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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.

Calculus, MET1 2023 VCAA 5

  1. Evaluate  \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\).   (1 mark)
  2. Hence, or otherwise, find all values of \(k\) such that \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx=\displaystyle \int_{0}^{\frac{\pi}{2}} \cos(x)\,dx\), where \(-3\pi<k<2\pi\).   (3 marks)
Show Answers Only

a.    \(\dfrac{1}{2}\)

b.    \(k=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Show Worked Solution
a.    \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\) \(=\left[-\cos x\right]_0^\frac{\pi}{3}\)
    \(=-\cos\dfrac{\pi}{3}+\cos 0\)
    \(=-\dfrac{1}{2}+1\)
    \(=\dfrac{1}{2}\)

 

b.    \(\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\) \(=\left[\sin x\right]_k^\frac{\pi}{2}\)
    \(=\sin\bigg(\dfrac{\pi}{2}\bigg)-\sin (k)\)
    \(=1-\sin (k)\)

 
\(\text{Using part (a):}\)

\(1-\sin (k)\) \(=\dfrac{1}{2}\)
\(\sin (k)\) \(=\dfrac{1}{2}\)
\(\therefore\ k\) \(=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Calculus, MET2 2019 VCAA 4 MC

`int_0^(pi/6) (a sin (x) + b cos(x))\ dx`  is equal to

  1. `((2 - sqrt 3)a - b)/2`
  2. `(b - (2 - sqrt 3) a)/2`
  3. `((2 - sqrt 3)a + b)/2`
  4. `((2 - sqrt 3) b - a)/2`
  5. `((2 - sqrt 3) b + a)/2`
Show Answers Only

`C`

Show Worked Solution

`int_0^(pi/6) (a sin (x) + b cos(x))\ dx`

`= [-a cos(x) + b sin(x)]_0^(pi/6)`

`= [-a ⋅ sqrt 3/2 + b/2 – (-a + 0)]`

`= (2a – sqrt 3 a + b)/2`

`= ((2 – sqrt 3) a + b)/2`

`=>   C`

Calculus, MET1 2014 VCAA 7

If  `f^{prime}(x) = 2cos(x)-sin(2x)`  and  `f(pi/2) = 1/2`,  find  `f(x)`.   (3 marks)

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`2sinx + 1/2cos(2x)-1`

Show Worked Solution
`f(x)` `= int(2cosx-sin2x)dx`
  `= 2sinx + 1/2cos(2x) + c`

  
`text(Substitute)\ \ f(pi/2) = 1/2:`

`1/2` `= 2sin(pi/2) + 1/2cos(pi) + c`
`1/2` `= 2-1/2 + c`
`c` `=-1`
`:. f(x)` `= 2sinx + 1/2cos(2x)-1`