+ 5 ( ) d D 4 4t 4 Sttionry point when Check for minimum: 4t 4 t d ( D ) t ( D ) d When t, Therefore D nd D minimum when t Alterntely: dd 4t 4 4 t 4t+ 5 Sttionry point when 4t 4 t Check for minimum: dd 4 When t < 5 dd When t dd 8 When t > () () ( ) () ( ) () r i j r i j The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge

14 (b) x t () y t+ () From () y y t sub in () ( y) y x ( x+ ) ( y ) 4( x+ ) y ± 4 x+ y ± x+ As t, y y x+ (, ) y 7, 4 (, ) x Shpe Endpoint nd Intercepts ( mrks) (c) Let the position vector of the second prticle be rb () t xi + yj. When t ra() i j rb( t) xi + yj ra() rb( t) x y () ( ) When t x + y x + y () 6 x From () y sub in () x 7 4x 7 x 7 x + x x x± When x, y When x, y Position of second prticle is t, or, The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 4

15 QUESTION dt () kt ( ) dt k T ( ) t dt k T t log e T + c k t log e ( T ) + c s T > k When t, T 7 loge 4 + loge 4 k k ( ) c c ( ) t loge( T ) + loge4 k k 4 t loge k T 4 kt loge T kt e 4 T 4 T kt e T + 4 kt e (b) T kt e The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 5

16 (c) Solving T + 4 kt nd T kt simultneously kt e kt e + 4e e 5 55e kt 5 e kt 55 kt 5 e 5 kt loge 5 t loge k t loge k 5 (d) If k.5 t loge 5.8 minutes.5 5 (e) T ( C) (. ) (. 7) ( ) T T 5 t (min) Shpe nd endpoints Asymptotes Point of Intersection Appropritely scled ( mrks) ( mrks) The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 6

17 QUESTION 4 () R F A W (b) Resolving verticlly: F R W R W R g N or 7 6 N (c) Resolving horizontlly: F m A F m 4 μn 4 ( g ) 4 g 4 g.5 ms (d) As the ccelertion of the stem engine is constnt v t s u v.5(4) 4 ms The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 7

18 (e) Let the ngle of the incline be θ Resolving perpendiculr to the incline: F R g cosθ R gcosθ As the stem engine is trvelling t constnt speed up the incline then Resolving prllel to the incline: F 4 gsinθ ( gcosθ) 4 gsinθ gcosθ (Solve using CAS) θ.5 QUESTION 5 () (m/s ) Note: FR μn t (sec) First stge Second stge Third stge The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 8

19 t (b), t t v t v t + c When t, v c t v t, t (c) t 4 t v t t Intercepts t t nd t 4 Mximum occurs hlf wy between the intercepts t t 4 Therefore vmx m/s (d) On the xes below sketch the velocity time grph for the prticle from the strt t Point A until it comes to rest gin t Point B. v (m/s) 4 45 t (sec) First stge Second stge Third stge Both xes lbelled (e) Find the totl distnce trvelled by the prticle s it moves from Point A to Point B. t Totl distnce t metres The School For Excellence 5 Mster Clss Specilist Mthemtics Week Exm Pge 9

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