smc-966-40-Log graphs

Functions, 2ADV 2024 MET2 1 MC

The asymptote(s) of the graph of \(y=\log _e(x+1)-3\) are

\(x=-1\) only
\(x=1\) only
\(y=-3\) only
\(x=-1\) and \(y=-3\)

Show Answers Only

\(A\)

Show Worked Solution

\(\text{Asymptotes occur when}\ \ x+1=0\)

\(\therefore\ \text{Only one asymptote at}\ \ x=-1\)

\(\Rightarrow A\)

Functions, 2ADV 2021 HSC 3 MC

Which of the following represents the domain of the function `f(x)=ln(1-x)`?

`[1,oo)`
`(1, oo)`
`(–oo, 1]`
`(–oo, 1)`

Show Answers Only

`D`

Show Worked Solution

Mean mark 52%!

`1-x`	`>0`
`x`	`<1`

`x ∈ (–oo, 1)`

`=> D`

Calculus, 2ADV C3 2020 HSC 29

The diagram shows the graph of `y = c ln x, \ c > 0`.

Show that the equation of the tangent to `y = c ln x` at `x = p`, where `p > 0`, is

`\ \ \ \ \ y = c/p x - c + c ln p`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Find the value of `c` such that the tangent from part (a) has a gradient of 1 and passes through the origin. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(See Worked Solutions)`
`c = e`

Show Worked Solution

a. `y = c ln x`

`(dy)/(dx) = c/x`

`text(At)\ x = p,`

`m_text(tang) = c/p`

`text(T)text(angent passes through)\ (p, c ln p)`

`:.\ text(Equation of tangent)`

`y – c ln p`	`= c/p (x – p)`
`y`	`= c/p x – c + c ln p`

b. `text(If)\ m_text(tang) = 1,`

♦ Mean mark part (b) 40%.

`c/p`	`= 1`
`c`	`= p`

`text(If tangent passes through)\ (0, 0)`

`0`	`= −c + c ln c`
`0`	`= c(ln c – 1)`

`ln c = 1\ \ (c > 0)`

`:. c = e`

Calculus, 2ADV C3 2019 MET1 4

Given the function `f(x) = log_e (x-3) + 2`,

State the domain and range of `f(x)`. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
i. Find the equation of the tangent to the graph of `f(x)` at `(4, 2)`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

ii. On the axes below, sketch the graph of the function `f(x)`, labelling any asymptote with its equation.

Also draw the tangent to the graph of `f(x)` at `(4, 2)`. (4 marks)

--- 0 WORK AREA LINES (style=lined) ---

Show Answers Only

`x >3`

`y in R`
i. `y = x-2`

ii. `text(See Worked Solutions)`

Show Worked Solution

a.	`text(Domain)`	`: \ x > 3`
	`text(Range)`	`: \ y in R`

b.i.	`g(x)`	`= log_e (x-3) + 2`
	`g^{\prime}(x)`	`= 1/(x-3)`
	`g^{\prime}(4)`	`= 1`

`text(Equation of tangent),\ m = 1\ \ text(through)\ (4, 2):`

`y-2`	`= 1(x-4)`
`y`	`= x-2`

b.ii.

Functions, 2ADV 2019 HSC 1 MC

What is the domain of the function `f(x) = ln(4-x)`?

`x < 4`
`x <= 4`
`x > 4`
`x >= 4`

Show Answers Only

`A`

Show Worked Solution

`4-x`	`> 0`
`-x`	`> -4`
`x`	`< 4`

`=> A`

Functions, 2ADV F2 SM-Bank 1

Draw the graph `y = ln x`. (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
Explain how the above graph can be transformed to produce the graph

`y = 3ln(x + 2)`

and sketch the graph, clearly identifying all intercepts. (3 marks)

--- 9 WORK AREA LINES (style=lined) ---

Show Answers Only

Show Worked Solution

ii. `text(Transforming)\ \ y = ln x => \ y = ln(x + 2)`

`y = ln x\ \ =>\ text(shift 2 units to left.)`

`text(Transforming)\ \ y = ln(x + 2)\ \ text(to)\ \ y = 3ln(x + 2)`

`=>\ text(increase each)\ y\ text(value by a factor of 3)`