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Algebra, MET1-NHT 2019 VCAA 5a

Let  `h:[-3/2, oo) -> R,\ h(x) = sqrt(2x + 3) - 2.`

Find the value(s) of `x` such that  `[h(x)]^2 = 1`.  (2 marks)

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`3 or\ text(−1)`

Show Worked Solution
  `[h(x)]^2 = 1` `\ => \ h(x) = +- 1`
  `sqrt(2x + 3)` `= 1` `sqrt(2x + 3) – 2` `= -1`
  `sqrt(2x + 3)` `= 3` `sqrt(2x + 3)` `= 1`
  `2x + 3` `= 9` `2x + 3` `= 1`
  `x` `= 3` `x` `= -1`

 
`:. x = 3 or -1\ \ \ text{(both in the domain of}\ h)`

Graphs, MET1 2016 VCAA 5

Let  `f : (0, ∞) → R`, where  `f(x) = log_e(x)`  and  `g: R → R`, where  `g (x) = x^2 + 1`.

  1.   i. Find the rule for `h`, where  `h(x) = f (g(x))`.   (1 mark)

    ii. State the domain and range of `h`.   (2 marks)

  2. iii. Show that  `h(x) + h(-x) = f ((g(x))^2 )`.   (2 marks)

  3. iv. Find the coordinates of the stationary point of `h` and state its nature.   (2 marks)

     

  4. Let  `k: (-∞, 0] → R`  where  `k (x) = log_e(x^2 + 1)`.
  5.  i. Find the rule for  `k^(-1)`.   (2 marks)

  6. ii. State the domain and range of  `k^(-1)`.   (2 marks)

Show Answers Only
  1.   i. `log_e(x^2 + 1)`
  2.  ii. `[0,∞)`
  3. iii. `text(See Worked Solutions)`
  4. iv. `(0, 0)`
  5.  i. `-sqrt(e^x-1)`
  6. ii. `text(Domain)\ (k) = (-∞,0]`
  7.    `text(Range)\ (k) = [0,∞)`
Show Worked Solution
a.i.    `h(x)` `= f(x^2 + 1)`
    `= log_e(x^2 + 1)`

 

a.ii.   `text(Domain)\ (h) =\ text(Domain)\ (g) = R`

♦♦ Mean mark part (a)(ii) 30%.
  `text(For)\ x ∈ R` `-> x^2 + 1 >= 1`
  `-> log_e(x^2 + 1) >= 0`

`:.\ text(Range)\ (h) = [0,∞)`

 

MARKER’S COMMENT: Many students were unsure of how to present their working in this question. Note the layout in the solution.
a.iii.   `text(LHS)` `= h(x) + h(−x)`
    `= log_e(x^2 _ 1) + log_e((-x)^2 + 1)`
    `= log_e(x^2 + 1) + log_e(x^2 + 1)`
    `= 2log_e(x^2 + 1)`

 

`text(RHS)` `= f((x^2 + 1)^2)`
  `= 2log_e(x^2 + 1)`

 

`:. h(x) + h(-x) = f((g(x))^2)\ \ text(… as required)`

 

a.iv.   `text(Stationary points when)\ \ h^{prime}(x) = 0`

♦ Mean mark part (a)(iv) 41%.
MARKER’S COMMENT: Solving a fraction is zero

   `text(Using Chain Rule:)`

`h^{prime}(x)` `= (2x)/(x^2 + 1)`

`:.\ text(S.P. when)\ \ x=0`

 

`text(Find nature using 1st derivative test:)`

`:.\ text{Minimum stationary point at (0, 0)}.`

 

b.i.   `text(Let)\ \ y = k(x)`

♦ Mean mark (b)(i) 49%.
MARKER’s COMMENT: Many students failed to consider the restrictions on the domain in `k(x)` and only select the negative root.

  `text(Inverse: swap)\ x ↔ y`

`x` `= log_e(y^2 + 1)`
`e^x` `= y^2 + 1`
`y^2` `= e^x-1`
`y` `= ±sqrt(e^x-1)`

 

`text(But range)\ \ (k^(-1)) =\ text(domain)\ (k)`

`:.k^(-1)(x) =-sqrt(e^x-1)`

♦ Mean mark part (b)(ii) 44%.

 

b.ii.   `text(Range)\ (k^(-1)) =\ text(Domain)\ (k) = (-∞,0]`

  `text(Domain)\ (k^(-1)) =\ text(Range)\ (k) = [0,∞)`