smc-1072-30-y = | f(x) |; y = f( |x| )

Functions, EXT1 F1 2024 HSC 6 MC

How many real value(s) of \(x\) satisfy the equation

\(\abs{b} = \abs{b\,\sin(4x)}\),

where \(x \in [0, 2\pi]\) and \(b\) is not zero?

\(1\)
\(2\)
\(4\)
\(8\)

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\(D\)

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\(\abs{b} = \abs{b\,\sin(4x)}\ \ \Rightarrow\ \ \abs{1} = \abs{\sin(4x)}\ \ (b \neq 0)\)

\(\sin(4x) = \pm 1\)

\(x \in [0, 2\pi]\ \ \Rightarrow\ \ 4x \in [0, 8\pi]\)

\(\therefore 8\ \text{real solutions}\)

\(\Rightarrow D\)

Functions, EXT1 F1 EQ-Bank 5

Given \(f(x)=x(x+2)(2-x)\)

On the graph, sketch the graphs of \(y=f(\abs{x})\) (2 marks)

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Functions, EXT1 F1 EQ-Bank 6

The diagram shows the graph of \(f(x)=x(x+2)(2-x)\)

On the graph, sketch the graph of \(y=\abs{f(x)}\) (2 marks)

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Functions, EXT1 F1 EQ-Bank 3

Using inequalities, define the shaded region in the diagram below. (2 marks)

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\(y \geqslant \abs{x+1}\ \cap \ y \leqslant 2 \)

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\(\text{Shaded area is defined by the inequalities:}\)

\(y \geqslant \abs{x+1}\ \cap \ y \leqslant 2 \)

Functions, EXT1 F1 EQ-Bank 2

Sketch the graph of \(y=\abs{x^2-5x+4}\). (2 marks)

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\(y=\abs{x^2-5 x+4}=\abs{(x-1)(x-4)}\)

\(x\text{-intercepts at} \ (1,0), (4,0)\)

\(y=(x-1)(x-4) \ \text{has low at} \ \ x=\dfrac{5}{2} \ \text{(by symmetry)}\)

\(\Rightarrow \ \text {Low at} \ \left(\dfrac{5}{2},\dfrac{-9}{4}\right)\)

Functions, EXT1 F1 EQ-Bank 1

Sketch the graph of \(y=\abs{x^2-1}\). (2 marks)

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\(y=\left|x^2-1\right|\)

\( x\text{-intercepts at} \ (-1,0), (1,0)\)

\(y=x^2-1 \ \ \text{has a low at} \ (0,-1)\)

Functions, EXT1 F1 2023 HSC 8 MC

The diagram shows the graph of a function.

Which of the following is the equation of the function?

\(y=\Big{|}1-\big{|}|x|-2\big{|}\Big{|}\)
\(y=\Big{|}2-\big{|}|x|-1\big{|}\Big{|}\)
\(y=\Big{|}1-\big{|}x-2\big{|}\Big{|}\)
\(y=\Big{|}2-\big{|}x-1\big{|}\Big{|}\)

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\(A\)

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\(\text{By elimination:}\)

\(\text{Even function}\ \rightarrow\ \text{Eliminate}\ C\ \text{and}\ D \)

\(\text{Graph passes through}\ (1, 0) \)

\(\text{Option}\ A:\ \ y=\Big{|}1-\big{|}1-2\big{|}\Big{|} =\Big{|}1-1\Big{|}=0\ \ \text{(lies on graph)} \)

\(\text{Option}\ B:\ \ y=\Big{|}2-\big{|}|1|-1\big{|}\Big{|} =\Big{|}2-0\Big{|}=2\ \ \text{(not on graph)} \)

\(\therefore\ \text{Eliminate}\ B \)

\(\Rightarrow A\)

Functions, EXT1 F1 2022 HSC 2 MC

The graph of `f(x)=(3)/(x-1)+2` is shown.

The graph of `f(x)` was transformed to get the graph of `g(x)` as shown.

What transformation was applied?

`g(x)=f(|x|)`
`g(x)=sqrt(f(x))`
`g(x)=f(-x)`
`g(x)=(1)/(f(x))`

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`A`

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`y=g(x)\ text{is an even function}`

`=> A`

Mean mark 56%.

Functions, EXT1′ F1 2008 HSC 3a

The following diagram shows the graph of `y = g(x)`.

Draw separate one-third page sketches of the graphs of the following:

`y = |g(x)|` (1 mark)

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`y = 1/(g(x))` (2 marks)

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Show Worked Solution

ii.

Functions, EXT1′ F1 2019 HSC 12d

Consider the function `f(x) = x^3 - 1`.

Sketch the graph `y = |\ f(x)\ |`. (1 mark)

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Sketch the graph `y = 1/(f(x))`. (2 marks)

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Without using calculus, sketch the graph `y = x/(f(x))`. (2 marks)

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`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`

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i. `y = |\ x^3 – 1\ |`

ii. `y = 1/(x^3 – 1)`

iii. `y = x/(x^3 – 1)`

Functions, EXT1 F1 SM-Bank 12

Given `f(x) = x^3 - x^2 - 2x`, without calculus sketch a separate half page graph of the following functions, showing all asymptotes and intercepts.

`y = |\ f(x)\ |` (1 mark)

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`y = f(|x|)` (2 marks)

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`y = 1/(f(x))` (2 marks)

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i.	`f(x)`	`= x^3 – x^2 – 2x`
		`= x(x^2 – x – 2)`
		`= x(x – 2)(x + 1)`

ii.

`y = f(|x|)\ text(is a reflection of)\ y = f(x)\ text(for)\ x > 0`

`text(is the)\ ytext(-axis.)`

iii.

Functions, EXT1′ F1 2018 HSC 12d

The diagram shows the graph of the function `f(x) = x/(x - 1)`.

Draw a separate half-page graph for each of the following functions, showing all asymptotes and intercepts.

`y = |\ f(x)\ |` (1 mark)

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`y = (f(x))^2` (2 marks)

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ii.

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ii.

Functions, EXT1 F1 2017 HSC 12b

Carefully sketch the graphs of `y = |\ x + 1\ |` and `y = 3 - |\ x - 2\ |` on the same axes, showing all intercepts. (3 marks)

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Using the graphs from part (i), or otherwise, find the range of values of `x` for which

`qquad qquad |\ x + 1\ | + |\ x - 2\ | = 3`. (1 mark)

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`−1 <= x <= 2`

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ii.	`\|\ x + 1\ \| + \|\ x – 2\ \|`	`= 3`
	`\|\ x + 1\ \|`	`= 3 – \|\ x – 2\ \|`

`:.\ text(Solution is intersection of the graphs)`

`:. −1 <= x <= 2`

Functions, EXT1′ F1 2016 HSC 1 MC

Which equation best represents the following graph?

`y = sqrt x`
`|\ y\ | = sqrt x`
`y = sqrt (|\ x\ |)`
`|\ y\ |\ = sqrt (|\ x\ |)`

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`=> C`

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`text(Graph is an even function.)`

`=> C`

Functions, EXT1′ F1 2007 HSC 3a

The diagram shows the graph of `y = f(x)`. The line `y = x` is an asymptote.

Draw separate one-third page sketches of the graphs of the following:

`f(-x).` (1 mark)

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`f(|\ x\ |).` (2 marks)

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`f(x) - x.` (2 marks)

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MARKER’S COMMENT: In part (ii), a significant number of students graphed `y=|f(x)|`.

ii.

iii.

Functions, EXT1′ F1 2012 HSC 11f

Sketch the following graphs, showing the `x`- and `y`-intercepts

`y = |\ x\ |- 1` (1 mark)
`y = x(|\ x\ | - 1)` (2 marks)

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`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

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i. & ii.

`text{Note for part (ii)}`

`=>y = x(|\ x\ | – 1)\ \ text(is an ODD function)`

Functions, EXT1′ F1 2014 HSC 12a

The diagram shows the graph of a function `f(x)`.

Draw a separate half-page graph for each of the following, showing all asymptotes and intercepts.

`y = f(|\ x\ |)` (2 marks)

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`y = 1/(f(x))` (2 marks)

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`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

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ii.

Functions, EXT1 F1 2008 HSC 3a

Sketch the graph of `y = |\ 2x - 1\ |`. (1 mark)

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Hence, or otherwise, solve `|\ 2x - 1\ | <= |\ x - 3\ |`. (3 marks)

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`-2 <= x <= 4/3`

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ii. `text(Solving for)\ \ |\ 2x – 1\ | <= |\ x – 3\ |`

`text(Graph shows the statement is TRUE)`

`text(between the points of intersection.)`

`=>\ text(Intersection occurs when)`

`(2x – 1)`	`= (x – 3)\ \ \ text(or)\ \ \ `	`-(2x – 1)`	`= x – 3`
`x`	`= -2`	`-2x + 1`	`= x – 3`
		`-3x`	`= -4`
		`x`	`= 4/3`

`:.\ text(Solution is)\ \ {x: -2 <= x <= 4/3}`

ii.	`\|\ x + 1\ \| + \|\ x – 2\ \|`	`= 3`
	`\|\ x + 1\ \|`	`= 3 – \|\ x – 2\ \|`