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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))`
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`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`

Algebra, MET2 2012 VCAA 19 MC

A function  `f` has the following two properties for all real values of  `theta.`

`f(pi - theta) = -f(theta)`  and  `f(pi - theta) = -f(- theta)`

A possible rule for  `f` is

  1. `f(x) = sin (x)`
  2. `f(x) = cos (x)`
  3. `f(x) = tan (x)`
  4. `f(x) = sin (x/2)`
  5. `f(x) = tan (2x)`
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`B`

Show Worked Solution

`text(Solution 1:)`

♦ Mean mark 45%.

`text(By elimination,)`

`f(pi – theta) = -f(theta)`

`text(Opposite signs in the 1st and 2nd quadrant)`

`:.\ text(Cannot be)\ sin\ \ text{(Eliminate}\ A\ text(and)\ D)`

 

`f(pi – theta) = -f(- theta)`

`text(Opposite signs in the 2nd and 4th quadrant)`

`:.\ text(Cannot be)\ tan\ \ text{(Eliminate}\ C\ text(and)\ E)`

`=> B`

 

`text(Solution 2:)`

`text(The given conditions imply the function is even.)`

`cos(x)\ \ text(is the only even function given.)`

`=>   B`