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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, SPEC2 2024 VCAA 3 MC

The graph of  \(f(x)=\dfrac{x-h}{(x+1)(x-4)}\),  where  \(h \in R\), will have no turning points when

  1. \( h<-1\)  and  \(h>4\)
  2. \(-4<h<1\)
  3. \(-1 \leq h \leq 4\)
  4. \(-4 \leq h \leq 1\)
Show Answers Only

\(C\)

Show Worked Solution

\(f(x)=\dfrac{x-h}{(x+1)(x-4)}\)

\(\text{Solve}\ \ f^{-1}(x)=0\ \ \text{(by CAS)}\)

\(x=h\pm \sqrt{h^2-3h-4}\)

\(\text{No turning points occur if}\ \ h^2-3h-4=(h-4)(h+1)<0\)

\(-1 \lt h \lt 4\)

\(\text{If}\ \ h=-1\ \ \text{or}\ \ 4, f(x)\ \text{is linear (no TPs)}\)

\(\therefore -1 \leq h \leq 4\)

\(\Rightarrow C\)

Calculus, SPEC2 2023 VCAA 1

Viewed from above, a scenic walking track from point \(O\) to point \(D\) is shown below. Its shape is given by

\(f(x)= \begin{cases}-x(x+a)^2, & 0 \leq x \leq 1 \\ e^{x-1}-x+b, & 1<x \leq 2 .\end{cases}\)

The minimum turning point of section \(O A B C\) occurs at point \(A\). Point \(B\) is a point of inflection and the curves meet at point \(C(1,0)\). Distances are measured in kilometres.
 

  1. Show that  \(a=-1\)  and  \(b=0\).  (1 mark)

  2. Verify that the two curves meet smoothly at point \(C\).  (2 marks)

  3.  i. Find the coordinates of point \(A\).  (1 mark)

  4. ii. Find the coordinates of point \(B\).  (1 mark)

The return track from point \(D\) to point \(O\) follows an elliptical path given by

\(x=2 \cos (t)+2, y=(e-2) \sin (t)\), where \(t \in\left[\dfrac{\pi}{2}, \pi\right]\).

  1. Find the Cartesian equation of the elliptical path.  (2 marks)

  2. Sketch the elliptical path from \(D\) to \(O\) on the diagram above.  (1 mark)

  3.  i. Write down a definite integral in terms of \(t\) that gives the length of the elliptical path from \(D\) to \(O\).  (1 mark)

  4. ii. Find the length of the elliptical path from \(D\) to \(O\).
  5.     Give your answer in kilometres correct to three decimal places.   (1 mark)

Show Answers Only

a.    \(\text{At}\ C(1,0): \)

\(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \)

    \(a=-1\)

\(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \)
 

b.    \(\text{Find derivative functions (by calc)}:\)

\(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \)

\(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \)

\(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\)
 

c.i.  \(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \)
 

c.ii.  \(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \)
 

d.   \(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\)
 

e.    
          
 

f.i.    \(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\)
 

f.ii.  \(\text{Length}\ = 2.255\)

Show Worked Solution

a.    \(\text{At}\ C(1,0): \)

\(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \)

    \(a=-1\)

\(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \)
 

b.    \(\text{Find derivative functions (by calc)}:\)

\(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \)

\(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \)

\(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\)
 

c.i.  \(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \Rightarrow \ \text{SP when}\ x=\dfrac{1}{3} \)

\(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \)
 

c.ii.  \(\text{POI at}\ B: \ f_1^{″}(x)=-6x+4=0\ \ \Rightarrow x=\dfrac{2}{3} \)

\(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \)
 

d.   \(\dfrac{x-2}{2}=\cos(t), \ \ \dfrac{y}{e-2}=\sin(t) \)

\(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\)
 

e.    
          
 

f.i.    \(\dfrac{dx}{dt} = -2\sin(t),\ \ \dfrac{dy}{dt} = (e-2)\cos(t) \)

\(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\)
 

f.ii.   \(\text{Evaluate the integral in part f.i.}\)

\(\Rightarrow \text{Length}\ = 2.255\)

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 2021 VCAA 9 MC

Which one of the following derivatives corresponds to a graph of  `f`  that has no points of inflection?

  1. `f′(x) = 2(x - 3)^2 + 5`
  2. `f′(x) = 2(x - 3)^3 + 5`
  3. `f′(x) = 5/2(x - 3)^2`
  4. `f′(x) = 1/2(x - 3)^2 - 5`
  5. `f′(x) = (x - 3)^3 - 12x`
Show Answers Only

`B`

Show Worked Solution

`text(If no POI,)\ \ f″(x)\ text(does not change sign).`

♦ Mean mark 38%.

`text(Consider option B)`

`f″(x) = 6(x – 3)^2 \ ->\ text(doesn’t change sign)`

`=>\ B`

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, SPEC2 2012 VCAA 11 MC

If  `(d^2y)/(dx^2) = x^2 - x`  and  `(dy)/(dx) = 0`  at  `x = 0`, then the graph of `y` will have

  1. a local minimum at  `x = 1/2`
  2. a local maximum at  `x = 0`  and a local minimum at  `x = 1`
  3. stationary points of inflection at  `x = 0`  and  `x = 1`, and a local minimum at  `x = 3/2`
  4. a stationary points of inflection at  `x = 0`, no other points of inflection and a local minimum at  `x = 3/2`
  5. a stationary point of inflection at  `x = 0`, a non-stationary point of inflection at  `x = 1`  and a local minimum at  `x = 3/2` 
Show Answers Only

`E`

Show Worked Solution

`(d^2y)/(dx^2)=x^2-x`

♦ Mean mark 51%.

`(d^2y)/(dx^2)=0\ \ text(when)\ \ x=0 or 1`

`(dy)/(dx)` ` = (x^3)/3 – (x^2)/2 + c`
  `= x^2(x/3 – 1/2) + c`

 
`text(Given)\ \ (dy)/(dx) = 0\ \ text(when)\ \ x = 0\ \ =>\ \ c=0`

`(dy)/(dx) = 0\ \ text(when)\ \ x = 0 or 3/2`

`(d^2y)/(dx^2)|_(x = 3/2) = 3/4>0`

`:.\ text(Local minimum at)\ \ x = 3/2`
 

`text(At)\ \ x=0, dy/dx<0\ \ text(either side)`

`:.\ text(Stationary point of inflection at)\ x=0`

`text(At)\ \ x=1, \ dy/dx!=0, \ dy/dx<0\ \ text(either side)`

`:.\ text(Non-stationary point of inflection at)\ \ x = 1`

`=> E`

Calculus, SPEC2 2017 VCAA 10 MC

A function  `f`, its derivative  `fprime` and its second derivative  `f″` are defined for  `x ∈ R`  with the following properties.

`f(a) = 1, f(−a) = −1`

`f(b) = −1, f(−b) = 1`

and  `f″(x) = ((x + a)^2(x - b))/(g(x))`, where  `g(x) < 0`

The coordinates of any points of inflection of   `|\ f(x)\ |`  are

  1.  `(−a,1) and (b,1)`
  2.  `(b,−1)`
  3.  `(−a,−1) and (b,−1)`
  4.  `(−a,1)`
  5.  `(b,1)`
Show Answers Only

`B`

Show Worked Solution

`text(P.O.I. requires:)`

♦♦♦ Mean mark 6%.

`f″(x) =0\ \ text(and)\ \ f″(x)\ \ text(to change sign)`

`f″(x)\ \ text(does not change sign around)\ \ x=a`

`(b, −1)\ text(is the only P.O.I. of)\ \ f(x)`

`:. (b, 1)\ text(is the only P.O.I. of)\ \ |\ f(x)\ |`

`=>B`

Calculus, SPEC2 2017 VCAA 8 MC

Let  `f(x) = x^3 - mx^2 + 4`, where  `m, x ∈ R`.

The gradient of  `f` will always be strictly increasing for

  1.  `x >= 0`
  2.  `x >= m/3`
  3.  `x <= m/3`
  4.  `x >= (2m)/3`
  5.  `x <= (2m)/3`
Show Answers Only

`B`

Show Worked Solution

  `f(x) = x^3 – mx^2 + 4`

♦♦ Mean mark 29%.

`fprime(x)` `= 3x^2 – 2mx`
`f″(x)` `= 6x – 2m`

 
`text(Gradient increasing when)\ \ f″(x)>=0`

`6x >= 2m`

 `x >= (2m)/6 = m/3`

 
`=>B`

Calculus, SPEC2 2017 VCAA 6 MC

Given that  `(dy)/(dx) = e^x\ text(arctan)(y)`, the value of  `(d^2y)/(dx^2)`  at the point  `(0,1)`  is

  1.      `1/2`
  2.     `(3pi)/8`
  3.  `−1/2`
  4.     `pi/4`
  5.  `−pi/8`
Show Answers Only

`B`

Show Worked Solution
`(d^2y)/(dx^2)` `= d/(dx)(e^x) xx text(arctan)(y) + d/(dx)(text(arctan)(y)) xx e^x`
  `= e^xtext(arctan)(y) + e^x (1/(1 + y^2))(dy)/(dx)`

♦ Mean mark 46%.

`(d^2y)/(dx^2)|_(text{(0, 1)})` `= e^0text(arctan)(1) + e^0(1/(1 + 1^2)) xxe^0text(arctan)(1)`
  `= 1 xx pi/4 + 1 xx 1/2 xx pi/4`
  `= (3pi)/8`

`=>B`