H4 2U Trial 2014 v24unitmaths.com/baulkham-hills-2u-2014.pdf · 1 2 d) Consider a parabola 2 L T 64...

1

BAULKHAM HILLS HIGH SCHOOL

TRIAL 2014 YEAR 12 TASK 4

Mathematics General Instructions Reading time ndash 5 minutes Working time ndash 180 minutes Write using black or blue pen Board-approved calculators may be used Show all necessary working in

Questions 11-16 Marks may be deducted for careless or

badly arranged work

Total marks ndash 100 Exam consists of 11 pages This paper consists of TWO sections Section 1 ndash Page 2-4 (10 marks) Questions 1-10 Attempt Question 1-10 Allow about 15 minutes for this section Section II ndash Pages 5-10 (90 marks) Attempt questions 11-16 Allow about 2 hours and 45 minutes for

this section Table of Standard Integrals is on page 11

2

Section I - 10 marks Use the multiple choice answer sheet for question 1-10

1 If3radic5 radic20 radic then is

(A) 5 (B) 25 (C) 125 (D) 15

2 What is the exact value of cos

(A) radic

(B) radic

(C) (D)

3 State which point on the diagram relates to the following

0 0 0

(A (B) (C) (D)

4 If 2then 2 3 is equal to (A) 2 (B) 13 (C) 7 (D) 10

5

The limrarr

4

(A) 1 (B) 4 (C) (D) 0

6 The function 3sin has a period of

(A) 3 (B) 10 (C) (D)

7

The sum of the series hellip is

(A) (B) (C) (D)

3

8

Part of the function is shown

The value of and respectively are (A) 2 3 2 (B) 2 3 2 (C) 2 3 2 (D) 2 3 2

9

The function y= shown below

Between and the function will have a (A) negative gradient (B) positive gradient (C) local minimum value (D) local maximum value

x

y

y = f (x)

a b

4

10 Consider the graph of 2 where is a real number The area of the shaded rectangles is used to find an approximation to the area of the region that is bounded by the graph the axis and the line 1 and 5

If the total area of the shaded rectangles is 44 then the value of is (A) 14

(B) -4

(C)

(D)

End of Section 1

5

Section II ndash Extended Response Attempt questions 11-16 All necessary working should be shown in every question

Question 11 (15 marks) (Answer on the appropriate page in your booklet)

a) Solve 1 leaving your answer in the exact form

2

b) Solve |3 4| 8

2

c) If tan and cos 0 find the exact value of sin

2

d) Solve for 9 27

2

e)

Differentiate with respect to

(i) cos

(ii)

2

2

f) Find the equation of the normal to 2 3 at a point 1

3

End of Question 11

6


a)

In the diagram is an arc of a circle with the centre The length of the arc is

The area of the sector is 4 Find the radius of the sector

2

b) The diagram shows the point 2 5 and 5 4

(i) Show that equation of line AB is 3 17 0

(ii) Find the coordinates of M the midpoint of AB

(iii) Show that the equation of the perpendicular bisector of AB is

3 6 0

(iv) The perpendicular bisector of AB cuts the -axis at C Find the coordinates of C

(v) Find the area of ∆

2 1 2

1 2

c)

Give the exact value of log1

radic3

1

d)

Find sin2

2

e) If the limiting sum of the series 1 2 4 hellip is Find the value of 2

End of Question 12

A

B

O

Not to scale

52

7


a) Consider the function given by 2 3 36 26

(i) Find the coordinates of the stationary points of the curve and determine their nature

(ii) Find the coordinates of any point of inflection

(iii) Hence sketch the graph of 2 3 36 26 by showing the above information

(iv) For what values of is the curve concave down and decreasing

3 2 2 2

b) Danny and Stella are pilots of two light planes which leave Rodeo Station at the same time Danny flies on a bearing of 330degT at a speed of 180kmh and Stella flies on a bearing of 080degT at a speed of 240kmh Copy the diagram below onto your answer page and mark the information on the diagram

(i) How far apart are Danny and Stella after 2 hours

(ii) What is the bearing of Stella from Danny after 2 hours (to the nearest degree)

2 2

c) There is one red and three green jelly beans in a jar One jellybean is selected at random eaten and then a second jellybean is selected at random and is also eaten Find the probability

(i) The two jellybeans eaten are both green

(ii) The red jellybean is the second one eaten

1 1

End of Question 13

D

S

R

N

8


a) The acceleration of a moving body is given by 212 msta If the body starts from rest find its velocity after 4 seconds

2

b) Find the area of the region in the diagram bounded by curve log axis and line =6

2

c) The velocity of an object is given by the equation 6 5 where is in seconds and velocity in m sfrasl It begins its motion at x 5 metres

(i) Find an equation for the displacement of the object

(ii) At what two times is the object stationary

(iii) Find the distance travelled by the object in first 3 seconds

2 1 2

d) Consider a parabola 2 4

(i) Find the coordinates of the focus

(ii) Find the equation of the directrix

2 1

e) Let A be the point 2 0 and 6 0 The point is such that

(i) Find the gradient of

(ii) Hence find the equation for the locus of

1 2

End of Question 14

x

y

y = lnxNot to scale

1 6

9


a) (i) Show that

cosec2 - cot2 - cos2

cos2 |tan |

(ii) Find the value of x if cosec2 - cot2 - cos2

cos2 = 1 for 0 x π

2 1

b)

The population of a species of bacteria P at time (minutes) grows such that 2000 where is the positive constant

(i) Show that the rate of increase of the population is proportional to the size of the population at that time

(ii) Given that the initial population doubles after 4 minutes calculate the value of correct to 3 significant figures

(iii) Find the population after 6 minutes( correct to nearest whole number)

1 2 2

c)

In the diagram ∥ is a right angled triangle is the midpoint of and is the midpoint of

(i) Explain why ang ang

(ii) Show that ∆ equiv ∆

(iii) Show that 2

1 3

3

End of Question 15

A B

DEF

C

Not to scale

10


a) Given that 2 3 Express in terms of

3

b) Sonia and Jenny want to save a deposit of $50 000 to buy a house They devise a savings plan to allow them to achieve this goal Beginning on 1 January 2014 they deposit $1000 on the first day of each month into an account which pays 9 pa compounded monthly

i Find the amount they have in the account on 31 January 2014 (ie at the end of the first month)

ii Show that the amount they have in the account at the end of months is

given by

iii Hence find the least number of months they need to save their deposit

1

2

2

c) A truncated cone is to be used as a part of a hopper for a grain harvester It has a height of

metres The radius is to be times greater than the bottom radius which is 2

metres

(i) If is the height of the removed section of the original cone show that

(ii) Show that the volume of the truncated cone is given by

1

(iii) If the upper radius plus the lower radius plus the height of the truncated cone

must total 12 metres calculate the maximum volume of the hopper

END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0

I t cisshy1- -=- 2shy

~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[

311 = gt -4gtgt1shy-xo q- X gt -1J

lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)

1I-t-3-tl=O (OR ~t g h-dU1IG c1P 1 ~uJ~ ~ ~hL-lt to s uJ c-llt u p-c3 +cent ~

2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy

_2 cn +shy- LcrS2 = -+- L

II- -

Ds~ DR-+ KS-- 2 x DRX 11 Unamp

- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It

v poundc- _S_t-l shy

l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S

5lt )L 6tl--t3gt_St- -t-J

3

~ Ij Skl- ()YQ vJ u2~ v 0

61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc

(Iii) ~ ~ IJ t -- t-JxI-I- Ga --t Uk D 3

~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~

~lG-gt-1 -JI -t~l-I-i)-LIL-IO-) 5 -t- (3 - - 2) == b ) ~ V

3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y

~ _gt)l-- ~lY -t1-J V VeAkv Cl--))

1laquogt-0( oO-Y1-shy

ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v

DJ-tvVQ Io =-1 t- L )

(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(

t _ tlvcct- 1+1shyM shy

= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy

Ilt -~l l shy0-11middot shy

= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L

CcJfvy-v-oJe L-lt V1r ftf II r D ) v

Cl~ -VIb cD e 4- b -lt=- 13 LC~D =- Lfh3( F~ oblll-(

L ( D =- L L iT j] LJ_H N~ L (IV

~ wF 1)) --W

L- ~ L-O (irv-L) - amp ~ c 1 ~) tgt LD F b Cfr8 - v

UILI ~Igt pound0 CI~V~ gdLoo-a- oM) ~ 0 0 P l b- ~ tt- ~d ~ 1- F0) v

ftpound - e + d 1tit)-1) M) II tshy

_ P-~ P 0 is rCl~~V v0

Ui0 50000 loolS ~ o ()D1J

o TtJl- =- Imiddot 0 OIS - I

-11 L =- I c-l) r Y

~Ctl ~111 - Y 1~~1middotlt)1-)

h loaltrl~ la-a)- an~r

4- 3li1shy

0tgtp~Dlt=I- ctGD

L eco LI-GO io (~tV)

I- j-il i) S Uz t

b ~Q D Il f e- LD

( Rc r C oa all MLLI-c~ ~V ~Iivl 0r )

_ e-oAJ vS( ~

b vC-t ~0 ~p -0 Ixltshy(OV~ampI

I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~

t hi )L 4- ~ YI j 1middotH 11lt -II 1

~ Ih )C + ~ IVto- 3

i VI 7j == l~-A~ 7lt-J =e

SJIos~J ~ ~ l01 cmcJJwl j ~

wvvk- SOI4- Sltgt-l~ ~U JR lI)

IIJ17V(iOOlr) J _ ~ i (]V r 0 L)~~-t nlcA d ~)

2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)

iltNl r cro 1 r - r- - - ltJ V ~ -----~~ ~ ~

f

ilJ VCJSvlL l -y H shy

V~~l ~ V~- Vs~ cnu

~ Ii l1-t-))lt+l) shy

=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-

V ~n(O-1-c-Jl1lt-b+~

- l 1- ~ LlS-- Ic)Lrlr t- +lj t 1 [s t - -t-~- t- -t-~- 1--t1-_tJ

8 [_t~tt-1-+(t-irJ dv _ ~ ~1l[ l shyok 3 -Jt +8t +iJ

rVY ~1A vef- Qy= 0 (lt shy

-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~

GuJ- t =0 ~ -t 2-ll 3middot oi 3

iVJ -t 1)

d l--~ ii1 tit-i--3J d-l- ltJ

--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

2

Section I - 10 marks Use the multiple choice answer sheet for question 1-10

1 If3radic5 radic20 radic then is

(A) 5 (B) 25 (C) 125 (D) 15

2 What is the exact value of cos

(A) radic

(B) radic

(C) (D)

3 State which point on the diagram relates to the following

0 0 0

(A (B) (C) (D)

4 If 2then 2 3 is equal to (A) 2 (B) 13 (C) 7 (D) 10

5

The limrarr

4

(A) 1 (B) 4 (C) (D) 0

6 The function 3sin has a period of

(A) 3 (B) 10 (C) (D)

7

The sum of the series hellip is

(A) (B) (C) (D)

3

8



9



x

y

y = f (x)

a b

4



(B) -4

(C)

(D)

End of Section 1

5




2

b) Solve |3 4| 8

2


2

d) Solve for 9 27

2

e)


(i) cos

(ii)

2

2


3

End of Question 11

6


a)



2





3 6 0



2 1 2

1 2

c)


radic3

1

d)

Find sin2

2


End of Question 12

A

B

O

Not to scale

52

7







3 2 2 2




2 2




1 1

End of Question 13

D

S

R

N

8



2


2





2 1 2




2 1




1 2

End of Question 14

x

y

y = lnxNot to scale

1 6

9


a) (i) Show that


cos2 |tan |


cos2 = 1 for 0 x π

2 1

b)





1 2 2

c)




(iii) Show that 2

1 3

3

End of Question 15

A B

DEF

C

Not to scale

10



3




given by


1

2

2



metres



1



END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

3

8



9



x

y

y = f (x)

a b

4



(B) -4

(C)

(D)

End of Section 1

5




2

b) Solve |3 4| 8

2


2

d) Solve for 9 27

2

e)


(i) cos

(ii)

2

2


3

End of Question 11

6


a)



2





3 6 0



2 1 2

1 2

c)


radic3

1

d)

Find sin2

2


End of Question 12

A

B

O

Not to scale

52

7







3 2 2 2




2 2




1 1

End of Question 13

D

S

R

N

8



2


2





2 1 2




2 1




1 2

End of Question 14

x

y

y = lnxNot to scale

1 6

9


a) (i) Show that


cos2 |tan |


cos2 = 1 for 0 x π

2 1

b)





1 2 2

c)




(iii) Show that 2

1 3

3

End of Question 15

A B

DEF

C

Not to scale

10



3




given by


1

2

2



metres



1



END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

4



(B) -4

(C)

(D)

End of Section 1

5




2

b) Solve |3 4| 8

2


2

d) Solve for 9 27

2

e)


(i) cos

(ii)

2

2


3

End of Question 11

6


a)



2





3 6 0



2 1 2

1 2

c)


radic3

1

d)

Find sin2

2


End of Question 12

A

B

O

Not to scale

52

7







3 2 2 2




2 2




1 1

End of Question 13

D

S

R

N

8



2


2





2 1 2




2 1




1 2

End of Question 14

x

y

y = lnxNot to scale

1 6

9


a) (i) Show that


cos2 |tan |


cos2 = 1 for 0 x π

2 1

b)





1 2 2

c)




(iii) Show that 2

1 3

3

End of Question 15

A B

DEF

C

Not to scale

10



3




given by


1

2

2



metres



1



END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

5




2

b) Solve |3 4| 8

2


2

d) Solve for 9 27

2

e)


(i) cos

(ii)

2

2


3

End of Question 11

6


a)



2





3 6 0



2 1 2

1 2

c)


radic3

1

d)

Find sin2

2


End of Question 12

A

B

O

Not to scale

52

7







3 2 2 2




2 2




1 1

End of Question 13

D

S

R

N

8



2


2





2 1 2




2 1




1 2

End of Question 14

x

y

y = lnxNot to scale

1 6

9


a) (i) Show that


cos2 |tan |


cos2 = 1 for 0 x π

2 1

b)





1 2 2

c)




(iii) Show that 2

1 3

3

End of Question 15

A B

DEF

C

Not to scale

10



3




given by


1

2

2



metres



1



END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

6


a)



2





3 6 0



2 1 2

1 2

c)


radic3

1

d)

Find sin2

2


End of Question 12

A

B

O

Not to scale

52

7







3 2 2 2




2 2




1 1

End of Question 13

D

S

R

N

8



2


2





2 1 2




2 1




1 2

End of Question 14

x

y

y = lnxNot to scale

1 6

9


a) (i) Show that


cos2 |tan |


cos2 = 1 for 0 x π

2 1

b)





1 2 2

c)




(iii) Show that 2

1 3

3

End of Question 15

A B

DEF

C

Not to scale

10



3




given by


1

2

2



metres



1



END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

7







3 2 2 2




2 2




1 1

End of Question 13

D

S

R

N

8



2


2





2 1 2




2 1




1 2

End of Question 14

x

y

y = lnxNot to scale

1 6

9


a) (i) Show that


cos2 |tan |


cos2 = 1 for 0 x π

2 1

b)





1 2 2

c)




(iii) Show that 2

1 3

3

End of Question 15

A B

DEF

C

Not to scale

10



3




given by


1

2

2



metres



1



END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

8



2


2





2 1 2




2 1




1 2

End of Question 14

x

y

y = lnxNot to scale

1 6

9


a) (i) Show that


cos2 |tan |


cos2 = 1 for 0 x π

2 1

b)





1 2 2

c)




(iii) Show that 2

1 3

3

End of Question 15

A B

DEF

C

Not to scale

10



3




given by


1

2

2



metres



1



END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

9


a) (i) Show that


cos2 |tan |


cos2 = 1 for 0 x π

2 1

b)





1 2 2

c)




(iii) Show that 2

1 3

3

End of Question 15

A B

DEF

C

Not to scale

10



3




given by


1

2

2



metres



1



END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

10



3




given by


1

2

2



metres



1



END OF EXAM

2 2 3

2t m

2 m

Nottoscale

x m

h m

A B

C

D

E

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

liD

l1i)

= E)l--~)lt shy 3 u

)--- gtlt - -1 0

vic

IL-Ie 2 (3

~c 1 f

qA1 13

IDmiddot D5c b 13

~ II d L-l = Itshy

2shyx-x~1 V

tl-_ )(-1 - 0


~-x-4)~b ~)t-i S 6 ~

)1-7 -~-[


lt l- ~) v-J

ISE) -c 4~

= l~ il

2gtt - 3

d ()) 3 1 v

-Ljlt - ~ = 3

~ =

-I) It A-Q ~ =-lj 0 5-) h

C- ~ ltd I 5

-)_ S = -) ( -2) v

j - IS= -XT)


2-~~)

1 -i3 8- -t) v I b ()eclW =- 3

1-Act ~

( C) 0) V

~ -= ~ f S 7 v- ~

) ~t-

v-c lOCl

(JJY ---Xishy


II- -


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf


- 3jD-+ ~H-_--3D)(lJD)((~HO

- Il1Qu L I~

f) $ ( q 2 +0 tkc ~+ k

(6111)

II)

lbO 611 )1-

)161 o XS 110

61()-1--

- O(SLl--b shy

Q- Omiddotlll- shy

-=

- 0

I~V~ ~D-4-)O = 01

0 T

c f (G c) =dK2 =J v - (~ 4- 3 shy

~ I-t -l - (~D f ( ~Jj It


l1) -x=- pound C - -Ie - - L l- J

at-t -0 - -S


3


61- -S-_tl u

G-S-)(t- -IJ == 0

--t = S-sc


~ 3tl-_ 5 -~1 + L3tl--_s-- -~J~


3 s

shy

d ) 1 ~ )t - ()L

)----)( 1-=- )~ y



ll) ~~()_L-) V

( r1A vI (~ )( 1J 1 = - 2t v


(=-] 3j

V (21-t1) 1-_1 ~ 1

4 - c-i 2--i=

t ~D V ok t- I

P(~ 1l ~Q k -= kG)(


= lQ150S--3

) I r ~ rlshy

V~ 1 ~ 1-) Y P1 =- )lt-t-l shyX-tl

(Ii) M 18 = ~ -4Ie

5 ce A-fl- tgt i3

MH)ltVIltpa=-l

~ )( ~ = - +1- -

1- -C)-)jl)lt-6)

)lt-_ -x -+ ~-=-- ll-V

D =

(10 -shy ~o -c=

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

l 0 ()O = 2- cJD 0 cl IL

~- eI-

ItA L ~ IL ltshy


= on] v

+-(

P 2 OOOe( IL

~ S( db f 1shy

=- S( 2~

~WLLDoLA3L




~ wF 1)) --W







-11 L =- I c-l) r Y



4- 3li1shy

0tgtp~Dlt=I- ctGD


I- j-il i) S Uz t

b ~Q D Il f e- LD


_ e-oAJ vS( ~


I C11 l l ) =- 3 + ( )- h~l~

2- 1 )+ 2-~ j ==- 1 T (n l( -I f1 ~


~ Ih )C + ~ IVto- 3





2

tn ()1S~ - 0 01 ~ J gt )11Ct cYD u-

(l1JV l 0-0 I r-J 11

~ ~C 001r) l IoU l l~+ -- - live 1

)


f


V~~l ~ V~- Vs~ cnu


=-1-nCt1-~-+ shy

~1(tgt--t-r ~ -~l- -t -)

~1hltl )3 -- _L --1 I V

=11~(tL--t ) J L t if

~ -1 Rl + -t - +1)

Ci~0 2 t- + lt -+ -0 I L

2-1- -+ 10

~ = jO-lt1-





-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf


-1t-~ampt--t-1 ~CJ 1-rl-_ ~+ -4 U

i =- Eplusmn~Iti (

=- plusmn )-0 ~


iVJ -t 1)


--= _ 8 q 2-- - 0

H4 2U Trial 2014

UntitledPDFpdf

H4 2U Trial 2014 v24unitmaths.com/baulkham-hills-2u-2014.pdf · 1 2 d) Consider a parabola 2 L T 64...

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Transcript of H4 2U Trial 2014 v24unitmaths.com/baulkham-hills-2u-2014.pdf · 1 2 d) Consider a parabola 2 L T 64...