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\lhead{\footnotesize\color{ibblue}\bfseries Quadratics \& Trigonometry Review}
\rhead{\footnotesize\color{accent}IBDP Prep $\cdot$ Worked Solutions}
\cfoot{\footnotesize\thepage}
\rfoot{\footnotesize J. Murphy}

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\begin{document}

\begin{center}
  {\color{ibblue}\Large\bfseries Quadratics \& Trigonometry Review}\\[2pt]
  {\color{accent}\footnotesize Worked Solutions $\cdot$ full method and final answers}
\end{center}
\vspace{2pt}
\small

%% ===================== Q1 =====================
\Q{1}{6}

The depth of water, $D$ metres, is modelled by $D(t) = 4.8 - 1.6\cos(30^{\circ} \times t)$.

\pt{a} The cosine function has range $[-1, 1]$.

Minimum depth (when $\cos = +1$):
\[
  D_{\min} = 4.8 - 1.6(+1) = 3.2~\text{m}
\]
Maximum depth (when $\cos = -1$):
\[
  D_{\max} = 4.8 - 1.6(-1) = 6.4~\text{m}
\]
\ans{D_{\min} = 3.2\text{ m},\quad D_{\max} = 6.4\text{ m.}} \mk{M1 for $\cos=\pm1$; A1 for each value}

\note{The period is $360^{\circ}/30^{\circ} = 12$ hours. Both values required.}

\pt{b} Set $D(t) = 5.6$:
\[
  4.8 - 1.6\cos(30^{\circ} \times t) = 5.6
  \;\Rightarrow\; \cos(30^{\circ} \times t) = \frac{4.8 - 5.6}{1.6} = -0.5
\]
\[
  30^{\circ} \times t = 120^{\circ} + 360^{\circ}k \;\text{ or }\; 30^{\circ} \times t = 240^{\circ} + 360^{\circ}k, \quad k \in \mathbb{Z}
\]
\[
  t = 4 + 12k \;\text{ or }\; t = 8 + 12k
\]
Values of $t$ greater than 10: from $t = 4+12k$ the first is $t = 16$; from $t = 8+12k$ the first is $t = 20$.

The first time after $t = 10$ is \ans{t = 16 \text{ hours.}} \mk{M1 for correct equation; M1 for general solution; A1 for $t = 16$}

\vspace{4pt}

%% ===================== Q2 =====================
\Q{2}{6}

Car hire model: $C = 45d + 120$.

\pt{a} Substitute $d = 5$:
\[
  C = 45(5) + 120 = 225 + 120 = 345
\]
\ans{345 \text{ EUR.}} \mk{A1}

\pt{b} The graph is a straight line, gradient 45 (EUR/day), $C$-intercept 120 EUR.

\begin{center}
\begin{tikzpicture}
\begin{axis}[
  width=8cm, height=5.5cm,
  axis lines=left,
  xmin=0, xmax=7.5, ymin=0, ymax=560,
  xtick={1,2,3,4,5,6,7}, ytick={100,200,300,400,500},
  tick label style={font=\scriptsize},
  xlabel={$d$ (days)}, ylabel={$C$ (EUR)},
  xlabel style={right}, ylabel style={above},
  grid=both, grid style={line width=0.3pt, draw=gray!30},
  major grid style={line width=0.5pt, draw=gray!50},
  clip=false]
  \addplot[ibblue,thick,domain=0:7] {45*x + 120};
  \addplot[accent,only marks,mark=*,mark size=1.5pt] coordinates {(0,120) (7,435)};
\end{axis}
\end{tikzpicture}
\end{center}
\mk{M1 for straight line; A1 for correct gradient and intercept}

\pt{c} Set $C = 435$:
\[
  45d + 120 = 435 \;\Rightarrow\; 45d = 315 \;\Rightarrow\; d = 7
\]
\ans{d = 7 \text{ days.}} \mk{A1}

\pt{d} PremierDrive: insurance $= 120 - 40 = 80$ EUR, daily rate $= 45 + 10 = 55$ EUR/day.
\[
  C_P = 55d + 80.
\]
EcoCars is cheaper when $C < C_P$:
\[
  45d + 120 < 55d + 80 \;\Rightarrow\; 40 < 10d \;\Rightarrow\; d > 4.
\]
At $d = 4$: both cost 300 EUR (equal). EcoCars is cheaper from $d = 5$ days onwards.

\ans{\text{Least number of days: } 5.} \mk{M1 for forming and solving inequality; A1 for 5 days}

\newpage

%% ===================== Q3 =====================
\Q{3}{6}

Graph shows $f(x) = 3x^2 + bx + c$ with $x$-intercepts at $x = -2$ and $x = 5$.

\pt{a} The $y$-intercept is at $x = 0$: $f(0) = c$. From the factored form $f(x) = 3(x+2)(x-5)$:
\[
  f(0) = 3(2)(-5) = -30.
\]
\ans{c = -30.} \mk{A1}

\pt{b} Expanding $f(x) = 3(x+2)(x-5) = 3(x^2 - 3x - 10) = 3x^2 - 9x - 30$,
\[
  \text{so } b = -9.
\]
\ans{b = -9,\quad f(x) = 3x^2 - 9x - 30.} \mk{M1 for factored form; A1 for $b$; A1 for $f(x)$}

\pt{c} The vertex $x$-coordinate:
\[
  x_V = -\frac{b}{2a} = -\frac{-9}{6} = 1.5.
\]
\[
  y_V = 3(1.5)^2 - 9(1.5) - 30 = 6.75 - 13.5 - 30 = -36.75.
\]
\ans{\text{Vertex} = (1.5,\;{-36.75}).} \mk{M1 for vertex method; A1 for correct coordinates}

\vspace{4pt}

%% ===================== Q4 =====================
\Q{4}{6}

Temperature model: $T(t) = a\cos(bt) + d$, with maximum $7^{\circ}$C and minimum $-15^{\circ}$C.

\pt{a}
\[
  a = \frac{7 - (-15)}{2} = \frac{22}{2} = 11,
  \qquad
  d = \frac{7 + (-15)}{2} = \frac{-8}{2} = -4.
\]
\ans{a = 11,\quad d = -4.} \mk{A1 for $a$; A1 for $d$}

\pt{b} The period is 24 hours:
\[
  b = \frac{360^{\circ}}{24} = 15.
\]
\ans{b = 15.} \mk{M1; A1}

\pt{c} Temperature is decreasing between its maximum (at $t = 0$) and its minimum (at $t = 12$).

$T = 3^{\circ}$C:
\[
  11\cos(15t) - 4 = 3 \;\Rightarrow\; \cos(15t) = \tfrac{7}{11}
  \;\Rightarrow\; 15t = \arccos\!\bigl(\tfrac{7}{11}\bigr) \approx 50.5^{\circ}
  \;\Rightarrow\; t \approx 3.37~\text{h}.
\]
$T = -9^{\circ}$C:
\[
  11\cos(15t) - 4 = -9 \;\Rightarrow\; \cos(15t) = -\tfrac{5}{11}
  \;\Rightarrow\; 15t = \arccos\!\bigl(-\tfrac{5}{11}\bigr) \approx 117.0^{\circ}
  \;\Rightarrow\; t \approx 7.80~\text{h}.
\]
\ans{3.37 < t < 7.80 \text{ hours.}} \mk{M1 for each equation; A1 for interval}

\newpage

%% ===================== Q5 =====================
\Q{5}{6}

Height of gondola: $H(t) = 35\sin(60^{\circ} \times t) + 40$.

\pt{a} The sine function has minimum value $-1$:
\[
  H_{\min} = 35(-1) + 40 = 5~\text{m.}
\]
\ans{5 \text{ m.}} \mk{M1; A1}

\pt{b} Substitute $t = 7$:
\[
  H(7) = 35\sin(420^{\circ}) + 40 = 35\sin(60^{\circ}) + 40
  = 35 \cdot \frac{\sqrt{3}}{2} + 40 \approx 70.3~\text{m.}
\]
\ans{H(7) = \tfrac{35\sqrt{3}}{2} + 40 \approx 70.3 \text{ m.}} \mk{M1 for reducing angle; A1 for value}

\pt{c}
\[
  \text{Period} = \frac{360^{\circ}}{60^{\circ}} = 6~\text{seconds.}
\]
\ans{6 \text{ seconds.}} \mk{M1; A1}

\vspace{4pt}

%% ===================== Q6 =====================
\Q{6}{6}

TalkDirect plan: $C = 25m + 15$.

\pt{a} Substitute $m = 4$:
\[
  C = 25(4) + 15 = 115~\text{EUR.}
\]
\ans{115 \text{ EUR.}} \mk{A1}

\pt{b} Straight line: gradient 25 (EUR/month), $C$-intercept 15 EUR.

\begin{center}
\begin{tikzpicture}
\begin{axis}[
  width=8cm, height=5.5cm,
  axis lines=left,
  xmin=0, xmax=8.5, ymin=0, ymax=230,
  xtick={1,2,3,4,5,6,7,8}, ytick={50,100,150,200},
  tick label style={font=\scriptsize},
  xlabel={$m$ (months)}, ylabel={$C$ (EUR)},
  xlabel style={right}, ylabel style={above},
  grid=both, grid style={line width=0.3pt, draw=gray!30},
  major grid style={line width=0.5pt, draw=gray!50},
  clip=false]
  \addplot[ibblue,thick,domain=0:8] {25*x + 15};
  \addplot[accent,only marks,mark=*,mark size=1.5pt] coordinates {(0,15) (6,165)};
\end{axis}
\end{tikzpicture}
\end{center}
\mk{M1 for straight line; A1 for correct gradient and intercept}

\pt{c} Set $C = 165$:
\[
  25m + 15 = 165 \;\Rightarrow\; 25m = 150 \;\Rightarrow\; m = 6.
\]
\ans{m = 6 \text{ months.}} \mk{A1}

\pt{d} DataPlus connection fee $= 15 + 35 = 50$ EUR; monthly rate $= 18$ EUR/month.
\[
  C_D = 18m + 50.
\]
DataPlus cheaper when $C_D < C$:
\[
  18m + 50 < 25m + 15 \;\Rightarrow\; 35 < 7m \;\Rightarrow\; m > 5.
\]
At $m = 5$: both cost 140 EUR (equal). DataPlus cheaper from $m = 6$ months onwards.

\ans{\text{Least number of months: } 6.} \mk{M1 for inequality; A1 for 6 months}

\newpage

%% ===================== Q7 =====================
\Q{7}{6}

Singapore Flyer model: $h(t) = 60\cos\!\left(\dfrac{\pi}{15}(t-15)\right) + 65$.

\pt{a} Maximum when $\cos = +1$:
\[
  h_{\max} = 60(1) + 65 = 125~\text{m.}
\]
\ans{125 \text{ m.}} \mk{A1}

\pt{b} Substitute $t = 12$:
\[
  h(12) = 60\cos\!\!\left(\frac{\pi}{15}(12-15)\right) + 65
         = 60\cos\!\!\left(-\frac{\pi}{5}\right) + 65
         = 60\cos(36^{\circ}) + 65.
\]
\[
  \cos(36^{\circ}) \approx 0.8090, \quad h(12) \approx 60(0.8090) + 65 \approx 113.5~\text{m.}
\]
\ans{h(12) \approx 113.5 \text{ m.}} \mk{A1}

\pt{c} Angular frequency $\omega = \dfrac{\pi}{15}$ rad/min, so:
\[
  T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/15} = 30~\text{minutes.}
\]
\ans{30 \text{ minutes.}} \mk{M1; A1}

\pt{d} Find values of $t \in [0,30]$ where $h(t) > 95$:
\[
  60\cos\!\!\left(\frac{\pi}{15}(t-15)\right) + 65 > 95
  \;\Rightarrow\; \cos\!\!\left(\frac{\pi}{15}(t-15)\right) > \frac{30}{60} = 0.5
\]
\[
  \frac{\pi}{15}(t-15) \in \left(-\frac{\pi}{3},\;\frac{\pi}{3}\right)
  \;\Rightarrow\; t-15 \in (-5,\;5)
  \;\Rightarrow\; t \in (10,\;20).
\]
Duration $= 20 - 10 = $ \ans{10 \text{ minutes.}} \mk{M1 for inequality; A1 for 10 min}

\note{$h(10) = h(20) = 60(0.5) + 65 = 95$, confirming the boundaries.}

\vspace{4pt}

%% ===================== Q8 =====================
\Q{8}{6}

Trajectory: $y = 6 + 4x - 0.4x^2$.

\pt{a} At $x = 0$: $y = 6$. \ans{\text{Height of launch} = 6 \text{ m.}} \mk{A1}

\pt{b} Complete the square on $y = -0.4x^2 + 4x + 6$:
\[
  y = -0.4\!\left(x^2 - 10x\right) + 6
    = -0.4\!\left[(x-5)^2 - 25\right] + 6
    = -0.4(x-5)^2 + 10 + 6
    = -0.4(x-5)^2 + 16.
\]
Maximum height $= 16$ m, at $x = 5$ m.

\ans{y_{\max} = 16 \text{ m, reached at } x = 5 \text{ m.}} \mk{M1 for completing square; A1 for $x=5$; A1 for $y=16$}

\pt{c} Set $y = 0$:
\[
  -0.4x^2 + 4x + 6 = 0 \;\Rightarrow\; x^2 - 10x - 15 = 0 \;\Rightarrow\;
  x = \frac{10 \pm \sqrt{100 + 60}}{2} = \frac{10 \pm \sqrt{160}}{2} = 5 \pm 2\sqrt{10}.
\]
Taking the positive root (since $x \ge 0$):
\[
  x = 5 + 2\sqrt{10} \approx 11.32~\text{m.}
\]
\ans{x = 5 + 2\sqrt{10} \approx 11.32 \text{ m.}} \mk{M1 for correct quadratic; A1 for positive root}

\newpage

%% ===================== Q9 =====================
\Q{9}{6}

Quadratic: $y = ax^2 + bx + c$. Given: $\mathrm{A}(0,18)$, $\mathrm{B}(3,0)$, vertex at $x = 6$.

\pt{a} At $x = 0$, $y = c$. From point A: \ans{c = 18.} \mk{A1}

\pt{b} By the symmetry of the parabola, the vertex lies midway between the two $x$-intercepts. With one root at $x = 3$ and vertex at $x = 6$:
\[
  x_2 = 2(6) - 3 = 9.
\]
Write in factored form $y = a(x-3)(x-9)$ and substitute A$(0,18)$:
\[
  18 = a(0-3)(0-9) = 27a \;\Rightarrow\; a = \frac{18}{27} = \frac{2}{3}.
\]
Expand to find $b$:
\[
  y = \tfrac{2}{3}(x-3)(x-9) = \tfrac{2}{3}(x^2 - 12x + 27)
    = \tfrac{2}{3}x^2 - 8x + 18.
\]
\ans{a = \tfrac{2}{3},\quad b = -8.} \mk{M1 for symmetry; M1 for factored form; A1 for $a$; A1 for $b$}

\pt{c} The second $x$-intercept is \ans{(9,\;0).} \mk{A1}

\vspace{30pt}

\begin{center}
{\color{ibblue!60}\rule{0.8\linewidth}{0.4pt}}\\[4pt]
{\footnotesize\color{ibblue!60}\itshape End of solutions --- 9 questions, 54 marks}
\end{center}

\end{document}