Consider the function $f\left(x\right)=\frac{27}{x^2}-<!--\; -\; -->16x,x\ne <!--\; \ne \; -->0$.

A water container is made in the shape of a cylinder with internal height $h$ cm and internal base radius $r$ cm.

The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.

The volume of the water container is $0.5{\text{m}}^3$.

The water container is designed so that the area to be coated is minimized.

One can of water-resistant material coats a surface area of $2000\text{ c}{\text{m}}^2$.

Let $f\left(x\right)=4-x^3$ and $g\left(x\right)=\ln x$, for $x>0$.

The cross-sectional view of a tunnel is shown on the axes below. The line $\left[\text{AB}\right]$ represents a vertical wall located at the left side of the tunnel. The height, in metres, of the tunnel above the horizontal ground is modelled by $y=-0.1x^3+ 0.8x^2, 2\le x\le 8$, relative to an origin $\text{O}$.

Point $\text{A}$ has coordinates $(2, 0)$, point $\text{B}$ has coordinates $(2, 2.4)$, and point $\text{C}$ has coordinates $(8, 0)$.

When $x=4$ the height of the tunnel is $6.4 \text{m}$ and when $x=6$ the height of the tunnel is $7.2 \text{m}$. These points are shown as $\text{D}$ and $\text{E}$ on the diagram, respectively.

Urvashi wants to model the height of a moving object. She collects the following data showing the height, $h$  metres, of the object at time $t$ seconds.

She believes the height can be modeled by a quadratic function, $h\left(t\right)=a{t}^2+bt+c$, where $a\text{,}b\text{,}c\in <!--\; \in \; -->\mathbb{R}$.

Hence find

A wind turbine is designed so that the rotation of the blades generates electricity. The turbine is built on horizontal ground and is made up of a vertical tower and three blades.

The point $\text{A}$ is on the base of the tower directly below point $\text{B}$ at the top of the tower. The height of the tower, $\text{AB}$, is $90 \text{m}$. The blades of the turbine are centred at $\text{B}$ and are each of length $40 \text{m}$. This is shown in the following diagram.

The end of one of the blades of the turbine is represented by point $\text{C}$ on the diagram. Let $h$ be the height of $\text{C}$ above the ground, measured in metres, where $h$ varies as the blade rotates.

Find the

The blades of the turbine complete $12$ rotations per minute under normal conditions, moving at a constant rate.

The height, $h$, of point $\text{C}$ can be modelled by the following function. Time, $t$, is measured from the instant when the blade $\text{[BC]}$ first passes $\text{[AB]}$ and is measured in seconds.

$h\left(t\right)=90-40 \cos \left(72t°\right), t\ge 0$

Looking through his window, Tim has a partial view of the rotating wind turbine. The position of his window means that he cannot see any part of the wind turbine that is more than $\mathbf{100}\mathbf{ }\mathbf{\text{m}}$ above the ground. This is illustrated in the following diagram.

The Texas Star is a Ferris wheel at the state fair in Dallas. The Ferris wheel has a diameter of $61.8 \text{m}$. To begin the ride, a passenger gets into a chair at the lowest point on the wheel, which is $2.7 \text{m}$ above the ground, as shown in the following diagram. A ride consists of multiple revolutions, and the Ferris wheel makes $1.5$ revolutions per minute.

The height of a chair above the ground, $h$, measured in metres, during a ride on the Ferris wheel can be modelled by the function $h\left(t\right)=-a \cos \left(bt\right)+d$, where $t$ is the time, in seconds, since a passenger began their ride.

Calculate the value of

A ride on the Ferris wheel lasts for $12$ minutes in total.

For exactly one ride on the Ferris wheel, suggest

Big Tex is a $16.7$ metre-tall cowboy statue that stands on the horizontal ground next to the Ferris wheel.

^{[Source: Aline Escobar., n.d. Cowboy. [image online] Available at: https://thenounproject.com/search/?q=cowboy&i=1080130
This file is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/deed.en [Accessed 13/05/2021]. Source adapted.]}

There is a plan to relocate the Texas Star Ferris wheel onto a taller platform which will increase the maximum height of the Ferris wheel to $65.2 \text{m}$. This will change the value of one parameter, $a$, $b$ or $d$, found in part (a).

A function $f$ is given by $f(x)=(2x+2)(5-<!--\; -\; -->x^2)$.

The graph of the function $g(x)=5^x+6x-<!--\; -\; -->6$ intersects the graph of $f$.

A pan, in which to cook a pizza, is in the shape of a cylinder. The pan has a diameter of 35 cm and a height of 0.5 cm.

A chef had enough pizza dough to exactly fill the pan. The dough was in the shape of a sphere.

The pizza was cooked in a hot oven. Once taken out of the oven, the pizza was placed in a dining room.

The temperature, $P$, of the pizza, in degrees Celsius, °C, can be modelled by

$$P(t)=a(2.06{)}^{-<!--\; -\; -->t}+19,t⩾<!--\; ⩾\; -->0$$

where $a$ is a constant and $t$ is the time, in minutes, since the pizza was taken out of the oven.

When the pizza was taken out of the oven its temperature was 230 °C.

The pizza can be eaten once its temperature drops to 45 °C.

Consider the curve y = 2x³ − 9x² + 12x + 2, for −1 < x < 3

At Grande Anse Beach the height of the water in metres is modelled by the function $h(t)=p\cos <!--\; \; -->(q\times <!--\; \times \; -->t)+r$, where $t$ is the number of hours after 21:00 hours on 10 December 2017. The following diagram shows the graph of $h$ , for $0⩽<!--\; ⩽\; -->t⩽<!--\; ⩽\; -->72$.

The point $\text{A}(6.25,0.6)$ represents the first low tide and $\text{B}(12.5,1.5)$ represents the next high tide.

Let $f(x)=x^2-<!--\; -\; -->1$ and $g(x)=x^2-<!--\; -\; -->2$, for $x\in <!--\; \in \; -->\mathbb{R}$.

Consider the function $f(x)=0.3x^3+\frac{10}{x}+2^{-<!--\; -\; -->x}$.

Consider a second function, $g(x)=2x-<!--\; -\; -->3$.

Consider the function $g(x)=x^3+k{x}^2-<!--\; -\; -->15x+5$.

The tangent to the graph of $y=g(x)$ at $x=2$ is parallel to the line $y=21x+7$.

The diagram shows the straight line $L_1$. Points $\text{A}\left(-9, -1\right)$, $\text{M}\left(-3, 2\right)$ and $\text{C}$ are points on $L_1$.

$\text{M}$ is the midpoint of $\text{AC}$.

Line $L_2$ is perpendicular to $L_1$ and passes through point $\text{M}$.

The point $\text{N}\left(k, 4\right)$ is on $L_2$.

Let $f\left(x\right)=2\text{sin}\left(3x\right)+4$ for $x\in <!--\; \in \; -->\mathbb{R}$.

Let $g\left(x\right)=5f\left(2x\right)$.

The function $g$ can be written in the form $g\left(x\right)=10\text{sin}\left(bx\right)+c$.

The graph of the quadratic function $f\left(x\right)=\frac{1}{2}\left(x-2\right)\left(x+8\right)$ intersects the $y$-axis at $\left(0, c\right)$.

The vertex of the function is $\left(-3, -12.5\right)$.

The equation $f\left(x\right)=12$ has two solutions. The first solution is $x=-10$.

Let $T$ be the tangent at $x=-3$.

The diagram below shows a circular clockface with centre $\text{O}$. The clock’s minute hand has a length of $10 \text{cm}$. The clock’s hour hand has a length of $6 \text{cm}$.

At $4:00$ pm the endpoint of the minute hand is at point $\text{A}$ and the endpoint of the hour hand is at point $\text{B}$.

Between $4:00$ pm and $4:13$ pm, the endpoint of the minute hand rotates through an angle, $\theta$, from point $\text{A}$ to point $\text{C}$. This is illustrated in the diagram.

A second clock is illustrated in the diagram below. The clock face has radius $10 \text{cm}$ with minute and hour hands both of length $10 \text{cm}$. The time shown is $6:00$ am. The bottom of the clock face is located $3 \text{cm}$ above a horizontal bookshelf.

The height, $h$ centimetres, of the endpoint of the minute hand above the bookshelf is modelled by the function

$h\left(\theta \right)=10 \cos \theta +13, \theta \ge 0,$

where $\theta$ is the angle rotated by the minute hand from $6:00$ am.

The height, $g$ centimetres, of the endpoint of the hour hand above the bookshelf is modelled by the function

$g\left(\theta \right)=-10 \cos \left(\frac{\theta }{12}\right)+13, \theta \ge 0,$

where $\theta$ is the angle in degrees rotated by the minute hand from $6:00$ am.

Boris recorded the number of daylight hours on the first day of each month in a northern hemisphere town.

This data was plotted onto a scatter diagram. The points were then joined by a smooth curve, with minimum point $(0, 8)$ and maximum point $(6, 16)$ as shown in the following diagram.

Let the curve in the diagram be $y=f\left(t\right)$, where $t$ is the time, measured in months, since Boris first recorded these values.

Boris thinks that $f\left(t\right)$ might be modelled by a quadratic function.

Paula thinks that a better model is $f\left(t\right)=a \cos \left(bt\right)+d$, $t\ge 0$, for specific values of $a, b$ and $d$.

For Paula’s model, use the diagram to write down

The true maximum number of daylight hours was $16$ hours and $14$ minutes.

The rate of change of the height $\left(h\right)$ of a ball above horizontal ground, measured in metres, $t$ seconds after it has been thrown and until it hits the ground, can be modelled by the equation

$\frac{dh}{dt}=11.4-9.8t$

The height of the ball when $t=0$ is $1.2 \text{m}$.

In the month before their IB Diploma examinations, eight male students recorded the number of hours they spent on social media.

For each student, the number of hours spent on social media ($x$) and the number of IB Diploma points obtained ($y$) are shown in the following table.

Use your graphic display calculator to find

Ten female students also recorded the number of hours they spent on social media in the month before their IB Diploma examinations. Each of these female students spent between 3 and 30 hours on social media.

The equation of the regression line y on x for these ten female students is

$$y=-<!--\; -\; -->\frac{2}{3}x+\frac{125}{3}.$$

An eleventh girl spent 34 hours on social media in the month before her IB Diploma examinations.

The braking distance of a vehicle is defined as the distance travelled from where the brakes are applied to the point where the vehicle comes to a complete stop.

The speed, $s\text{m}{\text{s}}^{-<!--\; -\; -->1}$, and braking distance, $d\text{m}$, of a truck were recorded. This information is summarized in the following table.

This information was used to create Model A, where $d$ is a function of $s$, $s$ ≥ 0.

Model A: $d\left(s\right)=p{s}^2+qs$, where $p$, $q\in <!--\; \in \; -->\mathbb{Z}$

At a speed of $6\text{m}{\text{s}}^{-<!--\; -\; -->1}$, Model A can be represented by the equation $6p+q=2$.

Additional data was used to create Model B, a revised model for the braking distance of a truck.

Model B: $d\left(s\right)=0.95s^2-<!--\; -\; -->3.92s$

The actual braking distance at $20\text{m}{\text{s}}^{-<!--\; -\; -->1}$ is $320\text{m}$.

Consider the function $f\left(x\right)=\frac{48}{x}+k{x}^2-<!--\; -\; -->58$, where x > 0 and k is a constant.

The graph of the function passes through the point with coordinates (4 , 2).

P is the minimum point of the graph of f (x).

A cafe makes $x$ litres of coffee each morning. The cafe’s profit each morning, $C$, measured in dollars, is modelled by the following equation

$C=\frac{x}{10}\left(k^2-\frac{3}{100}{x}^2\right)$

where $k$ is a positive constant.

The cafe’s manager knows that the cafe makes a profit of $\$426$ when $20$ litres of coffee are made in a morning.

The manager of the cafe wishes to serve as many customers as possible.

Let $f\left(x\right)=\frac{16}{x}$. The line $L$ is tangent to the graph of $f$ at $x=8$.

$L$ can be expressed in the form r $=\left(\begin{array}{c}8\\ 2\end{array}\right)+t$u.

The direction vector of $y=x$ is $\left(\begin{array}{c}1\\ 1\end{array}\right)$.

The depth of water in a port is modelled by the function $d(t)=p\cos <!--\; \; -->qt+7.5$, for $0⩽<!--\; ⩽\; -->t⩽<!--\; ⩽\; -->12$, where $t$ is the number of hours after high tide.

At high tide, the depth is 9.7 metres.

At low tide, which is 7 hours later, the depth is 5.3 metres.

Consider the function $f(x)=-<!--\; -\; -->x^4+a{x}^2+5$, where $a$ is a constant. Part of the graph of $y=f(x)$ is shown below.

It is known that at the point where $x=2$ the tangent to the graph of $y=f(x)$ is horizontal.

There are two other points on the graph of $y=f(x)$ at which the tangent is horizontal.

Let $f(x)=x^2+2x+1$ and $g(x)=x-<!--\; -\; -->5$, for $x\in <!--\; \in \; -->\mathbb{R}$.

Consider the function $f\left(x\right)=x^2+x+\frac{50}{x}, x\ne 0.$

The following diagram shows a water wheel with centre $\text{O}$ and radius $10$ metres. Water flows into buckets, turning the wheel clockwise at a constant speed.

The height, $h$ metres, of the top of a bucket above the ground $t$ seconds after it passes through point $\text{A}$ is modelled by the function

$h\left(t\right)=13+8 \cos \left(\frac{\mathrm{\pi }}{18}t\right)-6 \sin \left(\frac{\mathrm{\pi }}{18}t\right)$, for $t\ge 0$.

A bucket moves around to point $\text{B}$ which is at a height of $4.06$ metres above the ground. It takes $k$ seconds for the top of this bucket to go from point $\text{A}$ to point $\text{B}$.

The chord $\left[\text{AB}\right]$ is $17.0$ metres, correct to three significant figures.

Scott purchases food for his dog in large bags and feeds the dog the same amount of dog food each day. The amount of dog food left in the bag at the end of each day can be modelled by an arithmetic sequence.

On a particular day, Scott opened a new bag of dog food and fed his dog. By the end of the third day there were $115.5$ cups of dog food remaining in the bag and at the end of the eighth day there were $108$ cups of dog food remaining in the bag.

Find the number of cups of dog food

In $2021$, Scott spent $\$625$ on dog food. Scott expects that the amount he spends on dog food will increase at an annual rate of $6.4\%$.

Consider the function $f\left(x\right)=x^3-<!--\; -\; -->5x^2+6x-<!--\; -\; -->3+\frac{1}{x}$, $x>0$

The function $f\left(x\right)=x^3-<!--\; -\; -->5x^2+6x-<!--\; -\; -->3+\frac{1}{x}$, $x>0$, models the path of a river, as shown on the following map, where both axes represent distance and are measured in kilometres. On the same map, the location of a highway is defined by the function $g\left(x\right)=0.5{\left(3\right)}^{-<!--\; -\; -->x}+1$.

The origin, O(0, 0) , is the location of the centre of a town called Orangeton.

A straight footpath, $P$, is built to connect the centre of Orangeton to the river at the point where $x=\frac{1}{2}$.

Bridges are located where the highway crosses the river.

A straight road is built from the centre of Orangeton, due north, to connect the town to the highway.

Note: In this question, distance is in millimetres.

Let $f(x)=x+a\sin <!--\; \; -->\left(x-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}\right)+a$, for $x⩾<!--\; ⩾\; -->0$.

The graph of $f$ passes through the origin. Let ${\text{P}}_k$ be any point on the graph of $f$ with $x$-coordinate $2k\mathrm{\pi <!--\; \pi \; -->}$, where $k\in <!--\; \in \; -->\mathbb{N}$. A straight line $L$ passes through all the points ${\text{P}}_k$.

Diagram 1 shows a saw. The length of the toothed edge is the distance AB.

The toothed edge of the saw can be modelled using the graph of $f$ and the line $L$. Diagram 2 represents this model.

The shaded part on the graph is called a tooth. A tooth is represented by the region enclosed by the graph of $f$ and the line $L$, between ${\text{P}}_k$ and ${\text{P}}_{k+1}$.

The Voronoi diagram below shows four supermarkets represented by points with coordinates $\text{A}(0, 0)$, $\text{B}(6, 0)$, $\text{C}(0, 6)$ and $\text{D}(2, 2)$. The vertices $\text{X}$, $\text{Y}$, $\text{Z}$ are also shown. All distances are measured in kilometres.

The equation of $\text{(XY)}$ is $y=2-x$ and the equation of $\text{(YZ)}$ is $y=0.5x+3.5$.

The coordinates of $\text{Y}$ are $(-1, 3)$ and the coordinates of $\text{Z}$ are $(7, 7)$.

A town planner believes that the larger the area of the Voronoi cell $\text{XYZ}$, the more people will shop at supermarket $\text{D}$.

All lengths in this question are in metres.

Let $f(x)=-<!--\; -\; -->0.8x^2+0.5$, for $-<!--\; -\; -->0.5⩽<!--\; ⩽\; -->x⩽<!--\; ⩽\; -->0.5$. Mark uses $f(x)$ as a model to create a barrel. The region enclosed by the graph of $f$, the $x$-axis, the line $x=-<!--\; -\; -->0.5$ and the line $x=0.5$ is rotated 360° about the $x$-axis. This is shown in the following diagram.

Let $f\left(x\right)=12\text{cos}x-<!--\; -\; -->5\text{sin}x,-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}⩽<!--\; ⩽\; -->x⩽<!--\; ⩽\; -->2\mathrm{\pi <!--\; \pi \; -->}$, be a periodic function with $f\left(x\right)=f\left(x+2\mathrm{\pi <!--\; \pi \; -->}\right)$

The following diagram shows the graph of $f$.

There is a maximum point at A. The minimum value of $f$ is −13 .

A ball on a spring is attached to a fixed point O. The ball is then pulled down and released, so that it moves back and forth vertically.

The distance, d centimetres, of the centre of the ball from O at time t seconds, is given by

$d\left(t\right)=f\left(t\right)+17,0⩽<!--\; ⩽\; -->t⩽<!--\; ⩽\; -->5.$

Consider the function $f\left(x\right)=\frac{1}{3}{x}^3+\frac{3}{4}{x}^2-<!--\; -\; -->x-<!--\; -\; -->1$.

The following diagram shows the graph of $f(x)=a\sin <!--\; \; -->bx+c$, for $0⩽<!--\; ⩽\; -->x⩽<!--\; ⩽\; -->12$.

The graph of $f$ has a minimum point at $(3,5)$ and a maximum point at $(9,17)$.

The graph of $g$ is obtained from the graph of $f$ by a translation of $\left(\begin{array}{c}k\\ 0\end{array}\right)$. The maximum point on the graph of $g$ has coordinates $(11.5,17)$.

The graph of $g$ changes from concave-up to concave-down when $x=w$.

A theatre set designer is designing a piece of flat scenery in the shape of a hill. The scenery is formed by a curve between two vertical edges of unequal height. One edge is $2$ metres high and the other is $1$ metre high. The width of the scenery is $6$ metres.

A coordinate system is formed with the origin at the foot of the $2$ metres high edge. In this coordinate system the highest point of the cross‐section is at $\left(2, 3.5\right)$.

A set designer wishes to work out an approximate value for the area of the scenery $(A {\text{m}}^2 )$.

In order to obtain a more accurate measure for the area the designer decides to model the curved edge with the polynomial $h\left(x\right)=a{x}^3+b{x}^2+cx+d a, b, c, d\in \mathrm{ℝ}$ where $h$ metres is the height of the curved edge a horizontal distance $x \text{m}$ from the origin.

At an amusement park, a Ferris wheel with diameter 111 metres rotates at a constant speed. The bottom of the wheel is k metres above the ground. A seat starts at the bottom of the wheel.

The wheel completes one revolution in 16 minutes.

After t minutes, the height of the seat above ground is given by $h\left(t\right)=61.5+a\text{cos}\left(\frac{\mathrm{\pi <!--\; \pi \; -->}}{8}t\right)$, for 0 ≤ t ≤ 32.

A new concert hall was built with $14$ seats in the first row. Each subsequent row of the hall has two more seats than the previous row. The hall has a total of $20$ rows.

Find:

The concert hall opened in $2019$. The average number of visitors per concert during that year was $584$. In $2020$, the average number of visitors per concert increased by $1.2\%$.

The concert organizers use this data to model future numbers of visitors. It is assumed that the average number of visitors per concert will continue to increase each year by $1.2\%$.

The following diagram shows the graph of a function $y=f(x)$, for $-<!--\; -\; -->6⩽<!--\; ⩽\; -->x⩽<!--\; ⩽\; -->-<!--\; -\; -->2$.

The points $(-<!--\; -\; -->6,6)$ and $(-<!--\; -\; -->2,6)$ lie on the graph of $f$. There is a minimum point at $(-<!--\; -\; -->4,0)$.

A hollow chocolate box is manufactured in the form of a right prism with a regular hexagonal base. The height of the prism is $h \text{cm}$, and the top and base of the prism have sides of length $x \text{cm}$.