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Graphs, MET2 2020 VCAA 20 MC

Let  `f:R→R, \ f(x)=cos(ax)`, where  `a in R text(\{0})`, be a function with the property

`f(x)=f(x+h),` for all  `h in Z`

Let  `g:D rarr R, \ g(x)=log_(2)(f(x))`  be a function where the range of `g` is `[-1,0]`.

A possible interval for `D` is

  1. `[(1)/(4),(5)/(12)]`
  2. `[1,(7)/(6)]`
  3. `[(5)/(3),2]`
  4. `[-(1)/(3),0]`
  5. `[-(1)/(12),(1)/(4)]`
Show Answers Only

`B`

Show Worked Solution

`f(x)=cos(alpha x)=f(x+h)=cos(a(x+h))`

♦♦♦ Mean mark 18%.

`=>a=2pi`

`g(x)=log_(2)(f(x))=log_(2)(f(x+h))=log_(2)(cos(a(x+h)))`

`-1leqlog_(2)(cos(2pi x))leq0`

`(1)/(2)leq cos(2pi x)leq1`

`text(Sketch) \ y=cos(2pi x)\ \ text(by CAS.)`

`text(By inspection of graph,)\ \ (1)/(2)leq cos(2pi x)leq1\ \text{for}\ x in [1,(7)/(6)]`

`=>B`

Graphs, MET2 2020 VCAA 13 MC

The transformation  `T:R^(2)rarrR^(2)`  that maps the graph of  `y=cos(x)`  onto the graph of  `y=cos(2x+4)`  is

  1. `T([[x],[y]])=[[(1)/(2),0],[0,1]]([[x],[y]]+[[-4],[0]])`
  2. `T([[x],[y]])=[[(1)/(2),0],[0,1]][[x],[y]]+[[-4],[0]]`
  3. `T([[x],[y]])=[[(1)/(2),0],[0,1]]([[x],[y]]+[[-2],[0]])`
  4. `T([[x],[y]])=[[2,0],[0,1]]([[x],[y]]+[[2],[0]])`
  5. `T([[x],[y]])=[[2,0],[0,1]][[x],[y]]+[[2],[0]]`
Show Answers Only

`A`

Show Worked Solution

`y=cos(2x+4)=cos(2(x+2))`

♦♦♦ Mean mark 26%.

`text{Dilation of factor} \ 1/2 \ text{from} \ y text{-axis}.`

`text{Translation of 2 units to the left.}`
 

`T([[x],[y]])=[[(1)/(2),0],[0,1]][[x],[y]]+[[-2],[0]]=[[(1)/(2),0],[0,1]]([[x],[y]]+[[-4],[0]])`
 

`=>  A`

Functions, MET2-NHT 2019 VCAA 17 MC

The graph of the function `g` is obtained from the graph of the function `f` with rule  `f(x) = cos(x) - (3)/(8)`  by a dilation of factor  `(4)/(pi)`  from the `y`-axis, a dilation of factor  `(4)/(3)`  from the `x`-axis, a reflection in the  `y`-axis and a translation of  `(3)/(2)`  units in the positive `y` direction, in that order.

The range and period of `g` are respectively

  1. `[–(1)/(3) , (7)/(3)] \ text(and) \ 2`
  2. `[–(1)/(3) , (7)/(3)] \ text(and) \ 8`
  3. `[–(7)/(3) , (1)/(3)] \ text(and) \ 2`
  4. `[–(7)/(3) , (1)/(3)] \ text(and) \ 8`
  5. `[–(4)/(3) , (4)/(3)] \ text(and) \ (pi^2)/(2)`
Show Answers Only

`B`

Show Worked Solution

`text(Dilation of) \ \ (4)/(pi) \ \ text(from) \ y text(-axis:)`

`cos x -(3)/(8) \ => \ cos ((pi x)/(4)) – (3)/(8)`
 
`text(Dilation of) \ \ (4)/(3) \ \ text(from) \ x text(-axis:)`

`cos((pi x)/(4)) – (3)/(8) \ => \ (4)/(3) (cos({pi x}/{4}) -(3)/(8)) = (4)/(3) \ cos((pi x)/(4)) – (1)/(2)`
 
`text(Reflection in) \ y text(-axis and translation up) \ (3)/(2) :`

`(4)/(3) cos((pi x)/(4)) – (1)/(2) \ => \ (4)/(3) \ cos((–pi x)/(4)) + 1`

`:. y = (4)/(3) \ cos ((–pi x)/(4)) + 1`
 
`text(Range:) \ [1 – (4)/(3) , 1 + (4)/(3)] = [–(1)/(3) , (7)/(3)]`
 
`text(Period:) \ (2 pi)/(n) = (pi)/(4) \ => \ n = 8`

`=> \ B`

Graphs, MET2 2008 VCAA 18 MC

Let  `f: [0, pi/2] -> R,\ f(x) = sin(4x) + 1.` The graph of  `f`  is transformed by a reflection in the `x`-axis followed by a dilation of factor 4 from the `y`-axis.

The resulting graph is defined by

  1. `g: [0, pi/2] -> R\ \ \ \ \ \ g(x) = -1 - 4 sin (4x)`
  2. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = -1 - sin (16x)`
  3. `g: [0, pi/2] -> R\ \ \ \ \ \ g(x) = 1 - sin (x)`
  4. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = 1 - sin (4x)`
  5. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = -1 - sin (x)`
Show Answers Only

`E`

Show Worked Solution

`f(x) = sin(4x)+1`

♦♦ Mean mark 36%.

`text(Reflecting in the)\ x text(-axis,)`

`=> h(x) = – sin (4x) – 1`

 

`text(Dilation by a factor of 4 from the)\ \ y text(-axis,)`

`=> g(x) = -sin(x)-1`

`text(Dilation factor: adjust the domain from)\ \ [0, pi/2]`

`text(to)\ \ [0xx4, pi/2 xx 4]=[0,2pi].`

`=>   E`

Graphs, MET2 2009 VCAA 12 MC

A transformation  `T: R^2 -> R^2`  that maps the curve with equation  `y = sin (x)`  onto the curve with equation  `y = 1 - 3 sin(2x + pi)`  is given by

  1. `T [(x), (y)] = [(2, 0), (0, -3)] [(x), (y)] + [(pi), (1)]`
  2. `T [(x), (y)] = [(– 1/2, 0), (0, 3)] [(x), (y)] + [(pi/2), (1)]`
  3. `T [(x), (y)] = [(0, – 3), (2, 0)] [(x), (y)] - [(pi), (1)]`
  4. `T [(x), (y)] = [(1/2, 0), (0, – 3)] [(x), (y)] + [(– pi/2), (1)]`
  5. `T [(x), (y)] = [(1/2, 0), (0, 3)] [(x), (y)] + [(– pi/2), (– 1)]`
Show Answers Only

`D`

Show Worked Solution

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

`text(Let)\ \ g(x) = – 3 sin (2 (x + pi/2)) + 1`

`text(Find transformations taking)\ \ f -> g`

 

`3 f (2x) = 3 sin (2x) = h(x)`

`– h(x) = – 3 sin (2x) = k(x)`

`k (x + pi/2) + 1 = – 3 sin (2 (x + pi/2)) + 1 = g(x)`

 

`text(Dilate by factor 3 from)\ \ x text(-axis)`

`text(Dilate by factor)\ \ 1/2\ \ text(from)\ \ y text(-axis)`

`text(Reflect in)\ \ x text(-axis)`

`text(Translate left)\ \ pi/2\ \ text(up 1)`

`=>   D`

Calculus, MET2 2016 VCAA 1

Let  `f: [0, 8 pi] -> R, \ f(x) = 2 cos (x/2) + pi`.

  1. Find the period and range of `f`.   (2 marks)

  2. State the rule for the derivative function `f^{′}`.   (1 mark)

  3. Find the equation of the tangent to the graph of `f` at  `x = pi`.   (1 mark)

  4. Find the equations of the tangents to the graph of  `f: [0, 8 pi] -> R,\ \ f(x) = 2 cos (x/2) + pi`  that have a gradient of 1.   (2 marks)

  5. The rule of  `f^{′}` can be obtained from the rule of `f` under a transformation `T`, such that
      
    `qquad T: R^2 -> R^2,\ T([(x), (y)]) = [(1, 0), (0, a)] [(x), (y)] + [(−pi), (b)]`
     

     

    Find the value of `a` and the value of `b`.   (3 marks)

  6. Find the values of  `x, \ 0 <= x <= 8 pi`, such that  `f(x) = 2 f^{′} (x) + pi`.   (2 marks)

Show Answers Only
  1. `text(Period:)\ 4 pi; qquad text(Range:)\ [pi-2, pi + 2]`
  2. `f^{′} (x) =-sin (x/2)`
  3. `y =-x + 2 pi`
  4. `y = x-2 pi and y = x-6 pi`
  5. `a = 1/2 and b =-pi/2`
  6. `x = (3 pi)/2, (7 pi)/2, (11 pi)/2, (15 pi)/2`
Show Worked Solution

a.   `text(Period)= (2pi)/n = (2 pi)/(1/2) = 4pi`

MARKER’S COMMENT: Including round brackets rather than square ones was a common mistake.

`text(Range:)\ [pi-2, pi + 2]`
  

b.   `f^{′} (x) = text(−sin) (x/2)`
 

c.   `[text(CAS: tangentLine)\ (f(x), x, pi)]`

`y = -x + 2 pi`
 

d.   `text(Solve)\ \ f^{′} (x) = 1\ \ text(for)\ x in [0, 8 pi]`

♦ Mean mark part (d) 50%.

`-> x = 3 pi or 7 pi`

`:. y = x-2 pi and y = x-6 pi\ \ [text(CAS)]`

 

e.   `text(Using the transition matrix,)`

♦♦ Mean mark part (e) 27%.
`x_T` `=x-pi`
`x` `=x_T+pi`
`y_T` `=ay+b`
`y` `=(y_T-b)/a`
   

`f(x)= cos (x/2) + pi/2\ \ ->\ \ f{′}(x) = -sin(x/2)`

`(y_T-b)/a` `=2cos((x_T+pi)/2)+pi`
`y_T` `=2a cos((x_T+pi)/2)+a pi +b`
  `=-2a sin(x_T/2)+a pi + bqquad [text(Complementary Angles)]`
   
`-2a` `=-1`
`:. a` `=1/2`
`1/2 pi +b` `=0`
`:.b` `=-pi/2`

 

f.   `text(Solve)\ \ f(x) = 2 f^{′} (x) + pi\ \ text(for)\ \ x in [0, 8 pi]`

♦ Mean mark part (f) 50%.
`2 cos (x/2) + pi` `= -2 sin(x/2)+pi`
`tan(x/2)` `=-1`
`x/2` `=(3pi)/4, (7pi)/4, (11pi)/4, (15pi)/4`
`:.x` `= (3 pi)/2, (7 pi)/2, (11 pi)/2, (15 pi)/2`

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.`

  1. 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^{′}, P^{′}` and `C^{′}` be the image, under this transformation, of the points `X, P` and `C` respectively. 

     

    1. Find the values of `k` and `m`  if  `X^{′}P^{′} = 10`  and  `X^{′} C^{′} = 30.`   (2 marks)

    2. Find the coordinates of the point `P^{′}.`   (1 mark)

Show Answers Only
  1. `t = 0, or 16\ text(h)`
  2. `16\ text(hours)`
  3. `8/3`
  4. `8\ text(hours)`
  5.  i.  `-1/2`
    ii.  `text(See worked solution)`
  6.  i.  `k=(20sqrt3)/3, m=10sqrt3`
    ii.  `P^{′}((20pisqrt3)/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.   `X^{′} ((2pi)/3 m,0)qquadP^{′}((2pi)/3 m, sqrt3/2 k)qquadC^{′} ((sqrt3 + (2pi)/3)m, 0)`

`X^{′}P^{′}` `= 10`
`sqrt3/2 k` `= 10`
`:. k` `= 20/sqrt3`
  `=(20sqrt3)/3`
♦♦♦ Mean mark part (f)(i) 14%.

 

`X^{′}C^{′}=30`

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

 

f.ii.    `P^{′}((2pi)/3 m, sqrt3/2 k)` `= P^{′}((2pi)/3 xx 10sqrt3, sqrt3/2 xx 20/sqrt3)`
    `= P^{′}((20pisqrt3)/3,10)`