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Functions, MET1 2021 VCAA 9

Consider the unit circle  `x^2 + y^2 = 1`  and the tangent to the circle at the point `P`, shown in the diagram below.
 

  1. Show that the equation of the line that passes through the points `A` and `P` is given by  `y = -x/sqrt3 + 2/sqrt3`.  (2 marks)

Let  `T : R^2 -> R^2, T ([(x),(y)]) = [(1, 0),(0, q)][(x),(y)]`,  where  `q ∈ R text{\}{0}`, and let the graph of the function `h` be the transformation of the line that passes through the points `A` and `P` under `T`.

    1. Find the values of `q` for which the graph of `h` intersects with the unit circle at least once.  (1 mark)
    2. Let the graph of `h` intersect the unit circle twice.
    3. Find the values of `q` for which the coordinates of the points of intersection have only positive values.  (1 mark)
  2. For  `0 < q <= 1`, let  `P′` be the point of intersection of the graph of `h` with the unit circle, where  `P′` is always the point of intersection that is closest to `A`, as shown in the diagram below.
     

Let `g` be the function that gives the area of triangle `OAP′` in terms of `theta`.

    1. Define the function `g`.  (2 marks)
    2. Determine the maximum possible area of triangle `OAP′`.  (2 marks)
Show Answers Only
  1. `text(See Worked Solution)`
  2.  i. `q in [–1,1]text{\}{0}`
  3. ii. `q in (sqrt3/2, 1)`
  4. i. `g(theta) = sin theta,\ \ theta in (0, pi/3]`
  5. ii. `sqrt3/2\ text(u²)`
Show Worked Solution

a.   `text(In)\ ΔOAP, text(by Pythagoras:)`

♦♦♦ Mean mark part (a) 18%.

`AP=sqrt(2^2-1^2) = sqrt3`

`tan ∠OAP = 1/sqrt3`

`=> m_(AP) = – 1/sqrt3`

`text{Find equation of line}\ \ m=-1/sqrt3\ \ text{through (2, 0):}`

`y-0` `=- 1/sqrt3 (x-2)`  
`y` `=- x/sqrt3 + 2/sqrt3\ \ text{… as required}`  

 

♦♦♦ Mean mark part (b)(i) 6%.
bi.  `text{Transformation matrix → dilates line by a factor of}\ q\ text{from}\ xtext{-axis}`

`:. h(x)\ text{intersects circle for}\ \ q in [–1,1]text{\}{0}`

 

b.ii. `text(Positive coordinates → line can only intersect circle in top right quadrant.)`

 `text(Line)\ AP\ text{can move down until it cuts the circle at (0, 1)`

♦♦♦ Mean mark part (b)(ii) 4%.

`=>\ text(S)text(ince)\ AP\ text(cuts)\ ytext{-axis at}\ 2/sqrt3,`

`q=sqrt3/2\ \ text{dilates}\ AP\ text{where it cuts at (0, 1)`

`:. q in (sqrt3/2, 1)`
 

c.i.   `text(If)\ \ q=1, \ P′ = P and theta=pi/3`

♦♦♦ Mean mark part (c)(i) 10%.
`g(theta)` `=1/2 xx 1 xx 2 xx sin theta`  
  `=sin theta, \ \ theta in (0, pi/3]`  

 
c.ii.  `text{S}text{ince}\ \ g(theta) = sin theta\ \ text(for)\ \ theta in (0, pi/3],`

♦♦♦ Mean mark part (c)(ii) 12%.

`text(Maximum)\ g(theta)\ text(occurs when)\ \ theta = pi/3`

`:.g(theta)_text(max)` `=sin (pi/3)`  
  `=sqrt3/2\ \ text(u²)`  

Algebra, MET2 2017 VCAA 4

Let  `f : R → R :\  f (x) = 2^(x + 1)-2`. Part of the graph of  `f` is shown below.
 

  1. The transformation  `T: R^2 -> R^2, \ T([(x),(y)]) = [(x),(y)] + [(c),(d)]`  maps the graph of  `y = 2^x`  onto the graph of  `f`.

     

    State the values of `c` and `d`.  (2 marks)

  2. Find the rule and domain for  `f^(−1)`, the inverse function of  `f`.  (2 marks)
  3. Find the area bounded by the graphs of  `f` and  `f^(−1)`.  (3 marks)
  4. Part of the graphs of  `f` and  `f^(−1)` are shown below.
     

         
     
    Find the gradient of  `f` and the gradient of  `f^(−1)`  at  `x = 0`.  (2 marks)

The functions of  `g_k`, where  `k ∈ R^+`, are defined with domain `R` such that  `g_k(x) = 2e^(kx)-2`.

  1. Find the value of `k` such that  `g_k(x) = f(x)`.  (1 mark)
  2. Find the rule for the inverse functions  `g_k^(−1)` of  `g_k`, where  `k ∈ R^+`.  (1 mark)
    1. Describe the transformation that maps the graph of  `g_1` onto the graph of  `g_k`.  (1 mark)
    2. Describe the transformation that maps the graph of  `g_1^(–1)` onto the graph of  `g_k^(–1)`.  (1 mark)
  4. The lines `L_1` and `L_2` are the tangents at the origin to the graphs of  `g_k`  and  `g_k^(–1)`  respectively.
  5. Find the value(s) of `k` for which the angle between `L_1` and `L_2` is 30°.  (2 marks)
  6. Let `p` be the value of `k` for which  `g_k(x) = g_k^(−1)(x)`  has only one solution.
  7.  i. Find `p`.  (2 marks)
  8. ii. Let  `A(k)`  be the area bounded by the graphs of  `g_k`  and  `g_k^(–1)`  for all  `k > p`.
  9.     State the smallest value of `b` such that  `A(k) < b`.  (1 mark)
Show Answers Only

a.  `c = −1, \ d = −2`

b.  `f^(−1)(x) = log_2 (x + 2)-1, \ x ∈ (−2,∞)`

c.  `3-2/(log_e(2))\ text(units)²`

d. `f′(0)= 2log_e(2) and f^(-1′)(0)= 1/(2log_e(2))`

e.  `k = log_e(2)`

f.  `g_k^(−1)(x)== 1/k log_e((x + 2)/2)`

g.i.  `text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

g.ii.  `text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

h.  `k=sqrt3/6\ or\ sqrt3/2`

i.i.  `p = 1/2`

i.ii.  `b=4`

Show Worked Solution

a.   `text(Using the matrix transformation:)`

`x′` `=x+c\ \ => x=x′-c`
`y′` `=y+d\ \ =>y= y′-d`
   
`y′-d` `=2^(x′-c)`
`y′` `= 2^(x′-c)+d`

 

`:. c = −1, \ d = −2`

 

b.   `text(Let)\ \ y = f(x)`

`text(Inverse : swap)\ x\ ↔ \ y`

`x` `= 2^(y + 1)-2`
`x + 2` `= 2^(y + 1)`
`y + 1` `= log_2(x + 2)`
`y` `= log_2(x + 2)-1`

 

`text(dom)(f^(−1)) = text(ran)(f)`

`:. f^(−1)(x) = log_2 (x + 2)-1, \ x ∈ (−2,∞)`

 

c.   `text(Intersection points occur when)\ \ f(x)=f^(−1)(x)`

MARKER’S COMMENT: Specifically recommends using  `f^(-1)(x)-f(x)`  in this type of integral to avoid errors in stating the integral.

`x` `= −1, 0`

 

`text(Area)` `= int_(−1)^0 (f^(−1)(x)-f(x))\ dx`
  `= 3-2/(log_e(2))\ text(units)²`

 

d.    `f′(0)` `= 2log_e(2)`
  `f^(-1′)(0)` `= 1/(2log_e(2))`

 

e.   `g_k(x) = 2e^(kx)-2`

`text(Solve:)\ \ g_k(x) = f(x)quad text(for)\ k ∈ R^+`

`:. k = log_e(2)`

 

f.   `text(Let)\ \ y = g_k(x)`

`text(Inverse : swap)\ x\ text(and)\ y`

`x` `= 2e^(ky)-2`
`e^(ky)` `=(x+2)/2`
`ky` `=log_e((x+2)/2)`
`:. g_k^(−1)(x)` `= 1/k log_e((x + 2)/2)`

 

♦♦ Mean mark part (g)(i) 31%.

g.i.    `g_1(x)` `= 2e^x-2`
  `g_k(x)` `= 2e^(kx)-2`

 
`:. text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

 

♦♦ Mean mark part (g)(ii) 30%.

g.ii.    `g_1^(−1)(x)` `= log_e((x + 2)/2)`
  `g_k^(−1)(x)` `= 1/klog_e((x + 2)/2)`

 
`:. text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

 

h.   `text(When)\ \ x=0,`

♦♦♦ Mean mark part (h) 13%.

`g_k′(0)=2k\ \ => m_(L_1)=2k`

`g_k^(-1′)(0)=1/(2k)\ \ => m_(L_2)=1/(2k)`

 

`text(Using)\ \ tan 30^@=|(m_1-m_2)/(1+m_1m_2)|,`

`text(Solve:)\ \ 1/sqrt3` `=+- ((2k-1/(2k)))/2\ \ text(for)\ k`

 
`:. k=sqrt3/6\ or\ sqrt3/2`

 

i.i   `text(By inspection, graphs will touch once if at)\ \ x=0,`

♦♦♦ Mean mark part (i)(i) 4%.

`m_(L_1)` `=m_(L_2)`
`2k` `=1/(2k)`
`k` `=1/2,\ \ (k>0)`

 
`:. p = 1/2`

 

i.ii  `text(As)\ k→oo, text(the graph of)\ g_k\ text(approaches)\ \ x=0`

♦♦♦ Mean mark part (i)(ii) 2%.

`text{(vertically) and}\ \ y=–2\ \ text{(horizontally).}`

`text(Similarly,)\ \ g_k^(-1)\ \ text(approaches)\ \ x=–2`

`text{(vertically) and}\ \ y=0\ \ text{(horizontally).}`

 

`:. lim_(k→oo) A(k) = 4`

`:.b=4`

Graphs, MET2 2013 VCAA 1

Trigg the gardener is working in a temperature-controlled greenhouse. During a particular 24-hour time interval, the temperature  `(Ttext{°C})` is given by  `T(t) = 25 + 2 cos ((pi t)/8), \ 0 <= t <= 24`, where `t` is the time in hours from the beginning of the 24-hour time interval.

  1. State the maximum temperature in the greenhouse and the values of `t` when this occurs.  (2 marks)
  2. State the period of the function `T.`  (1 mark)
  3. Find the smallest value of `t` for which  `T = 26.`  (2 marks)
  4. For how many hours during the 24-hour time interval is  `T >= 26?`  (2 marks)

Trigg is designing a garden that is to be built on flat ground. In his initial plans, he draws the graph of  `y = sin(x)`  for  `0 <= x <= 2 pi`  and decides that the garden beds will have the shape of the shaded regions shown in the diagram below. He includes a garden path, which is shown as line segment `PC.`

The line through points  `P((2 pi)/3, sqrt 3/2)`  and  `C (c, 0)`  is a tangent to the graph of  `y = sin (x)`  at point `P.`

    1. Find  `(dy)/(dx)`  when  `x = (2 pi)/3.`  (1 mark)
    2. Show that the value of `c` is  `sqrt 3 + (2 pi)/3.`  (1 mark)

In further planning for the garden, Trigg uses a transformation of the plane defined as a dilation of factor `k` from the `x`-axis and a dilation of factor `m` from the `y`-axis, where `k` and `m` are positive real numbers.

  1. Let `X prime, P prime` and `C prime` be the image, under this transformation, of the points `X, P` and `C` respectively. 

     

    1. Find the values of `k` and `m`  if  `X prime P prime = 10`  and  `X prime C prime = 30.`  (2 marks)
    2. Find the coordinates of the point `P prime.`  (1 mark)
Show Answers Only
  1. `27^@Cquadtext(when)\ t = 0, 16`
  2. `text(16 hours)`
  3. `8/3`
  4. `text(8 hours)`
    1. `−1/2`
    2. `text(See Worked Solutions)`
    1. `k = 20/sqrt3, quadm = 30/sqrt3`
    2. `Pprime((20pi sqrt3)/3,10)`
Show Worked Solution

a.   `T_text(max)\ text(occurs when)\ \ cos((pit)/8) = 1,`

`T_text(max)= 25 + 2 = 27^@C`

`text(Max occurs when)\ \ t = 0, or 16\ text(h)`

 

b.    `text(Period)` `= (2pi)/(pi/8)`
    `= 16\ text(hours)`

 

c.   `text(Solve:)\ \ 25 + 2 cos ((pi t)/8)=26\ \ text(for)\ t,`

`t`  `= 8/3,40/3,56/3\ \ text(for)\ t ∈ [0,24]`
`t_text(min)` `= 8/3`

 

d.   `text(Consider the graph:)`

met2-2013-vcaa-sec1-answer1

`text(Time above)\ 26 text(°C)` `= 8/3 + (56/3 – 40/3)`
  `= 8\ text(hours)`

 

e.i.   `(dy)/(dx) = cos(x)`

`text(At)\ x = (2pi)/3,`

`(dy)/(dx)` `= cos((2pi)/3)= −1/2`

 

e.ii.  `text(Solution 1)`

`text(Equation of)\ \ PC,`

`y-sqrt3/2` `=-1/2(x-(2pi)/3)`
`y` `=-1/2 x +pi/3 +sqrt3/2`

 

`PC\ \ text(passes through)\ \ (c,0),`

`0` `=-1/2 c +pi/3 + sqrt3/2`
`c` `=sqrt3 + (2 pi)/3\ …\ text(as required)`

 

`text(Solution 2)`

`text(Equating gradients:)`

`- 1/2` `= (sqrt3/2 – 0)/((2pi)/3 – c)`
`-1` `= sqrt3/((2pi – 3c)/3)`
`3c – 2pi` `= 3sqrt3`
`3c` `= 3 sqrt3 + 2pi`
`:. c` `= sqrt3 + (2pi)/3\ …\ text(as required)`

 

f.i.   `Xprime ((2pi)/3 m,0)qquadPprime((2pi)/3 m, sqrt3/2 k)qquadCprime ((sqrt3 + (2pi)/3)m, 0)`

`XprimePprime` `= 10`
`sqrt3/2 k` `= 10`
`:. k` `= 20/sqrt3`
  `=(20sqrt3)/3`
♦♦♦ Mean mark part (f)(i) 14%.

 

`XprimeCprime=30`

`((sqrt3 + (2pi)/3)m) – (2pi)/3 m` `= 30`
`:. m` `= 30/sqrt3`
  `=10sqrt3`
♦♦♦ Mean mark part (f)(ii) 12%.

 

f.ii.    `Pprime((2pi)/3 m, sqrt3/2 k)` `= Pprime((2pi)/3 xx 10sqrt3, sqrt3/2 xx 20/sqrt3)`
    `= Pprime((20pisqrt3)/3,10)`

Graphs, MET2 2015 VCAA 11 MC

The transformation that maps the graph of  `y = sqrt(8x^3 + 1)`  onto the graph of  `y = sqrt(x^3 + 1)`  is a

  1. dilation by a factor of `2` from the `y`-axis.
  2. dilation by a factor of `2` from the `x`-axis.
  3. dilation by a factor of `1/2` from the `x`-axis.
  4. dilation by a factor of `8` from the `y`-axis.
  5. dilation by a factor of `1/2` from the `y`-axis.
Show Answers Only

`A`

Show Worked Solution
♦♦ Mean mark 24%.
`text(Let)\ f(x)` `= sqrt(8x^3 + 1)`
`f(1/2 x)` `= sqrt(8(1/2 x)^3 + 1)`
  `= sqrt(x^3 + 1)`

 
`f(1/2 x) -> text(dilate by factor of 2 from)\ ytext(-axis)`

`=>   A`

Algebra, MET2 2013 VCAA 18 MC

Let  `g(x) = log_2(x),\ \ x > 0`

Which one of the following equations is true for all positive real values of  `x?`

  1. `2g (8x) = g (x^2) + 8`
  2. `2g (8x) = g (x^2) + 6`
  3. `2g (8x) = (g (x) + 8)^2`
  4. `2g (8x) = g (2x) + 6`
  5. `2g (8x) = g (2x) + 64`
Show Answers Only

`B`

Show Worked Solution

`text(Consider Option)\ B:`

♦♦ Mean mark 35%.
`text(LHS)` `= 2g(8x)`
  `= 2log_2(8x)`
  `= 2log_2(8) + 2log_2(x)`
  `=2log_2 (2^3)+ 2log_2(x)`
  `= 6 + log_2(x^2)`
  `= g(x^2) + 6`

 
`=>   B`