VCAA Mathematical Methods Functions, relations and graphs

15 sample questions with marking guides and sample answers · Avg. score: 54.2%

Q5
2022
VCAA
Paper 2
1 mark
Q5
1 mark

The largest value of aa such that the function f:(,a]R,f(x)=x2+3x10f: (-\infty, a] \rightarrow R, f(x) = x^2 + 3x - 10, where ff is one-to-one, is

A

12.25-12.25

B

5-5

C

1.5-1.5

D

00

E

22

Reveal Answer
A

12.25-12.25

This is the y-coordinate of the vertex (the minimum value of the function), not the x-coordinate required for the domain.

B

5-5

This is one of the roots of the quadratic equation (x=5x=-5), not the x-coordinate of the vertex that determines where the function changes direction.

C

1.5-1.5

Correct Answer

A quadratic function is one-to-one on an interval ending at its vertex. The x-coordinate of the vertex is found using b/(2a)=3/2=1.5-b/(2a) = -3/2 = -1.5.

D

00

The interval (,0](-\infty, 0] includes the vertex at x=1.5x = -1.5, meaning the function decreases and then increases within this domain, so it is not one-to-one.

E

22

This is the other root of the quadratic equation (x=2x=2). The interval (,2](-\infty, 2] includes the vertex, so the function is not one-to-one.

Q1
2022
VCAA
Paper 2
1 mark
Q1
1 mark

The period of the function f(x)=3cos(2x+π)f(x) = 3 \cos(2x + \pi) is

A

2π2\pi

B

π\pi

C

2π3\frac{2\pi}{3}

D

22

E

33

Reveal Answer
A

2π2\pi

Incorrect. This is the standard period of y=cos(x)y = \cos(x). To find the period of this function, you must divide 2π2\pi by the coefficient of xx, which is 22.

B

π\pi

Correct Answer

Correct. The period of a cosine function y=Acos(Bx+C)y = A \cos(Bx + C) is 2πB\frac{2\pi}{|B|}. Here, B=2B = 2, so the period is 2π2=π\frac{2\pi}{2} = \pi.

C

2π3\frac{2\pi}{3}

Incorrect. This incorrectly divides the standard period 2π2\pi by the amplitude 33. The period depends on the coefficient of xx, not the amplitude.

D

22

Incorrect. This is the coefficient of xx (often denoted as BB), which represents the angular frequency, not the period itself.

E

33

Incorrect. This is the amplitude of the function, which determines its maximum and minimum values, not its period.

Q6
2020
SCSA
Paper 1
7 marks
Q6

Consider the function f(x)=ln(x)f(x) = \ln(x). The function g(x)=f(x)+ag(x) = f(x) + a is a vertical translation of ff by aa units.

Q6a
2 marks

Express the function g(x)=ln(4x)g(x) = \ln(4x) in terms of a vertical translation of ff (i.e. in the form g(x)=f(x)+ag(x) = f(x) + a), stating the number of units that ff is translated.

Reveal Answer

g(x)=ln(4x)=ln(4)+ln(x)=f(x)+ln(4)\begin{align*} g(x) &= \ln(4x)\\ &= \ln(4) + \ln(x)\\ &= f(x) + \ln(4) \end{align*}

ff is translated vertically (upward) by ln(4)\ln(4) units.

Marking Criteria
DescriptorMarks

expresses g(x)g(x) as a sum of logs

1

recognises a vertical translation by ln(4)\ln(4) units

1
Q6b
2 marks

The function h(x)=cf(x)h(x) = cf(x) is a vertical dilation of ff by a scale factor of cc.

Express the function h(x)=ln(x)h(x) = \ln(\sqrt{x}) in terms of a vertical dilation of ff, stating the scale factor.

Reveal Answer

h(x)=ln(x)=ln(x0.5)=0.5ln(x)=0.5f(x)\begin{align*} h(x) &= \ln(\sqrt{x})\\ &= \ln(x^{0.5})\\ &= 0.5 \ln(x)\\ &= 0.5 f(x) \end{align*}

ff is scaled vertically by a factor of 0.5.

Marking Criteria
DescriptorMarks

expresses hh as a product involving ln(x)\ln(x)

1

recognises a vertical scaling by a scale factor of 0.5

1
Q6c
3 marks

The function p(x)=f(bx)p(x) = f(bx) is a horizontal dilation of ff by a scale factor of 1b\frac{1}{b}.

Express the function p(x)=ln(x)+4p(x) = \ln(x) + 4 in terms of a horizontal dilation of ff, stating the scale factor.

Reveal Answer

p(x)=ln(x)+4=ln(x)+4ln(e)=ln(x)+ln(e4)=ln(e4x)=f(e4x)\begin{align*} p(x) &= \ln(x) + 4\\ &= \ln(x) + 4\ln(e)\\ &= \ln(x) + \ln(e^4)\\ &= \ln(e^4x)\\ &= f(e^4x) \end{align*}

ff is scaled horizontally by a scale factor of e4e^{-4}.

Marking Criteria
DescriptorMarks

expresses 4 as 4ln(e)4\ln(e)

1

expresses pp using a single logarithm

1

states horizontal scale factor

1
Q15
2025
VCAA
Paper 2
1 mark
Q15
1 mark

The graph of y=g(x)y = g(x) passes through the point (1,3)(1, 3).

The graph of y=1g(2x3)y = 1 - g(2x - 3) must pass through the point

A

(1,2)(-1, -2)

B

(2,2)(2, -2)

C

(1,2)(-1, 2)

D

(2,2)(2, 2)

Reveal Answer
A

(1,2)(-1, -2)

This incorrectly assumes the horizontal transformation is 2x+3=12x + 3 = 1, which would give x=1x = -1. However, the argument is 2x32x - 3.

B

(2,2)(2, -2)

Correct Answer

Since we know g(1)=3g(1) = 3, we set the argument 2x3=12x - 3 = 1 to find x=2x = 2. Substituting x=2x = 2 into the equation gives y=1g(1)=13=2y = 1 - g(1) = 1 - 3 = -2, resulting in the point (2,2)(2, -2).

C

(1,2)(-1, 2)

This option uses an incorrect xx-value by solving 2x+3=12x + 3 = 1 and an incorrect yy-value calculation.

D

(2,2)(2, 2)

While this correctly identifies x=2x = 2, it incorrectly calculates the yy-value as y=1+3=2y = -1 + 3 = 2 instead of y=13=2y = 1 - 3 = -2.

Q3
2023
VCAA
Paper 2
1 mark
Q3
1 mark

Two functions, pp and qq, are continuous over their domains, which are [2,3)[-2, 3) and (1,5](-1, 5], respectively.

The domain of the sum function p+qp + q is

A

[2,5][-2, 5]

B

[2,1)(3,5][-2, -1) \cup (3, 5]

C

[2,1)(1,3)(3,5][-2, -1) \cup (-1, 3) \cup (3, 5]

D

[1,3][-1, 3]

E

(1,3)(-1, 3)

Reveal Answer
A

[2,5][-2, 5]

This represents the union of the two domains. However, the domain of a sum function is the intersection of the individual domains, not the union.

B

[2,1)(3,5][-2, -1) \cup (3, 5]

This represents the intervals where only one of the functions is defined. For the sum function to be defined, both functions must be defined simultaneously.

C

[2,1)(1,3)(3,5][-2, -1) \cup (-1, 3) \cup (3, 5]

This represents the union of the domains excluding the points 1-1 and 33. The domain of p+qp + q requires finding the overlapping region (intersection) of both domains.

D

[1,3][-1, 3]

This incorrectly includes the endpoints 1-1 and 33. The value 1-1 is not in the domain of qq, and 33 is not in the domain of pp, so they cannot be in the domain of p+qp + q.

E

(1,3)(-1, 3)

Correct Answer

The domain of the sum function p+qp + q is the intersection of their individual domains. The intersection of [2,3)[-2, 3) and (1,5](-1, 5] is exactly (1,3)(-1, 3).

Q8
2023
QCAA
Paper 2
1 mark
Q8
1 mark

The number of koalas in a conservation park is modelled by N=15ln(7t+1)N = 15 \ln(7t + 1), t1t \geq 1, where tt represents the time (years) since the park opened. There were 20 koalas in the park when it opened.

Determine the approximate rate of change in the number of koalas when t=3t = 3.

A

46

B

26

C

25

D

5

Reveal Answer
A

46

This is the value of the function N(3)=15ln(22)46N(3) = 15 \ln(22) \approx 46. This represents the number of koalas (or the population increase) at year 3, rather than the rate at which the population is changing.

B

26

This value appears to be the result of calculating N(3)2026N(3) - 20 \approx 26. This subtracts the initial population from the model's value at t=3t=3, which does not represent the instantaneous rate of change.

C

25

This is an incorrect value. It does not correspond to the derivative at t=3t=3 or the function value, likely resulting from a calculation error.

D

5

Correct Answer

The rate of change is found by taking the derivative dNdt\frac{dN}{dt}. Using the chain rule, dNdt=1517t+17=1057t+1\frac{dN}{dt} = 15 \cdot \frac{1}{7t+1} \cdot 7 = \frac{105}{7t+1}. Evaluating at t=3t=3 gives 105224.77\frac{105}{22} \approx 4.77, which rounds to 5.

Q5
2021
VCAA
Paper 1
4 marks
Q5

Let f:RR,f(x)=x24f : R \rightarrow R, f(x) = x^2 - 4 and g:RR,g(x)=4(x1)24g : R \rightarrow R, g(x) = 4(x - 1)^2 - 4.

Q5a
2 marks

The graphs of ff and gg have a common horizontal axis intercept at (2,0)(2, 0).

Find the coordinates of the other horizontal axis intercept of the graph of gg.

Reveal Answer

4(x1)24=04(x-1)^2 - 4 = 0
(x1)2=1\therefore (x-1)^2 = 1
x1=±1x-1 = \pm 1
x=1+1,1+1x = -1+1, 1+1
x=0,2\therefore x = 0, 2
Therefore, the other xx-intercept is (0,0)(0,0)

Marking Criteria
DescriptorMarks

Sets g(x)=0g(x) = 0 or demonstrates a valid method to find the horizontal axis intercept (e.g., using symmetry around the turning point).

1

States the correct coordinates of the other horizontal axis intercept as (0,0)(0,0).

1
Q5b
2 marks

Let the graph of hh be a transformation of the graph of ff where the transformations have been applied in the following order:

  • dilation by a factor of 12\frac{1}{2} from the vertical axis (parallel to the horizontal axis)
  • translation by two units to the right (in the direction of the positive horizontal axis)

State the rule of hh and the coordinates of the horizontal axis intercepts of the graph of hh.

Reveal Answer

h(x)=f(2(x2))=4(x2)24h(x) = f(2(x-2)) = 4(x-2)^2 - 4 or 4x216x+124x^2 - 16x + 12
intercepts at (1,0),(3,0)(1,0), (3,0)

Marking Criteria
DescriptorMarks

States the correct rule for hh, e.g., h(x)=4(x2)24h(x) = 4(x-2)^2 - 4 or h(x)=4x216x+12h(x) = 4x^2 - 16x + 12.

1

States the correct coordinates of the horizontal axis intercepts as (1,0)(1,0) and (3,0)(3,0).

1
Q4
2022
VCAA
Paper 2
1 mark
Q4
1 mark

Which one of the following functions is not continuous over the interval x[0,5]x \in [0, 5]?

A

f(x)=1(x+3)2f(x) = \frac{1}{(x + 3)^2}

B

f(x)=x+3f(x) = \sqrt{x + 3}

C

f(x)=x13f(x) = x^{\frac{1}{3}}

D

f(x)=tan(x3)f(x) = \tan\left(\frac{x}{3}\right)

E

f(x)=sin2(x3)f(x) = \sin^2\left(\frac{x}{3}\right)

Reveal Answer
A

f(x)=1(x+3)2f(x) = \frac{1}{(x + 3)^2}

Incorrect. The function has a vertical asymptote at x=3x = -3, which is outside the interval [0,5][0, 5], meaning it is continuous on the given interval.

B

f(x)=x+3f(x) = \sqrt{x + 3}

Incorrect. The square root function is continuous for all values where its argument is non-negative (x3x \ge -3), which fully includes the interval [0,5][0, 5].

C

f(x)=x13f(x) = x^{\frac{1}{3}}

Incorrect. The cube root function is defined and continuous for all real numbers, so it is continuous on the interval [0,5][0, 5].

D

f(x)=tan(x3)f(x) = \tan\left(\frac{x}{3}\right)

Correct Answer

Correct. The tangent function has a vertical asymptote when its argument is π2\frac{\pi}{2}. Setting x3=π2\frac{x}{3} = \frac{\pi}{2} gives x=3π24.71x = \frac{3\pi}{2} \approx 4.71, which falls within the interval [0,5][0, 5] and creates a discontinuity.

E

f(x)=sin2(x3)f(x) = \sin^2\left(\frac{x}{3}\right)

Incorrect. The sine function is continuous for all real numbers, so its square is also continuous everywhere, including the interval [0,5][0, 5].

Q13
2020
VCAA
Paper 2
1 mark
Q13
1 mark

The transformation T:R2R2T : R^2 \rightarrow R^2 that maps the graph of y=cos(x)y = \cos(x) onto the graph of y=cos(2x+4)y = \cos(2x + 4) is

A

T([xy])=[12001]([xy]+[40])T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix} \left( \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -4 \\ 0 \end{bmatrix} \right)

B

T([xy])=[12001][xy]+[40]T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -4 \\ 0 \end{bmatrix}

C

T([xy])=[12001]([xy]+[20])T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix} \left( \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -2 \\ 0 \end{bmatrix} \right)

D

T([xy])=[2001]([xy]+[20])T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \left( \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 2 \\ 0 \end{bmatrix} \right)

E

T([xy])=[2001][xy]+[20]T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 2 \\ 0 \end{bmatrix}

Reveal Answer
A

T([xy])=[12001]([xy]+[40])T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix} \left( \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -4 \\ 0 \end{bmatrix} \right)

Correct Answer

To map y=cos(x)y = \cos(x) to y=cos(2x+4)y' = \cos(2x' + 4), we need y=yy' = y and x=2x+4x = 2x' + 4. Solving for xx' gives x=12(x4)x' = \frac{1}{2}(x - 4), which corresponds to a horizontal shift of 4-4 followed by a horizontal compression by 12\frac{1}{2}, exactly as this transformation matrix and vector addition describe.

B

T([xy])=[12001][xy]+[40]T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -4 \\ 0 \end{bmatrix}

This transformation results in x=12x4x' = \frac{1}{2}x - 4. Substituting x=2x+8x = 2x' + 8 into y=cos(x)y = \cos(x) would map the graph to y=cos(2x+8)y = \cos(2x + 8), not y=cos(2x+4)y = \cos(2x + 4).

C

T([xy])=[12001]([xy]+[20])T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix} \left( \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -2 \\ 0 \end{bmatrix} \right)

This transformation results in x=12(x2)=12x1x' = \frac{1}{2}(x - 2) = \frac{1}{2}x - 1. This would map the original graph to y=cos(2x+2)y = \cos(2x + 2), which is incorrect.

D

T([xy])=[2001]([xy]+[20])T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \left( \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 2 \\ 0 \end{bmatrix} \right)

This transformation results in x=2(x+2)=2x+4x' = 2(x + 2) = 2x + 4. This represents a horizontal stretch rather than a compression, mapping the graph to y=cos(12x2)y = \cos(\frac{1}{2}x - 2).

E

T([xy])=[2001][xy]+[20]T\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 2 \\ 0 \end{bmatrix}

This transformation results in x=2x+2x' = 2x + 2. This applies a horizontal stretch instead of a compression, mapping the graph to y=cos(12x1)y = \cos(\frac{1}{2}x - 1).

Q20
2025
VCAA
Paper 2
1 mark
Q20
1 mark

Let a>1a > 1, and consider the functions ff and gg defined below.

f:RR,  f(x)=axf : R \rightarrow R, \; f(x) = a^x g:RR,  g(x)=a2x+2g : R \rightarrow R, \; g(x) = a^{2x + 2}

Which one of the following sequences of transformations, when applied to f(x)f(x), does not produce g(x)g(x)?

A

dilation by a factor of 12\frac{1}{2} from the yy-axis, then translation by 11 unit in the negative direction of the xx-axis

B

dilation by a factor of 12\frac{1}{2} from the yy-axis, then dilation by a factor of a2a^2 from the xx-axis

C

dilation by a factor of aa from the xx-axis, then dilation by a factor of 12\frac{1}{2} from the yy-axis, then translation by 11 unit in the positive direction of the xx-axis

D

dilation by a factor of a3a^3 from the xx-axis, then translation by 11 unit in the positive direction of the xx-axis, then dilation by a factor of 12\frac{1}{2} from the yy-axis

Reveal Answer
A

dilation by a factor of 12\frac{1}{2} from the yy-axis, then translation by 11 unit in the negative direction of the xx-axis

Dilation by 1/21/2 from the yy-axis gives a2xa^{2x}, and translating left by 11 unit gives a2(x+1)=a2x+2a^{2(x+1)} = a^{2x+2}. This produces g(x)g(x), so it is not the correct answer.

B

dilation by a factor of 12\frac{1}{2} from the yy-axis, then dilation by a factor of a2a^2 from the xx-axis

Dilation by 1/21/2 from the yy-axis gives a2xa^{2x}, and dilating by a2a^2 from the xx-axis gives a2a2x=a2x+2a^2 \cdot a^{2x} = a^{2x+2}. This produces g(x)g(x), so it is not the correct answer.

C

dilation by a factor of aa from the xx-axis, then dilation by a factor of 12\frac{1}{2} from the yy-axis, then translation by 11 unit in the positive direction of the xx-axis

Correct Answer

Dilating by aa from the xx-axis gives ax+1a^{x+1}, dilating by 1/21/2 from the yy-axis gives a2x+1a^{2x+1}, and translating right by 11 unit gives a2(x1)+1=a2x1a^{2(x-1)+1} = a^{2x-1}. This does not equal g(x)g(x), making it the correct choice.

D

dilation by a factor of a3a^3 from the xx-axis, then translation by 11 unit in the positive direction of the xx-axis, then dilation by a factor of 12\frac{1}{2} from the yy-axis

Dilating by a3a^3 from the xx-axis gives ax+3a^{x+3}, translating right by 11 unit gives a(x1)+3=ax+2a^{(x-1)+3} = a^{x+2}, and dilating by 1/21/2 from the yy-axis gives a2x+2a^{2x+2}. This produces g(x)g(x), so it is not the correct answer.

Q2
2022
QCAA
Paper 2
1 mark
Q2
1 mark

Identify the correct features of the function f(x)=xexf(x) = xe^x

A

f(1)=0,f(1)<0f'(-1)=0, f''(-1)<0

B

f(1)=0,f(1)>0f'(-1)=0, f''(-1)>0

C

f(1)<0,f(1)<0f'(-1)<0, f''(-1)<0

D

f(1)<0,f(1)>0f'(-1)<0, f''(-1)>0

Reveal Answer
A

f(1)=0,f(1)<0f'(-1)=0, f''(-1)<0

This option is incorrect because while f(1)=0f'(-1)=0, the second derivative f(1)=e1(1+2)=e1f''(-1) = e^{-1}(-1+2) = e^{-1} is positive, not negative.

B

f(1)=0,f(1)>0f'(-1)=0, f''(-1)>0

Correct Answer

This is correct. Using the product rule, f(x)=ex(1+x)f'(x) = e^x(1+x) and f(x)=ex(x+2)f''(x) = e^x(x+2). Evaluating at x=1x=-1 gives f(1)=0f'(-1)=0 and f(1)=e1>0f''(-1) = e^{-1} > 0.

C

f(1)<0,f(1)<0f'(-1)<0, f''(-1)<0

This option is incorrect because the first derivative evaluates to zero at x=1x=-1, not a negative value, and the second derivative is positive.

D

f(1)<0,f(1)>0f'(-1)<0, f''(-1)>0

This option is incorrect because the first derivative f(1)f'(-1) equals 00, not a value less than 00.

Q9
2024
QCAA
Paper 1
1 mark
Q9
1 mark

At a certain location, the temperature (°C) can be modelled by the function T=5sin(π12x)+23T = 5\sin\left(\frac{\pi}{12}x\right) + 23, where xx is the number of hours after sunrise.

Determine the rate of change of temperature (°C/hour) when x=4x = 4

A

5π48\frac{5\pi}{48}

B

5π24\frac{5\pi}{24}

C

5π324\frac{5\pi\sqrt{3}}{24}

D

5π36\frac{5\pi\sqrt{3}}{6}

Reveal Answer
A

5π48\frac{5\pi}{48}

This incorrect value is half of the correct answer, which may result from a calculation error during the multiplication of fractions or evaluating the trigonometric ratio.

B

5π24\frac{5\pi}{24}

Correct Answer

The rate of change is the derivative T(x)=5π12cos(π12x)T'(x) = 5 \cdot \frac{\pi}{12}\cos\left(\frac{\pi}{12}x\right). Evaluating at x=4x=4 gives 5π12cos(π3)=5π12(12)=5π24\frac{5\pi}{12}\cos\left(\frac{\pi}{3}\right) = \frac{5\pi}{12}\left(\frac{1}{2}\right) = \frac{5\pi}{24}.

C

5π324\frac{5\pi\sqrt{3}}{24}

This answer results from evaluating sin(π3)=32\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} instead of cos(π3)\cos\left(\frac{\pi}{3}\right) in the derivative, or incorrectly assuming the derivative of sine is sine.

D

5π36\frac{5\pi\sqrt{3}}{6}

This option is incorrect and likely results from misapplying the chain rule or arithmetic errors when combining the constants.

Q14
2023
VCAA
Paper 2
1 mark
Q14
1 mark

A polynomial has the equation y=x(3x1)(x+3)(x+1)y = x(3x - 1)(x + 3)(x + 1).
The number of tangents to this curve that pass through the positive xx-intercept is

A

00

B

11

C

22

D

33

E

44

Reveal Answer
A

00

The tangent at the positive xx-intercept (1/3,0)(1/3, 0) itself is one such tangent, so there is at least one valid tangent line.

B

11

While the tangent at the xx-intercept is one solution, there are other points on the curve whose tangent lines also pass through this intercept.

C

22

Solving for the xx-coordinates of the points of tangency results in an equation with three distinct real roots, not two.

D

33

Correct Answer

The positive xx-intercept is (1/3,0)(1/3, 0). Equating the derivative f(a)f'(a) to the slope between the tangency point (a,f(a))(a, f(a)) and (1/3,0)(1/3, 0) yields 3(a1/3)2(3a2+8a+3)=03(a - 1/3)^2(3a^2 + 8a + 3) = 0, which has exactly 3 distinct real roots.

E

44

Although the original polynomial is of degree 4, the condition for the tangent passing through (1/3,0)(1/3, 0) simplifies to a cubic equation, yielding a maximum of 3 tangents.

Q5
2025
QCAA
Paper 2
1 mark
Q5
1 mark

Which statement best describes a feature of the graph of the exponential function y=exy = e^x, xRx \in R?

A

limx(ex)=e\lim_{x \to \infty} (e^x) = e

B

When x=0x = 0, y=ey = e

C

The graph has an asymptote with the equation x=0x = 0

D

The gradient of the graph has the same value as the function at all points on the graph.

Reveal Answer
A

limx(ex)=e\lim_{x \to \infty} (e^x) = e

Incorrect. As xx approaches infinity, the value of exe^x grows without bound, meaning the limit is infinity, not ee.

B

When x=0x = 0, y=ey = e

Incorrect. Any non-zero base raised to the power of 0 equals 1, so when x=0x = 0, y=e0=1y = e^0 = 1, rather than ee.

C

The graph has an asymptote with the equation x=0x = 0

Incorrect. The function y=exy = e^x has a horizontal asymptote at y=0y = 0 as xx approaches negative infinity, not a vertical asymptote at x=0x = 0.

D

The gradient of the graph has the same value as the function at all points on the graph.

Correct Answer

Correct. A unique property of the natural exponential function y=exy = e^x is that its derivative is also exe^x, meaning the gradient at any point equals the function's value.

Q13
2024
VCAA
Paper 2
1 mark
Q13
1 mark

The function f:(0,)R,f(x)=x2+2xf : (0, \infty) \rightarrow R, f(x) = \frac{x}{2} + \frac{2}{x} is mapped to the function gg with the following sequence of transformations:

  1. dilation by a factor of 3 from the yy-axis
  2. translation by 1 unit in the negative direction of the yy-axis.

The function gg has a local minimum at the point with the coordinates

A

(6,1)(6, 1)

B

(23,1)\left(\frac{2}{3}, 1\right)

C

(2,5)(2, 5)

D

(2,13)\left(2, -\frac{1}{3}\right)

Reveal Answer
A

(6,1)(6, 1)

Correct Answer

The transformed function is g(x)=f(x3)1=x6+6x1g(x) = f\left(\frac{x}{3}\right) - 1 = \frac{x}{6} + \frac{6}{x} - 1. Setting the derivative g(x)=166x2g'(x) = \frac{1}{6} - \frac{6}{x^2} to zero gives x=6x = 6, and evaluating g(6)g(6) yields a minimum value of 11.

B

(23,1)\left(\frac{2}{3}, 1\right)

This is the local minimum if the function was incorrectly dilated by a factor of 1/31/3 from the yy-axis, which would use f(3x)f(3x) instead of f(x3)f\left(\frac{x}{3}\right).

C

(2,5)(2, 5)

This is the local minimum if the function was dilated by a factor of 3 from the xx-axis (evaluating 3f(x)13f(x) - 1) rather than from the yy-axis.

D

(2,13)\left(2, -\frac{1}{3}\right)

This is the local minimum if the function was dilated by a factor of 1/31/3 from the xx-axis (evaluating 13f(x)1\frac{1}{3}f(x) - 1) rather than from the yy-axis.

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