How many QCAA Specialist Mathematics Alternative Sequence questions cover Integration techniques?

AusGrader has 12 QCAA Specialist Mathematics Alternative Sequence questions on Integration techniques, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics Alternative Sequence — Integration techniques Questions & Answers

Q11

2021

QCAA

Paper 1

5 marks

Q11

Let $f(x)=\tan^{-1}\!\left(\dfrac{x}{2}\right)$ for suitable values of $x$ where $f(x)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ .

Q11a

1 mark

Determine $f(2)$ .

Reveal Answer

$f(2) = \tan^{-1}(1)$
$= \frac{\pi}{4}$ as $f(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Marking Criteria

Descriptor	Marks
correctly determines the required value	1

Q11b

2 marks

Determine $f'(2)$ .

Reveal Answer

$f'(x) = \frac{d}{dx}\left(\tan^{-1}\left(\frac{x}{2}\right)\right)$
$= \frac{2}{4 + x^2}$
$f'(2) = \frac{1}{4}$

Marking Criteria

Descriptor	Marks
correctly determines the gradient function	1
determines gradient of the tangent	1

Q11c

2 marks

Use the results from Questions 11a) and 11b) to determine the equation of the tangent to the graph of $y=f(x)$ at $x=2$ .

Reveal Answer

Equation of the tangent has the form $y = mx + c$
From 11a) $x = 2, y = \frac{\pi}{4}$
From 11b) $m = \frac{1}{4}$
$\frac{\pi}{4} = \frac{1}{4}(2) + c \Rightarrow c = \frac{\pi}{4} - \frac{1}{2}$
Equation of the tangent is
$y = \frac{1}{4}x + \frac{\pi}{4} - \frac{1}{2}$

Marking Criteria

Descriptor	Marks
determines $y$ -intercept of the tangent	1
determines equation of the tangent	1

Q3

2022

QCAA

Paper 2

1 mark

Q3

1 mark

Determine the solution of the differential equation $\frac{dy}{dx} = \frac{\sin(2x)}{\cos(2x)}$ given $y = 0$ when $x = \frac{\pi}{5}$ .

A

$y = -2\ln|\cos(2x)| - 2.35$

B

$y = -2\ln|\cos(2x)| + 2.35$

C

$y = -\frac{1}{2}\ln|\cos(2x)| - 0.59$

D

$y = -\frac{1}{2}\ln|\cos(2x)| + 0.59$

Reveal Answer

A

$y = -2\ln|\cos(2x)| - 2.35$

This is incorrect because it results from multiplying by 2 instead of dividing by 2 when applying the chain rule during integration.

B

$y = -2\ln|\cos(2x)| + 2.35$

This is incorrect because it results from multiplying by 2 during integration and making a sign error when solving for the constant $C$ .

C

$y = -\frac{1}{2}\ln|\cos(2x)| - 0.59$

Correct Answer

This is correct. Integrating $\frac{\sin(2x)}{\cos(2x)}$ using u-substitution yields $-\frac{1}{2}\ln|\cos(2x)| + C$ . Substituting $x = \frac{\pi}{5}$ and $y = 0$ gives $C = \frac{1}{2}\ln|\cos(\frac{2\pi}{5})| \approx -0.59$ .

D

$y = -\frac{1}{2}\ln|\cos(2x)| + 0.59$

This is incorrect. While the integration is correct, a sign error was made when solving for the constant of integration $C$ , which should be negative.

Q15

2020

QCAA

Paper 1

4 marks

Q15

The motion of an object that moves in a straight line is given by $v(x) = \cos^{-1}(2x)$ where $v$ is the velocity ( $\text{m s}^{-1}$ ) and $x$ is the displacement (m) from the origin.

Q15a

2 marks

Determine $a(x)$ where $a$ is the acceleration ( $\text{m s}^{-2}$ ) of the object.

Reveal Answer

$v(x)=\cos^{-1}(2x)$
$a=v\frac{dv}{dx}$
$\frac{dv}{dx}=\frac{-1}{\sqrt{0.25-x^2}}$
$a=\frac{-\cos^{-1}(2x)}{\sqrt{0.25-x^2}}$

Marking Criteria

Descriptor	Marks
correctly determines $\frac{dv}{dx}$	1
determines an expression for acceleration as a function of displacement	1

Q15b

2 marks

Use the result from 15a) to determine $a(0)$ , given $-2\pi \le a(0) \le 0$ . Express your answer in simplest form.

Reveal Answer

When $x=0$
$a=\frac{-\cos^{-1}(0)}{0.5}$
$\cos^{-1}(0)=\frac{\pi}{2}$
$a(0)= -\frac{\left(\frac{\pi}{2}\right)}{0.5}$
$=-\pi\ (\text{m s}^{-2})$

Marking Criteria

Descriptor	Marks
determines a correct exact value for the inverse trigonometric expression on the numerator	1
determines a reasonable solution for $a(0)$ based on the given range ( $-2\pi\le a(0)\le 0$ )	1

Q15

2021

QCAA

Paper 1

4 marks

Q15

4 marks

Use partial fractions to determine $\int \dfrac{4x-17}{x^2-x-6}\,dx$ , where $x\in\mathbb{R}$ , $x\ne -2$ , $x\ne 3$ .

Express your answer in the form $\ln|f(x)|+c$ .

Reveal Answer

$\frac{4x - 17}{x^2 - x - 6} = \frac{A}{(x + 2)} + \frac{B}{(x - 3)}$
$= \frac{A(x - 3) + B(x + 2)}{(x + 2)(x - 3)}$
$x = 3: -5 = 5B \Rightarrow B = -1$
$x = -2: -25 = -5A \Rightarrow A = 5$
$\int \frac{4x - 17}{x^2 - x - 6} \, dx = \int \frac{5}{(x + 2)} + \frac{-1}{(x - 3)} \, dx$
$= 5\ln|x + 2| - \ln|x - 3|$
$= \ln|x + 2|^5 - \ln|x - 3|$
$= \ln\left|\frac{(x + 2)^5}{x - 3}\right| + c$

Marking Criteria

Descriptor	Marks
correctly factorises the denominator to establish the form of the partial fraction decomposition	1
determines values of A and B	1
determines indefinite integral of the fraction	1
determines expression in the form $\ln\|f(x)\| + c$	1

Q17

2021

QCAA

Paper 2

7 marks

Q17

7 marks

An object with a mass of 2 kg is released from rest at the top of a 1 metre long frictionless plane inclined at $30^\circ$ to the horizontal.

A force of $\boldsymbol{P}$ newtons acting parallel to the plane opposes the motion of the object as it travels down the plane.

When the object is $x$ metres from the top of the plane, its velocity is $v\text{ m s}^{-1}$ .

Given $|\boldsymbol{P}| = \frac{4}{\sqrt{4-x^2}}$ , determine $x$ when $v = 2$ .

Reveal Answer

Method 1
Resolving net forces along the plane
$F_{\text{net}} = 2g \sin(30^\circ) - \frac{4}{\sqrt{4 - x^2}}$

$F_{\text{net}} = ma$
$g - \frac{4}{\sqrt{4 - x^2}} = 2a$

$g - \frac{4}{\sqrt{4 - x^2}} = 2v \frac{dv}{dx}$

$v \frac{dv}{dx} = 4.9 - \frac{2}{\sqrt{4 - x^2}}$

$\int v \ dv = \int 4.9 - \frac{2}{\sqrt{4 - x^2}} \ dx$

$\frac{v^2}{2} = 4.9x - 2 \sin^{-1} \left(\frac{x}{2}\right) + c$

Given $v = 0$ when $x = 0$
$0 = 0 - 2 \sin^{-1}(0) + c$
$c = 0$
$\therefore v^2 = 9.8x - 4 \sin^{-1} \left(\frac{x}{2}\right)$

When $v = 2$
$4 = 9.8x - 4 \sin^{-1} \left(\frac{x}{2}\right)$

Solving for $x$ using GDC
$x = 0.51 \text{ m}$

Marking Criteria

Descriptor	Marks
correctly determines the net forces along the plane	1
determines equation for acceleration along the plane	1
determines differential equation in terms of velocity and displacement	1
determines general solution to a differential equation	1
determines value of arbitrary constant	1
establishes equation to solve for $x$ when $v = 2$	1
determines $x$	1

Q18

2021

QCAA

Paper 2

6 marks

Q18

The function $y = g(x)$ for $x \in \left(-\frac{\pi}{4}, \frac{\pi}{2}\right)$ is defined by the parametric equations

$x = \tan^{-1}(t-1)$
$y = t - 2 + 2t^{-1}$

Q18a

4 marks

Show that the area under the graph of $y = g(x)$ between $x = a$ and $x = b$ $\left(\text{where } -\frac{\pi}{4} < a < b < \frac{\pi}{2}\right)$ can be expressed as $\ln\left(\frac{\tan(b)+1}{\tan(a)+1}\right)$

Reveal Answer

Given $x = \tan^{-1}(t - 1)$
$t = \tan(x) + 1$

Substituting into $y = t - 2 + 2t^{-1}$
$y = \tan(x) + 1 - 2 + \frac{2}{\tan(x) + 1}$

$= \frac{(\tan(x) - 1)(\tan(x) + 1) + 2}{\tan(x) + 1}$

$= \frac{\tan^2(x) + 1}{\tan(x) + 1}$

$= \frac{\sec^2(x)}{\tan(x) + 1}$

Area $= \int_a^b \frac{\sec^2(x)}{\tan(x) + 1} \ dx$

Let $u = \tan(x) + 1 \Rightarrow \frac{du}{dx} = \sec^2(x)$

Area $= \int_{x=a}^{x=b} \frac{du}{u} = \ln|u| \ \Big|_{x=a}^{x=b} = \ln(\tan(x) + 1) \ \Big|_a^b$

$= [\ln(\tan(b) + 1) - \ln(\tan(a) + 1)]$

$= \ln \left( \frac{\tan(b) + 1}{\tan(a) + 1} \right)$

Marking Criteria

Descriptor	Marks
correctly expresses the parameter in terms of $x$	1
uses Pythagorean identity to determine a simplified Cartesian equation of $y$ in terms of $x$	1
demonstrates suitable trigonometric substitution method to integrate an expression representing the required area	1
provides evidence to show that the given expression represents the required area	1

Q18b

2 marks

Use the values of $a = 0$ and $b = 1$ to verify that $\ln\left(\frac{\tan(b)+1}{\tan(a)+1}\right)$ represents the area under the graph of $y = g(x)$ .

Reveal Answer

Given $a = 0$ and $b = 1$
Area $= \ln \left( \frac{\tan(1) + 1}{\tan(0) + 1} \right) \approx 0.94 \text{ units}^2$

Using GDC
Area $= \int_0^1 \frac{\sec^2(x)}{\tan(x) + 1} \ dx \approx 0.94 \text{ units}^2$

The result is verified for this example.

Marking Criteria

Descriptor	Marks
correctly determines the area using $\ln \left( \frac{\tan(b) + 1}{\tan(a) + 1} \right)$	1
verifies the result	1

Q14

2022

QCAA

Paper 2

5 marks

Q14

An object is moving in a straight line with an acceleration represented by the differential equation $\frac{dv}{dt} = -\left(4 + v^2\right)$ , where $v$ is the object's velocity $\left(\text{m s}^{-1}\right)$ over time, $t\text{ (s)}$ , where $t \ge 0$ , until it comes to rest.

Q14a

3 marks

Determine the general solution of the differential equation.

Reveal Answer

$\frac{dv}{dt} = -(4 + v^2)$
$\frac{1}{4 + v^2} \frac{dv}{dt} = -1$
$\int \frac{1}{4 + v^2} dv = \int -1 dt$
$\frac{1}{2} \int \frac{2}{4 + v^2} dv = \int -1 dt$
$\frac{1}{2} \tan^{-1} \left(\frac{v}{2}\right) = -t + c$

Marking Criteria

Descriptor	Marks
correctly establishes a suitable integration result based on the separation of variables technique	1
determines one side of the general solution in terms of $v$	1
determines the other side of the general solution in terms of $t$	1

Q14b

2 marks

The initial velocity of the object is $1.5 \text{ m s}^{-1}$ .

Determine the time when the particle comes to rest.

Reveal Answer

Given $v(0) = 1.5$
$\frac{1}{2} \tan^{-1} \left(\frac{1.5}{2}\right) = c$
$c \approx 0.32$

When $v = 0$ :
$\frac{1}{2} \tan^{-1}(0) \approx -t + 0.32$
$t \approx 0.32 \text{ s}$
The particle comes to rest after 0.32 s.

Marking Criteria

Descriptor	Marks
determines an expression that represents the integration constant	1
determines the time when the particle is at rest	1

Q12

2020

QCAA

Paper 2

9 marks

Q12

For a certain experiment, the number of yeast cells, $N$ , after $t$ hours in a test tube can be modelled by the differential equation

$\frac{dN}{dt} = \frac{1}{1000}N(1000 - N)$ for $t \ge 0$

Q12b

A scientist commenced this experiment at 9:00 am on a certain day and observed that 100 yeast cells were present at this time.

Q12d

The scientist predicted that the number of yeast cells would eventually exceed 1200.

Q12a

2 marks

Given $\frac{1000}{N(1000 - N)} = \frac{1}{N} + \frac{1}{1000 - N}$ , show that the general solution of the differential equation can be expressed as

$\ln\left|\frac{N}{1000 - N}\right| = t + c$

Reveal Answer

\frac{dN}{dt} = \frac{1}{1000}N(1000 - N)

\frac{1000}{N(1000-N)} \frac{dN}{dt} = 1

\left(\frac{1}{N} + \frac{1}{1000-N}\right) \frac{dN}{dt} = 1

\int \left(\frac{1}{N} + \frac{1}{1000-N}\right) dN = \int 1 dt

\ln|N| - \ln|1000 - N| = t + c

\ln\left|\frac{N}{1000 - N}\right| = t + c

Marking Criteria

Descriptor	Marks
correctly uses separation of variables technique and substitutes the given result into the differential equation	1
correctly develops the required general solution	1

Q12b

3 marks

Show that the solution of the differential equation can be expressed as

$N = \frac{1000}{1 + 9e^{-t}}$

Reveal Answer

\text{Let } t = \text{time after 9:00 am (in hours)}

\text{When } t = 0, N = 100

\ln\left|\frac{100}{1000-100}\right| = 0 + c

c = \ln\left(\frac{1}{9}\right)

\ln\left(\frac{N}{1000 - N}\right) = t + \ln\left(\frac{1}{9}\right)

\frac{N}{1000 - N} = e^{t + \ln(1/9)}

\frac{N}{1000 - N} = \frac{e^t}{9}

9N = 1000e^t - Ne^t

N(9 + e^t) = 1000e^t

N = \frac{1000e^t}{(9 + e^t)}

N = \frac{1000}{1 + 9e^{-t}}

Marking Criteria

Descriptor	Marks
correctly determines c	1
substitutes the value of c into the general equation and simplifies sufficiently to produce a function that includes the term e^t	1
develops a solution for N	1

Q12c

2 marks

Determine the time of day when 900 yeast cells were present.

Reveal Answer

\text{Given } N = 900

900 = \frac{1000}{1 + 9e^{-t}}

\text{Using solve facility of GDC}

t \approx 4.394 \text{ (hours)}

\text{The time of day is 1:24 pm.}

Marking Criteria

Descriptor	Marks
correctly determines the value of t when N = 900	1
communicates the time of day	1

Q12d

2 marks

Evaluate the reasonableness of the scientist's prediction.

Reveal Answer

\text{Given } N = 1200:

\text{As } t \to \infty, N \to 1000

\text{The number of yeast cells has a limit of 1000.}

\text{As } N \text{ will never reach 1200, the scientist's prediction is not reasonable.}

Marking Criteria

Descriptor	Marks
correctly recognises that the number of yeast cells will never exceed 1000	1
comments that the prediction is not reasonable	1

QCAA Specialist Mathematics Alternative Sequence Integration techniques

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