2013 Independent 2U

8/16/2019 2013 Independent 2U

1/32

Mathematics

eneral Instructions

• Reading time 5 minutes

• Working time 3 hours

• Board approved calculators may be

used.

• Write using black or blue pen

• A table of standard integrals is

provided at the back of the paper

• All necessary working should

be

shown in Question 16

• Write your student number and/or

name at the top

of

every page

2 13

Higher

School

Certificate

Trial Examination

Total marks 100

Section

I Pages 3 6

1

marks

Attempt Questions 1 10

Allow about 5 minutes for this section

Section

II Pages 7

3

90 marks

Attempt Questions 6

Allow about 2 hours 45 minutes for this

section

This paper MUST NOT be removed from the examination room

STUDENT NUMBER/NAME:


2/32

STUDENT NUMBERJNAME:

Section I

1 marks

Attempt Questions

1 10

Allow about 5 minutes for this section

Select the alternative A, B, C or D that best answers the question and indicate your choice with a

cross X) in the appropriate space

on

the grid below.

A B

c D

2

3

4

5

6

7

8

9

1

2


3/32

STUDENT NUMBER/NAME: . . . . . . . . . . . . . . . . . . . . . . . . .

1

What is 5.9893 correct to 2 significant figures?

A. 5.98

B 5.99

c

5.9

D

6.0

2

Which of the following is equivalent to J7

1

J

?

7 2 3

A

J7 2J3

J7 2J3

c

J7 2J3

-5

D

J7 2J3

-5

3.

· l fy x2

-5.xy

imp

1

2 2

•

-25y

A

x-5y

x Sy

c

I x

I-Sy

D

x 5y

x 25y

3


4/32


.........................

4 Which diagram indicates /

1

a >

0 and /

a <

0?

A

y

B

c

y

D

y

5 What is the solution to the equation 2 cos

x 1 =

0 in the domain 0

;

;

n?

A

Jr

2tr

3

3

B

2tr 4tr

3 3

c

Jr

Sn

?tr 1 ltr

6

-

6

-

6

6

D

Jr 2tr

4tr 5tr

3

-

3 3

4


5/32

STUDENT NUMBER/NAME: .........................

6

The diagram shows the region enclosed by

x y

= 2 and

y

=

e-x.

Which of the following pairs of inequalities describes the shaded region in the diagram?

A

x y:S:2 and y:S:e-x

B

x y:S:2 and

y ~ e x

c

x y ~ 2 and y ;; e x

D

x y

2 and

y e

x

7

What is the equation

of

the parabola with vertex

4,2)

and focus 3,2)?

A

x-4)

2

=4 y-2)

B

x-4)

2

=-4 y-2)

c

y-2)

2

=4 x-4)

D

y-2)

2

=-4 x-4)

5


6/32


8

What

is

the angle of inclination of the line 3x

y

= 7 with the positive direction of the

x axis?

A 33°41

B 56°19

c

123°41

D 146°19

9

What is the value ofx if log x

- log 3x =log

9?

A 3

B 9

c 18

D 27

5

10. The value of L n

is?

=2

A

50

B

108

c

2 5

1

32

D

3 6 ~

32

6


7/32


Section II

9 marks

Attempt Question 11 16

Allow about 2 hours 45 minutes for this section

Answer each question in a SEPARATE writing booklet. Extra writing booklets are available.

All necessary working should be shown in every question.

Question

15 marks). Use a SEPARATE writing booklet.

a)

b)

c)

d)

Solve

3

x =16 .

Solve

5

- 2xl ? 9.

Factorise 2x

9x 5

.

Find the equation of the tangent to the curve

y =2x

1

at the

point

where

x= 1.

e) Differentiate

with

respect to x:

f

i)

ii)

e

x sinx

cosx

4 x

Find

ec

2

3xdx .

7

2

2

2

3

2

2


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Question 12

15 marks). Use a

SEPARATE

writing booklet.

a

f

x

Evaluate

2 x-1

b)

Five

values of the function f

x

are

shown

in the table.

I

x

I

1

2

6

I

I

I : I I

18

15

Use Simpson s Rule with the five values given in the table to estimate

f

8

2

f x)dx

c) There are 2 piles of playing cards. One pile contains 7 red and 4 black cards

while

the other pile contains 3 red and 4 black cards.

Felicity

randomly

chooses

one

card from

each

pile.

i)

What is the probabil ity that both cards are black?

ii)

What

is the probability that at least

one

card is red?

iii) What is the probability that both cards are different colours?

Question 12 continues on following page

8

3

3

2


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.........................

Question 12 continued)

d) In the diagram the coordinates

of

the points and B are -2, -1) and 1 -3)

respectively. The line AD has equation y =2x 3 and the line CD has

equation

2x 3y =

17. BC

AD

y

x

NOTTO

SCALE

Copy the diagram into your examination booklet clearly labelling all given

information.

i)

Find the equation of the line BC

ii)

Find the perpendicular distance ofB 1

-3)

from the line AD

y = 2x 3.

iii) Find the coordinates

of

D the point intersection

of

y =

2x

3 and

2x 3y =17.

iv)

Hence,

or

otherwise, find the area

of

parallelogram ABCD

nd o Question 2

9

2


10/32

STUDENT NUMBER/NAME: ........................

Question 3

15 marks). Use a SEPARATE writing booklet.

a) A function is given

by x

=

12x-3x

- 2x

3

b)

i)

ii)

iii)

iv)

Find the coordinates

of

the stationary points of

x

and determine

their nature.

Hence sketch the graph

y

=

x

showing the stationary points and

y-intercept.

For

what

values

ofx

is the function decreasing?

For

what

values

of k

will

12x-

3x

- 2x

3

+k

=0 have 2 real

solutions?

Find the exact area bounded

by

the curve

y

= -

3

- the

x axis

and the lines

x 2

x=3

and

x=5.

c) Kevin has started an exercise program to lose weight.

When

he

started the

program

he

weighed 105 kg.

In the first

month

he lost 5 kg,

in

the second month

he

lost 4

kg

and

in

the third

month he lost 3.2 kg.

f

his weight loss trend continues

i)

how

much

will Kevin lose

in

the fourth month?

ii) what will

be

his ultimate weight?

10

3

2

2

3

3


11/32


Question 4 (15 marks). Use a SEPARATE writing booklet.

a)

y

x

The shaded area bounded by y

=

1 y

= 0

x =

0 and

y =

J

x 4 is rotated

about the x axis.

Find the

volume

generated.

b) The acceleration of a particle is given by

x

=4 sin

2t

where

xis

displacement

in metres and

t

is time in seconds.

Initially the particle is 1 metre to the left of the origin with a velocity of 2

metres per second.

(i)

Show that the velocity of the particle is given by x=4

2

cos 2t

(ii) Show the particle never comes to rest.

(iii) Find the equation for the displacement of the particle.

(iv)

Find the distance travelled by the particle in the first 4 seconds.

c)

Mrs Marsh

has just retired from work.

She

has received a large superannuation

payment and has worked out that she will require 4000 per month for the next

25 years.

She intends to invest a set amount

P)

at 3 p.a. interest compounding

monthly and withdraw the 4000 from this account at the end of each month.

(i)

(ii)

(iii)

Show that the amount A

1

remaining in the account after the first

withdrawal of 4000 is

A

1

=P(I.0025)-4000

Show

that the amount A

3

remaining

in the account after

the

third

withdrawal

of

4000 is

A

3

=

P I.0025)3

-12030.025

Find the

amount P that will need to

be

invested

if

she is to withdraw

4000

each month for

the next

25 years.

3

2

2

2

3


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STUDENT NUMBERJNAME: . . . . . . . . . . . . . . . . . . . . . . . . .

Question 5 (15 marks).

Use

a

SEPARATE

writing booklet.

a)

b

T

In the diagram PQRS is a rectangle and PQ = 3PS. The

point

Tlies on QR so

that RT = TQ The line PTmeets SQ at X.

(i)

(ii)

(iii)

Show

that triangles XSP and XQT are similar.

Show

that

2SQ

=

5QX.

Show

that 8(PS)

2

=

5 QX)2.

A die is biased so that the probability

of

rolling a six is 0.1.

What

is the minimum number

of

rolls of the die required to have a greater than

70%

chance of rolling at least

one

6?

c) Jeff and Barry have been offered jobs in a mine. The jobs are for 20 days.

Jeff agrees to be paid 5 cents for the

st day's

work, 10 cents for the

2nd

day's

work, 20 cents for the 3rd day's work with the company doubling his wage

each

day of the 20 day contract.

Barry agrees to be paid 500 for the

st

day's work, 550 for the 2nd day's

work, 600 for the

3rd

day's work

with

the

company

increasing his wage

by

50 a day for

each

day of the 20 day contract.

(i)

(ii)

(iii)

(iv)

How

much

will

Jeff

earn

on

the

6th

day

of

his contract?

How much will Barry earn on the 6th day of his contract?

Barry

finds the

job

too hard and only completes 10 days.

Show that

he

earns 7250 for the 10 day's work.

What is the minimum number of days Jeff will have to

work

to

ensure that

he

earns

more

than

Barry's wage

of 7250?

12

2

2

2

3

3


13/32


Marks

Question 6 15 marks). Use a SEPARATE writing booklet.

a)

Show that the quadratic equation in

x,

a

2

-

b

2

) x

2

+ b a

-

c x+ b

2

- c

2

)

= 0

has real and rational roots for all values of x, if a, b and c are rational.

3

b) A ball bearing is dropped into a vat

of

motor oil. The rate at which the ball

bearing decelerates is proportional to its velocity. i.e.

dv

= kv where vis the

dt

c)

velocity in centimetres

per

second, tis the time in seconds and

k

is a constant.

i)

ii)

Show that

v

=

Se-k

satisfies

dv

=

kv .

dt

The ball bearing is travelling at 120 centimetres

per

second when it

first enters the oil.

Find the value

of

S

iii) After travelling through the oil for 0.5 seconds the ball bearing has a

velocity

of

75 centimetres per second.

Show the value

of

k is 0.94, correct to 2 decimal places.

iv) Find the distance travelled through the oil in the first 2 seconds.

v)

f

he vat is 1.25 metres deep, how long will it take for the ball bearing

2

to reach the bottom of the vat?

The diagram shows a sector

of

a circle with radius

r

cm. The angle at the

centre is Bradians and the area is 18 cm

2

.

i)

ii)

iii)

Find an expression for r in terms

of B

Show that

P

the perimeter

of

the sector in cm, is given by

6 2+B)

P=

J

Find the minimum perimeter and the value

of

B for which this occurs.

13

2

3


14/32

dx

f _dx

x

f axdx

f

osax

dx

f in ax dx

f ec

2

ax

dx

f ec ax

tan ax dx

f

1 dx

.J

2 2

a

x

f ; dx

x2

a2

f

1

d

.Jx2

+a2

x

STUDENT NUMBERJNAME: ......................... .

STANDARD INTEGRALS

1

n I

1

f

= x

n :t= ; x-:t=O,

1 n


15/32

Section I

Question

1.

2.

3.

4.

S.

6.

7.

8.

9.

10.

Answer

D

c

B

A

c

B

D

c

D

B

NSW INDEPENDENT TRIAL EXAMS 2 13

MATHEMATICS YR12 TRIAL EXAM)

MARKING GUIDELINES

Solution

6.0

1 1

J7-2J3

= x

J7 2J3 J7 2J3 J7-2J3

J7-2J3

=

7-12

J7-2J3

=

-5

x

2

-5.iy

~

=

~ x + 5 y )

2

-25y

2

x

=

x 5y

Rising concave down

at

the

point x = a

2cos2x-1

=0

1

cos2x

= -

2

2

=TC TC 7TC

1 TC

3 3 3 3

T

TC

7TC

1

TC

x

6

6 6 6

x + y::; 2 and y : :

e x

(y-2)

2

=-4(x-4)

3

3

m

tanB=--

e= 123°41

2 2

logx

2

-log3x = log9

log ;:

J=

og9

~ =

3

x=27

5

2)n

=8+18+32+50

n=

=108

Outcome

P3

P3

P4

H6,H7

HS

P4

P4

P3

H3

HS

NSW

Independent Trial Exams 2 13 - HSC Trial Mathematics Examination: Marking Criteria - Page 1


16/32

Section II

Question 11

a

Outcome Assessed: P3 P4

•

Expresses both sides with a base 2.

•

Solves correctly

Answer:

23x-2 =

16

23x-2 =24

3x-2

=4

3x=6

x=2

b Outcome Assessed: P3 P4

•

x:S; 2

• x?:.7

Answer:

l5-2xl?:. 9

- 5-2x)?:.

9

-5+2x?:. 9

M ki G d

r

r

ng

m

e mes

Criteria

Marking Guidelines

Criteria

5-2x?:.

9

-2x?:.

4

x:S; 2

2x

?:.14

x?:

7

c

Outcome Assessed: P3 P4

Marking Guidelines

Criteria

•

Uses an acceptable factorization method and partially correct answer

•

Correct factorization

Answer:

2x

2

+9x-5 = 2x-l) x+5)

d Outcome Assessed: P4 P5

M ki G d

r

r

ng

m e mes

Criteria

•

Correct y value:- y

=

1

•

Correct gradient:- m = 4

•

Correct equation of tangent:- 4x y 3 =0

Answer:

At x

=

-1,

=

1,

=

2x

2

1

y =4x

y =-4

-1, 1

y-l=-4 x+l )

y 1 = 4 x 4

4x+

y+3

=0

Marks

Marks

Marks

Marks

NSW Independent Trial Exams 2013 - HSC Trial Mathematics Examination: Marking Criteria - Page 2


17/32

Question continued)

e) i Outcome Assessed: P7

M ki G .d

r

r

ng

m e mes

Criteria

•

Product rule and conect differentiation

of

either exponential or Trig function

•

CotTect solution

Answer:

y

=

e

2

x

sinx

dy

-·- =vu +uv

dx

=

2e

2

x

sin

x

+

e

2

x

cos

x

=

e

x 2 sin

x

+

cos x)

e)ii) Outcome Assessed: P7

•

Quotient rule

•

Conect solution

Answer:

cosx

y

4 x

2x

u =e

U =

2e

2

x

v =

smx

v =

cosx

M ki G .d r

r

ng

m e mes

Criteria

u

=cosx

dy 4 -

x). -

sin x

-

cos

x -1

dx =

4-x)

2

u =-sinx

-4sinx

xsin x+ cosx

4 x

)

2

f

Outcomes Assessed: H5

M ki G .d

I

r

n

m e mes

Criteria

•

Recognizing integral as tan

•

CotTect integral

+

c

Answer:

Jec

2

3xdx ~ t a n 3x

+

c

v=4 x

v =-1

Marks

1

1

Marks

1

1

Marks

1

1

NSW Independent Trial Exams 2013

- HSC

Trial Mathematics Examination: Marking Criteria

-

Page 3


18/32

uestion

12.

a) Outcome Assessed: H8

ar

ng

Ul

emes

k

G d I

Criteria

•

Correct expression for integral with correct index

•

Correct integration

•

Correct answer

Answer:

dx =

\ -1 ±

dx

2 ..Jx-1 2

~ [ 2 x - 1 ) i J :

[

2..Jx-lJ

=2[Fs-1]

b) Outcome Assessed: H8

M

ki G d

I

r

ng m e mes

Criteria

•

Attempt at Simpson s Rule with correct h

•

Correct substitution into Simpson s Rule

•

Correct solution

Answer:

8 4

x dx =-

16+

15) +4 14+ 21) +2 19))

2

3

=

± 209)3

836

3

=2783.

3

;:;:;279

c)(i) Outcome Assessed: HS

• Correct solution

Answer:

Pile

1.

P B) = ~

P R)

=}_

11

Pile2. P B)=_i, P R)=;?_

7 7

P BB)

=

P B)xP B)

4 4

X

7

16

77

Marking Guidelines

Criteria

Marks

1

1

1

Marks

1

1

1

Marks

1

NSW Independent Trial Exams 2013 - HSC Trial Mathematics Examination: Marking Criteria - Page 4


19/32


c) ii) Outcome Assessed: H5

Markin Guidelines


Answer:

P(atleast red)=l-P(BB)

=

1- . i

61

c) iii) Outcome Assessed: H5

Criteria

M ki G .d r

r

ng

m e mes

Criteria

.

28

12

•

Either - or -

77 77

•

Correct solution

Answer:

P(both

cards different)=

P RB)+

P BR)

7 4 4 3

= x x

11

7 11 7

28 12

=

77

40

77

d)

d) i) Outcome Assessed: P4

Markin Guidelines

Criteria


Answer:

Gradient

of

BC

is 2

y+3=2 x- l)

y+3 =

2x-2

y =

2x-5

x

NOTTO

SCALE

Marks

Marks

Marks

NSW



20/32


d) ii) Outcome Assessed: P4


Answer:

2x-y 3

=

0

=

12 1)- -3)+31

~ 2 ) 2

+ 1)2

2+3+3

JS

s s J

d) iii) Outcome Assessed: P4

•

Correct

x

value

•

C01Tect

y value

Answer:

y =2x 3

3y

=-2x l 7

4y=20

y=5

5 =2x 3

2x=2

x=l

Dis l ,5)

....................... 1)

....................... 2)

d) iv) Outcome Assessed: P4


Answer:

1

A=2x

2

8x3)

=24 u

2

Markin Guidelines

Criteria

Marks

ar

n

m e mes

ki G .d r

Criteria

Marks

Markin Guidelines

Criteria Marks

NSW



21/32

uestion13.

a) i) Outcome Assessed: H6

M ki G d

r

r

ng

Ul

e mes

Criteria

•

Derivative and

x

coordinates

of

stationary points

•

Correct minimum point with test

•

Correct maximum point with test

Answer:

f

(x) =

12x -

3x

2

- 2x

3

f (

x) =12-6x-6x

0 =12 -

6x

-

6x

2

0

=

x

2

x-2

0= x+2) x-1)

x =-2,1

f (

2)

=

-24

-12

+

16

=-20

f x = -6

-12x

f (-2) =-6+24

=18 > 0 concave up

: -2,-20) 1s a mm1mum

/ 1)

=12-3-2

=7

f

=

-18

<

0 concave down

: 1,

7 is a maximum

a) ii) Outcome Assessed: H6

ar

ng

m e mes

ki G d r

Criteria

•

Correct curve with maximum and minimum indicated

•

Correct y intercept

Answer:

x

-2, -20)

Marks

Marks

NSW

Independent Trial Exams 2013- HSC Trial Mathematics Examination: Marking Criteria - Page 7


22/32

Question 13 (continued)

a) iii) Outcome Assessed: H6

• Correct x values

Answer:

x < 2 and x

>

1

a) iv) Outcome Assessed: H6

• k = 7

• k=2

Answer:

Markin Guidelines

Criteria

Marking Guidelines

Criteria

Move function down

7

= 7

or up 20,

= 20


•

Correct integral

•

Correct substitution

•

Correct solution

Answer:

A f s_3_dx

3 x 2

= [

3 ln x - 2)J

=

3ln3 3ln

1

=

3ln3

u

2

c) i) Outcome Assessed:

HS


Answer:

5

4 3.2,

3.2

4

r

4 5

=3.2x

5

=2.56

kg

a

GP.

c ) ii) Outcome Assessed:

HS

M ki G d

I

r

ng m e mes

Criteria

Markin Guidelines

Criteria

M

ki G d

I

r

ng

m e mes

Criteria

•

Identifying limiting sum and correct fommla.

•

Correct limiting sum

of 2S

•

Correct final weight

Answer:

s

=

1 0.8

=25kg

Final weight 105 - 25

=

80kg

Marks

1

Marks

1

1

Marks

1

1

1

Marks

1

Marks

1

1

1

NSW Independent Trial Exams 2 13 HSC Trial Mathematics Examination: Marking Criteria - Page 8


23/32

uestion14.

a

Outcome Assessed: H8

M

ki G d I

r

ng

U e mes

Criteria

•

Volume of a cylinder and knowing to subtract area

•

Correct expression for volume generated by curve from

=

4 to 5

•

Correct solution

Answer:

V = ;r l)

2

x

5 nJ

4

5

x-4)

dx

=5J r

2

9;r 3

= U

2

b) i) Outcome Assessed:

H5

•

Correct integral

•

Correctly evaluates

c

Answer:

=4sin2t

x

=

-2 cos

2t

c

att

=

0,

x

2

2 =

-2cos2(

0) + c

c=4

x

=

4-2cos2t

b) ii) Outcome Assessed:

H5

• Correctly Shown

Answer:

Particle comes to rest when

x

= 0

x=

4-2cos2t

0 = 4-2cos2t

4

= 2cos2t

cos2t

= 2

2t

=cos- (2)

Marking Guidelines

Criteria

Marking Guidelines

Criteria

No solutions

:

Particle never comes to rest.

Marks

1

1

1

Marks

1

1

Marks

1

NSW Independent Trial Exams 2013

-

HSC Trial Mathematics Examination: Marking Criteria

-

Page 9


24/32

Question

4

(continued)

b )(iii) Outcome Assessed:

H5

•

Integral plus k

•

Correct answer

Answer:

i =4-2cos2t

x

=

4t - sin 2t k

When t =

0,

x =-1

-1

=k

x =

4t

- sin 2t - 1

b )(iv) Outcome Assessed: H5

•

Position after 4 seconds

•

Distance travelled

Answer:

t=4

x =16 - sin 8 - 1

=

14.01064175

::::::14

Distance travelled

14

- -1 =15m

c)(i) Outcome Assessed: H4, H5

• Correctly Shown

Answer:

Interest rate

=3 12

= 0.25 per

month

=

0.0025

A

1

=

P

0.0025)-4000

=P(l .0025)-4000

M ki G

d

r

r

ng

m e mes

Criteria

Marks

1

M ki G

d

r

r

ng

m e mes

Criteria Marks

Markin Guidelines

Criteria

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Question

4

continued)

c)(ii) Outcome Assessed: H4,

H5

Markin Guidelines

Criteria

• Correctly Shown

Answer:

A

2

=

A

1.0025)-4000

= (P l.0025)-4000 )1.0025-4000

=

P l.0025)

2

-4000 1.0025)-4000

=P l.0025)

2

-4000 1.0025+1)

A

3

=

A

2

1.0025)-

4000

= (

P l

.0025)

2

-4000 1.0025)-4000

)1.0025-4000

=

P l.0025)

3

-4000 1.0025)

2

-4000 1.0025)-4000

=P l.0025)

3

-4000 1.0025

2

+l.0025+1)

=

P l.0025)

3

-4000 3.00750625)

=

P l

.0025)

3

-12030.025

c)(iii) Outcome Assessed: H4,

H5

M ki G d r

r

ng

m e mes

Criteria

•

Co1Tect expressions for A

3

•

Correctly equating to 0 and recognizing sum

of

GP

•

Correct answer

Answer:

25

years

is

300 months

ll =P i.0025r-4000 1+i.0025+i.0025

2

+ ........

+i.0025

-

1

)

A

3

= P l .0025)

300

-4000 1+1.0025+1.0025

2

+ ........ +1.0025

299

)

A3oo 0

(

1(1 0025

300

-1)]

0 =

P l.0025)

300

-4000 ·

1.0025-1

1 0025

300

-1)]

P l.0025)

300

4000

- ---.

0.0025

P=4

(

(1.0025

300

-1 ]

( 0.0025) ( 1.0025)

300

p = 843505.81

Marks

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11


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uestion

15.

a) i) Outcome Assessed: H2

M

ki

G

d I

r

ng

m e mes

Criteria

•

2 angles and reasons

•

Conclusion

Answer:

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

R

p

In

b.SXP

and

b.QXT

LSXP=LQXT

LXSP=LXQT

LXPS=LXTQ

... b.SXP b.QXT

T

ve1iically opposite

angles=

alternate angles,

SP TQ

alternate angles, SP TQ

equiangular)

a) ii) Outcome Assessed: H2

M ki G d

I

r

ng

m e mes

Criteria

•

Correct proportionality statement

•

Correctly shown

Answer:

Let RT

=x

TQ=2x

PS=3x

PQ=9x

Given TQ =2RT)

PS =RQ opposite sides of rectangle =

Given

PQ

=3PS)

Marks

Marks

Let

QX

=

TQ

QS-QX SP

matching corresponding) sides in similar triangles proportional)

QX j

QS-QX 3/

3QX 2QS - 2QX

5QX=2QS

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a)(iii) Outcome Assessed: H2

Marking Guidelines

Criteria

•

Recognizing and applying Pythagoras Theorem

•

Correctly shown

Answer:

QS= 5QX

Apply Pythagoras Theorem in

1SPQ

(SQ)

2

=(PS)

2

+(PQ)

2

(s;x) ( P s )

+(3PS)

?

25 QXt

=(PS)2 +9(PS)2

4

25(

ix

2

= 10(PS)

2

4

25(QX)

2

=40(PS)

2

5 QX)

2

=8(PS)

2


•

•

•

Answer:

P(atleast

1

6 )=1-(0.9)

n > 11.42717266

Correct answer

P(6)=0.1

P 6)

=

0.9

P(

at least 1

11

6

11

)

=

1-P

all

6)

=1-(0.9)

1-(0.9)

>0.7

0.3

> (0.9)

ln(0.3) > ln(0.9)

ln(0.3) > nln(0.9)

ln(0.3)

~ ~ < n

ln(0.9)

n

> 11.42717266

n

=12

(from part (ii)

(Given PQ =

3PS)

M

ki G

d r

r

ng

m e mes

Criteria

Marks

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Question 15 (continued)

c)(i) Outcome Assessed:

H5

• Correct answer

Answer:

0.05, 0.10, 0.20, 0.40, 0.80, .....

a

=0.05,

r

=

2,

I i6

= (0.05) x 2

5

=$1638.40

T

=

ar

c)(ii) Outcome Assessed:

H5

• Correct answer

Answer:

500, 550, 600, 650, ..... .

a=500, d=50,

T =a+ n- l)d

I i

6

=

500+15x50

=$1250

c)(iii) Outcome Assessed:

H5

• Correctly shown

Answer:

S

11

= ~ 2 a + n - l ) d )

2

S O

=5 ( 1000 +

9

x 50)

=$7250

c)(iv) Outcome Assessed:

H5

•

Correct sum of GP

•

Correct inequality

•

Correct answer

Answer:

a r

-1

Sn =-- --- -

r -1

0.05(2 -1)

Sn=

Markin Guidelines

Criteria

Markin

Guidelines

Criteria

Markin Guidelines

Criteria

M ki G

d r

r

ng

m e mes

Criteria

0.05( 2

11

-1) > 7250

2

11

-1>145000

2

11

> 145001

nln(2) > ln(145001)

ln(145001)

n

ln(2)

n > 17.1457

18 days

Marks

1

Marks

1

Marks

1

Marks

1

1

1

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-HSC

Trial Mathematics Examination: Marking Criteria - Page 4


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uestion16.

a

Outcome Assessed: H2, HS

M

ki G d

I

r

ng

m e mes

Criteria

•

Correct expression for the discriminant

•

Correct expansion

•

Correct perfect square

Answer:

= b

2

-4ac

=4

b

4

-2ab

2

c

a

2

c

2

)

=4 b

2

-ac

2

=

2 b

2

- ac)

r

a perfect square : has real rational roots.

b) i) Outcome Assessed: H3, HS

• Correctly shown

Answer:

v

=

se-

1

dv S k

-k1

= e

dt

=-k se-kl)

=-kv

b) ii) Outcome Assessed: H3, HS

• Correct value

of S.

Answer:

t = 0,

v

=

120

v =se-

1

12 =Se

0

S=12

Markin Guidelines

Criteria

Marking Guidelines

Criteria

Marks

1

1

1

Marks

1

Marks

1

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b) iii) Outcome Assessed: H3,

HS

• Correctly shown

Answer:

t

=

0.5, v

=

75

v

=

20e-k

1

75

=

l20e-

05

k

0.625 = e-

0

·

5

k

0.5k

= ln 0.625)

k = ln 0.625)

-0.5

k

= 0.940007258

k = 0.94

b) iv) Outcome Assessed: H3,

HS


Answer:

s

= 120e 0.941

dt

=[EQ_ e 0.941

J

-0.94 0

=

EQ__[ e-0.94x2)-eo]

-0.94

120

=- - 0 . 1525 ..

-1)

-0.94

= 108.179

= 108 cm

b) v) Outcome Assessed: H3,

HS

Markin Guidelines

Criteria

Markin Guidelines

Criteria

M ki G •ctr

r

ng

m e mes

Criteria

•

Correct integral integration including

k.

•

C01Tect

solution

Answer:

125 = k l

20e 0·

t

dt

=[EQ_

e o.941 J

-0.94

0

-0.979166 ... =

[

0

·

k -e

0

J

= 0.94k ) - l

0.02083333 = e-

9

k

ln

0.0208333 ... ) = 0.94k

ln 0.0208333

...

)

k -

-0.94

k

= 4.12 seconds

Marks

Marks

Marks

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16


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c) i) Outcome Assessed: HS

• Correct value for r

Answer:

?

A = O r

2

?

8

=-or-

2

36 =

2

?

36

6

r= J

c) ii) Outcome Assessed:

HS

•

Correct expression for perimeter

•

Correctly shown

Answer:

P=2r r0

2 6

= JB

J

12+60

J

6 2+0)

J

Markin Guidelines

Criteria Marks

M ki G d r

r

ng

m e mes

Criteria Marks

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c) iii) Outcome Assessed: H5

M ki G

0

d

r

r

n

m e mes

Criteria

•

Equating correct first derivative to 0

•

Correct value of

B

•

Correct perimeter

Answer:

l l

p = 128

2

68

2

3 l

P' =-68

2

38

2

6

3

0=

8J8 JB

6 3

eJB JB

2JB =8J8

0=8J8 2JB

8 = o 2

but

8 o

5 3

P

=98

2

-

8-2

2

At 8

=

2,

5

3

3

P

=

9 2r2 -

2

2r2

=1.06

.

> 0 concave up

. . minimum at 8=

6 4)

P

J2

= 2J2

cm

Marks

2013 Independent 2U

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