How many SCSA Mathematics Methods questions cover The logarithmic function?

AusGrader has 129 SCSA Mathematics Methods questions on The logarithmic function, all with instant AI grading and detailed marking feedback.

SCSA Mathematics Methods The logarithmic function

5 sample questions with marking guides and sample answers · Avg. score: 61.1%

2024

QCAA

Paper 1

1 mark

Differentiate $y = \ln(x) \cos(x)$ with respect to $x$ .

$\frac{\cos(x)}{x}$

$-\frac{\sin(x)}{x}$

$\frac{\cos(x)}{x} + \ln(x) \sin(x)$

$\frac{\cos(x)}{x} - \ln(x) \sin(x)$

2024

QCAA

Paper 1

1 mark

Simplify $y = 2\ln(e^x)$

$y = 2x$

$y = 2^x$

$y = \frac{2}{x}$

$y = x^2$

2020

SCSA

Paper 1

7 marks

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

Q6a

2 marks

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

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

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

Marking Criteria

Descriptor	Marks
expresses $g(x)$ as a sum of logs	1
recognises a vertical translation by $\ln(4)$ units	1

Q6b

2 marks

The function $h(x) = cf(x)$ is a vertical dilation of $f$ by a scale factor of $c$ .

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

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

$f$ is scaled vertically by a factor of 0.5.

Marking Criteria

Descriptor	Marks
expresses $h$ as a product involving $\ln(x)$	1
recognises a vertical scaling by a scale factor of 0.5	1

Q6c

3 marks

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

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

\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*}

$f$ is scaled horizontally by a scale factor of $e^{-4}$ .

Marking Criteria

Descriptor	Marks
expresses 4 as $4\ln(e)$	1
expresses $p$ using a single logarithm	1
states horizontal scale factor	1

Q16

2021

SCSA

Paper 2

12 marks

Q16

An analyst was hired by a large company at the beginning of 2021 to develop a model to predict profit. At that time, the company’s profit was ＄4 million. The model developed by the analyst was:

$P(x)=\dfrac{20\ln(x+a)}{x+5}$ ,

where $P(x)$ is the profit in millions of dollars after $x$ weeks and $a$ is a constant.

Q16a

2 marks

Show that $a=e$ .

Descriptor	Marks
recognises $P(0) = 4$	1
obtains $\ln(a) = 1$	1

Q16b

1 mark

What does the model predict the profit will be after five weeks?

$P(5) = \frac{20 \ln(5 + e)}{5 + 5}$
$= 4.087 \text{ (3 d.p.)}$

Profit will be approximately ＄4 087 000

Marking Criteria

Descriptor	Marks
states the profit	1

Q16c

3 marks

Showing use of the quotient rule, determine an equation that, when solved, will give the time when the model predicts the profit will be maximised.

$P'(x) = \frac{(x + 5) \frac{20}{(x + e)} - 20 \ln(x + e)}{(x + 5)^2}$

For maximum profit we require :
$\frac{(x + 5) \frac{20}{(x + e)} - 20 \ln(x + e)}{(x + 5)^2} = 0$

Marking Criteria

Descriptor	Marks
demonstrates use of the quotient rule	1
writes correct expression for $P'(x)$	1
equates $P'(x)$ to zero	1

Q16d

2 marks

What is this maximum profit and during which week will it occur?

Maximum profit is approximately ＄4 436 000
This occurs when $x \approx 1.79$ , so during the second week.

Marking Criteria

Descriptor	Marks
states maximum profit	1
states it occurs in the second week	1

Q16e

1 mark

According to the model, during which week will the company’s profit fall below its value at the beginning of 2021?

$4 = \frac{20 \ln(x + e)}{x + 5}$
$x = 0 \text{ or } 5.581$

The model predicts during the $6^{\text{th}}$ week

Marking Criteria

Descriptor	Marks
determines when the profit falls below the 2021 value	1

Q16f

3 marks

The model proved accurate and after 10 weeks the company implemented some changes. From this time the analyst used a new model to predict the profit:

$N(y)=2e^{b(10+y)}$ ,

where $N(y)$ is the profit in millions of dollars $y$ weeks from this point in time and $b$ is a constant.

The company is projecting its profit to exceed ＄5 million. During which week does the new model suggest this will happen?

$P(10) = 3.39072 = N(0)$
$3.39072 = 2e^{10b}$
$b = 0.05279$
$5 = 2e^{b(10+y)}$
$y \approx 7.36$

Profit should exceed ＄5 million during the $8^{\text{th}}$ week after the changes.

Marking Criteria

Descriptor	Marks
determines $P(10)$	1
determines the value of the constant $b$	1
determines the week when the profit exceeds ＄5 million	1

2021

SCSA

Paper 1

9 marks

Q1a

3 marks

Differentiate $\frac{3x + 1}{x^3}$ and simplify your answer.

\begin{aligned} \frac{d}{dx}\left(\frac{3x+1}{x^3}\right) &= \frac{x^3(3) - 3x^2(3x+1)}{x^6} \\ &= \frac{3x^3 - 9x^3 - 3x^2}{x^6} \\ &= \frac{-6x - 3}{x^4} \end{aligned}

Marking Criteria

Descriptor	Marks
recognises the need for the quotient rule	1
correctly differentiate the expression	1
simplifies the result	1

Q1b

3 marks

Let $f'(x) = x\ln(2x)$ . Determine a simplified expression for the rate of change of $f'(x)$ .

\begin{aligned} f''(x) &= x \times \frac{2}{2x} + 1 \times \ln(2x) \\ &= 1 + \ln(2x) \end{aligned}

Marking Criteria

Descriptor	Marks
identifies the rate of change as $f''(x)$	1
correctly determines $f''(x)$	1
simplifies the expression for $f''(x)$	1

Q1c

3 marks

Given that $g'(x) = 4e^{2x}$ and $g(1) = 0$ , determine $g(5)$ .

\begin{aligned} g(x) &= 2e^{2x} + c \\ \text{Since } g(1) &= 0, \\ 0 &= 2e^2 + c \\ \therefore c &= -2e^2 \\ \therefore g(x) &= 2e^{2x} - 2e^2 \\ \therefore g(5) &= 2e^{10} - 2e^2 \end{aligned}

Marking Criteria

Descriptor	Marks
states an expression for $g(x)$ , including the constant of integration	1
correctly determines the constant	1
correctly determines $g(5)$ as the final solution	1

SCSA Mathematics Methods The logarithmic function

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