Year 2015 VCE Mathematical Methods Trial Examination 1 ... · Title: 2015 Kilbaha VCE Trial Examination - Mathematical Methods CAS Exam 1 - Suggested Answers Author: Kilbaha Multimedia

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Year 2015

VCE

Mathematical Methods

Trial Examination 1

Solutions

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Mathematical Methods CAS Trial Examination 1 2015 Solutions Page 3


Question 1

a. ( ) using the quot

loie

g 3nt

2rulee x

yx

=

( )log 3 2eu x v x= =

1 2du dv

dx x dx= = M1

( )

( )( )( )

2 2

12 2log 3 2 1 log 342

e ex x xdy x

dx xx

× − −= =

( )( )2

1 1 log 32 e

dy xdx x

= − A1

b. ( ) 24 9f x x x= +

2 using the product and chain4 9 rulesy x x= +

( )

( )

12 2 2

12 2

2

4 9 4 9

1 41 8 4 92 4 9

u x v x x

du dv xx xdx dx x

−

= = + = +

= = × × + =+

( )2

2

2

4 4 94 9

dv du xf x u v xdx dx x

′ = + = + ++ M1

( ) 16 162 25 5525

f ′ − = + = +

( ) 41 12 or 8 or 8.25 5

f ′ − = A1 Question 2

3 9 28 3 9 0x x× − × + = ( ) ( )22 2 2let 3 then 9 3 3 3

xx x x xu u= = = = =

( )( )

23 28 9 03 1 9 0

13 3 93

x x

u uu u

u u

− + =

− − =

= = = =

M1

1 or 2x x= − = A1



Question 3

( )8 4

3 2 1k x y k

x k y+ = −

+ − = ( )

( )( )

282 24 2 24

3 2

6 4

kk k k k

k

k k

Δ = = − − = − −−

Δ = − + M1

There is a unique solution 0 that is \{when } 4,6k RΔ ≠ ∈ − A1

When the equations beco e 4 mk = − 4 8 83 6 3

x yx y

− + = −− = these lines are parallel

with different -intercepts, therefore there is no solution when 4y k = −

the equations beWhen co 6 mek = 6 8 23 4 1

x yx y+ =+ = these lines are both the same line,

therefore we have an infinite number of solutions when 6k = A1 Question 4

a. 3 1

2y

x= −

+

( )crosses the -axis w 0 2 3 1h 0n 1,e y xx x= ⇒ + = =

crosses the -axis wh 3 1 10 1 0,n2

e2 2

x yy y ⎛ ⎞= ⇒ = − = ⎜ ⎟⎝ ⎠

is a vertical asymptote, is a horizontal asymptot2 e1x y= − = − A1 correct graph, shape, asymptotes axial intercepts G1

1 HAy = −

2 VAx = −

( )1,0

10,2

⎛ ⎞⎜ ⎟⎝ ⎠



b. 4

1

The area is below the -axis, the area i2

s 3 1A dx xx

⎛ ⎞= − −⎜ ⎟+⎝ ⎠⌠⎮⌡ A1

( )( )

4

13log 2

3log 6 4 3log 3 1e

e e

A x x

A

⎡ ⎤= − + +⎣ ⎦= − + + −

( ) ( )13log 3 or 3 3log 2 or 3 log 82e e eA ⎛ ⎞= + − −⎜ ⎟

⎝ ⎠ A1

c. 1 3into 1

2y y

x x= = −

+ the translations must come last.

dilate by a factor of 3 parallel to the y-axis ( or away from the x-axis )

translate 2 units to the left parallel to the x-axis ( or away from the y-axis )

translate 1 unit down parallel to the y-axis ( or away from the x-axis ) A1

d. 3: 1

2f y

x= −

+ swap x and y

1 3 1231

232

13 2

1

f xy

xy

yx

yx

− = −+

+ =+

+ =+

= −+

M1

1 1but dom \{ 2} ran and ran \{ 1} domf R f f R f− −= − = = − =

To state the function, we must state its domain

( )1 1 3: \{ 1} , 21

f R R f xx

− −− → = −+ A1

Question 5

( ) 24 1xf x e−′ = − ( ) ( )2 24 1 2x xf x e dx e x c− −= − = − − +∫ M1 ( )0 1f =

01 2 3e c c−⇒ = − + ⇒ = ( ) 23 2 xf x e x−= − − A1



Question 6

a. [ ] 2range 2 6 2,6 , period 12 , amplitude 4

6

y T ππ− ≤ ≤ = − = = A1

b. 12 4sin 0 4sin 2 or sin

6 6 6 2x x xπ π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = ⇒ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( )

( )

( ) ( )

1 11 12 sin or 2 1 sin6 2 6 2

2 or 2 16 6 6 6

12 1 or 12 56 6 6 6

x xn n

x xn n

x xn n

π ππ π

π π π ππ π

π π π π

− −⎛ ⎞ ⎛ ⎞= + = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + = + −

= + = +

M1

12 1 or 12 5 wherex n n n Z= + + ∈ A1

c. 0 or 1n n= = 1,5,13x⇒ = A1

d. correct graph and coordinates G2

x

y

-2 0 2 4 6 8 10 12 14

-6

-4

-2

2

4

6

e. ( ) ( ) ( ) ( )not including 1, 5domain 0,13 0,1 1, 3or 5 5,1∪ ∪ A1

( )13,0

( )5,0( )1,0

( )9,6

( )0,2 ( )3,2



Question 7

a. P pleasant weather, B uses BBQ, using a tree diagram

( ) ( )( )

1 3 1Pr 1 603 4 4Pr | 1 3 2 2 1 4 4 31Pr

3 4 3 5 4 15

P BP B

B

×∩= = = = ×

× + × + M1

( ) 15Pr |31

P B = A1

b. Since the probabilities sum to one. ( )Pr 1X x= =∑

( ) ( )( )( )

( )( )

( )( )

8 8

8

2

2

log 1 log 3 1

log 1 3 1

1 3 8

4 3 84 5 05 1 0

5 or 1 but 3

k k

k k

k k

k kk kk k

k k k

− + − =

⎡ ⎤− − =⎣ ⎦− − =

− + =

− − =

− + =

= = − >

M1

only solution 5k = A1

23

P

13

P

2|5

B P

1|4

B P

3|4

B P

3|5

B P



Question 8

a. using the product rule

( )( ) ( )( ) ( ) ( )

( ) ( )

cos 2 cos 2 cos 2

2 sin 2 cos 2

d d dx x x x x xdx dx dx

x x x

= +

= − + A1

b. [ ) ( ) ( ): 0, , sin 2f R f x x x∞ → =

( )The graph crosses the -axis wh sinen 2 0x x =

so that 2 0, 0 ,2

x x ππ= =

( )20

The area is sin 2A x x dxπ

= ∫ A1

( )( ) ( ) ( )Since cos 2 2 sin 2 cos 2 from a.d x x x x xdx

= − +

( ) ( )( ) ( )( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

2 sin 2 cos 2 cos 2

2 sin 2 cos 2 cos 2

2 sin 2 cos 2 cos 2

1 sin 2 cos 221 1sin 2 sin 2 cos 24 2

x x x dx x x

x x dx x dx x x

x x dx x dx x x

x x x

x x dx x x x c

− + =

− + =

= −

= −

= = − +

∫∫ ∫∫ ∫

∫

M1

( ) ( )

( ) ( ) ( )

2

0

1 1sin 2 cos 24 2

1 1 1sin cos sin 0 04 2 2 4

A x x x

A

π

ππ π

⎡ ⎤= −⎢ ⎥⎣ ⎦

= − × − −

2units

4A π= A1

Question 9

1The normal 4 0 44 4 4N Tx cy x c y m m− + = ⇒ = − ⇒ = = − M1

( )2 22 1 4 4 0 so 3 0,3x xdyy e e x y Pdx

− −= + ⇒ = − = − ⇒ = = M1

4 12c x y= − = − A1



Question 10 a. ( ) the total length o1 f4 3 strin36 gx y+ =

( )2 2 0 total area of square and equilater1 si al triangn 602

leA x y= +

( )2

2 324yA x= + A1

( ) ( ) ( ) ( )41 3 36 4 4 9 substitute 9 into 23

y x x y x⇒ = − = − = − But since the lengths cannot be negative, we require 0 9x≤ ≤ A1

( ) ( ) ( )

( )

22 2 2

2

3 4 4 39 81 184 3 9

4 3 1 8 3 36 3 for 0 99

A x x x x x x

A x x x x

⎛ ⎞= + − = + − +⎜ ⎟⎝ ⎠

⎛ ⎞= + − + ≤ ≤⎜ ⎟⎜ ⎟⎝ ⎠

b. For a maximum or a minimum 4 3we 2 1 8 3 0

9 require dA x

dx⎛ ⎞

= + − =⎜ ⎟⎜ ⎟⎝ ⎠

4 3 36 3 364 3 4 3 9 4 3 31

9

x⇒ = = =+ ++

54 36 3 72Now 1.5 3 2 so that 3 515 174 3 9

< < < < < <+

To find the maximum value investigate the end-points of the function

( ) ( )2

2 3 4 94 3

A x x x⎛ ⎞= + −⎜ ⎟⎝ ⎠

( ) ( )23 4 90 36 3 and 9 81

4 3A A×⎛ ⎞= × = =⎜ ⎟

⎝ ⎠ A1

The minimum value of the area occurs wh 36 39

en4 3

x =+

9.That is when only the square is formed.The maximum value of the area occurs when x =

A1

END OF SUGGESTED SOLUTIONS

yx

060

x

A

0 3 6 9

20

40

60

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Year 2015 VCE Mathematical Methods Trial Examination 1 ... · Title: 2015 Kilbaha VCE Trial Examination - Mathematical Methods CAS Exam 1 - Suggested Answers Author: Kilbaha Multimedia