[Methods] 2005 VCAA Exam 1 MAV Solutions PDF

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2005 Mathematical Methods Written Examination 1 Suggested Answers & Solutions Part 1 (Multiple choice) Answers 1. D 2. B 3. D 4. E 5. D 6. D 7. E 8. D 9. E 10. A 11. D 12. A 13. A 14. C 15. E 16. A 17. B 18. C 19. C 20. C 21. D 22. C 23. D 24. B 25. E 26. B 27. A

= 8.0023 Answer: 8 games Question 6 [D] Using Factor 7 (or 9) the factorised form is ( ─ 2)(( + 2)( + 3). So (x?3) is not a factor. Answer: ( ─ 3)

Part 1 (Multiple choice) Solutions Question 1 [D] At least two hours means 2 or 3 or 4 hours. The sum of the proportions is 7 10 4 21 + + = 30 30 30 30 Question 2 [B] Mean = ∑ ( × proportion) 3 6 4 = 0 × + 1 × + ...4 × 30 30 30 0 + 6 + 14 + 30 + 16 = 30 66 = 30

Question 7 [E] The turning point of this quadratic is at (3, 1). The function needs to be one?to? one for the inverse to exist. Since the domain extended to +∞, it starts from the  value of the turning point. The domain is [ 3, ∞) or  ≥ 3. Question 8 [D] 3 2 = 4 2  = 4/3 2 ln(  ) = ln(4/3) 2= ln(4/3) and so = ½ ln(4/3)  = 0.144 to 3 decimal places.

Question 3 [D] InvNorm ( 0.95, 0, 1) = 1.645 from calculator.

Question 9 [E] 2 log10 () + 3 = 5 log10( ) Therefore 3 = 5 log10( )─ 2 log10 () 3 = 3 log10 () 1 = log10 ()and so  = 10.

Question 4 [E] Hypergeometric distribution. The probability of at least one box with a prize is the same as taking the probability of no prizes from 1. 4  2×2  0 = 1─ 6 2 = 9/15

Question 10 [A] The amplitude is 2, eliminating alternatives C and D. 2π 5 2 = 1 ÷ = and so = 5π. Period: 2 5  This eliminates E. When  = 0 the value of  = 3. Out of the two alternatives remaining, A fulfils this requirement.

Question 5 [D] Binomial: Bi(n, 0.3) Pr(John wins no games) = 0.7 n Therefore 0.7n = 0.0576  log(0.7) = log(0.0576)

© 2006 The Mathematical Association of Victoria

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MAV Mathematical Methods Examination E 1 Solutions

Question 11 [D] The period of the graph  = tan (is π and so the period here iis π ÷ ¼ or  4π. Question 12 [A] Using the Graphics Calcullator with y1= cos2and y2 =√3 sin2, only one point of intersection is fou und within the domain.

Dividing this  value show wn on the screen by π gives 0.0833333, (the 1 decimal equivalent for ) and so the 12 π exact answer is . 12 Or, algebraically, Cos(2x) = √3sin(2x) 1 Tan(2x) = 3 π 2x = 6 π x= 12

Question 13 [A] If sin( +  < 0 then  < ─as the maximum value of sin(isi 1.

Question 14 [C] The  and  values interch ange for the inverse function. Because the original function had a horizontal aasymptote with equation  = 1, this will w become a vertical asymptote with eq quation  = 1 for the inverse. Only altern natives C and E display this feature. Alsoo, a function and its inverse are mirror images i in the line  =. Only C displayss this.

© 2006 The Mathematical Asso ociation of Victoria

Question 15 [E] A translation of ─2 paraallel to the  axis of  =  2 results in a graph with equation  = (+ 2)2 . This new graph is then ddilated by a factor of ½ from the  axxis. This means that the intercept of ( ─2, ─ 0) is now positioned at (─1, 0), i.e e. half the distance from the axis.. E is the only alternative to do this. Answer:  = (2+ 2)2 . Question 16 [A] A factor of (─ 1)2 is re equired because the graph touches the aaxis at = 1. This eliminates B and C . It also requires factors of (─3 3) and (+2), or the negative of both of o these. This eliminates D. If  = 0 thee value of is negative. A has this prop perty whereas E does not. Question 17 [B] The vertical asymptote hhas equation ─1 = 0 and so ─1 = + The value of is ─1. Only alternatives a A and B are possible answ ers. The graph’s horizontal asymptote has  is the equation = 2. But =  horizontal asymptote in the given equation. Hence  = 2. B satisfies these two requirements. Question 18 [C] 5 −2 = = 1− +3 +3 5 → 0 , so y→1 So as x→±∞, + 3 So asymptotes occur at y =1 and x= ?3. Question 19 [C] The gradient of  () shoould have the following features: It • is always positive • approaches zero as  → ∞ • is large positive as  → 0. Only C has these properrties.

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MAV Mathematical Methods Examination E 1 Solutions

Question 20 [C]  (1) −  (0) . 1− 0  (1) = 12 + 1 and  (0) = 0 2 + 0 = 1 1 +  −1 Average rate = = . 1

Average rate of change =

Question 21 [D]  9 (10 (1 − ) )    9 = (1─ ) 9 (10  ) + 10 (1 −  )     = 10(1─ )9 + 10 (9)(─1))(1─ )8 = 10(1─ )8 [1─  ─ 9] = 10(1─ )8 (1─ 10)

Using Product Rule:

Question 22 [C]       (2 ) = (  ( )) ×     where  = 2 = 2 2 ′( ) = 22 ′(2 )

Question 23 [D]  = 8  3 ─ 122 .   At  = 2, the values of a and are 0  and 16 respectively. The equation of the tangent is  ─ 0 = 16 (  ─ 2 ). This is alternative D.

= 2 4 ─ 43 and so

  The domain for this window is [0, 2π]. By observation there aree 4 points of intersection.

Question 26 [B]  1 1 +1 ∫ ( + )  =  ×  +1 ( +  ) + c. In this case = 2 , = ─ 1 , = ─3/2 and so the integral is: 1 1 3× × (2─ 11) ─1/2 +  2 − 1.5 + 1 = ─3(2─ 1) ─1/2 + 

Question 27 [A] If any of the sets for inncluded zero then the definite integrall could include 0

∫  ( ) which is zero annd so G() 0

Question 24 [B] The gradient of this curve is negative for > 2 and positive for all a other values of except at = 0 and = 2 where it is zero. Alternative B shows these features.

Question 25 [E] If () = sin 2+ cos theen the derivative is 2 cos 2─ sinn 

would not be positive. This T eliminates C and D. Alternative E iis readily discarded because if = /2 the area is clearly positive. Alternative B has certainnly got a positive value for G() but b the word prevents it from beiing a correct answer. Alternative A is correct. If < 0 then 

∫  ( )

will be the negative of a

0

negative number, i.e. positive. If > 0 then the integral is clearly positive. Alternative A is correct. © 2006 The Mathematical Asso ociation of Victoria

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MAV Mathematical Methods Examination E 1 Solutions

PART II Question 1 Normalcdf ( ─∞, 46, 41, 3 ) = 0.9522. Answer: 0.952 to three deccimal places. Select a point other thann (0,5) to indicate scale along the  axis. (1, 6.90) would suffice. Horizontal asymptote = 8. Do NOT sketch anything for < 0. b. Interchanging and : = 8 ─ 3 ─ 3  ─= 8 ─  8−  ─= 3 8−  ─ = loge ( ) 3

Question 2 The display from a graphiccs calculator is shown. The x?values of 0,1,2,3,4 need to be placed horizonttally and y values of 0.1, 0.2, 0.3, 0.4,, 0.5 vertically. Do NOT join the points.

1(x) = ? loge (

Question 3

8−  ) for 5 ≤  < 8 3

Question 6 a.  = ( + 2 )( 2 ─4  + 3) b.  = ( + 2 )( ─ 1)( ─ 3) and so the intercepts are 1, 3 aand ─ 2. 1

c. The vertical asymptotes occcur at = π and = ─π . 0) and (0,1). The intercepts are ( ─π/2,0

3

1

 4 2 3 5 2 − − + 6  ]1− 2 4 3 2 4 3 2 2 5  3 − + 6 ]1 ?[ − 3 2 4

=[







 

 − 2  2 − 5  + 6)

3 2 ─ ∫ (  − 2  − 5  + 6) 





3

−2

Question 4



∫ (

 







Question 5 a.

© 2006 The Mathematical Asso ociation of Victoria

16 16 20 1 2 5 − −12 ) =( − − + 6) ─( + 4 3 4 3 2 2 1 2 5 81 54 45 − + 18 ) +(( − − + 6 ) ─( − 4 3 2 4 3 2 = 21.08

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MAV Mathematical Methods Examination 1 Solutions

Question 7 a. 3 =  Taking logs to base of both sides: log e (3 )= loge ( ) log e 3 = so  = loge ( 3) for all . Answer:  = loge 3 b.     log 3 (3 ) = ( )   = loge(3) ×  log3 or loge(3) × 3 

© 2006 The Mathematical Association of Victoria

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