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Functions, EXT1 F1 2024 HSC 12e

The diagram shows the graph of  \(y=\dfrac{1}{\abs{x-5}}\).
 

For what values of \(x\) is  \(\dfrac{x}{6} \geq\dfrac{1}{\abs{x-5}}\) ?   (3 marks)

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\(x \in[2,3] \cup[6, \infty)\)

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\(\dfrac{x}{6} \geqslant \dfrac{1}{|x-5|}\)

\(x|x-5| \geqslant 6\)

\(\text{Case  1:}\)

\(x(x-5) \geqslant 6\)

\(x^2-5 x-6 \geqslant 0\)

\((x-6)(x+1) \geqslant 0\)

\(x \leqslant-1\ \ \text{or}\ \ x \geqslant 6\)

\(\text {By inspection of graph} \ \Rightarrow \ x \leqslant -1\ \text{is not a solution}\)

\(\Rightarrow x \geqslant 6\)
 

\(\text {Case 2: }\)

\(-x(x-5) \geqslant 6\)

\(-x^2+5 x-6 \geqslant 0\)

\(x^2-5 x+6 \leqslant 0\)

\((x-3)(x-2) \leqslant 0\)

\(\Rightarrow 2 \leqslant x \leqslant 3\)

\(\therefore x \in[2,3] \cup[6, \infty)\)

Functions, EXT1 F1 2020 SPEC1 4

Solve the inequality  `3 - x > 1/|x - 4|`  for `x`, expressing your answer in interval notation.  (3 marks)

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`x ∈ (– oo, (7 – sqrt 5)/2)`

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`3 – x > 1/|x – 4|`

`|x – 4| (3 – x) > 1`
 

`text(If)\ \ x – 4 > 0, x > 4`

`(x – 4) (3 – x)` `> 1`
`3x – x^2 – 12 + 4x` `> 1`
`-x^2 + 7x – 13` `> 0`

 
`Delta = 7^2 – 4 ⋅ 1 ⋅ 13 = -3 < 0`

`=>\ text(No Solutions)`
 

`text(If)\ \ x – 4 < 0, x < 4`

`-(x – 4) (3 – x)` `> 1`
`x^2 – 7x + 12` `> 1`
`x^2 – 7x + 11` `> 0`
`x` `= (7 +- sqrt(7^2 – 4 ⋅ 1 ⋅ 11))/2`
  `= (7 +- sqrt 5)/2`

`text(Combining solutions)`

`(x < (7 – sqrt 5)/2  ∪ x > (7 + sqrt 5)/2)  nn x < 4`

`x ∈ (– oo, (7 – sqrt 5)/2)`

Functions, EXT1 F1 SM-Bank 2

Solve  `3/(|\ x - 3\ |) < 3`.  (3 marks)

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`x < 2\ ∪\ x > 4`

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`text(Solution 1)`

`3/(|\ x – 3\ |) < 3`

`|\ x – 3\ |` `> 1`
`(x^2 – 6x + 9)` `> 1^2`
`x^2 – 6x + 8` `> 0`
`(x – 4)(x – 2)` `> 0`

 

 

`:. {x: \ x < 2\ ∪\ x > 4}`

 

`text(Solution 2)`

`|\ x – 3\ | > 1`

`text(If)\ \ (x – 3)` `> 0,\ text(i.e.)\ x >3`
`x – 3` `> 1`
`x` `> 4`

 
`=> x > 4\ (text(satisfies both))`
 

`text(If)\ \ (x – 3)` `< 0,\ text(i.e.)\ x <3`
`−(x – 3)` `> 1`
`−x + 3` `> 1`
`x` `< 2`

 
`=> x < 2\ (text(satisfies both))`

`:. {x: \ x < 2\ ∪\ x > 4}`

Functions, EXT1 F1 2004 HSC 1a

Indicate the region on the number plane satisfied by  `y ≥ |\ x + 1\ |.`  (2 marks) 

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`text(See Worked Solution)`

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 Real Functions, EXT1 2004 HSC 1a Answer

`y ≥ |\ x + 1\ |`

`text(Test)\ (0, 0)`

`0 ≥ |\ 0 + 1\ |`

`0 ≥ 1\ \ \ \ \ \ ⇒\ text(False)`

 

`:.\ text(Shaded area represents)`

`y ≥ |\ x + 1\ |`

Functions, EXT1* F1 2005 HSC 1e

Find the values of  `x`  for which  `|\ x − 3\ | ≤ 1`.  (2 marks)

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`2 ≤ x ≤ 4`

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`|\ x − 3\ | ≤ 1`

`text(Solution 1)`

`(x − 3)^2 ≤ 1`

`x^2 − 6x + 9 ≤ 1`

`x^2 − 6x +8 ≤ 0`

`(x − 4)(x − 2) ≤ 0`
 

Algebra, 2UA 2005 HSC 1e Answer  

 
`:. 2 ≤ x ≤ 4`

 

`text(Alternative Solution)`

`(x − 3)` `≤1` `–(x − 3)` ` ≤ 1`
`x` `≤4` `–x +3` `≤ 1`
    `–x` `≤−2`
    `x` `≥ 2`

`:. 2 ≤ x ≤ 4`

Functions, EXT1* F1 2004 HSC 1f

Find the values of  `x`  for which  `|\ x + 1\ |<= 5`.  (2 marks)

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`-6 <= x <= 4`

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`text(Solution 1)`

`|\ x + 1\ |<= 5`

`-5` `≤x+1≤5`
`:.-6` `≤x≤4` 

 

`text(Solution 2)`

`|\ x + 1\ |<= 5`

`(x+1)^2` `<= 5^2`
` x^2 + 2x + 1` `<= 25`
 `x^2 + 2x – 24` `<= 0`
`(x + 6)(x – 4)` `<= 0`

Algebra, 2UA 2004 HSC 1f Answer

`:.\ -6 <= x <= 4`

Functions, EXT1 F1 2008 HSC 3a

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

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

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  1.  
    Real Functions, EXT1 2008 HSC 3a Answer

  2. `-2 <= x <= 4/3`
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i.    Real Functions, EXT1 2008 HSC 3a Answer

 

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}`

Functions, EXT1 F1 2013 HSC 10 MC

Which inequality has the same solution as  `|\ x + 2\ | + |\ x- 3\ | = 5`?

  1. `5/(3 - x) >= 1`
  2. `1/(x - 3)\ - 1/(x + 2) <= 0`
  3. `x^2 - x - 6 <= 0`
  4. `|\ 2x - 1\ | >= 5`
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`C`

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♦♦ Mean mark 39%
COMMENT: Note that the quick elimination of `A, B` and `D` is sufficient to get to the correct answer without proving `C` (although we have done this in the solution).

`text(In)\ A\ text(and)\ B, \ x ≠ 3\ text(but when)\ x=3,`

`|\ 3 + 2\ | + |\ 3 – 3\ | = 5\ \ text(is correct.)`

`:.\ text(Not)\ \ A\ \ text(or)\ \ B.`

 

`text(Consider)\ D`

`x -> oo\ text(satisfies)\ |\ 2x – 1\ | >= 5,\ \ text(but)`

`text(obviously not)\ |\ x + 2\ | + |\ x – 3\ | = 5.`

 
`text(Consider)\ C`

`x^2 – x – 6` `<= 0`
`(x – 3)(x + 2)` `<= 0`

 

`text(True when)\ \ \ -2 <= x <= 3.`

 
`text(In this range,)`

`(x + 2) >= 0\ \ text(and)\ \ (x – 3)<= 0`

`:.\ text(We can write)`

`|\ x + 2\ | + |\ x – 3\ |` `= (x + 2)\ – (x – 3)`
  `= x + 2 – x + 3`
  `= 5`

 
`:. C\ text(has the same solution)`

`=>  C\ text(is correct.)`

Functions, EXT1* F1 2012 HSC 11b

 Solve  `|\ 3x -1\ | < 2`   (2 marks)

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 ` -1/3 < x < 1 `

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 MARKER’S COMMENT: Note that both conditions must be satisfied! Dealing with negative signs and division for inequalities produced many errors.

`|\ 3x -1\ | < 2`

`3x -1` `<2`  `\ \ \ \ \-(3x -1)` `< 2`
`3x`  `<3` `-3x + 1` `< 2`
`x` `< 1`  `3x` `> -1`
    `x` `> -1/3`

`:. -1/3 < x < 1`