2012 itute exam 2 solutions PDF

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1 has a vertical and a horizontal asymptotes a + bx + ax 2 when a + bx + ax 2 is a perfect square, i.e. ∆ = b2 − 4a 2 = 0 a 1 .: =± E 2 b Q6 y =

2012 Specialist Maths Trial Exam 2 Solutions Free download from www.itute.com © Copyright itute.com Section 1 1 A

2 C

3 E

4 D

5 E

6 E

7 A

8 E

9 C

10 A

11 B

1 1  Q7 Asymptotes: y = 2 ±  x + 1 , y − 2 = ± (x + 2 ) 2 2  .: the two asymptotes intersect at ( − 2,2 ) which is the centre of the hyperbola, also

12 D

13 B

14 B

15 B

16 B

17 A

18 C

19 C

20 B

21 D

22 E

Q1 A possible complex number is z = 1 − i, z = 2

b 1 = . a 2

A possible equation of the hyperbola is

A

( x + 2 )2 − ( y − 2)2 22

12

i.e. (x + 2 )2 − 4(y − 2)2 − 4 = 0

= 1, A

Q8 sec(a) + cosec (b) = 0 , sec( a ) = − cosec (b )

π  .: cos(a ) = − sin(b ), cos(a ) = − cos − b  2  π 3π π  3π −b and 0 < b < , .: a = π +  − b  = Since π < a < 2 2 2  2 3π .: a + b = E 2

Q2 z = −icis ( 2) is the clockwise rotation of cis( 2) about the origin O by

π 2

. .: Arg (z ) = 2 −

Q9 The domain of the inverse of f is the range of f .

π

C

2

Q3 Let z = x + yi .

(

2

) (

)

az 2 + b z − c = 0 , a x 2 − y 2 + 2xyi + b x 2 + y 2 − c = 0

[

]

.: (a + b) x2 + (b − a )y 2 − c + 2axyi = 0

C

Q10 tan−1( x − a + b) − tan−1 ( x − a) = π

.: (a + b) x2 + (b − a ) y2 − c = 0 and xy = 0

.: either x = 0 and y = ±

As x → 1 +, y → sec π + a π = aπ −1 5 4π + aπ = aπ − 2 At x = , y = sec 3 3 .: the domain of the inverse of f is [aπ − 2 , aπ − 1) .

tan −1 ( x − a + b) and tan −1( x − a ) have the same range

c if b > a b −a

c or y = 0 and x = ± b+ a

E

π π , 0 < Arg(1 + i) + Arg( z) < , 2 2 π π π π 0 < + Arg( z) < , − < Arg (z )< 4 2 4 4

 π π −1 −1  − ,  , .: tan ( x− a + b) − tan ( x − a) ≠ π for x ∈ R  2 2

A

~ ~ ~ Q11 Since a~ − b + c~ = 0 , .: unit vectors a~ , b and ~c form an equilateral triangle as shown below:

Q4 0 < Arg[(1 + i) z ] <

~ ~ a~.b − b .c~ + ~c .~ a = cos 60 ° − cos 60° + cos 120° 1 =− 2

~c D

~ b

a~

Q5

B

~ ~ ~ ~ ~ ~ Q12 Let j + ck = m k + a i + n i + bj where m and n are nonzero real numbers. ~ ~ ~ ~ ~ j + ck = (ma + n )i + nb j + mk

(

1 3

3

9

( y − k )2

= 1 or

2 ( x − h)2 + ( y − k )

9

=1

2012 Specialist Maths Trial Exam 2 Solutions

)

.: ma + n = 0 , nb = 1 and m = c .: ca + n = 0 , n = −ca .: abc = −1

1

( x − h)2 +

) (

E

D

~ ~ ~ Q13 ~p = ± i + j + k are vectors which make equal angle θ ~ ~ ~ with each of the orthogonal unit vectors i , j and k . 1 cosθ = ± , .: θ ≈ 55° and 125 ° B 1+1+1

(

)

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1

0 .2

Q14 Given f (0 ) = 1.2 , by CAS f ( 0.2) =

 0

0.4



f ( 0.4) =

0 0.8



f ( 0.8) =

0

1 dx +1.2 ≈1.6 , f (0.6 ) = cos 4 x 1 dx + 1.2 ≈ 2.6 , f (1) = cos 4 x

Average value ≈

1

 0

1 dx + 1.2 ≈ 1.4, cos 4 x

0.6

1

 cos 0

4

x

dx + 1.2 ≈ 2

(

)( )

Distance = loge 2 ≈ 0.7 m

1 dx + 1. 2 ≈ 4 cos4 x

1. 2 + 1.4 + 1. 6 + 2 + 2. 6 + 4 ≈2 6

dv = 2 − ex − ex = 0 , dx x e = 2 , x = log e 2 and v = 0 . .: the particle is moving in the same direction from x = 0 to x = log e 2 . Q19 Given v = 2 − e x , when a = v

C

+ + + Q20 a = 9. 8 sin 30°= 4.9 , s = 1.3 , u = 0 , v ? v 2 = u 2 + 2as , v ≈+ 3.57

B

Q15 At a point of inflection the gradient of the curve is a local maximum/minimum. y = 3 x5 + 5 x 4 −10 x 3 − 30 x 2 + 5 x − 10

dy = 15 x 4 + 20 x 3 − 30 x 2 − 60 x + 5 dx

Momentum = mv ≈ 1.3× + 3.57≈ + 4.6 kg m s -1

B

Q21 The particle exerts a downward force on the inclined plane equal in magnitude to the weight of the particle, .: the upward reaction force of the inclined plane on the particle is also equal in magnitude to the weight of the particle according to Newton’s D third law, mg = 2.04× 9.8 ≈ 20N upward Q22 Acceleration of the crate = acceleration of the chain Consider the forces on the chain: 1000 N

a= dy has a local minimum at x = 1. dx Q16 When y = 2 and 0 ≤ x ≤

B



R 1010 −1000 = 1 m s-2 = m 10

E

Section 2 2

1  =4 2

2

1 1 ( 2) → x 2 + 4 y −  = 4 , and in simplified form 2 2

tan2 xdx ≈ 0.8929

0

Area of the required region ≈ 4× 1.10715− 0.8929≈ 3.5

B

Q17 The quadratic equation has exactly one solution, .: ∆ = 0 k 2 − 4(1)(1 ) = 0 , .: k = 2

−

+

+

(

2

A

(

)

3 ,0 .

2

Q1ci x 2 + ( y − 1) = 4 , ( y − 1) = 4 − x2 , y − 1 = ± 4 − x 2 .: g( x) = 1 − 4 − x 2 ,

Q18 a = 2 , u = 10 , s = 16− 5= 11, v ? v2 = u2 + 2as, v = +12 v = 2t −10

2012 Specialist Maths Trial Exam 2 Solutions

)

The x-intercepts are − 3, 0 and

+

Total distance = 25 +36 = 61 m

x2 + ( y − 1)2 = 4 . Q1b Let y = 0 , x 2 = 3, x = ± 3

2

dy  dy  = −1 , y = −x + c + 1 = 0 , .:  dx  dx  (1,1), .: c = 2 , .: x + y = 2 +

1010 N

 (1) x 2 + 4 y − Q1a x2 + 4y2 = 4 → 

3 , tan2 x = 4 , x ≈ 1.10715 2

1.10715

By CAS,

10 kg

.: f (x )= g (x ) =

4 − x 2 − 1 for − 3 < x < x

Q1cii

C

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2

Q1d Volume V1 of solid by formed by rotating f ( x ): 1



2

2

y = 4 − x −1 ,

V1 =

2

4 − x = y +1 , x = 4 − ( y + 1)

2

1

(

)

πx2 dy = π 4 − ( y + 1)2 dy =

0

0

5π 3

=

u

du dx 2

+2



1+ x 2 dx = 1+ x 4

dx =

1

 2 +u

2

1 x2

 (x −

+1

)

1 2 x

+2

dx Let u = x −

1 x

du 1 =1 + 2 dx x

du

 u  1  + c tan −1 2  2  1  1 1  = tan −1  x −   + c x  2 2   =

4 32π 3 π 2 = 3 3 .: Volume of revolution of the required region 32π 5π 22π = − 2× = 3 3 3

Volume of the large sphere V2 =

Q2diii y =

( )

Q2div When x = 0. 5 , y = 0.5

 1  1  tan−1  0.5−   + c , c =1. 0762 0 .5  2  2  1  1  1 .: y = tan −1  x −   + 1. 0762 x  2  2 When x = 2 , y ≈ 1. 7

Q2a

0. 5 =

1

2 ~ ~ ~ a , P1P2 = P1P2 • P1P2 = b − a~ • b − a~ Q3a P1P2 = b − ~

(

)(

)

~ ~ ~ ~ ~ 2 = a~ • a~ + b • b − 2a~ • b = a~ + b − 2 a~ b cos α 2

Q3b Since the length of arc P1P2 equals the length of arc P2P3 , .: ∠P2OP3 = ∠P1OP2 = α 2 ~2 ~ 2 P2 P3 = b + ~ c − 2 b c~ cos α Q3c Since the length of arc P1 P2 equals the length of arc P2P3 , When x = 2 , y ≈ 1. 7

2

Q2b x = 0.5

y = 0.5

x = 1.0

y = 0. 5 + 0. 5 × 1. 1765 ≈ 1. 0882

x = 1.5

y = 1. 0882 + 0.5 × 1 ≈ 1. 5882

x = 2.0

y ≈ 1. 5882 + 0.5 × 0.5361 ≈ 1. 9 2

Q2c y =



0.5

2

.: P1P2 = P2P3

dy ≈ 1.1765 dx dy =1 dx dy ≈ 0.5361 dx

1 + x2 dx + 0.5 ≈ 1.7 by CAS 1 + x4

~2 ~ ~2 ~ 2 2 a + b − 2 a~ b cos α = b + c~ − 2 b c~ cos α .: ~ ~ 2 2 .: ~a − ~c = 2 b ( a~ − ~ c )cos α 2 2 a~ − c~ .: cosα = ~ a −~ c 2b (~

)

=

a~ + ~ c ~ b 2

Q3d Since OP1 is a diameter, .: ∠OP2 P1 and ∠OP3 P1 are right angles. ~ c .: b is the vector resolute of ~a in the direction of OP2 and ~ ~ is the vector resolute of a in the direction of OP . 3

Q2di

1+x 2 x2 1+x 4 x2

=

1 x2 1 x2

~ Q3e Given the length of OP1 = ~a =1 , .: b = 1 cos α = cos α and c~ = 1 cos 2α = cos 2α .

+1 + x2 2

1  b b2 2  + c = x + 2b + 2 + c , 2 =  x+ x x x  2 .: b = 1, 2b + c = 0 and c > 0 .: b = −1 and c = 2 Q2dii x2 +

.: x 2 +

1  = x − x2 

2

1  +2 x

2012 Specialist Maths Trial Exam 2 Solutions

Q3f cosα =

~ a +~ c 1 + cos 2α 2 ~ , cos α = 2cos α , .:1 + cos 2α = 2 cos α 2b

~ ~ b +d ~ Q3g cosα = and d = cos 3α by similar methods, 2 c~ cos α + cos 3α , 2 cos 2α .: cos α + cos 3 α = 2 cos α cos 2α .: cosα =

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3

Q4a

z9 + 1 z9 + 0 z6 + 0 z3 + 1 = 3 z −z +1 z 6 − z 3 +1

Q4dii The roots of P2 (iz ) = 0 are:  π   π  cis  ±  cis  ±  9  π   9  = cis  − 7 π  or cis  − 11π  iz = cis  ±  , z = = i π  9  18   18  cis

6

6

3

z − z +1

z3 + 1 z + 0 z + 0 z 3 +1

)

9

6

(

− z9 − z 6 + z 3

  2 

)

 5π   π   17π  iz = cis  ±  , z = cis  or cis   9   18   18 

z 6 −z 3 + 1 − z 6 − z 3 +1 0

(

.:

)

 7π   5π   13π  iz = cis  ±  , z = cis   or cis    9   18   18 

z9 + 1 = z 3 +1 z6 − z3 + 1

Q4b z 9 + 1 = 0 has a real root z = −1 , together with the other 8 roots they space out equally on the unit circle.

7π   11π  π   17 π   5π   13π   cis  −  cis   cis  cis   cis   18 18 18 18 18          18   

Q4e cis  −

7 π 11π π 17π 5π 13π  = cis  − − + + + +  = cis (π ) = − 1 18 18 18 18 18   18

Q5a

Nt Nc Rc

Rt

T

Fd

Wc

Wt

15000 5000 kg; mass of the trailer= kg 9. 8 9.8 At t = 0 , a = 0. 8 cos 2 0 = 0.8 Apply Newton’s second law to the car and the trailer: R = ma , 15000 + 5000 ×0.8 Fd − 300 − 200 − 15000sin 20° − 5000 sin 20° = 9. 8 .: Fd ≈ 9.0 ×103 N Q5b Mass of the car =

Q4c

2π 9

(

)(

) (

)

Q4di z9 + 1 = z3 + 1 z6 − z3 + 1 = z3 + 1 P2 (z) = 0 The roots of z + 1 = 0 (left) and the roots of P2 (z ) = 0 (right) are shown below: 3

Q5c Apply Newton’s second law to the trailer: 5000 T − 200 − 5000sin 20 ° = × 0.8 , T ≈ 2.3 ×103 N 9 .8 Q5d The car and the trailer accelerate for the first 2 s: 2

 πt  ∆v = 0.8 cos 2   dt = 0.8 ; there is no change in the velocity  4 0 in the next second. The car and the trailer start from rest, .: the exact speed of the trailer after the first 3 s is 0.8 m s-1.



Q5e Distance = 0.8(8 − 3 ) = 4 m.

The roots of P2( z ) = 0 are:

 π   5π   7π  z = cis ±  , z = cis  ±  and z = cis  ± .  9  9   9 

2012 Specialist Maths Trial Exam 2 Solutions

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