MATHEMATICAL METHODS (CAS)

The Mathematical Association of VictoriaMATHEMATICAL METHODS (CAS)SOLUTIONS: Trial Exam 2015Written Examination 2SECTION 11. A2. B3. D4. C5. D6. B7. B8. E9. D10. A 11. C12. E 13. C 14. A 15. E 16. B 17. C 18. E 19. D 20. D 21. B 22. AMULTIPLE CHOICE - WORKED SOLUTIONSQuestion 1x is in the fourth quadrant5212tan( x) 212AQuestion 2 x f : R R, f ( x) 3sin 1 1 2 The amplitude isThe period Question 3 3 32π 4π 1 2 g ( x) tan ( 3x 1) Bπ2πThe period 3πLet 3 x 1 2π 1x 6 3General solutionπ 1 πx k, k Z6 3 3(2k 1)π 2x ,k Z6D The Mathematical Association of Victoria, 2015

2015 MAV Mathematical Methods (CAS) Trial Exam 2 SolutionsQuestion 4A cos2 ( x) B cos( x) 0 , x [0, 2π ]cos( x) ( A cos( x) B ) 0Solve cos( x) 0π3π2 solutions2Solve A cos( x) B 0Bcos( x) AB 1 1ATo get no solutions B A Cx 2or x Question 51 2g(x) x 2x 1221 x 121 x 12()then g (a 1) 11a 1 1 a 2 ,22where a 2, g (a 1) 1(a 2)2D The Mathematical Association of Victoria, 20152

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutions3Question 6A possible equation for the graph shown is y 9 x 2 1It is a semicircle with radius 3 and centre ( 0, 1)Question 7g : R \{1} R, g ( x ) 22and f : [0, ) R, f ( x ) ( x 2) 1x 1The domain of f g is[0,1) (1, ) .The range is ( , 10] 12 4 10, The Mathematical Association of Victoria, 2015)BB

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutions4Question 8This is the graph of a one to many relation (horizontal:vertical line test).It is not a function as a vertical line hits the graph more than once for x 0.Options A – D are correctQuestion 942 x The image of the function g ( x ) 2 x is y 1 .3 3 4The transformations that have been applied are:reflection in the x-axisgives y1 2 x 4then a dilation from the y-axis by a factor of 3 x gives y2 2 3 4then a translation in the negative direction of the x axis by 3 x gives y3 2 1 3 4followed by a dilation from the x-axis by a factor of2 x gives y4 1 3 3 4D The Mathematical Association of Victoria, 201513E

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutions5Question 10 3x 23The inverse is f 1 ( x) 2xf : ( , 2 ) R, f ( x) Domain f 1 range f ( 0, )3f 1 : ( 0, ) R, f 1 ( x) 2xAQuestion 1122The graph with equation y ( x 1) is transformed to its image equation y 5 ( x 3) 2 .Option A is incorrect as the translation is incorrect.0 x 0 5 y 2 x 1T y 0Option B is incorrect as the dilation of a factor of 5 from the y-axis should be a factor of 5 from the x-axis. x 5 0 x 4 T y 0 1 y 2 Test Option C x 1 0 x 2 T y 2 5y0 x 2 x1 x 2 x1Gives5 y 2 y1 y y1 252In equation y ( x 1)We gety1 22 ( 2 x1 1)52y1 5 (3 x1 ) 22Giving y 5 ( x 3) 2C The Mathematical Association of Victoria, 2015

2015 MAV Mathematical Methods (CAS) Trial Exam 2 SolutionsQuestion 1233f ( x) f '( x) 2xx f '(3) 13EQuestion 13h( x) x 2 2 x 1 h '( x) 2 x 2For {x : h '( x) 1}Solve 2x 2 1Gives x 12CQuestion 14 x 1, x [ 2, ) f ( x) 2 x 1, x ( 0, 2 ) 2 x 3 , x ( , 0] 1x . Hence differentiate the middle rule of the function.2ddx()2x 1 1 1f ' 2 2 12x 1A The Mathematical Association of Victoria, 20156

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutions7Question 15Area f (0) f (1) f (2)By symmetry this is the same asf (4) f (5) f (6)as the asymptote is at x 3EQuestion 16The average value 1π0 π ( e2cos( x π )) dx 2.3 correct to one decimal place Question 17 3 1h( x)dx 5 1 (1 5h( x) ) dx (1 5h( x) ) dx (1) dx ( 5h( x) ) dx (1) dx 5 h( x)dx33 133 13 13 1 13 [ x] 5 5 1 (3 1) 25 4 25 29C The Mathematical Association of Victoria, 2015B

2015 MAV Mathematical Methods (CAS) Trial Exam 2 SolutionsQuestion 18 E ( X ) ( x e x ) dx 102 SD ( X ) ( x e ) dx ( x e x ) dx 10 0 2 xLet median aaSolve1 ( e ) dx 2 for a x0a loge (2)EQuestion 190.1 a b c 1a b c 0.9 B TrueUsing E ( X )0.1 3a 5b 7c 43a 5b 7c 3.9 C TrueUsingVar ( X )0.1 9a 25b 49c 42 2Hence 9a 25b 49c 1.9 D falseDSolving the three equations shows A is correct0.1 0.325 0.425 The median is 5. E is correct. The Mathematical Association of Victoria, 20158

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutions9Question 20Using z scores, z x µσPhysics is the highest.DQuestion 21Pr ( X 135 X 130) Pr (130 x 135)Pr( x 130)Pr (130 x 135)12 2Pr (130 X 135) 2Pr (125 X 130 )BQuestion 22A score of 15 can be obtained by a 5 then 10 or a 10 then 5.Area of the inner triangle 4 3Area of the middle section 25 3 4 3 21 3Area of the board 100 3Probability of a score of 15 2 4 321 3421 2 100 100100 3 100 3 The Mathematical Association of Victoria, 2015A

2015 MAV Mathematical Methods (CAS) Trial Exam 2 SolutionsSECTION 2EXTENDED RESPONSE QUESTIONSQuestion 1 (13 marks)a. Shape1A(2, 0) closed circle(8, 6 ) open circle1A1A(2, 0)b.f (4) c. Solvex 942x 2 1A121Ad. Equation of the tangent at x 4 is y The Mathematical Association of Victoria, 20152x41A10

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutions(Note that this tangent goes through the origin)24 2 2 xdx x x 2 dx 0 4 2 4 2 2 0 4 x dx 4 2 2 4 x x 2 dx e. Area 1A1A1AOR4!42 xdx #& 4 %0"2Area 4!240" #4 (Area )x 2 dx1A x &dx%1A)1Ax 2 dx2f.(2 23 The Mathematical Association of Victoria, 20151A11

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutionsg. Equation of normal is y 2 2 x 9 2 .The normal intersects the x-axis at x 4.5.4Area ()2Area 4.5x 2 dx ( 2)2 x 9 2 dx4121A1A19 2square units 1A12Question 2 (15 marks)a.5 2 x 0, x a 52b. Solvex c.521Ag ( x) loge (5 2 x ) 1 05 e21Ag (0) 1 loge (5) 1A The Mathematical Association of Victoria, 2015

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutions13These are now the axial intercepts.d. Shape1AAsymptote 1AAxial InterceptsGraph of5 e , 0 2 (0,1 loge (5) ), 1Ag ( x) loge (5 2 x ) 1e. Shape (cusp)Asymptotes x Axial intercepts( ) 521Ax 2.51A5 e 5 e , 0 , , 0 2 2 (0,1 loge (5) ) cusp, (Graph of g x 1 loge 5 2 x) The Mathematical Association of Victoria, 20151A

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutionsx 2.5f. 5 2 x 2.55 2 differentiable for domain x , 0 0, 5 ( )2 21Ag '( x ) 5 2x 5 For x , 0 , g ( x ) 1 loge (5 2 x ) 2 21Ag '( x ) 5 2x g. For x 0, , g x 1 loge (5 2 x )Correct domain1A 2 5 5 2 x , x 2 , 0 g ' ( x ) 2 , x 0, 5 2 5 2 x The Mathematical Association of Victoria, 20151A14

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutionsh. 3 correct transformations in a correct order 1AThe remaining correct and in a correct order 1AThere are other possibilities. Translation of 1 unit in the negative direction of the y-axis Translation of 2.5 units in the negative direction of the x-axis Dilation by a factor of 2 units from the y-axis Reflection over the y-axis Reflection over the x-axis The Mathematical Association of Victoria, 201515

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutions16Question 3a. This is at the point of inflection.x π1A35πb. x 4 5π S 160 cm to the nearest cm 4 S1 is 160 cm below the waterc.1A1A 1 π π x h : 0, R, h( x) tan x 6 3cos 3 4 2 2 d. The minimum value occurs when x 0.701 1Ah( x) 336 cm The Mathematical Association of Victoria, 20151A Rule1M1A Domain

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutionse.πwhen B1 ends4 π B 2.7716. 4 x SolveS ( x ) 2.7716. for xx 3.588.1AHorizontal distance is 3.588. f.Period 2π 6 2π 3 300 50 waves6g.17π4 280 cm to the nearest cm1A1Aw(t ) 3 for the slide to be completely under water. The Mathematical Association of Victoria, 20151M1A

2015 MAV Mathematical Methods (CAS) Trial Exam 2 SolutionsSolve w(t ) 3 for t1 5t ,2 2B1 will be under water for 2 seconds per cycle (6 seconds)1of the time1A3Question 4 (17 marks)Let P pass and F faila.Pr ( PPP ) 0.3 0.6 0.6271A250b. Pr( PFF ) Pr( FPF ) Pr( FFP)1M 0.3 0.4 0.7 0.7 0.3 0.4 0.7 0.7 0.31A 0.315Pr ( P 3 P 1)c. Pr ( P 3 P 1) 1MPr ( P 1) 0.108 orPr ( P 3)1 Pr( F 3)0.3 0.6 0.6 1 0.73121A 73 The Mathematical Association of Victoria, 201518

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutions 0.7 0.4 d. 0.3 0.6 0.572 0.428 5 1 0 191MThe probability he will pass the 6th test is 0.428 correct to 3 decimal places1A0.330.44 fail, pass0.3 0.4 70.3 0.4 7Jeremy is more likely to fail a test in the long term,e. Steady states1A34as the probability of Jeremy passing in the long term isand failing is .772f. X : N µ , σ()Pr ( X 86) 0.1, Pr ( X 95) 0.0186 µ1A 1.281. (1)σ95 µ1A 2.326. (2)σ1Aµ 74.96σ 8.61g.1APr ( X a ) 0.85a 83.891A The Mathematical Association of Victoria, 20151A

2015 MAV Mathematical Methods (CAS) Trial Exam 2 Solutionsh.1Ap3 p 2 p 1p 0.543.Pr ( S R ) Pr(S ) Pr( R)Let1Ma Pr ( S R )Solve a (a p) p 3 for a1Aa 0.10 The Mathematical Association of Victoria, 201520

MATHEMATICAL METHODS (CAS) SOLUTIONS: Trial Exam 2015 Written Examination 2 SECTION 1 1. A 2. B 3. D 4. C 5. D 6. B 7. B 8. E 9. D 10. A 11. C 12. E 13. C 14. A 15. E 16. B 17. C 18. E 19. D 20. D 21. B 22. A MULT