Question
This question asks you to investigate the throwing of darts at a target.
Whenever you throw a dart at a circular target of radius 1 , you hit a random point on the target.
Let $X=$ distance from dart’ $s$ landing point to centre of target.
(a) (i) Find the median value of $X$.
(ii) By considering $\mathrm{P}(X<x)$, or otherwise, show that the probability density function of $X$ is $f(x)=2 x$.
(iii) Find $\mathrm{E}(X)$.
Define “score” as $S=\frac{1}{X}$.
(b) (i) Find the median value of $S$.
(ii) By considering $\mathrm{P}(X<x)$, or otherwise, show that the probability density function of $S$ is $g(s)=2 s^{-3}$.
(iii) Find $\mathrm{E}(S)$.
Your IB Higher Level Mathematics teacher, whose accuracy is also random, throws a dart. You then throw darts until you get a score that is higher than your teacher’s score.
Let $N=$ number of your throws.
(c) (i) Show that $\mathrm{P}(N=5)=\left(\frac{1}{6}\right)\left(\frac{1}{5}\right)$.
(ii) Verify that the sum of probabilities of all $N$ equals 1.
(iii) Show that $\mathrm{E}(N)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots$.
(iv) By comparing $\mathrm{E}(N)$ with $1\left(\frac{1}{2}\right)+2\left(\frac{1}{4}\right)+4\left(\frac{1}{8}\right)+\cdots$, or otherwise, show that $\mathrm{E}(N)=\infty$.
(v) Is the result $\mathrm{E}(N)=\infty$ realistic? Explain.
(d) Your teacher asks you to design a non-circular target so that $S$ is approximately normally distributed. Is this possible? If yes, draw such a target; if no, explain why not.
▶️Answer/Explanation
(a) (i)
$$
\begin{aligned}
& \pi m_X^2=\frac{1}{2} \pi 1^2 \\
& m_X=\frac{1}{\sqrt{2}}
\end{aligned}
$$
(ii)
$$
\begin{aligned}
& \mathrm{P}(X<x)=\frac{\pi x^2}{\pi 1^2}=x^2 \\
& f(x)=\frac{\mathrm{d}}{\mathrm{d} x} x^2=2 x
\end{aligned}
$$
(iii) $\mathrm{E}(X)=\int_0^1 x(2 x) \mathrm{d} x=\frac{2}{3}$
(b) (i) $m_S=\frac{1}{m_X}=\sqrt{2}$
(ii)
$$
\begin{aligned}
& \mathrm{P}(S<s)=\mathrm{P}\left(\frac{1}{X}<s\right)=\mathrm{P}\left(X>\frac{1}{s}\right)=1-\mathrm{P}\left(X<\frac{1}{s}\right)=1-\int_0^{\frac{1}{s}} 2 x \mathrm{~d} x=1-s^{-2} \\
& g(s)=\frac{\mathrm{d}}{\mathrm{d} s}\left(1-s^{-2}\right)=2 s^{-3}
\end{aligned}
$$
(iii) $\mathrm{E}(S)=\int_1^{\infty} s\left(2 s^{-3}\right) \mathrm{d} s=\lim _{b \rightarrow \infty}\left[-2 s^{-1}\right]_1^b=\lim _{b \rightarrow \infty}\left(-\frac{2}{b}+2\right)=2$
(c) (i)
$$
\begin{aligned}
& \mathrm{P}(N=5) \\
& =\mathrm{P}(\text { in } 6 \text { throws (including teacher’s), highest score is 6th, and 2nd highest score is } 1 \mathrm{st} \text { ) } \\
& =\left(\frac{1}{6}\right)\left(\frac{1}{5}\right)
\end{aligned}
$$
(ii) $\frac{1}{2}+\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)\left(\frac{1}{4}\right)+\left(\frac{1}{4}\right)\left(\frac{1}{5}\right)+\cdots=\frac{1}{2}+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\cdots=1$
(iii) $\mathrm{E}(N)=1\left(\frac{1}{2}\right)\left(\frac{1}{1}\right)+2\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)+3\left(\frac{1}{4}\right)\left(\frac{1}{3}\right)+\cdots=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$
(iv)
$$
\begin{aligned}
& \left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots>1\left(\frac{1}{2}\right)+2\left(\frac{1}{4}\right)+4\left(\frac{1}{8}\right)+\cdots=\infty \\
& \mathrm{E}(N)=\infty
\end{aligned}
$$
(v) Realistic. In real life, distance measurements have limited precision, so there is a top score. If the teacher gets the top score, then $N=\infty$. So $\mathrm{E}(N)=\infty$.
\text { (d) Example: }
Question 2
\text { This question asks you to investigate a ladder of length } l \mathrm{~m} \text { sliding down a wall and across the floor. }
(a) Find the equation of the path of the midpoint of the ladder, in terms of $x, y$ and $l$.
(b) The speed, in $\mathrm{m} \mathrm{s}^{-1}$, of the top end of the ladder is equal to the distance, in $\mathrm{m}$, between the other end and the wall. The ladder is gently nudged from the vertical position and begins to slide. Find the time it takes to become horizontal.
(c) Now assume that the bottom end of the ladder has constant speed $1 \mathrm{~m} \mathrm{~s}^{-1}$.
(i) Find the speed of the top end just before it hits the floor.
(ii) Is your answer to (c)(i) realistic? Explain.
As the ladder slides, an envelope curve appears. The ladder is always tangent to the envelope curve.
(d) (i) On the same set of axes, in the first quadrant, sketch and label $x^n+y^n=l^n$ for $n=\frac{1}{2}, 1,2,3$.
(ii) Show that the equation of the tangent to $x^n+y^n=l^n, n \neq 0$, at $(p, q)$ is $y=-p^{n-1} q^{1-n} x+q^{1-n} l^n$.
(iii) Find the equation of the envelope curve, in terms of $x, y$ and $l$.
▶️Answer/Explanation
(a) Draw another ladder that makes an ” $\mathrm{X}$ ” with the first ladder.
As the ladders slide, their midpoint moves along the circle $x^2+y^2=\left(\frac{l}{2}\right)^2$.
(b)
$$
\begin{aligned}
& -\frac{\mathrm{d} b}{\mathrm{~d} t}=a=\sqrt{l^2-b^2} \\
& \int \frac{1}{\sqrt{l^2-b^2}} \mathrm{~d} b=\int-\mathrm{d} t \\
& \arcsin \frac{b}{l}=-t+c \\
& t=0, \quad b=l \quad \Rightarrow \quad c=\frac{\pi}{2} \\
& b=0 \quad \Rightarrow \quad t=\frac{\pi}{2} \mathrm{~s}
\end{aligned}
$$
(c) (i)
$$
\begin{aligned}
& a^2+b^2=l^2 \\
& 2 a \frac{\mathrm{d} a}{\mathrm{~d} t}+2 b \frac{\mathrm{d} b}{\mathrm{~d} t}=0 \\
& \frac{\mathrm{d} b}{\mathrm{~d} t}=-\frac{a}{b} \\
& \text { As } b \rightarrow 0, \quad\left|\frac{\mathrm{d} b}{\mathrm{~d} t}\right| \rightarrow \infty \mathrm{m} \mathrm{s}^{-1}
\end{aligned}
$$
(ii) Not realistic. In real life, the ladder will separate from the wall before it hits the floor.
d(i)
(ii)
$$
\begin{aligned}
& n x^{n-1}+n y^{n-1} y^{\prime}=0 \quad \Rightarrow \quad y^{\prime}=-p^{n-1} q^{1-n} \\
& y-q=-p^{n-1} q^{1-n}(x-p) \\
& y=-p^{n-1} q^{1-n} x+q^{1-n}\left(p^n+q^n\right) \\
& y=-p^{n-1} q^{1-n} x+q^{1-n} l^n
\end{aligned}
$$
(iii) Assume the equation of the envelope curve is of the form $x^n+y^n=l^n$.
Make the distance between the tangent’s axes intercepts equal to $l$.
$$
\begin{aligned}
& \left(p^{1-n} l^n\right)^2+\left(q^{1-n} l^n\right)^2=l^2 \\
& p^{2-2 n}+q^{2-2 n}=l^{2-2 n} \\
& p^n+q^n=l^n \\
& n=2-2 n \quad \Rightarrow \quad n=\frac{2}{3}
\end{aligned}
$$
There is a solution for $n$ that is independent of $p$ and $q$, so the assumption is correct.
$$
x^{\frac{2}{3}}+y^{\frac{2}{3}}=l^{\frac{2}{3}}
$$
Question
This question asks you to investigate the packing density of circles and spheres.
(a) An infinitely large table is covered by non-overlapping circular disks of equal radii. Find the maximum proportion of the table that is covered.
(b) (i) Expand $(k+1)^3$.
(ii) By summing each side of (b)(i) from $k=1$ to $k=n$, show that $\sum_{k=1}^n k^2=\frac{1}{3} n^3+\frac{1}{2} n^2+\frac{1}{6} n$.
The diagram below shows spherical balls of diameter 1 arranged in a triangular pyramid with $n$ layers.
(c) (i) Find the number of balls in the $k$ th layer from the top, simplifying your answer.
(ii) Show that the total number of balls in the pyramid is $\frac{1}{6} n^3+\frac{1}{2} n^2+\frac{1}{3} n$.
Let $A, B, C, D$ be the centres of the balls at the vertices of the pyramid.
(d) Show that the volume of tetrahedron $A B C D$ is $\frac{\sqrt{2}}{12}(n-1)^3$.
(e) Find the exact value of the proportion of tetrahedron $A B C D$ that is occupied by the balls as $n$ approaches infinity.
▶️Answer/Explanation
(a)
$$
\begin{aligned}
\text { Maximum proportion } & =\frac{A_{\text {sectors in triangle }}}{A_{\text {triangle }}}(M 1) \\
& =\frac{\frac{1}{2} \pi r^2}{\frac{1}{2}(2 r)^2 \sin \frac{\pi}{3}}=\frac{\pi}{2 \sqrt{3}}
\end{aligned}
$$
(b) (i) $(k+1)^3=k^3+3 k^2+3 k+1$
$$
\text { (ii) } \begin{aligned}
\sum_{k=1}^n(k+1)^3 & =\sum_{k=1}^n k^3+3 \sum_{k=1}^n k^2+3 \sum_{k=1}^n k+\sum_{k=1}^n 1 \\
\sum_{k=1}^n k^2 & =\frac{1}{3}\left(\sum_{k=1}^n(k+1)^3-\sum_{k=1}^n k^3-3 \sum_{k=1}^n k-\sum_{k=1}^n 1\right)=\frac{1}{3}\left((n+1)^3-1-\frac{3 n(n+1)}{2}-n\right) \\
& =\frac{1}{3}\left(n^3+\frac{3}{2} n^2+\frac{1}{2} n\right)=\frac{1}{3} n^3+\frac{1}{2} n^2+\frac{1}{6} n
\end{aligned}
$$
(c) (i) $1+2+3+\cdots+k=\frac{k(k+1)}{2}$
(ii)
$$
\begin{aligned}
\sum_{k=1}^n \frac{k(k+1)}{2} & =\frac{1}{2}\left(\sum_{k=1}^n k^2+\sum_{k=1}^n k\right) \\
& =\frac{1}{2}\left(\frac{1}{3} n^3+\frac{1}{2} n^2+\frac{1}{6} n+\frac{1}{2} n(n+1)\right)=\frac{1}{6} n^3+\frac{1}{2} n^2+\frac{1}{3} n
\end{aligned}
$$
(d) Edge length $=n-1$
$$
\begin{aligned}
& \text { Area of base }=\frac{1}{2}(n-1)^2 \sin \frac{\pi}{3}=\frac{\sqrt{3}}{4}(n-1)^2 \\
& \text { Height }=\sqrt{(n-1)^2-\left(\frac{2}{\sqrt{3}}\left(\frac{n-1}{2}\right)\right)^2}=\sqrt{\frac{2}{3}}(n-1) \\
& \text { Volume }=\frac{1}{3}\left(\frac{\sqrt{3}}{4}\right)(n-1)^2 \sqrt{\frac{2}{3}}(n-1)=\frac{\sqrt{2}}{12}(n-1)^3
\end{aligned}
$$
(e)
$$
\begin{aligned}
& \text { Volume of each ball }=\frac{4 \pi}{3}\left(\frac{1}{2}\right)^3=\frac{\pi}{6} \\
& \text { Proportion }=\lim _{n \rightarrow \infty} \frac{\frac{\pi}{6}\left(\frac{1}{6} n^3+\frac{1}{2} n^2+\frac{1}{3} n\right)}{\frac{\sqrt{2}}{12}(n-1)^3}=\frac{\pi}{3 \sqrt{2}}
\end{aligned}
$$
Question
\text { This question asks you to investigate a circle inscribed in a sector that is inscribed in another circle. }
$$
\begin{array}{llll}
r=\text { radius of small circle } & L=\text { radius of sector } & R=\text { radius of big circle } & \theta=\frac{\text { sector angle }}{2} \\
A_1=\text { area of small circle } & A_2=\text { area of sector } & A_3=\text { area of big circle }
\end{array}
$$
(a) (i) Show that $\frac{A_2}{A_3}=\frac{4 \theta}{\pi} \cos ^2 \theta$.
(ii) Find the values of $\frac{A_2}{A_3}$ when $\theta=\frac{\pi}{6}$ and $\theta=\frac{\pi}{4}$.
Naim guesses that $\frac{A_2}{A_3}$ is maximized when $\theta=\frac{1}{2}\left(\frac{\pi}{6}+\frac{\pi}{4}\right)=\frac{5 \pi}{24}$.
(b) (i) Find, to four decimal places, the value of $\frac{A_2}{A_3}$ when $\theta=\frac{5 \pi}{24}$.
(ii) Determine if Naim’s guess is correct or not.
(c) (i) Show that $\frac{A_1}{A_3}=\left(\frac{2 \cos \theta \sin \theta}{1+\sin \theta}\right)^2$.
(ii) Find, to four decimal places, the maximum value of $\frac{A_1}{A_3}$.
Renee guesses that $\frac{A_1}{A_3}$ is maximized when the circles are concentric.
(d) Determine if Renee’s guess is correct or not.
▶️Answer/Explanation
(a)(i)
$$
\cos \theta=\frac{\frac{L}{2}}{R} \quad \Rightarrow \quad L=2 R \cos \theta \quad \Rightarrow \quad \frac{A_2}{A_3}=\frac{\frac{1}{2} L^2(2 \theta)}{\pi R^2}=\frac{4 \theta}{\pi} \cos ^2 \theta
$$
(ii) $\theta=\frac{\pi}{6} \Rightarrow \frac{A_2}{A_3}=\frac{1}{2} \quad \theta=\frac{\pi}{4} \Rightarrow \frac{A_2}{A_3}=\frac{1}{2}$
(b) (i) 0.5245
(ii) GDC: At $\left(\frac{A_2}{A_3}\right)_{\max }, \theta=0.65327 \ldots \neq \frac{5 \pi}{24}$.
Naim’s guess is incorrect.
(c) (i)
$$
\begin{aligned}
& L=\frac{r}{\sin \theta}+r \\
& 2 R \cos \theta=\frac{r}{\sin \theta}+r \Rightarrow \frac{r}{R}=\frac{2 \cos \theta}{\frac{1}{\sin \theta}+1}=\frac{2 \cos \theta \sin \theta}{1+\sin \theta} \Rightarrow \frac{A_1}{A_3}=\left(\frac{2 \cos \theta \sin \theta}{1+\sin \theta}\right)^2
\end{aligned}
$$
(ii) GDC: $\left(\frac{A_1}{A_3}\right)_{\max }=0.3607$
(d)
If circles are concentric, $L=2 \sqrt{R^2-r^2}=R+r$
$$
\begin{aligned}
& 4 R^2-4 r^2=R^2+2 r R+r^2 \Rightarrow 3 R^2-2 r R-5 r^2=0 \quad \Rightarrow \quad R=\frac{2 r+\sqrt{4 r^2+60 r^2}}{6}=\frac{5}{3} r \\
& \frac{A_1}{A_3}=\left(\frac{r}{R}\right)^2=\frac{9}{25} \neq 0.3607
\end{aligned}
$$
Renee’s guess is incorrect.
Question
This question asks you to investigate the Maclaurin series of $\mathrm{e}^x$.
(a) Write down the Maclaurin series of $\mathrm{e}^x$ using sigma notation.
(b) Hence show that e $>2.5$.
(c) (i) Prove by induction that $2^{n-1}<n$ ! for $n \in \mathbb{Z}, n \geq 3$.
(ii) By comparing with a suitable geometric series, or otherwise, show that e $<3$.
(d) The probability density function of the nomal distribution is $\frac{1}{\sigma \sqrt{2 \pi}} \mathrm{e}^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$. Given $Z \sim \mathrm{N}(0,1)$, show that $\mathrm{P}(0<Z<a) \approx \frac{1}{\sqrt{2 \pi}}\left(a-\frac{1}{6} a^3\right)$ for small $a$.
(e) Show that $f(x)=\sum_{k=0}^{\infty} \frac{x^k}{k !}>0$ for $x \in \mathbb{R}$.
(f) Determine, with justification, whether $g(x)=\sum_{k=0}^n \frac{x^k}{k !}>0$ for $x \in \mathbb{R}, n \in \mathbb{Z}^{+}$.
Let $h(x)=\sum_{k=0}^{2 n} \frac{x^k}{k !}$ for $n \in \mathbb{Z}^{+}$.
(g) (i) By considering $\lim _{x \rightarrow \infty} h(x)$ and $\lim _{x \rightarrow-\infty} h(x)$, or otherwise, show that $h(x)$ has a minimum point.
(ii) Show that, at this minimum point, $h(x)=\frac{x^{2 n}}{(2 n) !}$.
(h) Show that $h(x)>0$ for $x \in \mathbb{R}$.
▶️Answer/Explanation
(a) $\sum_{k=0}^{\infty} \frac{x^k}{k !}$
(b) $\mathrm{e}=1+1+\frac{1^2}{2 !}+\frac{1^3}{3 !}+\cdots>1+1+\frac{1^2}{2 !}=2.5$
(c) (i) $H_n: 2^{n-1}<n$ !
When $n=3$, LHS $=4$, RHS $=6 \Rightarrow H_3$ is true.
Assume $H_k$ is true for some $k \geq 3$.
$$
\begin{aligned}
H_k & \Rightarrow 2^{k-1}<k ! \\
& \Rightarrow 2^{(k+1)-1}<2 k !<(k+1) k !=(k+1) ! \\
& \Rightarrow H_{k+1}
\end{aligned}
$$
$H_3$ is true, and $H_k \Rightarrow H_{k+1}$.
$H_n$ is true for $n \in \mathbb{Z}, n \geq 3$.
(ii) $\mathrm{e}=2.5+\sum_{k=3}^{\infty} \frac{1}{k !}<2.5+\sum_{k=3}^{\infty} \frac{1}{2^{k-1}}=2.5+\frac{\frac{1}{4}}{1-\frac{1}{2}}=3$
(d)
$$
\begin{aligned}
\mathrm{P}(0<Z<a) & =\frac{1}{\sqrt{2 \pi}} \int_0^a \mathrm{e}^{-\frac{1}{2} x^2} \mathrm{~d} x=\frac{1}{\sqrt{2 \pi}} \int_0^a\left(1-\frac{1}{2} x^2+\cdots\right) \mathrm{d} x=\frac{1}{\sqrt{2 \pi}}\left[x-\frac{1}{6} x^3+\cdots\right]_0^a \\
& =\frac{1}{\sqrt{2 \pi}}\left(a-\frac{1}{6} a^3+\cdots\right) \approx \frac{1}{\sqrt{2 \pi}}\left(a-\frac{1}{6} a^3\right) \text { for small } a
\end{aligned}
$$
(e) $f(x)=\mathrm{e}^x>0$
(f) Not always true. E.g. when $n=1$ and $x=-2, g(x)=1-2=-1<0$
(g) (i)
$$
\begin{aligned}
& \lim _{x \rightarrow-\infty} h(x)=\lim _{x \rightarrow-\infty}\left(x^{2 n} \sum_{k=0}^{2 n} \frac{x^{k-2 n}}{k !}\right)=\lim _{x \rightarrow-\infty}\left(\frac{x^{2 n}}{(2 n) !}\right)=\infty \\
& \lim _{x \rightarrow \infty} h(x)=\infty \\
& h(x) \text { is continuous. } \\
& \therefore h(x) \text { has a minimum point. }
\end{aligned}
$$
(ii) At this minimum point, $h(x)=h^{\prime}(x)+\frac{x^{2 n}}{(2 n) !}=\frac{x^{2 n}}{(2 n) !}$.
(h) Assume $h(x)$ has a real root.
At the minimum point, $h(x)=\frac{x^{2 n}}{(2 n) !}$ cannot be negative, so $h(x)=0$ and so $x=0$.
But $h(0)=1$, contradiction.
$h(x)$ has no real root.
$h(0)=1>0$.
$h(x)>0$ for $x \in \mathbb{R}$.
Question
This question asks you to investigate the area of a triangle on a clock.
The lengths of the hour, minute and second hands are 3, 4, 4, respectively. $H, M, S$ are the tips of the hour, minute and second hands, respectively.
$A=$ area of $\triangle H M S$
$t=$ time in seconds after midnight
(a) Show that $A=|6 \sin \alpha+6 \sin \beta-8 \sin (\alpha+\beta)|$.
(b) Find the angle (in radians) swept between midnight and $t$ by the:
(i) second hand
(ii) minute hand
(iii) hour hand
(c) Express $A$ in terms of $t$.
(d) (i) Let $I=$ area of $\triangle H M S$ if the hands can be positioned independently (e.g. battery is dead). Find the maximum value of $I$ to two decimal places.
(ii) Find the maximum value of $A$ to two decimal places.
▶️Answer/Explanation
(a)
$$
\begin{aligned}
& A=\left|\frac{1}{2}(3)(4) \sin \alpha+\frac{1}{2}(3)(4) \sin \beta+\frac{1}{2}(4)(4) \sin (2 \pi-\alpha-\beta)\right| \\
& A=|6 \sin \alpha+6 \sin \beta-8 \sin (\alpha+\beta)|
\end{aligned}
$$
(b) (i) $\frac{t \pi}{30}$
(ii) $\frac{t \pi}{1800}$
(iii) $\frac{t \pi}{21600}$
(c) $\alpha=\frac{t \pi}{1800}-\frac{t \pi}{21600}=\frac{11 \pi}{21600} t$
$$
\begin{aligned}
& \beta=\frac{t \pi}{21600}-\frac{t \pi}{30}=-\frac{719 \pi}{21600} t \\
& A=\left|6 \sin \left(\frac{11 \pi}{21600} t\right)-6 \sin \left(\frac{719 \pi}{21600} t\right)+8 \sin \left(\frac{59 \pi}{1800} t\right)\right|
\end{aligned}
$$
(d) (i) At $I_{\max }, \alpha=\beta$.
$$
I=|12 \sin \alpha-8 \sin (2 \alpha)|
$$
GDC: $I_{\max }=17.37$ to two decimal places
(ii) GDC: $A$ exceeds 17.365 , e.g. $A(1361.5773) \approx 17.368247$
$$
17.365<A_{\max }<17.375
$$
$A_{\max }=17.37$ to two decimal places
Question
This question asks you to investigate the horizontal distance from the airport, $l \mathrm{~m}$, at which an airplane should start its descent, given the following conditions:
I. As the airplane descends, it follows a cubic curve, $y=a x^3+b x^2+c x+d$.
$y$ is the airplane’s vertical distance in $\mathrm{m}$ from the horizontal ground.
$x$ is the airplane’s horizontal distance in $\mathrm{m}$ from the point where it begins its descent.
II. The airplane’s cruising altitude is $6 \mathrm{~km}$.
III. The airplane’s horizontal velocity is a constant $100 \mathrm{~m} \mathrm{~s}^{-1}$.
IV. For passenger comfort, vertical acceleration ranges from $-0.1 \mathrm{~m} \mathrm{~s}^{-2}$ to $0.1 \mathrm{~m} \mathrm{~s}^{-2}$.
(a) Find $d$ and $c$.
(b) By considering the point at which $x=l$, or otherwise, show that $a l^3+b l^2+6000=0$ and $3 a l+2 b=0$.
(c) (i) Sketch a graph of the airplane’s vertical velocity against time.
(ii) Sketch a graph of the airplane’s vertical acceleration against time.
(iii) Determine the airplane’s vertical acceleration at the beginning of its descent.
(iv) Express the airplane’s vertical acceleration in terms of $a, b$ and $x$.
(v) Show that $b=-5 \times 10^{-6}$.
(d) Find $l$.
(e) If the airplane can follow any curve, find the minimum value of $l$.
▶️Answer/Explanation
(a)
$$
\begin{aligned}
& d=6000 \\
& y^{\prime}=3 a x^2+2 b x+c \\
& c=0
\end{aligned}
$$
(b) At $x=l, y=0 \Rightarrow 0=a l^3+b l^2+6000$
At $x=l, y^{\prime}=0 \quad \Rightarrow \quad 0=3 a l^2+2 b l \quad \Rightarrow \quad 0=3 a l+2 b$
(c) 
(iii) $-0.1 \mathrm{~m} \mathrm{~s}^{-2}$
(iv)
$$
\begin{aligned}
& \text { vertical velocity }=\frac{\mathrm{d} y}{\mathrm{~d} t}=\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)=\left(3 a x^2+2 b x\right) 100 \\
& \text { vertical acceleration }=\frac{\mathrm{d} v}{\mathrm{~d} t}=\left(\frac{\mathrm{d} v}{\mathrm{~d} x}\right)\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)=(6 a x+2 b) 100^2
\end{aligned}
$$
(v)
$$
\begin{aligned}
& x=0, \text { vertical acceleration }=-0.1: \quad-0.1=(2 b) 100^2 \\
& b=-5 \times 10^{-6}
\end{aligned}
$$
(d)
$$
\begin{aligned}
& 0=3 a l+2\left(-5 \times 10^{-6}\right) \\
& a=\frac{1}{\left(3 \times 10^5\right) l} \\
& 0=\frac{1}{\left(3 \times 10^5\right) l}\left(l^3\right)+\left(-5 \times 10^{-6}\right) l^2+6000 \\
& l=60000
\end{aligned}
$$
(e) On the graph of vertical velocity against time, the area enclosed by the curve and the $t$-axis equals vertical distance travelled $(6 \mathrm{~km})$ and is therefore constant.
Vertical acceleration ranges from -0.1 to 0.1 , so $l$ is minimized when the graph has this shape:
So for $0<x<\frac{l}{2}$ the plane’ s path is $y=6000-k x^2$.
$$
\begin{aligned}
& \left(\frac{l}{2}, 3000\right) \Rightarrow k=\frac{12000}{l^2} \Rightarrow y=6000-\frac{12000}{l^2} x^2 \\
& \text { vertical velocity }=-\frac{24000}{l^2} x(100) \\
& \text { vertical acceleration }=-\frac{24000}{l^2}(100)^2=-0.1 \\
& l_{\min }=20000 \sqrt{6}
\end{aligned}
$$
Question
This question asks you to investigate the number of throws of a die up to and including a certain result.
Throw an unbiased 6-sided die until you get the first 6. Let $X=$ number of throws.
(a) (i) Find $\mathrm{P}(X=1)$.
(ii) Explain why $\mathrm{P}(X=4)=\left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right)$.
(iii) Find $\mathrm{P}(X<9)$.
$[5]$
Let $p=\frac{5}{6}, \quad q=\frac{1}{6}$.
(b) (i) Express $\mathrm{E}(X)$ as an infinite series with $p$ and $q$.
(ii) By using the fact that $q=1-p$, or otherwise, show that $\mathrm{E}(X)=6$.
(c) (i) Show that $\operatorname{Var}(X)=2 p\left(1+2 p+3 p^2+\cdots\right)+\left(1+p+p^2+p^3+\cdots\right)-36$.
(ii) By considering $\frac{\mathrm{d}}{\mathrm{d} t}\left(t+t^2+t^3+\cdots\right)$, or otherwise, find $\operatorname{Var}(X)$.
Throw the die until you get the same score two consecutive times. Let $Y=$ number of throws.
(d) (i) Find $\mathrm{P}(Y=2)$.
(ii) Find $\mathrm{P}(Y=5)$.
(iii) Find $\mathrm{E}(Y)$.
Throw the die until you get each score at least once. Let $Z=$ number of throws.
(e) (i) Find $\mathrm{P}(Z=6)$.
(ii) Find $\mathrm{P}(Z=7)$.
(iii) Find $\mathrm{E}(Z)$.
▶️Answer/Explanation
(a) (i) $\mathrm{P}(X=1)=\frac{1}{6}$
(ii) If $X=4$, the first three throws are not 6 , and the 4 th throw is 6 .
(iii) $\mathrm{P}(X<9)=1-\mathrm{P}(X>8)=1-\left(\frac{5}{6}\right)^8 \approx 0.767$
(b) (i) $\mathrm{E}(X)=q+2 p q+3 p^2 q+4 p^3 q+\cdots$
(ii)
$$
\begin{aligned}
\mathrm{E}(X) & =(1-p)+2(1-p) p+3(1-p) p^2+4(1-p) p^3+\cdots \\
& =1-p+2 p-2 p^2+3 p^2-3 p^3+4 p^3-4 p^4+\cdots \\
& =1+p+p^2+p^3+\cdots \\
& =\frac{1}{1-p} \\
& =6
\end{aligned}
$$
(c) (i)
$$
\begin{aligned}
\operatorname{Var}(X) & =q+4 p q+9 p^2 q+16 p^3 q+\cdots-6^2 \\
& =(1-p)+4(1-p) p+9(1-p) p^2+16(1-p) p^3+\cdots-36 \\
& =1-p+4 p-4 p^2+9 p^2-9 p^3+16 p^3-16 p^4+\cdots-36 \\
& =1+3 p+5 p^2+7 p^3+\cdots-36 \\
& =2 p\left(1+2 p+3 p^2+\cdots\right)+\left(1+p+p^2+p^3+\cdots\right)-36
\end{aligned}
$$
(ii)
$$
\begin{aligned}
& 1+2 t+3 t^2+\cdots=\frac{\mathrm{d}}{\mathrm{d} t}\left(t+t^2+t^3+\cdots\right)=\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{t}{1-t}\right)=\frac{1}{(1-t)^2} \\
& \operatorname{Var}(X)=2 p\left(\frac{1}{(1-p)^2}\right)+\frac{1}{1-p}-36 \\
& =60+6-36 \\
& =30 \\
&
\end{aligned}
$$
(d) (i) $\mathrm{P}(Y=2)=\frac{1}{6}$
(ii) $\mathrm{P}(Y=5)=\left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right)=\frac{125}{1296}$
(iii) $\mathrm{E}(Y)=1+\mathrm{E}(X)=7$
(e) (i) $\mathrm{P}(Z=6)=\frac{6 !}{6^6}=\frac{5}{324}$
(ii)
$$
\begin{aligned}
P(Z=7) & =\frac{6\left(\frac{6 !}{2 !}\right) 5}{6^7} \\
& =\frac{25}{648}
\end{aligned}
$$
(iii) $\mathrm{E}(Z)=\mathrm{E}$ (number of throws to get 1st new score)
$$
\begin{aligned}
& +E(\text { number of throws to get } 2 \text { nd new score })+\cdots \\
& +E(\text { number of throws to get } 6 \text { th new score) } \\
= & 1+\frac{1}{\frac{5}{6}}+\frac{1}{\frac{4}{6}}+\frac{1}{\frac{3}{6}}+\frac{1}{\frac{2}{6}}+\frac{1}{\frac{1}{6}}=\frac{147}{10}
\end{aligned}
$$
Question
This question asks you to investigate the hyperbolic functions cosh and sinh.
Draw line segments from $(0,0)$ to $(p, q)$ and $(p,-q)$ on the right side of the hyperbola $x^2-y^2=1$. Call the area of the enclosed region $A$ if $q \geq 0$, or $-A$ if $q<0$.
Define: $\quad \cosh A=p \quad \sinh A=q$
(a) Show that $A=2 \int_0^q\left(\sqrt{y^2+1}-\frac{\sqrt{q^2+1}}{q} y\right) \mathrm{d} y$.
(b) (i) Differentiate $f(y)=y \sqrt{y^2+1}+\ln \left(y+\sqrt{y^2+1}\right)$ with respect to $y$ and simplify.
(ii) Hence, or otherwise, show that $A=\ln \left(q+\sqrt{q^2+1}\right)$.
(c) Show that $\sinh A=\frac{\mathrm{e}^A-\mathrm{e}^{-A}}{2}$ and $\cosh A=\frac{\mathrm{e}^A+\mathrm{e}^{-A}}{2}$.
(d) (i) Show that $\cosh ^2 x-\sinh ^2 x=1$.
(ii) Show that $\frac{\mathrm{d}}{\mathrm{d} x} \cosh x=\sinh x$ and $\frac{\mathrm{d}}{\mathrm{d} x} \sinh x=\cosh x$.
(iii) Given that $\tanh x=\frac{\sinh x}{\cosh x}$, find $\frac{\mathrm{d}}{\mathrm{d} x} \tanh x$ and simplify your answer.
The inverse of $\sinh$, denoted by $\sinh ^{-1}$, is defined by $\sinh ^{-1}(\sinh x)=x$.
(e) Show that $\int \frac{1}{\sqrt{1+x^2}} \mathrm{~d} x=\sinh ^{-1} x+c$.
▶️Answer/Explanation
(a) The line from $(0,0)$ to $(p, q)$ has equation $x=\frac{p}{q} y \quad \Rightarrow \quad x=\frac{\sqrt{q^2+1}}{q} y$ The right side of $x^2-y^2=1$ has equation $x=\sqrt{y^2+1}$.
$$
A=2 \int_0^q\left(\sqrt{y^2+1}-\frac{\sqrt{q^2+1}}{q} y\right) \mathrm{d} y
$$
(b) (i)
$$
\begin{aligned}
f^{\prime}(y) & =y \frac{2 y}{2 \sqrt{y^2+1}}+\sqrt{y^2+1}+\frac{1+\frac{2 y}{2 \sqrt{y^2+1}}}{y+\sqrt{y^2+1}} \\
& =\frac{y^2\left(y+\sqrt{y^2+1}\right)+\left(y^2+1\right)\left(y+\sqrt{y^2+1}\right)+\sqrt{y^2+1}+y}{\sqrt{y^2+1}\left(y+\sqrt{y^2+1}\right)} \\
& =\frac{2\left(y^3+y^2 \sqrt{y^2+1}+y+\sqrt{y^2+1}\right)}{\sqrt{y^2+1}\left(y+\sqrt{y^2+1}\right)}=\frac{2\left(y\left(y^2+1\right)+\sqrt{y^2+1}\left(y^2+1\right)\right)}{\sqrt{y^2+1}\left(y+\sqrt{y^2+1}\right)}=2 \sqrt{y^2+1}
\end{aligned}
$$
(ii)
$$
\begin{aligned}
A & =\left[y \sqrt{y^2+1}+\ln \left(y+\sqrt{y^2+1}\right)-\frac{\sqrt{q^2+1}}{q} y^2\right]_0^q \\
& =q \sqrt{q^2+1}+\ln \left(q+\sqrt{q^2+1}\right)-q \sqrt{q^2+1}=\ln \left(q+\sqrt{q^2+1}\right)
\end{aligned}
$$
(c)
$$
\begin{gathered}
\frac{\mathrm{e}^A-\mathrm{e}^{-A}}{2}=\frac{1}{2}\left(q+\sqrt{q^2+1}-\frac{1}{q+\sqrt{q^2+1}}\left(\frac{q-\sqrt{q^2+1}}{q-\sqrt{q^2+1}}\right)\right) \\
=\frac{1}{2}\left(q+\sqrt{q^2+1}+q-\sqrt{q^2+1}\right)=q=\sinh A \\
\cosh A=p=\sqrt{q^2+1}=\sqrt{\left(\frac{\mathrm{e}^A-\mathrm{e}^{-A}}{2}\right)^2+1}=\frac{\mathrm{e}^A+\mathrm{e}^{-A}}{2}
\end{gathered}
$$
(d) (i)
$$
\cosh ^2 x-\sinh ^2 x=\left(\frac{\mathrm{e}^x+\mathrm{e}^{-x}}{2}\right)^2-\left(\frac{\mathrm{e}^x-\mathrm{e}^{-x}}{2}\right)^2=\frac{1}{4}\left(\mathrm{e}^{2 x}+2+\mathrm{e}^{-2 x}-\mathrm{e}^{2 x}+2-\mathrm{e}^{-2 x}\right)=1
$$
(ii)
$$
\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{d} x} \cosh x=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{\mathrm{e}^x+\mathrm{e}^{-x}}{2}\right)=\frac{\mathrm{e}^x-\mathrm{e}^{-x}}{2}=\sinh x \\
& \frac{\mathrm{d}}{\mathrm{d} x} \sinh x=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{\mathrm{e}^x-\mathrm{e}^{-x}}{2}\right)=\frac{\mathrm{e}^x+\mathrm{e}^{-x}}{2}=\cosh x
\end{aligned}
$$
(iii) $\frac{\mathrm{d}}{\mathrm{d} x} \tanh x=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{\sinh x}{\cosh x}\right)=\frac{\cosh ^2 x-\sinh ^2 x}{\cosh ^2 x}=\frac{1}{\cosh ^2 x}$
(e)
$$
\begin{aligned}
& q=\sinh A \quad \Rightarrow \quad \sinh ^{-1} q=A=\ln \left(q+\sqrt{q^2+1}\right) \\
& \frac{\mathrm{d}}{\mathrm{d} q} \sinh ^{-1} q=\frac{1+\frac{2 q}{2 \sqrt{q^2+1}}}{q+\sqrt{q^2+1}}=\frac{1}{\sqrt{q^2+1}} \Rightarrow \int \frac{1}{\sqrt{1+x^2}} \mathrm{~d} x=\sinh ^{-1} x+c
\end{aligned}
$$
Question
This question asks you to investigate the number of lattice points enclosed by a circle.
(a) A circle of radius $r$ encloses exactly four lattice points. Find the possible values of $r$.
(b) Prove that $\sqrt{2}$ is irrational.
You are given that $\sqrt{3}$ and $\sqrt{6}$ are also irrational.
(c) By squaring both sides, or otherwise, show that the only rational solution to $\sqrt{2} x+\sqrt{3} y=z$ is $x=y=z=0$.
(d) Show that $(\sqrt{2}, \sqrt{3})$ cannot be equidistant to two distinct lattice points $(a, b)$ and $(c, d)$ where $a, b, c, d \in \mathbb{Z}$.
(e) Show that, for any positive integer $n$, there exists a circle that encloses exactly $n$ lattice points.
▶️Answer/Explanation
(a) $\frac{\sqrt{2}}{2}<r<\frac{\sqrt{10}}{2}$
(b) Assume $\sqrt{2}$ is rational, i.e. $\sqrt{2}=\frac{p}{q}$ where $p$ and $q$ are coprime integers.
$$
p^2=2 q^2
$$
$p$ is even, so $p=2 k, k \in \mathbb{Z}$.
$(2 k)^2=2 q^2$
$$
q^2=2 k^2
$$
$q$ is even.
$p$ and $q$ are not coprime, contradiction.
$\sqrt{2}$ is irrational.
(c) Assume $x, y, z$ are rational and not all 0 .
Square both sides: $2 x^2+2 \sqrt{6} x y+3 y^2=z^2$
If $x \neq 0$ and $y \neq 0$, then $\sqrt{6}=\frac{-z^2-2 x^2-3 y^2}{2 x y} \in \mathbb{Q}$, contradiction
If $x=0$ and $y \neq 0$, then $\sqrt{3}=\frac{z}{y} \in \mathbb{Q}$, contradiction
If $y=0$ and $x \neq 0$, then $\sqrt{2}=\frac{z}{x} \in \mathbb{Q}$, contradiction
$\therefore x=y=z=0$ is the only rational solution.
(d) Assume $(\sqrt{2}, \sqrt{3})$ is equidistant to two distinct lattice points $(a, b)$ and $(c, d)$.
$$
\begin{aligned}
& (\sqrt{2}-a)^2+(\sqrt{3}-b)^2=(\sqrt{2}-c)^2+(\sqrt{3}-d)^2 \\
& 2-2 \sqrt{2} a+a^2+3-2 \sqrt{3} b+b^2=2-2 \sqrt{2} c+c^2+3-2 \sqrt{3} d+d^2 \\
& \sqrt{2}(c-a)+\sqrt{3}(d-b)=\frac{c^2+d^2-a^2-b^2}{2} \\
& \text { From (c), } c-a=d-b=0 \quad \Rightarrow \quad a=c, \quad b=d
\end{aligned}
$$
From (c), $c-a=d-b=0 \quad \Rightarrow \quad a=c, \quad b=d$
$(a, b)$ and $(c, d)$ are not distinct, contradiction
$(\sqrt{2}, \sqrt{3})$ cannot be equidistant to two distinct lattice points $(a, b)$ and $(c, d)$.
(e) Let $(\sqrt{2}, \sqrt{3})$ be the centre of a circle.
Increase the radius. The circle will never touch two lattice points at the same time.
So the radius can be increased until the circle encloses exactly $n$ lattice points.
Question
This question asks you to investigate the roots of a polynomial.
$f(x)=a x^3+b x^2+c x+d=0$ has roots $\alpha, \beta, \gamma$.
Let $S_n=\alpha^n+\beta^n+\gamma^n$.
(a) Factorise $a x^3+b x^2+c x+d$, then expand your result with $\alpha, \beta, \gamma$.
(b) (i) Show that $S_1 a+b=0$.
(ii) By expanding $(\alpha+\beta+\gamma)^2$, or otherwise, show that $S_2 a+S_1 b+2 c=0$.
(iii) By considering $\alpha^{n-3} f(\alpha), \beta^{n-3} f(\beta), \gamma^{n-3} f(\gamma)$, or otherwise, show that $S_n a+S_{n-1} b+S_{n-2} c+S_{n-3} d=0$.
Now you are given that $x^3-x+d=0, d \neq 0$, has roots $\alpha, \beta, \gamma$.
(c) Find the value of:
(i) $\alpha+\beta+\gamma$
(ii) $\alpha^2+\beta^2+\gamma^2$
(iii) $\alpha^3+\beta^3+\gamma^3$
(iv) $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$
(v) $\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\alpha \gamma}$
(d) Find an equation with roots $\alpha+\beta, \beta+\gamma, \alpha+\gamma$.
▶️Answer/Explanation
(a)
$$
\begin{aligned}
& a x^3+b x^2+c x+d \\
& =a(x-\alpha)(x-\beta)(x-\gamma) \\
& =a\left(x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\alpha \gamma) x-\alpha \beta \gamma\right)
\end{aligned}
$$
(b) (i)
$$
\begin{aligned}
& b=-a(\alpha+\beta+\gamma) \\
& b=-a S_1 \\
& S_1 a+b=0
\end{aligned}
$$
(ii)
$$
\begin{aligned}
& (\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha \beta+\beta \gamma+\alpha \gamma) \\
& S_1{ }^2=S_2+2\left(\frac{c}{a}\right) \\
& S_1\left(-\frac{b}{a}\right) a=S_2 a+2 c \\
& S_2 a+S_1 b+2 c=0
\end{aligned}
$$
(iii) Add the equations $\alpha^{n-3} f(\alpha)=0, \quad \beta^{n-3} f(\beta)=0, \quad \gamma^{n-3} f(\gamma)=0$
$$
S_n a+S_{n-1} b+S_{n-2} c+S_{n-3} d=0
$$
(c) (i) $S_1=0$
(ii) $S_2+0 S_1-2=0 \quad \Rightarrow \quad S_2=2$
(iii) $S_3+0 S_2-S_1+S_0 d=0 \Rightarrow S_3=-3 d$
(iv) $S_2+0 S_1-S_0+S_{-1} d=0 \quad \Rightarrow \quad S_{-1}=\frac{1}{d}$
(v) $\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\alpha \gamma}=\frac{\alpha+\beta+\gamma}{\alpha \beta \gamma}=0$
(d) $\alpha+\beta+\gamma=0$, so roots are $-\alpha,-\beta,-\gamma$.
Reflect in $y$-axis.
$$
\begin{aligned}
& (-x)^3-(-x)+d=0 \\
& x^3-x-d=0
\end{aligned}
$$
Question
This question asks you to investigate the envelope curve of projectiles.
A projectile is launched from a point on flat ground with speed $v_0$ at an angle $\theta$ with the ground. The acceleration due to gravity is $g$. You may ignore air resistance.
(a) (i) Express the projectile’s horizontal displacement $x$, in terms of $v_0, \theta$ and time $t$.
(ii) Show that the projectile’ s vertical displacement is $y=-\frac{g}{2} t^2+v_0(\sin \theta) t$.
(iii) Show that the equation of the projectile’ s path is $y=-\frac{g}{2 v_0^2}\left(1+\tan ^2 \theta\right) x^2+(\tan \theta) x$.
The diagram below shows several projectile paths and their envelope (dotted curve), which encloses all the projectile paths.
For any point $(x, y)$ on the envelope, there is only one value of $\theta$ that would cause a projectile to reach that point.
(b) (i) Express the projectile’ s path as a quadratic equation in $\tan \theta$.
(ii) Show that the envelope of projectiles has equation $y=\frac{v_0^2}{2 g}-\frac{g}{2 v_0^2} x^2$.
The “danger zone” is the region in which a bird could be hit by a projectile.
(c) Find the volume of the danger zone.
(d) Verify that, at the apex of a projectile path, $x^2=-4 y^2+\frac{2 v_0^2}{g} y$.
(e) A bird is randomly located in the danger zone. Find the probability that it could be hit from below. [6]
▶️Answer/Explanation
(a) (i) $x=v_0(\cos \theta) t$
(ii)
$$
\begin{aligned}
& a_y=-g \\
& v_y=-g t+v_0(\sin \theta) \\
& y=-\frac{g}{2} t^2+v_0(\sin \theta) t
\end{aligned}
$$
(iii)
$$
\begin{aligned}
& t=\frac{x}{v_0(\cos \theta)} \\
& y=-\frac{g}{2}\left(\frac{x}{v_0(\cos \theta)}\right)^2+v_0(\sin \theta) \frac{x}{v_0(\cos \theta)}=-\frac{g}{2 v_0^2}\left(1+\tan ^2 \theta\right) x^2+(\tan \theta) x
\end{aligned}
$$
(b)
(i)
$$
\frac{g x^2}{2 v_0^2} \tan ^2 \theta-x \tan \theta+y+\frac{g x^2}{2 v_0^2}=0
$$
(ii)
$$
\begin{aligned}
& \Delta=x^2-4\left(\frac{g x^2}{2 v_0^2}\right)\left(y+\frac{g x^2}{2 v_0^2}\right)=0 \\
& y+\frac{g x^2}{2 v_0^2}=\frac{v_0^2}{2 g x^2} x^2 \\
& y=\frac{v_0^2}{2 g}-\frac{g}{2 v_0^2} x^2
\end{aligned}
$$
(c)
$$
\begin{aligned}
V & =\pi \int_0^{\frac{v_0^2}{2 g}} x^2 \mathrm{~d} y=\pi \int_0^{\frac{v_0^2}{2 g}}\left(\frac{v_0^4}{g^2}-\frac{2 v_0^2}{g} y\right) \mathrm{d} y \\
& =\pi\left[\frac{v_0^4}{g^2} y-\frac{v_0^2}{g} y^2\right]_0^{\frac{v_0^2}{2 g}}=\pi\left(\frac{v_0^4}{g^2}\left(\frac{v_0^2}{2 g}\right)-\frac{v_0^2}{g}\left(\frac{v_0^2}{2 g}\right)^2\right)=\frac{v_0^6 \pi}{4 g^3}
\end{aligned}
$$
(d)
$$
\begin{aligned}
& \text { At apex, } v_y=-g t+v_0(\sin \theta)=0 \quad \Rightarrow \quad t=\frac{v_0(\sin \theta)}{g} \\
& x=\frac{v_0^2(\cos \theta)(\sin \theta)}{g} \\
& y=\frac{v_0^2 \sin ^2 \theta}{2 g} \\
& \text { LHS }=\left(\frac{v_0^2(\cos \theta)(\sin \theta)}{g}\right)^2=\frac{v_0^4}{g^2}\left(\cos ^2 \theta\right)\left(\sin ^2 \theta\right) \\
& \text { RHS }=-4\left(\frac{v_0^2 \sin ^2 \theta}{2 g}\right)^2+\frac{2 v_0^2}{g}\left(\frac{v_0^2 \sin ^2 \theta}{2 g}\right)=\frac{v_0^4}{g^2}\left(-\sin ^4 \theta+\sin ^2 \theta\right)=\frac{v_0^4}{g^2}\left(\cos ^2 \theta\right)\left(\sin ^2 \theta\right) \\
& \text { LHS }=\text { RHS }
\end{aligned}
$$
(e) Let $V_B=$ volume of region in which bird could be hit from below
$$
\begin{aligned}
V_B & =\pi \int_0^{\frac{v_0^2}{2 g}}\left(-4 y^2+\frac{2 v_0^2}{g} y\right) \mathrm{d} y=\pi\left[-\frac{4}{3} y^3+\frac{v_0^2}{g} y^2\right]_0^{\frac{v_0{ }^2}{2 g}} \\
& =\pi\left(-\frac{4}{3}\left(\frac{v_0^6}{8 g^3}\right)+\frac{v_0^2}{g}\left(\frac{v_0^2}{2 g}\right)^2\right)=\frac{v_0{ }^6 \pi}{12 g^3} \\
\frac{V_B}{V} & =\frac{1}{3}
\end{aligned}
$$
Question
This question asks you to investigate the probability of getting consecutive outcomes when tossing a coin.
Toss an unbiased coin 10 times to get a sequence of heads $(H)$ and tails $(T)$.
(a) (i) Find the number of possible sequences.
(ii) Find the number of sequences with $5 T$ and $5 H$ and no consecutive $H$.
(iii) Find the number of sequences with $6 T$ and $4 H$ and no consecutive $H$.
(iv) Find the probability of getting at least two consecutive $H$.
[10]
Toss an unbiased coin $n$ times. Let $C_n=$ number of sequences with at least three consecutive $H$.
(b) Find $C_3$.
A sequence can have three consecutive $H$ before the last toss, or can have three consecutive $H$ for the first time with the last toss.
(c) Explain why $C_n=2 C_{n-1}+2^{n-4}-C_{n-4}$ for $n \geq 4$.
(d) Find $C_{10}$.
(e) Find the probability of getting at least three consecutive $H$ in 10 tosses.
(f) Find the probability of getting at least three consecutive $H$ or $T$ in 10 tosses.
▶️Answer/Explanation
(a) (i) $2^{10}(=1024)$
(ii) Put $5 T$ down, then put each $H$ separately between $T$ s or at ends: $6 C 5=6$
(iii) Put $6 T$ down, then put each $H$ separately between $T$ s or at ends: $7 C 4=35$
(iv) $1-\frac{6 C 5+7 C 4+8 C 3+9 C 2+10 C 1+11 C 0}{2^{10}}=\frac{55}{64}$
(b) $C_3=1$
(c) $2 C_{n-1}=$ number of sequences with 3 consecutive $H$ before last toss and last is either $H$ or $T$
$\boldsymbol{R} 1 R 1$ $2^{n-4}-C_{n-4}=$ number of sequences with 3 consecutive $H$ for first time with last toss, so last 4 are $\mathrm{THHH}$ and before that there were no 3 consecutive $H$.
$R 1 R 1$
(d)
$$
\begin{aligned}
& C_4=2 C_3+2^0-C_0=2+1-0=3 \\
& C_5=2 C_4+2^1-C_1=6+2-0=8 \\
& C_6=2 C_5+2^2-C_2=16+4-0=20 \\
& C_7=2 C_6+2^3-C_3=40+8-1=47 \\
& C_8=2 C_7+2^4-C_4=94+16-3=107 \\
& C_9=2 C_8+2^5-C_5=214+32-8=238 \\
& C_{10}=2 C_9+2^6-C_6=476+64-20=520
\end{aligned}
$$
(e) $\frac{C_{10}}{2^{10}}=\frac{65}{128}(M 1) A 1$
(f) For $n$ tosses, let $D_n=$ number of sequences with at least 3 consecutive $H$ or $T$.
For $n \geq 7$, use $D_n=2 D_{n-1}+2\left(2^{n-4}-D_{n-4}-\left(2^{n-7}-D_{n-7}\right)\right)$.
For $n \leq 6$, use $D_n=2 C_n-$ number of sequences with 3 consecutive $H$ and 3 consecutive $T$ \}.
$$
\begin{aligned}
& D_7=2 D_6+2\left(2^3-D_3-\left(2^0-D_0\right)\right)=76+2(8-2-(1-0))=86 \\
& D_8=2 D_7+2\left(2^4-D_4-\left(2^1-D_1\right)\right)=172+2(16-6-(2-0))=188 \\
& D_9=2 D_8+2\left(2^5-D_5-\left(2^2-D_2\right)\right)=376+2(32-16-(4-0))=400 \\
& D_{10}=2 D_9+2\left(2^6-D_6-\left(2^3-D_3\right)\right)=800+2(64-38-(8-2))=840 \\
& \frac{D_{10}}{2^{10}}=\frac{105}{128}
\end{aligned}
$$
Question
This question asks you to investigate definite integrals of powers of $\left(1+x^2\right)^{-1}$.
(a) On the same set of axes, sketch and label $y=\left(1+x^2\right)^{-1}, y=\left(1+x^2\right)^{-2}$ and $y=\left(1+x^2\right)^{-3}$ for $0<x<1$.
(b) Find the exact value of $\int_0^1\left(1+x^2\right)^{-1} \mathrm{~d} x$.
(c) By substituting $x=\tan \theta$, or otherwise, find the exact value of $\int_0^1\left(1+x^2\right)^{-2} \mathrm{~d} x$.
Let $I_n=\int_0^1\left(1+x^2\right)^{-n} \mathrm{~d} x$
(d) (i) By expressing $\left(1+x^2\right)^{-n}$ as $\left(1+x^2\right)^{-n}(1)$, or otherwise, show that $I_{n+1}=\left(1-\frac{1}{2 n}\right) I_n+\frac{2^{-n-1}}{n}$ for $n \geq 1$.
(ii) Find the exact value of $\int_0^1\left(1+x^2\right)^{-3} \mathrm{~d} x$.
(e) Find the exact value of $\int_0^1\left(x^2-2 x+2\right)^{-3} \mathrm{~d} x$.
▶️Answer/Explanation
(a)
(b)
$$
\int_0^1\left(1+x^2\right)^{-1} \mathrm{~d} x=[\arctan x]_0^1=\frac{\pi}{4}
$$
(c)
$$
\begin{aligned}
& x=\tan \theta \Rightarrow \frac{\mathrm{d} x}{\mathrm{~d} \theta}=\sec ^2 \theta \\
& \int\left(1+x^2\right)^{-2} \mathrm{~d} x=\int\left(1+\tan ^2 \theta\right)^{-2} \sec ^2 \theta \mathrm{d} \theta \\
&=\int \cos ^2 \theta \mathrm{d} \theta \\
&=\int \frac{\cos 2 \theta+1}{2} \mathrm{~d} \theta \\
&=\frac{1}{2}\left(\frac{\sin 2 \theta}{2}+\theta\right)+c
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1\left(1+x^2\right)^{-2} \mathrm{~d} x & =\frac{1}{2}\left[\frac{\sin 2 \theta}{2}+\theta\right]_0^{\frac{\pi}{4}} \\
& =\frac{1}{2}\left(\frac{1}{2}+\frac{\pi}{4}\right) \\
& =\frac{\pi}{8}+\frac{1}{4}
\end{aligned}
$$
(d) (i)
$$
\begin{aligned}
& I_n=\int_0^1\left(1+x^2\right)^{-n}(1) \mathrm{d} x \\
& =\left[x\left(1+x^2\right)^{-n}\right]_0^1+2 n \int_0^1 x^2\left(1+x^2\right)^{-n-1} \mathrm{~d} x \\
& =2^{-n}+2 n \int_0^1\left(1+x^2-1\right)\left(1+x^2\right)^{-n-1} \mathrm{~d} x \\
& =2^{-n}+2 n\left(I_n-I_{n+1}\right) \\
& 2 n I_{n+1}=(2 n-1) I_n+2^{-n} \\
& I_{n+1}=\left(1-\frac{1}{2 n}\right) I_n+\frac{2^{-n-1}}{n} \\
&
\end{aligned}
$$
(ii) $I_3=\left(1-\frac{1}{4}\right) I_2+\frac{2^{-3}}{2}=\frac{3}{4}\left(\frac{\pi}{8}+\frac{1}{4}\right)+\frac{1}{16}=\frac{3 \pi}{32}+\frac{1}{4}$
(e)
$$
\begin{aligned}
\int_0^1\left(x^2-2 x+2\right)^{-3} \mathrm{~d} x & =\int_0^1\left(1+(x-1)^2\right)^{-3} \mathrm{~d} x=\int_{-1}^0\left(1+x^2\right)^{-3} \mathrm{~d} x \\
& =\int_0^1\left(1+x^2\right)^{-3} \mathrm{~d} x=\frac{3 \pi}{32}+\frac{1}{4}
\end{aligned}
$$
Question
This question asks you to investigate the Pearson correlation coefficient.
Given a set of points $\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right), \ldots,\left(x_n, y_n\right)$, the Pearson correlation coefficient is defined as
$$
r=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\left(\sum(x-\bar{x})^2\right)\left(\sum(y-\bar{y})^2\right)}} .
$$
(a) (i) Express $r$ in terms of $\boldsymbol{u}=\left(\begin{array}{c}x_1-\bar{x} \\ x_2-\bar{x} \\ \vdots \\ x_n-\bar{x}\end{array}\right)$ and $\boldsymbol{v}=\left(\begin{array}{c}y_1-\bar{y} \\ y_2-\bar{y} \\ \vdots \\ y_n-\bar{y}\end{array}\right)$.
(ii) Show that $-1 \leq r \leq 1$.
$[4]$
(b) (i) For the points $(1,1),(2,2),(3,3), \ldots,(n, n)$, show that $r=1$.
(ii) For the points $(1, n),(2, n-1),(3, n-2), \ldots,(n, 1)$, show that $r=-1$.
$[4]$
Seven IB students’ Maths grades, $x$, were, in random order, $1,2,3,4,5,6,7$.
Their Physics grades, $y$, were, in random order, $1,2,3,4,5,6,7$.
(c) Find the number of possible distributions of their grades.
[2]
(d) (i) Find the value of $\sqrt{\left(\sum(x-\bar{x})^2\right)\left(\sum(y-\bar{y})^2\right)}$.
(ii) Show that there are no more than 57 possible values of $r$.
[6]
(e) (i) Show that if $(a, b)$ and $(c, d)$ are replaced with $(a, d)$ and $(c, b)$, then the value of $\sum(x-\bar{x})(y-\bar{y})$ is reduced by $(a-c)(b-d)$.
(ii) Find a distribution of the grades such that $r=0$.
▶️Answer/Explanation
(a) (i) $r=\frac{\boldsymbol{u} \cdot \boldsymbol{v}}{|\boldsymbol{u} \| \boldsymbol{v}|}$
(ii)
$$
\begin{aligned}
& r=\cos \theta A 1 \\
& -1 \leq \cos \theta \leq 1 \\
& -1 \leq r \leq 1
\end{aligned}
$$
(b) (i) $\boldsymbol{u}=\boldsymbol{v} \quad \Rightarrow \quad r=\cos 0=1$
(ii) $\boldsymbol{u}=-\boldsymbol{v} \quad \Rightarrow \quad r=\cos \pi=-1$
(c) $7 !^2(=25401600)$
(d) (i) $\sqrt{\left(\sum(x-\bar{x})^2\right)\left(\sum(y-\bar{y})^2\right)}=\sum(x-\bar{x})^2=2\left(3^2+2^2+1^2\right)+0^2=28$
(ii) $-28 \leq \sum(x-\bar{x})(y-\bar{y}) \leq 28$
A1
$$
\bar{x}, \bar{y} \in \mathbb{Z} \quad \Rightarrow \quad \sum(x-\bar{x})(y-\bar{y}) \in \mathbb{Z}
$$
$\boldsymbol{R} 1 R 1$
The number of possible $r$ values cannot exceed $28+1+28=57$.
(e) (i)
$$
\begin{aligned}
\text { Reduction } & =((a-4)(b-4)+(c-4)(d-4))-((a-4)(d-4)+(c-4)(b-4)) \\
& =a b+c d-a d-b c-4(a+b+c+d-a-d-c-b) \\
& =(a-c)(b-d)
\end{aligned}
$$
(ii) Example:
$(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7) \Rightarrow \sum(x-\bar{x})(y-\bar{y})=28$
$(1,6),(2,2),(3,3),(4,4),(5,5),(6,1),(7,7) \Rightarrow \sum(x-\bar{x})(y-\bar{y})=28-(1-6)(1-6)=3$
$(1,6),(2,4),(3,3),(4,2),(5,5),(6,1),(7,7) \Rightarrow \sum(x-\bar{x})(y-\bar{y})=3-(2-4)(2-4)=-1$
$(1,6),(2,3),(3,4),(4,2),(5,5),(6,1),(7,7) \Rightarrow \sum(x-\bar{x})(y-\bar{y})=-1-(2-3)(4-3)=0$
$r=0$ for $(1,6),(2,3),(3,4),(4,2),(5,5),(6,1),(7,7)$.
Question
This question asks you to investigate possible applications of the curve $y=\frac{\ln x}{x}$.
(a) For the curve $y=\frac{\ln x}{x}$, find the
(i) $x$-intercept(s)
(ii) asymptote(s)
(iii) coordinates of the stationary point(s) and their nature
(b) Sketch the curve $y=\frac{\ln x}{x}$.
(c) By considering two points on the curve $y=\frac{\ln x}{x}$, show that $\pi^e<\mathrm{e}^\pi$.
(d) Given that $a^b=b^a$ where $a, b \in \mathbb{R}$ and $0<a<b$, find the ranges of $a$ and $b$.
(e) Given that $\log _a b=\log _b a$ where $a, b \in \mathbb{R}$ and $a<b$, find the ranges of $a$ and $b$.
▶️Answer/Explanation
(a) (i) $x=1$
(ii) $x=0$
$$
\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{\ln x}{x}=\lim _{x \rightarrow \infty} \frac{\frac{1}{x}}{1}=0 \\
& y=0
\end{aligned}
$$
(iii) $y^{\prime}=\frac{x\left(\frac{1}{x}\right)-\ln x}{x^2}=0$
$$
\begin{aligned}
& x=\mathrm{e} \\
& y^{\prime \prime}=\frac{x^2\left(-\frac{1}{x}\right)-2 x(1-\ln x)}{x^4}=\frac{-\mathrm{e}}{\mathrm{e}^4}<0 \\
& \left(\mathrm{e}, \frac{1}{\mathrm{e}}\right) \text { maximum }
\end{aligned}
$$
(b)
$$
\begin{aligned}
& \text { (c) } \frac{\ln \pi}{\pi}<\frac{\ln \mathrm{e}}{\mathrm{e}} \\
& \mathrm{e} \ln \pi<\pi \ln \mathrm{e} \\
& \ln \pi^{\mathrm{e}}<\ln \mathrm{e}^\pi \\
& \pi^{\mathrm{e}}<\mathrm{e}^\pi \\
& \text { (d) } \ln a^b=\ln b^a \\
& b \ln a=a \ln b \\
& \frac{\ln a}{a}=\frac{\ln b}{b} \\
& 1<a<\mathrm{e}, \quad b>\mathrm{e}
\end{aligned}
$$
(d)
$$
\begin{aligned}
& \ln a^b=\ln b^a \\
& b \ln a=a \ln b \\
& \frac{\ln a}{a}=\frac{\ln b}{b} \\
& 1<a<\mathrm{e}, \quad b>\mathrm{e}
\end{aligned}
$$
(e)
$$
\begin{aligned}
& \frac{\ln b}{\ln a}=\frac{\ln a}{\ln b} \\
& (\ln a)^2=(\ln b)^2 \\
& \ln a=-\ln b \\
& a=\frac{1}{b} \\
& a<b \quad \Rightarrow \quad a<\frac{1}{a}, \quad b>\frac{1}{b} \\
& a^2<1, \quad b^2>1 \\
& 0<a<1, \quad b>1
\end{aligned}
$$
Question
This question asks you to investigate maximum binomial probabilities.
Toss a biased coin 10 times. Let $H=$ number of heads.
(a) (i) Find the maximum value of $\mathrm{P}(H=10)$, and the corresponding value of $\mathrm{P}($ head).
(ii) Find the maximum value of $\mathrm{P}(H=0)$, and the corresponding value of $\mathrm{P}($ head).
(iii) Deduce the value of $0^{\circ}$.
Let $\mathrm{P}(n, r)=\left(\begin{array}{l}n \\ r\end{array}\right) x^r(1-x)^{n-r}$, where $r \neq 0, n$.
(b) (i) Show that $\ln u<u-1$ for $u>0, u \neq 1$.
(ii) Show that $y=\left(1-\frac{1}{x}\right)^{x-1}$ is decreasing for $x>1$.
(c) Show that $\mathrm{P}(n, n-1)_{\max }=\mathrm{P}(n, 1)_{\max }=\frac{1}{2}$.
(d) (i) Express $\mathrm{P}(n, r)$ as a linear combination of $\mathrm{P}(n-1, r-1)$ and $\mathrm{P}(n-1, r)$.
(ii) Show that either $\mathrm{P}(n, r) \leq \mathrm{P}(n-1, r-1)$ or $\mathrm{P}(n, r) \leq \mathrm{P}(n-1, r)$.
(iii) Show that, for $1<r<n-1, \mathrm{P}(n, r) \leq \frac{1}{2}$.
(e) Find $\mathrm{P}(n, r)_{\max }$.
▶️Answer/Explanation
(a) (i) $\mathrm{P}(H=10)_{\max }=1, \mathrm{P}($ head $)=1$
(ii) $\mathrm{P}(H=0)_{\max }=1, \mathrm{P}($ head $)=0$
(iii) When $\mathrm{P}($ head $)=0, \mathrm{P}(H=0)=\left(\begin{array}{c}10 \\ 0\end{array}\right) 0^0(1-0)^{10}=1$
$$
0^0=1
$$
(b) (i) $u-1$ is tangent to $\ln u$ at $u=1$, and $\ln u$ is concave down.
$$
\therefore \ln u<u-1 \text { for } u>0, u \neq 1 \text {. }
$$
(ii)
$$
\begin{aligned}
& \ln y=(x-1) \ln \left(1-\frac{1}{x}\right) \\
& y^{\prime}=\left(\frac{1}{x}+\ln \left(1-\frac{1}{x}\right)\right)\left(1-\frac{1}{x}\right)^{*-1} \\
& x>1 \\
& 1-\frac{1}{x}>0, \quad 1-\frac{1}{x} \neq 1 \\
& \ln \left(1-\frac{1}{x}\right)<\left(1-\frac{1}{x}\right)-1 \\
& \frac{1}{x}+\ln \left(1-\frac{1}{x}\right)<0 \\
& y^{\prime}<0
\end{aligned}
$$
(c)
$$
\begin{aligned}
& \mathrm{P}(n, n-1)=\left(\begin{array}{c}
n \\
n-1
\end{array}\right) x^{n-1}(1-x)^1=n x^{n-1}-n x^n \\
& \frac{\mathrm{d}}{\mathrm{d} x} \mathrm{P}(n, n-1)=n(n-1) x^{n-2}-n^2 x^{n-1}=n x^{n-2}(n-1-n x)=0 \\
& x=1-\frac{1}{n} \\
& r \neq 0, n \Rightarrow n_{\min }=2 \\
& \mathrm{P}(n, n-1)=n\left(1-\frac{1}{n}\right)^{n-1}\left(\frac{1}{n}\right)=\left(1-\frac{1}{n}\right)^{n-1} \text { is decreasing. } \\
& \mathrm{P}(n, n-1)_{\max }=\left(1-\frac{1}{2}\right)^{2-1}=\frac{1}{2} \\
&
\end{aligned}
$$
By symmetry, $\mathrm{P}(n, 1)_{\max }=\frac{1}{2}$
(d) (i) $\mathrm{P}(n, r)=x \mathrm{P}(n-1, r-1)+(1-x) \mathrm{P}(n-1, r)$
(ii) Assume $\mathrm{P}(n-1, r-1)<P(n, r)$ and $\mathrm{P}(n-1, r)<\mathrm{P}(n, r)$. $\mathrm{P}(n, r)<x \mathrm{P}(n, r)+(1-x) \mathrm{P}(n, r)=\mathrm{P}(n, r)$, contradiction. $\therefore \mathrm{P}(n, r) \leq \mathrm{P}(n-1, r-1)$ or $\mathrm{P}(n, r) \leq \mathrm{P}(n-1, r)$.
(iii) If $1<r<n-1$ then $\mathrm{P}(n, r) \leq \mathrm{P}(k, 1)$ or $\mathrm{P}(n, r) \leq \mathrm{P}(k, k-1)$ for some $k$.
$$
\mathrm{P}(n, r) \leq \frac{1}{2}
$$
(e) $\mathrm{P}(n, r)_{\max }=\frac{1}{2}$
Question
This question asks you to investigate the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$.
(a) By considering the graph below, or otherwise, show that $\sum_{n=1}^{\infty} \frac{1}{n}=\infty$.
(b) (i) By considering the Maclaurin series of $\frac{\sin x}{x}$, or otherwise, solve
$$
1-\frac{x^2}{3 !}+\frac{x^4}{5 !}-\frac{x^6}{7 !}+\cdots=0
$$
Euler claimed that $1-\frac{x^2}{3 !}+\frac{x^4}{5 !}-\frac{x^6}{7 !}+\cdots=\left(1-\frac{x^2}{(\pi)^2}\right)\left(1-\frac{x^2}{(2 \pi)^2}\right)\left(1-\frac{x^2}{(3 \pi)^2}\right) \cdots$
Bob thinks Euler was right because both sides have the same complex roots and non-zero $y$-intercept.
(ii) By considering $\sin (\cos x)$ and $(\sin 1)(\cos x)$, or otherwise, show that Bob’s reasoning is flawed.
Euler’s claim is in fact correct (but not for the reason given by Bob).
(iii) By equating the coefficients of $x^2$ in Euler’s claim, or otherwise, find the exact value of $\sum_{n=1}^{\infty} \frac{1}{n^2}$.
(iv) Find the exact value of $\sum_{n=2}^{\infty} \frac{1}{n^2-1}$.
(c) (i) By considering the graphs below, or otherwise, show that $\frac{1}{98}<\sum_{n=7}^{\infty} \frac{1}{n^3}<\frac{1}{72}$.
(ii) Approximate \( \sum_{n=1}^{\infty} \frac{1}{n^3} \)to three significant figures.
▶️Answer/Explanation
(a) $\sum_{n=1}^{\infty} \frac{1}{n}>\int_1^{\infty} \frac{1}{x} \mathrm{~d} x=\lim _{b \rightarrow \infty}[\ln x]_1^b=\lim _{b \rightarrow \infty}(\ln b)=\infty$
$$
\sum_{n=1}^{\infty} \frac{1}{n}=\infty
$$
(b) (i) $\frac{\sin x}{x}=1-\frac{x^2}{3 !}+\frac{x^4}{5 !}-\frac{x^6}{7 !}+\cdots=0$
$$
x=k \pi, k \in \mathbb{Z}, k \neq 0
$$
(ii) $\sin (\cos x)$ and $(\sin 1)(\cos x)$ have the same complex roots and non-zero $y$-intercept but they are not the same.
(iii) $-\frac{1}{3 !}=-\left(\frac{1}{(\pi)^2}+\frac{1}{(2 \pi)^2}+\frac{1}{(3 \pi)^2}+\cdots\right)$
$$
\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}
$$
(iv)
$$
\begin{aligned}
\sum_{n=2}^{\infty} \frac{1}{n^2-1} & =\sum_{n=2}^{\infty} \frac{1}{(n+1)(n-1)} \\
& =\sum_{n=2}^{\infty}\left(\frac{1}{2}\right)\left(\frac{1}{n-1}-\frac{1}{n+1}\right) \\
& =\frac{1}{2}\left(\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\cdots\right) \\
& =\frac{3}{4}
\end{aligned}
$$
(c) (i) $\int_7^{\infty} \frac{1}{x^3} \mathrm{~d} x<\sum_{n=7}^{\infty} \frac{1}{n^3}<\int_6^{\infty} \frac{1}{x^3} \mathrm{~d} x$
$$
\begin{aligned}
& \lim _{b \rightarrow \infty}\left[-\frac{1}{2 x^2}\right]_7^b<\sum_{n=7}^{\infty} \frac{1}{n^3}<\lim _{b \rightarrow \infty}\left[-\frac{1}{2 x^2}\right]_6^b \\
& \frac{1}{98}<\sum_{n=7}^m \frac{1}{n^3}<\frac{1}{72}
\end{aligned}
$$
$$
\text { (ii) } \begin{aligned}
& \sum_{n=1}^6 \frac{1}{n^3}+\frac{1}{98}<\sum_{n=1}^6 \frac{1}{n^3}+\sum_{n=7}^{\infty} \frac{1}{n^3}<\sum_{n=1}^6 \frac{1}{n^3}+\frac{1}{72} \\
& 1.200496 \ldots<\sum_{n=1}^{\infty} \frac{1}{n^3}<1.204180 \ldots \\
& \sum_{n=1}^{\infty} \frac{1}{n^3}=1.20 \text { to } 3 \text { sf }
\end{aligned}
$$
Question
This question asks you to investigate the pigeonhole principle (PHP), which states that if $n$ pigeons are put in $m$ pigeonholes, where $n>m$, then at least one pigeonhole must contain more than one pigeon.
(a) An equilateral triangle of side length $2 \mathrm{~m}$ contains five points. By dividing the triangle into four equilateral triangles, or otherwise, show that two of the points must be within $1 \mathrm{~m}$ of each other.
$[2]$
(b) (i) A circle contains four points. By drawing a diameter through one of the points, or otherwise, show that three of the points must be in the same closed semicircle.
(ii) A circle contains four randomly located points. Find the probability that all of the points are in the same closed semicircle.
[7]
(c) A sphere contains five points. Show that four of the points must be in the same closed hemisphere.
$[3]$
(d) (i) List the possible remainders when a positive integer is divided by $n$, where $n \in Z^{+}$.
(ii) By considering the numbers $7,77,777, \ldots, 7 \sum_{k=0}^n 10^k$, or otherwise, show that,
for any $n \in \mathbf{Z}^{+}$, there is a number whose digits are limited to 0 and 7 that is divisible by $n$. [5]
(e) Given real numbers $x_1, x_2, x_3, x_4$, by letting $x=\tan \theta$, or otherwise, show that two of them must satisfy $0 \leq \frac{x_t-x_j}{1+x_i x_j}<\sqrt{3}$.
▶️Answer/Explanation
(a)
By PHP, at least one small triangle must contain two points.
Those two points must be within $1 \mathrm{~m}$ of each other.
(b) (i) Draw a diameter through one of the points to create two closed semicircles.
By PHP, one of the semicircles must contain two of the other three points.
That semicircle contains three of the four points.
(ii) Call the points $A, B, C, D$.
Draw a radius through $A$ and rotate the radius clockwise $\pi$ to form a closed semicircle.
The probability all four points lie in a semicircle with $A$ as the first one clockwise is $\left(\frac{1}{2}\right)^3$. Likewise for $B, C, D$.
So the total probability is $4\left(\frac{1}{2}\right)^3=\frac{1}{2}$.
(c) Draw a plane through two of the points and the centre of the sphere to create two closed hemispheres.
By PHP, one of the hemispheres must contain two of the other three points.
That hemisphere contains four of the five points.
(d) (i) $0,1,2, \ldots, n-1$
(ii) By PHP, two of the numbers $7,77,777, \ldots, 7 \sum_{k=0}^n 10^k$ must have the same remainder when divided by $n$.
The difference between them has digits limited to 0 and 7 , and is divisible by $n$.
(e) Let $x=\tan \theta, 0 \leq \theta<\pi$.
Divide the interval 0 to $\pi$ into three equal width intervals (each closed on left and open on right). By PHP, at least one interval must contain two $\theta$, call them $\theta_i$ and $\theta_j$ where $\theta_i \geq \theta_j$.
$$
\begin{aligned}
& 0 \leq \theta_i-\theta_j<\frac{\pi}{3} \\
& 0 \leq \tan \left(\theta_i-\theta_j\right)<\sqrt{3} \\
& 0 \leq \frac{\tan \theta_i-\tan \theta_j}{1+\tan \theta_i \tan \theta_i}<\sqrt{3} \\
& 0 \leq \frac{x_i-x_j}{1+x_i x_j}<\sqrt{3}
\end{aligned}
$$
Question
(a) In each part below, show that $y=f(x)$ goes through the origin and is concave up, then sketch
$$
y=\frac{f(x)}{x} .
$$
(i) $f(x)=\mathrm{e}^x-1$
(ii) $f(x)=-\ln (x+1)$
(iii) $f(x)=\sec x-1,-\frac{\pi}{2}<x<\frac{\pi}{2}$
(b) Make a conjecture about $\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{f(x)}{x}\right), x \neq 0$.
Let the tangent to $y=f(x)$ at $x \neq 0$ be $y=m x+c$.
(c) (i) Show that $\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{f(x)}{x}\right)=-\frac{c}{x^2}$ for $x \neq 0$.
(ii) Show that $y=\frac{f(x)}{x}$ is increasing for $x \neq 0$.
(d) Show that $y=\frac{\arcsin x}{x}$ is increasing for $0<x<1$.
(e) (i) By expressing $\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$ as $\mathrm{e}^{g(x)}$, or otherwise, show that $y=\left(\frac{a^x+b^x}{2}\right)^{\frac{2}{x}}, 0<a<b$, is increasing for $x \neq 0$.
(ii) Find the range of $y=\left(\frac{a^x+b^x}{2}\right)^{\frac{4}{x}}, 0<a<b$, in terms of $a$ and $b$.
(f) Determine, with justification, whether $y=x f(x)$ is increasing for all $x$ in its domain.
▶️Answer/Explanation
\text { (a) (i) } \begin{array}{ll}
& f(0)=\mathrm{e}^0-1=0 \\
& f^{\prime}(x)=\mathrm{e}^x \\
& f^{\prime \prime}(x)=\mathrm{e}^x>0
\end{array}
\text { (ii) } \begin{aligned}
f(0) & =-\ln (0+1)=0 \\
f^{\prime}(x) & =-\frac{1}{x+1} \\
f^{\prime \prime}(x) & =\frac{1}{(x+1)^2}>0
\end{aligned}
\text { (iii) } \begin{aligned}
& f(0)=\sec 0-1=0 \\
f^{\prime}(x) & =(\sec x)(\tan x) \\
f^{\prime \prime}(x) & =\sec ^3 x+(\sec x)\left(\tan ^2 x\right)>0
\end{aligned}
(b) $\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{f(x)}{x}\right)>0$
(c) (i)
$$
\begin{aligned}
\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{f(x)}{x}\right) & =\frac{x f^{\prime}(x)-f(x)}{x^2} \\
& =\frac{x m-(m x+c)}{x^2} \\
& =-\frac{c}{x^2}
\end{aligned}
$$
(ii) $y=m x+c$ is tangent to $y=f(x)$ at $x \neq 0$ and $y=f(x)$ is concave up.
$y=m x+c$ is below $y=f(x)$
$$
\begin{aligned}
& c<0 \\
& \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{f(x)}{x}\right)=-\frac{c}{x^2}>0
\end{aligned}
$$
(e) (i)
$$
\begin{aligned}
&\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\mathrm{e}^{\ln \left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}}=\mathrm{e}^{\frac{\ln \left(\frac{a^x+b^x}{2}\right)}{x}} \\
& \ln \left(\frac{a^0+b^0}{2}\right)=0 \\
& \frac{\mathrm{d}}{\mathrm{d} x} \ln \left(\frac{a^x+b^x}{2}\right)=\frac{a^x \ln a+b^x \ln b}{a^x+b^x} \\
& \frac{\mathrm{d}^2}{\mathrm{~d} x^2} \ln \left(\frac{a^x+b^x}{2}\right)=\frac{\left(a^x+b^x\right)\left(a^x(\ln a)^2+b^x(\ln b)^2\right)-\left(a^x \ln a+b^x \ln b\right)^2}{\left(a^x+b^x\right)^2} \\
&=\frac{a^x b^x(\ln a-\ln b)^2}{\left(a^x+b^x\right)^2}>0
\end{aligned}
$$
$\frac{\ln \left(\frac{a^x+b^x}{2}\right)}{x}$ is increasing for $x \neq 0$.
$\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$ is increasing for $x \neq 0$.
(ii)
$$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\lim _{x \rightarrow \infty} \mathrm{e}^{\frac{\ln \left(\frac{a^x+b^x}{2}\right)}{x}}=\lim _{x \rightarrow \infty} \mathrm{e}^{\left(\frac{a^x \ln a+b^x \ln b}{a^x+b^x}\right)}=\lim _{x \rightarrow \infty} \mathrm{e}^{\left(\frac{\left(\frac{a}{b}\right)^x \ln a+\ln b}{\left(\frac{a}{b}\right)^x+1}\right)}=b \\
& \lim _{x \rightarrow 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\lim _{x \rightarrow 0} \mathrm{e}^{\left(\frac{\left(\frac{a}{b}\right)^x \ln a+\ln b}{\left(\frac{a}{b}\right)^x+1}\right)}=\sqrt{a b} \\
& \lim _{x \rightarrow-\infty}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\lim _{x \rightarrow-\infty} \mathrm{e}^{\left(\frac{\ln a+\left(\frac{b}{a}\right)^x \ln b}{1+\left(\frac{b}{a}\right)^x}\right)}=a \\
& a<\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}<b, \quad\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}} \neq \sqrt{a b}
\end{aligned}
$$
(f) No. E.g. $\frac{\mathrm{d}}{\mathrm{d} x}\left(x\left(\mathrm{e}^x-1\right)\right)=x \mathrm{e}^x+\mathrm{e}^x-1=-1<0$ when $x=-1$.
Question
This question asks you to investigate the application of complex nur
(a) Show that, when a complex number $z$ is multiplied by i, the r plane is rotated $\frac{\pi}{2}$ anticlockwise about the origin.
The diagram below shows an arbitrary quadrilateral with squares $v$ each side. Van Aubel’s theorem states that $P R \perp O S$ and $P R=O S$
In the diagram below, the quadrilateral and squares are superimposed on the complex plane.
(b) (i) Explain why $B=2 a+b+b \mathrm{i}$.
(ii) Express $A, C, D$ in terms of $a, b, c, d$.
(c) Show that $P R \perp Q S$ and $P R=Q S$.
(d) Determine whether $P R$ and $Q S$ necessarily bisect each other.
▶️Answer/Explanation
(a)
$$
\begin{aligned}
& |\mathrm{i} z|=|\mathrm{i}||z|=|z| \\
& \arg (\mathrm{iz})=\arg (\mathrm{i})+\arg (z)=\arg (z)+\frac{\pi}{2}
\end{aligned}
$$
(b) (i) To move from 0 to $B$, we move from 0 to $2 a$, then move half the displacement to $2(a+b)$, which is $b$, then move $b$ rotated anticlockwise $\frac{\pi}{2}$, which is $b \mathrm{i}$.
(ii)
$$
\begin{aligned}
& A=a+a \mathrm{i} \\
& C=2 a+2 b+c+c \mathrm{i} \\
& D=2 a+2 b+2 c+d+d \mathrm{i}
\end{aligned}
$$
(c)
$$
\begin{aligned}
C-A & =a+2 b+c+c \mathrm{i}-a \mathrm{i} \\
& =b-d+c \mathrm{i}-a \mathrm{i} \\
B-D & =-b+b \mathrm{i}-2 c-d-d \mathrm{i} \\
& =a-c+b \mathrm{i}-d \mathrm{i} \\
& =\mathrm{i}(b-d+c \mathrm{i}-a \mathrm{i}) \\
& =\mathrm{i}(C-A)
\end{aligned}
$$
$B-D$ is $C-A$ rotated $\frac{\pi}{2}$ anticlockwise about the origin $\Rightarrow \quad P R \perp Q S$ and $P R=Q S$
(d)
$$
\begin{aligned}
& \frac{1}{2}(A+C)=\frac{1}{2}(3 a+2 b+c+\mathrm{i}(a+c)) \\
& \frac{1}{2}(B+D)=\frac{1}{2}(4 a+3 b+2 c+d+\mathrm{i}(b+d))=\frac{1}{2}(3 a+2 b+c+\mathrm{i}(b+d))
\end{aligned}
$$
$a+c$ is not necessarily equal to $b+d$.
$P R$ and $Q S$ do not necessarily bisect each other.
Question
This question asks you to investigate sums and products of cosines and sines.
For this question, answers obtained only by technology will not receive marks.
(a) Let $z_1=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7}$.
(i) Show that $z_1$ is a root of $z^7-1=0$.
(ii) Show that $z_1$ is a root of $z^6+z^5+z^4+z^3+z^2+z+1=0$.
(iii) Simplify $\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$.
(iv) Simplify $\cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}+\cos \frac{7 \pi}{7}$.
(b) (1) Solve $\sin 7 \theta=0$.
(ii) Show that $\frac{\sin 7 \theta}{\sin \theta}=64 \cos ^6 \theta-80 \cos ^4 \theta+24 \cos ^2 \theta-1$ for $\sin \theta \neq 0$.
(iii) Solve $64 x^6-80 x^4+24 x^2-1=0$.
(iv) Deduce the value of $\left(\cos \frac{\pi}{7}\right)\left(\cos \frac{2 \pi}{7}\right)\left(\cos \frac{3 \pi}{7}\right) \ldots\left(\cos \frac{7 \pi}{7}\right)$.
(c) (i) Find the value of $\left(\sin \frac{\pi}{100}\right)\left(\sin \frac{2 \pi}{100}\right)\left(\sin \frac{3 \pi}{100}\right) \ldots\left(\sin \frac{100 \pi}{100}\right)$.
(ii) Approximate $\sin \frac{\pi}{100}+\sin \frac{2 \pi}{100}+\sin \frac{3 \pi}{100}+\cdots+\sin \frac{100 \pi}{100}$.
▶️Answer/Explanation
(a) (i) $z_1^7-1=\cos 2 \pi+i \sin 2 \pi-1=0$
(ii) $z_1^6+z_1^5+z_1^4+\cdots+1=\frac{z_1^7-1}{z_1-1}=0$
(iii)
$$
\begin{aligned}
& \operatorname{Re}\left(z_1^6+z_1^5+z_1^4+\cdots+1\right)=0 \\
& \cos \frac{12 \pi}{7}+\cos \frac{10 \pi}{7}+\cos \frac{8 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{2 \pi}{7}+1=0 \\
& \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{2 \pi}{7}+1=0 \\
& \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=-\frac{1}{2}
\end{aligned}
$$
(iv) $\sum_{k=0}^7 \cos \frac{k \pi}{7}=0 \Rightarrow \cos \frac{\pi}{7}+\cos \frac{3 \pi}{7}+\cos \frac{5 \pi}{7}+\cos \frac{7 \pi}{7}=-\frac{1}{2}$
(b) (i) $\theta=\frac{k \pi}{7}, k \in \mathbb{Z}$
$$
\text { (ii) } \begin{aligned}
\frac{\sin 7 \theta}{\sin \theta} & =\frac{\operatorname{lm}(\cos \theta+\mathrm{i} \sin \theta)^7}{\sin \theta} \\
& =\frac{1}{s}\left(7 c^6 s-35 c^4 s^3+21 c^2 s^5-s^7\right) \\
& =7 c^6-35 c^4\left(1-c^2\right)+21 c^2\left(1-c^2\right)^2-\left(1-c^2\right)^3 \\
& =7 c^6-35 c^4\left(1-c^2\right)+21 c^2\left(1-2 c^2+c^4\right)-\left(1-3 c^2+3 c^4-c^6\right) \\
& =64 \cos ^6 \theta-80 \cos ^4 \theta+24 \cos ^2 \theta-1
\end{aligned}
$$
(iii) $x=\cos \frac{k \pi}{7}, k-1,2,3,4,5,6$
(iv) $\left(\cos \frac{\pi}{7}\right)\left(\cos \frac{2 \pi}{7}\right)\left(\cos \frac{3 \pi}{7}\right) \ldots\left(\cos \frac{7 \pi}{7}\right)=($ product of roots $)\left(\cos \frac{7 \pi}{7}\right)=\left(-\frac{1}{64}\right)(-1)=\frac{1}{64}$
(c) (i) 0
(ii) $\sum_{k=1}^{100} \sin \frac{k \pi}{100} \approx 100 \times$ (average value of $\sin x$ from $x=0$ to $x=\pi$ )
$$
\begin{aligned}
& =100 \frac{\int_0^\pi \sin x d x}{\pi} \\
& =\frac{200}{\pi}
\end{aligned}
$$
Question
This question asks you to investigate the radii of “stacked” circles.
It can be shown that the radii of the circles on the left are in a geometric sequence, and the radii of the circles on the right are in an arithmetic sequence. In this question, we are looking for the sequence of radii of the circles in the middle.
(a) Using a geometrical argument, or otherwise, show that $\lim _{\frac{r_2}{r_1} \rightarrow 1}\left(\frac{r_1+r_2}{r_2^{1.5}-r_1^{1.5}}\right)=1$
(b) Show that $\lim _{\frac{r_2}{r_1} \rightarrow 1}\left(\sqrt{r_2}-\sqrt{r_1}\right)=\frac{2}{3}$.
(c) Show that, for large values of $r, r_n \approx\left(\frac{2}{3}(n-1)+\sqrt{r_1}\right)^2$.
(d) (i) Show that $c_1=t_1^{1.5}+\frac{2}{3} t_1{ }^{0.5}$.
(ii) Show that if $t_1=100$ then $r_1 \approx 100.222$ and $r_2 \approx 114.010$.
(iii) Comment on the closeness of the approximation in (c).
(e) On the same set of axes, sketch graphs of $r_n$ against $n$ for stacked circles on $y=|x|^1, y=|x|^{1.5}$ and $y=|x|^2$, assuming that their $r_1$ are equal and large.
▶️Answer/Explanation
(a) As $\frac{r_2}{r_1} \rightarrow 1, r_1+r_2=c_2-c_1 \approx t_2{ }^{1.5}-t_1{ }^{1.5} \approx r_2{ }^{1.5}-r_1{ }^{1.5}$
$$
\frac{r_1+r_2}{r_2^{1.5}-r_1^{1.5}} \approx 1
$$
(b)
$$
\begin{aligned}
& \lim _{\frac{r_2}{r_1} \rightarrow 1}\left(\sqrt{r_2}-\sqrt{r_1}\right)=\lim _{\frac{r_2}{r_1} \rightarrow 1}\left(\sqrt{r_2}-\sqrt{r_1}\right)\left(\frac{r_1+r_2}{r_2^{1.5}-r_1^{1.5}}\right)=\lim _{\frac{r_2}{r_1} \rightarrow 1}\left(\frac{r_1+r_2}{r_1}\right)\left(\frac{r_1 \sqrt{r_2}-r_1 \sqrt{r_1}}{r_2^{1.5}-r_1^{1.5}}\right) \\
= & 2 \lim _{\frac{r_2}{r_1} \rightarrow 1}\left(\frac{\left(\frac{r_2}{r_1}\right)^{0.5}-1}{\left(\frac{r_2}{r_1}\right)^{1.5}-1}\right)=2 \lim _{\frac{r_2 \rightarrow 1}{r_1} \rightarrow 1} \frac{0.5\left(\frac{r_2}{r_1}\right)^{-0.5}}{1.5\left(\frac{r_2}{r_1}\right)^{0.5}}=\frac{2}{3}
\end{aligned}
$$
(c)
$$
\begin{aligned}
& \sqrt{r_n} \approx \frac{2}{3}+\sqrt{r_{n-1}} \approx \frac{2}{3}+\frac{2}{3}+\sqrt{r_{n-2}} \approx \frac{2}{3}+\frac{2}{3}+\frac{2}{3}+\sqrt{r_{n-3}} \approx \cdots \approx \frac{2}{3}(n-1)+\sqrt{r_1} \\
& r_n \approx\left(\frac{2}{3}(n-1)+\sqrt{r_1}\right)^2
\end{aligned}
$$
(d)
$$
\begin{aligned}
& \text { (i) } \frac{\mathrm{d}}{\mathrm{d} x} x^{1.5}=1.5 x^{0.5} \\
& c_1=t_1^{1.5}+\frac{t_1}{1.5 t_1{ }^{0.5}}=t_1^{1.5}+\frac{2}{3} t_1{ }^{0.5}
\end{aligned}
$$
(ii)
$$
\begin{aligned}
& r_1=\sqrt{t_1{ }^2+\left(\frac{2}{3} t_1{ }^{0.5}\right)^2} \approx 100.2219759 \\
& c_2-c_1=r_1+r_2 \\
& \left(t_2^{1.5}+\frac{2}{3} t_2{ }^{0.5}\right)-\left(t_1^{1.5}+\frac{2}{3} t_1{ }^{0.5}\right)=100.2219759+\sqrt{t_2{ }^2+\left(\frac{2}{3} t_2{ }^{0.5}\right)^2} \\
& \text { GDC: } t_1=100 \Rightarrow t_2=113.7876 \\
& r_2=\sqrt{t_2{ }^2+\left(\frac{2}{3} t_2{ }^{0.5}\right)^2} \approx 114.010
\end{aligned}
$$
(iii) From (c), letting $n=2$, LHS $=114.010$, RHS $=114.015$.
Approximation is close.
(e)
Question
This question asks you to investigate the shortest distance between a Christmas tree and a bird flying past it.
The Christmas tree can be modelled by a cone with base perimeter $x^2+z^2=1$ and tip $(0,5,0)$.
Line $l_1$ passes through the tip and base perimeter of the cone.
(a) Show that the equation of $l_1$ is $r=\left(\begin{array}{l}0 \\ 5 \\ 0\end{array}\right)+\lambda\left(\begin{array}{c}\cos \theta \\ -5 \\ \sin \theta\end{array}\right)$.
The bird’s path can be modelled by line $l_2: r=\left(\begin{array}{l}0 \\ 0 \\ 5\end{array}\right)+t\left(\begin{array}{c}1 \\ 1 \\ -2\end{array}\right)$.
(b) (i) Find a vector that is perpendicular to $l_1$ and the surface of the cone.
(ii) Show that the distance between $l_1$ and $l_2$ is minimized when $\theta \approx 0.55321$.
(c) Find the coordinates of the point on the tree that is closest to the bird.
(d) Find the shortest distance between the bird and the tree.
A second bird’ s path can be modelled by line $l_3: r=\left(\begin{array}{c}10 \\ 0 \\ -10\end{array}\right)+t\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)$.
(e) Find the shortest distance between the second bird and the tree.
▶️Answer/Explanation
(a) General point on base permimeter is $(\cos \theta, 0, \sin \theta)$.
Direction of $l_1:\left(\begin{array}{c}\cos \theta \\ 0 \\ \sin \theta\end{array}\right)-\left(\begin{array}{l}0 \\ 5 \\ 0\end{array}\right)=\left(\begin{array}{c}\cos \theta \\ -5 \\ \sin \theta\end{array}\right)$
$$
l_1: \quad \mathbf{r}=\left(\begin{array}{l}
0 \\
5 \\
0
\end{array}\right)+\lambda\left(\begin{array}{c}
\cos \theta \\
-5 \\
\sin \theta
\end{array}\right)
$$
(b) (i) $\left(\begin{array}{c}5 \cos \theta \\ 1 \\ 5 \sin \theta\end{array}\right)$
(ii)
$$
\begin{aligned}
& \left(\begin{array}{c}
5 \cos \theta \\
1 \\
5 \sin \theta
\end{array}\right) \cdot\left(\begin{array}{c}
1 \\
1 \\
-2
\end{array}\right)=0 \\
& 5 \cos \theta-10 \sin \theta+1=0
\end{aligned}
$$
Bird flies on right side of tree, so $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$.
GDC: $\theta \approx 0.55321$
(c) Common perpendicular to $l_1$ and $l_2:\left(\begin{array}{c}\lambda \cos \theta \\ 5-5 \lambda \\ \lambda \sin \theta\end{array}\right)-\left(\begin{array}{c}t \\ t \\ 5-2 t\end{array}\right)=\left(\begin{array}{c}\lambda \cos \theta-t \\ -5 \lambda-t+5 \\ \lambda \sin \theta+2 t-5\end{array}\right)$
$$
\begin{aligned}
& \left(\begin{array}{c}
\lambda \cos \theta-t \\
-5 \lambda-t+5 \\
\lambda \sin \theta+2 t-5
\end{array}\right) \cdot\left(\begin{array}{c}
\cos \theta \\
-5 \\
\sin \theta
\end{array}\right)=0 \\
& 26 \lambda+(-\cos \theta+5+2 \sin \theta) t-25-5 \sin \theta=0 \\
& 26 \lambda+5.2 t-27.62711=0 \quad[\mathrm{~A}] \\
& \left(\begin{array}{c}
\lambda \cos \theta-t \\
-5 \lambda-t+5 \\
\lambda \sin \theta+2 t-5
\end{array}\right) \cdot\left(\begin{array}{c}
1 \\
1 \\
-2
\end{array}\right)=0 \\
& (\cos \theta-5-2 \sin \theta) \lambda-6 t+15=0 \\
& -5.2 \lambda-6 t+15=0 \quad[\mathrm{~B}] \\
& 6[\mathrm{~A}]+5.2[\mathrm{~B}]: \quad 128.96 \lambda-87.76264=0 \\
& \lambda=0.68054 \\
& (0.57903,1.59729,0.35757) \\
& (0.579,1.60,0.358) \text { to } 3 \text { sf } \\
& t=1.91020 \\
& (1.91020,1.91020,1.17960) \\
& d=\sqrt{(1.91020-0.57903)^2+(1.91020-1.59729)^2+(1.17960-0.35757)^2}=1.60
\end{aligned}
$$
[A]
(d) $t=1.91020$
$(1.91020,1.91020,1.17960)$
$$
d=\sqrt{(1.91020-0.57903)^2+(1.91020-1.59729)^2+(1.17960-0.35757)^2}=1.60
$$
(e) Shortest distance is when $t=0$.
Shortest distance $=10 \sqrt{2}-1$
Question
This question asks you to investigate arithmetic and geometric sequences that have three matching terms. For example, the arithmetic sequence $1,-2,-5,-8$ and the geometric sequence $1,-2,4,-8$ have the same $u_1, u_2$ and $u_4$.
(a) Find a geometric sequence that has three matching terms with the arithmetic sequence
$1, \frac{5}{8}, \frac{1}{4},-\frac{1}{8}$.
Suppose an arithmetic sequence with common difference $d$, and a geometric sequence with common ratio $r$, have the same $u_1, u_{p+1}$ and $u_{q+1}$ where $0<p<q$.
(b) (i) Sketch graphs of $y=u_1 r^{x-1}$ for $x>0$, for the cases $r>1, r=1$ and $0<1<r$.
(ii) Find the possible positive values of $r$, justifying your answer.
(c) Show that $p\left(r^q-1\right)-q\left(r^D-1\right)=0$.
Let $f(r)=p\left(r^q-1\right)-q\left(r^D-1\right)$, where $p$ is even and $q$ is odd.
(d) (i) Show that $f(r)$ has a root between -1 and 0 .
(ii) Show that $f(r)$ cannot have more than one negative root.
[10]
(e) (i) Evaluate $\lim _{q \rightarrow \infty} \frac{r^q-1}{r^2-1}$ for the negative root $r$.
(ii) Show that the negative root $r$ can be arbitrarily close to -1 .
▶️Answer/Explanation
(a)1,\(-\frac{1}{2}\),\( \frac{1}{4}\),\(-\frac{1}{8}\)
(b)(i)
(ii) If $0<r<1$ or $r>1$, a straight line cannot intersect $y=u_1 r^{x-1}$ at three points.
$\therefore$ The only possible positive value of $r$ is 1 .
(c)
$$
\begin{aligned}
& u_{p+1}=u_1+p d=u_1 r^p \\
& u_{q+1}=u_1+q d=u_1 r^q \\
& d=\frac{u_1 r^p-u_1}{p}=\frac{u_1 r^q-u_1}{q} \\
& p\left(r^q-1\right)-q\left(r^p-1\right)=0
\end{aligned}
$$
(d) (i)
$$
\begin{aligned}
& f(-1)=-2 p<0 \\
& f(0)=q-p>0 \\
& f(r) \text { is continuous. } \\
& \therefore f(r) \text { has a root between }-1 \text { and } 0 .
\end{aligned}
$$
$f(r)$ is continuous.
$\therefore f(r)$ has a root between -1 and 0 .
(ii) Assume $f(r)$ has more than one negative root.
$$
\begin{aligned}
& f^{\prime}(r)=0 \text { for some } r<0 . \\
& f^{\prime}(r)=p q r^{q-1}-p q r^{p-1}=p q r^{p-1}\left(r^{q-p}-1\right)=0 \\
& r=0,1 \text {, contradiction } \\
& \therefore f(r) \text { cannot have more than one negative root. }
\end{aligned}
$$
(e) (i)
$$
\begin{aligned}
& \frac{r^q-1}{r^2-1}=\frac{q}{2} \\
& \lim _{q \rightarrow \infty} \frac{r^q-1}{r^2-1}=\lim _{q \rightarrow \infty} \frac{q}{2}=\infty
\end{aligned}
$$
(ii)
$$
\begin{aligned}
& -2 \leq \lim _{q \rightarrow \infty}\left(r^q-1\right) \leq-1 \\
& \lim _{q \rightarrow \infty}\left(r^2-1\right)=0 \\
& \lim _{q \rightarrow \infty} r=-1
\end{aligned}
$$
$r$ can be arbitrarily close to -1 .
Question
This question asks you to investigate arrowheads, which are bounded by a circle and tangents to the circle.