Calculus — Free Maths AA SL Practice Questions

Consider the function $$g(x) = \begin{cases} 2x + 1, & x < 2 \\ 5, & x = 2 \\ x^2 - 3, & x > 2 \end{cases}$$

Consider the function $f(x) = \dfrac{x^2 - 4}{x - 2}$ for $x \neq 2$.

A function $g$ is defined by $g(x) = \dfrac{x^2 - 4}{|x-2|}$ for $x \in \mathbb{R}$, $x \neq 2$.

Consider the function $f(x) = \dfrac{x^2 - 1}{x - 1}$, for $x \neq 1$.

The function $f$ is defined by $f(x) = \dfrac{3}{x-1}$, for $x \neq 1$.

The graph of $y = f(x)$ for $-2 \leq x \leq 4$ is shown below. The tangent to the curve at point $P$, where $x = 1$, is also drawn.

A water tank has a small hole at its base. The volume of water, $V$ litres, remaining in the tank $t$ minutes after the hole is unplugged is given by $$V(t) = 100\left(1 - \frac{t}{20}\right)^2, \quad 0 \leq t \leq 20.$$

A particle moves along a straight line such that its displacement, $s$ metres, from a fixed point $O$ at time $t$ seconds is given by $$s(t) = t^3 - 6t^2 + 9t + 2, \quad t \geq 0.$$

A company models the profit $P(x)$, in thousands of dollars, from selling $x$ hundred units of a product as $$P(x) = (2x + 3)(x - 1)^2, \quad x \geq 1.$$ -

The graph of a function $f$ is shown below. The points $A(1,\,2)$ and $B(4,\,8)$ lie on the curve $y = f(x)$. The tangent to the curve at $A$ has equation $y = 3x - 1$.

A particle moves along a straight line such that its velocity $v\,\text{m s}^{-1}$ at time $t$ seconds is given by $v(t) = 3t^2 - 4t + 5$.

A particle moves along a straight line such that its velocity $v\,\text{m s}^{-1}$ at time $t$ seconds is given by $v(t) = 3t^2 - 12t + 9$, for $t \geq 0$.

A manufacturer produces a decorative rod whose cross-sectional area (in $\text{cm}^2$) at distance $x$ metres from one end is given by $A(x) = 4xe^{-2x} + 3$, for $0 \leq x \leq 3$.

Consider the function $f(x) = \dfrac{1}{x^2}$ for $x > 0$.

Consider the function $f(x) = \dfrac{12}{(2x+1)^2}$, for $x > -\dfrac{1}{2}$.

The number of bacteria, $N$, in a culture grows at a rate proportional to $N$. This modelled by the differential equation $\dfrac{dN}{dt} = kN$, where $t$ is the time in hours and $k$ is a positive constant.

The population of a colony of ants is modelled by the differential equation $$\frac{dP}{dt} = \frac{1}{50}P(500 - P)$$ where $P$ is the population at time $t$ years, $t \geq 0$. Initially, the population is $100$ ants.

A cylindrical water tank has a constant cross-sectional area of $2\,\text{m}^2$ and a small hole at its base. Water leaks such that the rate of change of volume $V$ (in $\text{m}^3$) at time $t$ (in hours) is proportional to the square root of the depth $h$ (in metres). The tank is initially full with depth $4\,\text{m}$. After $1$ hour, the depth has decreased to $3\,\text{m}$. Using $V = 2h$, the differential equation for the depth is $$\frac{dh}{dt} = -k\sqrt{h}, \quad k > 0.$$

Consider the differential equation $\dfrac{dy}{dx} = \dfrac{3x^2}{y}$, where $y > 0$.

A chemical reaction is modelled by the differential equation $$\frac{dx}{dt} = k(10 - x)(5 - x)$$ where $x$ grams is the amount of product formed after $t$ minutes and $k$ is a positive constant. Initially, $x = 0$.

The function $f$ is defined by $f(x) = \dfrac{x^2 - 2x - 3}{x^2 - 9}$, for $x \neq \pm 3$.

The graph of a function $h$ is shown below. The function has vertical asymptotes at $x = -2$ and $x = 2$, and a horizontal asymptote at $y = 1$. Using the graph, determine the value of each of the following limits. If a limit does not exist, state this clearly.

A function is defined as $f(x) = \dfrac{2x^2 - 5x - 3}{x - 3}$ for $x \neq 3$.

The graph of a function $y = p(x)$ is shown below.

The graph of $f(x) = x^3 - 6x^2 + 9x + 1$ is shown below for $0 \leq x \leq 4$.

The function $f(x) = \dfrac{1}{x}$ is defined for $x > 0$. The graph of $f$ passes through points $P(1,1)$ and $Q(2, 0.5)$.

The volume $V$ of water in a cylindrical tank, measured in litres, is given by $V(t) = 2000 + 120t - 3t^2$, where $t$ is the time in minutes after a valve is opened.

The height $h$ metres of a ball thrown vertically upwards from ground level is given by $h(t) = 24t - 4.9t^2$, where $t$ is the time in seconds after release.

The population $P$ of a town, measured in thousands, is modelled by $P(t) = 50 + 8t - 0.2t^2$, where $t$ is the number of years after the year 2000.

A company manufactures decorative tiles. The curved edge of a tile is modelled by the function $f(x) = 8 - 2\ln(x+1)$, for $0 \leq x \leq 6$, where $x$ is the horizontal distance from the left edge of the tile and $f(x)$ is the vertical height above the base, both measured in centimetres.

The graph of $f(x) = x^3 - 2x$ is shown for $x \in \mathbb{R}$. The shaded region is bounded by the curve, the $x$-axis, and the vertical lines $x = -1$ and $x = 2$.

The graph shows $f'(x)$, the derivative of $f(x)$, for $0 \leq x \leq 6$. The graph consists of three line segments joining the points $(0,0)$, $(2,4)$, $(4,0)$, and $(6,-2)$ in order. Given that $f(0) = 1$:

The rate of change of the volume of water in a reservoir, measured in thousands of cubic metres per day, is modelled by $$R(t) = 12t^2 - 120t + 300$$ where $t$ is the time in days after 1 January 2020. The reservoir initially contained $5000$ thousand cubic metres of water.

The acceleration of a particle moving along a straight line is given by $a(t) = 6t - 4$, where $t \geq 0$ is time in seconds and $a(t)$ is in $\text{m\,s}^{-2}$. The initial velocity of the particle is $v(0) = 10\,\text{m\,s}^{-1}$ and the initial displacement is $s(0) = 0$.

A population of rabbits on an island is modelled by the differential equation $$\frac{dP}{dt} = \frac{P(500 - P)}{1000}$$ where $P$ is the number of rabbits at time $t$ years, $t \geq 0$.

The rate of cooling of a cup of coffee is modelled by Newton's Law of Cooling: $$\frac{dT}{dt} = -k(T - T_s)$$ where $T$ is the temperature of the coffee in $^\circ\text{C}$ at time $t$ minutes, $T_s = 20\,^\circ\text{C}$ is the constant room temperature, and $k > 0$ is a constant.

A population of bacteria grows at a rate proportional to its size. The population $P$, measured in thousands, satisfies the differential equation $$\frac{dP}{dt} = kP$$ where $t$ is time in hours and $k$ is a positive constant. At $t = 0$ the population is $1000$ bacteria. After $3$ hours the population is $2000$ bacteria.

A tank contains $100$ litres of water with a dissolved pollutant concentration of $0.5\,\text{kg}$ per litre. Fresh water enters the tank at a rate of $2$ litres per minute, and the well-mixed solution leaves at the same rate. Let $x(t)\,\text{kg}$ be the mass of pollutant in the tank at time $t$ minutes.

A population of bacteria in a laboratory culture grows according to the differential equation $\dfrac{dP}{dt} = kP$, where $P$ is the population size at time $t$ days and $k$ is a positive constant. Initially, the population is 200 bacteria. After 3 days, the population has grown to 600 bacteria.