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Calculus, MET2 2022 VCAA 5

Consider the composite function `g(x)=f(\sin (2 x))`, where the function `f(x)` is an unknown but differentiable function for all values of `x`.

Use the following table of values for `f` and `f^{\prime}`.

`\quad x \quad` `\quad\quad 1/2\quad\quad` `\quad\quad(sqrt{2})/2\quad\quad` `\quad\quad(sqrt{3})/2\quad\quad`
`f(x)` `-2` `5` `3`
`\quad\quad f^{prime}(x)\quad\quad` `7` `0` `1/9`

 

  1. Find the value of `g\left(\frac{\pi}{6}\right)`.   (1 mark)

The derivative of `g` with respect to `x` is given by `g^{\prime}(x)=2 \cdot \cos (2 x) \cdot f^{\prime}(\sin (2 x))`.

  1. Show that `g^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{9}`.   (1 mark)

  2. Find the equation of the tangent to `g` at `x=\frac{\pi}{6}`.   (2 marks)

  3. Find the average value of the derivative function `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`.   (2 marks)

  4. Find four solutions to the equation `g^{\prime}(x)=0` for the interval `x \in[0, \pi]`.   (3 marks)

Show Answers Only

a.    `3`

b.    `1/9`

c.    `y=1/9x+3-pi/54`

d.    `-48/pi`

e.    ` x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`

Show Worked Solution
 

a.  `g(pi/6)`
 `= f(sin(pi/3))`  
  `= f(sqrt3/2)`  
  `= 3`  

 

b.  `g\ ^{prime}(x)` `= 2\cdot\ cos(pi/3)\cdot\ f\ ^{prime}(sin(pi/3))`  
`g\ ^{prime}(pi/6)` `= 2 xx 1/2 xx f\ ^{prime}(sqrt3/2)`  
  `= 1/9`  

 

c.   `m = 1/9`  and  `g(pi/6) = 3`

`y  –  y_1` `= m(x-x_1)`  
`y  –  3` `= 1/9(x-pi/6)`  
`y` `= 1/9x + 3-pi/54`  

♦♦ Mean mark (c) 45%.
MARKER’S COMMENT: Some students did not produce an equation as required.

  
d.   The average value of `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`

Average `= \frac{1}{\frac{\pi}{6}-\frac{\pi}{8}}\cdot\int_{\frac{\pi}{8}}^{\frac{\pi}{6}} g^{\prime}(x) d x`  
  `=24/pi \cdot[g(x)]_{\frac{\pi}{8}}^{\frac{\pi}{\6}}`  
  `= 24/pi \cdot(f(sqrt3/2)-f(sqrt2/2))`  
  `= 24/pi (3-5) = -48/pi`  

♦♦ Mean mark (d) 30%.
MARKER’S COMMENT: Those who used the Average Value formula were generally successful.
Some students substituted `g^{\prime}(x)`, not `g(x)`.

e.   `2 \cos (2 x) f^{\prime}(\sin (2 x)) = 0`

`:.\   2 \cos (2 x) = 0\ ….(1)`  or  ` f^{\prime}(\sin (2 x)) = 0\ ….(2)`

(1):   ` 2 \cos (2 x)`  `= 0`      `x \in[0, \pi]`
`\cos (2 x)` `= 0`      `2 x \in[0,2 \pi]`
`2x` `= pi/2 , (3pi)/2`  
`x` `= pi/4 , (3pi)/4`  
     
(2): ` f^{\prime}(\sin (2 x)) ` `= sqrt2/2`  
`2x` `= pi/4 , (3pi)/4`  
`x` `= pi/8 , (3pi)/8`  

  
`:. \  x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`


♦♦ Mean mark (e) 30%.
MARKER’S COMMENT: Some students were able to find `pi/4, (3pi)/4`. Some solved `2 cos(2x)=0` or `f^{\prime}(sin(2x))=0` but not both.

Functions, MET1 2021 VCAA 3

Consider the function  `g: R -> R, \ g(x) = 2sin(2x).`

  1. State the range of `g`.   (1 mark)

  2. State the period of `g`.   (1 mark)

  3. Solve  `2 sin(2x) = sqrt3`  for  `x ∈ R`.   (3 marks)

Show Answers Only
  1. `[-2,2]`
  2. `pi`
  3. `x= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`
Show Worked Solution

a.   `text(S)text(ince)  -1<sin(2x)<1,`

`text(Range)\  g(x) = [-2,2]`
 

b.   `text(Period) = (2pi)/n = (2pi)/2 = pi`
 

c.    `2sin(2x)` `=sqrt3`
  `sin(2x)` `=sqrt3/2`
  `2x` `=pi/3, (2pi)/3, pi/3 + 2pi, (2pi)/3 + 2pi, …`
  `x` `=pi/6, pi/3, pi/6+pi, pi/3+pi, …`

 
`:.\ text(General solution)`

`= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`

Functions, MET1 2011 VCAA 3b

Solve the equation

`qquad sin (2x + pi/3) = 1/2\ \ text(for)\ \ x in [0, pi].`   (2 marks)

Show Answers Only

`x = pi/4, (11 pi)/12`

Show Worked Solution

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

`=>\ text(Base angle is)\ \ pi/6`

`(2x + pi/3)` `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, …`
`2x` `= -pi/6, pi/2, (11pi)/6, (15pi)/6, …`
`x` `= -pi/12, pi/4, (11pi)/12, (15pi)/12, …`

 

`:. x = pi/4, (11 pi)/12,\ \ x in [0, pi]`

Algebra, MET2 2017 VCAA 12 MC

The sum of the solutions of  `sin(2x) = (sqrt3)/2`  over the interval `[−pi,d]`  is  `−pi`.

The value of `d` could be

  1. `0`
  2. `pi/6`
  3. `(3pi)/4`
  4. `(7pi)/6`
  5. `(3pi)/2`
Show Answers Only

`C`

Show Worked Solution
`sin(2x)` `=sqrt3/2`
`2x` `=pi/3, (2pi)/3, ..`
`x` `=pi/6, pi/3, ..`

 

`text(Expand and test possible solutions to solve:)`

♦ Mean mark 45%.

`x=-(5pi)/6, -(2pi)/3, pi/6, pi/3, ..`

`-(5pi)/6+ -(2pi)/3+ pi/6+ pi/3=-pi`

 

`=> C`

Algebra, MET2 2009 VCAA 4 MC

The general solution to the equation  `sin (2x) = -1`  is

  1. `x = n pi - pi/4,\ n in Z`
  2. `x = 2n pi + pi/4 or x = 2n pi - pi/4,\ n in Z`
  3. `x = (n pi)/2 + (-1)^n pi/2,\ n in Z`
  4. `x = (n pi)/2 + (-1)^n pi/4,\ n in Z`
  5. `x = n pi + pi/4 or x = 2n pi + pi/4,\ n in Z`
Show Answers Only

`A`

Show Worked Solution
`2x` `= 2n pi – pi/2,\ \ n in Z`
`x` `= n pi – pi/4,\ \ n in Z`

 
`=>   A`

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)

  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)

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`

Functions, MET1 2013 VCAA 4

Solve the equation  `sin (x/2) =-1/2`  for  `x in [2 pi, 4 pi].`   (2 marks)

Show Answers Only

`x = (7 pi)/3, (11 pi)/3`

Show Worked Solution

`sin (x/2) = -1/2`

`=>\ text(Base angle is)\ \ pi/6`

`x/2` `=pi/6 + pi, 2pi-pi/6, 2pi + (pi/6 +pi), …`
  `=(7pi)/6, (11pi)/6, (19pi)/6, …`
`:. x` `=(7pi)/3, (11pi)/3, (19pi)/3, …`
   

`text(Given)\ \ x in [2 pi, 4 pi],`

`:. x = (7 pi)/3, (11 pi)/3`

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

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])`