Band 5

Financial Maths, STD2 2025 HSC 34 (Adapted)

The table shows future value interest factors for an annuity of $1.

Larry invests a single amount of $18 000 for 5 years at 9% per annum, compounding monthly.

Tobias wants to end up with the same amount as Larry by using an annuity. He will pay a fixed sum into an account at the end of each month for 5 years, with the account also paying 9% per annum, compounding monthly.

Using the table, work out how much Tobias must deposit each month. (3 marks)

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$\$373.65$

Show Worked Solution

$r=\dfrac{0.09}{12}=0.0075,\ \ n=12\times 5=60$

$\text{Larry’s investment:}$

$FV=18\,000(1+0.0075)^{60}=28\,182.26$

$\text{Tobias’s investment:}$

$\text{Annuity factor:}\ 75.42414$

$\text{Annuity}\times 75.42414$	$=\$28\,182.26$
$\text{Annuity}$	$=\dfrac{28\,182.26}{75.42414}=\$373.65$

Financial Maths, STD2 2021 HSC 31 (Adapted)

Present value interest factors for an annuity of $1 for various interest rates $(r)$ and numbers of periods $(N)$ are given in the table.

A bank lends Paula $600 000 to buy an apartment, with interest charged at 1.8% per annum compounding monthly. She agrees to repay the loan in equal monthly repayments over a 25-year period.

What monthly repayment is needed to repay the loan in 25 years? Give your answer correct to the nearest cent. (2 marks)

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$\text{Monthly repayment}=\$2485.11$

Show Worked Solution

$\text{Monthly interest rate}\ (r)=\dfrac{1.8}{12}\%=0.15\%=0.0015$

$N=25\times 12=300$

$\Rightarrow\ \text{PV annuity factor}=241.43789$

$\therefore\ \text{Monthly repayment}=\dfrac{600\,000}{241.43789}=\$2485.11$

Financial Maths, STD2 2021 HSC 21 (Adapted)

Trevor opens a savings account with $8000. The account pays interest at a fixed monthly rate. At the end of each month the interest is added, and Trevor then deposits a further $400.

The spreadsheet below records the first six months of the account, together with the start of the seventh month.

By first finding the monthly interest rate, complete the row for month 7. (3 marks)

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$\text{Monthly rate}=0.2\%$

$\text{Month 7: beginning}=\$10\,508.52,\ \text{interest}=\$21.02,\ \text{end}=\$10\,929.54$

Show Worked Solution

$\text{Monthly interest rate}=\dfrac{16.00}{8000}=0.002=0.2\%$

$\text{Row 7 calculations:}$

$\text{Beginning balance}=\$10\,508.52$

$\text{Monthly interest}=10\,508.52\times 0.002=\$21.02$

$\text{End of month balance}$	$=10\,508.52+21.02+400$
	$=\$10\,929.54$

Financial Maths, STD2 2024 HSC 20 (Adapted)

A table of future value interest factors for an annuity of $1 is shown.

Callum invests $300 at the end of each year for 6 years into an account earning 4% per annum, compounded annually. Using the table, calculate the future value of Callum’s investment. (1 mark)
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Devi wants to have saved $6200 in 4 years. She will make equal payments at the end of every six months into an account paying 6% per annum, compounded six-monthly.
Using the table, find the minimum amount Devi must pay each six months. Give your answer to the nearest $10 and support it with calculations. (2 marks)
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a. $\$1989.90$

b. $$700\ \text{(nearest \$10)}$

Show Worked Solution

a. $\text{6 annual periods at 4% p.a.}\Rightarrow\text{Factor}=6.6330$

$FV=300\times 6.6330=\$1989.90$

b. $r=\dfrac{6\%}{2}=3\%\ \text{per 6 months}$

$\text{Compounding periods}=4\times 2=8$

$\Rightarrow\ \text{Factor}=8.8923$

$\text{Find minimum payment:}$

$6200$	$=\text{Payment}\times 8.8923$
$\text{Payment}$	$=\dfrac{6200}{8.8923}=697.23\ldots=$700\ \text{(nearest \$10)}$

Financial Maths, STD2 EO-Bank 28

Leon opens a superannuation account to build up savings for retirement. At the end of each year he pays in $4000, and the account earns 5% per annum, compounded annually.

The spreadsheet below models the first 4 years of the account.

Write down the formula used in cell C9, using appropriate grid references. (1 mark)

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Determine the value that belongs in cell C9. (1 mark)
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Starting from the end of year 4, Leon lifts his yearly payment from $4000 to $7000. Find the balance in the account at the end of year 7, and state how much larger this is than if he had stayed with $4000 payments. (3 marks)
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a. $\text{=E8*B3}$

b. $\text{C9}=\$410.00$

c. $\text{Balance at end of year 7}=\$42\,025.54$

$\text{Leon has}\ \$9457.50\ \text{more than at the standard contribution.}$

Show Worked Solution

a. $\text{Formula: =E8*B3}$

b. $\text{C9 (Year 3 interest)}=\text{balance at start}\times\text{rate}$

$\text{C9}=8200\times 0.05=\$410.00$

c. $\text{Using}\ \ P+I+C\ \ \text{from end of year 4 balance}\ \$17\,240.50:$

$\text{Increased contributions of}\ \$7000\ \text{from year 5:}$

$\text{Year 5:}\ 17\,240.50+17\,240.50\times 0.05+7000=\$25\,102.53$

$\text{Year 6:}\ 25\,102.53+25\,102.53\times 0.05+7000=\$33\,357.66$

$\text{Year 7:}\ 33\,357.66+33\,357.66\times 0.05+7000=\$42\,025.54$

$\text{Standard contributions of}\ \$4000\ \text{from year 5:}$

$\text{Year 5:}\ 17\,240.50+17\,240.50\times 0.05+4000=\$22\,102.53$

$\text{Year 6:}\ 22\,102.53+22\,102.53\times 0.05+4000=\$27\,207.66$

$\text{Year 7:}\ 27\,207.66+27\,207.66\times 0.05+4000=\$32\,568.04$

$\text{Difference}=42\,025.54-32\,568.04=\$9457.50$

$\therefore\ \text{Leon has}\ \$9457.50\ \text{more by increasing his contributions.}$

Financial Maths, STD2 2013 23 MC (Adapted)

Elias opens a savings account that pays 4% per annum, compounded quarterly. At the end of every quarter he deposits $800, beginning one quarter after the account is opened, and continues for the following 18 months.

How much is in Elias’s account at the end of the 18 months?

$4800.00
$4921.61
$4970.83
$5095.30

Show Answers Only

$B$

Show Worked Solution

$\text{Interest: 4% p.a.}\ \ \Rightarrow\ \ \text{1% per quarter}$

$\text{18 months}=6\ \text{end-of-quarter deposits}$

$\text{Value of 1st deposit}=800(1.01)^5=840.81$

$\text{Value of 2nd deposit}=800(1.01)^4=832.48$

$\text{Value of 3rd deposit}=800(1.01)^3=824.24$

$\text{Value of 4th deposit}=800(1.01)^2=816.08$

$\text{Value of 5th deposit}=800(1.01)^1=808.00$

$\text{Value of 6th deposit}=800$

$\therefore\ \text{Amount in account}$

$=840.81+832.48+824.24+816.08+808.00+800.00$

$=\$4921.61$

$\Rightarrow B$

Financial Maths, STD2 EO-Bank 34

Farah wants to buy a commercial espresso machine for her cafe at a cost of $4200. She has decided to pay off whatever she owes in a single amount 75 days after the purchase, and is weighing up two ways to fund it.

A credit card that compounds interest daily at 17% per annum, but does not charge interest during the first 25 days after a purchase.
A personal loan on which simple interest is charged at 8.9% per annum.

Advise Farah on the cheaper option, backing up your recommendation with the relevant calculations. (4 marks)

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$\text{Credit card interest}=\$98.93$

$\text{Personal loan interest}=\$76.81$

$\text{As }\$76.81\lt\$98.93,\text{ Farah should choose the personal loan.}$

Show Worked Solution

$\text{Days accruing interest}=75-25=50$

$\text{Credit card option (interest compounds daily):}$

$\text{Amount owing}$	$=4200\left(1+\dfrac{0.17}{365}\right)^{50}$
	$=4298.931\ldots$
	$=\$4298.93\ \text{(nearest cent)}$

$\therefore\ \text{Interest}=4298.93-4200=\$98.93$

$\text{Personal loan option (simple interest applies):}$

$I=Prn$	$=4200\times 0.089\times\dfrac{75}{365}$
	$=76.808\ldots$
	$=\$76.81\ \text{(nearest cent)}$

$\therefore\ \text{As }\$76.81\lt\$98.93,\text{ the personal loan}$

$\text{is cheaper, so Farah should choose it.}$

Financial Maths, STD2 2023 HSC 32 (Adapted)

Theo’s credit card charges interest at 15.9% per annum, compounded daily, on any balance owing and has no interest-free period.

His only transaction for the month was a single purchase of $680, which he paid off completely 28 days afterwards.

Assuming interest applied across all 28 days, calculate the amount of interest to be charged to the purchase. (2 marks)
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Express this interest as a percentage of the total amount Theo repaid, rounding to two decimal places. (2 marks)
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a. $\text{Interest}=\$8.34$

b. $1.21\%$

Show Worked Solution

a. $\text{Daily interest rate}\ (r)=\dfrac{0.159}{365}$

$n=28\ \text{days}$

$FV$	$=PV(1+r)^n$
	$=680\left(1+\dfrac{0.159}{365}\right)^{28}$
	$=\$688.34$

$\therefore\ \text{Interest}=688.34-680=\$8.34$

b. $\text{Interest as % of total repaid}$

$=\dfrac{8.34}{688.34}\times 100$

$=1.2116\ldots$

$=1.21\%\ \text{(to 2 d.p.)}$

Financial Maths, STD2 2016 HSC 17 MC (Adapted)

A credit card balance of $920 is outstanding for 20 days. Compound interest is charged at 0.041% per day, with no interest-free period.

Which calculation gives the amount of interest charged on this balance?

$920\times 0.00041\times 20$
$920(1.00041)^{20}$
$920(1.00041)^{20}-920$
$920(1+0.00041\times 20)$

Show Answers Only

$C$

Show Worked Solution

C is correct: compound interest $=$ total owing $-$ principal $=920(1.00041)^{20}-920$.

Other options:

A is incorrect: this is simple interest $(Prn)$, not compound interest.
B is incorrect: this gives the total amount owing, not the interest.
D is incorrect: this gives the total owing using simple interest.

$\Rightarrow C$

Financial Maths, STD2 2019 HSC 27 (Adapted)

Bianca has a credit card that offers no interest-free period. At the end of each month, interest is added to the account at 19.5% per annum, compounded daily. Interest is worked out from the day a purchase is made (included) through to the last day of the month (included).

Part of Bianca’s June statement is shown below, with two figures left blank.

For this account, the minimum payment is set at 3% of the closing balance on 30 June.

Find the minimum payment, giving your answer to the nearest cent. (3 marks)

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$\text{Minimum payment}=\$78.42$

Show Worked Solution

$\text{Days of interest}\ (n)=\ \text{21 June – 30 June}=10$

$\text{Daily interest rate}\ (r)=\dfrac{0.195}{365}=0.0005342\ldots$

$\text{Closing balance}=2600(1+0.0005342\ldots)^{10}=\$2613.92$

$\text{Minimum payment}$	$=\dfrac{3}{100}\times 2613.92$
	$=78.4176\ldots$
	$=\$78.42\ \text{(nearest cent)}$

Financial Maths, STD2 EO-Bank 26

Kiara wants to purchase a used motorbike selling for $2500 and will be in a position to settle the whole debt in a single payment 40 days after the purchase.

Two methods of financing the purchase are open to her.

Using a credit card, where interest of 20.5% per annum is compounded daily. There is no interest-free period, so interest is charged from the day after the purchase.
Taking out a 40-day personal loan, where simple interest is charged at 12% per annum.

Determine the interest that would build up under each method across the 40 days. (2 marks)
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A friend insists that the personal loan will leave Kiara better off by more than $25 compared with the credit card. Decide whether the friend is correct, justifying your answer with calculations. (2 marks)
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a. $\text{Credit card: }\$56.78\ \text{, Personal loan: }\$32.88$

b. $\text{Difference}=\$23.90$

$\text{As }\$23.90\lt\$25,\text{ the friend’s claim is incorrect.}$

Show Worked Solution

a. $\text{Credit card option (compounding daily):}$

$\text{Amount owing}$	$=2500\left(1+\dfrac{0.205}{365}\right)^{40}$
	$=2556.784\ldots$
	$=\$2556.78\ \text{(nearest cent)}$

$\therefore\ \text{Interest}=2556.78-2500=\$56.78$

$\text{Personal loan option (simple interest):}$

$I=Prn$	$=2500\times 0.12\times\dfrac{40}{365}$
	$=32.876\ldots$
	$=\$32.88\ \text{(nearest cent)}$

b. $\text{Difference}=56.78-32.88=\$23.90$

$\text{As }\$23.90\lt\$25,\text{ the friend’s claim is incorrect.}$

v1 Measurement, STD2 M1 2008 HSC 28b*

A tunnel is excavated with a cross-section as shown.

Find an expression for the area of the cross-section using the Trapezoidal rule. (2 marks)

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The area of the cross-section must be 600 m². The tunnel is 80 m wide.

If the value of `a` increases by 2 metres, by how much will `b` change? (2 marks)

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a. `h(2a + b)`

b. `b\ text(decreases by 4.)`

Show Worked Solution

`A`	`~~ h/2[0 + 2(a + b + a) + 0]`
	`~~ h/2(4a + 2b)`
	`~~ h(2a + b)`

b. `A = 600\ text(m²)`

`text(If tunnel is 80 metres wide)`

`4h=80\ \ =>\ \ h=20`

`text{Using part (a):}`

`600`	`=20(2a+b)`
`2a + b`	`= 30`
`b`	`= 30-2a`

`:.\ text(If)\ a\ text(increases by 2,)\ b\ text(must decrease by 4.)`

Financial Maths, STD2 EQ-Bank 29

Mei sets up a superannuation account to save for retirement. She contributes $5000 at the end of each year into the account which earns interest at 6% per annum, compounded annually.

The spreadsheet shown models the first 4 years of the account.

Calculate the value in cell C9. (1 mark)
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From the end of year 4, Mei increases her annual contribution from $5000 to $8000. Calculate the balance in her superannuation account at the end of year 7, and determine how much more this is than if she had continued contributing $5000. (3 marks)
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a. $\text{C9} = \$618.00$

b. $\text{Balance at end of year 7} = \$51\,519.99$

$\text{Mei has}\ \$9550.80\ \text{more than at the standard contribution.}$

Show Worked Solution

a. $\text{C9 (Year 3 interest)} = \text{balance at start} \times \text{rate}$

$\text{C9} = 10\,300 \times 0.06 = \$618.00$

b. $\text{Using}\ \ P+I+C\ \ \text{from end of year 4 balance}\ \$21\,873.08:$

$\text{Increased contributions of}\ \$8000\ \text{from year 5:}$

$\text{Year 5:}\ 21\,873.08+21\,873.08 \times 0.06+8000=\$31\,185.46$

$\text{Year 6:}\ 31\,185.46+31\,185.46 \times 0.06+8000=\$41\,056.59$

$\text{Year 7:}\ 41\,056.59+41\,056.59 \times 0.06+8000=\$51\,519.99$

$\text{Standard contributions of}\ \$5000\ \text{from year 5:}$

$\text{Year 5:}\ 21\,873.08+21\,873.08 \times 0.06+5000=\$28\,185.46$

$\text{Year 6:}\ 28\,185.46+28\,185.46 \times 0.06+5000=\$34\,876.59$

$\text{Year 7:}\ 34\,876.59+34\,876.59 \times 0.06+5000=\$41\,969.19$

$\text{Difference}=51\,519.99-41\,969.19=\$9550.80$

$\therefore\ \text{Mei has}\ \$9550.80\ \text{more by increasing her contributions.}$

Financial Maths, STD2 EQ-Bank 32

Patricia borrows $20 000 as a reducing balance loan at an interest rate of 6% per annum. She is comparing two repayment options, each repaying the loan over 2 years.

Option A: equal monthly repayments
Option B: equal quarterly repayments

A table of present value interest factors for an annuity of $1 is shown.

Using the table, calculate Patricia's repayment under each option. (2 marks)
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Determine which option costs less in total interest, and by how much. (3 marks)
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a. $\text{Monthly} = \$886.41,\ \ \text{Quarterly} = \$2671.69$

b. $\text{Monthly repayments cost}\ \$99.68\ \text{less in interest.}$

Show Worked Solution

a. $\text{Monthly interest rate} = \dfrac{6}{12 \times 100}\ \ \Rightarrow\ \ r=0.005$

$\text{Quarterly interest rate} = \dfrac{6}{4 \times 100}\ \ \Rightarrow\ \ r=0.015$

$\text{Using Repayment} = \dfrac{\text{loan amount}}{\text{PV factor}}:$

$\text{Monthly repayment}= \dfrac{20\,000}{22.5629} = \$886.41$

$\text{Quarterly repayment}= \dfrac{20\,000}{7.4859} = \$2671.69$

b. $\text{Total paid} = \text{repayment} \times \text{number of repayments}$

$\text{Monthly:}\ 886.41 \times 24 = \$21\,273.84$

$\text{Quarterly:}\ 2671.69 \times 8 = \$21\,373.52$

$\text{Interest (monthly)} = 21\,273.84-20\,000 = \$1273.84$

$\text{Interest (quarterly)} = 21\,373.52-20\,000 = \$1373.52$

$\text{Difference} = 1373.52-1273.84 = \$99.68$

$\therefore\ \text{Monthly repayments cost}\ \$99.68\ \text{less in interest.}$

Financial Maths, STD2 EO-Bank 29

Dao takes out a reducing balance loan of $5000. The loan has an interest rate of 12% per annum, compounded monthly, and Dao makes monthly repayments of $900.

The spreadsheet shown models the first 4 months of the loan.

Calculate the value in cell C9. (1 mark)
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By continuing the spreadsheet, determine the number of months it takes Dao to repay the loan in full, and calculate the value of the final repayment. (3 marks)
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a. $\text{C9} = \$41.50$

b. $\text{The loan is repaid in}\ 6\ \text{months.}$

$\text{Final repayment} = \$670.78$

Show Worked Solution

a. $\text{Monthly interest rate} = \dfrac{12\%}{12} = 1\%$

$\text{C9 (Month 2 interest)} = 4150 \times 0.01 = \$41.50$

b. $\text{Continue the schedule using}\ \ P+I-R\ \ \text{each month:}$

$\text{Month 5:}\ 1548.65+1548.65 \times 0.01-900=\$664.14$

$\text{Month 6:}\ 664.14+664.14 \times 0.01=\$670.78$

$\text{The Month 6 balance owing}\ (\$670.78)\ \text{is less than the}$

$\text{usual}\ \$900\ \text{repayment, so this final repayment clears the loan.}$

$\therefore\ \text{The loan is repaid in}\ 6\ \text{months, with a final}$

$\text{repayment of}\ \$670.78.$

Financial Maths, STD2 EO-Bank 28

Talia takes out a reducing balance loan of $12 000 to buy a boat. The loan has an interest rate of 7.2% per annum and Talia makes monthly repayments of $300.

The spreadsheet shown models the first 3 months of the loan.

Calculate the value in cell C9. (1 mark)
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From the end of month 3, Talia increases her monthly repayment from $300 to $500. Calculate the balance owing at the end of month 6, and determine how much less Talia owes compared to keeping repayments at $300. (3 marks)
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a. $\text{C9} = \$70.63$

b. $\text{Balance at end of month 6} = \$10\,007.71$

$\text{Talia owes}\ \$603.61\ \text{less than at the standard repayment.}$

Show Worked Solution

a. $\text{Monthly interest rate} = \dfrac{7.2\%}{12} = 0.6\%$

$\text{C9 (Month 2 interest)} = 11\,772 \times 0.006 = \$70.63$

b. $\text{Using}\ \ P+I-R\ \ \text{from end of month 3 balance}\ \$11\,311.89:$

$\text{Increased repayments of}\ \$500\ \text{from month 4:}$

$\text{Month 4:}\ 11\,311.89+11\,311.89 \times 0.006-500=\$10\,879.76$

$\text{Month 5:}\ 10\,879.76+10\,879.76 \times 0.006-500=\$10\,445.04$

$\text{Month 6:}\ 10\,445.04+10\,445.04 \times 0.006-500=\$10\,007.71$

$\text{Standard repayments of}\ \$300\ \text{from month 4:}$

$\text{Month 4:}\ 11\,311.89+11\,311.89\times 0.006-300=\$11\,079.76$

$\text{Month 5:}\ 11\,079.76+11\,079.76\times 0.006-300=\$10\,846.24$

$\text{Month 6:}\ 10\,846.24+10\,846.24\times 0.006-300=\$10\,611.32$

$\text{Difference}=10\,611.32-10\,007.71=\$603.61$

$\therefore\ \text{Talia owes}\ \$603.61\ \text{less by increasing her repayments.}$

v1 Financial Maths, STD2 F1 2007 HSC 26bii

Isla is in her third year working as a freelance photographer.

Isla purchased photography equipment for $4200.

Isla earns $760 per week. Calculate her taxable income for this year if the only allowable deduction is the amount of depreciation of her photography equipment in the third year of use. (1 mark)

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Use this tax table to calculate Isla’s tax payable. (2 marks)

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Taxable income} \rule[-1ex]{0pt}{0pt}& \textit{Tax on this income} \\
\hline
\rule{0pt}{2.5ex}0 - \$18\,200 \rule[-1ex]{0pt}{0pt}& \text{Nil} \\
\hline
\rule{0pt}{2.5ex}\$18 \, 201 - \$45\,000 \rule[-1ex]{0pt}{0pt}& \text{16 cents for each \$1 over \$18 200} \\
\hline
\rule{0pt}{2.5ex}\$45\,001 - \$135\,000 \rule[-1ex]{0pt}{0pt}& \$4288 \text{ plus 30 cents for each \$1 over \$45 000} \\
\hline
\rule{0pt}{2.5ex}\$135\,001 - \$190\,000 \rule[-1ex]{0pt}{0pt}& \$31 \, 288 \text{ plus 37 cents for each \$1 over \$135 000} \\
\hline
\rule{0pt}{2.5ex}\$190\,001 \text{ and over} \rule[-1ex]{0pt}{0pt}& \$51 \, 638 \text{ plus 45 cents for each \$1 over \$190 000} \\
\hline
\end{array}

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a. $\$38\,790$

b. $\$3294.40$

Show Worked Solution

a. $\text{Income per year}=52 \ 760= $39\,520$

$\text{Taxable income}= 39\ 520-350= $38\,790$

b.	$\text{Tax payable}$	$= 0.16 \times (38\,790-18\,200)$
		$= 0.16 \times 20\,590$
		$= $3294.40$

Calculus, EXT1 EQ-Bank 32

The polynomial $h(x)=x^3+2x+1$ passes through the point $(1,4)$.

Find the gradient of the tangent to $f(x)=x h^{-1}(x)$ at the point where $x=4$. (3 marks)

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Show Answers Only

$\dfrac{9}{5}$

Show Worked Solution

$h(1)=4 \ \ \Rightarrow\ \ h^{-1}(4)=1$

$f(x)=x h^{-1}(x)$

$\text{Using the product rule:}$

$f^{\prime}(x)=h^{-1}(x)+x\cdot \dfrac{d}{dx}\left(h^{-1}(x)\right)$

$\text{Find}\ \dfrac{d}{dx}\left(h^{-1}(x)\right):$

$\text{Let}\ \ y=h^{-1}(x)\ \ \Rightarrow\ \ x=h(y)$

$\dfrac{dx}{dy}=h^{\prime}(y)\ \ \Rightarrow\ \ \dfrac{dy}{dx}=\dfrac{1}{h^{\prime}(y)}$

$\dfrac{d}{dx}\left(h^{-1}(x)\right)=\dfrac{1}{h^{\prime}\left(h^{-1}(x)\right)}$

$f^{\prime}(x)=h^{-1}(x)+\dfrac{x}{h^{\prime}\left(h^{-1}(x)\right)}$

$f^{\prime}(4)=h^{-1}(4)+\dfrac{4}{h^{\prime}\left(h^{-1}(4)\right)}=1+\dfrac{4}{h^{\prime}(1)}$

$h(x)=x^3+2x+1\ \ \Rightarrow\ \ h^{\prime}(x)=3x^2+2$

$f^{\prime}(4)=1+\dfrac{4}{3(1)^2+2}=\dfrac{9}{5}$

$\therefore\ \text{Gradient of tangent}=\dfrac{9}{5}$

Financial Maths, STD2 EQ-Bank 32

Yasmin is planning to purchase a Newcastle Knights corporate box for $5000. She intends to repay the entire amount 90 days after making the purchase.

She is comparing two credit options.

Credit card: compound interest charged daily at 18% per annum, with a 30-day interest-free period from the date of purchase.
Personal loan: simple interest charged at 9.5% per annum.

Determine which option Yasmin should choose, showing calculations to support your answer. (4 marks)

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$\text{Credit card (daily compound interest):}$

$\text{Days accruing interest}\ =90-30=60$

$\text{Amount owing}$	$= 5000\left(1+\dfrac{0.18}{365}\right)^{60}$
	$= 5150.118\ldots$
	$= \$5150.12\ \text{(nearest cent)}$

$\text{Interest} = 5150.12-5000 = \$150.12$

$\text{Personal loan (simple interest):}$

$I = Prn$	$=5000 \times 0.095 \times \dfrac{90}{365}$
	$= 117.123\ldots$
	$= \$117.12\ \text{(nearest cent)}$

$\therefore\ \text{Since \$117.12 < \$150.12, Yasmin should choose the personal loan.}$

Show Worked Solution

$\text{Days accruing interest}\ =90-30=60$

$\text{Credit card (daily compound interest):}$

$\text{Amount owing}$	$= 5000\left(1+\dfrac{0.18}{365}\right)^{60}$
	$= 5150.118\ldots$
	$= \$5150.12\ \text{(nearest cent)}$

$\text{Interest} = 5150.12-5000 = \$150.12$

$\text{Personal loan (simple interest):}$

$I = Prn$	$=5000 \times 0.095 \times \dfrac{90}{365}$
	$= 117.123\ldots$
	$= \$117.12\ \text{(nearest cent)}$

$\therefore\ \text{Since \$117.12 < \$150.12, Yasmin should choose the personal loan.}$

Financial Maths, STD2 EQ-Bank 28

Sam needs to borrow $3000 and plans to repay it in full after 30 days. He is comparing two options.

Credit card: compound interest charged daily at 19.9% per annum, with no interest-free period.
Personal loan: simple interest charged at 11.5% per annum.

Calculate the interest Sam would be charged on the credit card over the 30 days. (2 marks)

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How much does Sam save by choosing the personal loan. (2 marks)

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a. $\$49.46$

b. $\text{Saving =}\ \$21.10$

Show Worked Solution

a. $\text{Daily interest rate} = \dfrac{0.199}{365}$

$\text{Amount owing}$	$= 3000\left(1+\dfrac{0.199}{365}\right)^{30}$
	$= 3049.458\ldots$
	$= \$3049.46\ \text{(nearest cent)}$

$\therefore \text{Interest} = 3049.46-3000 = \$49.46$

b. $\text{Personal loan (simple interest):}$

$I = Prn$	$=3000 \times 0.115 \times \dfrac{30}{365}$
	$= 28.356\ldots$
	$= \$28.36\ \text{(nearest cent)}$

$\therefore \text{Saving} = 49.46-28.36 = \$21.10$

Financial Maths, STD2 EQ-Bank 26

A credit card has an interest-free period of 55 days from and including the date of purchase. Interest is charged on purchases, compounding daily at a rate of 16.8% per annum. Interest is charged from the day following the interest-free period.

Furniture was purchased for $1200 using this credit card. Full payment was made on the 78th day after the date of purchase. There were no other purchases on this credit card.

For how many days is interest charged on the purchase? Give your answer to the nearest cent. (1 mark)

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Calculate the total interest charged when the account was paid in full. (2 marks)

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a. $23 \text{ days}$

b. $\$12.77$

Show Worked Solution

a. $\text{Interest is charged after the 55-day interest-free period.}$

$\text{Days charged} = 78-55 = 23 \text{ days}$

b. $\text{Daily interest rate} = \dfrac{0.168}{365}$

$\text{Amount owing} $	$= 1200\left(1+\dfrac{0.168}{365}\right)^{23}$
	$= 1212.768\ldots$
	$= \$1212.77\ \text{(nearest cent)}$

$\therefore\ \text{Interest} = 1212.77-1200 = \$12.77$

Financial Maths, STD2 EQ-Bank 27

A credit card account has no interest-free period. Compound interest is calculated daily at a rate of 18.5% per annum and charged to the account up to and including the statement date.

A single purchase of $1500 was made and interest accrued on the purchase for 28 days, including the statement date. There were no other purchases on the account.

Calculate the closing balance on the statement date. (2 marks)

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The minimum payment is calculated as $30 or 3% of the closing balance, whichever is greater. Calculate the minimum payment due. (2 marks)

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a. $\$1521.43$

b. $\$45.64$

Show Worked Solution

a. $\text{Daily interest rate} = \dfrac{0.185}{365}$

$\text{Closing balance}$	$= 1500\left(1+\dfrac{0.185}{365}\right)^{28}$
	$= 1521.433\ldots$
	$= \$1521.43$

b. $3\% \text{ of closing balance} = 0.03 \times 1521.43= \$45.64$

$\text{Since } \$45.64 > \$30, \text{ the minimum payment due is } \$45.64.$

Networks, STD2 EQ-Bank 28

Lena is opening a new café. The project requires the completion of 7 activities, $A$ to $G$. The project is due to be completed in 15 days.

The directed network diagram shows these activities with their completion times in days.

Use the information from the network diagram to complete the Gantt chart, including the critical path and all other activities required for the project. (3 marks)
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List all activities, not on the critical path, which could be occurring at midday on day 7. (1 mark)
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b. $B\text{, }D\text{ and }E$

Show Worked Solution

b. $\text{By inspection of the Gantt chart.}$

$\text{Activities which could be occurring at }\approx 6.5\ \text{on chart (midday day 7):}$

$B\text{, }D\text{ and }E$

Networks, STD2 EQ-Bank 27

A project requires the completion of 8 activities, $A$ to $H$. The project is due to be completed in 16 days.

The directed network diagram shows these activities with their completion times in days.

Use the information from the network diagram to complete the Gantt chart, including the critical path and all other activities required for the project. (3 marks)
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List all activities, not on the critical path, which could be occurring at midday on day 9. (1 mark)
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b. $C, D\ \text{and}\ F$

Show Worked Solution

b. $\text{By inspection of the Gantt chart.}$

$\text{Activities which could be occurring at}\ \approx 8.5\ \text{on chart (midday day 9):}$

$C, D\ \text{and}\ F$

Networks, STD2 EQ-Bank 25

The construction of a new reptile exhibit is a project involving nine activities, $A$ to $I$. The network diagram below shows the activities and their completion times in weeks. Some values are missing.

The Gantt chart below has been created for this project.

Using the Gantt chart, identify the critical path. (1 mark)
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Use the Gantt chart to determine the missing values in the network diagram. (2 marks)
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Activity $E$ is delayed by 8 weeks. Using the Gantt chart, explain whether this will affect the minimum completion time of the project. (2 marks)
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a. $ACDFGI$

b. $\text{B} = 5 \text{ weeks, H} = 7 \text{ weeks}$

c. $\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}$

$\text{extension from week 10 to 16).}$

$\text{The delay of 8 weeks exceeds the float of 6 weeks.}$

$\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}$

$\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}$

Show Worked Solution

a. $\text{The critical path is the continuous solid bar on row 1 of the Gantt chart.}$

$\text{Critical path:}\ ACDFGI$

b. $\text{Activities B and H are not labelled in the network diagram.}$

$\text{From the Gantt chart:}$

$\text{B starts at week 0, ends at week 5} \to \text{duration} = 5 \text{ weeks}$

$\text{H starts at week 7, ends at week 14} \to \text{duration} = 7 \text{ weeks}$

c. $\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}$

$\text{extension from week 10 to 16).}$

$\text{The delay of 8 weeks exceeds the float of 6 weeks.}$

$\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}$

$\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}$

Statistics, EXT1 EQ-Bank 32

A company accountant has previously found that the mean amount owed on any individual unpaid invoice is $\$800$ with a standard deviation of $\$200$.

Using the normal distribution table (included), determine the probability that, in a random sample of 60 unpaid invoices, the total amount owed is more than $\$47\,500$. Give your answer as a percentage correct to one decimal place. (3 marks)

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$62.6\%$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}$

$\bar{X} \sim N\left(\mu, \dfrac{\sigma^2}{n} \right) \sim N\left(800, \dfrac{200^2}{60} \right)$

$P(\Sigma X) > 47\,500 = P\left(\bar{X}>\dfrac{47\,500}{60}\right)=P(\bar{X})>791.67 $

$\text{By the central limit theorem:}$

$Z=\dfrac{\bar{X}-800}{\frac{200}{\sqrt{60}}}\sim N(0,1)$

$Z=\dfrac{791.67-800}{\frac{200}{\sqrt{60}}}=-0.32 \ \text{(2 d.p.)}$

$\text{Using Normal Distribution Table of Values:}$

$P(\Sigma X > 47\,500)$	$=P(Z>-0.32)$
	$=1-P(Z\leq-0.32)$
	$=1-0.3745$
	$=62.6\%$

Probability, STD2 EQ-Bank 31

A survey of 60 people found the following information about their exercise habits.

35 enjoy hiking $(H)$
28 enjoy cycling $(C)$
8 enjoy neither hiking nor cycling

Draw a Venn diagram to represent this information. (2 marks)

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One person is selected at random. What is the probability that the person enjoys hiking or cycling? (1 mark)

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b. $\dfrac{13}{15}$

Show Worked Solution

a. $\text{Number in either }H\text{ or }C = 60-8 = 52$

$\text{Number in both }H\text{ and }C = 35+28-52 = 11$

$H\text{ only} = 35-11 = 24$

$C\text{ only} = 28-11 = 17$

b. $P(H \text{ or } C) = \dfrac{24+11+17}{60} = \dfrac{52}{60} = \dfrac{13}{15}$

$\text{OR, using the complement:}$

$P(H \text{ or } C) = 1-\dfrac{8}{60} = \dfrac{13}{15}$

Algebra, STD2 A4 EQ-Bank 27

SunPower Solutions is a business that installs solar panels. Fixed costs are $1200. Each panel costs $150.00 to install and generates revenue of $350.00.

The spreadsheet below models the business's costs and revenue for different numbers of panels installed.

Calculate the spreadsheet values for the installation of 4 panels (cells B9, C9, D9) and 6 panels (cells B10, C10, D10). (2 marks)
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Using the spreadsheet, identify the break-even point and explain what it means for the business. (2 marks)
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SunPower Solutions has received a large order that will see them make a profit of $4200. Calculate the number of solar panels $(x)$ they will be installing. (2 marks)
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a. $\text{4 panels: TC (B9) = \$1800.00}$

$\text{Revenue (C9) = \$1400.00, Profit/Loss (D9) = }-\$400.00$

$\text{6 panels: TC (B10) = \$2100.00}$

$\text{Revenue (C10) = \$2100.00, Profit/Loss (D10) = \$0.00}$

b. $\text{Break-even = 6 panels, See worked solution}$

c. $27 \text{ panels}$

Show Worked Solution

a. $\text{4 panels:}$

$\text{Total Cost (B9)} = \$1200+4\times\$150 = \$1800.00$

$\text{Revenue (C9)} = 4\times\$350 = \$1400.00$

$\text{Profit/Loss (D9)} = \$1400.00-\$1800.00 = -\$400.00$

$\text{6 panels:}$

$\text{Total Cost (B10)} = \$1200+6\times\$150 = \$2100.00$

$\text{Revenue (C10)} = 6\times\$350 = \$2100.00$

$\text{Profit/Loss (D10)} = \$2100.00-\$2100.00 = \$0.00$

b. $\text{When 6 panels are installed:}$

$\text{Revenue = Total costs = \$2100.00 (breakeven)}$

$\text{This is the point at which the business covers all of its costs and}$

$\text{begins to make a profit.}$

$\text{OR}$

$\text{If fewer than 6 panels are installed the business will make a loss.}$

c. $\text{Let } x = \text{the number of solar panels to be installed.}$

$\text{Profit}$	$ = \text{Revenue}-\text{Total costs}$
$4200$	$ = 350x-(1200+150x)$
$4200$	$= 200x-1200$
$5400$	$=200x$
$x$	$=27$

$\text{SunPower Solutions will install 27 solar panels.}$

Mechanics, EXT2 EQ-Bank 36

An experimental rocket is at a height of 5000 m, ascending at a speed of $50\sqrt{2}$ m s$^{-1}$ at an angle of 45° to the horizontal, when its engine stops. The rocket is then subject to gravity and to air resistance proportional to its velocity. Take $g$ = 10 m s$^{-2}$.

The velocity vector of the rocket, $t$ seconds after the engine stops, is

$\mathbf{v}(t) = 50e^{-0.2t}\,\mathbf{i} + (100e^{-0.2t}-50)\mathbf{j}.$ (Do NOT prove this.)

Show that the rocket reaches its greatest height when $t =5\ln 2$ seconds, and calculate its greatest height. (3 marks)

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The pilot can only operate the ejection seat while the rocket is descending at an angle between 45° and 60° to the horizontal. Find the earliest and latest times at which the pilot can eject. (3 marks)

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As the rocket continues to fall, its speed approaches a limiting value. Find this terminal speed, justifying your answer. (1 mark)

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a. $\text{See Worked Solutions}$

b. $\text{Pilot can eject between 5.5 and 6.6 seconds.}$

c. $\text{Terminal speed}=50 \ \text{ms}^{-1}$

Show Worked Solution

a. $\mathbf{v}(t)=50 e^{-0.2 t}\,\mathbf{i}+\left(100 e^{-0.2 t}-50\right)\mathbf{j}$

$\text{Max height occurs when} \ \ \dot{y}=0:$

$100 e^{-0.2 t}-50$	$=0$
$e^{-0.2 t}$	$=\dfrac{1}{2}$
$-0.2 t$	$=-\ln 2$
$t$	$=5 \ln 2$

$\text{Find} \ y \ \text{when}\ \ t=5 \ln 2:$

$y(t)=\displaystyle \int 100 e^{-0.2 t}-50\, d t=-500 e^{-0.2 t}-50 t+c$

$\text{When} \ \ t=0, y=5000:$

$5000=-500 e^{\circ}+c \ \Rightarrow \ c=5500$

$y=5500-500 e^{-0.2 t}-50 t$

$\text{At} \ \ t=5\ln 2:$

$y=5500-500 e^{-\ln 2}-50 \times 5 \ln 2=5076.71 \ldots=5077 \ \text{m}$.

b. $\text {On descent,} \ \ \dot{y}<0.$

$\text{Let} \ \ \theta=\text{angle below the horizontal}$

$\tan \theta=\dfrac{\abs{\dot{y}}}{\dot{x}}$

$\text{Let}\ \ a=e^{-0.2 t}$

$\tan \theta=\dfrac{50-100 a}{50 a}=\dfrac{1}{a}-2 \ \Rightarrow \ \theta=\tan ^{-1}\left(\dfrac{1}{a}-2\right)$

$\text{When} \ 45^{\circ} \ \text {is reached:}$

$\dfrac{1}{a}-2=1 \ \Rightarrow \ a=3$

$e^{-0.2 t}=\dfrac{1}{3} \ \Rightarrow \ t=\dfrac{\ln 3}{0.2} \approx 5.5 \ \text{s (1 d.p.)}$

$\text{When} \ 60^{\circ} \ \text {is reached:}$

$\dfrac{1}{a}-2=\sqrt{3} \ \Rightarrow \ a=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}$

$e^{-0.2 t}$	$=2-\sqrt{3}$
$-0.2 t$	$=\ln (2-\sqrt{3})$
$t$	$=-5\ln (2-\sqrt{3}) \approx 6.6 \ \text{s (1 d.p.)}$

$\therefore \ \text{Pilot can eject between 5.5 and 6.6 seconds.}$

c. $\text{As} \ \ t \rightarrow \infty:$

$e^{-0.2 t} \rightarrow 0\ \ \Rightarrow\ \ \dot{x} \rightarrow 0, \ \ \dot{y} \rightarrow -50$

$\therefore \ \text{Terminal speed}=\sqrt{0^2+50^2}=50 \ \text{ms}^{-1}$

Mechanics, EXT2 EQ-Bank 31

A particle is projected from the origin with speed, $V$, at an angle of $\alpha$ above the horizontal. It is subject to both gravity and an air resistance proportional to its velocity, so that its horizontal and vertical components of acceleration while it is rising are given by

$\ddot{x}=-k\dot{x}$ and $\ddot{y} = -g-k\dot{y}$

Show that $\dot{x} = V\cos\,\alpha\ e^{-kt}$ and $\dot{y} = \left( \dfrac{g}{k} + V\sin\,\alpha\right)e^{-kt}-\dfrac{g}{k} $ (2 marks)

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Show that when the particle reaches its greatest height, it has travelled a horizontal distance of
$\dfrac{V^2\sin\,2\alpha}{2(g+Vk\,\sin\,\alpha)}$. (3 marks)

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a. $\text{See Worked Solutions}$

b. $\text{See Worked Solutions}$

Show Worked Solution

a. $\ddot{x}=-k \dot{x} \ \Rightarrow \ \dfrac{d \dot{x}}{d t}=-k \dot{x}$

$\text{Solving 1st order differential equation:}$

$\dot{x}=A e^{-k t}$

$\text{Since}\ \ \dot{x}=V \cos \alpha\ \ \text{when} \ \ t=0\ \ \Rightarrow\ \ A=V \cos \alpha$

$\dot{x}=V \cos \alpha e^{-k t}$

$\ddot{y}=-g-k \dot{y}\ \Rightarrow \ \dfrac{d \dot{y}}{d t}+k \dot{y}=-g$

$\text{Solving 1st order differential equation:}$

$\dot{y}=B e^{-k t}-\dfrac{g}{k}$

$\text{Since} \ \ \dot{y}=V \sin \alpha \ \ \text{when} \ \ t=0:$

$V \sin \alpha=B e^{-k t}-\dfrac{g}{k} \ \Rightarrow \ B=\dfrac{g}{k}+V \sin \alpha$

$\dot{y}=\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}$

b. $\text{At max height,} \ \ \dot{y}=0$

$\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}=0 \ \Rightarrow \ e^{-k t}=\dfrac{g}{g+V k \, \sin \alpha}\ \ldots\ (1)$

$\text{Find horizontal distance} \ (x):$

$x$	$=\displaystyle \int_0^t V \cos \alpha\, e^{-k t}\, d t$
	$=-\dfrac{V \cos \alpha}{k}\big[e^{-k t}\big]_0^t$
	$=\dfrac{V \cos \alpha}{k}\left(1-e^{-k t}\right)$
	$=\dfrac{V \cos \alpha}{k}\left(1-\dfrac{g}{g+Vk\, \sin \alpha}\right)\ \ \ \text{(using (1) above)}$
	$=\dfrac{V \cos \alpha}{k}\left(\dfrac{g+Vk\, \sin \alpha-g}{g+Vk\, \sin \alpha}\right)$
	$=\dfrac{V^2 \sin \alpha\, \cos \alpha}{g+Vk\, \sin \alpha}$
	$=\dfrac{V^2 \sin 2 \alpha}{2(g+Vk\, \sin \alpha)}$

Proof, EXT2 EQ-Bank 25

Use mathematical induction to prove De Moivre's theorem:

$(\cos \theta+i \,\sin \theta)^n=\cos n \theta+i \, \sin n \theta$

for all integers $n \geq 1$. (3 marks)

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$\text{Proof (See Worked Solutions)}$

Show Worked Solution

$\text{Prove} \ \ (\cos \theta+i \, \sin \theta)^n=\cos (n \theta)+i \sin (n \theta) \ \text {for} \ \ n \geq 1$

$\text{If} \ \ n=1:$

$(\cos \theta+i \, \sin \theta)^1=\cos \theta+i \, \sin \theta=\cos (1 \theta)+i \, \sin (1 \theta)$

$\therefore \ \text{True for} \ \ n=1$

$\text{Assume true for} \ \ n=k:$

$(\cos \theta+i \, \sin \theta)^k=\cos k \theta+i \,\sin k \theta\ \ldots\ (1)$

$\text{Prove true for} \ \ n=k+1:$

$\text{i.e.}\ \ (\cos \theta+i \,\sin \theta)^{k+1}=\cos (k+1) \theta+i \, \sin (k+1) \theta.$

$\text{LHS}$	$=(\cos \theta+i \, \sin \theta)^{k+1}$
	$=(\cos \theta+i \, \sin \theta)^k(\cos \theta+i \, \sin \theta)$
	$=(\cos k \theta+i \, \sin k \theta)(\cos \theta+i \, \sin \theta) \ \ \text{(using}\ (1)\ \text{above)}$
	$=\cos k \theta \, \cos \theta+i \, \cos k \theta \,\sin \theta+i \, \sin k \theta \,\cos \theta+i^2 \, \sin k \theta \,\sin \theta$
	$=(\cos k \theta \, \cos \theta-\sin k \theta \,\sin \theta)+i(\sin k \theta \,\cos \theta+\cos k \theta \,\sin \theta)$
	$=\cos (k+1) \theta+i \, \sin (k+1) \theta$

$\Rightarrow \text{True for} \ \ n=k+1$

$\therefore \ \text{Since true for \(n=1$, by PMI, true for integers $n \geqslant 1$.}\)

Proof, EXT2 EQ-Bank 34

Consider a sequence of rectangles with side lengths $a_{ n }$ and $b_{ n }$.

The first rectangle has $a_1=2$ and $b_1=1$.

For integers $n \geq 1,\ \ a_{n+1}=\dfrac{a_n+b_n}{2}$ and $b_{n+1}=\dfrac{2}{a_{n+1}}.$

Show that every rectangle in the sequence has an area of 2 square units. (1 mark)

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Use the relationship between the arithmetic mean and the geometric mean to prove that $a_n \geq \sqrt{2}$ for any integer $n \geq 1$. (2 marks)

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Use mathematical induction to prove that $a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2})$ for any integer $n \geq 1$. (4 marks)

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Use the squeeze theorem to show that the rectangles approach a square as $n$ approaches infinity. (2 marks)

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Show Worked Solution

a. $\text{Since}\ \ a_1 b_1=2\ \ \text{and}\ \ a_{n+1} b_{n+1}=a_{n+1} \times \dfrac{2}{a_{n+1}}=2$

$\Rightarrow\ \text{Each rectangle has area 2.}$

b. $a_1=2\ \ \Rightarrow\ \ a_1 \geq \sqrt{2}\ \ \text{(true for 1st rectangle)}$

$\text{AM/GM inequality:}\ \ \dfrac{a_n+b_n}{2} \geq \sqrt{a_nb_n} $

$a_nb_n=2\ \ \text{(from part (a))}$

$\dfrac{a_n+b_n}{2} \geq \sqrt{2}\ …\ (1)$

$\text{Since}\ \ a_{n+1}=\dfrac{a_n+b_n}{2}:$

$\ a_{n+1} \geq \sqrt{2}\ \ \ \text{(using (1) above)}$

$\therefore a_n \geq \sqrt{2}$

c. $\text{Prove}\ \ a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}),\ \ \text{for}\ \ n \geq 1$

$\text{If}\ \ n=1:$

$\ a_1-\sqrt{2}=2-\sqrt{2} \leq \dfrac{1}{2^{0}}(2-\sqrt{2})$.

$\Rightarrow\ \text{True for}\ \ n=1.$

$\text{Assume true for}\ \ n=k:$

$a_k-\sqrt{2} \leq \dfrac{1}{2^{k-1}}(2-\sqrt{2})\ …\ (1) $

$\text{Prove true for}\ \ n=k+1:$

$\text{i.e.}\ \ a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})$

$\text{By definition,} \ \ a_{k+1}=\dfrac{1}{2}\left(a_k+b_k\right)$

$a_{k+1}-\sqrt{2}$	$=\dfrac{1}{2}\left(a_k-\sqrt{2}\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) $
	$\leq \dfrac{1}{2}\left(\dfrac{1}{2^{k-1}}(2-\sqrt{2})\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right)\ \ \text{(see (1) above)}\ $

$\text{Since}\ \ a_k \geq \sqrt{2}\ \ \text{and}\ \ a_kb_k=2:$

$\ b_k \leq \sqrt{2}\ \ \text{and}\ \ b_k-\sqrt{2} \leq 0$

$a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) \leq \dfrac{1}{2^k}(2-\sqrt{2})$

$\Rightarrow\ \text{True for}\ \ n=k+1.$

$\therefore\ \text{Since true for}\ \ n=1,\ \text{by PMI, true for integers}\ \ n \geq 1.$

d. $\text {Combining parts (b) and (c):}$

$0 \leq a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}).$

$\text{As}\ \ n \rightarrow \infty, \ \dfrac{1}{2^{n-1}} \rightarrow 0$

$\Rightarrow a_n-\sqrt{2} \rightarrow 0\ \ \text{(by squeeze theorem)}$

$\text{Since rectangles have an area = 2:}$

$\text{As}\ \ n \rightarrow \infty,\ b_n \rightarrow \sqrt{2}$

$\text{i.e. rectangles approach a square.}$

Mechanics, EXT2 EQ-Bank 30

Luggage at an airport is delivered to its owners via a ramp that is inclined at 30° to the horizontal. A 20 kg suitcase, initially at rest at the top of the ramp, slides down the ramp against a resistance of `v` newtons per kilogram, where `v\ text(ms)^(-1)` is the speed of the suitcase.

By resolving forces parallel to the ramp, show that the magnitude of the acceleration, `ddot{x}\ text(ms)^(-2)`, of the suitcase down the ramp is given by `ddot{x} = (g-2v)/2`. (2 marks)

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Using 9.8 `text(ms)^(-2)` as the acceleration due to gravity, find the distance `x` metres that the suitcase has slid as a function of `v`. Give your answer in the form `x = bv + c\ log_e(c/(c-v))`, where `b, c in R`. (3 marks)

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a. `text(Proof)\ \ text{(See Worked Solutions)}`

b. `x = -v + 4.9 ln ((4.9)/(4.9-v))`

Show Worked Solution

`sumF`	`=20g sin 30^@-20v`
`m ddot{x}`	`=10g-20v`
`ddot{x}`	`= (g-2v)/2`

b. `text{Using}\ \ ddot{x}=v *(dv)/(dx):`

`(dv)/(dx)`	`= (g-2v)/(2v)`
`(dx)/(dv)`	`= (2v)/(g-2v)`
`(dx)/(dv)`	`= -(2v)/(2v-g)=-((2v-g + g))/(2v-g)= -1-g/(2v-g)`

`text{Find the distance travelled:}`

`x`	`= int_0^v-1-g/(2v-g)\ dv`
	`= int_0^v-1-g/2 (2/(2v-g))\ dv`
	`= [-v-4.9 xx ln\ \|2v-g\|]_0^v`

`text(When)\ \ x=0, v=0:`

`2v-g < 0\ \ =>\ \ |2v-g| = g-2v`

`x`	`= [-v-4.9 ln (g-2v)]_0^v`
	`= -v-4.9 ln (g-2v)-(0-4.9 ln (g))`
	`= -v + 4.9 ln (g/(g-2v))`
	`=-v + 4.9 ln(9.8/(9.8-2v))`
	`= -v + 4.9 ln ((4.9)/(4.9-v))`

Complex Numbers, EXT2 N2 2023 14a*

Let $z$ be the complex number $z=\text{cis}\dfrac{\pi}{6} $ and $w$ be the complex number $w=\text{cis}\dfrac{3\pi}{4} $.

By first writing $z$ and $w$ in Cartesian form, or otherwise, show that
$|z+w|^2=\dfrac{4-\sqrt{6}+\sqrt{2}}{2}$. (3 marks)

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The complex numbers $z, w$ and $z+w$ are represented in the complex plane by the vectors $\overrightarrow{O A},\overrightarrow{O B}$ and $\overrightarrow{O C}$ respectively, where $O$ is the origin.
Show that $\angle A O C=\dfrac{7 \pi}{24}$. (2 marks)

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Deduce that $\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}$. (1 mark)

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i. $\text{See Worked Solutions}$

ii. $\text{See Worked Solutions}$

iii. $\text{See Worked Solutions}$

Show Worked Solution

i. $z= \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i $

$w= \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} $

$\|z+w\|^2$	$=\Bigg{\|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{\|}$
	$=\Bigg{\|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{\|}$
	$=\Bigg{\|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{\|}$
	$= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}$
	$= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}$
	$=\dfrac{16-4\sqrt6+4\sqrt2}{8} $
	$=\dfrac{4-\sqrt6+\sqrt2}{2} $

ii.

$\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} $

$ |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} $

$\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC $

$\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB $

$\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} $

iii. $\text{In}\ \triangle AOC: $

$ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} $

$\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. $

$\text{Using the cos rule in}\ \triangle AOC: $

$\cos\,\dfrac{7\pi}{24}$	$=\dfrac{\|z\|^2+\|z+w\|^2-\|w\|^2}{2\|z\|\|z+w\|}$
	$=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1 \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} $
	$=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} $
	$=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} $
	$=\dfrac{8-2\sqrt6+2\sqrt2}{4} $

♦♦ Mean mark (iii) 26%.

Vectors, EXT2 EQ-Bank 34

A sphere of radius $r=3$ is centred at $C(6,-3,2)$.

A line passes through $A(3,-1,6)$ and $B(5,-1,-5)$.

Show that this line is a tangent to the sphere. (4 marks)

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$\text{See Worked Solutions}$

Show Worked Solution

$\text{The sphere centred at} \ \ C(6,-3,2) \ \ \text {with radius} \ \ r=3$

$(x-6)^2+(y+3)^2+(z-2)^2=3^2=9$

$\text{The line}\ A B\ \text{has direction}$

$\overrightarrow{O B}-\overrightarrow{O A}=\left[\begin{array}{c}5-3 \\ -1-(-1) \\ -5-6\end{array}\right]=\left[\begin{array}{c}2 \\ 0 \\ -11\end{array}\right]$

$\text{Line \(A B$ has parametric equation:}\)

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\overrightarrow{O A}+\lambda \overrightarrow{A B}=\left[\begin{array}{c}3 \\ -1 \\ 6\end{array}\right]+\lambda\left[\begin{array}{c}2 \\ 0 \\ -11\end{array}\right]=\left[\begin{array}{c}3+2 \lambda \\ -1 \\ 6-11 \lambda\end{array}\right]$

$\text{Substitute into the equation for the sphere:}$

$(3+2 \lambda-6)^2+(-1+3)^2+(6-11 \lambda-2)^2$	$=9$
$(2 \lambda-3)^2+4+(4-11 \lambda)^2$	$=9$
$9-12 \lambda+4 \lambda^2+4+16-88 \lambda+121 \lambda^2$	$=9$

$125 \lambda^2-100 \lambda+20$	$=0$
$5\left(25 \lambda^2-20 \lambda+4\right)$	$=0$
$5(5 \lambda-2)^2$	$=0$

$\text{There is only one}\ \lambda\ \text{that solves this equation.}$

$\text{i.e. one common point on the line and the sphere.}$

$\therefore\ \text{The line is a tangent to the sphere.}$

Mechanics, EXT2 EQ-Bank 7 MC

Particles of mass 3 kg and $m$ kg are attached to the ends of a light inextensible string that passes over a smooth pulley, as shown.

If the acceleration of the 3 kg mass is 4.9 m s$^{-2}$ upwards, then

$m= 4.5$
$m = 6.0$
$m= 9.0$
$m= 13.5$

Show Answers Only

$C$

Show Worked Solution

$\text{Using}\ \ \Sigma F=m \ddot{x}:$

$mg-3g$	$= (m + 3) \times 4.9$
$9.8m-3 \times 9.8$	$= 4.9m + 14.7$
$m$	$= 9$

$\Rightarrow C$

Mechanics, EXT2 2019 SPEC1 9

A light inextensible string is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs in equilibrium from a smooth ring on the string, as shown in the diagram below. The string makes an angle `alpha` with the ceiling.

Express the tension, `T` newtons, in the string in terms of `m`, `g` and `alpha`. (1 mark)

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A different light inextensible sting is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs from a smooth ring on the string. A horizontal force of `F` newtons is applied to the ring. The tension in the sting has a constant magnitude and the system is in equilibrium. At one end the string makes an angle `beta` with the ceiling and at the other end the string makes an angle `2beta` with the ceiling, as shown in the diagram below.

Show that `F = mg((1-cos(beta))/(sin(beta)))`. (3 marks)

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a. `T = (mg)/(2sinalpha)`

b. `text(See Worked Solutions)`

Show Worked Solution

`2 xx Tsinalpha`	`= mg`
`:.T`	`= (mg)/(2sinalpha)`

`text(Resolving forces vertically:)`

`Tsin(beta) + Tsin(2beta)`	`= mg`
`T`	`= (mg)/(sin(beta) + sin(2beta))`

`text(Resolving forces horizontally:)`

`F + Tcos(2beta)`	`= Tcos(beta)`
`F`	`= Tcos(beta)-Tcos(2beta)`
	`= T(cos(beta)-cos(2beta))`
	`= T[cos(beta)-(2cos^2beta-1)]`
	`= T(−2cos^2(beta) + cos(beta) + 1)`
	`= T(−2cos(beta)-1)(cos(beta)-1)`
	`= (mg(1 -cos(beta))(2cosbeta + 1))/(sin(beta) + 2sin(beta)cos(beta))`
	`= (mg(1-cos(beta))(2cos(beta) + 1))/(sin(beta)(1+2cos(beta)))`
	`= mg((1-cos(beta))/(sin(beta)))`

Mechanics, EXT2 EQ-Bank 30

A mass is acted upon by three forces (as shown in the diagram below). Express the resultant force using the standard unit vectors $\mathbf{i}$ and $\mathbf{j}$ in the horizontal and vertical directions, and hence calculate the size of the resultant force acting on the mass. Describe the direction of this resultant force. (4 marks)

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Show Worked Solution

$\text{Resolving forces into \(\mathbf{i}$ and $\mathbf{j}$ components}\)

$\text{Resultant}$

$\text{Force:}$

$F$	$=(-20+15 \cos 45+10 \cos 60) \mathbf{i}+(15 \sin 45-10 \sin 60)\mathbf{j}$
	$=\left(\dfrac{15 \sqrt{2}}{2}-15\right)\mathbf{i}+\left(\dfrac{15 \sqrt{2}}{2}-5 \sqrt{3}\right)\mathbf{j}$
$\abs{F}$	$=\sqrt{\left(\dfrac{3 \sqrt{2}}{2}-15\right)^2+\left(\dfrac{5 \sqrt{2}}{2}-5 \sqrt{3}\right)^2} =4.805 \cdots=4.81 \ N$

$\text{Direction}$:

$\tan \theta$	$=\dfrac{\frac{15 \sqrt{2}}{2}-5 \sqrt{3}}{\frac{15 \sqrt{2}}{2}-15}=-0.443 \ldots$
$\theta$	$=180-23.89=156^{\circ}$

$\text{i.e.} \ 156^{\circ} \ \text{direction measured from the \(x$-axis}\)

Mechanics, EXT2 EQ-Bank 30

In a circus act, an 8 kg cannon ball is projected from the origin into the air with an initial velocity of 26 m s$^{-1}$ and at an angle of 67.4° to the horizontal. The ball is caught at the top of its trajectory by a performer who is at the position $(A, B)$.

The velocity vector, $\mathbf{v} (t)$, of the ball at time $t$ seconds after launch is given by

$\mathbf{v}(t)=10 e^{-0.8 t} \mathbf{i} +\left[36.5 e^{-0.8 t}-12.5\right] \mathbf{j}$. (Do NOT Prove this.)

Show that the ball reaches the performer at $t=1.339$ (to three decimal places). (2 marks)

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Find the values of $A$ and $B$. (3 marks)

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What is the speed of the ball when it reaches the performer? (1 mark)

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What is the magnitude of the force on the ball when it reaches the performer? (2 marks)

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a. $t=1.339 \ \text{s}$

b. $A=8.217\ \text{m},\ \ B=13.257\ \text{m}$

c. $\text{Speed}\ = 3.426\ \text{m s}^{-1}$

d. $82.951 \ \text{N}$

Show Worked Solution

a. $\text{At top of trajectory:}$

$36.5 e^{-0.8 t}-12.5$	$=0$
$36.5 e^{-0.8 t}$	$=12.5$
$e^{-0.8 t}$	$=\dfrac{12.5}{36.5}$
$-0.8 t$	$=\ln \dfrac{12.5}{36.5}$
$ t$	$=\dfrac{\ln\frac{12.5}{36.5}}{-0.8}=1.339 \ \text{s (3 d.p.)}$

b. $\text{Horizontal velocity}\ =10 e^{-0.8 t}$

$x=\displaystyle \int 10 e^{-0.8 t}\,d t=\dfrac{10}{-0.8} e^{-0.8 t}+c_1=-12.5 e^{-0.8 t}+c_1$

$\text{When} \ \ t=0, \ x=0:$

$0=-12.5 e^0+c_1\ \ \Rightarrow\ \ c_1=12.5$

$x=-12.5 e^{-0.8 t}+12.5$

$\text{When } t=1.339, \ x=A:$

$A=-12.5 e^{-0.8 \times 1.339}+12.5=8.217\ \text{m (3 d.p.)}$

$\text {Vertical velocity}\ =36.5 e^{-0.8 t}-12.5$

$y=\displaystyle \int\left(36.5 e^{-0.8 t}-12.5\right)\,d t=\dfrac{36.5}{-0.8} e^{-0.8 t}-12.5 t+c_2=-45.625 e^{-0.8 t}-12.5 t+c_2$

$\text{When} \ \ t=0, \ y=0:$

$0=-45.625+c_2\ \ \Rightarrow\ \ c_2=45.625$

$y=-45.625 e^{-0.8 t}-12.5 t+45.625$

$\text{When}\ \ t=1.339, \ y=B:$

$B=-45.625 e^{-0.8 \times 1.339}-12.5 \times 1.339+45.625=13.257\ \text{m (3 d.p.)}$

c. $\text{At top of trajectory,}\ \mathbf{j} \text{-component of velocity = 0.}$

$\Rightarrow\ \text{Speed at top is the}\ \mathbf{i} \text{-component of velocity at}\ t=1.339:$

$\text{Speed}\ =10e^{-0.8 \times 1.339} = 3.426\ \text{m s}^{-1}\ \text{(3 d.p.)}$

d. $\text{Using}\ \ F=m \ddot{x}:$

$\mathbf{v}(t)=10 e^{-0.8 t}\mathbf{i} +\left(36.5 e^{-0.8 t}-12.5\right) \mathbf{j}$

$\mathbf{a} =\dfrac{d v }{d t}=-8 e^{-0.8 t} \mathbf{i} -29.2 e^{-0.8 t} \mathbf{j}$

$\text{When} \ \ t=1.339:$

$\mathbf{a}=-8 e^{-0.8 \times 1.339}\mathbf{i}-29.2 e^{-0.8 \times 1.339}\mathbf{j}=-2.74 1\,\mathbf{i} -10\, \mathbf{j}$

$\text{Magnitude of acceleration}$

$=\sqrt{(2.741)^2+(10)^2}=10.36885 \ldots\ \text{ms}^{-2}$

$\therefore \ \text{Magnitude of force}$

$=8 \times 10.36885 \ldots =82.951 \ \text{N (3 d.p.)}$

Mechanics, EXT2 2021 SPEC2 16 MC

An object of mass `m` kilograms slides down a smooth slope that is inclined at an angle of `theta^@` to the horizontal, where `0^@ < theta^@ < 45^@`. The acceleration of the object down the slope is `a\ text(ms)^(-2), a > 0`.

If the angle of inclination of the slope is doubled to `2theta^@`, then the acceleration of the object down the slope, in `text(ms)^(-2)`, is

`2a`
`(2a)/gsqrt(g^2-a^2)`
`(2a^2-g^2)/g`
`a/g sqrt(g^2-a^2)`

Show Answers Only

`B`

Show Worked Solution

`ma = mg\ sintheta`

`sintheta`	`= a/g`
`cos^2theta`	`= 1-(a^2)/(g^2)`
`costheta`	`= sqrt(1-(a^2)/(g^2))`

`text(If incline angle) = 2theta:`

`ma`	`= mgsin(2theta)`
`a`	`= g*2sinthetacostheta`
	`= g 2 a/g sqrt(1-(a^2)/(g^2))`
	`= (2a)/g sqrt(g^2-a^2)`

`=>\ B`

Financial Maths, STD2 EQ-Bank 29

Shown below is part of the output from a spreadsheet used to model a 25-year reducing balance loan with equal monthly repayments and a constant interest rate, $r$, expressed as a decimal.

Note that many rows and values have been removed from the output shown.

Show that the amount owing at the end of month 120 is $407 860.99. (4 marks)

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$\text{See worked solution}$

Show Worked Solution

$\text{Find } r:$

$\text{Interest}$	$= r \times \text{amount owing}$
$3500$	$= r \times 500\,000$
$r$	$= 0.007$

$\text{Find monthly repayment using PVIF table:}$

$\text{At } N = 300\ \text{ and }\ r = 0.007,\ \text{ PVIF} = 125.23492$

$\text{Monthly repayment}= \dfrac{500\,000}{125.23492}= \$3992.50$

$\text{Amount owing at start of month 120:}$

$\text{Amount owing at end of month 119} = \$408\,990.56$

$\therefore\ \text{Interest charged in month 120}$	$= 0.007 \times \$408\,990.56$
	$= \$2862.93$

$\text{Amount owing at end of month 120:}$

$= \$408\,990.56+\$2862.93-\$3992.50$

$= \$407\,860.99 \quad \checkmark$

Trigonometry, EXT1 EQ-Bank 7 MC

Which curve best represents the graph of the function $f(x)=-a \sin x+b \cos x$ given that the constants $a$ and $b$ are both positive?

Show Answers Only

$D$

Show Worked Solution

$\text{Method 1 (non-calculus)}$

$f(x)=-a \sin x+b \cos x$

$\text{At}\ \ x=0, \ f(x)=b \gt 0\ \ \text{(Eliminate A and C)}$

$\text{Express as auxiliary angle:}$

$f(x)$	$=-a \sin x+b \cos x$
	$=b \cos x-a \sin x$
	$=R \cos (x+\alpha)$

$\text{where}\ \ R=\sqrt{a^2+b^2}\ \ \text{and}\ \ \tan \alpha = \dfrac{a}{b}.$

$\text{Since}\ a, b \gt 0,\ \ 0 \lt \alpha \lt \dfrac{\pi}{2}.$

$\text{So}\ f(x)\ \text{is}\ \ y=\cos x\ \ \text{dilated by factor}\ R\ \text{from}\ x\text{-axis and shifted left by}\ \alpha.$

$\Rightarrow D$

$\text{Method 2 (with calculus)}$

$f(x)=-a \sin x+b \cos x,\ \ f^{′}(x)=-a \cos x-b \sin x$

$\text{By elimination:}$

$\text{At}\ \ x=0, \ f(x)=b \gt 0\ \ \text{(Eliminate A and C)}$

$\text{At}\ \ x=0, \ f^{′}(x)=-a \lt 0$

$\text{By inspection of graphs, option B gradient > 0 at}\ \ x=0\ \text{(Eliminate B)}$

$\Rightarrow D$

Vectors, EXT1 EQ-Bank 34

The position vector of a particle at time $t$ is given by $\mathbf{r}(t)=n e^{-2 t}\,\mathbf{i}-t^2\,\mathbf{j}$, where $n$ is a positive constant.

Determine $n$ if the particle's acceleration is perpendicular to its velocity when $t=\dfrac{1}{2}$. (3 marks)

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$n=\dfrac{e}{2}$

Show Worked Solution

$\underset{\sim}{r}(t)$	$=n e^{-2 t} \underset{\sim}{i}-t^2 \underset{\sim}{j}$
$\underset{\sim}{v}(t)$	$=-2n e^{-2t} \underset{\sim}{i}-2 t\underset{\sim}{i} \ \ \Rightarrow \ \ v\left(\frac{1}{2}\right)=-2 ne^{-1} \underset{\sim}{i}-\underset{\sim}{j}$
$\underset{\sim}{a}(t)$	$=4 ne^{-2t} \underset{\sim}{i}-2 \underset{\sim}{j} \ \ \Rightarrow \ \ a\left(\frac{1}{2}\right)=4ne^{-1} \underset{\sim}{i}-2 \underset{\sim}{j}$

$\text{Velocity}\perp \text{acceleration at}\ \ t=\dfrac{1}{2}:$

$\displaystyle \binom{-\tfrac{2 n}{e}}{-1}\binom{\tfrac{4 n}{e}}{-2}=0$

$-\dfrac{8 n^2}{e^2}+2=0 \ \ \Rightarrow \ \ n^2=\dfrac{e^2}{4} \ \ \Rightarrow \ \ n=\dfrac{e}{2}\ \ (n\gt 0)$

Vectors, EXT1 EQ-Bank 40

Determine the component of $\textbf{a} = 2\textbf{i}-\textbf{j} + 3\textbf{k}$ that is perpendicular to $\textbf{b} = \textbf{i} + \textbf{j}-\textbf{k}.$ (3 marks)

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$\textbf{a}-\operatorname{proj}_{\textbf{b}}\textbf{a}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)+\dfrac{2}{3}\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)=\left(\begin{array}{c}2 \frac{2}{3} \\ -\frac{1}{3} \\ 2 \frac{1}{3}\end{array}\right)$

Show Worked Solution

$\textbf{a}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right), \ \textbf{b}=\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)$

$\textbf{a} \cdot \textbf{b}=2-1-3=-2$

$\abs{\textbf{b}}^2=1^2+1^2+(-1)^2=3$

$\text{Projection of} \ \textbf{a} \ \text{in the direction of} \ \textbf{b}$:

$\operatorname{proj}_{\textbf{b}} \textbf{a}=\left(\dfrac{\textbf{a} \cdot \textbf{b}}{\abs{\textbf{b}}^2}\right) \textbf{b}=-\dfrac{2}{3}\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)$

$\text {Component of} \ \textbf{a} \ \text {that is perpendicular to} \ \textbf{b}$:

Vectors, EXT1 EQ-Bank 38

Let $\underset{\sim}{a}=2 \underset{\sim}{i}-3 j+\underset{\sim}{k}$ and $\underset{\sim}{b}=\underset{\sim}{i}+m j-\underset{\sim}{k}$, where $m$ is an integer.

The vector resolute of $\underset{\sim}{a}$ in the direction of $\underset{\sim}{b}$ is $-\dfrac{11}{18}(\underset{\sim}{i}+m\underset{\sim}{j}-\underset{\sim}{k})$.

Find the value of $m$. (3 marks)

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Find the component of $\underset{\sim}{a}$ that is perpendicular to $\underset{\sim}{b}$. (1 mark)

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a. $m=4$

b. $\left(\begin{array}{c}2 \frac{11}{18} \\ -\frac{5}{9} \\ \frac{7}{18}\end{array}\right)$

Show Worked Solution

a. $\underset{\sim}{a}=\left(\begin{array}{c}2 \\ -3 \\ 1\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}1 \\ m \\ -1\end{array}\right)$

$\underset{\sim}{b} \cdot \underset{\sim}{a}=2-3 m-1=1-3 m$

$\abs{\underset{\sim}{b}}=\sqrt{1^2+m^2+(-1)^2}=\sqrt{2+m^2}$

$\operatorname{proj}_{\underset{\sim}{b}}\underset{\sim}{a}=\left(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{b}^2}\right) \underset{\sim}{b}=\dfrac{1-3 m}{2+m^2}\, \underset{\sim}{b}$

$\text{Equating projection vectors:}$

$\dfrac{1-3 m}{m^2+2}$	$=-\dfrac{11}{18}$
$18-54 m$	$=-11 m^2-22$

$11 m^2-54 m+40$	$=0$
$(11 m-10)(m-4)$	$=0$

$\therefore m=4\ \left(m \neq \frac{10}{11}, m \in Z\right)$

b. $\text{Component of \(\underset{\sim}{a}$ perpendicular to $\underset{\sim}{b}$:}\)

$\underset{\sim}{a}-\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{a}=\left(\begin{array}{c}2 \\ -3 \\ 1\end{array}\right)+\dfrac{11}{18}\left(\begin{array}{c}1 \\ 4 \\ -1\end{array}\right)=\left(\begin{array}{c}2 \frac{11}{18} \\ -\frac{5}{9} \\ \frac{7}{18}\end{array}\right)$

Vectors, EXT1 2014 SPEC1 1

Consider the vector `underset ~a = sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k`, where `underset ~i, underset ~j` and `underset ~k` are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

Find the unit vector in the direction of `underset ~a`. (1 mark)

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Find the acute angle that `underset ~a` makes with the positive direction of the `x`-axis. (2 marks)

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The vector `underset ~b = 2 sqrt 3 underset ~i + m underset ~j-5 underset ~k`.
Given that `underset ~b` is perpendicular to `underset ~a,` find the value of `m`. (2 marks)

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a. `1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`

b. `theta = 45^@`

c. `m = 6 + 5 sqrt 2`

Show Worked Solution

a. `|underset ~a|= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)= sqrt 6`

`hat underset ~a= underset ~a/|underset ~a|= 1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`

b. `x text{-axis vectors include}\ (1,0,0).`

`underset ~a ⋅ underset ~i = ((\sqrt3),(-1),(-\sqrt2))((1),(0),(0))=\sqrt3`

	`underset ~a ⋅ underset ~i`	`= \|underset ~a\|\|underset ~i\| cos theta= sqrt 6 cos theta`
	`sqrt 3`	`= sqrt 6 cos theta`
	`cos theta`	`=1/sqrt 2`
	`:. theta`	`= 45^@`

c. `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6-m + 5 sqrt 2`	`=0`
`:. m`	`=6 + 5 sqrt 2`

Calculus, EXT1 EQ-Bank 28

Given the differential equation $\dfrac{d y}{d x}=-\dfrac{x}{y e^{x^2}}$, determine the particular solution that passes through the point $(0,1 )$. (3 marks)

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$y=e^{-\tfrac{x^2}{2}}$

Show Worked Solution

$\dfrac{d y}{d x}$	$=-\dfrac{x}{y e^{x^{2}}}$
$\displaystyle \int y\, d y$	$=-\displaystyle \int x e^{-x^2}\, d x$
$\dfrac{y^2}{2}$	$=\displaystyle \dfrac{1}{2} \int(-2 x) e^{-x^2}\, d x$
$y^2$	$=e^{-x^2}+c$

$\text{Given the solution passes through}\ (0,1):$

$1^2=e^0+c \ \ \Rightarrow \ \ c=0$

$y^2=e^{-x^2}$

$y=\left(e^{-x^2}\right)^{\tfrac{1}{2}}=e^{-\tfrac{x^2}{2}}$

Vectors, EXT1* V1 2025 HSC 11d

Force ${\underset{\sim}{F}}_1$ has magnitude 12 newtons in the direction of vector $2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}$.
Show that ${\underset{\sim}{F}}_1=8 \underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}$. (1 mark)

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Force ${\underset{\sim}{F}}_1$ from part (i) and a second force, ${\underset{\sim}{F}}_2=-6 \underset{\sim}{i}+12 \underset{\sim}{j}+4 \underset{\sim}{k}$, both act upon a particle.
Show that the resultant force acting on the particle is given by:
${\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}.$ (1 mark)

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Calculate ${\underset{\sim}{F}}_3 \cdot \underset{\sim}{d}$, where ${\underset{\sim}{F}}_3$ is the resultant force from part (ii) and $\underset{\sim}{d}=\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k}$. (1 mark)

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i. $\text{Unit vector of the direction vector:}$

$\dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}} = \dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\sqrt{2^2+(-2)^2 + 1^2}} = \dfrac{1}{3} \left( 2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k} \right)$

$\text{Since \({\underset{\sim}{F}}_1$ has magnitude 12:}\)

${\underset{\sim}{F}}_1=12 \times \dfrac{1}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)$

${\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}$

ii. ${\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)$

${\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}$

iii. ${\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22$

Show Worked Solution

i. $\text{Unit vector of the direction vector:}$

$\text{Since \({\underset{\sim}{F}}_1$ has magnitude 12:}\)

${\underset{\sim}{F}}_1=12 \times \dfrac{1}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)$

${\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}$

${\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}$

iii. ${\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22$

Calculus, EXT1 C2 EQ-Bank 30

Solve $ \displaystyle\int_0^1 \dfrac{1}{(1+x^2)^\tfrac{3}{2}}\, dx $ using the substitution $x=\tan \theta$. (3 marks)

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$\dfrac{1}{\sqrt{2}}$

Show Worked Solution

$\text {Let} \ \ x=\tan \theta$

$\dfrac{d x}{d \theta}=\sec ^2 \theta \ \Rightarrow \ d x=\sec ^2 \theta \, d \theta$

$\text{Also,} \ \ 1+x^2=1+\tan ^2 \theta=\sec ^2 \theta$

$\left(1+x^2\right)^{\tfrac{3}{2}}=(\sec ^2 \theta)^{\tfrac{3}{2}}=\sec ^3 \theta$

$\text{Adjust limits:}$

$x=0 \Rightarrow \theta=0, \ \ x=1 \Rightarrow \theta=\dfrac{\pi}{4}$

$\displaystyle\int_0^1 \dfrac{1}{\left(1+x^2\right)^{\tfrac{3}{2}}} \, d x$	$=\displaystyle\int_0^{\tfrac{\pi}{4}} \frac{\sec ^2 \theta}{\sec ^3 \theta} \, d \theta$
	$=\displaystyle \int_0^{\tfrac{\pi}{4}} \cos \theta \, d \theta$
	$=\Big[\sin \theta\Big]_0^{\tfrac{\pi}{4}}$
	$=\sin \dfrac{\pi}{4}-\sin 0$
	$=\dfrac{1}{\sqrt{2}}$

Calculus, 2ADV EQ-Bank 22

The diagram shows the graph of $y=\log _e(x+1)$

Express $x$ as a function of $y$. (1 mark)

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Hence, or otherwise, find the exact area of the shaded region bounded by the curve $y=\log _e(x+1)$, the $x$-axis, and the line $x=3$. (3 marks)

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a. $x=e^y-1$

b. $A=4 \ln 4-3 \ \text{u}^2$

Show Worked Solution

a. $y=\ln (x+1) \ \Rightarrow \ x+1=e^y \ \Rightarrow \ x=e^y-1$

b. $\text{Area of rectangle}=\ln 4 \times 3=3 \ln 4$

$\text{Find the area between curve and \(y$-axis from $\ y=0\ $ to $\ y=\ln 4$:}\)

$A$	$=\displaystyle \int_0^{\ln 4} e^y-1\, d y$
	$=\Big[e^y-y\Big]_0^{\ln 4}$
	$=\left(e^{\ln 4}-\ln 4\right)-(1)$
	$=4-\ln 4-1$
	$=3-\ln 4$

$\text{Shaded Area}$	$=3 \ln 4-(3-\ln 4)$
	$=4 \ln 4-3 \ \text{u}^2$

Calculus, 2ADV EQ-Bank 29

The diagram shows the graph of $y=\log _2 2 x$

Determine the exact value of the shaded area bounded by the $x$-axis, the $y$-axis, and the curve $y=\log _2 2 x$.

Express your answer is the form $\dfrac{a}{\ln b}$, where $a$ and $b$ are integers. (4 marks)

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$A=\dfrac{7}{\ln 4}\ \text{u}^2$

Show Worked Solution

$y=\log _2(2 x) \ \Rightarrow \ 2 x=2^y \ \Rightarrow \ x=\dfrac{1}{2} \times 2^y$

$A$	$=\dfrac{1}{2} \displaystyle \int_0^3 2^y\, d y$
	$=\dfrac{1}{2}\left[\dfrac{2^y}{\ln 2}\right]_0^3$
	$=\dfrac{1}{2}\left[\dfrac{2^3}{\ln 2}-\dfrac{1}{\ln 2}\right]$
	$=\dfrac{7}{2 \ln 2}$
	$=\dfrac{7}{\ln 4}\ \text{u}^2$

Statistics, 2ADV EQ-Bank 34

All the students in a class of 30 did a test.

The marks, out of 10, are shown in the dot plot.

Find the median test mark. (1 mark)

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The mean test mark is 5.4. The standard deviation of the test marks is 4.22.
Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean. (2 marks)
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A student states that for any data set, 68% of the scores should lie within one standard deviation of the mean. With reference to the dot plot, explain why the student’s statement is NOT relevant in this context. (1 mark)

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i. `6`

ii. `text(43%)`

iii. `text(The statement assumes the data is normally distributed which is incorrect.)`

Show Worked Solution

♦ Mean mark (i) 50%.
♦♦Mean mark (ii) 34%.

i. `text(Median)= text(15th + 16th score)/2= (4 + 8)/2= 6`

ii. `text(Lower limit) = 5.4-4.22 = 1.18`

`text(Upper limit) = 5.4 + 4.22 = 9.62`

`:.\ text(Percentage in between)`

`= 13/30 xx 100`

`= 43.33…`

`= 43text{% (nearest %)}`

iii. `text(The statement assumes the data is normally distributed.)`

♦♦♦ Mean mark (iii) 13%.

`text(This is incorrect in this case.)`

Statistics, 2ADV EQ-Bank 9 MC

The scores on an examination are normally distributed with a mean of 70 and a standard deviation of 6. Michael received a score on the examination between the lower quartile and the upper quartile of the scores.

Which shaded region most accurately represents where Michael's score lies?

A.		B.
C.		D.

Show Answers Only

`A`

Show Worked Solution

`text{68% of marks lie between 64 and 76 (mean ± 1 σ).}`

`text(50% of marks lie between)\ Q_1\ text(and)\ Q_3.`

`=> A`

Statistics, 2ADV S3 EQ-Bank 24

A continuous random variable $X$ has probability density function $f(x)$ given by

\begin{align*}
f(x)=\left\{\begin{array}{cl}
k x(1-x)^5, & \text { for } 0 \leq x \leq 1 \\
0, & \text { for all other values of } x
\end{array}\ \ \ , \text { where } k\right. \text { is a constant. }
\end{align*}

It is given that

$\displaystyle \int_0^a x(1-x)^5\, d x=\frac{1}{42}+\frac{(1-a)^7}{7}-\frac{(1-a)^6}{6}$

and $\displaystyle\int_0^1 x^m(1-x)^5\, d x=\dfrac{120}{(m+1)(m+2)(m+3)(m+4)(m+5)(m+6)}$

where $a>0$ and $m>0$.

Show that $k=42$. (1 mark)

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Show that $E (X)=0.25$. (2 marks)

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Show that the median of $X$ is less than the expected value of $X$. (3 marks)

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a. $k \displaystyle \int_0^1 x(1-x)^5 d x=1$

$k\left[\dfrac{1}{42}+\dfrac{(1-1)^7}{7}-\dfrac{(1-1)^6}{6}\right]=1$

$\dfrac{k}{42}$	$=1$
$k$	$=42$

b.	$E (X)$	$=\displaystyle \int_0^1 x \times f(x)\, d x$
		$=\displaystyle \int_0^1 42 x^2(1-x)^5\, d x$
		$=42 \times \dfrac{120}{3 \times 4 \times 5 \times 6 \times 7 \times 8}$
		$=0.25$

c. $\text{Let}\ m =\text{ median}$

$P(X\leqslant m) = 0.5\ \ \Rightarrow\ \ \displaystyle \int_0^m 42 x(1-x)^5\, d x=0.5 $

$E(X)=0.25$

$\text{Calculate }\ P(X\leqslant 0.25):$

$\displaystyle \int_0^{0.25} 42 x(1-x)^5\, d x$	$=42\left[\frac{1}{42}+\dfrac{(1-0.25)^7}{7}-\dfrac{(1-0.25)^6}{6}\right]$
	$=0.555 \ldots\ \text{(3 dp)}$

$\therefore \displaystyle \int_0^m 42 x(1-x)^5 d x=0.5 \ \ \text{requires the median to be less than 0.25.}$

Show Worked Solution

a. $k \displaystyle \int_0^1 x(1-x)^5 d x=1$

$k\left[\dfrac{1}{42}+\dfrac{(1-1)^7}{7}-\dfrac{(1-1)^6}{6}\right]=1$

$\dfrac{k}{42}$	$=1$
$k$	$=42$

b.	$E (X)$	$=\displaystyle \int_0^1 x \times f(x)\, d x$
		$=\displaystyle \int_0^1 42 x^2(1-x)^5\, d x$
		$=42 \times \dfrac{120}{3 \times 4 \times 5 \times 6 \times 7 \times 8}$
		$=0.25$

c. $\text{Let}\ m =\text{ median}$

$P(X\leqslant m) = 0.5\ \ \Rightarrow\ \ \displaystyle \int_0^m 42 x(1-x)^5\, d x=0.5 $

$E(X)=0.25$

$\text{Calculate }\ P(X\leqslant 0.25):$

$\displaystyle \int_0^{0.25} 42 x(1-x)^5\, d x$	$=42\left[\frac{1}{42}+\dfrac{(1-0.25)^7}{7}-\dfrac{(1-0.25)^6}{6}\right]$
	$=0.555 \ldots\ \text{(3 dp)}$

$\therefore \displaystyle \int_0^m 42 x(1-x)^5 d x=0.5 \ \ \text{requires the median to be less than 0.25.}$

Calculus, 2ADV C1 EQ-Bank 32

The graph of $y=x^4$ is dilated horizontally by a factor of $k$ where $k>0$. The normal to this dilated graph at $x=k$ intersects the $y$-axis at $(0,5)$.

Find the value of $k$. (4 marks)

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$k=4$

Show Worked Solution

$\text{Transformed graph has equation} \ y=\dfrac{x^4}{k^4}$

$\dfrac{d y}{d x}=\dfrac{4 x^3}{k^4}$

$\text{At }\ x=k:$

$m_{\text{tang}} = \dfrac{4 k^3}{k^4}=\dfrac{4}{k}$

$m_{\text{norm}} =-\dfrac{k}{4}\ (m_1m_2=-1)$

$\text{Equation of normal:}$

$y-1=-\dfrac{k}{4}(x-k)\ \ \Rightarrow\ \ y=-\dfrac{k}{4}x+1+\dfrac{k^2}{4}$

$\text{When }\ x=0, \ y=1+\dfrac{k^2}{4}$

$\text{\(y$-intercept at }\left(0,1+\dfrac{k^2}{4}\right)\)

$1+\dfrac{k^2}{4}=5\ \ \Rightarrow\ \ k^2=16$

$\therefore k=4\ \ (k>0)$.

Statistics, 2ADV EQ-Bank 37

The probability density function for the normal distribution with mean $\mu$ and standard deviation $\sigma$ is

$f(x)=\dfrac{1}{\sigma \sqrt{2 \pi}} e^{-\tfrac{(x-\mu)^2} {2 \sigma^2}}$

The graph of $y=e^{-\tfrac{1}{2}(x-1.5)^2}$ is shown. The point $M$ is a local maximum.

Using a $z$-score table of values, calculate the area of the shaded region. Give your answer correct to three decimal places. (4 marks)

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$\text{Comparing the graph to the normal distribution PDF:}$

$\mu=1.5, \ \sigma=1$

$e^{-\tfrac{1}{2}(x-1.5)^2} = \sqrt{2 \pi} \times f(x)$

$\Rightarrow\ \text{Total area under the curve} = \sqrt{2 \pi}\ \text{u}^2$

$\text{Convert the \(x$-values to $z$-scores:}\)

$\text{When }\ x=0:\ \ z=\dfrac{0-1.5}{1}=-1.5 $

$\text{When }\ x=1.5:\ \ z=\dfrac{1.5-1.5}{1}=0 $

$P(Z \leqslant 0)=0.5000$

$P(Z \leqslant 1.5)=0.9332\ \ \text{(from table)}$

$P(Z \leqslant -1.5)=1-0.9332=0.0668\ \ \text{(by symmetry)}$

$P(-1.5 \leqslant Z \leqslant 0)=0.5000-0.0668=0.4332$

$\text{Area under curve} = 0.4332 \times \sqrt{2\pi} \approx 1.08587$

$\text {Shaded area}$	$=\ \text{Area of rectangle}-\text{Area under curve}$
	$=(1.5 \times 1)-1.08587 \ldots$
	$=0.414 \ \text{u}^2 \ \text{(3 d.p.)}$

Show Worked Solution

$\text{Comparing the graph to the normal distribution PDF:}$

$\mu=1.5, \ \sigma=1$

$e^{-\tfrac{1}{2}(x-1.5)^2} = \sqrt{2 \pi} \times f(x)$

$\Rightarrow\ \text{total area under the curve} = \sqrt{2 \pi}\ \text{u}^2$

$\text{Convert the \(x$-values to $z$-scores:}\)

$\text{When }\ x=0:\ \ z=\dfrac{0-1.5}{1}=-1.5 $

$\text{When }\ x=1.5:\ \ z=\dfrac{1.5-1.5}{1}=0 $

$P(Z \leqslant 0)=0.5000$

$P(Z \leqslant 1.5)=0.9332\ \ \text{(from table)}$

$P(Z \leqslant -1.5)=1-0.9332=0.0668\ \ \text{(by symmetry)}$

$P(-1.5 \leqslant Z \leqslant 0)=0.5000-0.0668=0.4332$

$\text{Area under curve} = 0.4332 \times \sqrt{2\pi} \approx 1.08587$

$\text {Shaded area}$	$=\ \text{Area of rectangle}-\text{Area under curve}$
	$=(1.5 \times 1)-1.08587 \ldots$
	$=0.414 \ \text{u}^2 \ \text{(3 d.p.)}$

Probability, STD2 EQ-Bank 28

History and Geography are two of the subjects students may decide to study. For a group of 40 students, the following is known.

- 7 students study neither History nor Geography
- 20 students study History
- 18 students study Geography

Draw a Venn diagram that represents the information given, and hence find the number of students that study both History and Geography. (2 marks)

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A student is chosen at random. Determine the probability that the students studies Geography only. (1 mark)

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a. $\text{Venn diagram:}$

$n\text{(study H and G)}= 5$

b. $P(\text{study G only}) = \dfrac{13}{40}=32.5\% $

Show Worked Solution

a. $\text{Venn diagram:}$

$n\text{(study H and G)}= 5$

b. $P(\text{study G only}) = \dfrac{13}{40}=32.5\% $

Probability, STD2 EQ-Bank 30

In a workplace of 25 employees, each employee speaks either French or German, or both.

If 36% of the employees speak German, and 20% speak both French and German.

Draw a Venn diagram that represents the information given. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
If one person is chosen at random, what is the probability they can speak French but cannot speak German? (2 marks)

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a. `text(Venn diagram:)`

b. $\text{Number who can speak French but not German = 16}$

$P(F\ \text{but not}\ G)=\dfrac{16}{25} = 64\%$

Show Worked Solution

a. `text(Venn diagram:)`

b. $\text{Number who can speak French but not German = 16}$

$P(F\ \text{but not}\ G)=\dfrac{16}{25} = 64\%$

Algebra, STD2 EQ-Bank 39

The graph of the parabola $y=a(x+1)(x+7)$ for some value of $a$ is shown.

By first finding the value of $a$, find the coordinates of the vertex. (3 marks)

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$\text{Vertex:}\ (-4,-27)$

Show Worked Solution

$\text{Since graph passes through}\ (0,21):$

$21$	$=a(0+1)(0+7)$
$21$	$=7a$
$a$	$=3$

$\text{Vertex is halfway between \(x$-intercepts.}\)

$\Rightarrow \ x=\dfrac{-7+(-1)}{2}=-4$

$y$	$=3(-4+1)(-4+7)$
	$=3 \times (-3) \times 3$
	$=-27$

$\therefore\ \text{Vertex at}\ (-4,-27).$

Algebra, STD2 EQ-Bank 38

The graph of the parabola $y=a(x+2)(x-6)$ for some value of $a$ is shown.

By first finding the value of $a$, find the coordinates of the vertex. (3 marks)

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$\text{Vertex:}\ (2,32)$

Show Worked Solution

$\text{Since graph passes through}\ (0,24):$

$24$	$=a(0+2)(0-6)$
$24$	$=-12a$
$a$	$=-2$

$\text{Vertex is halfway between \(x$-intercepts.}\)

$\Rightarrow \ x=\dfrac{-2+6}{2}=2$

$y$	$=-2(2+2)(2-6)$
	$=-2 \times 4 \times (-4)=32$

$\therefore\ \text{Vertex at}\ (2,32).$

\(\text{Annuity}\times 75.42414\)	\(=\$28\,182.26\)
\(\text{Annuity}\)	\(=\dfrac{28\,182.26}{75.42414}=\$373.65\)

\(6200\)	\(=\text{Payment}\times 8.8923\)
\(\text{Payment}\)	\(=\dfrac{6200}{8.8923}=697.23\ldots=$700\ \text{(nearest \$10)}\)

\(\text{Amount owing}\)	\(=4200\left(1+\dfrac{0.17}{365}\right)^{50}\)
	\(=4298.931\ldots\)
	\(=\$4298.93\ \text{(nearest cent)}\)

\(I=Prn\)	\(=4200\times 0.089\times\dfrac{75}{365}\)
	\(=76.808\ldots\)
	\(=\$76.81\ \text{(nearest cent)}\)

\(FV\)	\(=PV(1+r)^n\)
	\(=680\left(1+\dfrac{0.159}{365}\right)^{28}\)
	\(=\$688.34\)

\(\text{End of month balance}\)	\(=10\,508.52+21.02+400\)
	\(=\$10\,929.54\)

\(\text{Minimum payment}\)	\(=\dfrac{3}{100}\times 2613.92\)
	\(=78.4176\ldots\)
	\(=\$78.42\ \text{(nearest cent)}\)

\(I=Prn\)	\(=2500\times 0.12\times\dfrac{40}{365}\)
	\(=32.876\ldots\)
	\(=\$32.88\ \text{(nearest cent)}\)

b.	\(\text{Tax payable}\)	\(= 0.16 \times (38\,790-18\,200)\)
		\(= 0.16 \times 20\,590\)
		\(= $3294.40\)