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Algebra, MET2 2018 VCAA 7 MC

Let  `f: R^+ -> R,\ f(x) = k log_2(x),\ k in R`.

Given that  `f^(-1)(1) = 8`, the value of `k` is

  1. `0`
  2. `1/3`
  3. `3`
  4. `8`
  5. `12`
Show Answers Only

`B`

Show Worked Solution

`y = k log_2(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= k log_2 (y)`
`log_2(y)` `= x/k`
`y` `= 2^(x/k)`

 
`text(Given)\ \ f^(-1)(1) = 8,`

`8` `= 2^(1/k)`
`1/k` `= 3`
`:. k` `= 1/3`

 
`=>   B`

Algebra, MET1 SM-Bank 24

The rule for  `f`  is  `f(x) = e^x − e^(-x)`.

Show that the inverse function is given by

    `f^(-1)(x) = log_e((x + sqrt(x^2 + 4))/2)` (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions.)}`

Show Worked Solution

`y = e^x − e^(-x)`

`text(Inverse: swap)\  x harr y`

`x` `= e^y − 1/(e^y)`
`xe^y` `= e^(2y) − 1`
`e^(2y) − xe^y − 1` `= 0`

 

`text(Let)\ \ A = e^y`

`:.A^2 − xA − 1 = 0`
 

`text(Using the quadratic formula)`

`A` `=(x ± sqrt((-x)^2 − 4 · 1 · (-1)))/(2 · 1)`
  `=(x ± sqrt(x^2 + 4))/2`

 

`text(S)text(ince)\ \ (x – sqrt(x^2 + 4))/2<0\ \ text(and)\ \ e^y>0,`

`:.e^y` `= (x + sqrt(x^2 + 4))/2`
`log_e e^y` ` = log_e((x + sqrt(x^2 + 4))/2)`
`y` `= log_e((x + sqrt(x^2 + 4))/2)`
`:.f^(-1)(x)` `= log_e((x + sqrt(x^2 + 4))/2)\ \  …\ text(as required)`

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)
  2.  ii. State the domain and range of `h`.  (2 marks)
  3. iii. Show that  `h(x) + h(−x) = f ((g(x))^2 )`.  (2 marks)
  4. iv. Find the coordinates of the stationary point of `h` and state its nature.  (2 marks)
     
  5. Let  `k: (– ∞, 0] → R`  where  `k (x) = log_e(x^2 + 1)`.
  6.  i. Find the rule for  `k^(−1)`.  (2 marks)
  7. ii. State the domain and range of  `k^(– 1)`.  (2 marks)
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  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′(x) = 0`

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

   `text(Using Chain Rule:)`

`h′(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,∞)`

Algebra, MET2 2009 VCAA 16 MC

The inverse of the function  `f: R^+ -> R,\ f(x) = e^(2x + 3)`  is

  1. `{:(f^-1:\ R^+ -> R, qquad qquad qquad qquad f^-1 (x) = e^(-2x - 3)):}`
  2. `{:(f^-1:\ R^+ -> R, qquad qquad qquad qquad f^-1 (x) = e^((x - 3)/2)):}`
  3. `{:(f^-1:\ (e^3, oo) -> R, qquad qquad f^-1 (x) = log_e (sqrt x) - 3/2):}`
  4. `{:(f^-1:\ (e^3, oo) -> R, qquad qquad f^-1 (x) = e^((x - 3)/2)):}`
  5. `{:(f^-1:\ (e^3, oo) -> R, qquad qquad f^-1 (x) = -log_e (2x - 3)):}`
Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ y = f(x)`

`text(Inverse:  swap)\ \ x harr y`

`x` `=e^(2y + 3)`
`log_e x` `=2y+3`
`2y` `=log_e x -3`
`y` `=1/2 log_e (x) – 3/2`

 
`:. f^-1 (x) = log_e (sqrt x) – 3/2`

`=>   C`

Functions, MET1 2006 VCAA 2

For the function  `f: R -> R, f(x) = 3e^(2x)-1,`

  1. find the rule for the inverse function  `f^-1`  (2 marks)
  2. find the domain of the inverse function  `f^-1`  (1 mark)
Show Answers Only
  1. `f^-1 (x) = 1/2 log_e ((x + 1)/3)`
  2. `x in (– 1, oo)`
Show Worked Solution

a.    `text(Let)\ \ y = f(x)`

`text(For inverse, swap)\ \ x harr y`

`x` `= 3e^(2y)-1`
`(x + 1)/3` `= e^(2y)`
`2y` `= log_e ((x + 1)/3)`
`y` `= 1/2 log_e ((x + 1)/3)`

 

`:. f^-1 (x) = 1/2 log_e ((x + 1)/3)`

 

♦ Mean mark 45%.

b.   `text(Domain)\ (f^-1) = text(Range)\ f(x),`

`:.x in (– 1, oo)`