L.C. MATHS

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Past Papers

## Question 1

(a) Write the function $$f(x)=2x^2-7x-10$$, where $$x\in\mathbb{R}$$, in the form $$a(x+h)^2+k$$, where $$a$$, $$h$$ and $$k\in\mathbb{Q}$$.

$$f(x)=2\left(x-\dfrac{7}{4}\right)^2-\dfrac{129}{8}$$

Solution

\begin{align}f(x)&=2x^2-7x-10\\&=2\left(x^2-\frac{7}{2}x-5\right)\\&=2\left[\left(x-\frac{7}{4}\right)^2-\frac{129}{16}\right]\\&=2\left(x-\frac{7}{4}\right)^2-\frac{129}{8}\end{align}

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(b) Hence, write the minimum point of ݂$$f$$.

$$\left(\dfrac{7}{4},-\dfrac{129}{8}\right)$$

Solution

$$\left(\dfrac{7}{4},-\dfrac{129}{8}\right)$$

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(c)

(i) Explain why ݂$$f$$ must have two real roots.

(ii) Write the roots of ݂$$f(x)=0$$ in the form $$p\pm q$$,where $$p$$ and $$q\in\mathbb{Q}$$.

(i) The discriminant is greater than zero.

(ii) $$\dfrac{7}{4}\pm\sqrt{\dfrac{129}{16}}$$

Solution

(i)

\begin{align}b^2-4ac&=(-7)^2-4(2)(-10)\\&=49+80\\&=129\end{align}

As the discriminant is greater than zero, there must be two real roots.

(ii)

\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(-10)}}{2(2)}\\&=\frac{7\pm\sqrt{129}}{4}\\&=\frac{7}{4}\pm\sqrt{\frac{129}{16}}\end{align}

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## Question 2

$$z=-\sqrt{3}+i$$, where $$i^2=-1$$.

(a) Use De Moivre’s Theorem to write $$z^4$$ in the form ܽ$$a+b\sqrt{c}i$$, where ܽ$$a$$, $$b$$, and $$c\in\mathbb{Z}$$.

$$z^4=-8-8\sqrt{3}i$$

Solution

Modulus

\begin{align}r&=\sqrt{(-\sqrt{3})^2+1^2}\\&=\sqrt{3+1}\\&=2\end{align}

$\,$

Argument

\begin{align}\tan\theta=-\frac{1}{\sqrt{3}}\end{align}

Reference Angle:

\begin{align}\tan A=\frac{1}{\sqrt{3}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A=\frac{\pi}{6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta&=\pi-\frac{\pi}{6}\\&=\frac{5\pi}{6}\end{align}

$\,$

Polar Form

\begin{align}z=2(\cos\left(\frac{5\pi}{6}\right)+i\sin\left(\frac{5\pi}{6}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z^4&=2^4(\cos\left(\frac{5(4)\pi}{6}\right)+i\sin\left(\frac{5(4)\pi}{6}\right)\\&=-8-8\sqrt{3}i\end{align}

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(b) The complex number $$w$$ is such that $$|w|=3$$ and $$w$$ makes an angle of $$30^{\circ}$$ with the positive sense of the real axis. If $$t=zw$$, write $$t$$ in its simplest form.

$$t=-6$$

Solution

\begin{align}w=3\left[\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right]\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=zw\\&=2\left[\cos\left(\frac{5\pi}{6}\right)+i\sin\left(\frac{5\pi}{6}\right)\right]\times3\left[\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right]\\&=6(\cos\pi+i\sin\pi)\\&=6(-1+0i)\\&=-6\end{align}

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## Question 3

(a) Differentiate $$\dfrac{1}{3}x^2-x+3$$ from first principles with respect to $$x$$.

$$\dfrac{2}{3}x-1$$

Solution

\begin{align}f(x)=\frac{1}{3}x^2-x+3\end{align}

and

\begin{align}f(x+h)&=\frac{1}{3}(x+h)^2-(x+h)+3\\&=\frac{1}{3}x^2+\frac{2}{3}xh+\frac{1}{3}h^2-x-h+3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{df}{dx}&=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\\&=\lim_{h\rightarrow0}\frac{\frac{1}{3}x^2+\frac{2}{3}xh+\frac{1}{3}h^2-x-h+3-(\frac{1}{3}x^2-x+3)}{h}\\&=\lim_{h\rightarrow0}\frac{\frac{2}{3}xh+\frac{1}{3}h^2-h}{h}\\&=\lim_{h\rightarrow0}\frac{2}{3}x+\frac{1}{3}h-1\\&=\frac{2}{3}x-1\end{align}

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(b) ݂$$f(x)=\ln(3x^2+2)$$ and $$g(x)=x+5$$, where $$x\in\mathbb{R}$$.
Find the value of the derivative of ݂$$f(g(x))$$ at $$x=\dfrac{1}{4}$$.
Give your answer correct to $$3$$ decimal places.

$$0.372$$

Solution

\begin{align}f(g(x))&=\ln[3(x+5)^2+2]\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f'(g(x))&=\frac{1}{3(x+5)^2+2}[6(x+5)]\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f’\left(g\left(\frac{1}{4}\right)\right)&=\frac{1}{3\left(\frac{1}{4}+5\right)^2+2}\left[6\left(\frac{1}{4}+5\right)\right]\\&\approx0.372\end{align}

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## Question 4

(a) The amount of a substance remaining in a solution reduces exponentially over time.
An experiment measures the percentage of the substance remaining in the solution.
The percentage is measured at the same time each day. The data collected over the first $$4$$ days are given in the table below. Based on the data in the table, estimate which is the first day on which the percentage of the substance in the solution will be less than $$0.01\%$$.

Day $$1$$ $$2$$ $$3$$ $$4$$

Percentage of substance (%)

$$95$$

$$42.75$$

$$19.2375$$

$$8.6569$$

The $$12$$th day.

Solution

\begin{align}r&=\frac{42.75}{95}\\&=\frac{9}{20}\end{align}

\begin{align}T_n<0.01\end{align}

\begin{align}\downarrow\end{align}

\begin{align}ar^{n-1}<0.01\end{align}

\begin{align}\downarrow\end{align}

\begin{align}95\left(\frac{9}{20}\right)^{n-1}<0.01\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\left(\frac{9}{20}\right)^{n-1}<\frac{0.01}{95}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\ln\left(\frac{9}{20}\right)^{n-1}<\ln\left(\frac{0.01}{95}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(n-1)\ln\left(\frac{9}{20}\right)<\ln\left(\frac{0.01}{95}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n>1+\frac{\ln\left(\frac{0.01}{95}\right)}{\ln\left(\frac{9}{20}\right)}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n>12.47\end{align}

\begin{align}\downarrow\end{align}

$$12$$th day

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(b) A square has sides of length $$2\mbox{ cm}$$. The midpoints of the sides of this square are joined to form another square. This process is continued.
The first three squares in the process are shown below.
Find the sum of the perimeters of the squares if this process is continued indefinitely.
Give your answer in the form $$a+b\sqrt{c}\mbox{ cm}$$, where $$a$$, $$b$$ and $$c\in\mathbb{N}$$.

$$16+8\sqrt{2}\mbox{ cm}$$

Solution

\begin{align}S=4(2)+4\sqrt{2}+4+…\end{align}

\begin{align}a=8&&r=\frac{1}{\sqrt{2}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{8}{1-\frac{1}{\sqrt{2}}}\\&=\frac{8}{1-\frac{1}{\sqrt{2}}}\times\frac{1+\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}\\&=\frac{8+\frac{8}{\sqrt{2}}}{1-\frac{1}{2}}\\&=16+\frac{16}{\sqrt{2}}\\&=16+8\sqrt{2}\mbox{ cm}\end{align}

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## Question 5

The function $$f$$ is such that $$f(x)=2x^3+5x^2-4x-3$$, where $$x\in\mathbb{R}$$.

(a) Show that $$x=-3$$ is a root of ݂$$f(x)$$ and find the other two roots.

$$x=-3$$, $$x=-\dfrac{1}{2}$$ and $$x=1$$

Solution

\begin{align}f(-3)&=2(-3)^3+5(-3)^2-4(-3)-3\\&=-54+45+12-3\\&=0\end{align}

Therefore, $$x=-3$$ is a root.

$\require{enclose} \begin{array}{rll} 2x^2-x-1\phantom{000000}\, \\[-3pt] x+3 \enclose{longdiv}{\,2x^3+5x^2-4x-3} \\[-3pt] \underline{2x^3+6x^2\phantom{00000000}\,\,} \\[-3pt] -x^2-4x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-x^2-3x\phantom{00}\,}\\[-3pt]\phantom{00}-x-3\\[-3pt]\phantom{00}\underline{-x-3}\\[-3pt]0 \end{array}$

\begin{align}\downarrow\end{align}

\begin{align}f(x)&=(x+3)(2x^2-x-1)\\&=(x+3)(2x+1)(x-1)\end{align}

\begin{align}\downarrow\end{align}

$$x=-3$$, $$x=-\dfrac{1}{2}$$ and $$x=1$$

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(b) Find the co-ordinates of the local maximum point and the local minimum point of the function $$f$$.

Maximum: $$(-2,9)$$

Minimum: $$\left(\dfrac{1}{3},-\dfrac{100}{27}\right)$$

Solution

\begin{align}f'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x^2+10x-4=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x^2+5x-2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+2)(3x-1)=0\end{align}

\begin{align}\downarrow\end{align}

$$x=-2$$ and $$x=\dfrac{1}{3}$$

$$f(-2)=9$$ and $$f\left(\dfrac{1}{3}\right)=-\dfrac{100}{27}$$

\begin{align}\downarrow\end{align}

Maximum: $$(-2,9)$$

and

Minimum: $$\left(\dfrac{1}{3},-\dfrac{100}{27}\right)$$

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(c) $$f(x)+a$$, where $$a$$ is a constant, has only one real root.
Find the range of possible values of $$a$$.

$$a<-9$$ or $$a>\dfrac{100}{27}$$

Solution

$$a<-9$$ or $$a>\dfrac{100}{27}$$

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## Question 6

The graph of the function $$g(x)=e^x$$, $$x\in\mathbb{R}$$, $$0\leq x\leq 1$$, is shown on the diagram below.

(a) On the same diagram, draw the graph of $$h(x)=e^{-x}$$, $$x\in\mathbb{R}$$, in the domain $$0\leq x\leq 1$$.

Solution
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(b) Find the area enclosed by $$g(x)=e^x$$, $$h(x)=e^{-x}$$ and the line $$x=0.75$$.
Give your answer correct to $$4$$ decimal places.

$$0.5894$$

Solution

\begin{align}A&=\int_0^{0.75}(g(x)-h(x)\,dx\\&=\int_0^{0.75}(e^x-e^{-x})\,dx\\&=\left.e^x+e^{-x}\right|_0^{0.75}\\&=(e^{0.75}+e^{-0.75})-(e^0+e^0)\\&\approx0.5894\end{align}

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## Question 7

Sometimes it is possible to predict the future population in a city using a function.
The population in Sapphire City, over time, can be predicted using the following function:

\begin{align}p(t)=Se^{0.1t}\times10^6\end{align}

The population in Avalon, over time, can be predicted using the following function:

\begin{align}q(t)=3.9e^{kt}\times10^6\end{align}

In the functions above, $$t$$ is time, in years; $$t=0$$ is the beginning of $$2010$$; and both $$S$$ and $$k$$ are constants.

(a) The population in Sapphire City at the beginning of $$2010$$ is $$1{,}100{,}000$$ people.
Find the value of $$S$$.

$$1.1$$

Solution

\begin{align}p(0)=1{,}100{,}000\end{align}

\begin{align}\downarrow\end{align}

\begin{align}Se^{0.1(0)}\times10^6=1{,}100{,}000\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S&=\frac{1{,}100{,}000}{10^6}\\&=1.1\end{align}

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(b) Find the predicted population in Sapphire City at the beginning of $$2015$$.

$$1{,}813{,}593$$

Solution

\begin{align}p(5)&=1.1e^{0.1(5)}\times10^6\\&\approx1{,}813{,}593\end{align}

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(c) Find the predicted change in the population in Sapphire City during $$2015$$.

$$190{,}737$$

Solution

\begin{align}p(6)-p(5)&=1.1e^{0.1(6)}\times10^6-1.1e^{0.1(5)}\times10^6\\&\approx190{,}737\end{align}

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(d) The predicted population in Avalon at the beginning of $$2011$$ is $$3{,}709{,}795$$ people.
Write down and solve an equation in $$k$$ to show that $$k=-0.05$$, correct to $$2$$ decimal places.

Solution

\begin{align}q(1)=3{,}709{,}795\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3.9e^{k(1)}\times10^6=3{,}709{,}795\end{align}

\begin{align}\downarrow\end{align}

\begin{align}e^k=\frac{3{,}709{,}795}{3.9\times10^6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k&=\ln\left(\frac{3{,}709{,}795}{3.9\times10^6}\right)\\&\approx-0.05\end{align}

as required.

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(e) Find the year during which the populations in both cities will be equal.

$$2018$$

Solution

\begin{align}p(t)=q(t)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.1e{0.1t}\times10^6=3.9e^{-0.05t}\times 10^6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}e^{0.15t}=\frac{39}{11}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.15t=\ln\left(\frac{39}{11}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{1}{0.15}\ln\left(\frac{39}{11}\right)\\&=8.44…\end{align}

The populations will be equal in $$2018$$.

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(f) Find the predicted average population in Avalon from the beginning of $$2010$$ to the beginning of $$2025$$.

$$2{,}743{,}694$$

Solution

\begin{align}\mbox{Average}&=\frac{1}{b-a}\int_a^bq(t)\,dt\\&=\frac{1}{15-0}\int_0^{15}(3.9e^{-0.05t}\times10^6)\,dt\\&=-\frac{3.9\times10^6}{15(0.05)}\left.e^{-0.05t}\right|_0^{15}\\&=-\frac{3.9\times10^6}{15(0.05)}(e^{-0.05(15)}-e^{-0.05(0)})\\&\approx2{,}743{,}694\end{align}

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(g) Use the function $$q(t)=3.9e^{-0.05t}\times 10^6$$ to find the predicted rate of change of the population in Avalon at the beginning of $$2018$$.

$$-130{,}712$$

Solution

\begin{align}q'(t)=-0.05(3.9e^{-0.05t})\times10^6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q'(8)&=-0.05(3.9e^{-0.05(8))}\times10^6\\&=-130{,}712\end{align}

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## Question 8

(a) When a loan of €$$P$$ is repaid in equal repayments of amount €$$A$$ ,at the end of each of $$t$$ equal periods of time, where $$i$$ is the periodic compound interest rate (expressed as a decimal), the formula below can be used to find the amount of each repayment.

\begin{align}A=P\frac{i(1+i)^t}{((1+i)^t-1)}\end{align}

Show how this formula is derived. You may use the formula for the sum of a finite geometric series.

Solution

\begin{align}S_n=\frac{a(1-r^n)}{1-r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P=\frac{a(1-r^t)}{1-r}\end{align}

\begin{align}a=\frac{A}{1+i}&&r=\frac{1}{1+i}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P&=\frac{\frac{A}{1+i}\left(1-\left(\frac{1}{1+i}\right)^t\right)}{1-\frac{1}{1+i}}\\&=\frac{A\left(1-\left(\frac{1}{1+i}\right)^t\right)}{1+i-1}\\&=\frac{A[(1+i)^t-1]}{i(1+i)^t}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A=P\frac{i(1+i)^t}{(1+i)^t-1}\end{align}

as required.

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(b) Alex has a credit card debt of €$$5000$$. One method of clearing this debt is to make a fixed repayment at the end of each month. The amount of this repayment is $$2.5\%$$ of the original debt.

(i) What is the fixed monthly repayment, €$$A$$, required to pay the debt of €$$5000$$?

(ii) The annual percentage rate (APR) charged on debt by the credit card company is $$21.75\%$$, fixed for the term of the debt. Find as a percentage, correct to $$3$$ significant figures, the monthly interest rate that is equivalent to an APR of $$21.75\%$$.

(iii) Assume Alex pays the fixed monthly repayment, €$$A$$,each month and does not have
any further transactions on that card. Complete the table below to show how the balance of the debt of €$$5000$$ is reducing each month for the first three months, assuming an APR of $$21.75\%$$, charged and compounded monthly.

Payment Number Fixed Monthly Payment Interest Previous balance reduced by New balance of debt

$$0$$

\

\

\

$$5{,}000$$

$$1$$

$$42.50$$

$$4{,}957.50$$

$$2$$

$$3$$

(iv) Using the formula you derived on the previous page, or otherwise, find how long it would take to pay off a credit card debt of €$$5000$$, using the repayment method outlined at the beginning of part (b) above.

(v) Alex decides to borrow €$$5000$$ from the local Credit Union to pay off this credit card debt of €$$5000$$. The APR charge for the Credit Union loan is $$8.5\%$$ fixed for the term of the loan. The loan is to be repaid in equal weekly repayments, at the end of each week, for $$156$$ weeks. Find the amount of each weekly repayment.

(vi) How much will Alex save by paying off the credit card debt using the loan from the Credit Union instead of paying the fixed repayment from part (b)(i) each month to the credit card company?

(i) $$125\mbox{ euro}$$

(ii) $$1.65\%$$.

(iii)

Payment Number Fixed Monthly Payment Interest Previous balance reduced by New balance of debt

$$0$$

\

\

\

$$5{,}000$$

$$1$$

$$125$$

$$82.50$$

$$42.50$$

$$4{,}957.50$$

$$2$$

$$125$$

$$81.80$$

$$43.20$$

$$4{,}914.30$$

$$3$$

$$125$$

$$81.09$$

$$43.91$$

$$4{,}870.39$$

(iv) $$66\mbox{ months}$$

(v) $$36.16\mbox{ euro}$$

(vi) $$2{,}609.04\mbox{ euro}$$

Solution

(i)

\begin{align}A&=5{,}000\times\left(\frac{2.5}{100}\right)\\&=125\mbox{ euro}\end{align}

(ii)

\begin{align}(1+i)^{1/12}&=(1.2175)^{1/12}\\&=1.01653…\end{align}

Therefore, the monthly interest rate is $$1.65\%$$.

(iii)

Payment Number Fixed Monthly Payment Interest Previous balance reduced by New balance of debt

$$0$$

\

\

\

$$5{,}000$$

$$1$$

$$125$$

$$82.50$$

$$42.50$$

$$4{,}957.50$$

$$2$$

$$125$$

$$81.80$$

$$43.20$$

$$4{,}914.30$$

$$3$$

$$125$$

$$81.09$$

$$43.91$$

$$4{,}870.39$$

(iv)

\begin{align}A=P\frac{i(1+i)^t}{(1+i)^t-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A[(1+i)^t-1]=Pi(1+i)^t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A(1+i)^t-A=Pi(1+i)^t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(1+i)^t(A-Pi)=A\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(1+i)^t=\frac{A}{A-Pi}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\ln(1+i)^t=\ln\left(\frac{A}{A-Pi}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t\ln(1+i)=\ln\left(\frac{A}{A-Pi}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{1}{\ln(1+i)}\times\ln\left(\frac{A}{A-Pi}\right)\\&=\frac{1}{\ln\left(1+\frac{1.65}{100}\right)}\times\ln\left(\frac{125}{125-5000\left(\frac{1.65}{100}\right)}\right)\\&\approx66\mbox{ months}\end{align}

(v)

\begin{align}A&=P\frac{i(1+i)^t}{(1+i)^t-1}\\&=(5000)\frac{(1.085^{1/52}-1)(1.085)^3}{(1.085)^3-1}\\&\approx36.16\mbox{ euro}\end{align}

(vi)

\begin{align}125\times66-(36.16)(156)=2{,}609.04\mbox{ euro}\end{align}

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## Question 9

The depth of water, in metres, at a certain point in a harbour varies with the tide and can be modelled by a function of the form

\begin{align}f(t)=a+b\cos ct\end{align}

where $$t$$ is the time in hours from the first high tide on a particular Saturday and $$a$$, $$b$$ and $$c$$ are constants. (Note: $$ct$$ is expressed in radians.)

On that Saturday, the following were noted:

• The depth of the water in the harbour at high tide was $$5.5\mbox{ m}$$
• The depth of the water in the harbour at low tide was $$1.7\mbox{ m}$$
• High tide occurred at $$02:00$$ and again at $$14:34$$.

(a) Use the information you are given to add, as accurately as you can, labelled and scaled axes to the diagram below to show the graph of ݂ over a portion of that Saturday.
The point $$P$$ should represent the depth of the water in the harbour at high tide on that Saturday morning.

Solution
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(b)

(i) Find the value of $$a$$ and the value of $$b$$.

(ii) Show that $$c=0.5$$, correct to $$1$$ decimal place.

(i) $$a=3.6$$ and $$b=1.9$$

Solution

(i)

\begin{align}a+b=5.5\end{align}

\begin{align}a-b=1.7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2a=7.2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a=3.6\end{align}

and

\begin{align}b&=a-1.7\\&=3.6-1.7\\&=1.9\end{align}

(ii)

\begin{align}T=\frac{2\pi}{f}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12\frac{34}{60}=\frac{2\pi}{c}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}c&=\frac{2\pi}{12\frac{34}{60}}\\&\approx0.5\end{align}

as required.

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(c) Use the equation$$f(t)=a+b\cos ct$$ to find the times on that Saturday afternoon when the depth of the water in the harbour was exactly $$5.2\mbox{ m}$$.
Give each answer correct to the nearest minute.

$$13\colon26$$ and $$15\colon42$$

Solution

\begin{align}f(t)=5.2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a+b\cos ct=5.2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3.6+1.9\cos0.5t=5.2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\cos0.5t=\frac{5.2-3.6}{1.9}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.5t=\cos^{-1}\left(\frac{5.2-3.6}{1.9}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=2\cos^{-1}\left(\frac{5.2-3.6}{1.9}\right)\\&=1.139…\mbox{ hrs}\\&\approx1\mbox{ hr }8\mbox{ min}\end{align}

Therefore, the times are $$14\colon34\pm1\mbox{ hr }8\mbox{ min}$$, i.e. $$13\colon26$$ and $$15\colon42$$.

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