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Calculus, SPEC2 2024 VCAA 1

Consider the function \(f\) with rule  \(f(x)=\dfrac{x^4-x^2+1}{1-x^2}\).

  1. Sketch the graph of  \(y=f(x)\)  on the set of axes below. Label the vertical asymptotes with their equations and label the stationary points with their coordinates.   (3 marks)
     


  1. The region bounded by the graph of  \(y=f(x)\)  and the lines  \(y=1\)  and  \(y=6\)  is rotated about the \(y\)-axis to form a solid of revolution.
    1. Write down a definite integral involving only the variable \(y\), that when evaluated, will give the volume of the solid.   (2 marks)

    2. Find the volume of the solid, correct to one decimal place.   (1 mark)

  1. Now consider the function \(g\) with rule  \(g(x)=\dfrac{x^4+b}{1-x^2}\), where \(b \in R\).  
  2. For what value of \(b\) will the graph of \(g\) have no asymptotes?    (1 mark)

  3. The gradient function of \(g\) is given by  \(g^{\prime}(x)=\dfrac{-2 x\left(\left(x^2-1\right)^2-(b+1)\right)}{\left(1-x^2\right)^2}\).
  4. For what values of \(b\) will the graph of \(g\) have exactly
    1. one stationary point?   (1 mark)

    2. three stationary points?   (1 mark)

    3. five stationary points?   (1 mark)

Show Answers Only

a.   

b.i.   \(V=\pi \displaystyle\int_1^6 \frac{1-y+\sqrt{y^2+2y-3}}{2} \, dy\)

b.ii.  \(V=11.2\ \text{u}^3\)

c.   \(b=-1\)

d.i.   \(b \leqslant-1\)

d.ii.  \(b \geqslant 0\)

d.iii. \(-1<b<0\)

Show Worked Solution

a.   \(\text{Using CAS (set domain, range to match image):}\)

\(1-x^2 \neq 0 \ \Rightarrow \ \text {Vertical asymptotes at}\ \  x= \pm 1\)
 

b.i    \(V\) \(=\pi \displaystyle \int_1^6 x^2 \ dy\)
    \(=\pi \displaystyle\int_1^6 \frac{1-y+\sqrt{y^2+2y-3}}{2} \, dy \ \ \text{(by CAS)}\)

 
b.ii.  \(V=11.2\ \text{u}^3 \ \text{(1 d.p.)}\)
 

c.    \(g(x)=\dfrac{x^4+b}{1-x^2}=-x^2-1+\dfrac{b+1}{1-x^2} \ \ \text{(by polynomial division)}\)

\(\text{No asymptotes when}\ \ b+1=0\ \ \Rightarrow\ \ b=-1\)
  

d.i. \(\text{Since  \(g^{\prime}(0)=0\)  provides 1 SP, no solutions are required for}\)

\(\dfrac{(x^2-1)^2-(b+1)}{(1+x^2)^2}=0\)

\(b+1<0 \ \Rightarrow \ b<-1\)

\(\text{Consider} \ \ b=-1:\)

\(\dfrac{\left(x^2-1\right)^2-0}{\left(1-x^2\right)^2}=1 \neq 0 \ \text{(no solution)}\)

\(\therefore b \leqslant-1\)

♦♦♦ Mean mark (d.i.) 27%.
♦♦♦ Mean mark (d.ii.) 27%.
♦♦♦ Mean mark (d.iii.) 25%.
 

d.ii.  \(\text{3 SPs:} \ \left(x^2-1\right)^2=b+1 \ \ \text{has two non-zero, real solutions}\)

\(x^2-1\) \(= \pm \sqrt{b+1}\)
\(x\) \(=\pm \sqrt{1 \pm \sqrt{b+1}}\)

  

 \(\Rightarrow \sqrt{b+1} \geqslant 1 \ \text{for 2 solution}\)

\(\Rightarrow b \geqslant 0\)
 

d.iii  \(\text{5 SPs:} \ \left(x^2-1\right)^2 = b+1 \ \text{has four non-zero real solutions}\)

\(x=\pm\sqrt{1 \pm \sqrt{b+1}} \ \ \text{has 4 solutions if}\)

\(b+1 >\) \(0\) \(\text{and}\)    \(\sqrt{b+1}<1\)
\(b >\) \( -1\)   \(b<0\)

 
\(\therefore -1<b<0\)

Calculus, SPEC1 2024 VCAA 3

Let  \(f: R \backslash\{-1\} \rightarrow R, f(x)=\dfrac{(x-1)^2}{(x+1)^2}\)

The rule \(f(x)\) can be written in the form  \(f(x)=A+\dfrac{B}{x+1}+\dfrac{C}{(x+1)^2}\),  where \(A, B, C \in Z\).

  1. Show that  \(A=1, B=-4\)  and  \(C=4\).   (1 mark)

  2. The graph of \(f\) has one turning point.  
  3. Find the coordinates of this turning point.  (2 marks)

  4. Sketch the graph of  \(y=f(x)\)  on the set of axes below. Label the asymptotes with their equations and the axial intercepts with their coordinates.   (3 marks)
     

Show Answers Only

a.   \(\text{See worked solutions.}\)

b.   \(\text{Turning point at}\ (1,0)\)

c.   
       

Show Worked Solution

a.    \(\dfrac{(x-1)^2}{(x+1)^2}\) \(=\dfrac{x^2-2x+1}{x^2+2x+1}\)
    \(=\dfrac{x^2+2x+1}{x^2+2x+1}-\dfrac{4x}{x^2+2x+1}\)
    \(=1-\dfrac{4(x+1)}{(x+1)^2}+\dfrac{4}{(x+1)^2}\)
    \(=1-\dfrac{4}{(x+1)}+\dfrac{4}{(x+1)^2}\)

 

b.   \(f^{′}(x)=\dfrac{4}{(x+1)^2}-\dfrac{8}{(x+1)^3}\)

\(\text{SP’s when}\ \ f^{′}(x)=0:\)

\(\dfrac{4}{(x+1)^2}\) \(=\dfrac{8}{(x+1)^3}\)  
\(4(x+1)\) \(=8\)  
\(x\) \(=1\)  

 
\(f(1)=0\)

\(\therefore \text{Turning point at}\ (1,0)\)
 

c.    \(\text{Vertical asymptote at}\ \ x=-1\)

\(f(0)=\dfrac{(-1)^2}{(1)^2}=1\ \ \Rightarrow \ \ y\text{-intercept at}\ (0,1)\)

\(\text{As}\ x\rightarrow \infty, \ 1-\dfrac{4}{(x+1)}+\dfrac{4}{(x+1)^2} \rightarrow 1^{-} \)

\(\text{As}\ x\rightarrow -\infty, \ 1-\dfrac{4}{(x+1)}+\dfrac{4}{(x+1)^2} \rightarrow 1^{+} \)

\(\text{Horizontal asymptote at}\ \ y=1 \)

♦♦ Mean mark (c) 31%.
 
        

Calculus, SPEC2 2021 VCAA 1

Let  `f(x) = ((2x-3)(x + 5))/((x-1)(x + 2))`.

  1. Express `f(x)` in the form  `A + (Bx + C)/((x-1)(x + 2))`, where `A`, `B` an `C` are real constants.   (1 mark)

  2. State the equation of the asymptotes of the graph of `f`.   (2 marks)

  3. Sketch the graph of `f` on the set of axes below. Label the asymptotes with their equations, and label the maximum turning point and the point of inflection with their coordinates, correct to two decimal places. Label the intercepts with the coordinate axes.   (3 marks)

     

     

  4. Let  `g_k(x) = ((2x-3)(x + 5))/((x-k)(x + 2))`, where `k` is a real constant.
  5.  i. For what values of `k` will the graph of `g_k`, have two asymptotes?   (2 marks)

  6. ii. Given that the graph of `g_k` has more than two asymptotes, for what values of `k` will the graph of `g_k` have no stationary points?   (2 marks)

Show Answers Only
  1. `f(x) = 2 + (5x-11)/((x-1)(x + 2))`
  2. `text(Horizontal asymptote:)\ y = 2`

  3.  
  4.  i. `k = 3/2 => g_k(x) = 2 + 6/(x + 2)`
  5. ii. `k , -5\ \ text(or)\ \ k > 3/2`
Show Worked Solution

a.   `text{By CAS  (prop Frac}\ f(x)):`

`f(x) = 2 + (5x-11)/((x-1)(x + 2))`
 

b.   `text(Vertical asymptotes:)\ x = 1, x = –2`

`text(As)\ \ x -> ∞, y -> 2`

`text(Horizontal asymptote:)\ y = 2`

♦ Mean mark part (c) 48%.

 
c.   

 

d.i.   `text(Two asymptotes only when:)`

♦♦ Mean mark part (d)(i) 24%.

`k = -2 \ => \ g_k(x) = 2-(23 + x)/((x + 2)^2)`

`k = -5 \ => \ g_k(x) = 2-7/(x + 2)`

`k = 3/2 \ => \ g_k(x) = 2 + 6/(x + 2)`

 

d.ii.   `text(By CAS, solve)\ \ d/(dx)(g_k(x)) = 0\ \ text(for)\ \ x:`

♦♦♦ Mean mark part (d)(ii) 16%.

`x = (-4k + 15 ± sqrt(-21(2k^2 + 7k-15)))/(2k + 3)`
 

`text(No solutions occur when:)`

`k = -3/2\ \ text(or)`

`2k^2 + 7k-15 < 0`

`=> k < -5\ \ text(or)\ \ k > 3/2`

Calculus, SPEC2 2020 VCAA 3

Let  `f(x) = x^2e^(−x)`.

  1. Find an expression for  `f′(x)`  and state the coordinates of the stationary points of  `f(x)`.  (2 marks)
  2. State the equation(s) of any asymptotes of  `f(x)`.  (1 mark)
  3. Sketch the graph of  `y = f(x)`  on the axes provided below, labelling the local maximum stationary point and all points of inflection with their coordinates, correct to two decimal places.  (3 marks)
     
         

Let  `g(x) = x^n e^(−x)`, where  `n ∈ Z`.

  1. Write down an expression for  `g″(x)`.  (1 mark)
  2.  i. Find the non-zero values of `x` for which  `g″(x) = 0`.  (1 mark)
  3. ii. Complete the following table by stating the value(s) of `n` for which the graph of  `g(x)`  has the given number of points of inflection.  (2 marks)
     
       
         
Show Answers Only
  1. `f(0) = 0; f(2) = 4e^(−2) text(and SP’s at)\ (0, 0), (2, 4e^(−2))`
  2. `y = 0`
  3.   
  4. `g″(x) = x^(n – 2) e^(−x)(x^2 – 2xn + n^2 – n)`
  5.  i. `x = x ± sqrtn`
  6. ii.
       
Show Worked Solution

a.   `f′(x) = 2xe^(−x) – x^2e^(−x)`

`text(SP’s when)\ \ f′(x) = 0:`

`x^2e^(−x)` `= 2xe^(−x)`
`x` `= 2\ \ text(or)\ \ 0`

 
`f(0) = 0; \ f(2) = 4e^(−2)`

`:. text(SP’s at)\ \ (0, 0) and  (2, 4e^(−2))`

 

b.   `text(As)\ \ x -> ∞, \ f(x) -> 0^+`

`:. text(Horizontal asymptote at)\ \ y = 0`

 

c.   

`text(POI when)\ \ f″(x) = 0`

`:. text(POI’s:)\ (0.59, 0.19), \ (3.41, 0.38)`

 

d.   `g′(x) = x^(n – 1) e^(−x)(n – x)`

`g″(x) = x^(n – 2) e^(−x)(x^2 – 2xn + n^2 – n)`

 

e.i.   `text(Solve:)\ \ x^2 – 2xn + n^2 – n = 0`

`x = n ± sqrtn`

♦♦♦ Mean mark (e)(ii) 9%.

 

e.ii.   

Calculus, SPEC1 2020 VCAA 6

Let  `f(x) = arctan (3x - 6) + pi`.

  1. Show that  `f^{\prime}(x) = 3/(9x^2 - 36x + 37)`.  (1 mark)

  2. Hence, show that the graph of  `f`  has a point of inflection at  `x = 2`.  (2 marks)

  3. Sketch the graph of  `y = f(x)`  on the axes provided below. Label any asymptotes with their equations and the point of inflection with its coordinates.   (2 marks)

Show Answers Only

  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`

Show Worked Solution

a.    `f^{\prime}(x)` `= (d/(dx) (3x – 6))/(1 + (3x – 6)^2)`
    `= 3/(9x^2 – 36x + 37)`

 

b.   `f^{\prime\prime}(x) = (3(18x – 36))/(9x^2 – 36x + 37)^2`

♦ Mean mark part (b) 42%.

`f^{\prime\prime}(x) = 0\ \ text(when)\ \ 18x – 36 = 0 \ => \ x = 2`

`text(If)\ \ x < 2, 18x – 36 < 0 \ => \ f^{\prime\prime}(x) < 0`

`text(If)\ \ x > 2, 18x – 36 > 0 \ => \ f^{\prime\prime}(x) > 0`

`text(S) text(ince)\ \ f^{\prime\prime}(x)\ \ text(changes sign about)\ \ x = 2,`

`text(a POI exists at)\ \ x = 2`

 

c.   

Calculus, SPEC2-NHT 2019 VCAA 2

Consider the function `f` with rule  `f(x) = (x^2 + x + 1)/(x^2-1)`.

  1. State the equations of the asymptotes of the graph of `f`.  (2 marks)

  2. State the coordinates of the stationary points and the point of inflection. Give your answers correct to two decimal places.  (2 marks)

  3. Sketch the graph of `f` from  `x = -6`  to  `x = 6`  (endpoint coordinates are not required) on the set of axes below, labeling the turning points and the point of inflection with their coordinates correct to two decimal places. Label the asymptotes with their equations.  (3 marks)

Consider the function `f_k` with rule  `f_k(x) = (x^2 + x + k)/(x^2-1)`  where  `k ∈ R`.

  1. For what values of `k` will `f_k` have no stationary points?  (2 marks)

  2. For what value of `k` will the graph of `f_k` have a point of  inflection located on the `y`-axis?  (1 marks)

Show Answers Only

i.     `x = 1, x = -1, y = 1`

 ii.   `(-3.73, 0.87) \ text(and) \ (-0.27, -0.87)`

`(-5.52, 0.88)`
iii.

iv.  `-2`

v.  `-1`

Show Worked Solution

i.    `f(x) = 1 + (x + 2)/((x + 1)(x-1))`

`:. \ text(Asymptotes at)\ \ x = 1, x = -1, y = 1`
  

ii.   `f′(x) = (-(x^2 + 4x + 1))/(x^2-1)^2`

`text(Solve:) \ \ f^{′}(x) = 0 \ => \ x = -2 ± sqrt3`

`text(SP’s at) \ (-3.73, 0.87) \ \ text(and)\ \ (-0.27, -0.87)`
 
`f^{″}(x) = (-2x^3-12x^2-6x-4)/(x^2-1)^3`

`text(Solve:) \ \ f^{″}(x) = 0 \ => \ x = -5.52`

`text(POI at) \ \ (-5.52, 0.88)`
 

iii.

 

iv.    `f_k^{′}(x) = (-x^2-2(k+1) x-1)/(x^2-1)^2`

`text(If no SP’s,) \ \ Δ < 0`

`[-2( k + 1)]^2-4(-1)(-1)` `< 0`
`4k^2 + 8k + 4-4` `< 0`
`4k(k + 2)` `< 0`

 
`:. \ -2 < k < 0`
 

v.    `f_k^{″}(x) = (2(x^3 + 3(k + 1) x^2 + 3x + k + 1))/((x^2-1)^3)`

`text(Solve:)\ \ f_k^{″}(x) = 0`

`k = -1`

Calculus, SPEC1 2019 VCAA 5

The graph of  `f(x) = cos^2(x) + cos(x) + 1`  over the domain  `0 <= x <= 2pi`  is shown below.

  1.  i.  Find `f^{′}(x)`.  (1 mark)

  2. ii. Hence, find the coordinates of the turning points of the graph in the interval  `(0, 2pi)`.  (2 marks)

  3. Sketch the graph of  `y = 1/(f(x))`  on the set of axes above. Clearly label the turning points and endpoints of this graph with their coordinates.  (3 marks)
Show Answers Only
  1. i.  `−sin(x)(2cos(x) + 1)`
  2. ii. `(pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`
  3.  

Show Worked Solution
a.i.    `f(x)` `= cos^2(x) + cos(x) + 1`
  `f^{′}(x)` `= -2sin(x)cos(x)-sin(x)`
    `= -sin(x)(2cos(x) + 1)`

 

a.ii.   `text(SP when)\ \ sin(x) = 0\ \ text(or)\ \ 2cos(x) + 1 = 0`

`sin(x) = 0 \ => \ x = pi\ \ (x = 0\ \ text{not in domain})`

`2cos(x) + 1` `= 0`
`cos(x)` `= -1/2`
`x` `= (2pi)/3, (4pi)/3`

 
`text(When)\ \ cos(x) = −1/2 \ => \ f(x) = 1/4-1/2 + 1 = 3/4`

`:.\ text(Turning Points:)\ (pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`

 

b.   

Graphs, SPEC1 2013 VCAA 4

  1. State the maximal domain and the range of  `y = arccos(1-2x).`   (2 marks)

  2. Sketch the graph of  `y = arccos(1-2x)`  over its maximal domain. Label the endpoints with their coordinates.   (2 marks)

     

     
              VCAA 2013 spec 4b
     

  3. Find the gradient of the tangent to the graph of  `y = arccos (1 – 2x)`  at  `x = 1/4.`   (2 marks)

Show Answers Only
  1. `text(Maximal domain)\ \ [0, 1];\ \ \ text(Range)\ \ [0, pi]`
  2.  

  3. `4/sqrt3`
Show Worked Solution

a.  `text(Maximal domain:)`

`1-2x` `∈ [−1, 1]`
`-2x` `∈ [−2, 0]`
`x` `∈ [0, 1]`

 
`text(When)\ \ x = 0,\ \ y = cos^-1 1= 0;`

`text(When)\ \ x = 1,\ \ y = cos^-1 (-1)= pi`

`:.\ text(Range is)\ \ [0, pi]`

 

b.   

 

c.   `y = cos^-1 (1-2x),`

`(dy)/(dx)` `= (-1(-2))/sqrt(1-(1-2x)^2)`
  `= 2/sqrt(1-(1-2x)^2)`

 
`text(At)\ \ x = 1/4,`

`m_T` `= 2/sqrt(1-(1-1/2)^2)`
  `=2/sqrt(3/4)`
  `= 4/sqrt 3`
  `= (4 sqrt 3)/3`

Calculus, SPEC2 2017 VCAA 1

Let  `f:D ->R, \ f(x) = x/(1 + x^3)`, where `D` is the maximal domain of  `f`.

  1.   i. Find the equations of any asymptotes of the graph of  `f`.   (1 mark)

  2.  ii. Find  `f′(x)`  and state the coordinates of any stationary points of the graph of  `f`, correct to two decimal places.  (2 marks)

  3. iii. Find the coordinates of any points of inflection of the graph of  `f`, correct to two decimal places.  (2 marks)

  4. Sketch the graph of  `f(x) = x/(1 + x^3)`  from  `x=–3`  and  `x = 3`  on the axes provided below, marking all stationary points, points of inflection and intercepts with axes, labelling them with their coordinates. Show any asymptotes and label them with their equations.  (3 marks)

     

     

  5. The region `S`, bounded by the graph of  `f`, the `x`-axis and the line  `x = 3`, is rotated about the `x`-axis to form a solid of revolution. The line  `x = a`, where  `0 < a < 3`, divides the region `S` into two regions such that, when the two regions are rotated about the `x`-axis, they generate solids of equal volume.
  6. i.  Write down an equation involving definite integrals that can be used to determine `a`.  (2 marks)

  7. ii. Hence, find the value of `a`, correct to two decimal places.  (1 mark)

Show Answers Only

  1. i.  `text(vertical asymptote:)\ \ x = −1“text(horizontal asymptote:)\ \ y = 0`
  2. ii. `text(S.P.)\ \ (0.79, 0.53)`
  3. iii. `text(P.O.I.)\ (1.26, 0.42)`
  4.  
  5. i.  `pi int_a^3 (x^2)/((1 + x^3)^2) dx`
  6. ii. `~~ 0.98`

Show Worked Solution

a.i.   `text(Graphing the function on CAS:`

♦♦ Mean mark 36%.

`text(Vertical asymptote:)\ x = −1`

`text(Horizontal asymptote:)\ \ y = 0`
 

a.ii.   `u = x, \ u′ = 1,\ \ v = 1 + x^3, \ v′ = 3x^2\ \ \ text{(manual or by CAS)}`

`f′(x)` `= (1(1 + x^3)-x(3x^2))/((1 + x^3)^2)`
  `= (1-2x^3)/((1 + x^3)^2)`

 
`text(S.P. when)\ \ f′(x)=0:\ `

`=>  (0.79, 0.53)\ \ \ text{(by CAS)}`

 

a.iii.  `text(When)\ \ f″(x)=0,\ \ \ text{(by CAS)}`

`=> x = 0, \ x = -1, \ x = 2`

`x != -1`

`text(Check concavity changes:)`

`f″(−1/2) = −1632/343`

`f″(1) = −3/4`

`f″(3) = 675/(10\ 976)`
 

`text(P.O.I. at)\ \ x = sqrt2 ~~ 1.26\ \ text{(concavity changes)}`

`=> f(sqrt2) ~~ 0.42`

`:. text(P.O.I.)\ (1.26, 0.42)`

 

b.   

 

c.i.    `V_1` `= pi int_0^a y^2\ dx`
  `V_2` `= pi int_a^3 y^2\ dx`

 
`:.\ text(Equation to solve for)\ a:`

`int_0^a x^2/(1 + x^3)^2\ dx = int_a^3 x^2/(1 + x^3)^2\ dx`

 

c.ii.  `a=0.98\ \ \ text{(by CAS)}`