12 Maths b Kap Revision Term 3 Solutions[1]

8/4/2019 12 Maths b Kap Revision Term 3 Solutions[1]

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12 Maths B Kap Revision Term 3

Question 1a ) Find how long it would take for an initial investment of $22 000 to grow into $25 000 if the

money was invested in a capital growth fund returning 8.6 % pa interest with fortnightly rests.What is the effective interest rate for this investment.

(1 0.086/26)

(1 )

0.08625000 22000(1 )

26

25log ( )

22

38.71

n

n

FV PV i

n

n

+

= +

= +

=

=

ANSWER: 38.71 fortnights

Effective Rate:Invest $1 for 1 year (26 fortnights):

261(1 0.086 / 26)

1.08965

FV = +

=

Interest earned on $1 for 1 year = 1.08965 1 = $0.08965

ANSWER: Effective interest rate = 8.965%

b) Compare the effective interest rate of these two investments. A simple interest investment of14 % over 4 years AND an investment paying 11.25 % pa compounding weekly over the sametime period.

SIMPLE INTEREST INVESTMENTAssume $1 is invested @ 14% simple interest for 4 years.

1 0.14 4

0.56

SI PRT =

=

=

1 0.56 1.56A = + =

Determine rate of annual compound interest that would return the 1.56 after 4 years:

4

41.56 1(1 )

1.56 1

0.1176

ii

i

= +

=

=


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ANSWER: effective rate = 11.76% compound interest p.a

COMPOUND INTEREST INVESTMENT

Invest $1 @ 11.25% compounding weekly

520.11251(1 ) 1.11894

52A = + =

I = 1.11894 1 = 0.11894

ANSWER: effective rate = 11.89%

c ) What is the current value of 3 future payments you owe the bank if the payments are:Payment 1 $ 5500 in 8 weeks from now

(Assume a discount rate of 1.7 % pa compounding daily applies.)

8 7

56

(1 )

5500 (1 0.017 / 365)

5500

(1 0.017 / 365)

5485.67

nFV PV i

PV

PV

PV

= +

= +

=+

=

Payment 2 $ 3800 in 12 months time

365

3800

(1 0.017 / 365)3735.95

PV

PV

=

+=

Payment 3 $ 4200 in 2 years time.

365 2

4200

(1 0.017 / 365)

4059.60

PV

PV

=

+

=

PV of all 3 payments is:

5485.67 3735.95 4059.60

13281.22

PV = + +

=

ANSWER: PV=13281.22


3/12

Question 2a) The first term of a GP is 10 and the fifth term is 160. Find the geometric sequence and hence

( i ) the 4th term. And ( ii ) the Sum of the first 8 terms.

1n

nt ar

=

1 10t = 10a = 5 1 4

5 160 10 160 16 2t r r r

= = = =

I will assume r=2 (Strictly speaking r could also be -2)1

10 2n

nt =

i) 4 14

10 2 80t= =

ii)

8

8

( 1)

1

10(2 1)2550

2 1

n

n

a rS

r

S

=

= =

b) Find the sum of the following geometric sequence: 1 + 0.8 + 0.64 + for 10 terms

2

1

1

0.80.8

1

a

tr

t

=

= = =

10

10

( 1)

( 1)

(0.8 1)1 4.4631

0.8 1

n

n

a rS

r

S

=

= =

Question 3

a) Calculate the following present value annuity factors.

i)06.0/10

a ii)075.0/12

a

1 (1 )n

n i

ia

i

+=

10

10 0.061 (1.06) 7.360

0.06a

= = 12

12 0.0751 (1.075) 7.735

0.075a

= =

b) Find the present value of an immediate annuity which offers payments of $350 a month for 8years where an interest rate of 9% pa applies. Clearly state the significance of the value youhave just calculated.

12 81 (1 0.09 /12)

3500.09 /12

23890.45

n iPV p a

=

+=

=

There are several ways of interpreting this result.


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1. $23890.45 is the amount you could borrow today and repay over 8 years with monthlyrepayments of 350 @ 9% compounding monthly. After 8 years of payments the loan wouldhave been repaid.

2. $23890.45 is the amount you could invest today @ 9% compounding monthly and make awithdrawal of $350 at the end of each month. There would be $0 left after 8 years of regular

withdrawals.

3. You could invest a single sum of $23890.45 today @ 9% compounding monthly for 8 yearsOR start with $0 in the account and make regular deposits of $350 at the end of each monthfor 8 years. You will end up having exactly the same balance in your account after 8 yearswith either approach.

8 1223890.45(1 0.09/ 12) 48949.66FV = + =

OR96

96 0.09/12

(1 0.09 / 12) 1350 350 48949.65

0.09 /12FV S

+ = = =

c) Mary has $250 000 to invest in an annuity at 8% pa. She is 40yrs old now. Assuming she livesto be 80 yrs old, then, how much per month would she get for this investment.

40 12

40 12

1 (1 0.08 /12)250000

0.08 /12

2500001 (1 0.08 /12)

0.08 /12

1738.28

n iPV p a

p

p

p

=

+=

= +

=

Mary can make a monthly withdrawal of $1738.28. The account would be $0 after 40 years ofpayments.

d) Jason wants an annuity of $500 per fortnight over 10 years. If the interest rate is 6.8 %, what

would he expect to pay for the annuity.

(Note: I assume the payment occurs atthe end of each fortnight)

10 261 (1 0.068 / 26)500

0.068 / 26

94237.21

n iPV p a

PV

PV

=

+=

=


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(If Annuity Due i.e payment atthe start of each fortnight)

10 26

(1 )

1 (1 0.068 / 26)500 (1 0.068 / 26)

0.068 / 26

94483.68

n iPV p a i

PV

PV

= +

+= +

=

Question 4

a) Calculate the following sum of annuity actors.

i)06.0/10

S ii)075.0/12

S

(1 ) 1n

n i

iS

i

+ =

10

10 0.06

(1 0.06) 113.181

0.06S

+ = =

12

12 0.075

(1.075) 118.424

0.075S

= =

b) Find the value in 10 years time of 120 monthly payments of$220 made into an account paying 7.5 % paif the first payment is made immediately.

Annuity Due ( i.e payment at start so multiply by extra period of interest (1 )i + )

120

(1 )

(1 0.075 / 12) 1220 (1 0.075 / 12)

0.075 / 12

39389.33

n iFV p S i= +

+ = +

=


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c) Find the value of a superannuation plan with payments of $5000 pa over 25 yearsat 9% interest.

(I will assume payment occurs at the end of each year)

25(1 0.09) 1

50000.09

423504.48

n iFV p S=

+ =

=

d) Jane wants to retire in 30 years with a lump sum payment of $250 000 from asuperannuation investment at 8% pa. What monthly contributions are required?

12 30

250000

(1 0.08 / 12) 1

0.08/ 12

167.74

n i

FVp

S

=

=+

=


7/12

Question5.

What is the domain and range of (i) y= (x+3)/(x-4), (ii) y = (x-4)/( x-1), (iii) y=(x +7x+12) 2/1

Skipped

Question 6.

Which of the following are functions? (i)y = 4 (ii) x=3 (iii) y= x + 2x+1(iv) y=1/x(v) x + y=9 (vi) xy +2x =10(vii) y=4x-2x+1

Skipped

Question 7.(a)Find the inverse of f(x) = 3 (x-2)+4. Is the inverse a function? Why or why not?

23( 2) 4y x= +

Swap x and y:2

0.5

3( 2) 4

42

3

42

3

x y

xy

OR

xy

= +

= +

= +

Not a function. Its graph can be cut at more than 1 point by a vertical line.

(b)Find the inverse of x+y =9.Swap x and y:

2 29y x+ =

i.e circle is its own inverse.

(c)Find the inverse of y = 3x-9x +4.


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23 9 4 y x x= +

Complete the square:

2

2

( 1.5 )

2 2 2

2 2

2

3( 3 ) 4

3( 1.5 1.5 1.5 1.5 ) 4

3( 1.5) 3 1.5 4

3( 1.5) 2.75

x

y x x

y x x x

y x

y x

= +

= + +

= +

=

Now swap x and y:

23( 1.5) 2.75x y=

Rearrange to make y the subject of the eqn:

0.52.75

1.53

xy

+ = +

Question 8. A function has the equation y=)2(

1

+x+3

State (i) the domain (ii) the range (iii) any asymptotes of the function. Now sketch the function.

Domain: All real number except 2x = Vertical Asymptote: 2x = Range: All real numbers except 3y =

Horizontal Asymptote: 3y =

y-int (set x=0)

13 3.5

2y = + =

x-int( set y=0)

10 3 0 1 3( 2) 2.33

2x x

x= + = + + =

+

Note: The graph SHOULD NOT cross the asymptotes. If they do you will lose some marks.


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Question 9 For the relation y= 2 (x-4)^ +3, state the turning point. Sketch the relation.

Note: I will assume 2 4 3y x= + is wanted.

Turning pt = (4,3)

y-int ( set x=0)0.5

2( 4) 3y = + is undefined. No y-int.

x-int (set y=0)0.5

2

0 2( 4) 3

( 1.5) 4

2.25 4

6.25

x

x

x

x

= +

=

= +

=

Question 10 f(x) = x 2 and g(x) =2

)1(

1

x+4.

Sketch f(x). What transformations of f(x) result in g(x)? Sketch g(x).

Skipped. Not required to know about Truncus.

Question 11. Write the equation of a circle with (a) centre at (-2, 5) and a radius of 3. (b) centre(4,5) and radius of 6.

2 2 2( ) ( ) x B y C A + =

a) 2 2 22 2

( 2) ( 5) 3

( 2) ( 5) 9

x y

x y

+ =

+ + =

b) 2 2( 4) ( 5) 36x y + =


10/12

Question 12. Find the centre and radius of the circle with an equation (a) x+y-2x + 4y -9=0(b) 3x+3y-6x -9y +6=0

2 2

2 2 2 2 2 2

2 2

2 2

2 4 9 0

1 1 1 1 2 2 2 2 9 0

( 1) ( 2) 1 4 9 0

( 1) ( 2) 14

x x y y

x x x y y y

x y

x y

+ + =

+ + + + + =

+ + =

+ + =

Centre = (1, -2) and radius = 14

2 2

2 2 2 2 2 2

2 2

2 2

2 2

3 3 6 9 6 0

3( 2 ) 3( 3 ) 6 0

3( 1 1 1 1 ) 3( 1.5 1.5 1.5 1.5 ) 6 03( 1) 3( 1.5) 3 6.75 6 0

3( 1) 3( 1.5) 3.75

( 1) ( 1.5) 1.25

x y x y

x x y y

x x x y y yx y

x y

x y

+ + =

+ + =

+ + + + =

+ + =

+ =

+ =

Centre = (1, 1.5) and radius = 1.25 5 / 4 5 / 2or =

Question 13. For each of the followingState (a) the vertices (b) the value of e (c)the foci (d) thelength of the major axis (e) the length of the minor axis for each of the following ellipses.

(i)49

2

x +36

2

y =1

2 2

2 2

( 0) ( 0)1

7 6

x y + =

49 > 36 Therefore a Horizontal Ellipse i.e Major axis is horizontal and Minor axis is vertical

7 & 6a b = =

Centre: (0,0)Vertices: (7, 0) , (-7, 0) and (0, 6) and (0, -6)Length of major axis = 2 2 7 14a = = Length of minor axis 2 2 6 12b= = =

Eccentricity2

2

361 1 0.515

49

be

a= = =

Foci: (always lie on the major axis)(0 , 0) (0 7 0.515) ( 3.605, 0)ae = =


11/12

(ii)25

)3( 2+x+

4

)2( 2y=1

2 2

2 2

( 3) ( 2)1

5 2

x y + =

25 > 4 Horizontal Ellipse i.e Major axis is horizontal and Minor axis is vertical

5a = and 2b = Major axis length = 10Minor axis length = 4Centre: (-3, 2)

Vertices:(-3 + 5, 2)= (2, 2)(-3 -5, 2)= (-8, 2)

(-3, 2+2) = (-3, 4)(-3, 2-2) = (-3, 0)

41 0.917

25e = =

Foci (Again Horizontal ellipse, so foci lie on the Major axis)( 3 , 2) ( 3 5 0.917, 2) (1.58, 2) ( 7.58, 2)ae and = =

Major axis = 14

Minor axis = 12


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AnswersQ1 a 78 weeks, 8.96% b compound = 11.89 % simple = 11.75 % compound investment better.c 5485.67 + 3735.94 + 4059.61 = 13281.22 Q2 a Term 4 = 80 Sum = 2550 b sum = 4.4631Q3 a 7.360087 7.735278 b 23 890.46 c 1 738.28 per month d 94 237.21 Q4 a 13.180818.42 b $39389.33 c $423504.48 d. $176.75Q5 (i) domain reals except x=2, range real numbers (ii) reals except x=1:range y cannot bebetween 1 and 4(iii)x not between -3 and-4 :rangey0Q6(i) yes (ii) no (iii) yes (iv) no (v) no (vi)no (vii)yes Q7(a)y=

3

)4(2/1x +2, no some x values

have 2 y values (b) x+y =9. (c) y=3

)4/11( 2/1+x +1.5

Q8(i) reals except x=-2,(ii) range.. reals except y=3 (iii) x=-2 and y=3 Check sketch on calculator.Q9. (4,3) Check sketch on calculator ..show x intercept of 6.25Q10.check calculator for f(x).vertical asymptote moved 1 unit to right, ie at x =1 horizontalasymptote now at y=4 so up 4 units. Check calculator for sketch.Q11(a) (x+2) +(y-5)=9 (b) (x-4) +(y-5)=6 Q12(a) centre(1,-2) r=14 (b) Centre(1,1.5) r=5/2Q13(i) vertices at (7,0) and (0,6) e=13/7 , foci (13,0), major axis is 14 units long and minoraxis is 12 units long (ii) vertices at (-8, 2),(2, 2) and(-3,0) (-3,4), e=(21/25),foci at (-7.58, 2)and(1.58, 2) major axis is 10 units and minor axis is 4 units long

Major axis = 10

Minor axis = 10

12 Maths b Kap Revision Term 3 Solutions[1]

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