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Functions, 2ADV F1 SM-Bank 42

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

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`x=2 or 4`

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`|\ x − 3\ | = 1`

`text(Method 1)`

`(x − 3)^2 = 1`

`x^2 − 6x + 9 = 1`

`x^2 − 6x +8 = 0`

`(x − 4)(x − 2) = 0`

`:. x=2 or 4`
 

`text(Method 2)`

`(x − 3)` `=1` `–(x − 3)` ` = 1`
`x` `=4` `–x +3` `=1`
    `x` `=−2`

Functions, 2ADV F1 SM-Bank 37

Find all values of  `x`  for which  `|\ x - 4\ | = x/2 + 7`.  (3 marks)

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`x = 22\ \ text(or)\ −2`

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`x – 4` `= x/2 + 7` `text(or)` `-(x-4)` `= x/2 + 7`
`2x – 8` `= x + 14`   `−2x + 8` `= x + 14`
`x` `= 22`   `3x` `= −6`
      `x` `= −2`

`:. x = 22\ \ text(or)\ −2`

Functions, 2ADV F1 2019 HSC 13e

  1. Sketch the graph of  `y = |\ x - 1\ |`  for  `-4 <= x <= 4`.  (1 mark)

  2. Using the sketch from part i, or otherwise, solve  `|\ x - 1\ | = 2x + 4`.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `(-1, 2)`
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(i)   

 

ii.    `text(By inspection, intersection when)\ x = -1`

`text(Test:)`

`|-1-1|` `= -2 + 4`
`2` `= 2`

 
`:.\ text(Intersection at)\ (-1, 2)`

Functions, 2ADV F1 2017 HSC 11g

Solve  `|\ 3x - 1\ | = 2`.  (2 marks)

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`x = 1 or -1/3`

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

`:. x = 1 or -1/3`