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Calculus, MET2 2023 VCAA 1

Let \(f:R \rightarrow R, f(x)=x(x-2)(x+1)\). Part of the graph of \(f\) is shown below.

  1. State the coordinates of all axial intercepts of \(f\).   (1 mark)
  2. Find the coordinates of the stationary points of \(f\).   (2 marks)
    1. Let \(g:R\rightarrow R, g(x)=x-2\).
    2. Find the values of \(x\) for which \(f(x)=g(x)\).   (1 mark)
    1. Write down an expression using definite integrals that gives the area of the regions bound by \(f\) and \(g\).  (2 marks)
    2. Hence, find the total area of the regions bound by \(f\) and \(g\), correct to two decimal places.   (1 mark)
  1. Let \(h:R\rightarrow R, h(x)=(x-a)(x-b)^2\), where \(h(x)=f(x)+k\) and \(a, b, k \in R\).
  2. Find the possible values of \(a\) and \(b\).   (4 marks)
Show Answers Only

a.    \((-1, 0), (0, 0), (2, 0)\)

b.    \(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)

c.i.  \(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

c.ii. \(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

 \(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)

c.iii. \(5.95\)

d.   \(\text{1st case }\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case }\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

Show Worked Solution

a.    \((-1, 0), (0, 0), (2, 0)\)
  

b.    \(\text{Using CAS solve for}\ x:\)

\(\dfrac{d}{dx}(x(x-2)(x+1))=0\)

\(\therefore\ x=\dfrac{1-\sqrt{7}}{3}\ \text{and }x=\dfrac{1+\sqrt{7}}{3}\)

\(\text{Substitute }x\ \text{values into }f(x)\ \text{using CAS to get}\ y\ \text{values}\)

\(\text{The stationary points of }f\ \text{are}:\)

\(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)
  

ci    \(\text{Given }f(x)=g(x)\)

\(x(x-2)(x+1)\) \(=x-2\)
\(x(x-2)(x+1)(x-2)\) \(=0\)
\((x-2)(x(x+1)-1)\) \(=0\)
\((x-2)(x^2+x-1)\) \(=0\)

  
\(\therefore\ \text{Using CAS: } \)

\(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

cii  \(\text{Area of bounded region:}\)

\(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

\(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)
  

ciii  \(\text{Solve the integral in c.ii above using CAS:}\)
  \(\text{Total area}=5.946045..\approx 5.95\)

  

d.   \(\text{Method 1 – Equating coefficients}\)

\((x-a)(x-b)^2=x(x-2)(x+1)+k\)

\(x^3-2bx^2-ax^2+b^2x+2abx-ab^2=x^3-x^2-2x+k\)

\((x^3-(a+2b)x^2+(2ab+b^2)x-ab^2=x^3-x^2-2x+k\)

\(\therefore\ -(a+2b)=-1\ \to\ a=1-2b …(1)\)

\(2ab+b^2=-2\ \ …(2)\)

\(\text{Substitute (1) into (2) and solve for }b.\)

\(2b(1-2b)+b^2\) \(=-2\)
\(3b^2-2b-2\) \(=0\)
\(b\) \(=\dfrac{1\pm \sqrt{7}}{3}\)
\(\text{When }b\) \(=\dfrac{1+\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1+\sqrt{7}}{3}\Bigg)=\dfrac{-2\sqrt{7}+1}{3}\)
\(\text{When }b\) \(=\dfrac{1-\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1-\sqrt{7}}{3}\Bigg)=\dfrac{2\sqrt{7}+1}{3}\)

  

\(\text{Method 2 – Using transformations}\)

\(\text{The squared factor in }(x-a)(x-b)^2=x(x-2)(x+1)+k,\)

\(\text{shows that the turning point is on the }x\ \text{axis}.\)

\(\therefore\ \text{Lowering }f(x)\ \text{by }\dfrac{2(7\sqrt{7}-10)}{27}\ \text{and raising }f(x)\ \text{by }\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\text{will give the 2 possible sets of values for }a\ \text{and}\ b.\)

\(\text{1st case – lowering using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)-\dfrac{2(7\sqrt{7}-10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case – raising using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)+\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

 

Calculus, MET2 2019 VCAA 5

Let  `f: R -> R, \ f(x) = 1 - x^3`. The tangent to the graph of  `f` at  `x = a`, where  `0 < a < 1`, intersects the graph of  `f`  again at `P` and intersects the horizontal axis at `Q`. The shaded regions shown in the diagram below are bounded by the graph of  `f`, its tangent at  `x = a`  and the horizontal axis.
 


 

  1. Find the equation of the tangent to the graph of  `f`  at  `x = a`, in terms of  `a`.   (1 mark)
  2. Find the `x`-coordinate of `Q`, in terms of  `a`.  (1 mark)
  3. Find the `x`-coordinate of `P`, in terms of  `a`. (2 marks)

Let  `A`  be the function that determines the total area of the shaded regions.

  1. Find the rule of  `A`, in terms of  `a`. (3 marks)
  2. Find the value of  `a`  for which  `A`  is a minimum. (2 marks)

Consider the regions bounded by the graph of  `f^(-1)`, the tangent to the graph of  `f^(-1)`  at  `x = b`, where  `0 < b < 1`, and the vertical axis.

  1. Find the value of  `b`  for which the total area of these regions is a minimum.  (2 marks)
  2. Find the value of the acute angle between the tangent to the graph of  `f`  and the tangent to the graph of  `f^(-1)`  at  `x = 1`.  (1 mark)
Show Answers Only
  1. `y = 2a^3 – 3a^2x + 1`
  2. `(2a^3 + 1)/(3a^2)`
  3. `-2a`
  4. `(80a^6 + 8a^3 – 9a^2 + 2)/(12a^2)`
  5. `10^(-1/3)`
  6. `9/10`
  7. `tan^(-1)(1/3)`
Show Worked Solution

a.   `f(x) = 1 – x^3,\ \ f^{\prime}(x) = -3x^2`

`m_text(tang) = -3a^2\ \ text(through)\ \ (a, 1 – a^3)`

`y = 2a^3 – 3a^2x + 1`

 

b.   `text(Solve for)\ \x:`

`2a^3 – 3a^2x + 1 = 0`

`x = (2a^3 + 1)/(3a^2)`

 

c.   `text(Solve for)\ \ x:`

`2a^3 – 3a^2x + 1 = 1 – x^3`

`x=-2a`

`:. x text(-coordinate of)\ P = -2a`

 

d.   `P(-2a, 8a^3 + 1),\ \ Q((2a^3 + 1)/(3a^2), 0)`

`A` `= text(Area of triangle)-int_(-2a)^1 f(x)\ dx`
  `= 1/2((2a^3 + 1)/(3a^2) + 2a)(8a^3 + 1)-int_(-2a)^1 1-x^3\ dx`
  `= (80a^6 + 8a^3 – 9a^2 + 2)/(12a^2)\ \ text{(by CAS)}`

 

e.   `text(Solve for)\ a: \ (dA)/(da) = 0\ \ text{(by CAS)}`

`a = 10^(-1/3)`

 

f.   `text(Consider)\ f(x):\ \ f(10^(-1/3)) = 9/10`

`A_text(min)\ text(for)\ \ f(x)\ \ text(occurs at)\ \ (10^(-1/3), 9/10)`

`=> A_text(min)\ text(for)\ \ f^(-1)(x)\ \ text(occurs when)\ \ x=9/10`

`:. b = 9/10`

 

g.   `f^{\prime} (1)  = -3`

`text(Gradient of)\ \  f^(-1)(x)\ \ text(at)\ \ x = 1\ \ text(is vertical line.)`

`tan theta` `= 1/3`
`theta` `= tan^(-1)(1/3)`
  `=18.4°\ \ text{(to 1 d.p.)}`

Calculus, MET2 2017 VCAA 1

Let  `f : R → R,\  f (x) = x^3 - 5x`. Part of the graph of  `f` is shown below.

 

  1. Find the coordinates of the turning points.  (2 marks)
  2. `A(−1, f (−1))`  and  `B(1, f (1))`  are two points on the graph of  `f`.

     

    1. Find the equation of the straight line through `A` and `B`.  (2 marks)
    2. Find the distance `AB`.  (1 mark)

Let  `g : R → R, \ g(x) = x^3 - kx, \ k ∈ R^+`.

  1. Let  `C(–1, g(−1))` and `D(1, g(1))` be two points on the graph of `g`.

     

    1. Find the distance `CD` in terms of `k`.  (2 marks)
    2. Find the values of `k` such that the distance `CD` is equal to  `k + 1`.  (1 mark)
  2. The diagram below shows part of the graphs of `g` and  `y = x`. These graphs intersect at the points with the coordinates `(0, 0)` and `(a, a)`.
  3.  
  4.  
    1. Find the value of `a` in terms of `k`.  (1 mark)
    2. Find the area of the shaded region in terms of `k`.  (2 marks)
Show Answers Only
  1. `(- sqrt15/3,(10sqrt15)/9)\ text(and)\ (sqrt15/3,-(10sqrt15)/9)`
    1. `y =-4x`
    2. `2sqrt17`
    1. `2sqrt(k^2 – 2k + 2)`
    2. `k = 1quadtext(or)quadk = 7/3`
    1. `sqrt(k + 1)`
    2. `((k + 1)^2)/4\ text(units)²`
Show Worked Solution
a.   
`text(Solve)\ \ f′(x)` `= 0\ \ text(for)\ x:`
`x` `= ± sqrt15/3`

 

`f(sqrt15/3) = -(10sqrt15)/9`

`f(−sqrt15/3) = (10sqrt15)/9`

 

`:.\ text(Turning points:)`

`(- sqrt15/3,(10sqrt15)/9)\ text(and)\ (sqrt15/3,-(10sqrt15)/9)`

 

b.i.   `A(-1,4),\ \ B(1,–4)`

`m_(AB) = (4 – (−4))/(−1 – (1)) = −4`

 

`text(Equation of)\ AB, m=-4, text(through)\ \ (-1,4):`

` y – 4` `= −4(x – (−1))`
`:. y` `= −4x`

 

MARKER’S COMMENT: Students skilled in the use of technology will be much more efficient and minimise errors here.
b.ii.    `d_(text(AB))` `=sqrt((x_2-x_1)^2+(f(x_2)-f(x_1))^2)`
    `= sqrt((1- (- 1))^2 + (f(1) – f(−1))^2)`
    `= 2sqrt17`

 

c.i.    `d_(text(CD))` `=sqrt((x_2-x_1)^2+(g(x_2)-g(x_1))^2)`
    `=sqrt((1-(-1))^2+(g(1)-g(-1))^2)`
    `= 2sqrt(k^2 – 2k + 2)`

 

c.ii.   `text(Solve:)quad2sqrt(k^2 – 2k + 2) = k + 1quadtext(for)quadk > 0`

`:. k = 1quadtext(or)quadk = 7/3`

 

d.i.   `text(Solve:)quada^3 – ka = a\ \ \ text(for)quad a,\ \ (a > 0)`

`:. a = sqrt(k + 1)`

 

d.ii.    `text(Area)` `= int_0^(sqrt(k + 1))(x – g(x))\ dx`
    `= ((k + 1)^2)/4\ text(units)²`

Calculus, MET1 2011 VCAA 9

Parts of the graphs of the functions

`f: R -> R, \ f(x)` `= x^3 - ax\ \ \ \ \ ` `a > 0`
`g: R -> R, \ g(x)` `= ax` `a > 0`
are shown in the diagram below.

The graphs intersect when  `x = 0`  and when  `x = m.`

vcaa-2011-meth-9a

The area of the shaded region is 64.

Find the value of `a` and the value of `m.`  (4 marks)

Show Answers Only

`a = 8,\ \ m = 4`

Show Worked Solution

`text(Intersection between)\ f(x) and g(x):`

♦ Mean mark 41%.
MARKER’S COMMENT: Few students correctly used the intersection to achieve the final answer.
`f(x)` `= g(x)`
`x^3 – ax` `= ax`
`x^3 – 2ax` `= 0`
`x (x^2 – 2a)` `= 0`

`:. x = 0,\ \ x = +- sqrt (2a)`

`:. m = sqrt (2a),\ \ m>0`

 

`text{Shaded Area = 64  (given)},`

`:. int_0^(sqrt(2a)) (ax – (x^3 – ax))\ dx` `=64`
`int_0^(sqrt(2a)) (2ax – x^3)\ dx` `=64`
`[ax^2 – 1/4 x^4]_0^(sqrt(2a))` `=64`
`(a (2a) – 1/4 (4a^2)) – (0)` `=64`
`2a^2 – a^2` `=64`
`a^2` `=64`
`:. a` `=8,\ \ \ a > 0`

 

`:. m` `= sqrt (2 xx 8)=4`

 `:. a = 8,\ \ m = 4`

Calculus, MET1 2014 VCAA 5

Consider the function  `f:[−1,3] -> R`,  `f(x) = 3x^2 - x^3`.

  1. Find the coordinates of the stationary points of the function.  (2 marks)
  2. On the axes  below, sketch the graph of  `f`.

     

    Label any end points with their coordinates.  (2 marks)

     

     
    met1-2014-vcaa-q5
     

  3. Find the area enclosed by the graph of the function and the horizontal line given by `y = 4`.  (3 marks)
Show Answers Only
  1. `(0, 0) and (2, 4)`
  2.  
    met1-2014-vcaa-q5-answer3
  3. `27/4\ text(u²)`
Show Worked Solution
a.    `text(SP’s occur when)\ \ f′(x)` `= 0`
`6x – 3x^2`  `= 0` 
 `3x(2 – x)` `=0`

`x = 0,\ \ text(or)\ \ 2`

 

`:.\ text{Coordinates are (0, 0) and (2, 4)}`

 

b.    met1-2014-vcaa-q5-answer3

 

♦ Part (c) mean mark 48%.
c.    met1-2014-vcaa-q5-answer4

`text(Solution 1)`

`text(Area)` `= int_(−1)^2 4 – (3x^2 – x^3)dx`
  `= int_(−1)^2 4 – 3x^2 + x^3dx`
  `= [4x – x^3 + 1/4x^4]_(−1)^2`
  `= (8 – 8 + 4) – (−4 – (−1) + 1/4)`
   
`:.\ text(Area)` `= 27/4 text(units²)`

 

`text(Solution 2)`

`text(Area)` `= 12 – int_(−1)^2(3x^2 – x^3)dx`
  `= 12 – [x^3 – 1/4 x^4]_(−1)^2`
  `= 12 – [(8 – 4) – (−1 – 1/4)]`
  `= 27/4\ text(units²)`

Calculus, MET2 2015 VCAA 1

Let  `f: R -> R,\ \ f(x) = 1/5 (x - 2)^2 (5 - x)`. The point  `P(1, 4/5)`  is on the graph of  `f`, as shown below.

The tangent at `P` cuts the y-axis at `S` and the x-axis at `Q.`

VCAA 2015 1ai

  1. Write down the derivative  `f prime (x)` of `f (x)`.  (1 mark)
    1. Find the equation of the tangent to the graph of  `f` at the point  `P(1, 4/5)`.  (1 mark)
    2. Find the coordinates of points `Q` and `S`.  (2 marks)
  3. Find the distance `PS` and express it in the form  `sqrt b/c`, where `b` and `c` are positive integers.  (2 marks)

VCAA 2015 1di

  1. Find the area of the shaded region in the graph above.   (3 marks)
Show Answers Only
  1. `−3/5(x – 4)(x – 2)`
    1. `y = – 9/5x + 13/5`
    2. `S (0, 13/5)`

       

      `Q (13/9, 0)`

  3. `sqrt 106/5`
  4. `108/5\ text(units²)`
Show Worked Solution
a.    `f(x)` `= 1/5 (x – 2)^2 (5 – x)`
  `fprime(x)` `=1/5 xx 2(x-2)(5-x) – 1/5 (x-2)^2`
    `= −3/5(x – 4)(x – 2)`

 

b.i.   `text(Solution 1)`

`y = −9/5x + 13/5qquad[text(CAS:tangentLine)\ (f(x),x,1)]`

 

`text(Solution 2)`

`m_text(tan) = -9/5,\ \ text(through)\ \ (1, 4/5)`

`y-4/5` `=-9/5 (x-1)`
`y` `=-9/5 x +13/5`

 

b.ii.   `text(At)\ S,\ \ x=0`

`y = −9/5 xx 0 + 13/5 = 13/5`

`:. S(0,13/5)`

 

`text(At)\ Q,\ \ y=0`

`0` `= −9/5x + 13/5`
`x` `=13/5 xx 5/9=13/9`

 
`:. Q(13/9,0)`

 

c.   `P(1, 4/5),\ \ S(0,13/5)`

`text(dist)\ PS` `=sqrt((x_2-x_1)^2 + (y_2-y_1)^2)`
  `= sqrt((1 – 0)^2 + (4/5 – 13/5)^2)`
  `= (sqrt106)/5`

 

d.   `text(Find intersection pts of)\ SQ\ text(and)\ f(x),`

MARKER’S COMMENT: Many students complicated their answer by splitting up the area.

`text{Solve (using technology):}`

`1/5 (x – 2)^2 (5 – x) = −9/5x + 13/5`

`x = 1,7`

`:.\ text(Area)` `= int_1^7(f(x) – (−9/5x + 13/5))dx`
  `= 108/5\ text(u²)`

Calculus, MET2 2014 VCAA 3 MC

The area of the region enclosed by the graph of  `y = x (x + 2) (x − 4)`  and the `x`-axis is

A.   `128/3`

B.   `20/3`

C.   `236/3`

D.   `148/3`

E.   `36`

Show Answers Only

`D`

Show Worked Solution

met2-2014-vcaa-3-mc-answer

`text(Area)` `= int_-2^0 x(x + 2)(x – 4)\ dx – int_0^4 x(x + 2)(x – 4)\ dx`
  `= 148/3`

 
`=>   D`