% =============================================================
% Y11 EXTENDED — END OF YEAR REVIEW BOOKLET
% Cumulative review: Probability, Functions, Exp/Log, Trigonometry
% Style: matches Trig_Review_Source structure (4 packets, Parts A/B/C)
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\fancyfoot[L]{\small May 2026}
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  \fancyfoot[R]{\small\itshape J.\ Murphy --- EOY Test Preparation}
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\begin{document}
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% TITLE PAGE
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\begin{center}
{\LARGE\bfseries\color{tealmain} End of Year (Cumulative) Review}\\[3pt]
{\large\color{tealdark} Year 11 Extended}\\[2pt]
{\small\itshape May / June 2026 --- Assessment Preparation}\\[10pt]
\ifsol
  {\large\bfseries\color{sectionblue} WORKED SOLUTIONS}
\else
  {\large\bfseries\color{sectionblue} QUESTION BOOKLET}
\fi
\end{center}
\vspace{6pt}
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\noindent\begin{tabular}{@{}p{4cm}p{6cm}p{4cm}@{}}
\textbf{Name:} \rule{3cm}{0.4pt} & \textbf{Class:} \rule{4.5cm}{0.4pt} & \textbf{Date:} \rule{2.5cm}{0.4pt}
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\vspace{6pt}
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\begin{packinfo}
\textbf{\textcolor{tealdark}{About this booklet.}} Four review packets covering the Year~11 Extended cumulative content: \textbf{Probability}, \textbf{Functions}, \textbf{Exponentials \& Logarithms}, and \textbf{Trigonometry}. Each packet starts with warm-up practice (Part~A), works through core skills (Part~B), and ends with applied or unfamiliar contexts (Part~C). Total recommended time: approximately 5--6 hours, split across multiple sessions.

\medskip
Diagrams are not drawn to scale --- sketch your own where helpful. Give answers exactly or to 3~significant figures unless told otherwise. Show all working with each step on a new line. Pay attention to angle units (DEG vs RAD) and to which paper a question would belong to (a calculator may or may not be available).

\medskip
{\small\textbf{Codes used in this booklet.} \dfm{384a} = Dr~Frost Maths skill code (search the code at \url{drfrostmaths.com}). \haese{10E:17H} = Haese Mathematics 10 Extended, chapter~17, section~H.}
\end{packinfo}
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\newpage

% =============================================================
% PACKET 1 --- PROBABILITY
% =============================================================
\section{Packet 1: Probability (Sets, Venn \& Probability)}
\begin{packinfo}
\textbf{Topic:} Probability and set notation. \textbf{Suggested time:} approx.\ 75 minutes. \textbf{Questions:} 9.

\medskip
\textbf{Haese chapter references:} \haese{10E:3} (Sets \& set notation), \haese{10E:17} (Probability and tree diagrams), \haese{10E:17H} (Conditional probability --- Extended only).

\medskip
\textbf{Skills covered (DFM):} \dfm{384}, \dfm{395}, \dfm{396}, \dfm{397}, \dfm{398}.

\medskip
\textbf{Key formulas.}
$P(A') = 1 - P(A)$;\quad
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$;\quad
$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$.\\
$A$ and $B$ \emph{mutually exclusive} iff $P(A \cap B) = 0$.\quad
$A$ and $B$ \emph{independent} iff $P(A \cap B) = P(A)\,P(B)$.
\end{packinfo}
\subsection{Part A --- Warm-up: simple probability \& Venn}
\question{1}\level{1--2}
A fair die is rolled. Let $A$ = "score is even" and $B$ = "score is greater than~3". Find:
\begin{enumerate}[label=(\alph*)]
  \item $P(A)$;
  \item $P(B)$;
  \item $P(A \cap B)$;
  \item $P(A \cup B)$.
\end{enumerate}
\workspace{4cm}
\begin{soln}
$A = \{2,4,6\}$, $B = \{4,5,6\}$, $A\cap B = \{4,6\}$, $A\cup B = \{2,4,5,6\}$.
(a)~$\tfrac{1}{2}$.\quad (b)~$\tfrac{1}{2}$.\quad (c)~$\tfrac{1}{3}$.\quad (d)~$\tfrac{2}{3}$.
\end{soln}
\question{2}\level{1--2}
In a class of $30$ students, $18$ play tennis, $12$ play hockey, and $5$ play both.
\begin{enumerate}[label=(\alph*)]
  \item Draw a Venn diagram.
  \item How many play neither tennis nor hockey?
  \item Find $P(\text{plays tennis only})$.
\end{enumerate}
\workspace{5cm}
\begin{soln}
(a)~Tennis only: $13$; both: $5$; hockey only: $7$; neither: $5$.\quad
(b)~$\mathbf{5}$.\quad (c)~$\dfrac{13}{30}$.
\end{soln}
\question{3}\level{3--4}
A bag contains $4$ red and $3$ blue counters. Two are drawn one at a time \emph{with} replacement.
\begin{enumerate}[label=(\alph*)]
  \item Draw a tree diagram.
  \item Find $P(\text{both red})$.
  \item Find $P(\text{exactly one red})$.
\end{enumerate}
\workspace{5cm}
\begin{soln}
(a)~Tree with $P(R) = \tfrac{4}{7}$, $P(B) = \tfrac{3}{7}$ on each branch.
(b)~$\left(\tfrac{4}{7}\right)^2 = \tfrac{16}{49}$.\quad
(c)~$2 \cdot \tfrac{4}{7}\cdot\tfrac{3}{7} = \tfrac{24}{49}$.
\end{soln}
\subsection{Part B --- Core skills}
\question{4}\level{3--4}
A bag contains $5$ red and $4$ blue marbles. Two are drawn \emph{without} replacement. Find:
\begin{enumerate}[label=(\alph*)]
  \item $P(\text{both blue})$;
  \item $P(\text{one of each colour})$.
\end{enumerate}
\workspace{5cm}
\begin{soln}
(a)~$\dfrac{4}{9}\cdot\dfrac{3}{8} = \dfrac{12}{72} = \mathbf{\dfrac{1}{6}}$.\quad
(b)~$\dfrac{5}{9}\cdot\dfrac{4}{8} + \dfrac{4}{9}\cdot\dfrac{5}{8} = \dfrac{40}{72} = \mathbf{\dfrac{5}{9}}$.
\end{soln}
\question{5}\level{3--4}
The universal set is $U = \{1, 2, 3, \ldots, 15\}$. Let $A = \{x \in U : x \text{ is a multiple of } 3\}$ and $B = \{x \in U : x \text{ is a factor of } 12\}$.
\begin{enumerate}[label=(\alph*)]
  \item List the elements of $A \cap B$ and of $A \cup B$.
  \item Find $P(A' \cap B)$ when an element of $U$ is chosen at random.
  \item State, with reasoning, whether $A$ and $B$ are mutually exclusive.
\end{enumerate}
\workspace{6cm}
\begin{soln}
$A = \{3,6,9,12,15\}$, $B = \{1,2,3,4,6,12\}$.\\
(a)~$A\cap B = \{3,6,12\}$;\;\; $A\cup B = \{1,2,3,4,6,9,12,15\}$.\\
(b)~$A' \cap B = \{1,2,4\}$, so $P = \mathbf{\tfrac{3}{15} = \tfrac{1}{5}}$.\\
(c)~\textbf{Not} mutually exclusive, since $A \cap B = \{3,6,12\} \neq \emptyset$.
\end{soln}
\question{6}\level{3--4}
A diagnostic test is used to detect a rare condition. In the population:
\begin{itemize}[nosep, leftmargin=1.4em, topsep=2pt]
  \item $2\%$ of people have the condition.
  \item Among those who have the condition, the test correctly returns positive in $90\%$ of cases.
  \item Among those who do not have the condition, the test incorrectly returns positive in $5\%$ of cases (a false positive).
\end{itemize}
A person is selected at random from the population and tested.
\begin{enumerate}[label=(\alph*)]
  \item Draw a tree diagram showing the four outcomes (with/without condition; positive/negative test).
  \item Find the probability that the test returns positive.
  \item Given that the test returns positive, find the probability that the person actually has the condition. Comment briefly on what this tells you about the test.
\end{enumerate}
\workspace{8cm}
\begin{soln}
Let $C$ = "has the condition" and $T^+$ = "tests positive".\\
(a)~Branches: $C$~(0.02) $\to$ $T^+$~(0.90), $T^-$~(0.10); $C'$~(0.98) $\to$ $T^+$~(0.05), $T^-$~(0.95).\\
(b)~$P(T^+) = (0.02)(0.90) + (0.98)(0.05) = 0.018 + 0.049 = \mathbf{0.067}$.\\
(c)~$P(C \mid T^+) = \dfrac{0.018}{0.067} \approx \mathbf{0.269}$, i.e.\ about $27\%$.\;
Despite the test being $90\%$ accurate at detecting the condition, only about $27\%$ of those who test positive actually have it. This is because the condition is rare ($2\%$), so even a small false-positive rate produces many more false positives than true positives.
\end{soln}
\subsection{Part C --- Applied / Unfamiliar Contexts}
\question{7}\level{5--6}
At a school of $100$ students: $55$ enjoy Maths, $48$ enjoy Science, $30$ enjoy Art. $22$ enjoy Maths \& Science, $10$ enjoy Science \& Art, $15$ enjoy Maths \& Art, and $6$ enjoy all three.
\begin{enumerate}[label=(\alph*)]
  \item Find the number of students who enjoy at least one of the three subjects.
  \item Find the number who enjoy exactly one of the three.
  \item Find the probability that a randomly chosen student enjoys none of the three.
\end{enumerate}
\workspace{7cm}
\begin{soln}
(a)~$|M\cup S\cup A| = 55+48+30 - 22 - 10 - 15 + 6 = \mathbf{92}$.\\
(b)~Pair-only counts (subtract all-three): $16, 4, 9$. Singles: $24, 22, 11$. Total exactly one $= \mathbf{57}$.\\
(c)~None: $100 - 92 = 8$, so $P = \mathbf{0.08}$.
\end{soln}
\question{8}\level{5--6}
At a school, $40\%$ of students study French ($F$) and $25\%$ study Spanish ($S$). Of those who study French, $30\%$ also study Spanish.
\begin{enumerate}[label=(\alph*)]
  \item Find $P(F \cap S)$.
  \item Find $P(F \cup S)$.
  \item Determine, with reasoning, whether $F$ and $S$ are independent.
\end{enumerate}
\workspace{6cm}
\begin{soln}
(a)~$P(F \cap S) = P(F)\,P(S \mid F) = 0.40 \times 0.30 = \mathbf{0.12}$.\\
(b)~$P(F \cup S) = 0.40 + 0.25 - 0.12 = \mathbf{0.53}$.\\
(c)~Test: $P(F)\,P(S) = 0.40 \times 0.25 = 0.10 \neq 0.12$, so \textbf{not independent}.
\end{soln}
\question{9}\level{7--8}\;\textit{(Y10 carryover --- combinatorial reasoning)}
A four-digit code is formed using the digits $1$, $2$, $3$, $4$, $5$ \emph{without} repetition.
\begin{enumerate}[label=(\alph*)]
  \item How many different codes are possible?
  \item Find the probability that a randomly formed code is even.
  \item Find the probability that a randomly formed code starts with~$1$ \emph{and} ends with~$5$.
\end{enumerate}
\workspace{6cm}
\begin{soln}
(a)~$5\times 4\times 3\times 2 = \mathbf{120}$ codes.\\
(b)~Even iff last digit is $2$ or $4$. $2 \times 4! / (5!/4!\cdot\text{ish}) = $ alternatively: $2 \cdot 4\cdot 3\cdot 2 = 48$ favourable. $P = \tfrac{48}{120} = \mathbf{\tfrac{2}{5}}$.\\
(c)~First $=1$, last $=5$: $1 \cdot 3 \cdot 2 \cdot 1 = 6$ codes. $P = \tfrac{6}{120} = \mathbf{\tfrac{1}{20}}$.
\end{soln}
\newpage

% =============================================================
% PACKET 2 --- FUNCTIONS
% =============================================================
\section{Packet 2: Functions (Domain, Range, Composite, Inverse, Quadratic, Transformations)}
\begin{packinfo}
\textbf{Topic:} Functions, domain \& range, composition, inverses, quadratics in vertex form, transformations of graphs. \textbf{Suggested time:} approx.\ 90 minutes. \textbf{Questions:} 9.

\medskip
\textbf{Haese chapter references:} \haese{10E:15A--E,G} (Composite, Inverse), \haese{10E:20} (Functions Review --- relations, domain, range), \haese{10E:9} \& \haese{10E:21} (Quadratics, vertex form, factoring), \haese{10E:6} (Lines \& gradients --- Y10 carryover).

\medskip
\textbf{Skills covered (DFM):} \dfm{225}, \dfm{227}, \dfm{228}, \dfm{279}, \dfm{281}, \dfm{198}, \dfm{256}.

\medskip
\textbf{Key facts.}
$f^{-1}$: domain of $f^{-1}$ = range of $f$ (and vice versa). Graph of $f^{-1}$ is reflection of $f$ in $y = x$.\\
Vertex form: $y = a(x - h)^2 + k$ has vertex $(h, k)$. If $a > 0$, minimum; if $a < 0$, maximum.\\
Transformations: $y = f(x) + a$ shifts up by $a$;\; $y = f(x - a)$ shifts right by $a$;\; $y = -f(x)$ reflects in $x$-axis;\; $y = f(ax)$ stretches horizontally by factor $\tfrac{1}{a}$.
\end{packinfo}
\subsection{Part A --- Warm-up: notation, domain, evaluation}
\question{1}\level{1--2}
Let $f(x) = 2x + 3$. Find:
\begin{enumerate}[label=(\alph*)]
  \item $f(5)$;
  \item $f(-1)$;
  \item the value of $x$ for which $f(x) = 11$.
\end{enumerate}
\workspace{4cm}
\begin{soln}
(a)~$13$.\quad (b)~$1$.\quad (c)~$2x+3 = 11 \Rightarrow x = \mathbf{4}$.
\end{soln}
\question{2}\level{3--4}
State the largest possible (natural) domain of each function over the real numbers.
\begin{enumerate}[label=(\alph*)]
  \item $f(x) = \dfrac{1}{x - 3}$;
  \item $g(x) = \sqrt{x - 1}$;
  \item $h(x) = \dfrac{1}{\sqrt{4 - x}}$.
\end{enumerate}
\workspace{4cm}
\begin{soln}
(a)~$x \in \mathbb{R}, x \neq 3$.\quad
(b)~$x \geq 1$.\quad
(c)~$4 - x > 0$, so $\mathbf{x < 4}$.
\end{soln}
\question{3}\level{3--4}
Solve each quadratic by factorising, then state the solutions.
\begin{enumerate}[label=(\alph*)]
  \item $x^2 - 7x + 12 = 0$;
  \item $2x^2 - 5x - 3 = 0$.
\end{enumerate}
\workspace{4cm}
\begin{soln}
(a)~$(x-3)(x-4) = 0$, so $x = 3$ or $x = 4$.\quad
(b)~$(2x+1)(x-3) = 0$, so $x = -\tfrac{1}{2}$ or $x = 3$.
\end{soln}
\subsection{Part B --- Core skills}
\question{4}\level{3--4}
Let $f(x) = 2x + 5$ and $g(x) = x^2 - 3$.
\begin{enumerate}[label=(\alph*)]
  \item Find $f(g(3))$.
  \item Find a simplified expression for $(f\circ g)(x)$.
  \item Find $f^{-1}(x)$, and state its domain.
\end{enumerate}
\workspace{7cm}
\begin{soln}
(a)~$g(3) = 6$; $f(6) = \mathbf{17}$.\\
(b)~$(f\circ g)(x) = 2(x^2 - 3) + 5 = \mathbf{2x^2 - 1}$.\\
(c)~$y = 2x + 5 \Rightarrow x = \tfrac{y-5}{2}$. So $f^{-1}(x) = \dfrac{x-5}{2}$.\;
Domain of $f^{-1}$ = range of $f$ = $\mathbf{\mathbb{R}}$.
\end{soln}
\question{5}\level{5--6}
Let $f(x) = x^2 + 8x + 10$.
\begin{enumerate}[label=(\alph*)]
  \item By completing the square, or otherwise, express $f(x)$ in vertex form.
  \item Hence, or otherwise, solve $f(x) = 0$, giving exact answers in surd form.
  \item State the range of $f$.
  \item Describe the single transformation that maps $y = x^2$ onto $y = (x+5)^2$.
\end{enumerate}
\workspace{8cm}
\begin{soln}
(a)~$f(x) = (x+4)^2 - 16 + 10 = \mathbf{(x+4)^2 - 6}$.\\
(b)~$(x+4)^2 = 6 \Rightarrow x = -4 \pm \sqrt{6}$.\\
(c)~Vertex $(-4,-6)$ minimum, so $\mathbf{y \geq -6}$.\\
(d)~\textbf{Translation 5 units left} (vector $\binom{-5}{0}$).
\end{soln}
\question{6}\level{5--6}
The graph of $y = f(x)$ has a minimum point at $(-1, 4)$ and crosses the $y$-axis at $(0, 7)$. State the coordinates of the corresponding point on each transformed graph:
\begin{enumerate}[label=(\alph*)]
  \item $y = f(x) - 5$ (state the new minimum);
  \item $y = f(x + 3)$ (state the new minimum);
  \item $y = -f(x)$ (state the new $y$-intercept and whether the turning point becomes a max or min);
  \item $y = f(x/2)$ (state the new minimum and the new $y$-intercept).
\end{enumerate}
\workspace{7cm}
\begin{soln}
(a)~$(-1, -1)$.\quad
(b)~$(-4, 4)$.\\
(c)~$(0, -7)$;\; turning point is now a \textbf{maximum} at $(-1, -4)$.\\
(d)~$y = f(x/2)$ is a horizontal stretch by factor~$2$. Min at $(-2, 4)$; $y$-intercept at $(0, 7)$ unchanged (since $f(0) = 7$).
\end{soln}
\subsection{Part C --- Applied / Unfamiliar Contexts}
\question{7}\level{5--7}
A water jet from a fountain follows the path $h(x) = -(x - 4)^2 + 9$, where $h$ is the height of the jet (in metres) above ground level and $x$ is the horizontal distance (in metres) from the nozzle.
\begin{enumerate}[label=(\alph*)]
  \item State the maximum height of the jet, and the horizontal distance at which it is reached.
  \item Find $h(0)$, and explain in context what this value tells you about the position of the nozzle.
  \item Find the values of $x$ at which the jet returns to ground level. Give exact (surd or integer) answers.
\end{enumerate}
\workspace{6cm}
\begin{soln}
(a)~Max $9$ m at $x = 4$ m.\\
(b)~$h(0) = -16 + 9 = -7$. The nozzle is $7$ m \emph{below} ground level (e.g.\ in a basin).\\
(c)~$(x-4)^2 = 9 \Rightarrow x - 4 = \pm 3 \Rightarrow x = 1$ or $x = 7$.
\end{soln}
\question{8}\level{7--8}
The points $A(-2, 2)$ and $B(4, 8)$ both lie on the parabola $y = \tfrac{1}{2}x^2$.
\begin{enumerate}[label=(\alph*)]
  \item Find the midpoint $M$ of $AB$, and the gradient of the chord $AB$.
  \item Find the equation of the perpendicular bisector of $AB$.
  \item The perpendicular bisector meets the parabola at points $P$ and $Q$. Show that the $x$-coordinates satisfy $x^2 + 2x - 12 = 0$.
  \item Hence find the $x$-coordinates of $P$ and $Q$ in exact (surd) form.
\end{enumerate}
\workspace{8cm}
\begin{soln}
(a)~$M = (1, 5)$;\; gradient of $AB = \dfrac{8-2}{4-(-2)} = 1$.\\
(b)~Perpendicular gradient $= -1$. Through $M$: $y - 5 = -1(x - 1) \Rightarrow y = -x + 6$.\\
(c)~Set equal: $\tfrac{1}{2}x^2 = -x + 6 \Rightarrow x^2 = -2x + 12 \Rightarrow x^2 + 2x - 12 = 0$.\\
(d)~$x = \dfrac{-2 \pm \sqrt{4 + 48}}{2} = \dfrac{-2 \pm \sqrt{52}}{2} = -1 \pm \sqrt{13}$.
\end{soln}
\question{9}\level{7--8}\;\textit{(Y10 carryover: lines + functions)}
The line $y = 2x + 1$ meets the curve $y = x^2 + k$ at exactly one point.
\begin{enumerate}[label=(\alph*)]
  \item Show that $x^2 - 2x + (k - 1) = 0$.
  \item Find the value of $k$ for which the line is tangent to the curve. Find also the coordinates of the point of tangency.
\end{enumerate}
\workspace{6cm}
\begin{soln}
(a)~Set equal: $x^2 + k = 2x + 1 \Rightarrow x^2 - 2x + (k-1) = 0$.\\
(b)~Tangent $\Leftrightarrow$ discriminant $=0$: $4 - 4(k - 1) = 0 \Rightarrow k = \mathbf{2}$. Then $x^2 - 2x + 1 = 0$, so $x = 1$, $y = 2(1) + 1 = 3$. Point of tangency $\mathbf{(1, 3)}$.
\end{soln}
\newpage

% =============================================================
% PACKET 3 --- EXP/LOG
% =============================================================
\section{Packet 3: Exponentials \& Logarithms}
\begin{packinfo}
\textbf{Topic:} Index laws (Y10 carryover), exponential graphs, logarithm definitions, log laws, exponential / logarithmic equations, growth and decay. \textbf{Suggested time:} approx.\ 90 minutes. \textbf{Questions:} 9.

\medskip
\textbf{Haese chapter references:} \haese{10E:15A--C} (Index laws --- Y10 carryover prerequisite), \haese{10E:18} (Exponential and logarithmic functions, including all of: graphs, log definitions, equations, growth and decay).

\medskip
\textbf{Skills covered (DFM):} \dfm{305}, \dfm{306}, \dfm{310}, \dfm{311}, \dfm{312}, \dfm{313}, \dfm{120}.

\medskip
\textbf{Key facts.}
$y = a^x$ passes through $(0, 1)$ with horizontal asymptote $y = 0$.\\
$\log_a x = y \Leftrightarrow a^y = x$ (with $a > 0, a \neq 1$, $x > 0$).\\
Log laws: $\log(MN) = \log M + \log N$;\; $\log(M/N) = \log M - \log N$;\; $\log(M^k) = k\log M$;\; $\log_a x = \dfrac{\ln x}{\ln a}$.
\end{packinfo}
\subsection{Part A --- Warm-up: indices, log/exp form}
\question{1}\level{1--2}
Simplify, leaving each as a single power.
\begin{enumerate}[label=(\alph*)]
  \item $x^5 \cdot x^3$;
  \item $\dfrac{y^7}{y^2}$;
  \item $\left(z^4\right)^3$;
  \item $\left(\dfrac{1}{2}\right)^{-3}$.
\end{enumerate}
\workspace{4cm}
\begin{soln}
(a)~$x^8$.\quad (b)~$y^5$.\quad (c)~$z^{12}$.\quad (d)~$2^3 = 8$.
\end{soln}
\question{2}\level{1--2}
Convert between exponential and logarithmic form.
\begin{enumerate}[label=(\alph*)]
  \item Write $2^5 = 32$ in logarithmic form.
  \item Write $\log_3 81 = 4$ in exponential form.
  \item Evaluate $\log_5 125$.
\end{enumerate}
\workspace{3cm}
\begin{soln}
(a)~$\log_2 32 = 5$.\quad (b)~$3^4 = 81$.\quad (c)~$\log_5 125 = 3$ (since $5^3 = 125$).
\end{soln}
\question{3}\level{3--4}
Sketch the graph of $y = 2^x$ for $-2 \leq x \leq 4$. Mark the $y$-intercept and the value at $x = 3$. State the equation of the horizontal asymptote.
\workspace{6cm}
\begin{soln}
$y$-intercept $(0, 1)$; $y(3) = 8$. Horizontal asymptote $y = 0$. Sketch increasing exponential through these points, approaching but never crossing the $x$-axis.
\end{soln}
\subsection{Part B --- Core skills}
\question{4}\level{3--4}
Without a calculator, simplify each expression to a single logarithm or numerical value.
\begin{enumerate}[label=(\alph*)]
  \item $\log 8 + \log 5 - \log 4$;
  \item $\log 2 + 2\log 5$;
  \item $\log_a(p^3 q^2) - \log_a(p\,q^3)$ in terms of $\log_a p$ and $\log_a q$.
\end{enumerate}
\workspace{5cm}
\begin{soln}
(a)~$\log\tfrac{40}{4} = \log 10 = 1$.\quad
(b)~$\log 2 + \log 25 = \log 50$.\\
(c)~$\log_a\tfrac{p^3 q^2}{p\,q^3} = \log_a\tfrac{p^2}{q} = 2\log_a p - \log_a q$.
\end{soln}
\question{5}\level{5--7}
Solve $\log_3(x) + \log_3(x - 2) = 1$. Show all working, and explain why one of the algebraic solutions must be rejected.
\workspace{6cm}
\begin{soln}
$\log_3[x(x-2)] = 1 \Rightarrow x(x-2) = 3$, so $x^2 - 2x - 3 = 0$.\\
$(x-3)(x+1) = 0 \Rightarrow x = 3$ or $x = -1$.\\
For both logs to be defined, $x > 2$. Reject $x = -1$. So $\mathbf{x = 3}$.
\end{soln}
\question{6}\level{5--6}
An investment of CHF\,$5000$ is placed in a savings account paying $4.5\%$ interest, compounded annually. Let $V$ (CHF) be the value after $t$ years.
\begin{enumerate}[label=(\alph*)]
  \item Write a formula for $V$ in terms of $t$.
  \item Find the value of the investment after $8$ years, correct to the nearest franc.
  \item Find the smallest integer $t$ for which the investment has at least \emph{doubled}.
  \item By what percentage has the investment grown after $25$ years?
\end{enumerate}
\workspace{7cm}
\begin{soln}
(a)~$V = 5000(1.045)^t$.\\
(b)~$V(8) = 5000(1.045)^8 \approx \mathbf{7106}$ CHF.\\
(c)~$(1.045)^t \geq 2 \Rightarrow t \geq \tfrac{\ln 2}{\ln 1.045} \approx 15.75$, so $\mathbf{t = 16}$ years.\\
(d)~$V(25)/5000 = (1.045)^{25} \approx 3.005$, so a growth of $\approx \mathbf{200\%}$ (i.e.\ tripled).
\end{soln}
\subsection{Part C --- Applied / Unfamiliar Contexts}
\question{7}\level{7--8}
Solve the equation $\;4 \cdot 2^{2x} - 9 \cdot 2^x + 2 = 0$.
\begin{enumerate}[label=(\alph*)]
  \item Show that the substitution $u = 2^x$ converts this into a quadratic in $u$.
  \item Solve the quadratic for $u$.
  \item Hence find all real values of $x$.
\end{enumerate}
\workspace{6cm}
\begin{soln}
(a)~$2^{2x} = (2^x)^2 = u^2$, so $4u^2 - 9u + 2 = 0$.\\
(b)~$(4u - 1)(u - 2) = 0$, so $u = \tfrac{1}{4}$ or $u = 2$.\\
(c)~$2^x = \tfrac{1}{4} = 2^{-2} \Rightarrow x = -2$;\; $2^x = 2 \Rightarrow x = 1$. Solutions $\mathbf{x = -2, \, 1}$.
\end{soln}
\question{8}\level{7--8}
A caf\'e records the number of weekly customers $C$ (in hundreds) at two times: at $t = 1$ week, $C = 6$;\; at $t = 4$ weeks, $C = 162$. The number of customers is to be modelled by $C(t) = a \cdot b^{\,t}$, where $a$ and $b$ are positive constants and $t$ is measured in weeks.
\begin{enumerate}[label=(\alph*)]
  \item Use the two observations to set up two equations in $a$ and $b$, and hence find the values of $a$ and $b$.
  \item Hence state the initial number of weekly customers $C(0)$.
  \item Sketch the graph of $C(t)$ for $0 \leq t \leq 4$, marking the $C$-intercept and the value at $t = 4$.
  \item Find, using the laws of logarithms, the time at which $C$ first reaches $1000$ (hundred). Give your answer correct to 3~s.f.
\end{enumerate}
\workspace{8cm}
\begin{soln}
(a)~$a \cdot b = 6$ \,(i),\quad $a \cdot b^4 = 162$ \,(ii). Divide (ii)/(i): $b^3 = 27$, so $b = \mathbf{3}$. From (i): $a \cdot 3 = 6$, so $a = \mathbf{2}$.\\
(b)~$C(0) = a = \mathbf{2}$ hundred (i.e.\ $200$ initial weekly customers).\\
(c)~Increasing exponential through $(0, 2)$; $C(4) = 2 \cdot 81 = \mathbf{162}$.\\
(d)~$2 \cdot 3^t = 1000 \Rightarrow 3^t = 500 \Rightarrow t = \log_3 500 = \tfrac{\ln 500}{\ln 3} \approx \mathbf{5.66}$ weeks.
\end{soln}
\question{9}\level{7--8}\;\textit{(Mixed: log + change-of-base)}
Solve $5^{x} = 17$, giving the answer correct to 3 significant figures. State the law of logarithms you used.
\workspace{4cm}
\begin{soln}
Take logs: $x\log 5 = \log 17$ (power law), so $x = \dfrac{\log 17}{\log 5} \approx \mathbf{1.76}$.
\end{soln}
\newpage

% =============================================================
% PACKET 4 --- TRIGONOMETRY
% =============================================================
\section{Packet 4: Trigonometry (Cumulative essentials)}
\begin{packinfo}
\textbf{Topic:} Exact trigonometric values, sine/cosine rule, triangle area, sectors and arcs (radians), 3D and bearings. \textbf{Suggested time:} approx.\ 75 minutes. \textbf{Questions:} 8.

\medskip
\textbf{Haese chapter references:} \haese{10E:12A--D} (Right-angle trig and bearings, Y10 carryover), \haese{10E:12E--I} (Sine rule, cosine rule, area), \haese{10E:21A--D,H} (Unit circle, exact values, equations), \haese{DP Core 6A} (Radians, arcs, sectors --- introduced in Y11).

\medskip
\textbf{Skills covered (DFM):} \dfm{327}, \dfm{465}, \dfm{466}, \dfm{467}, \dfm{318}, \dfm{319}, \dfm{584}, \dfm{412}, \dfm{503}.

\medskip
\textbf{Note.} For deeper trigonometry practice see the four-packet \textbf{Trigonometry Review Booklet} and its dedicated checklist. The pack here covers the cumulative essentials required for the EOY paper.

\medskip
\textbf{Key formulas.}\;
Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$;\quad
Cosine rule: $a^2 = b^2 + c^2 - 2bc\cos A$;\quad
Area: $\tfrac{1}{2}ab\sin C$.\;
Arc $s = r\theta$;\quad sector area $A = \tfrac{1}{2}r^2\theta$ (with $\theta$ in radians).
\end{packinfo}
\subsection{Part A --- Warm-up: exact values \& basic SOHCAHTOA}
\question{1}\level{1--2}
State the exact value of:
\begin{enumerate}[label=(\alph*)]
  \item $\sin 60\degree$;
  \item $\cos 30\degree$;
  \item $\tan 45\degree$;
  \item $\sin\dfrac{\pi}{6}$.
\end{enumerate}
\workspace{3cm}
\begin{soln}
(a)~$\tfrac{\sqrt{3}}{2}$.\quad (b)~$\tfrac{\sqrt{3}}{2}$.\quad (c)~$1$.\quad (d)~$\tfrac{1}{2}$.
\end{soln}
\question{2}\level{1--2}
Find the exact value of $\cos\!\left(\dfrac{5\pi}{6}\right)$.
\workspace{3cm}
\begin{soln}
Reference angle $\tfrac{\pi}{6}$ in Q\,II; cosine negative: $\mathbf{-\dfrac{\sqrt{3}}{2}}$.
\end{soln}
\subsection{Part B --- Core skills (sine/cosine rule, sector, equations)}
\question{3}\level{5--6}
Triangle $ABC$ has $AB = 12$~m, $AC = 9$~m and $\angle BAC = 75\degree$.
\begin{enumerate}[label=(\alph*)]
  \item Find the area of the triangle.
  \item Find the length $BC$.
  \item Find the angle $\angle ABC$.
\end{enumerate}
\workspace{7cm}
\begin{soln}
(a)~Area $= \tfrac{1}{2}(12)(9)\sin 75\degree \approx \mathbf{52.2}$~m\textsuperscript{2}.\\
(b)~$BC^2 = 144 + 81 - 2(12)(9)\cos 75\degree \approx 169.1$, so $BC \approx \mathbf{13.0}$~m.\\
(c)~$\dfrac{\sin B}{9} = \dfrac{\sin 75\degree}{13.0} \Rightarrow \angle ABC \approx \mathbf{42.0\degree}$.
\end{soln}
\question{4}\level{5--6}
A circular fan blade of radius $9$~cm sweeps out a sector with central angle $\theta = \dfrac{4\pi}{9}$ radians.
\begin{enumerate}[label=(\alph*)]
  \item Find the length of the arc swept.
  \item Find the area swept by the sector.
  \item Find the perimeter of the swept sector.
\end{enumerate}
\workspace{6cm}
\begin{soln}
(a)~$s = r\theta = 9 \cdot \tfrac{4\pi}{9} = 4\pi \approx \mathbf{12.6}$~cm.\\
(b)~$A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(81)\tfrac{4\pi}{9} = 18\pi \approx \mathbf{56.5}$~cm\textsuperscript{2}.\\
(c)~$P = 2r + s = 18 + 4\pi \approx \mathbf{30.6}$~cm.
\end{soln}
\question{5}\level{5--6}
In triangle $ABC$, $AB = 4$, $AC = 6$, $\angle BAC = 60\degree$. Give all answers exactly.
\begin{enumerate}[label=(\alph*)]
  \item Find $BC$.
  \item Find the area of the triangle.
\end{enumerate}
\workspace{5cm}
\begin{soln}
(a)~$BC^2 = 16 + 36 - 2(4)(6)\cos 60\degree = 52 - 24 = 28$, so $BC = \sqrt{28} = \mathbf{2\sqrt{7}}$.\\
(b)~Area $= \tfrac{1}{2}(4)(6)\sin 60\degree = 12 \cdot \tfrac{\sqrt{3}}{2} = \mathbf{6\sqrt{3}}$.
\end{soln}
\question{6}\level{5--6}
Solve each equation exactly in the given range.
\begin{enumerate}[label=(\alph*)]
  \item $2\cos x = -1$ for $0 \leq x \leq 2\pi$;
  \item $\sin x = \dfrac{\sqrt{2}}{2}$ for $0\degree \leq x \leq 360\degree$.
\end{enumerate}
\workspace{5cm}
\begin{soln}
(a)~$\cos x = -\tfrac{1}{2}$, reference $\tfrac{\pi}{3}$, cos negative in Q\,II, III: $x = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}$.\\
(b)~Reference $45\degree$, sin positive in Q\,I, II: $x = 45\degree, 135\degree$.
\end{soln}
\subsection{Part C --- Applied / Unfamiliar Contexts}
\question{7}\level{7--8}
A vertical flagpole stands at point $F$ on horizontal ground. Surveyor $S_1$ stands $80$~m due south of surveyor $S_2$. From $S_1$, the flagpole is on a bearing of $045\degree$. From $S_2$, the flagpole is on a bearing of $115\degree$. The angle of elevation of the top of the flagpole from $S_2$ is $28\degree$.
\begin{enumerate}[label=(\alph*)]
  \item Find the interior angles of the ground triangle $S_1\,F\,S_2$ at $S_1$ and $S_2$.
  \item Use the sine rule to find the distance $S_2\,F$.
  \item Hence find the height of the flagpole, correct to 3~s.f.
\end{enumerate}
\workspace{8cm}
\begin{soln}
(a)~$S_1S_2$ runs north (direction $S_1\to S_2$ is bearing $000\degree$). At $S_1$, $F$ is on bearing $045\degree$, so $\angle F\,S_1\,S_2 = 45\degree$. At $S_2$, $S_1$ is due south (bearing $180\degree$), $F$ is on bearing $115\degree$, so $\angle F\,S_2\,S_1 = 180\degree - 115\degree = 65\degree$.\\
(b)~$\angle S_1\,F\,S_2 = 180\degree - 45\degree - 65\degree = 70\degree$. Sine rule: $\dfrac{S_2 F}{\sin 45\degree} = \dfrac{80}{\sin 70\degree}$, so $S_2 F \approx \mathbf{60.2}$~m.\\
(c)~Height $= S_2 F \cdot \tan 28\degree \approx 60.2 \cdot 0.5317 \approx \mathbf{32.0}$~m.
\end{soln}
\question{8}\level{7--8}\;\textit{(Y10 carryover: bearings + cosine rule --- new scenario)}
A hiker starts at base camp $C$ and walks $5$~km on a bearing of $050\degree$ to a viewpoint $V$. From $V$ they continue onto a bearing of $145\degree$ for $7$~km to reach a lake $L$.
\begin{enumerate}[label=(\alph*)]
  \item Sketch the route, marking north arrows at each turning point, and find the interior angle at $V$.
  \item Find the straight-line distance $CL$, correct to 3~s.f.
  \item Find the bearing the hiker would walk to return directly from $L$ to $C$, correct to 3~s.f.
\end{enumerate}
\workspace{8cm}
\begin{soln}
(a)~At $V$: turning through external angle $145\degree - 50\degree = 95\degree$, so interior angle of triangle $C\,V\,L$ at $V$ is $180\degree - 95\degree = 85\degree$.\\
(b)~$CL^2 = 25 + 49 - 70\cos 85\degree \approx 74 - 6.10 \approx 67.9$, so $CL \approx \mathbf{8.24}$~km.\\
(c)~Sine rule: $\dfrac{\sin(\angle VCL)}{7} = \dfrac{\sin 85\degree}{8.24}$, so $\angle VCL \approx 57.9\degree$. Bearing of $L$ from $C$ $= 50\degree + 57.9\degree = 107.9\degree$. Return bearing $= 107.9\degree + 180\degree = \mathbf{288\degree}$ (3~s.f.).
\end{soln}
% =============================================================
\vspace{6pt}
\begin{center}
{\itshape End of Review.}
\end{center}
\end{document}
