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Mock Exams

HL Mock Exam A
Paper 1

Section A

Question 1

\begin{align}z^2-2z+1-2i=0\end{align}

has a root $$z_1=2+i$$, where $$i^2=-1$$.

(i) “According to the Conjugate Roots theorem, the other solution to this equation is $$z_2=2-i$$.”
Explain why this statement is incorrect.

(ii) Find the other root to this equation.

(i) The Conjugate Roots theorem cannot be used as the coefficients of the equation are not all real.

(ii) $$z_2=-i$$

(i) The Conjugate Roots Theorem only applies if the coefficients are real. In this case, this equation is of the form $$az^2+bz+c=0$$, but $$c$$ is not real.

(ii)

\begin{align}[z-(2+i)][z-(a+bi)]=z^2-2z+1-2i=0\end{align}

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

\begin{align}z[z-(a+bi)]-(2+i)[z-(a+bi)]=z^2-2z+1-2i\end{align}

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

\begin{align}z^2-(a+bi)z-2z+2(a+bi)-iz+i(a+bi)=z^2-2z+1-2i\end{align}

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

\begin{align}z^2-az-biz-2z+2a+2bi-iz+ai-b=z^2-2z+1-2i\end{align}

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

\begin{align}z^2+(-a-bi-2-i)z+(2a+2bi+ai-b)=z^2-2z+1-2i\end{align}

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

\begin{align}z^2+[(-a-2)+(-b-1)i]z+[(2a-b)+(2b+a)i]=z^2-2z+1-2i\end{align}

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

\begin{align}2a-b&=1\\2b+a&=-2\end{align}

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

\begin{align}b&=2a-1\\2b+a&=-2\end{align}

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

\begin{align}2(2a-1)+a=-2\end{align}

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

\begin{align}4a-2+a=-2\end{align}

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

\begin{align}5a=0\end{align}

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

\begin{align}a=0\end{align}

and

\begin{align}b&=2a-1\\&=2(0)-1\\&=-1\end{align}

Therefore, the other root is $$z_2=-i$$.

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(b) Write $$\sqrt{1+i\sqrt{3}}$$ in the form $$a+ bi$$, where $$a,b\in\mathbb{R}$$.

$$\pm\left(\dfrac{\sqrt{6}}{2}+\dfrac{1}{\sqrt{2}}i\right)$$

\begin{align}\sqrt{1+i\sqrt{3}}=a+bi\end{align}

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

\begin{align}1+i\sqrt{3}=(a+bi)^2\end{align}

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

\begin{align}1+i\sqrt{3}=a^2+2abi-b^2\end{align}

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

\begin{align}1+i\sqrt{3}=(a^2-b^2)+(2ab)i\end{align}

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

\begin{align}a^2-b^2&=1\\2ab&=\sqrt{3}\end{align}

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

\begin{align}a^2-b^2&=1\\a&=\frac{\sqrt{3}}{2b}\end{align}

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

\begin{align}\left(\frac{\sqrt{3}}{2b}\right)^2-b^2&=1=\end{align}

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

\begin{align}\frac{3}{4b^2}-b^2=1\end{align}

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

\begin{align}3-4b^4=4b^2\end{align}

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

\begin{align}4b^4+4b^2-3=0\end{align}

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

$$b^2=\dfrac{1}{2}$$ or $$b^2=-\dfrac{3}{2}$$

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

$$b=\pm\dfrac{1}{\sqrt{2}}$$

(as $$b\in\mathbb{R}$$)

and

\begin{align}a&=\frac{\sqrt{3}}{2b}\\&=\pm\frac{\sqrt{6}}{2}\end{align}

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

\begin{align}\sqrt{1+i\sqrt{3}}=\pm\left(\frac{\sqrt{6}}{2}+\frac{1}{\sqrt{2}}i\right)\end{align}

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Question 2

(a) Consider the following quadratic function:

\begin{align}h(x)=x^2-12x+50,\,x\in\mathbb{R}\end{align}

(i) Express $$h(x)$$ in the form $$(x-a)^2+b$$, where $$a,b\in\mathbb{Z}$$.

(ii) Hence, or otherwise, find the minimum value of $$\sqrt{h(x)}$$.

(i) $$(x-6)^2+14$$

(ii) $$\sqrt{14}$$

(i)

\begin{align}h(x)&=x^2-12x+50\\&=x^2-12x+36+14\\&=(x-6)^2+14\end{align}

(ii) The minimum value of $$\sqrt{(x-6)^2+14}$$ occurs when $$x=6$$, i.e. when $$\sqrt{h(x)}=\sqrt{14}$$.

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(b) Consider the following cubic function:

\begin{align}f(x)=2x^3-9x^2-11x+30\end{align}

(i) Using the factor theorem, or otherwise, show that $$(x-5)$$ is a factor of $$f(x)$$.

(ii) Hence, or otherwise, find the remaining two factors of $$f(x)$$.

(ii) $$(2x-3)$$ and $$(x+2)$$

(i)

\begin{align}f(5)&=2(5^3)-9(5^2)-11(5)+30\\&=250-225-55+30\\&=0\end{align}

as required.

(ii)

\begin{align}(x-5)(2x^2+bx+c)=2x^3-9x^2-11x+30\end{align}

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

\begin{align}2x^3+bx^2+cx-10x^2-5bx-5c=2x^3-9x^2-11x+30\end{align}

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

\begin{align}2x^3+(b-10)x^2+(c-5b)x-5c=2x^3-9x^2-11x+30\end{align}

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

\begin{align}b-10&=-9\\c-5b&=-11\\-5c&=30\end{align}

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

$$b=1$$ and $$c=-6$$

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

\begin{align}2x^3-9x^2-11x+30=(x-5)(2x^2+x-6)\end{align}

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

\begin{align}2x^3-9x^2-11x+30=(x-5)(2x-3)(x+2)\end{align}

Therefore, the other factors are $$(2x-3)$$ and $$(x+2)$$.

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Question 3

(a) Differentiate $$f(x)=5x^2-7$$ with respect to $$x$$, from first principles.

$$10x$$

\begin{align}\frac{df}{dx}&=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\\&=\lim_{h\rightarrow0}\frac{5(x+h)^2-7-(5x^2-7)}{h}\\&=\lim_{h\rightarrow0}\frac{5x^2+10xh+5h^2-7-5x^2+7}{h}\\&=\lim_{h\rightarrow0}\frac{10xh+5h^2}{h}\\&=\lim_{h\rightarrow0}(10x+5h)\\&=10x\end{align}

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(b) Consider the following linear function:

\begin{align}f^\prime(x)=4x+7\end{align}

where $$f^\prime(x)$$ is the derivative with respect to $$x$$ of the function $$f(x)$$.

Given that $$f(3)=f^{\prime\prime}(x)$$, write $$f(x)$$ in the form $$ax^2+bx+c$$, where $$a,b,c\in\mathbb{R}$$.

$$f(x)=2x^2+7x-35$$

\begin{align}f(x)&=\int(4x+7)dx\\&=2x^2+7x+C\end{align}

and

\begin{align}f(3)=f^{\prime\prime}(x)\end{align}

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

\begin{align}2(3^2)+7(3)+C=4\end{align}

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

\begin{align}C=-35\end{align}

and therefore

\begin{align}f(x)=2x^2+7x-35\end{align}

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Question 4

(a) Find the value of $$C$$ in the following equation:

\begin{align}\int_3^C(C-5x)\,dx=0\end{align}

where $$C\in\mathbb{R}$$ and $$C \neq 3$$.

$$C=-5$$

\begin{align}\int_3^C(C-5x)\,dx=0\end{align}

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

\begin{align}\left.Cx-\frac{5x^2}{2}\right|_3^C=0\end{align}

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

\begin{align}C^2-\frac{5C^2}{2}-3C+\frac{45}{2}=0\end{align}

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

\begin{align}2C^2-5C^2-6C+45=-16\end{align}

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

\begin{align}-3C^2-6C+45=0\end{align}

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

\begin{align}C^2+2C-15=0\end{align}

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

\begin{align}(C+5)(C-3)=0\end{align}

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

$$C=-5$$ or $$C=3$$

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

\begin{align}C=-5\end{align}

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(b) Consider the following quadratic functions:

\begin{align}p(x)&=-4x^2+10x+11\\q(x)&=x^2+3x+5\end{align}

(i) Find the intersection points of the two functions.

(ii) Hence, or otherwise, use integration to find the area bounded by these two functions, correct to two decimal places.

(i) $$(-0.6,3.56)$$ and $$(2,15)$$

(ii) $$14.65\mbox{ units}^2$$

(i)

\begin{align}-4x^2+10x+11=x^2+3x+5\end{align}

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

\begin{align}5x^2-7x-6=0\end{align}

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

\begin{align}a=5&&b=-7&&c=-6\end{align}

\begin{align}x&=\frac{-(-7)\pm\sqrt{(-7)^2-4(5)(-6)}}{2(5)}\\&=\frac{7\pm\sqrt{169}}{10}\\&=\frac{7\pm13}{10}\end{align}

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

$$x=-0.6$$ or $$x=2$$

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

\begin{align}p(-0.6)=q(-0.6)=3.56\end{align}

and

\begin{align}p(2)=q(2)=15\end{align}

Therefore, the intersection points are $$(-0.6,3.56)$$ and $$(2,15)$$.

(ii)

\begin{align}A&=\int_{-0.6}^2(-4x^2+10x+11)dx-\int_{-0.6}^2(x^2+3x+5)dx\\&=\int_{-0.6}^2(-4x^2+10x+11-x^2-3x-5)dx\\&=\int_{-0.6}^2(-5x^2+7x+6)\\&=\left.-\frac{5x^3}{3}+\frac{7x^2}{2}+6x\right|_{-0.6}^2\\&=-\frac{5(2^3)}{3}+\frac{7(2^2)}{2}+6(2)+\frac{5(-0.6)^3}{3}-\frac{7(-0.6)^2}{2}-6(-0.6)\\&\approx14.65\mbox{ units}^2\end{align}

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Question 5

(a) The first two terms of an arithmetic sequence are shown below:

\begin{align}2\log_2 a,\log_2 (ab)^2,… \end{align}

where $$a,b>0$$.

(i) Show that the common difference of this sequence is $$2\log_2 b$$.

(ii) $$ab^4=32$$. Using this, together with part (i), or otherwise, find the sum of the first $$9$$ terms of this sequence.

(ii) $$90$$

(i)

\begin{align}d&=\log_2(ab)^2-2\log_2a\\&=2\log (ab)-2\log_2a\\&=2\log\left(\frac{ab}{a}\right)\\&=2\log_2b\end{align}

as required.

(ii)

\begin{align}S_n=\frac{n}{2}[2a+(n-1)d]\end{align}

\begin{align}a=2\log_2a&&d=2\log_2b&&n=9\end{align}

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

\begin{align}S_{9}&=\frac{9}{2}[2(2\log_2 a)+(9-1)(2\log_2b)]\\&=9(2\log_2a+8\log_2b)\\&=9(2\log_2a+2\log_2b^4)\\&=18(\log_2a+\log_2b^4)\\&=18\log_2(ab^4)\\&=18\log_2(32)\\&=18(5)\\&=90\end{align}

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(b) Consider the following recurring decimal:

\begin{align}0.1\dot{2}\dot{3}=0.1232323…\end{align}

By rewriting this decimal as a geometric series, or otherwise, express this decimal in the form $$\dfrac{a}{b}$$, where $$a,b\in\mathbb{N}$$.

$$\dfrac{61}{495}$$

\begin{align}0.1\dot{2}\dot{3}&=0.1232323…\\&=0.1+0.023+0.00023+0.0000023+…\\&=0.1+\frac{23}{1{,}000}+\frac{23}{100{,}000}+\frac{23}{10{,}000{,}000}+…\end{align}

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

\begin{align}a=\frac{23}{1{,}000}&&r=\frac{1}{100}\end{align}

\begin{align}0.1\dot{2}\dot{3}&=0.1+\frac{\frac{23}{1{,}000}}{1-\frac{1}{100}}\\&=0.1+\frac{23\times100}{99\times1{,}000}\\&=0.1+\frac{23}{990}\\&=\frac{1}{10}+\frac{23}{990}\\&=\frac{99+23}{990}\\&=\frac{122}{990}\\&=\frac{61}{495}\end{align}

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Question 6

(a) Prove using induction that $$9^n-5^n$$ is divisible by $$4$$ for all $$n\in\mathbb{N}$$.

Let $$f(n)=9^n-5^n$$, where $$n\in\mathbb{N}$$.

$\,$

$$\mathbf{n=1}$$

$$f(1)=9-5=4$$ is indeed divisible by 4.

$\,$

$$\mathbf{n=k}$$

Assume that the statement is true for $$n=k$$, i.e. that $$f(k)=4p$$, where $$p\in\mathbb{N}$$.

$\,$

$$\mathbf{n=k+1}$$

Using this assumption, we now need to prove that the statement is true for $$n=k+1$$, i.e. that $$f(k+1)=4q$$, where $$q\in\mathbb{N}$$.

\begin{align}f(k+1)-f(k)=(9^{k+1}-5^{k+1})-(9^{k}-5^{k})\end{align}

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

\begin{align}f(k+1)-f(k)=9(9^{k})-5(5^{k})-9^{k}+5^{k}\end{align}

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

\begin{align}f(k+1)-f(k)=8(9^{k})-4(5^{k})\end{align}

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

\begin{align}f(k+1)-4p=8(9^{k})-4(5^{k})\end{align}

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

\begin{align}f(k+1)&=4p+8(9^{k})-4(5^{k})\\&=4[p+2(9^{k})-5^{k}]\\&=4q\end{align}

where $$q=p+2(9^{k})-5^{k}$$, as required.

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(b) When Gerard eventually retires, he wants to ensure that he has enough money saved such that he can spend $$12{,}000\mbox{ euro}$$ per year for the first $$15$$ years of his retirement.

For this to be the case, how much money does Gerard need to have in his bank account at the start of his retirement (correct to the nearest euro) assuming that the AER is $$3\%$$?

$$126{,}974\mbox{ euro}$$

\begin{align}P&=\frac{12{,}000}{1.03}+\frac{12{,}000}{1.03^2}+…+\frac{12{,}000}{1.03^{15}}\end{align}

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

\begin{align}a=\frac{12{,}000}{1.03}&&r=\frac{1}{1.05}&&n=15\end{align}

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

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

\begin{align}S_{15}&=\frac{12{,}000}{1.03}\times\frac{\left(1-\frac{1}{1.05}^{15}\right)}{1-\frac{1}{1.05}}\\&\approx126{,}974\mbox{ euro}\end{align}

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Section B

Question 7

A particular video game console, GameStation $$6$$, is reaching the end of its life cycle and is due to be replaced by a new, more powerful console.

The new console is due to be released at the beginning of next year (Year $$1$$).

Therefore, in Year $$1$$, the manufacturer of these consoles will start “winding down” how many GameStation $$6$$ consoles they will produce.

Specifically, they will produce $$80{,}000$$ GameStation $$6$$ consoles in January of Year $$1$$, reducing this production amount by $$25\%$$ each month thereafter.

(a) How many Gamestation $$6$$ consoles will be produced in July of Year $$1$$?

$$14{,}238$$

\begin{align}a=80{,}000 && r=0.75 && n=7\end{align}

\begin{align}T_n=ar^{n-1}\end{align}

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

\begin{align}T_7&=(80{,}000)(0.75^{7-1})\\&\approx14{,}238\end{align}

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(b) In total, how many Gamestation $$6$$ consoles will be produced in Year $$1$$?

$$309{,}864$$

\begin{align}a=80{,}000 && r=0.75 && n=12\end{align}

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

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

\begin{align}S_{12}&=\frac{80{,}000(1-0.75^{12})}{1-0.75}\\&\approx309{,}864\end{align}

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(c) If this company continues to produce these consoles at this rate indefinitely, how many consoles will they produce in total from the start of Year $$1$$?

$$320{,}000$$

\begin{align}a=80{,}000 && r=0.75\end{align}

\begin{align}S_\infty&=\frac{a}{1-r}\end{align}

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

\begin{align}S_\infty&=\frac{80{,}000}{1-0.75}\\&=320{,}000\end{align}

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The new console, GameStation $$7$$, will begin production at the start of Year $$1$$.

The company will produce $$20{,}000$$ GameStation $$7$$ consoles in January of Year $$1$$ and increase the production amount by $$10\%$$ each month thereafter.

(d) How many GameStation $$7$$ consoles will be produced in total in Year $$1$$?

$$427{,}686$$

\begin{align}a=20{,}000 && r=1.1 && n=12\end{align}

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

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

\begin{align}S_n&=\frac{20{,}000(1-1.1^{12})}{1-1.1}\\&\approx427{,}686\end{align}

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Eventually, the number of Gamestation $$7$$ consoles being produced in a particular month will surpass the number of Gamestation $$6$$ consoles being produced in that same month.

(e) Show that, if this occurs in the $$k$$th month, that $$k$$ satisfies:

\begin{align}\left(\frac{1.1}{0.75}\right)^{k-1}>4\end{align}

\begin{align}(20{,}000)(1.1^{k-1})>(80{,}000)(0.75^{k-1})\end{align}

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

\begin{align}\left(\frac{1.1}{0.75}\right)^{k-1}>4\end{align}

as required.

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(f) Using part (e), or otherwise, find the value of $$k$$.

$$k=5$$

\begin{align}\left(\frac{1.1}{0.75}\right)^{k-1}>4\end{align}

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

\begin{align}(k-1)\ln\left(\frac{1.1}{0.75}\right)>\ln 4\end{align}

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

\begin{align}k-1>\frac{\ln 4}{\ln\left(\frac{1.1}{0.75}\right)}\end{align}

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

\begin{align}k-1>3.6196…\end{align}

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

\begin{align}k>4.6196…\end{align}

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

\begin{align}k=5\end{align}

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Question 8

The figure below shows the blueprint of a running track that a university is planning to build.

The track consists of two semicircles of radius $$r$$ connected to a rectangle of width $$w$$, as shown.

As with all modern competitive running tracks, this track has a perimeter of $$400\mbox{ metres}$$.

(a) Show that the area $$A$$ of the track is given by:

\begin{align}A(r)=400r-\pi r^2\end{align}

where the area is measured in $$\mbox{metres}^2$$.

\begin{align}\pi r+w+\pi r+w=400\end{align}

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

\begin{align}2\pi r+2w=400\end{align}

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

\begin{align}\pi r+w=200\end{align}

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

\begin{align}w=200-\pi r\end{align}

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

\begin{align}A&=\pi r^2+2rw\\&=\pi r^2+2r(200-\pi r)\\&=\pi r^2+400r-2\pi r^2\\&=400r-\pi r^2\end{align}

as required.

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(b)

(i) How many maximum values does $$A(r)$$ have? Explain your reasoning.

(ii) How many minimum values does $$A(r)$$ have? Explain your reasoning.

(i) $$1$$ as this is a quadratic function with a negative coefficient in front of the squared term.

(ii) $$0$$ as this is a quadratic function with a negative coefficient in front of the squared term.

(i) $$1$$ as this is a quadratic function with a negative coefficient in front of the squared term.

(i) $$0$$ as this is a quadratic function with a negative coefficient in front of the squared term.

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“We wish to make the area of the track as large as possible in order to make it viewable to as many spectators as possible.”

(c) Using calculus, or otherwise, find the value of $$r$$ (in terms of $$\pi$$) that will make the area of the track as large as possible.

$$\dfrac{200}{\pi}\mbox{ m}$$

\begin{align}\frac{dA}{dr}=0\end{align}

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

\begin{align}400-2\pi r=0\end{align}

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

\begin{align}\end{align}

\begin{align}r=\frac{200}{\pi}\mbox{ m}\end{align}

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(d) What is the maximum area of the track (in terms of $$\pi$$)?

$$\dfrac{40{,}000}{\pi}\mbox{ m}^2$$

\begin{align}A(r_{\mbox{max}})&=400r_{\mbox{max}}-\pi r^2_{\mbox{max}}\\&=400\left(\frac{200}{\pi}\right)-\pi\left(\frac{200}{\pi}\right)^2\\&=\frac{80{,}000}{\pi}-\frac{40{,}000}{\pi}\\&=\frac{40{,}000}{\pi}\mbox{ m}^2\end{align}

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“No no, this won’t do! If we maximise the area, the shape of the track is a circle! Competitive running tracks aren’t allowed to be circles!”

(e) Show that this claim by the university representative regarding the shape of the running track is indeed accurate.

\begin{align}w&=200-\pi r\\&=200-\pi\left(\frac{200}{\pi}\right)\\&=0\end{align}

Therefore, there would be no rectangular section of the track and it would instead be just a circle, as required.

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Question 9

Newton’s Law of Cooling states that, when a hot object of temperature $$T$$ is placed into a room of constant temperature $$T_R$$, that object’s temperature will decrease over time according to the following equation:

\begin{align}T(t)=T_R+(T_0-A)e^{-kt}\end{align}

where $$T_0=T(0)$$, $$A\in\mathbb{R}$$, $$k>0,k\in\mathbb{R}$$ and $$t$$ is measured in minutes.

(a) Show that $$A=T_R$$.

\begin{align}T(0)=T_0\end{align}

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

\begin{align}T_R+(T_0-A)=T_0\end{align}

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

\begin{align}T_R+T_0-A=T_0\end{align}

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

\begin{align}T_R-A=0\end{align}

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

\begin{align}A=T_R\end{align}

as required.

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A cup of tea, of initial temperature $$75^{\circ}\mbox{C}$$, is placed into a room of temperature $$23^{\circ}\mbox{C}$$.

After $$30$$ seconds have passed, the temperature of the tea has decreased to $$73^{\circ}\mbox{C}$$.

(b) Using this information, find the value of $$k$$ correct to two decimal places.

$$0.08$$

\begin{align}T(t)=T_R+(T_0-T_R)e^{-kt}\end{align}

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

\begin{align}73=23+(75-23)e^{-k(0.5)}\end{align}

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

\begin{align}50=52e^{-0.5k}\end{align}

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

\begin{align}e^{-0.5k}=\frac{50}{52}\end{align}

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

\begin{align}-0.5k=\ln\left(\frac{50}{52}\right)\end{align}

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

\begin{align}k&=-2\ln\left(\frac{50}{52}\right)\\&\approx0.08\end{align}

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“According to the equation, the temperature of the tea gets closer and closer to the temperature of the room as time passes.”

(c) Using Newton’s Law of Cooling, or otherwise, show that the temperature of the tea is converging towards the temperature of the room as the time approaches infinity.

\begin{align}T(t)=23+(75-23)e^{-0.08t}\end{align}

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

\begin{align}T(\infty)&=23+(75-23)(0)\\&=23^{\circ}\mbox{C}\end{align}

as required.

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The person who poured the tea prefers drinking it when it has a temperature of $$70^{\circ}\mbox{C}$$.

(d) How many seconds does this person have to wait before drinking the tea (to the nearest second)?

$$76\mbox{ s}$$

\begin{align}70=23+(75-23)e^{-0.08t}\end{align}

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

\begin{align}47=52e^{-0.08t}\end{align}

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

\begin{align}e^{-0.08t}=\frac{47}{52}\end{align}

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

\begin{align}-0.08t=\ln\left(\frac{47}{52}\right)\end{align}

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

\begin{align}t&=-\frac{1}{0.08}\ln\left(\frac{47}{52}\right)\\&=1.2637…\mbox{ min}\\&\approx76\mbox{ s}\end{align}

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(e) When this person begins drinking their tea, what is the rate of change of the temperature of the tea at that time?

$$-3.76^{\circ}\mbox{C}/\mbox{min}$$

\begin{align}T(t)=T_R+(T_0-T_R)e^{-kt}\end{align}

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

\begin{align}\frac{dT}{dt}=-k(T_0-T_R)e^{-kt}\end{align}

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

\begin{align}\frac{dT}{dt}(t=1.2637…)&=-0.08(75-23)e^{-0.08(1.2637)}\\&\approx-3.76^{\circ}\mbox{C}/\mbox{min}\end{align}