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Functions, MET2 2023 VCAA 9 MC

The function \(f\) is given by
 

\(f(x) = \begin {cases}
\tan\Bigg(\dfrac{x}{2}\Bigg)         &\ \ 4 \leq x \leq 2\pi \\
\sin(ax) &\ \ \ 2\pi\leq x\leq 8
\end{cases}\)

 
The value of \(a\) for which \(f\) is continuous and smooth at  \( x\) = \(2\pi\)  is

  1. \(-2\)
  2. \(-\dfrac{\pi}{2}\)
  3. \(-\dfrac{1}{2}\)
  4. \(\dfrac{1}{2}\)
  5. \(2\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Solve for}\ a\ \text{given}\ \ x=2\pi:\)

\(\tan\Bigg(\dfrac{2\pi}{2}\Bigg)=\sin(a2\pi)=0\)

\(a=\pm \dfrac{1}{2}\)
 

\(\text{For smoothness, solve for}\ a\ \text{given}\ \ x=2\pi:\)

\(\dfrac{d}{dx}\tan\Bigg(\dfrac{x}{2}\Bigg)=\dfrac{d}{dx}\sin(ax)\)

\(\therefore a=-\dfrac{1}{2}\)

 
\(\Rightarrow C\)


♦ Mean mark 46%.

Calculus, MET1 2021 VCAA 8

The gradient of a function is given by  `(dy)/(dx) = sqrt(x + 6) - x/2 - 3/2`.

The graph of the function has a single stationary point at  `(3, 29/4)`.

  1. Find the rule of the function.  (3 marks)
  2. Determine the nature of the stationary point.  (2 marks)
Show Answers Only
  1. `y=2/3(x+6)^(3/2)-x^2/4-3/2x-4`
  2. `text{Maximum}`
Show Worked Solution
a.    `dy/dx` `= sqrt(x + 6) – x/2 – 3/2`
  `y` `=int sqrt(x + 6) – x/2 – 3/2\ dx`
    `=2/3(x+6)^(3/2)-x^2/4-3/2x + c`

 
`text{Graph passes through}\ (3, 29/4)`

`29/4` `=2/3*9^(3/2) – 3^2/4 – 3/2 * 3+c`  
`29/4` `=18-9/4-9/2+c`  
`c` `=29/4-18+9/4+9/2`  
  `=-4`  

 
`:. y=2/3(x+6)^(3/2)-x^2/4-3/2x-4`

♦ Mean mark part (b) 37%.

b. 
      

`:.(3,29/4)\ text{is a maximum}.`

Graphs, MET2 2015 VCAA 7 MC

The range of the function  `f:\ text{(−1, 2]} -> R,\ \ f(x) = -x^2 + 2x-3` is

  1. `R`
  2. `text{(−6, −3]}`
  3. `text{(−6, −2]}`
  4. `text{[−6, −3]}`
  5. `text{[−6, −2]}`
Show Answers Only

`C`

Show Worked Solution

`text(A sketch of the equation:)`

`y` `=-x^2 + 2x-3`
 `dy/dx` `=-2x+2`

 

`=>\ text(Concave down with turning point when)\ \ x=1,`
 

vcaa-2015-7mc-answer1
 

`:. text(Range) = (-6,-2]`

`=>   C`