2020 Mathematical Methods Trial Exam 2 Solutions 1 PDF

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2020 Mathematical Methods Trial Exam 2 Solutions © itute 2020

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Q11 y = (1 − x ) x 2 + bx + c has three x-intercepts. 2

.: ∆ = b − 4c ≥ 0 , b ≤ −2 c or b ≥ 2 c .

D

Q12 f (t + 10 ) = f (t) has a period of 10. Given f (5 + a) = − f (5 − a) for 0 < a < 5 , .: f (6 ) = − f (4) Hence f( 26 ) = − f (34 ) .

A

SECTION A – Multiple-choice questions 1 B

2 B

3 D

4 C

5 E

6 B

7 A

8 B

9 A

10 C

11 D

12 A

13 C

14 D

15 B

16 A

17 E

18 C

19 E

20 A

b 2 Q1 y = ax + , ay = (ax ) + b a 2

B

Q13 If the common tangent touches y = f ( x) at (a , b) , then it touches y = f ( x + h) + k at ( a − h, b + k ) . Gradient of the common tangent =

Q2 a 0 = 1 , a1 = −1 , a 2 = 1 , a3 = −1 , …, a n−1 = 1 .: sum = 1 Q3 α = β , .: α

−1

log 5 b

b

=β

−1

(

Q4 5loga = 5log5 a = 5log5

B

D

)

1

b log5 a

=b

1 log 5 a

(

= b log 5

a )− 1

Q5

C E

(b + k ) − b = − k (a − h ) − a h

Q14 Let P(x , y ) on the curve be the point closest to O . y 1 Gradient OP = , gradient of tangent to curve at P = . x x y 1 2 2 At P , × = − 1 , x + y = 0, .: x + log e x = 0 x x

C

D

4

 b −a  Q15 a × 1 +    x + 2 a − b  dx = 1 , .: 2a + b = 1 2     2



Q16 Pr( B′ ) = Pr( A∩ B′ )+ Pr( A′ ∩ B′ ) 2 7 1 Pr ( A ∩ B′) = Pr ( B′) − Pr ( A′ ∩ B′) = − = 3 12 12 1 ′ Pr ( A′ ∪ B )  = Pr (A ∩ B ′) =   12

B

A

1

 (x + 1) dx n

2 n− 1 = Q6 1− 0 1

0

Q17

, n ≈ 1.53

1 −0

1 4   Q7 A= 4 ×  x − x 4 x2 −1  dx = 3 3   0



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)

B

A

x Q8 y = sin  is an odd function. The number of intersections it m  B makes with y = mx cannot be 2.

1 Q9 y = a cos( nx) is an even function. When b > 1 , 0 < < 1 . b If p is a solution, then −p is also a solution. .: Sum= 0 A Q10 g ′(b) =

1 1 = = b−1 f ′( a) b

E

Q18 Sum of hidden numbers= 63 − 58 = 5 Possibilities: (113 ), (131), (311), (122 ), ( 212), ( 221) 6 1 Pr(58) = = 6× 6 × 6 36

C

ˆ = 0.20 ×0.80 = 0.04 Q19 E Pˆ = 0.20 , sd P 100 Pr Pˆ < 0.22 ≈ 0. 6915

()

()

(

)

Pr(at least one ) = 1 − Pr (none ) ≈ 1 − 0.0028 = 0. 9972 Q20 Pr (X ≤ 1) < 0.1 Pr( X = 0 ) + Pr( X = 1) < 0.1

(0.2 )n + n(0.2 n−1 )(0.8 ) < 1 , .:

C

n≥4

A

1 2020 Mathematical Methods Trial Exam 2 Solutions

E

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SECTION B 2  10  − 1  = 0 , t = 9 , time taken= 9.000 min. Q1a h =  9  t +1 

2 10  dV = − 2π × 10 Q1b V (t ) = π × 12 ×  − 1 , 9 t + 1  dt 9 (t + 1)2 20π dV = m3 min-1 2 dt 9(t + 1)

20π 2π 3 m 3 min-1, min = m min-1 Q1c Max = 9 90 Q1di At lowest level,

Q2di

dA = 2π r + 1− r 2 − 2r cos− 1 r , 0 < r < 1 dr

Q2dii 0 < r < 1 , .: 0 < θ <

π 2

,

dA dA < 2π r + 1 − r2 , .: > 0, dr dr .: A( r) is a strictly increasing function in 0 < r < 1

.: π r + 1− r 2 <

Q2e

dV 20π 20 π = 1− = 0 , (t + 1)2 = 2 dt 9 9 (t + 1)

t ≈1.642 min.

2π  10  −1 ,  9  t +1  2π  10  − 1  ≈ 3. 586 m 3 V (1. 642) ≈ 1. 642 +  9  1. 642 + 1  Q1dii V (t ) = 1× t +

Q1diii V( t) = 1× t +

2π  10  − 1 = 2π , t ≈ 5.981 min.  9 t +1 

Q1div Let r m 3min -1 be the constant rate, 2π  10 2π  − 1 = 2π , r = r ×9 + ≈ 0. 698  9  9+ 1  9

Q3a

2π =12 , m ≈ 0.5236 m

Q3b TP = 12 ×

34 2π = 17 , n ≈ 0. 3696 = 17 , 24 n

π Q3c h P (t ) = 17 sin (0.3696t )+ 20 = 17 cos − 0.3696t  + 20 2  = 17 cos (0.3696 (4.25 − t ))+ 20

Q2a

Q3d hQ (5 ) = 12 sin (0.5236 × 5 ) + 34.95 ≈ 40.95 Q3e hP (t ) = 17 sin (0.3696 t ) + 20 = 30

t=

Q3f t ≈ 7.03 ( ±0.02 )

PQ = 1 − r 2 , Area E = area ∆ POQ − sector F =

1 10  sin −1  ≈ 1.70 0.3696 17 

1 1 r 1 − r 2 − r 2θ 2 2

A = area of circle + 2 ×area E =π r2 + r 1 − r2 − r2θ Q2b 0 < r < 1 Q2ci limθ = 0 r →1

Q2cii lim A = π × 1 2 + 1 1 − 1 2 − 1 2 × 0 = π r→ 1

2 2020 Mathematical Methods Trial Exam 2 Solutions

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Q3g LCM (TP ,TQ ) = LCM (12,17) = 204 s or by graphs (see below) First time Second time

Q5a k × area under graph =1

4 1  1 k  (2 × 0.5 + 3× 3) + ( 0.5 + 3)3 = 1, k = 41 2  2 Q5b Pr (bus A late by > 5 ) =

4 1   × 3 × 3  ≈ 0. 4390 41  2 

Q5c Pr (miss either bus)

= Pr( bus A late by < 7) + Pr( bus B late by< 2) − Pr( bus A late by < 7 and bus B late by < 2 )

x− 2 y − 1= e 3 5 Sequence of transformations: Right 2, up 1, dilate from y-axis by 3, dilate from x-axis by 5. c = −2 , d = − 1 , a = 3 , b = 5 x −2

Q4a y = 5e 3

 4 1   4 1 = 1 −  × 1× 1  +  × 0.5× 2 41 2      41  2

+5,

Q4b Equation of curve B:

y −2 y x  x − 1 = e 3 , − 2 = log e  − 1 5 3 5 

 4 1   4 1 − 1 −  ×1 ×1   ×  × 0.5 × 2  ≈ 0.9536   41 2  41  2 Q5d Binomial: p ≈ 0. 4390244 , n = 5 Pr( X ≥ 2) ≈ 0.7271 Q5e

x  y = 3 loge  − 1 + 6 5 

0≤ t≤ 2  4t  5t− 7 2< t≤ 5  f (t ) =  6 t 8 5< t ≤ 8 −   0 elsewhere

Q4c Range of A: When x = 0 , y = 5e −2 + 5

[

]

Domain of B is the range of A, i.e. 5e −2 + 5, 40

Mean

 40  − 1  + 6 = 3 log e 7 + 6 Curve B: When x = 40 , y = 3 log e   5  Range of B is [0, 3 log e 7 + 6] .

x dy 5 3x −2 3 = e = 1 , − 2 = log e 3 3 5 dx 3 3   x = 3 loge + 6 , y = 3 + 5 = 8 , 3 log e + 6, 8 5 5   3   Curve B:  8, 3 log e + 6  5   Shortest distance

2

=



5

t× kf ( t) dt +

0



8



t× kf ( t) dt + t × kf (t) dt

2

5

8 81 108 575 = + + = ≈ 4. 674797 123 41 41 123

x−

2

Q4d Curve A: y = 5e 3 + 5 . Let

2

 kf ( t) dt ≈ 0.5299 4.674797

Q5g Binomial: n = 20 , p ≈ 0. 529907 Pr (X = 12 ) ≈ 0. 1473

2

3 3 3      = 8 − 3 log e − 6 +  3 loge + 6 − 8  = 2  2 − 3 log e  5 5 5      Q4e Area between curve A and curve B 40

= 40 2 − 2

8

Q5f

 x   2 3 log e  5 − 1 + 6 dx ≈ 977.30 m     − 2 5 e +5



dy 5 3x −2 5 a3 −2 = e = e , gradient from ( p, p) to dx 3 3 p −b p− b 3 2− a 5 a3 −2 p − b (a, b ) = × = −1 , =− e 3 , .: e p −a p− a p− a 5 3

Q4f At (a, b ),

11 11  11 ×9  ×9 11 +1. 96 20 20  Q5h  − 1. 96 20 20 ,  20 20  20 20   ≈ (0.3320, 0. 7680 ) Let p be the long term proportion of week days the 7:35 am bus of

575 min. 123 If many similar surveys were carried out and the 95% confidence interval calculated in each survey, 95% of them would have p in the interval. Company B is late for more than

Please inform [email protected] re conceptual and/or mathematical errors

Q4g p= 25.17 3 2020 Mathematical Methods Trial Exam 2 Solutions

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