2007 nsw hsc mathematics exam worked solutions€¦ · 2007 nsw hsc mathematics exam worked...

2007 NSW HSC Mathematics Exam Worked Solutions

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Question 1 (a)

!

" 2 + 5

!

= 3.8561126...

!

" 3.86 (2 d.p.) (b)

!

2x " 5 > "32x > 2x >1

2 310-1 4

(c)

!

13 "1

!

=13 "1

#3 +13 +1

!

=3 +1

3( )2"12

!

=3 +13"1

!

=3 +12

(d)

!

S" =

!

34"

# $ %

& '

1( 14"

# $ %

& '

=

!

34

÷34

= 1 (e)

!

2x 2 + 5x "12

!

= 2x " 3( ) x + 4( )

(f) First find the gradient of

!

2x + y + 4 = 0 , by writing it in the form

!

y = mx + b.

!

2x + y + 4 = 0 y =

!

"2x " 4 So the gradient of

!

2x + y + 4 = 0 is

!

"2. A line perpendicular to

!

2x + y + 4 = 0

would therefore have a gradient of

!

12

.

The equation of the line with gradient of

!

12

,

passing through

!

"1,3( ) will be given by:

!

y " 3 =

!

12x ""1( )

!

2 y " 3( ) =

!

x +1

!

2y " 6 =

!

x +1 0 =

!

x " 2y +1+ 6

!

x " 2y + 7 = 0



Question 2 (a) (i)

!

ddx

2xex +1"

# $ %

& ' =

!

2 ex +1( ) " ex .2xex +1( )2

=

!

2ex + 2 " 2xex

ex +1( )2

=

!

2 ex " xex +1( )ex +1( )2

(ii)

!

ddx

1+ tan x( )10[ ]

=

!

10 1+ tan x( )9 " ddx1+ tan x[ ]

=

!

10 1+ tan x( )9 " sec2 x =

!

10sec2 x 1+ tan x( )9 (b) (i)

!

1+ cos3x( )dx" =

!

x +13sin3x + c

(ii)

!

8x 21

4" dx =

!

8x"21

4# dx

=

!

8 " x#1

#1$

% &

'

( ) 1

4

=

!

8 " #1x

$

% & '

( ) 1

4

=

!

8 " #14##11

$

% &

'

( )

=

!

8 " #14

+1$

% &

'

( )

=

!

8 " 34

= 6

(c) y =

!

x sin x

!

dydx

=

!

x " ddx

sin x[ ]#

$ %

&

' ( + sin x " d

dxx[ ]

#

$ %

&

' (

=

!

x " cos x( ) + sin x "1( ) =

!

x cos x + sin x If

!

x = " , then

!

dydx

=

!

" cos" + sin"

=

!

" #$1+ 0 =

!

"# So we want the equation of the line with gradient =

!

"# , passing through the point

!

" ,0( ).

!

y " 0 =

!

"# x "#( ) y =

!

"#x + # 2

!

"x + y #" 2 = 0 Question 3 (a) (i) AC =

!

10 " 2( )2 + 5 "11( )2

=

!

82 + "6( )2 =

!

64 + 36 =

!

100 = 10 (ii)

midpoint =

!

10 + 22

,5 +112

"

# $

%

& '

=

!

122,162

"

# $

%

& '

=

!

6,8( )



Question 3 (a) – continued. (iii)

!

mOB =

!

1612

=

!

43

!

mAC =

!

11" 52 "10

=

!

6"8

=

!

"34

!

mOB "mAC =

!

43"#

34

=

!

"1

!

"OB#AC (product of gradients =

!

"1). (iv)

midpoint =

!

0 +122

,0 +162

"

# $

%

& '

=

!

6,8( ) Since the midpoints of OB and AC are both

!

6,8( ), the diagonals of the quadrilateral bisect each other. Since

!

OB"AC , the diagonals bisect each other at right angles. Therefore, the quadrilateral OABC is a rhombus. (v)

Area =

!

12" AC "OB

=

!

12"10 " 122 +162

=

!

5 " 144 + 256 =

!

5 " 400 =

!

5 " 20 = 100 units2.

(b) (i) distance =

!

750 + n "1( ) #100 metres. (ii)

!

T10 =

!

750 + 9 "100 =

!

750 + 900 =

!

1650 metres. (iii)

!

S10 =

!

102750 +1650( )

=

!

5 " 2400 = 12000 metres (iv)

!

Sn =

!

n22 " 750 + n #1( ) "100[ ]

34000 =

!

n21500 +100n "100[ ]

68000 =

!

n 1400 +100n[ ] 68000 =

!

1400n +100n2 0 =

!

100n2 +1400n " 68000 0 =

!

100 n2 +14n " 680( )

!

n2 +14n " 680 = 0

!

n " 20( ) n + 34( ) = 0 n = 20 OR n =

!

"34 Clearly n can not be negative, so Heather would complete a total of 34 kilometres after 20 days.

or use the quadratic formula



Question 4 (a)

!

2 sin x = 1

!

sin x =

!

12

x =

!

"4

OR x =

!

" #"4

x =

!

3"4

(b)

1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12

(i) P(10) =

!

336

=

!

112

(ii) P(Not 10) =

!

1" 112

=

!

1112

(c) (i)

!

OA( )2 + AC( )2 =

!

12 + 3( )2

=

!

1+ 3 = 4 =

!

22 =

!

OC( )2

!

"

!

"OAC is right angled with hypotenuse OC.

!

"

!

"OAC =#2

(c) (ii)

!

sin"ACO =

!

AOOC

and

!

sin"AOC =

!

ACOC

=

!

12

=

!

32

!

" #ACO =

!

"6

and

!

"AOC =

!

"3

(iii) Area of

!

AOBC = 2

!

" Area of

!

"AOC

=

!

2 " 12" AO" AC

=

!

1"1" 3 =

!

3 metres2. (iv) reflex

!

"ACB =

!

2" #$ACB =

!

2" # 2 $%ACO( )

=

!

2" # 2 $ "6

%

& '

(

) *

=

!

2" # "3

=

!

5"3

Area of major sector ACB

=

!

12" AC( )2 " reflex#ACB

=

!

12" 3( )

2"5#3

=

!

5"2

metres2.

(v) Total area = Area

!

AOBC + Area major sector ACB + Area major sector AOB

=

!

3 +5"2

+12#12 # 2" $ 2 #%AOC[ ]

&

' (

)

* +

=

!

3 +5"2

+12# 2 # " $

"3

%

& ' (

) * +

, -

.

/ 0

=

!

3 +5"2

+2"3

=

!

3 +19"6

#

$ %

&

' ( metres2.



Question 5 (a) (i) The interior angle of a regular n sided polygon

is given by

!

180 n " 2( )n

.

For a pentagon, this angle will be given by

!

180 " 5 # 2( )5

=

!

180 " 35

=

!

5405

= 108° (ii) In triangle BAC:

!

AB = BC (equal sides of a regular pentagon)

!

"#BAC is isosceles and

!

"BAC ="BCA .

!

"ABC +"BAC +"BCA = 180° (angle sum of

!

"BAC )

!

108° +"BAC +"BAC = 180°

!

2"BAC = 72°

!

"BAC = 36° (iii) Since

!

"BAC = 36°, similarly

!

"CBD = 36°.

!

"ABF =

!

"ABC #"CBD

!

"#ABF =

!

108°" 36° = 72°

!

"AFB +"BAC +"ABF = 180° (angle sum of

!

"ABF )

!

"AFB + 36° + 72° = 180°

!

"AFB = 72°

!

"#AFB =

!

"ABF (both = 72°)

!

"#ABF is isosceles (two equal angles)

(b) (i) Initial velocity occurs when

!

t = 0.

ie. v =

!

2 " 016 + 02

= 016

= 0 m/s.

(ii) a =

!

dvdt

=

!

ddt

2t16 + t 2"

# $ %

& '

=

!

2 16 + t 2( ) " 2t # 2t16 + t 2( )2

=

!

32 + 2t 2 " 4t 2

16 + t 2( )2

=

!

32 " 2t 2

16 + t 2( )2

(iii) If a = 0, then

!

32 " 2t 2

16 + t 2( )2 = 0

!

2 16 " t 2( ) = 0

!

4 " t( ) 4 + t( ) = 0

t = 4 OR t =

!

"4

but t must be positive, therefore the acceleration of the particle is zero when t = 4 seconds.



Question 5 (b) – continued.

(iv) displacement x =

!

v dt"

=

!

2t16 + t 2

dt"

=

!

loge 16 + t 2( ) + c

since the particle is initially at the origin, when t = 0, x = 0 also.

ie. 0 =

!

loge 16 + 02( ) + c

c =

!

"loge16

!

" x =

!

loge 16 + t 2( ) " loge16

=

!

loge16 + t 2

16"

# $

%

& '

If t = 4, x =

!

loge16 + 42

16"

# $

%

& '

=

!

loge3216"

# $

%

& '

=

!

loge 2

So when t = 4, the position of the particle is

!

loge 2 metres to the right of the origin.

Question 6 (a)

!

2e2x " ex = 0

!

2 ex( )2" ex = 0

let

!

m = ex

!

2m2 "m = 0

!

m 2m "1( ) = 0 m = 0 OR

!

2m "1 = 0

!

2m = 1 m =

!

12

but

!

m = ex , therefore

!

ex = 0 OR

!

ex =

!

12

(impossible) x =

!

loge 12 =

!

"loge 2

(b) (i) For y-intercepts, let

!

x = 0 . ie. y =

!

04 " 4 # 03 = 0 For x-intercepts, let

!

f x( ) = 0. ie. 0 =

!

x 4 " 4x 3 0 =

!

x 3 x " 4( ) x = 0 OR x = 4 So the curve crosses the axes at the points

!

0,0( ) and

!

4,0( ). (ii) For stationary points, let

!

" f x( ) = 0 . ie.

!

4x 3 "12x 2 = 0

!

4x 2 x " 3( ) = 0 x = 0 OR x = 3 Using the first derivative test to check the

gradients either side of x = 0:

x

!

0" 0

!

0+

!

" f x( ) -ve 0 -ve So when x = 0, there is a horizontal point of

inflexion. (note: we could use the second derivative test instead, but in this case it would have given a value of zero, which would not be enough to know the nature of the turning point. Then we would have had to resort to the first derivative test anyway, or check concavity either side)

Using the first derivative test to check the

gradients either side of x = 3:

x

!

3" 3

!

3+

!

" f x( ) -ve 0 +ve So when x = 3, there is a minimum turning point. When x = 0, y = 0, as shown in part (i). When x = 3, y =

!

34 " 4 # 33 =

!

81"108 =

!

"27 So there is a horizontal point of inflexion at

!

0,0( ) and a minimum turning point at

!

3,"27( ) .



Question 6 (continued) (iii) Points of inflexion only occur when

!

" " f x( ) = 0 ie

!

12x 2 " 24x = 0

!

12x x " 2( ) = 0 x = 0 OR x = 2 We have already shown that there is a point of inflexion at

!

0,0( ). If x = 2, there may also be a point of inflexion. Checking concavity either side:

x

!

2" 2

!

2+

!

" " f x( ) -ve 0 +ve So when x = 2, there IS a change in concavity, and therefore a point of inflexion. When x = 2, y =

!

24 " 4 # 23 =

!

16 " 32 =

!

"16 So there are points of inflexion at

!

0,0( ) and

!

2,"16( ) . (iv)

Question 7 (a) (i) y =

!

x 2 + 4

!

x 2 =

!

y " 4

!

x 2 =

!

4 " 14y # 4( )

which is a concave up parabola, vertex

!

0,4( ),

and focal length

!

14

.

Therefore the focus is the point

!

0,4 14( ) .

(ii) Solving simultaneously:

!

x 2 + 4 =

!

x + k

!

x 2 + 4 " x " k = 0

!

x 2 " x + 4 " k = 0 (iii) Only one point of intersection implies that

this equation has only one solution (ie. equal roots)

!

"# = 0 ie.

!

"1( )2 " 4 #1# 4 " k( ) = 0

!

1" 4 4 " k( ) = 0

!

1"16 + 4k = 0

!

4k = 15

k =

!

154

(iv)

!

x 2 " x + 4 " k = 0

!

x 2 " x + 4 " 154 = 0

!

4x 2 " 4x +16 "15 = 0

!

4x 2 " 4x +1 = 0

!

2x "1( )2 = 0

!

2x "1 = 0

!

2x = 1 x =

!

12

y =

!

x 2 + 4 y =

!

12( )2 + 4

=

!

4 14

So the point P is

!

12 ,4 1

4( ) .



Question 7 (a) – continued (v) Since the y-coordinates of S and P are both equal to

!

4 14 , the line SP is horizontal.

Since the parabola is concave up, its directrix is also horizontal.

!

" the line SP is parallel to the directrix of the parabola. (b) (i)

!

3cos x =

!

sin x

!

3cos xcos x

=

!

sin xcos x

!

3 =

!

tan x

x =

!

"3

OR x =

!

" +"3

=

!

4"3

So the x-coordinate of A is

!

"3

, and the

x-coordinate of B is

!

4"3

.

(ii)

Area =

!

sin x " 3cos x( )#3

4#3

$ dx

=

!

"cos x " 3sin x[ ]#3

4#3

=

!

"cos 4#3

$

% &

'

( ) " 3sin 4#

3$

% &

'

( )

*

+ ,

-

. / " "cos #

3$

% &

'

( ) " 3sin #

3$

% &

'

( )

*

+ ,

-

. /

=

!

cos "3

#

$ %

&

' ( + 3sin "

3#

$ %

&

' ( + cos

"3

#

$ %

&

' ( + 3sin "

3#

$ %

&

' (

=

!

2 cos "3

#

$ %

&

' ( + 3sin "

3#

$ %

&

' (

)

* +

,

- .

=

!

2 12

+ 3 " 32

#

$ %

&

' (

=

!

2 12

+32

"

# $

%

& '

= 4 units2.



Question 8 (a) (i) When

!

t =1:

!

1600 "106 =

!

Aek When

!

t = 2:

!

2600 "106 =

!

Ae2k Dividing the second equation by the first gives:

!

2600 "106

1600 "106 =

!

Ae2k

Aek

!

2616

=

!

e2k

ek

!

e2k"k = 1.625

!

ek = 1.625 k =

!

loge1.625 Substituting k into the first equation gives:

!

1600 "106 =

!

Ae loge 1.625( )

!

1.625 " A =

!

1600 "106

A =

!

1600 "106

1.625

!

"

!

984.6 "106 (ii) If the number of phones exceeds 4000 million, then N > 4000 million. ie

!

Aekt >

!

4000 "106

!

984.6 "106( )e loge 1.625( )t >

!

4000 "106

!

e loge 1.625( ) t >

!

4000 "106

984.6 "106

!

loge1.625( )t >

!

loge4000984.6"

# $

%

& '

t >

!

loge4000984.6"

# $

%

& '

loge1.625

t > 2.887… which is approximately 34.6 months, or 2 years and 10.6 months. ie the number of phones will exceed 4000 million in November 2010.

(b)

(i) RTP:

!

"ABE |||"BCD In

!

"s ABE and BCD:

!

"ABE ="BCD (given)

!

"EAB ="DBC (corresponding

!

"s on parallel lines AE and BD).

!

" #ABE |||#BCD (equiangular) (ii) RTP:

!

"EDB |||"BCD

!

"EBD =

!

"BEA (alternate

!

"s on parallel lines AE and BD) but

!

"BEA =

!

"CDB (corresponding

!

"s in similar

!

"s ABE and BCD)

!

" #EBD =#CDB (both =

!

"BEA) Also in

!

"s EDB and BCD,

!

"EDB ="BCD (given)

!

" #EDB |||#BCD (equiangular)



Question 8 (b) – continued (iii)

!

p8

=qp

(corresponding sides in similar

!

"s EDB and BCD)

!

q8

=27p

(corresponding sides in similar

!

"s ABE and BCD)

!

p8

=27q

(multiplying both sides by

!

pq

)

So

!

p8

=qp

=27q

!

" 8, p, q and 27 are in geometric progression, since the ratios of each consecutive pair of terms are equal.

(iv) Let r be the common ratio. Then 27 =

!

8r3

!

r3 =

!

278

r =

!

278

3

r =

!

32

p =

!

8 " 32

= 12

and q =

!

12 " 32

= 18

Question 9

(a) V =

!

" x 2 +1( )2dx

0

1

#

=

!

" x 4 + 2x 2 +1( ) dx0

1

#

=

!

"x 5

5+2x 3

3+ x

#

$ %

&

' ( 0

1

=

!

"15

+23

+1#

$ %

&

' ( )

05

+03

+ 0#

$ %

&

' (

*

+ ,

-

. /

=

!

"315

+1015

+1515

#

$ %

&

' ( ) 0

*

+ ,

-

. /

=

!

28"15

So the volume of the solid of revolution is

!

28"15

units3.

(b)

(i) P(different suits) =

!

3951

.

(ii) P(different suits) =

!

3951

"2650

"1349

=

!

219720825



Question 9 (continued) (c) (i) Let

!

An be the amount the investment has grown to after n years.

!

A1 =

!

1000 "1# 06

!

A2 =

!

A1 +1000( ) "1# 06

=

!

1000 "1# 06{ } +1000( ) "1# 06

=

!

1000 "1# 062 +1000 "1# 06 =

!

1000 "1# 06 +1000 "1# 062 . .

!

A18 =

!

1000 "1# 06 +1000 "1# 062 + ...+1000 "1# 0618 which is a G.P. with a = 1060, r = 1.06 and n = 18

So

!

A18 =

!

1060 1" 0618 #1( )1" 06 #1

!

" 32 759.99 ie. the total value of Mr Caine’s investment on his son’s eighteenth birthday is approximately

$32 759.99 (correct to the nearest cent). (ii) Mrs Caine’s first investment grows to:

!

1000 "1# 063. Mrs Caine’s second investment grows to:

!

1000 "1# 06( ) "1# 062 =

!

1000 "1# 063. Mrs Caine’s third investment also grows to

!

1000 "1# 063. So the total value on her son’s third birthday will be

!

3"1000 "1# 063

!

" $3 573.05 (iii)

!

18 "1000 "1# 0618 $ $51 378.10



Question 10

(a) (i) Distance

!

"230 + 4 #1+ 5( )

= 6 units. (ii) When

!

t > 5 .

(iii) The distance travelled away from the origin between t = 4 and t = 5 is equal to the distance travelled back towards the origin between t = 5 and t = 6. So the object will again be 6 units from the origin when t = 6. After t = 6 it will take

!

6 ÷ 5 =1.2 seconds to travel the 6 units back to the origin. So the object will return to the origin when t = 7.2 seconds.

(iv)



Question 10 (continued)

(b) (i)

!

N =L1x 2

+L2

m " x( )2

(ii) To minimise N we need to find when

the function is a minimum.

!

dNdx

=

!

ddx

L1x 2

+L2

m " x( )2#

$ %

&

' (

=

!

ddx

L1x"2 + L2 m " x( )"2[ ]

=

!

"2L1x"3 " 2L2 m " x( )"3 # d

dxm " x[ ]

=

!

"2L1x 3

"2L2m " x( )3

# "1

=

!

2L2m " x( )3

"2L1x 3

If

!

dNdx

= 0 then

!

2L2m " x( )3

"2L1x 3

= 0

!

2L2m " x( )3

=

!

2L1x 3

!

x 3

m " x( )3 =

!

L1L2

!

xm " x

=

!

L1L2

3

x =

!

L1L2

3 m " x( )

x =

!

mL1L2

3 " xL1L2

3

!

x + xL1L2

3 =

!

m L1L2

3

!

x 1+L1L2

3"

# $

%

& ' =

!

m L1L2

3

x =

!

mL1L2

3

1+L1L2

3"

# $

%

& '

(

L2L1

3

L2L1

3

x =

!

mL2L1

3 +1

!

d2Ndx 2

=

!

6L2m " x( )4

+6L1x 4

which is always +ve, so the graph of the function N will be concave up.

!

" when x =

!

mL2L1

3 +1 the point P will be

placed such that the sum of the noise levels is a minimum.

2007 nsw hsc mathematics exam worked solutions€¦ · 2007 nsw hsc mathematics exam worked...

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