SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Algebra, MET2-NHT 2019 VCAA 20 MC

Let  `f(x) = (ax + b)^5`  and let  `g`  be the inverse function of  `f`.

Given that  `f(0) = 1`, what is the value of  `g' (1)`

  1.  `(5)/(a)`
  2.  `1`
  3.  `(1)/(5a)`
  4.  `5a(a + 1)^4`
  5.  `0`
Show Answers Only

`C`

Show Worked Solution

`f(x) = (ax + b)^5`

`text(Given) \ \ f(0) = 1\ \ =>\ \ b^5 = 1 \ => \ b = 1`
  
`f ‘(x) = 5a(ax + 1)^4`

`f(0) = 1 \ => \ g(1) = 0 \ \ text{(Inverse: swap} \ x ↔ y )`
 
`text(Gradient of) \ \ f(x) \ \ text(at) \ \ x =0 \ \ text(will be the reciprocal of the gradient of) \ \ g(x) \ \ text(at) \ \ x = 1`

`f ‘(0) = 5a`

`:. \ g'(1) = (1)/(5a)`
 
`=> \ C`

Algebra, MET2-NHT 2019 VCAA 11 MC

The function  `f : D → R, \ f(x) = 5x^3 + 10x^2 + 1`  will have an inverse function for

  1.  `D = R`
  2.  `D = (–2, ∞)`
  3.  `D = (–∞ , (1)/(2)]`
  4.  `D = (–∞ , –1]`
  5.  `D = [0 , ∞)`
Show Answers Only

`E`

Show Worked Solution

`text(Graph:) \ \ y = 5x^3 + 10x^2 + 1 \ \ text{(by CAS)}`
 


 

`text(Max when) \ \ x = -(4)/(3)`

`text(Min when) \ \ x = 0`

`=> \ E`

Graphs, MET2 2008 VCAA 20 MC

The function  `f: B -> R`  with rule  `f(x) = 4x^3 + 3x^2 + 1`  will have an inverse function for

  1. `B = R`
  2. `B = (1/2, oo)`
  3. `B = (text{−∞}, 1/2]`
  4. `B = (text{−∞}, 1/2)`
  5. `B = [−1/2, oo)`
Show Answers Only

`B`

Show Worked Solution

`text(Inverse exists if)\ \ f(x)\ \ text(is)\ \ 1 – 1:`

 

vcaa-2008-20i

`text(Option B’s domain ensures)\ \ f(x)\ \ text(is)\ \ 1- 1`

`=>   B`

Graphs, MET1 SM-Bank 20

The rule for  `f` is  `f(x) = x-1/2 x^2`  for  `x <= 1`.  This function has an inverse,  `f^(-1) (x)`.

  1. Sketch the graphs of  `y = f(x)`  and  `y = f^(-1) (x)`  on the same set of axes. (Use the same scale on both axes.)  (2 marks
  2. Find the rule for the inverse function  `f^(-1) (x)`.   (2 marks)
  3. Evaluate  `f^(-1) (3/8)`.   (1 mark)
Show Answers Only
  1.  
    Inverse Functions, EXT1 2008 HSC 5a Answer
  2. `y = 1-sqrt(1-2x)`
  3. `1/2`
Show Worked Solution
a. 

Inverse Functions, EXT1 2008 HSC 5a Answer
b.   `y = x-1/2 x^2,\ \ \ x <= 1`

 

`text(For the inverse function, swap)\ \ x↔y,`

`x` `= y-1/2 y^2,\ \ \ y <= 1`
`2x` `= 2y-y^2`
`y^2-2y + 2x` `= 0`

 

`text(Using quadratic formula,)`

`y` `= (2 +- sqrt( (-2)^2-4 * 1 * 2x) )/2`
  `= (2 +- sqrt(4-8x))/2`
  `= (2 +- 2 sqrt(1-2x))/2`
  `= 1 +- sqrt (1-2x)`

 

`:. y = 1-sqrt(1-2x), \ \ (y <= 1)`

 

c.    `f^(-1) (3/8)` `= 1-sqrt(1 -2(3/8))`
    `= 1-sqrt(1-6/8)`
    `= 1-sqrt(1/4)`
    `= 1-1/2`
    `= 1/2`

Calculus, MET2 2011 VCAA 3

  1. Consider the function  `f: R -> R, f(x) = 4x^3 + 5x-9`.

     

    1. Find  `f^{′}(x)`  (1 mark)
    2. Explain why  `f^{′}(x) >= 5` for all `x`.  (1 mark)
  2. The cubic function `p` is defined by  `p: R -> R, p(x) = ax^3 + bx^2 + cx + k`, where `a`, `b`, `c` and `k` are real numbers.

     

    1. If `p` has `m` stationary points, what possible values can `m` have?  (1 mark)
    2. If `p` has an inverse function, what possible values can `m` have?  (1 mark)
  3. The cubic function `q` is defined by  `q:R -> R, q(x) = 3-2x^3`.

     

    1. Write down a expression for  `q^(−1)(x)`.  (2 marks)
    2. Determine the coordinates of the point(s) of intersection of the graphs of  `y = q(x)`  and  `y = q^(−1)(x)`.  (2 marks)
  4. The cubic function `g` is defined by  `g: R -> R, g(x) = x^3 + 2x^2 + cx + k`, where `c` and `k` are real numbers.

     

    1. If `g` has exactly one stationary point, find the value of `c`.  (3 marks)
    2. If this stationary point occurs at a point of intersection of  `y = g(x)`  and  `g^(−1)(x)`, find the value of `k`.  (3 marks)
Show Answers Only
    1. `f^{′}(x) = 12x^2 + 5`
    2. `text(See Worked Solutions)`
    1. `m = 0, 1, 2`
    2. `m = 0, 1`
    1. `q^(−1)(x) = root(3)((3-x)/2), x ∈ R`
    2. `(1, 1)`
    1. `4/3`
    2. `−10/27`
Show Worked Solution

a.i.   `f^{′}(x) = 12x^2 + 5`

 

a.ii.  `text(S)text(ince)\ \ x^2>=0\ \ text(for all)\ x,`

♦ Mean mark 47%.
` 12x^2` `>= 0`
`12x^2 + 5` `>=  5`
`f^{′}(x)` `>=  5\ \ text(for all)\ x`

 

b.i.   `p(x) = text(is a cubic)`

♦♦♦ Mean mark part (b)(i) 9%, and part (b)(ii) 20%.
MARKER’S COMMENT: Good exam strategy should point students to investigate earlier parts for direction. Here, part (a) clearly sheds light on a solution!

`:. m = 0, 1, 2`

`text{(Note: part a.ii shows that a cubic may have no SP’s.)}`

 

b.ii.   `text(For)\ p^(−1)(x)\ text(to exist)`

`:. m = 0, 1`

 

c.i.   `text(Let)\ y = q(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= 3-2y^3`
`y^3` `= (3-x)/2`

`:. q^(−1)(x) = root(3)((3-x)/2), \ x ∈ R`

 

c.ii.  `text(Any function and its inverse intersect on)`

   `text(the line)\ \ y=x.`

`text(Solve:)\ \ 3-2x^3` `= xqquadtext(for)\ x,`
`x` `= 1`

 

`:.\ text{Intersection at (1, 1)}`

 

♦ Mean mark part (d)(i) 44%.
d.i.    `g^{′}(x)` `= 0`
  `3x^2 + 4x + c` `= 0`
  `Delta` `= 0`
  `16-4(3c)` `= 0`
  `:. c` `= 4/3`

 

d.ii.   `text(Define)\ \ g(x) = x^3 + 2x^2 + 4/3x + k`

♦♦♦ Mean mark part (d)(ii) 14%.

  `text(Stationary point when)\ \ g^{′}(x)=0`

`g^{′}(x) = 3x^2+4x+4/3`

`text(Solve:)\ \ g^{′}(x)=0\ \ text(for)\ x,`

`x = −2/3`

`text(Intersection of)\ g(x)\ text(and)\ g^(−1)(x)\ text(occurs on)\ \ y = x`

`text(Point of intersection is)\  (−2/3, −2/3)`

`text(Find)\ k:`

`g(−2/3)` `= −2/3\ text(for)\ k`
`:. k` ` = −10/27`

Algebra, MET2 2010 VCAA 9 MC

The function  `f:\ (–oo, a] -> R`  with rule  `f(x) = x^3 - 3x^2 + 3`  will have an inverse function provided

  1. `a <= 0`
  2. `a >= 2`
  3. `a >= 0`
  4. `a <= 2`
  5. `a <= 1`
Show Answers Only

`A`

Show Worked Solution
`f(x)` `= x^3-3x^2 + 3`
`f′(x)` `=3x^2-6x`
  `=3x(x-2)`

 

`text(Stationary points at)\ \ x=0 and 2.`

`text{Local max at (0,3) and local min at (2,-1).}`

`text(Sketch the graph:)`

`text(Inverse exists if)\ \ f(x)\ \ text(is)\ \ 1-1.`

`:. x <= 0`

`=>   A`

Graphs, MET2 2011 VCAA 8 MC

Consider the function  `f: R -> R, \ f(x) = x(x - 4)`  and the function

`g: [3/2,5) -> R, \ g(x) = x + 3`.

If the function  `h = f + g`, then the domain of the inverse function of `h` is

  1. `[0,13)`
  2. `[−3/4,10]`
  3. `(−3/4,15/4]`
  4. `[3/4,13)`
  5. `[3/2,13)` 
Show Answers Only

`=> D`

Show Worked Solution
`h(x)` `= x^2 – 3x + 3, \ x ∈[3/2,5)`
`h′(x)` `=2x-3`

 
`:.\ text(Minimum at)\ \ (3/2, 3/4)`

`h(5)=5^2 -3 xx 5 +3=13`

 

`text(Sketch the graph:)` 

met1-2011-vcaa-8-mc-answer

`text(Domain) (h^(−1)(x))` `= text(Range)\ (h)`
  `= [3/4,13)`

`=> D`

Functions, MET1 2010 VCAA 3

Let  `f: R^+ -> R`  where  `f(x) = 1/x^2.`

  1. Find  `g(x) = f(f(x))`.  (1 mark)
  2. Evaluate  `g^-1 (16)`,  where `g^-1` is the inverse function of `g`.  (1 mark)
Show Answers Only
  1. `x^4,\ \ x in R^+`
  2. `g^-1(16) = 2`
Show Worked Solution
a.   `g(x)` `= f(1/x^2)`
    `= 1/((1/x^2)^2)`
    `=1/(1/x^4)`
     
  `:. g(x)` `= x^4,\ \ x in R^+`

 

b.   `text(Let)\ \ y = f(x)`

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

`x` `= y^4`
`y` `= +- x^(1/4),`
  `=x^(1/4)\ \ \ text{(Domain of}\ g(x)\ text(is)\ R^+ text{)}`

 

`:. g^-1 (16)` `= 16^(1/4)`
  `= 2`

Functions, MET1 2012 VCAA 3

The rule for function  `h`  is  `h(x) = 2x^3 + 1.`  Find the rule for the inverse function  `h^-1.`  (2 marks)

Show Answers Only

`h^-1 (x) = root 3 ((x – 1)/2),\ \ x in R`

Show Worked Solution

`h(x) = 2x^3+1,\ \ text(let)\ \ y = 2x^3+1,`

MARKER’S COMMENT: “It is important” students do not proceed directly from  `y-2x^3+1`  to  `x=2y^3+1`. See approach shown in working.

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

`x` `= 2y^3 + 1`
`y^3` `= (x – 1)/2`
`y` `= root 3 ((x – 1)/2)`
`:. h^-1 (x)` `= root 3 ((x – 1)/2),\ \ x in R`