Test category

Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`

Part of the graph of `y=f(x)` is shown below.

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

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1. Find `f^{\prime}(0)`. (2 marks)
  
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2. State the maximal domain over which `f` is strictly increasing. (1 mark)
  
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Show that `f(x)+f(-x)=0`. (1 mark)

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Find the domain and the rule of `f^{-1}`, the inverse of `f`. (3 marks)

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Let `h` be the function `h:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, h(x)=\frac{1}{k}\left(\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)\right)`, where `k \in R` and `k>0`.

The inverse function of `h` is defined by `h^{-1}: R \rightarrow R, h^{-1}(x)=\frac{e^{k x}-1}{2\left(e^{k x}+1\right)}`.
The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
The graph below shows the relevant area shaded.

You are not required to find or define `A(k)`.

Determine the range of values of `k` such that `A(k)>0`. (1 mark)

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Explain why the domain of `A(k)` does not include all values of `k`. (1 mark)

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Show Answers Only

a.	`R`
b.i	`f^{\prime}(0)=4`
b.ii	`\left(-\frac{1}{2}, \frac{1}{2}\right)`
c.	`0`
d.	`x \in \mathbb{R}`
e.i	` k > 4`
e.ii	No bounded area for `0<k \leq 4`

Show Worked Solution

a. `R` is the range.

b.i	`f(x)`	`= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)`
	`f^{\prime}(x)`	`= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}`
		`= \frac{2}{2 x+1}-\frac{2}{2 x-1}`
	`f^{\prime}(0)`	`= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}`
		`= 4`

b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`

c. `f(x)+f(-x)`	`= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)`
	`= 0`

d. To find the inverse swap `x` and `y` in `y=f(x)`

`x`	`= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)`
`x`	`= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)`
`e^x`	`=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}`
`y+\frac{1}{2}`	`= e^x\left(-y+\frac{1}{2}\right)`
`y+\frac{1}{2}`	`= -e^x y+\frac{e^x}{2}`
`y\left(e^x+1\right)`	`= \frac{e^x-1}{2}`
`:.\ f^(-1)(x)`	`= \frac{e^x-1}{2(e^x + 1)}`

`:.` Domain: `x \in \mathbb{R}`

e.i The vertical dilation factor of `f(x)` is `1/k`

For `A(k)>=0` , `h^{\prime}(0)<1`

`\frac{1}{k}(4)<1` [Using CAS]

`:.\ k > 4`

♦♦♦♦ Mean mark (e.i) 10%.
MARKER’S COMMENT: Incorrect responses included `k>0` and `4<k<33`.

e.ii When `h \geq h^{-1}` for `x>0` (or `h \leq h^{-1}` for `x<0`) there is no bounded area.

`:.` There will be no bounded area for `0<k \leq 4`.

♦♦♦♦ Mean mark (e.ii) 10%.