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Calculus, SPEC2 2022 VCAA 1

Consider the family of functions \(f\) with rule  \(f(x)=\dfrac{x^2}{x-k}\), where \(k \in R \backslash\{0\}\).

  1. Write down the equations of the two asymptotes of the graph of \(f\) when \(k=1\).   (2 marks)

  2. Sketch the graph of  \(y=f(x)\)  for  \(k=1\)  on the set of axes below. Clearly label any turning points with their coordinates and label any asymptotes with their equations.   (3 marks)
     

  1.  i. Find, in terms of \(k\), the equations of the asymptotes of the graph of  \(f(x)=\dfrac{x^2}{x-k}\).   (1 mark)

  2. ii. Find the distance between the two turning points of the graph of  \(f(x)=\dfrac{x^2}{x-k}\) in terms of \(k\).   (2 marks)

  3. Now consider the functions \(h\) and \(g\), where  \(h(x)=x+3\)  and  \(g(x)=\abs{\dfrac{x^2}{x-1}}\).
  4. The region bounded by the curves of \(h\) and \(g\) is rotated about the \(x\)-axis.
    1. Write down the definite integral that can be used to find the volume of the resulting solid.   (2 marks)

    2. Hence, find the volume of this solid. Give your answer correct to two decimal places.   (1 mark)

Show Answers Only

a.  \(\text {Asymptotes: } x=1,\  y=x+1\)

b.   
       

c.i.   \(\text {Asymptotes: } x=k,\  y=x+k\)

c.ii.  \(\text {Distance }=2 \sqrt{5}|k|\)

d.i.  \(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)

d.ii.  \(V=51.42\ \text{u}^3 \)

Show Worked Solution

a.    \(\text {When } k=1 :\)

\(f(x)=\dfrac{x^2}{x-1}=\dfrac{(x+1)(x-1)+1}{(x-1)}=x+1+\dfrac{1}{x-1}\)

\(\text {Asymptotes: } x=1,\  y=x+1\)
 

b.    
       

 

c.i. \(f(x)=\dfrac{x^2}{x-k}=\dfrac{(x+k)(x-k)+k^2}{x-k}=x+k+\dfrac{k^2}{x-k}\)

\(\text {Using part a.}\)

\(\text {Asymptotes: } x=k,\  y=x+k\)
 

c.ii.  \(f^{\prime}(x)=1-\left(\dfrac{k}{x-k}\right)^2\)

\(\text {TP’s when } f^{\prime}(x)=0 \text { (by CAS):}\)

\(\Rightarrow(2 k, 4 k),(0,0)\)

\(\text {Distance }\displaystyle=\sqrt{(2 k-0)^2+(4 k-0)^2}=\sqrt{20 k^2}=2 \sqrt{5}|k|\)
 

d.i  \(\text {Solve for intersection of graphs (by CAS):}\)

\(\displaystyle x+3=\left|\frac{x^2}{x-1}\right|\)

\(\displaystyle \Rightarrow x=\frac{3}{2}, x=\frac{-1 \pm \sqrt{7}}{2}\)

\(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)
 

d.ii. \(V=51.42\ \text{u}^3 \text{ (by CAS) }\)

♦♦ Mean mark (d)(ii) 37%.

Graphs, SPEC2 2022 VCAA 1 MC

For the interval  `\frac{1}{2} \le x \le3`, the graph of  `y=|2 x-1|-|x-3|`  is the same as the graph of

  1. `y=-x-2`
  2. `y=3x-4`
  3. `y=x+2`
  4. `y=3x+2`
  5. `y=x-4`
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`B`

Show Worked Solution

`text{Consider the graph (CAS):}`

`text{Interval:}\ \ \frac{1}{2} \le x \le3`

`text{Endpoints:}\ \ (1/2 , -2\frac{1}{2}),\ \ (3 , 5)`

`m = frac{7\frac{1}{2}}{2\frac{1}{2}} = 3`

`y-y_1` `= m(x-x_1)`  
`y-5` `= 3(x-3)`  
`y` `= 3x-4`  

 
`=>B`

Algebra, SPEC1 2020 VCAA 4

Solve the inequality  `3 - x > 1/|x - 4|`  for `x`, expressing your answer in interval notation.  (4 marks)

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`x ∈ (– oo, (7 – sqrt 5)/2)`

Show Worked Solution

`3 – x > 1/|x – 4|`

♦ Mean mark 46%.

`|x – 4| (3 – x) > 1`
 

`text(If)\ \ x – 4 > 0, x > 4`

`(x – 4) (3 – x)` `> 1`
`3x – x^2 – 12 + 4x` `> 1`
`-x^2 + 7x – 13` `> 0`

 
`Delta = 7^2 – 4 ⋅ 1 ⋅ 13 = -3 < 0`

`=>\ text(No Solutions)`
 

`text(If)\ \ x – 4 < 0, x < 4`

`-(x – 4) (3 – x)` `> 1`
`x^2 – 7x + 12` `> 1`
`x^2 – 7x + 11` `> 0`
`x` `= (7 +- sqrt(7^2 – 4 ⋅ 1 ⋅ 11))/2`
  `= (7 +- sqrt 5)/2`

`text(Combining solutions)`

`(x < (7 – sqrt 5)/2  ∪ x > (7 + sqrt 5)/2)  nn x < 4`

`x ∈ (– oo, (7 – sqrt 5)/2)`