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VCE Specialist Mathematics — Functions, relations and graphs Questions & Answers

Q2

2024

QCAA

Paper 1

1 mark

Q2

1 mark

Given that $\frac{A}{x-2} + \frac{3}{x} = \frac{x-6}{x(x-2)}$ , determine the value of $A$ .

A

$-4$

B

$-2$

C

$2$

D

$4$

Reveal Answer

A

$-4$

Substituting $A=-4$ and combining the fractions yields a numerator of $-4x + 3(x-2) = -x-6$ , which does not match the required numerator $x-6$ .

B

$-2$

Correct Answer

Multiplying the equation by the common denominator $x(x-2)$ gives $Ax + 3(x-2) = x-6$ . Expanding and comparing coefficients of $x$ leads to $A+3=1$ , so $A=-2$ .

C

$2$

Substituting $A=2$ results in a combined numerator of $2x + 3(x-2) = 5x-6$ , which is incorrect.

D

$4$

Substituting $A=4$ results in a combined numerator of $4x + 3(x-2) = 7x-6$ , which is incorrect.

Q2

2024

VCAA

Paper 2

1 mark

Q2

1 mark

Consider the function $f$ with rule $f(x) = \begin{cases} \frac{x^2 + 3x - 10}{x - 2}, & x \in R \setminus \{2\} \\ 7, & x = 2 \end{cases}$
Which of the following statements is correct?

A

The function $f$ is continuous.

B

The graph of $y = f(x)$ has a vertical asymptote.

C

The graph of $y = f(x)$ has a horizontal asymptote.

D

The graph of $y = f(x)$ has a point of discontinuity.

Reveal Answer

A

The function $f$ is continuous.

Correct Answer

For $x \neq 2$ , the function simplifies to $f(x) = x + 5$ . Since $\lim_{x \to 2} (x + 5) = 7$ , which equals $f(2)$ , the function is continuous everywhere.

B

The graph of $y = f(x)$ has a vertical asymptote.

The factor $(x - 2)$ in the denominator cancels with the numerator, creating a removable singularity rather than a vertical asymptote.

C

The graph of $y = f(x)$ has a horizontal asymptote.

As $x \to \pm\infty$ , the function behaves like the line $y = x + 5$ , meaning it approaches infinity rather than a horizontal asymptote.

D

The graph of $y = f(x)$ has a point of discontinuity.

Although the rational expression is undefined at $x = 2$ , the piecewise definition explicitly fills this "hole" by defining $f(2) = 7$ , making it continuous.

Q3

2025

VCAA

Paper 2

1 mark

Q3

1 mark

The graph of $y = \frac{x^2 + a}{bx + c}$ has an asymptote given by $y = -\frac{1}{2}x + \frac{1}{4}$ and a $y$ -intercept of $-2$ .

The values of $a$ , $b$ and $c$ are

A

$a = 2, b = -2, c = -1$

B

$a = 2, b = 2, c = -1$

C

$a = -2, b = -2, c = 1$

D

$a = -2, b = -2, c = -1$

Reveal Answer

A

$a = 2, b = -2, c = -1$

Correct Answer

Correct. The $y$ -intercept gives $\frac{a}{c} = -2$ . Polynomial division reveals the asymptote is $y = \frac{1}{b}x - \frac{c}{b^2}$ , so $\frac{1}{b} = -\frac{1}{2}$ and $-\frac{c}{b^2} = \frac{1}{4}$ , yielding $b = -2$ , $c = -1$ , and $a = 2$ .

B

$a = 2, b = 2, c = -1$

Incorrect. This option incorrectly sets $b = 2$ , which would result in an asymptote with a positive slope of $\frac{1}{2}$ instead of $-\frac{1}{2}$ .

C

$a = -2, b = -2, c = 1$

Incorrect. With $c = 1$ and $b = -2$ , the $y$ -intercept of the asymptote would be $-\frac{c}{b^2} = -\frac{1}{4}$ , which contradicts the given value of $\frac{1}{4}$ .

D

$a = -2, b = -2, c = -1$

Incorrect. With $a = -2$ and $c = -1$ , the $y$ -intercept of the function would be $\frac{a}{c} = 2$ , which contradicts the given $y$ -intercept of $-2$ .

Q15

2021

QCAA

Paper 1

4 marks

Q15

4 marks

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

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 $\ln\|f(x)\|$	1

Q3

2024

VCAA

Paper 2

1 mark

Q3

1 mark

The graph of $f(x) = \frac{x - h}{(x + 1)(x - 4)}$ , where $h \in R$ , will have no turning points when

A

$h < -1$ and $h > 4$

B

$-4 < h < 1$

C

$-1 \le h \le 4$

D

$-4 \le h \le 1$

Reveal Answer

A

$h < -1$ and $h > 4$

This range makes the discriminant of the derivative's numerator positive, which would result in the derivative changing signs and the function having turning points.

B

$-4 < h < 1$

This results from incorrectly factoring the discriminant inequality $h^2 - 3h - 4 \le 0$ as $(h+4)(h-1) < 0$ instead of $(h-4)(h+1) \le 0$ .

C

$-1 \le h \le 4$

Correct Answer

Setting the derivative $f'(x) = 0$ yields a quadratic numerator $-x^2 + 2hx - 3h - 4$ . For no turning points, its discriminant must be $\le 0$ , which simplifies to $h^2 - 3h - 4 \le 0$ , giving $-1 \le h \le 4$ .

D

$-4 \le h \le 1$

This results from incorrectly factoring the discriminant inequality $h^2 - 3h - 4 \le 0$ as $(h+4)(h-1) \le 0$ .

Q3

2025

SCSA

Paper 1

6 marks

Q3

6 marks

Determine $\int \frac{x+6}{x^2+2x} \, dx$ .

Reveal Answer

Using the method of partial fractions:

\begin{align*} \frac{x+6}{x(x+2)} &= \frac{a}{x} + \frac{b}{x+2}\\ &= \frac{a(x+2)+bx}{x(x+2)} = \frac{(a+b)x + 2a}{x(x+2)} \quad ... (1) \end{align*}

Equating coefficients: $a+b=1$ and $2a=6$
Solving gives $a=3, b=-2$

\begin{align*} \therefore \int \frac{x+6}{x^2+2x} dx &= \int \left(\frac{3}{x} - \frac{2}{x+2}\right) dx\\ &= 3\ln|x| - 2\ln|x+2| + c \end{align*}

Marking Criteria

Descriptor	Marks
factorises the denominator correctly	1
forms the numerator correctly using partial fractions (statement 1)	1
solves correctly for $a$ and $b$	1
re-writes the integrand in terms of the partial fractions correctly	1
anti-differentiates the reciprocal terms correctly using the absolute value	1
uses a constant of integration	1

Q1

2020

QCAA

Paper 1

1 mark

Q1

1 mark

The indefinite integral $\int \frac{3x-4}{1-x^2} dx$ can be determined using the partial fractions $\frac{-1}{1+x} + \frac{2}{1-x}$

The value of $A$ is

A

$-3$

B

$-1$

C

$1$

D

$3$

Reveal Answer

A

$-3$

This value is incorrect. Based on the provided partial fractions, the numerator for the term with denominator $1+x$ is $-1$ , not $-3$ .

B

$-1$

Correct Answer

The problem provides the partial fraction decomposition $\frac{-1}{1+x} + \frac{2}{1-x}$ . By comparing this to the standard form $\frac{A}{1+x} + \frac{B}{1-x}$ , we can identify that $A$ corresponds to the numerator of the first term, which is $-1$ .

C

$1$

This value has the wrong sign. The numerator given in the partial fraction decomposition is $-1$ .

D

$3$

This value is incorrect. It does not match the numerator $-1$ found in the term $\frac{-1}{1+x}$ .

Q4

2022

SCSA

Paper 1

8 marks

Q4a

3 marks

Function $f(x) = \frac{5(x + 1)}{(x - 1)(x^2 + 3x + 1)}$ can be expressed in the form $\frac{a}{x - 1} + \frac{bx + c}{x^2 + 3x + 1}$ .

Determine the value of the constants $a$ , $b$ and $c$ .

Reveal Answer

\begin{aligned} \frac{5x+5}{(x-1)(x^2+3x+1)} &= \frac{a(x^2+3x+1) + (x-1)(bx+c)}{(x-1)(x^2+3x+1)} \\ &= \frac{(a+b)x^2 + (3a-b+c)x + (a-c)}{(x-1)(x^2+3x+1)} \end{aligned}

Equating coefficients:

\begin{align*} a+b &= 0\\ 3a-b+c &= 5\\ a-c &= 5 \end{align*}

Solving gives $a=2$ , $b=-2$ , $c=-3$

\begin{align*} \text{i.e. }\frac{5(x+1)}{(x-1)(x^2+3x+1)} = \frac{2}{x-1} - \frac{(2x+3)}{x^2+3x+1} \end{align*}

Marking Criteria

Descriptor	Marks
forms the correct expression for the equivalent numerator	1
equates coefficients correctly to form 3 linear equations	1
solves correctly to determine $a$ , $b$ and $c$	1

Q4b

5 marks

Hence determine $\int \frac{10x + 10}{(x - 1)(x^2 + 3x + 1)} \, dx$ .

Reveal Answer

\begin{aligned} \int \frac{10x+10}{(x-1)(x^2+3x+1)} dx &= 2 \int \frac{5x+5}{(x-1)(x^2+3x+1)} dx \\ &= \int \frac{4}{x-1} - \frac{2(2x+3)}{x^2+3x+1} dx \\ &= 4\ln|x-1| - 2\ln|x^2+3x+1| + k \\ &= \ln\left( \frac{(x-1)^4}{(x^2+3x+1)^2} \right) + k \end{aligned}

Marking Criteria

Descriptor	Marks
expresses the given integrand as double $f(x)$	1
writes the integrand correctly in terms of the partial fractions	1
anti-differentiates $\frac{a}{x-1}$ correctly using the absolute value of a natural logarithm	1
anti-differentiates $\frac{bx+c}{x^2+3x+1}$ correctly	1
uses a constant of integration	1

Q4

2024

SCSA

Paper 1

6 marks

Q4a

2 marks

Given that $\frac{a}{x + 1} + \frac{b}{(x + 1)^2} = \frac{5x + 3}{(x + 1)^2}$ , determine the values for $a$ and $b$ .

Reveal Answer

$\frac{a}{x+1} + \frac{b}{(x+1)^2} = \frac{a(x+1)+b}{(x+1)^2} = \frac{ax+(a+b)}{(x+1)^2}$
Equating coefficients:
$a+b = 3$
$a = 5$
$b = -2$
Solving gives $a = 5, \ b = -2$

Marking Criteria

Descriptor	Marks
forms the equivalence of numerators correctly	1
solves for $a, b$ correctly	1

Q4b

4 marks

Hence determine $\int \frac{5x + 3}{(x + 1)^2} \, dx$ .

Reveal Answer

\begin{align*} \int \frac{5x+3}{(x+1)^2} \, dx &= \int \frac{5}{x+1} - \frac{2}{(x+1)^2} \, dx\\ &= 5\ln|x+1| + \frac{2}{x+1} + k \end{align*}

Marking Criteria

Descriptor	Marks
re-writes the integrand correctly in terms of the partial fractions	1
anti-differentiates the $(x+1)^{-1}$ term correctly using the absolute value of a natural logarithm	1
anti-differentiates the $2(x+1)^{-2}$ term correctly	1
uses a constant of integration	1

Q13

2022

QCAA

Paper 1

6 marks

Q13a

4 marks

Use partial fractions to determine $\int \frac{22}{(2x-3)(x+4)} dx$

Reveal Answer

$\frac{22}{(2x-3)(x+4)} = \frac{A}{2x-3} + \frac{B}{x+4}$
$\therefore 22 = A(x+4) + B(2x-3)$
Let $x = \frac{3}{2}: A(\frac{3}{2}+4) = 22 \Rightarrow A = 4$
Let $x = -4: B(2 \times -4 - 3) = 22 \Rightarrow B = -2$
$\int \frac{22}{2x^2+5x-12} dx = \int \frac{4}{2x-3} dx + \int \frac{-2}{x+4} dx$
$= 2 \int \frac{2}{2x-3} dx - 2 \int \frac{1}{x+4} dx$
$= 2 \ln|2x-3| - 2 \ln|x+4| + c$

Marking Criteria

Descriptor	Marks
correctly sets up the partial fractions	1
determines value of A	1
determines value of B	1
determines an expression for the indefinite integral	1

Q13b

2 marks

Use the result from Question 13a) to determine $\int_{-3}^{0} \frac{22}{(2x-3)(x+4)} dx$
Express your answer in simplest form.

Reveal Answer

$\int_{-3}^{0} \frac{22}{(2x-3)(x+4)} dx$
$= [2 \ln|2x-3| - 2 \ln|x+4|]_{-3}^{0}$
$= ((2 \ln|-3| - 2 \ln|4|) - (2 \ln|-9| - 2 \ln|1|))$
$= 2(\ln(3) - \ln(4) - \ln(9)) = 2 \ln(\frac{3}{9 \times 4})$
$= 2 \ln(\frac{1}{12})$

Marking Criteria

Descriptor	Marks
substitutes limits of integration into the result from 13a)	1
expresses a definite integral value in simplest form	1

Q5

2021

SCSA

Paper 1

5 marks

Q5a

2 marks

Given that $\frac{7x^2 - 12x + 2}{(x - 2)(x^2 + 2)} = \frac{a}{x - 2} + \frac{bx}{x^2 + 2}$ determine the values of $a$ and $b$ .

Reveal Answer

\begin{align*} \frac{a}{x-2} + \frac{bx}{x^2+2} = \frac{a(x^2+2)+bx(x-2)}{(x-2)(x^2+2)} = \frac{(a+b)x^2-2bx+2a}{(x-2)(x^2+2)} \end{align*}

Equating coefficients: $a+b=7$ , $-2b=-12$ , and $2a=2$

Solving gives $a=1, \ b=6$

Marking Criteria

Descriptor	Marks
forms the equivalence of numerators correctly	1
solves for $a,b$ correctly	1

Q5b

3 marks

Hence determine $\int \frac{7x^2 - 12x + 2}{(x - 2)(x^2 + 2)} \, dx$ .

Reveal Answer

\begin{align*} \int \frac{7x^2-12x+2}{(x-2)(x^2+2)} \, dx &= \int \left( \frac{1}{x-2} + \frac{6x}{x^2+2} \right) dx\\ &= \int \frac{1}{x-2} \, dx + 3 \int \frac{2x}{x^2+2} \, dx\\ &= \ln|x-2| + 3\ln(x^2+2) + k \end{align*}

Marking Criteria

Descriptor	Marks
re-writes the integrand correctly in terms of the partial fractions	1
anti-differentiates the $(x-2)^{-1}$ term correctly using the absolute value of a natural logarithm AND uses an integration constant	1
anti-differentiates the $6x(x^2+2)^{-1}$ term correctly (absolute value not required)	1

Q3

2023

VCAA

Paper 1

3 marks

Q3

A particle moves along a straight line. When the particle is $x$ m from a fixed point $O$ , its velocity, $v$ m s $^{-1}$ , is given by

$v = \frac{3x + 2}{2x - 1}\text{, where } x \ge 1.$

Q3a

2 marks

Find the acceleration of the particle, in m s $^{-2}$ , when $x = 2$ .

Reveal Answer

$a = v\frac{dv}{dx} = \frac{3x+2}{2x-1} \times \frac{-7}{(2x-1)^2}$

When $x=2$ , $a = -\frac{56}{27}$ .

Marking Criteria

Descriptor	Marks
Correctly differentiates $v$ with respect to $x$ to find $\frac{dv}{dx}$ , or sets up a correct expression for acceleration such as $v\frac{dv}{dx}$ or $\frac{d}{dx}\left(\frac{1}{2}v^2\right)$	1
Correctly calculates the final acceleration when $x=2$ as $-\frac{56}{27}$	1

Q3b

1 mark

Find the value that the velocity of the particle approaches as $x$ becomes very large.

Reveal Answer

$\frac{3}{2}$

Marking Criteria

Descriptor	Marks
States the correct limiting value of $\frac{3}{2}$	1

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