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% Y11 EXTENDED — END OF YEAR ASSESSMENT (June 2026)
% PAPER 1: NON CALCULATOR (70 minutes, 60 marks)
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  \lhead{\textbf{MYP 11E} \;\textbar\; EOY Paper 1 --- Non-Calculator}%
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\begin{document}
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\noindent
\begin{tabular}{@{}p{8.2cm}p{8.2cm}@{}}
\textbf{Name:} \rule{5.5cm}{0.4pt} & \textbf{MYP Criteria:}\quad A\;\;\rule{1cm}{0.4pt}/8\quad C\;\;\rule{1cm}{0.4pt}/8 \\
\textbf{Date:} \rule{5.5cm}{0.4pt} & \textbf{Time:} 70 minutes \;\textbar\; \textbf{Total: 60 marks} \\
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\vspace{6pt}

\begin{infobox}
\textbf{\color{mypdark}Instructions for candidates}
\begin{itemize}[nosep, leftmargin=1.4em, topsep=3pt, itemsep=2pt]
  \item Do not open this examination paper until instructed to do so.
  \item A graphic display calculator is \textbf{not} permitted for this paper.
  \item All questions are to be answered in the space provided on this paper.
  \item Unless otherwise stated, all numerical answers should be given \textbf{exactly}.
  \item The maximum mark for this examination paper is \textbf{[60 marks]}.
  \item Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations.
\end{itemize}
\end{infobox}

\ifsol\else
\vspace{4pt}
\noindent{\small\textit{\textcolor{markgray}{\textbf{A note on levels.}\; Each question is labelled \textbf{[Level X--Y]}. Questions marked \textbf{[Level 7--8]} are \emph{unfamiliar contexts}: they combine multiple Y11 topics with Y10 carryover skills (coordinate geometry, surds, indices, 3D thinking, perpendicular bisectors), require you to choose your own approach without method hints, and may ask you to define your own variables or recognise mathematical structure. Try the question even if the path isn't immediately obvious --- set up what you know and look for connections.}}}
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\ifsol
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\begin{notebox}
\textbf{\color{mypamber}Mark scheme notes.}\; Award method marks (\textbf{M}) for correct strategy; accuracy marks (\textbf{A}) for correct answer; follow-through (\textbf{ft}) where an earlier error has been carried forward correctly. Equivalent forms accepted unless otherwise stated.
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% =============================================================
\question{1}{5}{1--2}
Find the exact value of each:
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item $\log_2 32$ \hfill\partmarks{1}
    \workspace{1.8cm}
  \item $\sin\!\left(\dfrac{2\pi}{3}\right)$ \hfill\partmarks{2}
    \workspace{2.5cm}
  \item $\left(\dfrac{1}{4}\right)^{-3/2}$ \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) $32 = 2^5$, so $\mathbf{5}$. \,\textit{[A1]}\\
(b) Reference $\pi/3$ in Q\,II; $\mathbf{\dfrac{\sqrt{3}}{2}}$. \,\textit{[M1, A1]}\\
(c) $\left(\tfrac{1}{4}\right)^{-3/2} = 4^{3/2} = (4^{1/2})^3 = 2^3 = \mathbf{8}$. \,\textit{[M1, A1]}
\end{solution}

% =============================================================
\question{2}{5}{3--4}
The universal set is $U = \{1, 2, 3, \ldots, 12\}$. Let $A = \{$prime numbers in $U\}$ and $B = \{$even numbers in $U\}$.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item List the elements of $A \cap B$ and of $A \cup B$. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find $P(A' \cap B)$ when an element of $U$ is chosen at random. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item State, with reasoning, whether $A$ and $B$ are mutually exclusive. \hfill\partmarks{1}
    \workspace{1.8cm}
\end{enumerate}
\begin{solution}
$A = \{2,3,5,7,11\}$, $B = \{2,4,6,8,10,12\}$.\\
(a) $A\cap B = \{2\}$ \,\textit{[A1]};\;\; $A\cup B = \{2,3,4,5,6,7,8,10,11,12\}$ \,\textit{[A1]}.\\
(b) $A' \cap B = \{4,6,8,10,12\}$, so $P = \mathbf{5/12}$. \,\textit{[M1, A1]}\\
(c) \textbf{Not mutually exclusive}, since $A\cap B = \{2\} \neq \emptyset$. \,\textit{[A1]}
\end{solution}

% =============================================================
\question{3}{6}{3--4}
Let $f(x) = 3x - 2$ and $g(x) = x^2 + 1$.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Find $f(g(2))$. \hfill\partmarks{1}
    \workspace{1.8cm}
  \item Find a simplified expression for $(f\circ g)(x)$. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find $f^{-1}(x)$, and state its domain. \hfill\partmarks{3}
    \workspace{3.5cm}
\end{enumerate}
\begin{solution}
(a) $g(2) = 5$; $f(5) = \mathbf{13}$. \,\textit{[A1]}\\
(b) $(f\circ g)(x) = 3(x^2 + 1) - 2 = \mathbf{3x^2 + 1}$. \,\textit{[M1, A1]}\\
(c) $y = 3x - 2 \Rightarrow x = \tfrac{y+2}{3}$. So $\mathbf{f^{-1}(x) = \dfrac{x+2}{3}}$. \,\textit{[M1, A1]}\;
Domain of $f^{-1}$ = range of $f$ = $\mathbf{x \in \mathbb{R}}$. \,\textit{[A1]}
\end{solution}

\newpage

% =============================================================
\question{4}{5}{3--4}
Without using a calculator, simplify each expression to a single logarithm or numerical value.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item $\log 6 + \log 2 - \log 4$ \hfill\partmarks{2}
    \workspace{2.5cm}
  \item $2\log 5 + \log 4$ \hfill\partmarks{2}
    \workspace{2.5cm}
  \item $\log_a(p^2 q) - \log_a(pq^2)$, in terms of $\log_a p$ and $\log_a q$. \hfill\partmarks{1}
    \workspace{1.8cm}
\end{enumerate}
\begin{solution}
(a) $\log\!\left(\dfrac{12}{4}\right) = \mathbf{\log 3}$. \,\textit{[M1, A1]}\\
(b) $\log 25 + \log 4 = \log 100 = \mathbf{2}$. \,\textit{[M1, A1]}\\
(c) $\log_a\!\left(\dfrac{p^2 q}{pq^2}\right) = \mathbf{\log_a p - \log_a q}$. \,\textit{[A1]}
\end{solution}

% =============================================================
\question{5}{7}{5--6}
Let $f(x) = x^2 - 6x + 4$.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item By completing the square, or otherwise, express $f(x)$ in vertex form. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Hence, or otherwise, solve $f(x) = 0$, giving exact answers in surd form. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item State the range of $f$. \hfill\partmarks{1}
    \workspace{1.8cm}
  \item Describe the single transformation that maps $y = x^2$ onto $y = (x-3)^2$. \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) $f(x) = (x-3)^2 - 9 + 4 = \mathbf{(x-3)^2 - 5}$. \,\textit{[M1, A1]}\\
(b) From (a): $(x-3)^2 = 5 \Rightarrow \mathbf{x = 3 \pm \sqrt{5}}$. \,\textit{[M1, A1]}\\
\textit{Or otherwise:} formula gives $x = \dfrac{6 \pm \sqrt{20}}{2} = 3 \pm \sqrt{5}$. \,\textit{[A1 either method]}\\
(c) Vertex $(3, -5)$ is a minimum, so range $\mathbf{y \geq -5}$. \,\textit{[A1]}\\
(d) \textbf{Translation 3 units to the right} (i.e.\ shift by $\binom{3}{0}$). \,\textit{[A1, A1]}
\end{solution}

% =============================================================
\question{6}{5}{5--6}
In triangle $ABC$, $AB = 3$, $AC = 5$ and $\angle BAC = 60\degree$. Give all answers exactly.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Find the length $BC$. \hfill\partmarks{3}
    \workspace{3.5cm}
  \item Find the area of triangle $ABC$. \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) $BC^2 = 9 + 25 - 2(3)(5)\cos 60\degree = 34 - 15 = 19$. \,\textit{[M1, A1]}\\
$BC = \mathbf{\sqrt{19}}$. \,\textit{[A1]}\\
(b) Area $= \tfrac{1}{2}(3)(5)\sin 60\degree = \mathbf{\dfrac{15\sqrt{3}}{4}}$. \,\textit{[M1, A1]}
\end{solution}

% =============================================================
\question{7}{5}{5--6}
Solve each equation, giving answers exactly in the given range.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item $2\sin x = \sqrt{3}$ for $0 \leq x \leq 2\pi$. \hfill\partmarks{3}
    \workspace{3.5cm}
  \item $\cos x = -\dfrac{1}{2}$ for $0\degree \leq x \leq 360\degree$. \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) $\sin x = \dfrac{\sqrt{3}}{2}$. Reference $\pi/3$; positive in Q\,I, Q\,II.\\
$\mathbf{x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}}$. \,\textit{[M1, A1, A1]}\\
(b) Reference $60\degree$; cosine negative in Q\,II, Q\,III. $\mathbf{x = 120\degree, 240\degree}$. \,\textit{[A1, A1]}
\end{solution}

\newpage

% =============================================================
\question{8}{6}{5--7}
Solve $\log_2(x) + \log_2(x - 6) = 4$. Show all working, and explain why one of the algebraic solutions must be rejected.
\workspace{6cm}
\begin{solution}
Combine: $\log_2[x(x-6)] = 4 \Rightarrow x(x-6) = 16$. \,\textit{[M1]}\\
$x^2 - 6x - 16 = 0 \Rightarrow (x-8)(x+2) = 0 \Rightarrow x = 8$ or $x = -2$. \,\textit{[M1, A1, A1]}\\
For both logs to be defined, $x > 6$. Reject $x = -2$; $\mathbf{x = 8}$. \,\textit{[M1, A1]}
\end{solution}

% =============================================================
\question{9}{6}{7--8}
Solve the equation $\quad 3^{2x} - 4(3^x) + 3 = 0.$
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item By introducing a suitable substitution, reduce the equation to a quadratic. State your substitution clearly. \hfill\partmarks{3}
    \workspace{4cm}
  \item Hence find all real values of $x$. \hfill\partmarks{3}
    \workspace{4.5cm}
\end{enumerate}
\begin{solution}
(a) Let $u = 3^x$. Then $3^{2x} = (3^x)^2 = u^2$, so the equation becomes $u^2 - 4u + 3 = 0$. \,\textit{[M1 substitution stated, M1 reduction, A1 quadratic]}\\
(b) Factor: $(u-1)(u-3) = 0$, so $u = 1$ or $u = 3$. \,\textit{[M1]}\\
$3^x = 1 \Rightarrow \mathbf{x = 0}$;\;\; $3^x = 3 \Rightarrow \mathbf{x = 1}$. \,\textit{[A1, A1]}
\end{solution}

% =============================================================
\question{10}{10}{7--8}
The points $A(-1, 1)$ and $B(3, 9)$ both lie on the parabola $y = x^2$.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Find the midpoint $M$ of $AB$, and the gradient of the chord $AB$. \hfill\partmarks{3}
    \workspace{3.5cm}
  \item Find the equation of the perpendicular bisector of $AB$. \hfill\partmarks{3}
    \workspace{3.5cm}
  \item The perpendicular bisector meets the parabola $y = x^2$ at two points $P$ and $Q$. Show that the $x$-coordinates of $P$ and $Q$ satisfy
  \[ 2x^2 + x - 11 = 0. \] \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Hence find the $x$-coordinates of $P$ and $Q$ in exact (surd) form. \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) Midpoint $M = \left(\dfrac{-1+3}{2}, \dfrac{1+9}{2}\right) = \mathbf{(1, 5)}$. \,\textit{[A1, A1]}\\
Gradient of $AB = \dfrac{9 - 1}{3 - (-1)} = \mathbf{2}$. \,\textit{[A1]}\\
(b) Perpendicular gradient $= -\tfrac{1}{2}$. Through $M(1, 5)$:
\[ y - 5 = -\tfrac{1}{2}(x - 1) \Rightarrow \mathbf{y = -\tfrac{1}{2}x + \tfrac{11}{2}}. \]
\textit{[M1, M1, A1]}\\
(c) $x^2 = -\tfrac{1}{2}x + \tfrac{11}{2}$. Multiply by $2$: $2x^2 + x - 11 = 0$. \,\textit{[M1, A1]}\\
(d) Quadratic formula:
\[ x = \dfrac{-1 \pm \sqrt{1 + 88}}{4} = \mathbf{\dfrac{-1 \pm \sqrt{89}}{4}}. \]
\textit{[M1, A1]}
\end{solution}

\par\vspace*{\fill}
\begin{center}
\textit{End of Paper 1.}
\end{center}

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