One-to-One Grinds
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2018 Higher Level - Paper One
Section A
Question 1
(a) Solve the simultaneous equations.
\begin{align}2x+3y-z&=-4\\3x+2y+2z&=14\\x-3z&=-13\end{align}
\(x=2\) and \(y=-1\) and \(z=5\)
\begin{align}2x+3y-z&=-4\\3x+2y+2z&=14\\x-3z&=-13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4x+6y-2z&=-8\\9x+6y+6z&=42\\x-3z&=-13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x+8z=50\\x-3z&=-13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x+8z=50\\5x-15z&=-65\end{align}
\begin{align}\downarrow\end{align}
\begin{align}23z=115\end{align}
\begin{align}\downarrow\end{align}
\begin{align}z&=\frac{115}{23}\\&=5\end{align}
and
\begin{align}x-3z=-13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=3z-13\\&=3(5)-13\\&=2\end{align}
and
\begin{align}2x+3y-z=-4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y&=\frac{-4-2x+z}{3}\\&=\frac{-4-2(2)+5}{3}\\&=-1\end{align}
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(b) Solve the inequality \(\dfrac{2x-3}{x+2}\geq3\), where \(x\in\mathbb{R}\) and \(x\neq-2\).
\(-9\leq x<2\)
\begin{align}\frac{2x-3}{x+2}\geq3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(2x-3)(x+2)\geq3(x+2)^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2x^2+4x-3x-6\geq3x^2+12x+12\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-x^2-11x-18\geq0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2+11x+18\leq0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x+9)(x+2)\leq0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-9\leq x<-2 \end{align}
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Question 2
(a) The first three terms of a geometric series are \(x^2\), \(5x-8\), and \(x+8\) where \(x\in\mathbb{R}\).
Use the common ratio to show that \(x^3-17x^2+80x-64=0\).
The answer is already in the question!
\begin{align}\frac{5x-8}{x^2}=\frac{x+8}{5x-8}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(5x-8)^2=x^2(x+8)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}25x^2-80x+64=x^3+8x^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^3-17x^2+80x-64=0\end{align}
as required.
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(b) If ݂\(f(x)=x^3-17x^2+80x-64\), \(x\in\mathbb{R}\), show that ݂\(f(1)=0\), and find another value of \(x\) for which ݂\(f(x)=0\).
\(x=8\)
\begin{align}f(1)&=1^3-17(1^2)+80(1)-64\\&=1-17+80-64\\&=0\end{align}
as required.
\[
\require{enclose}
\begin{array}{rll}
x^2-16x+64\phantom{000000}\, \\[-3pt]
x-1 \enclose{longdiv}{\,x^3-17x^2+80x-64} \\[-3pt]
\underline{x^3-x^2\phantom{00000000000}\,\,} \\[-3pt]
-16x^2+80x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-16x^2+16x\phantom{00}\,}\\[-3pt]\phantom{00}64x-64\\[-3pt]\phantom{00}\underline{64x-64}\\[-3pt]0
\end{array}
\]
\begin{align}\downarrow\end{align}
\begin{align}f(x)&=x^3-17x^2+80x-64\\&=(x-1)(x^2-16x+64)\\&=(x-1)(x-8)^2\end{align}
Therefore, another value of \(x\) in which \(f(x)=0\) is \(x=8\).
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(c) In the case of one of the values of \(x\) from part (b), the terms in part (a) will generate a geometric series with a finite sum to infinity.
Find this value of \(x\) and hence find the sum to infinity.
\(x=8\) and \(S_{\infty}=128\)
For \(x=1\), the series \(1^2,5(1)-8,1+8…=1,-3,9\) doesn’t have a sum to infinity (as \(|r|>1\)).
For \(x=8\), the series \(8^2,5(8)-8,8+8…=64,32,16\) does have a sum to infinity (as \(|r|<1\)).
\begin{align}\downarrow\end{align}
\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{64}{1-\frac{1}{2}}\\&=128\end{align}
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Question 3
(a) Let \(h(x)=\cos(2x)\), where \(x\in\mathbb{R}\).
A tangent is drawn to the graph of \(h(x)\) at the point where \(x=\dfrac{\pi}{3}\).
Find the angle that this tangent makes with the positive sense of the \(x\)-axis.
\(\dfrac{2\pi}{3}\mbox{ rad}\)
\begin{align}h'(x)=-2\sin(2x)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h’\left(\frac{\pi}{3}\right)&=-2\sin\left(2\frac{\pi}{3}\right)\\&=-\sqrt{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan\theta=-\sqrt{3}\end{align}
\begin{align}\theta&=\pi-\frac{\pi}{3}\\&=\frac{2\pi}{3}\mbox{ rad}\end{align}
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(b) Find the average value of \(h(x)\) over the interval \(0\leq x \leq \dfrac{\pi}{4}\), \(x\in\mathbb{R}\).
Give your answer in terms of \(\pi\).
\(\dfrac{2}{\pi}\)
\begin{align}\mbox{Average}&=\frac{1}{b-a}\int_a^bh(x)\,dx\\&=\frac{1}{\frac{\pi}{4}-0}\int_0^{\pi/4}\cos(2x)\,dx\\&=\frac{4}{\pi}\left(\frac{1}{2}\right)\left.\sin(2x)\right|_0^{\pi/4}\\&=\frac{2}{\pi}\left[\sin\left(\frac{2\pi}{4}\right)-\sin(2(0))\right]\\&=\frac{2}{\pi}(1-0)\\&=\frac{2}{\pi}\end{align}
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Question 4
(a) Prove, using induction, that if \(n\) is a positive integer then
\((\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)\), where \(i^2=-1\).
The answer is already in the question!
For \(n=1\), the formula states that
\begin{align}(\cos\theta+i\sin\theta)^1=\cos(1\theta)+i\sin(1\theta)\end{align}
\begin{align}\cos\theta+i\sin\theta=\cos\theta+i\sin\theta\end{align}
and therefore it is true for \(n=1\).
Hence, let us assume that it is true for \(n=k\), i.e. that
\begin{align}(\cos\theta+i\sin\theta)^k=\cos(k\theta)+i\sin(k\theta)\end{align}
Using this assumption, we now wish to prove that it is also true for \(n=k+1\), i.e. that
\begin{align}(\cos\theta+i\sin\theta)^{k+1}=\cos[(k+1)\theta]+i\sin[(k+1)\theta]\end{align}
Let us look at the left hand side:
\begin{align}(\cos\theta+i\sin\theta)^{k+1}&=(\cos\theta+i\sin\theta)^{k}(\cos\theta+i\sin\theta)^{1}\\&=[\cos(k\theta)+i\sin(k\theta)](\cos\theta+i\sin\theta)\theta]\\&=[\cos(k\theta)-\sin(k\theta)\sin\theta]+i[\cos(k\theta)\sin\theta+\cos\theta\sin(k\theta)]\\&=\cos[(k+1)\theta]+i\sin[(k+1)\theta]\end{align}
which is indeed the right hand side.
Therefore, the formula is true for all positive integers \(n\).
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(b) Hence, or otherwise, find \(\left(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right)^3\) in its simplest form.
\(1\)
Modulus
\begin{align}r&=\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\\&=\sqrt{\frac{1}{4}+\frac{3}{4}}\\&=\sqrt{1}\\&=1\end{align}
\[\,\]
Argument
\begin{align}\tan\theta&=-\frac{\sqrt{3}/2}{1/2}\\&=-\sqrt{3}\end{align}
Reference Angle:
\begin{align}\tan A=\sqrt{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\frac{\pi}{3}\mbox{ rad}\end{align}
As \(\theta\) is in the second quadrant:
\begin{align}\theta&=\pi-A\\&=\pi-\frac{\pi}{3}\\&=\frac{2\pi}{3}\end{align}
\[\,\]
Polar Form
\begin{align}-\frac{1}{2}+\frac{\sqrt{3}}{2}i&=1\left[\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)\right]\\&=\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)^3&=\left[\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)\right]^3\\&=\cos(3)\left(\frac{2\pi}{3}\right)+i\sin(3)\left(\frac{2\pi}{3}\right)\\&=\cos(2\pi)+i\sin(2\pi)\\&=1+0i\\&=1\end{align}
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Question 5
(a) The Sieve of Sundaram is an infinite table of arithmetic sequences.
The terms in the first \(4\) rows and the first \(4\) columns of the table are shown below.
(i) Find the difference between the sums of the first \(45\) terms in the first two rows.
(ii) Find the number which is in the \(60\)th row and \(70\)th column of the table.
(i) \(2{,}115\)
(ii) \(8{,}530\)
(i)
\begin{align}S_{2,45}-S_{1,45}&=\frac{45}{2}[2(7)+(45-1)(5)]-\frac{45}{2}[2(4)+(45-1)(3)]\\&=2{,}115\end{align}
(ii)
\begin{align}T_{60,1}&=4+(60-1)(3)\\&=181\end{align}
and
\begin{align}T_{60,2}&=7+(60-1)(5)\\&=302\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_{60,70}&=181+(70-1)(302-181)\\&=8{,}530\end{align}
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(b) The first two terms of a sequence are ܽ\(a_1=4\) and ܽ\(a_2=2\).
The general term is defined by ܽ\(a_n=a_{n-1}-a_{n-2}\), when \(n\geq3\).
Write out the next \(6\) terms of the sequence and hence find the value of ܽ\(a_{2019}\).
\(-2,-4,-2,2,4,2\) and \(a_{2019}=-2\)
\begin{align}a_3&=a_2-a_1\\&=2-4\\&=-2\end{align}
and
\begin{align}a_4&=a_3-a_2\\&=-2-2\\&=-4\end{align}
and
\begin{align}a_5&=a_4-a_3\\&=-4-(-2)\\&=-2\end{align}
and
\begin{align}a_6&=a_5-a_4\\&=-2-(-4)\\&=2\end{align}
and
\begin{align}a_7&=a_6-a_5\\&=2-(-2)\\&=4\end{align}
and
\begin{align}a_8&=a_7-a_6\\&=4-2\\&=2\end{align}
The following pattern therefore repeats
\begin{align}4,2,-2,-4,-2,2,…\end{align}
As \(2019=2016+3\), and as \(2016\) is a multiple of \(6\):
\begin{align}a_{2019}=-2\end{align}
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Question 6
Parts of the graphs of the functions \(h(x)=x\) and \(k(x)=x^3\), \(x\in\mathbb{R}\) are shown in the diagram below.
(a) Find the co-ordinates of the points of intersection of the graphs of the two functions.
\((-1,-1)\), \((0,0)\) and \((1,1)\)
\begin{align}x^3=x\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^3-x=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x(x^2-1)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x(x+1)(x-1)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=-1\), \(x=0\) and \(x=1\)
\begin{align}\downarrow\end{align}
\((-1,-1)\), \((0,0)\) and \((1,1)\)
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(b)
(i) Find the total area enclosed between the graphs of the two functions.
(ii) On the diagram on the previous page, using symmetry or otherwise, draw the graph of \(k^{-1}\), the inverse function of ݇\(k\).
(i) \(\dfrac{1}{2}\mbox{ units}^2\)
(ii)
(i)
\begin{align}A&=2\int_0^1(h(x)-k(x)\,dx\\&=2\int_0^1(x-x^3)\,dx\\&=2\left(\frac{x^2}{2}-\frac{x^4}{4}\right)_0^1\\&=2\left(\frac{1^2}{2}-\frac{1^4}{4}\right)\\&=\frac{1}{2}\mbox{ units}^2\end{align}
(ii)
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Section B
Question 7
The time, in days of practice, it takes Jack to learn to type \(x\) words per minute (wpm) can be modelled by the function:
\(t(x)=k\left[\ln\left(1-\dfrac{x}{80}\right)\right]\), where \(0\leq x\leq70\), \(x\in\mathbb{R}\), and \(k\) is a constant.
(a) Based on the function \(t(x)\), Jack can learn to type \(35\) wpm in \(35.96\) days. Write the function above in terms of ݇\(k\) and hence show that ݇\(k=-62.5\), correct to \(1\) decimal place.
The answer is already in the question!
\begin{align}35.96=k\left[\ln\left(1-\frac{35}{80}\right)\right]\end{align}
\begin{align}\downarrow\end{align}
\begin{align}35.96=k\ln\left(\frac{45}{80}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k&=\frac{35.96}{\ln\left(\frac{45}{80}\right)}\\&\approx-62.5\end{align}
as required.
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(b) Find the number of wpm that Jack can learn to type with \(100\) days of practice.
Give your answer correct to the nearest whole number.
\(64\mbox{ wpm}\)
\begin{align}100=-62.5\ln\left(1-\frac{x}{80}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\ln\left(1-\frac{x}{80}\right)=-\frac{100}{62.5}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1-\frac{x}{80}=\exp\left(-\frac{100}{62.5}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{80-x}{80}=\exp\left(-\frac{100}{62.5}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=80-80\exp\left(-\frac{100}{62.5}\right)\\&\approx64\mbox{ wpm}\end{align}
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(c) Complete the table below, correct to the nearest whole number and hence draw the graph of \(t(x)\) for \(0\leq x\leq70\), \(x\in\mathbb{R}\).
| \(x\) (wpm) | \(0\) | \(10\) | \(20\) | \(30\) | \(40\) | \(50\) | \(60\) | \(70\) |
|---|---|---|---|---|---|---|---|---|
|
\(t(x)\) days |
|
|
|
|
|
|
|
|
| \(x\) (wpm) | \(0\) | \(10\) | \(20\) | \(30\) | \(40\) | \(50\) | \(60\) | \(70\) |
|---|---|---|---|---|---|---|---|---|
|
\(t(x)\) days |
\(0\) |
\(8\) |
\(18\) |
\(29\) |
\(43\) |
\(61\) |
\(87\) |
\(130\) |
| \(x\) (wpm) | \(0\) | \(10\) | \(20\) | \(30\) | \(40\) | \(50\) | \(60\) | \(70\) |
|---|---|---|---|---|---|---|---|---|
|
\(t(x)\) days |
\(0\) |
\(8\) |
\(18\) |
\(29\) |
\(43\) |
\(61\) |
\(87\) |
\(130\) |
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(d) A simpler function that could also be used to model the number of days needed to attain \(x\) wpm is \(p(x)=1.5x\).
Draw, on the diagram above, the graph of \(p(x)\) for \(0\leq x\leq 70\), \(x\in\mathbb{R}\).
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(e) Let \(h(x)=p(x)-t(x)\).
(i) Use your graphs above to estimate the solution to \(h(x)=0\) for \(x>0\).
(ii) Use calculus to find the maximum value of \(h(x)\) for \(0\leq x\leq 70\), \(x\in\mathbb{R}\).
Give your answer correct to the nearest whole number.
(i) \(\approx62\mbox{ wpm}\)
(ii) \(17\mbox{ days}\)
(i) \(\approx62\mbox{ wpm}\)
(ii)
\begin{align}h'(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p'(x)-t'(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1.5+62.5\left(-\frac{1}{80}\right)\left(1-\frac{x}{80}\right)^{-1}=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1.5+62.5\left(\frac{1}{x-80}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=80-\frac{62.5}{1.5}\\&=\frac{115}{3}\mbox{ words}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h\left(\frac{115}{3}\right)&=1.5\left(\frac{115}{3}\right)+62.5\ln\left[1-\frac{\frac{115}{3}}{80}\right]\\&\approx17\mbox{ days}\end{align}
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Question 8
The graph of the symmetric function \(f(x)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\) is shown below.
(a) Find the co-ordinates of \(A\), the point where the graph intersects the \(y\)-axis.
Give your answer in terms of \(\pi\).
\(\left(0,\dfrac{1}{\sqrt{2\pi}}\right)\)
\begin{align}f(0)&=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(0^2)}\\&=\frac{1}{\sqrt{2\pi}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x,y)=\left(0,\frac{1}{\sqrt{2\pi}}\right)\end{align}
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(b) The co-ordinates of \(B\) are \(\left(-1,\frac{1}{\sqrt{2\pi e}}\right)\). Find the area of the shaded rectangle in the diagram above. Give your answer correct to \(3\) decimal places.
\(0.484\mbox{ units}^2\)
\begin{align}A&=l\times w\\&=\left(\frac{1}{\sqrt{2\pi e}}\right)\times(1+1)\\&\approx0.484\mbox{ units}^2\end{align}
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(c) Use calculus to show that ݂\(f(x)\) is decreasing at \(C\).
The answer is already in the question!
\begin{align}f'(x)&=\frac{1}{\sqrt{2\pi}}(-x)e^{-\frac{1}{2}x^2}\\&=-\frac{x}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f'(1)&=-\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(1^2)}\\&<0\end{align}
Therefore, \(f(x)\) is decreasing at \(C\).
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(d) Show that the graph of ݂\(f(x)\) has a point of inflection at \(B\).
The answer is already in the question!
\begin{align}f'(x)=-\frac{x}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f^{\prime\prime}(x)&=-\frac{x}{\sqrt{2\pi}}(-x)e^{-\frac{1}{2}x^2}-\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\\&=\frac{x^2-1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f^{\prime\prime}(-1)=\frac{(-1)^2-1}{\sqrt{2\pi}}e^{-\frac{1}{2}(-1)^2}\\&=0\end{align}
Therefore, \(f(x)\) has a point of inflection at \(B\).
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Question 9
The diagram below shows the first \(4\) steps of an infinite pattern which creates the Sierpinski Triangle. The sequence begins with a black equilateral triangle. Each step is formed by removing an equilateral triangle from the centre of each black triangle in the previous step, as shown. Each equilateral triangle that is removed is formed by joining the midpoints of the sides of a black triangle from the previous step.
(a) The table below shows the number of black triangles at each of the first \(4\) steps and the fraction of the original triangle remaining at each step. Complete the table.
| Step | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
|
Number of black triangles |
\(1\) |
|
|
|
|
Fraction of the original triangle remaining |
\(1\) |
|
\(\dfrac{9}{16}\) |
|
| Step | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
|
Number of black triangles |
\(1\) |
\(3\) |
\(9\) |
\(27\) |
|
Fraction of the original triangle remaining |
\(1\) |
\(\dfrac{3}{4}\) |
\(\dfrac{9}{16}\) |
\(\dfrac{27}{64}\) |
| Step | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
|
Number of black triangles |
\(1\) |
\(3\) |
\(9\) |
\(27\) |
|
Fraction of the original triangle remaining |
\(1\) |
\(\dfrac{3}{4}\) |
\(\dfrac{9}{16}\) |
\(\dfrac{27}{64}\) |
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(b)
(i) Write an expression in terms of \(n\) for the number of black triangles in step \(n\) of the pattern.
(ii) Step \(k\) is the first step of the pattern in which the number of black triangles exceeds one thousand million (i.e. \(1\times10^9\)) for the first time. Find the value of \(k\).
(i) \(3^n\)
(ii) \(k=19\)
(i) \(3^n\)
(ii)
\begin{align}3^k>1\times10^9\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k>\log_3(1\times10^9)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k>18.863…\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k=19\end{align}
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(c)
(i) Step \(h\) is the first step of the pattern in which the fraction of the original triangle remaining is less than \(\dfrac{1}{100}\) of the original triangle. Find the value of \(h\).
(ii) What fraction of the original triangle remains after an infinite number of steps of the
pattern?
(i) \(h=17\)
(ii) None of the triangle remains.
(i)
\begin{align}\left(\frac{3}{4}\right)^h<\frac{1}{100}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\ln\left(\frac{3}{4}\right)^h<\ln\left(\frac{1}{100}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h\ln\left(\frac{3}{4}\right)<\ln\left(\frac{1}{100}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h>\frac{\ln\left(\frac{1}{100}\right)}{\ln\left(\frac{3}{4}\right)}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h>16.007…\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h=17\end{align}
(ii) \(\left(\dfrac{3}{4}\right)^n\) approaches zero as \(n\) approaches infinity.
Therefore, none of the triangle remains.
Our Video Walkthroughs are in the final stages of editing and will go live during Summer 2023.
In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!
(d)
(i) The side length of the triangle in Step \(0\) is \(1\) unit. The table below shows the total perimeter of all the black triangles in each of the first \(5\) steps.
Complete the table below.
| Step | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
|
Perimeter |
\(3\) |
|
\(\dfrac{27}{4}\) |
|
|
(ii) Find the total perimeter of the black triangles in step \(35\) of the pattern.
Give your answer correct to the nearest unit.
(iii) Use your answers to part (c)(ii) and part (d)(ii) to comment on the total area and the total perimeter of the black triangles in step ݊\(n\) of the pattern, as ݊\(n\) tends to infinity.
(i)
| Step | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
|
Perimeter |
\(3\) |
\(\dfrac{9}{2}\) |
\(\dfrac{27}{4}\) |
\(\dfrac{81}{8}\) |
\(\dfrac{243}{16}\) |
(ii) \(4{,}368{,}329\)
(iii) The area tends to zero and the perimeter tends to infinity.
(i)
| Step | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
|
Perimeter |
\(3\) |
\(\dfrac{9}{2}\) |
\(\dfrac{27}{4}\) |
\(\dfrac{81}{8}\) |
\(\dfrac{243}{16}\) |
(ii)
\begin{align}P_n=\frac{3^{n+1}}{2^n}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}P_{35}&=\frac{3^{35+1}}{2^35}\\&=4{,}368{,}329\end{align}
(iii) The area tends to zero and the perimeter tends to infinity.
