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Functions, MET2 2020 VCAA 12 MC

A clock has a minute hand that is 10 cm long and a clock face with a radius of 15 cm, as shown below.
 


 

At 12.00 noon, both hands of the clock point vertically upwards and the tip of the minute hand is at its maximum distance above the base of the clock face.

The height, `h` centimetres, of the tip of the minute hand above the base of the clock face `t` minutes after 12.00 noon is given by

  1. `h(t)=15+10 sin((pi t)/(30))`
  2. `h(t)=15-10 sin((pi t)/(30))`
  3. `h(t)=15+10 sin((pi t)/(60))`
  4. `h(t)=15+10 cos((pi t)/(60))`
  5. `h(t)=15+10 cos((pi t)/(30))`
Show Answers Only

`E`

Show Worked Solution

`text(By elimination)`

♦ Mean mark 45%.

`text(At)\ \ t=0,\ text(height = 25 cm)`

`=>\ text(Eliminate)\ A, B, C`
 

`text(Period = 60)`

`(2pi)/n` `=60`  
`60n` `=2pi`  
`n` `=pi/30`  

 
`=>\ text(Eliminate)\ D`

`=>E`

Filed Under: Trig Equations Tagged With: Band 5, smc-725-45-Unknown trig ratio, smc-725-60-Max/min (non-calc)

Functions, MET1 2007 VCAA 8

Let  `f: R -> R`,  `f(x) = sin((2pix)/3)`.

  1. Solve the equation  `sin((2pix)/3) = -sqrt3/2`  for  ` x ∈ [0,3]`.   (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

  2. Let  `g: R -> R`,  `g(x) = 3f(x-1) + 2`.
  3. Find the smallest positive value of `x` for which `g(x)` is a maximum.   (2 marks)

    --- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a.   `x = 2, 5/2`

b.   `7/4`

Show Worked Solution

a.    `sin((2pix)/3) = -sqrt3/2`

`=>\ text(Base angle)\ = pi/3`

`(2 pi x)/3` `=(4pi)/3, (5pi)/3, (10pi)/3, …` 
`:.x` `=2 or 5/2, \ \ \ text(for)\ x ∈ [0,3]`

  

b.   `g(x) = 3sin[(2pi)/3 (x-1)] + 2`

♦♦ Mean mark 29%.
STRATEGY: Max/min questions involving trig functions can often use the powerful identity, `-1 <= sin theta <=1` to solve efficiently and reduce errors.

`text(Maximum occurs when:)`

`sin[(2pi)/3 (x-1)]` `= 1`
`(2pi)/3 (x-1)` `= pi/2`
`x-1` `= pi/2 xx 3/(2pi)`
`:. x` `=7/4`

Filed Under: Trig Equations Tagged With: Band 4, Band 5, smc-725-10-Sin, smc-725-60-Max/min (non-calc)

Functions, MET1 2015 VCAA 5

On any given day, the depth of water in a river is modelled by the function

`h(t) = 14 + 8sin((pit)/12),\ \ 0 <= t <= 24`

where `h` is the depth of water, in metres, and  `t`  is the time, in hours, after 6 am. 

  1. Find the minimum depth of the water in the river.   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  2. Find the values of  `t`  for which  `h(t) = 10`.   (2 marks)

    --- 6 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `6\ text(m)`
  2. `14quadtext(or)quad22`
Show Worked Solution

a.   `h_(text(min))\ text(occurs when)\ \ sin((pit)/12)=-1`

MARKER’S COMMENT: Students who used calculus to find the minimum were less successful.
`:. h_(text(min))` `= 14-8`
  `= 6\ text(m)`

 

b.    `14 + 8sin(pi/12t)` `= 10`
  `sin(pi/12t)` `=-1/2`

 

`text(Solve in general:)`

`pi/12t` `=(7pi)/6 + 2pi n\ \ \ \ text(or)\ \ \ `  `pi/12t` `= (11t)/6 + 2pi n,`
`t` `= 14 + 24n` `t` `=22 + 24n`

 

`text(Substitute integer values for)\ n,`

`:. t = 14quadtext(or)quad22,\ \ \ (t ∈ [0,24])`

Filed Under: Trig Equations Tagged With: Band 4, smc-725-10-Sin, smc-725-60-Max/min (non-calc)

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