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CORE, FUR1 2017 VCAA 18 MC

The first five terms of a sequence are 2, 6, 22, 86, 342 …

The recurrence relation that generates this sequence could be

  1. `P_0 = 2,qquadP_(n + 1) = P_n + 4`
  2. `P_0 = 2,qquadP_(n + 1) = 2 P_n + 2`
  3. `P_0 = 2,qquadP_(n + 1) = 3 P_n`
  4. `P_0 = 2,qquadP_(n + 1) = 4 P_n – 2`
  5. `P_0 = 2,qquadP_(n + 1) = 5 P_n – 4`
Show Answers Only

`D`

Show Worked Solution

`text(Using trial and error,)`

`text(Consider option)\ D:`

`P_1 = 4 xx 2 – 2 = 6`

`P_2 = 4 xx 6 – 2 = 22`

`P_3 = 4 xx 22 – 2 = 86\ \ text(etc…)`

`=> D`

CORE*, FUR1 2015 VCAA 3 MC

A town has a population of 200 people when a company opens a large mine.

Due to the opening of the mine, the town’s population is expected to increase by 50% each year.

Let `P_n` be the population of the town `n` years after the mine opened.

The expected growth in the town’s population can be modelled by

A.   `P_(n + 1) = P_n + 100` `\ \ \ \ \ P_0 = 200`
B.   `P_(n + 1) = P_n + 100` `\ \ \ \ \ P_1= 300`
C.   `P_(n + 1) = 0.5P_n` `\ \ \ \ \ P_0 = 200`
D.   `P_(n + 1) = 1.5P_n` `\ \ \ \ \ P_0 = 300`
E.   `P_(n + 1) = 1.5P_n` `\ \ \ \ \ P_1 = 300`
Show Answers Only

`E`

Show Worked Solution

`text(After 1 year,)`

`P_1` `= 1.5 xx P_0`
  `= 1.5 xx 200`
  `=300`

 
`=> E`

CORE*, FUR1 2006 VCAA 7 MC

The values of the first five terms of a sequence are plotted on the graph shown below.
 

 
The first order difference equation that could describe the sequence is

A.   `t_(n+1) = t_n + 5,` `\ \ \ \ \ t_1 = 4`
B.   `t_(n+1) = 2t_n + 1,` `\ \ \ \ \ t_1 = 4`
C.   `t_(n+1) = t_n - 3,` `\ \ \ \ \ t_1 = 4`
D.   `t_(n+1) = t_n + 3,` `\ \ \ \ \ t_1 = 4`
E.   `t_(n+1) = 3t_n,` `\ \ \ \ \ t_1 = 4`
Show Answers Only

`B`

Show Worked Solution

`text(By elimination,)`

`text(There is no common difference between terms,)`

`:.\ text(Cannot be A, C or D.)`

`text(The equation in B has)\ \ t_2=9,\ \ text(while the equation)`

`text(in C has)\ \ t_2=12.`

`rArr B`

CORE*, FUR1 2006 VCAA 3-4 MC

The following information relates to Parts 1 and 2.

A farmer plans to breed sheep to sell.

In the first year she starts with 50 breeding sheep.

During the first year, the sheep numbers increase by 84%.

At the end of the first year, the farmer sells 40 sheep.

Part 1

How many sheep does she have at the start of the second year?

A.     2

B.   42

C.   52

D.   84

E.   92

 

Part 2

If  `S_n`  is the number of sheep at the start of year `n`, a difference equation that can be used to model the growth in sheep numbers over time is

A.   `S_(n+1) = 1.84S_n - 40` `\ \ \ \ \ text(where)\ \ S_1 = 50`  
B.   `S_(n+1) = 0.84S_n - 50` `\ \ \ \ \ text(where)\ \ S_1 = 40`  
C.   `S_(n+1) = 0.84S_n - 40` `\ \ \ \ \ text(where)\ \ S_1 = 50`  
D.   `S_(n+1) = 0.16S_n - 50` `\ \ \ \ \ text(where)\ \ S_1 = 40`  
E.   `S_(n+1) = 0.16S_n - 40` `\ \ \ \ \ text(where)\ \ S_1 = 50`  
Show Answers Only

`text (Part 1:)\ C`

`text (Part 2:)\ A`

Show Worked Solution

`text (Part 1)`

`text(Number at the start of the 2nd year)`

`=50 + (84text(%) xx 50) – 40`

`=52`

`rArr C`

 

`text (Part 2)`

`text(If sheep numbers increase by 84% from the start of)`

`text(each year, and 40 sheep are then sold,)`

`S_(n+1)` `=S_n + 0.84S_n – 40`
  `=1.84 S_n – 40`

`rArr A`

CORE*, FUR1 2007 VCAA 8 MC

The first four terms of a sequence are

`12, 18, 30, 54`

A difference equation that generates this sequence is

A.   `t_(n+1)` `= t_n + 6` `\ \ \ \ t_1 = 12`
B.   `t_(n+1)` `= 1.5t_n` `\ \ \ \ t_1 = 12`
C.   `t_(n+1)` `= 0.5t_n + 12` `\ \ \ \ t_1 = 12`
D.   `t_(n+1)` `= 2t_n - 6` `\ \ \ \ t_1 = 12`
E.   `t_(n+2)` `= t_(n+1) + t_n` `\ \ \ \ t_1 = 12, t_2 = 18`
Show Answers Only

`D`

Show Worked Solution

`text(Calculating)\ t_3\ \ text(in each given option)`

`text(eliminates)\ A, B\ text(and)\ C.`

`text(Consider)\ E,`

`t_4 != t_3 + t_2\ \ text(which eliminates)\ E.`

`rArr D`

CORE*, FUR1 2007 VCAA 4-5 MC

The following information relates to Parts 1 and 2.

The number of waterfowl living in a wetlands area has decreased by 4% each year since 2003.

At the start of 2003 the number of waterfowl was 680.
 

Part 1

If this percentage decrease continues at the same rate, the number of waterfowl in the wetlands area at the start of 2008 will be closest to

A.   532

B.   544

C.   554

D.   571

E.   578
 

Part 2

`W_n` is the number of waterfowl at the start of the `n`th year.

Let  `W_1 = 680.`

The rule for a difference equation that can be used to model the number of waterfowl in the wetlands area over time is

A.   `W_(n+1) = W_n - 0.04n`

B.   `W_(n+1) = 1.04 W_n` 

C.   `W_(n+1) = 0.04 W_n` 

D.   `W_(n+1) = -0.04 W_n` 

E.   `W_(n+1) = 0.96 W_n`

Show Answers Only

`text (Part 1:)\ C`

`text (Part 2:)\ E`

Show Worked Solution

`text (Part 1)`

`text(After 1 year, number of waterfowls)`

`=680 – 4/100 xx 680`

`=680\ (0.96)^1`

`text(After 2 years)\ = 680\ (0.96)^2`

`vdots`

`text{After 5 years (in 2008)}`

`=680\ (0.96)^5 =554.45…`

`rArr C`

 

`text (Part 2)`

`text(Sequence is geometric where)\ \ r=0.96`

`:. W_(n+1)/W_n` `=0.96`
`W_(n+1)` `=0.96 W_n`

`rArr E`

CORE*, FUR1 2008 VCAA 4 MC

In 2008, there are 800 bats living in a park.

After 2008, the number of bats living in the park is expected to increase by 15% per year.

Let `Β_n` represent the number of bats living in the park `n` years after 2008.

A difference equation that can be used to determine the number of bats living in the park `n` years after 2008 is

A.   `B_n=1.15B_(n-1)-800` `\ \ \ \ \ B_0=2008`
B.   `B_n=B_(n-1)+1.15xx800` `\ \ \ \ \ B_0=2008`
C.   `B_n=B_(n-1)-0.15xx800` `\ \ \ \ \ B_0=800`
D.   `B_n=0.15B_(n-1)` `\ \ \ \ \ B_0=800`
E.    `B_n=1.15B_(n-1)` `\ \ \ \ \ B_0=800`
Show Answers Only

`E`

Show Worked Solution

`B_0=800`

`B_1= B_0 + 15 text(%) xx B_0=1.15 B_0`

`B_2= 1.15B_1`

`vdots`

`B_n=1.15 B_(n-1)`

`=> E`

CORE*, FUR1 2009 VCAA 8 MC

A patient takes 15 milligrams of a prescribed drug at the start of each day.

Over the next 24 hours, 85% of the drug in his body is used. The remaining 15% stays in his body.

Let  `D_n`  be the number of milligrams of the drug in the patient’s body immediately after taking the drug at the start of the `n`th day.

A difference equation for determining  `D_(n+1)`, the number of milligrams in the patient’s body immediately after taking the drug at the start of the `n+1`th day, is given by

A.   `D_(n + 1) = 85 D_n + 15` `D_1 = 15`
B.   `D_(n + 1) = 0.85 D_n + 15` `D_1 = 15`
C.   `D_(n + 1)= 0.15 D_n + 15` `D_1 = 15`
D.   `D_(n + 1)= 0.15 D_n + 0.85` `D_1 = 15`
E.   `D_(n + 1)= 15 D_n + 85` `D_1 = 15`
Show Answers Only

`C`

Show Worked Solution

`D_1=15`

♦♦ Mean mark 34%.

`text(85% of the drug is used up before the second dose.)`

`D_2` `=0.15 D_1 + 15\ \ \ text{(drug left from 1st day + new dose)}`
`D_3` `= 0.15 D_2 + 15\ \ \ text{(drug left from 2nd day + new dose)}`
  `vdots`
`D_(n+1)` `=0.15 D_n+15`

 
`=>  C`

CORE*, FUR1 2009 VCAA 6 MC

The `n`th term of a sequence is given by  `t_n = 100 − 20n`, where  `n = 1, 2, 3, 4\ . . .`

A difference equation that generates the same sequence is

A.   `t_(n +1)= 100 - 20t_n` `t_1 = 80`
B.   `t_(n+1) = 100t_n - 20` `t_1 = 1`
C.   `t_(n+1) = 80t_n` `t_1 = 80`
D.   `t_(n+1) = 100 - t_n` `t_1 = 20`
E.   `t_(n+1) = t_n - 20` `t_1 = 80`

 

Show Answers Only

`E`

Show Worked Solution

`text(By elimination,)`

♦♦ Mean mark 33%.
MARKERS’ COMMENT: Many students clearly did not reason their way through the solution by generating a few terms. This is an efficient and effective strategy.
`t_n` `= 100 − 20_n text( for ) n = 1, 2, …`
`t_1` `= 100 \ – 20 × 1`
  `= 80`

 
`∴\ text(Eliminate)\ B\ text(and)\ D`

`t_2` `= 100 − 20 × 2`
  `= 60`

 
`∴\ text(Eliminate)\ A\ text(and)\ C`

`=>  E`

CORE*, FUR1 2011 VCAA 7 MC

Let `P_2011` be the number of pairs of shoes that Sienna owns at the end of 2011.

At the beginning of 2012, Sienna plans to throw out the oldest 10% of pairs of shoes that she owned in 2011.

During 2012 she plans to buy 15 new pairs of shoes to add to her collection.

Let `P_2012` be the number of pairs of shoes that Sienna owns at the end of 2012.

A rule that enables `P_2012` to be determined from `P_2011` is

A.   `P_2012 = 1.1 P_2011 + 15`

B.   `P_2012 = 1.1 (P_2011 + 15)`

C.   `P_2012 = 0.1 P_2011 + 15`

D.   `P_2012 = 0.9 (P_2011 + 15)`

E.   `P_2012 = 0.9 P_2011 + 15`

Show Answers Only

`E`

Show Worked Solution

`text(By throwing out 10%, Sienna keeps 90% of her)`

`text{her 2011 shoes (or 0.9} \ P_2011 text{) and then adds 15.}`

`:. P_(2012) = 0.9\ P_2011 + 15`

`=> E`

CORE*, FUR1 2014 VCAA 6 MC

Consider the following sequence.

`2,\ 1,\ 0.5\ …`

Which of the following difference equations could generate this sequence?

A. `t_(n + 1) = t_n - 1` `t_1 = 2`
B. `t_(n + 1) = 3 - t_n` `t_1 = 2`
C. `t_(n + 1) = 2 × 0.5^(n – 1)` `t_1 = 2`
D. `t_(n + 1) = - 0.5t_n + 2` `t_1 = 2`
E. `t_(n + 1) = 0.5t_n` `t_1 = 2`
Show Answers Only

`E`

Show Worked Solution

`text(Sequence is)\ \ 2, 1, 0.5, …`

NOTE: “GP” is used as an abbreviation of “geometric sequence”.

`=>\ text(Geometric sequence where common ratio = 0.5)`

`∴\ text(Difference equation is)`

`t_(n + 1) = 0.5t_n`

`=>  E`

CORE*, FUR1 2014 VCAA 4 MC

On day 1, Vikki spends 90 minutes on a training program.

On each following day, she spends 10 minutes less on the training program than she did the day before.

Let  `t_n`  be the number of minutes that Vikki spends on the training program on day  `n`.

A difference equation that can be used to model this situation for  `1 ≤ n ≤ 10`  is

A.   `t_(n + 1) = 0.90t_n` `t_1 = 90`
B.   `t_(n + 1) = 1.10 t_n` `t_1 = 90`
C.   `t_(n + 1) = t_n - 0.10` `t_1 = 90`
D.   `t_(n + 1) = 1 - 10 t_n` `t_1 = 90`
E.   `t_(n + 1) = t_n - 10` `t_1 = 90`

 

Show Answers Only

`E`

Show Worked Solution

`text(Difference equation where each term is 10 minutes)`

`text(less than the preceding term.)`

`∴\ text(Equation)\ \ \t_(n+1) = t_n-10, t_1 = 90`

`=>  E`

CORE*, FUR1 2010 VCAA 7 MC

Each trading day, a share trader buys and sells shares according to the rule

 `T_(n+1)=0.6 T_n + 50\ 000` 

where `T_n` is the number of shares the trader owns at the start of the `n`th trading day.

From this rule, it can be concluded that each day

  1. the trader sells 60% of the shares that she owned at the start of the day and then buys another 50 000 shares.
  2. the trader sells 40% of the shares that she owned at the start of the day and then buys another 50 000 shares.
  3. the trader sells 50 000 of the shares that she owned at the start of the day.
  4. the trader sells 60% of the 50 000 shares that she owned at the start of the day.
  5. the trader sells 40% of the 50 000 shares that she owned at the start of the day.
Show Answers Only

`B`

Show Worked Solution

`T_(n+1)=0.6\ \T_n + 50\ 000`

`text(The difference equation describes a rule)`

`text(where a trader sells 40% of shares owned on)`

`text{the day before (left with 60% or 0.6}T_n text{)} `

`text(and then buys another 50 000 each day.)`

`=> B`

CORE*, FUR1 2012 VCAA 2 MC

A poultry farmer aims to increase the weight of a turkey by 10% each month.

The turkey’s weight, `T_n`, in kilograms, after `n` months, would be modelled by the rule

A.  `T_(n + 1) = T_n + 10`

B.  `T_(n + 1) = 1.1T_n + 10`

C.  `T_(n + 1) = 0.10T_n`

D.  `T_(n + 1) = 10T_n`

E.  `T_(n + 1) = 1.1T_n`

 

Show Answers Only

`E`

Show Worked Solution
`T_2` `=1.1T_1`  
`T_3` `= 1.1T_2` 
`vdots`   
`T_(n+1)` `= 1.1T_n`

 
`rArr E`