SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

EXAMCOPY MattTest Indenting

Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`

Part of the graph of `y=f(x)` is shown below.
 

  1. State the range of `f(x)`.   (1 mark)

  2. Matt 1
    1. lskdjflsdkfj 
    2. sldkjfsldkfj 
    1. Something
    1. abc
    2. def
  5. slkflskdfj
    1. slkdfjlsdkfj
    2. sdlkjfsdlkfj
    1. lskjdflksd
    2. sdkjflsdkfj
  7. See the items below
    1. first
    2. second
    3. third
    4. fourth
    5. fifth
    1. Find `f^{\prime}(0)`.   (2 marks)

    2. State the maximal domain over which `f` is strictly increasing.   (1 mark)
    1. slkdjflskdfj 
    2. slkdfjlsdkfj
  10. lsksdlkfj
    1. sdflkjsd
    2. sdlfkj
    1. sdfsdlkf
  12. slkdfsldkfj
    1. slkdfjsldkfj
  13. Show that `f(x)+f(-x)=0`.   (1 mark)

  14. Find the domain and the rule of `f^{-1}`, the inverse of `f`.   (3 marks)
  15. Let `h` be the function `h:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, h(x)=\frac{1}{k}\left(\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)\right)`, where `k \in R` and `k>0`.
  1. The inverse function of `h` is defined by `h^{-1}: R \rightarrow R, h^{-1}(x)=\frac{e^{k x}-1}{2\left(e^{k x}+1\right)}`.
  2. The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
  3. The graph below shows the relevant area shaded.
     

  1. You are not required to find or define `A(k)`.
  1. Determine the range of values of `k` such that `A(k)>0`.   (1 mark)

  2. Explain why the domain of `A(k)` does not include all values of `k`.   (1 mark)

Show Answers Only
a.     `R`
b.i `f^{\prime}(0)=4`
b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`
c. `0`
d. `x \in \mathbb{R}`
e.i  ` k > 4`
e.ii No bounded area for `0<k \leq 4`
Show Worked Solution

a.   `R` is the range.

b.i    `f(x)`
 `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)`  
  `f^{\prime}(x)` `= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}`  
    `= \frac{2}{2 x+1}-\frac{2}{2 x-1}`  
  `f^{\prime}(0)` `= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}`  
    `= 4`  

 
b.ii  `\left(-\frac{1}{2}, \frac{1}{2}\right)`

c.   `f(x)+f(-x)` `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)`  
  `= 0`  

 
d.   To find the inverse swap `x` and `y` in `y=f(x)`

`x` `= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)`  
`x` `= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)`  
`e^x` `=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}`  
`y+\frac{1}{2}` `= e^x\left(-y+\frac{1}{2}\right)`  
`y+\frac{1}{2}` `= -e^x y+\frac{e^x}{2}`  
`y\left(e^x+1\right)` `= \frac{e^x-1}{2}`  
`:.\ f^(-1)(x)` `= \frac{e^x-1}{2(e^x + 1)}`  

 
  `:.`  Domain: `x \in \mathbb{R}`
  

e.i   The vertical dilation factor of  `f(x)` is  `1/k`

For `A(k)>=0` , `h^{\prime}(0)<1`

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


♦♦♦♦ Mean mark (e.i) 10%.
MARKER’S COMMENT: Incorrect responses included `k>0` and `4<k<33`.

e.ii  When `h \geq h^{-1}` for  `x>0` (or `h \leq h^{-1}` for  `x<0`) there is no bounded area.

`:.`  There will be no bounded area for `0<k \leq 4`.


♦♦♦♦ Mean mark (e.ii) 10%.

EXAMCOPY Functions, MET2 2022 VCAA 4

Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`

Part of the graph of `y=f(x)` is shown below.
 

  1. State the range of `f(x)`.   (1 mark)

  2.  i. Find `f^{\prime}(0)`.   (2 marks)

  3. ii. State the maximal domain over which `f` is strictly increasing.   (1 mark)

  4. Show that `f(x)+f(-x)=0`.   (1 mark)

  5. Find the domain and the rule of `f^{-1}`, the inverse of `f`.   (3 marks)

  6. Let `h` be the function `h:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, h(x)=\frac{1}{k}\left(\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)\right)`, where `k \in R` and `k>0`.
  7. The inverse function of `h` is defined by `h^{-1}: R \rightarrow R, h^{-1}(x)=\frac{e^{k x}-1}{2\left(e^{k x}+1\right)}`.
  8. The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
  9. The graph below shows the relevant area shaded.
     

  1. You are not required to find or define `A(k)`.
  1. Determine the range of values of `k` such that `A(k)>0`.   (1 mark)

  2. Explain why the domain of `A(k)` does not include all values of `k`.   (1 mark)

Show Answers Only
a.     `R`
b.i `f^{\prime}(0)=4`
b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`
c. `0`
d. `x \in \mathbb{R}`
e.i  ` k > 4`
e.ii No bounded area for `0<k \leq 4`
Show Worked Solution

a.   `R` is the range.

b.i    `f(x)`
 `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)`  
  `f^{\prime}(x)` `= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}`  
    `= \frac{2}{2 x+1}-\frac{2}{2 x-1}`  
  `f^{\prime}(0)` `= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}`  
    `= 4`  

 
b.ii  `\left(-\frac{1}{2}, \frac{1}{2}\right)`

c.   `f(x)+f(-x)` `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)`  
  `= 0`  

 
d.   To find the inverse swap `x` and `y` in `y=f(x)`

`x` `= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)`  
`x` `= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)`  
`e^x` `=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}`  
`y+\frac{1}{2}` `= e^x\left(-y+\frac{1}{2}\right)`  
`y+\frac{1}{2}` `= -e^x y+\frac{e^x}{2}`  
`y\left(e^x+1\right)` `= \frac{e^x-1}{2}`  
`:.\ f^(-1)(x)` `= \frac{e^x-1}{2(e^x + 1)}`  

 
  `:.`  Domain: `x \in \mathbb{R}`
  

e.i   The vertical dilation factor of  `f(x)` is  `1/k`

For `A(k)>=0` , `h^{\prime}(0)<1`

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


♦♦♦♦ Mean mark (e.i) 10%.
MARKER’S COMMENT: Incorrect responses included `k>0` and `4<k<33`.

e.ii  When `h \geq h^{-1}` for  `x>0` (or `h \leq h^{-1}` for  `x<0`) there is no bounded area.

`:.`  There will be no bounded area for `0<k \leq 4`.


♦♦♦♦ Mean mark (e.ii) 10%.

Functions, MET2 2022 VCAA 4

Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`

Part of the graph of `y=f(x)` is shown below.
 

  1. State the range of `f(x)`.   (1 mark)

  2.  i. Find `f^{\prime}(0)`.   (2 marks)

  3. ii. State the maximal domain over which `f` is strictly increasing.   (1 mark)

  4. Show that `f(x)+f(-x)=0`.   (1 mark)

  5. Find the domain and the rule of `f^{-1}`, the inverse of `f`.   (3 marks)

  6. Let `h` be the function `h:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, h(x)=\frac{1}{k}\left(\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)\right)`, where `k \in R` and `k>0`.
  7. The inverse function of `h` is defined by `h^{-1}: R \rightarrow R, h^{-1}(x)=\frac{e^{k x}-1}{2\left(e^{k x}+1\right)}`.
  8. The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
  9. The graph below shows the relevant area shaded.
     

  1. You are not required to find or define `A(k)`.
  2.  i. Determine the range of values of `k` such that `A(k)>0`.   (1 mark)
  3. ii. Explain why the domain of `A(k)` does not include all values of `k`.   (1 mark
Show Answers Only
a.     `R`
b.i `f^{\prime}(0)=4`
b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`
c. `0`
d. `x \in \mathbb{R}`
e.i  ` k > 4`
e.ii No bounded area for `0<k \leq 4`
Show Worked Solution

a.   `R` is the range.

b.i    `f(x)`
 `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)`  
  `f^{\prime}(x)` `= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}`  
    `= \frac{2}{2 x+1}-\frac{2}{2 x-1}`  
  `f^{\prime}(0)` `= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}`  
    `= 4`  

 
b.ii  `\left(-\frac{1}{2}, \frac{1}{2}\right)`

c.   `f(x)+f(-x)`

`= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)`  
`= 0`  

 
d.   To find the inverse swap `x` and `y` in `y=f(x)`

`x` `= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)`  
`x` `= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)`  
`e^x` `=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}`  
`y+\frac{1}{2}` `= e^x\left(-y+\frac{1}{2}\right)`  
`y+\frac{1}{2}` `= -e^x y+\frac{e^x}{2}`  
`y\left(e^x+1\right)` `= \frac{e^x-1}{2}`  
`:.\ f^(-1)(x)` `= \frac{e^x-1}{2(e^x + 1)}`  

 
  `:.`  Domain: `x \in \mathbb{R}`
  

e.i   The vertical dilation factor of  `f(x)` is  `1/k`

For `A(k)>=0` , `h^{\prime}(0)<1`

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


♦♦♦♦ Mean mark (e.i) 10%.
MARKER’S COMMENT: Incorrect responses included `k>0` and `4<k<33`.

e.ii  When `h \geq h^{-1}` for  `x>0` (or `h \leq h^{-1}` for  `x<0`) there is no bounded area.

`:.`  There will be no bounded area for `0<k \leq 4`.


♦♦♦♦ Mean mark (e.ii) 10%.

Calculus, MET2 2022 VCAA 1

The diagram below shows part of the graph of `y=f(x)`, where `f(x)=\frac{x^2}{12}`.
 

  1. State the equation of the axis of symmetry of the graph of `f`.   (1 mark)

  2. State the derivative of `f` with respect to `x`.   (1 mark)

The tangent to `f` at point `M` has gradient `-2` .

  1. Find the equation of the tangent to `f` at point `M`.   (2 marks)

The diagram below shows part of the graph of `y=f(x)`, the tangent to `f` at point `M` and the line perpendicular to the tangent at point `M`.
 

 

  1.  i. Find the equation of the line perpendicular to the tangent passing through point `M`.   (1 mark)

  2. ii. The line perpendicular to the tangent at point `M` also cuts `f` at point `N`, as shown in the diagram above.
  3.     Find the area enclosed by this line and the curve `y=f(x)`.   (2 marks)

  4. Another parabola is defined by the rule `g(x)=\frac{x^2}{4 a^2}`, where `a>0`.
  5. A tangent to `g` and the line perpendicular to the tangent at `x=-b`, where `b>0`, are shown below.

  1. Find the value of `b`, in terms of `a`, such that the shaded area is a minimum.   (4 marks)

Show Answers Only

a.    `x=0`

b.    ` f^{\prime}(x)=1/6x`

c.    `x=-12`

di.    `y=1/2x + 18`

dii.   Area`= 375` units²

e.    `b = 2a^2`

Show Worked Solution

a.   Axis of symmetry:  `x=0`
  

b.    `f(x)`  `=\frac{x^2}{12}`  
  ` f^{\prime}(x)` `= 1/6x`  

  
c.   At `M` gradient `= -2`

`1/6x` `= -2`  
`x` `= -12`  

 
When  `x = -12`
 

`f(x) = (-12)^2/12 = 12`
 
Equation of tangent at `(-12 , 12)`:

`y-y_1` `=m(x-x_1)`  
`y-12` `= -2(x + 12)`  
`y` `= -2x-12`  


d.i   Gradient of tangent `= -2`

`:.`  gradient of normal `= 1/2` 

Equation at `M(- 12 , 12)`

`y -y_1` `=m(x-x_1)`  
`y-12` `= 1/2(x + 12)`  
`y` `=1/2x + 18`  


d.ii  Points of intersection of `f(x)` and normal are at  `M` and `N`.

So equate ` y = x^2/12` and  `y = 1/2x + 18` to find `N`

`x^2/12` `=1/2x + 18`  
`x^2-6x-216` `=0`  
`(x + 12)(x-18)` `=0`  

   
`:.\  x = -12` or `x = 18`

Area `= \int_{-12}^{18}\left(\frac{1}{2} x+18-\frac{x^2}{12}\right) d x`  
  `= [x^2/4 + 18x-x^3/36]_(-12)^18`  
  `= [18^2/4 +18^2-18^3/36] – [12^2/4 + 18 xx (-12)-(-12)^3/36]`  
  `= 375` units²  

 

e.   `g(x) = x^2/(4a^2)`   `a > 0`

At `x = -b`   `y = (-b)^2/(4a^2) = b^2/4a^2`

`g^{\prime}(x) = (2x)/(4a^2) = x/(2a^2)`

Gradient of tangent `= (-b)/(2a^2)`

Gradient of normal `= (2a^2)/b`

Equation of normal at `(- b , b^2/(4a^2))`

`y-y_1` `= m(x-x_1)`  
`y-b^2/(4a^2)` `= (2a^2)/b(x-(-b))`  
`y` `= (2a^2x)/b + 2a^2 + b^2/(4a^2)`  
`y` `= (2a^2x)/b +(8a^4 + b^2)/(4a^2)`  

 
Points of intersection of normal and parabola (Using CAS)

solve `((2a^2x)/b +(8a^4 + b^2)/(4a^2) = x^2/(4a^2),x)`

`x =-b`  or  `x = (8a^4+b^2)/b`

 
Calculate area using CAS

`A = \int_{-b}^{(8a^4+b^2)/b}\left(\frac{2a^2x}{b} +\frac{8a^4 + b^2}{4a^2}-frac{x^2}{4a^2} \right) dx`

`A = frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3}`
 
Using CAS Solve derivative of `A = 0`  with respect to `b` to find `b`

solve`(d/(db)(frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3})=0,b)`

 
`b =-2a^2`   and  `b = 2a^2`

Given `b > 0`

`b = 2a^2`


♦♦ Mean mark (e) 33%.
MARKER’S COMMENT: Common errors: Students found `\int_{-b}^{\frac{8 a^4+b^2}{b}}\left(y_n\right) d x` and failed to subtract `g(x)` or had incorrect terminals.

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, MET1 2023 VCAA 7

Consider \(f:(-\infty, 1]\rightarrow R, f(x)=x^2-2x\). Part of the graph of  \(y=f(x)\)  is shown below.
 

  1. State the range of \(f\).   (1 mark)
  2. Sketch the graph of the inverse function  \(y=f^{-1}(x)\) on the axes above. Label any endpoints and axial intercepts with their coordinates.   (2 marks)
  3. Determine the equation of the domain for the inverse function  \(f^{-1}\).   (2 marks)
  4. Calculate the area of the regions enclosed by the curves of \(f,\ f^{-1}\)  and  \(y=-x\).   (2 marks)
Show Answers Only

a.    \([-1, \infty)\)

b.   

c.    \(f^{-1}(x)=1-\sqrt{x+1}\)

\(\text{Domain}\ [-1, \infty)\)

d.    \(A=\dfrac{1}{3}\)

Show Worked Solution

a.    \([-1, \infty)\)

b.   

c.    \(\text{When }f(x)\ \text{is written in turning point form}\)

\(y=(x-1)^2-1\)
 

\(\text{Inverse function: swap}\ x \leftrightarrow y\)

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

 
\(\text{Domain}\ [-1, \infty)\)


♦ Mean mark (c) 48%.
MARKER’S COMMENT: Common error → writing the function as \(f^{-1}(x)=1+\sqrt{x+1}\).

d.     \(\text{One strategy of many possibilities:}\)

  \(A\) \(=2\displaystyle \int_{0}^{1} \big(-x-(x^2-2x)\big)\,dx\)
    \(=2\displaystyle \int_{0}^{1} \big(x-x^2\big)\,dx\)
    \(=2\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1\)
    \(=2\bigg(\dfrac{1}{2}-\dfrac{1}{3}-(0)\bigg)\)
    \(=\dfrac{1}{3}\)

♦♦♦ Mean mark (d) 24%.
MARKER’S COMMENT: Using the symmetry properties of the graph and its inverse helped answer this question efficiently.

Calculus, MET2 2020 VCAA 2

An area of parkland has a river running through it, as shown below. The river is shown shaded.

The north bank of the river is modelled by the function  `f_(1):[0,200]rarr R, \ f_(1)(x)=20 cos((pi x)/(100))+40`.

The south bank of the river is modelled by the function  `f_(2):[0,200]rarr R, \ f_(2)(x)=20 cos((pi x)/(100))+30`.

The horizontal axis points east and the vertical axis points north.

All distances are measured in metres.
 

A swimmer always starts at point `P`, which has coordinates  (50, 30).

Assume that no movement of water in the river affects the motion or path of the swimmer, which is always a straight line.

  1. The swimmer swims north from point `P`.
  2. Find the distance, in metres, that the swimmer needs to swim to get to get to the north bank of the river.   (1 mark)

  3. The swimmer swims east from point `P`.
  4. Find the distance, in metres, that the swimmer needs to swim to get to the north bank of the river.   (2 marks)

  5. On another occasion, the swimmer swims the minimum distance from point `P` to the north bank of the river.
  6. Find this minimum distance. Give your answer in metres, correct to one decimal place.   (2 marks)

  7. Calculate the surface area of the section of the river shown on the graph in square metres.   (1 mark)

  8. A horizontal line is drawn through point `P`. The section of the river that is south of the line is declared a no "no swimming" zone.
  9. Find the area of the "no swimming" zone, correct to the nearest square metre.   (3 marks)

  10. Scientists observe that the north bank of the river is changing over time. It is moving further north from its current position. They model its predicted new location using the function with rule  `y=kf_(1)(x)`, where `k >= 1`.
  11. Find the values of  `k` for which the distance north across the river, for all parts of the river, is strictly less than 20 m.   (2 marks)

Show Answers Only
  1. `10\ text{m}`
  2. `16 2/3\ text{m}`
  3. `8.5\ text{m}`
  4. `2000\ text{m}^2`
  5. `837\ text{m²}`
  6. `k in [1, 7/6)`
Show Worked Solution

a.   `text{Since swimmer swims due north,}`

`text{Distance}\ = 40-30=10\ text{m}`

 

b.  `text{Solve} \ f_(1)(x)=30 \ text{or} \ x in[50,100]`

`=> x=200/3`

`:.\ text{Distance to swim (east) to reach north bank}`

`=200/3-50`

`=16 2/3\ text{m}`
 

c.   `text{Let swimmer arrive at north bank at the point}\ \ (x,f_(1)(x))`

`text{By Pythagoras,}`

♦ Mean mark part (c) 39%.

`d(x)=sqrt((x-50)^(2)+(f_(1)(x)-30)^(2))`

`text{Solve} \ d/dx(d(x))=0 \ text{for} \ x:`

`x=54.47…`

`:. d_min=8.5\ text{m (to 1 d.p.)}`

 

d.   `text{Shaded Area}`

`=int_(0)^(200)(f_(1)(x)-f_(2)(x))\ dx`

`=2000\ text{m}^2`

 

e.   `text{Find}\ \ f_(1)(x) = 30 \ text{for} x in [50,150]:`

♦♦ Mean mark part (e) 35%.

`=>x=200/3, 400/3`

`text{Find}\ \ f_(2)(x) = 30 \ text{for} \ x in [50,150]:`

`=>x=50, 150`

`:.\ text{Area}` `=int_(50)^(150)(30-f_(2)(x))\ dx-int_((200)/(3))^((400)/(3))(30-f_(1)(x))\ dx`  
  `=837\ text{m² (to nearest m²)}`  

♦♦♦Mean mark part (f) 15%.

 

f.   `text{Let}\ \ D(x)=\ text{vertical distance between banks}`

`D(x)` `=kf_(1)(x)-f_(2)(x)`  
  `=20k cos((pi x)/(100))+40k-(20 cos((pi x)/(100))+30)`   
  `=(20k-20)cos((pi x)/(100)) +40k-30`  

  
`text{Given}\ \ D(x)<20 \ text{for}\ x in[0,200]`

`text{Maximum} \ cos((pi x)/(100)) = 1\  text{when}\ \ x=0, 200`

`text{Solve} \ 20k-20+40k-30<20\ \ text{for}\ k:`

`=> k<7/6`

`:. k in [1,7/6)`

Calculus, MET2 2020 VCAA 1

Let  `f:R rarr R, \ f(x)=a(x+2)^(2)(x-2)^(2)`, where `a in R`. Part of the graph of `f` is shown below. 
 

  1. Show that  `a = 1/4`.    (1 mark)

  2. Express  `f(x)=(1)/(4)(x+2)^(2)(x-2)^(2)`  in the form  `f(x)=(1)/(4)x^(4)+bx^(2)+c`  where `b` and `c` are integers.   (1 mark)

Part of the graph of the derivative function  `f^{′}` is shown below.
 
     
 

  1.  i. Write the rule for `f^{′}` in terms of `x`.   (1 mark)

  2. ii. Find the minimum value of the graph of `f^{′}` on the interval  `x in (0, 2)`.   (2 marks)  

Let  `h:R rarr R, \ h(x)=-(1)/(4)(x+2)^(2)(x-2)^(2)+2`. Parts of the graph of `f` and `h` are shown below.
 
 
       
 

  1. Write a sequence of two transformations that map the graph of `f` onto the graph of `h`.   (1 mark)

     
       
     

  2.   i. State the values of `x` for which the graphs of `f`and `h` intersect.   (1 mark)

  3.  ii. Write down a definite integral that will give the total area of the shaded regions in the graph above.   (1 mark)

  4. iii. Find the total area of the shaded regions in the graph above. Give your answer correct to two decimal places.   (1 mark)

  5. Let `D` be the vertical distance between the graphs of `f`and`h`.
  6. Find all values of `x` for which `D` is at most 2 units. Give your answers correct to two decimal places.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `f(x)=(1)/(4)x^(4)-2x^(2)+4`
  3.  i. `f^{′}(x)=x(x-2)(x+2)=x^(3)-4x`
  4. ii. `-(16sqrt3)/(9)`
  5. `text{Reflect in the x-axis then translate 2 units up, or translate 2 units down then reflect in the x-axis.}`
  6.   i. `x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
  7.  ii. `2int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx” or “int_(-sqrt6)^(-sqrt2)(h(x)-f(x))\ dx+int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx`
  8. iii. `text(2.72 u²)`
  9. `-2.61 <= x <= -1.08,1.08 <= x <= 2.61`
Show Worked Solution

a.   `text(S)text(ince)\ f(x)\ text{passes through (0, 4):}`

`4` `=a(2)^2(-2)^2`  
`4` `=16a`  
`:.a` `=1/4\ \ text(… as required)`  

 

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

 
c.i.  `f^{′}(x)=x(x-2)(x+2)=x^(3)-4x`
 

c.ii.  `text{Solve}\ \ f^{″}(x)=0\ \ \text{for}\ x:`

`3x^2-4=0`

`=> \ x=(2sqrt3)/(3)`

`f^{′}((2sqrt3)/(3))=-(16sqrt3)/(9)`
 

d.  `text{Reflect in the x-axis then translate 2 units up, or}`

`text{translate 2 units down then reflect in the x-axis.}`
 

e.i.  `text{By CAS, solve}\ \ f(x)=h(x)\ \ text{for}\ x:}`

`x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
 

e.ii.  `2int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx” or”`

`int_(-sqrt6)^(-sqrt2)(h(x)-f(x))\ dx+int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx`
 

e.iii.  `text(Area = 2.72 u²)`

♦♦ Mean mark part (f) 26%.

 

f.   `text{By CAS, solve}\ -2 <= h(x)-f(x) <= 2\ \ \text{for}\ x:`

`-2.61 <= x <= -1.08, \ 1.08 <= x <= 2.61,” or”`

`-sqrt(4+2sqrt2) <= x <= -sqrt(4-2sqrt2), \ sqrt(4-2sqrt2) <= x <= sqrt(4+2sqrt2)`

Calculus, MET2 2021 VCAA 3

Let  `q(x) = log_e (x^2-1)-log_e (1-x)`.

  1. State the maximal domain and the range of `q`.   (2 marks)

  2.  i. Find the equation of the tangent to the graph of `q` when  `x =-2`.   (1 mark)

  3. ii. Find the equation of the line that is perpendicular to the graph of `q` when  `x =-2`  and passes through the point  (-2, 0).   (1 mark)

Let  `p(x) = e^{-2x}-2e^-x + 1.`

  1. Explain why `p` is not a one-to one function.   (1 mark)

  2. Find the gradient of the tangent to the graph of `p` at  `x = a`.   (1 mark)

The diagram below shows parts of the graph of `p` and the line  `y = x + 2`.
 
 
                     
 
The line  `y = x + 2`  and the tangent to the graph of `p` at  `x = a`  intersect with an acute angle of `theta` between them.

  1. Find the value(s) of `a` for which  `theta = 60^@`. Give your answer(s) correct to two decimal places.   (3 marks)

  2. Find the `x`-coordinate of the point of intersection between the line  `y = x + 2` and the graph of `p`, and hence find the area bounded by  `y = x + 2`, the graph of `p` and the `x`-axis, both correct to three decimal places.   (3 marks)

Show Answers Only
  1. `text{Domain:} \ x ∈ (-∞, -1)`
    `text{Range:} \ y ∈ R`
  2. i.  `-x-2`
    ii. `x + 2`
  3. `text{Not a one-to-one function as it fails the horizontal line test.}`
  4. `-2e^{-2a} + 2e^{-a}`
  5. `-0.67`
  6. `1.038\ text(u)^2`
Show Worked Solution

a.    `text(Method 1)`

`x^2-1 > 0 \ \ => \ \ x > 1 \ \ ∪ \ \ x < -1`

`1-x > 0 \ \ => \ \ x < 1`
 
`:. \ x ∈ (-∞, -1)`
 

`text(Method 2)`

`text{Sketch graph by CAS}`

`text{Asymptote at} \ x = -1`

`text{Domain:} \ x ∈ (-∞, -1)`

`text{Range:} \ y ∈ R`
 
 

b.     i.   `text{By CAS (tanLine} \ (q (x), x, -2)):`

`y = -x-2`
 

ii.   `text{By CAS (normal} \ (q (x), x, -2)):`

`y = x + 2`
  

c.    `text{Sketch graph by CAS.}`

`p(x) \ text{is not a one-to-one function as it fails the horizontal line test}`

`text{(i.e. it is a many-to-one function)}`
 

d.   `p^{′}(x) = -2e^{-2x} + 2e^-x`

`p^{prime}(a) = -2e^{-2a} + 2e^{-a}`
 
 

e. 

 
`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a} -2e^{-2a} =-tan (15^@) \ \ text{for}\ a:`

`a = -0.11`
 

`text{Case 2}`

`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a}-2e^{-2a} = -tan 75^@\ \ text{for}\ a:`

`a = -0.67`
 

f.     `text{At intersection,}`

`x + 2 = e^{-2x} -2e^{-x} + 1`

`x = -0.750`
 

`text{Area}` `= int_{-2}^{-0.750} x + 2\ dx + int_{-0.750}^0 e^{-2x}-2e^{-x} + 1\ dx`
  `= 1.038 \ text(u)^2`

Calculus, MET2 2021 VCAA 2

Four rectangles of equal width are drawn and used to approximate the area under the parabola  `y = x^2`  from  `x = 0`  to  `x = 1`.

The heights of the rectangles are the values of the graph of  `y = x^2` at the right endpoint of each rectangle, as shown in the graph below.
 

  1. State the width of each of the rectangles shown above.  (1 mark)
  2. Find the total area of the four rectangles shown above.  (1 mark)
  3. Find the area between the graph of  `y = x^2`, the `x`-axis and the line  `x=1`.  (2 marks)
  4. The graph of `f` is shown below.
     
         

    Approximate  `int_(-2)^2 f(x)\ dx`  using four rectangles of equal width and the right endpoint of each rectangle.  (1 mark)

Parts of the graphs of  `y = x^2`  and  `y = sqrtx`  are shown below.
 
     

  1. Find the area of the shaded region.  (1 mark)
  2. The graph of  `y=x^2` is transformed to the graph of  `y = ax^2`, where  `a ∈ (0, 2]`.
  3. Find the values of `a` such that the area defined by region(s) bounded by the graphs of  `y = ax^2`  and  `y = sqrtx`  and the lines  `x = 0`  and  `x = a`  is equal to `1/3`. Give your answer correct to two decimal places.  (4 marks)
Show Answers Only
  1. `0.25`
  2. `15/32\ text(u)^2`
  3. `1/3\ text(u)^2`
  4. `-2`
  5. `1/3\ text(u)^2`
  6. `1.13`
Show Worked Solution

a.   `0.25`

 

b.    `text{Area}` `= 1/4 (1/16 + 1/4 + 9/16 + 1)`
    `= 15/32\ text(u)^2`

 

c.    `text{Area}` `= int_0^1 x^2 dx`
    `= 1/3\ text(u)^2`

 

d.

♦♦♦ Mean mark part (d) 16%.

`int_-2^2 f(x)\ dx` `≈ 6 + 2 – 4 – 6`
  `≈ -2`

 

e.   `text{Area}` `= int_0^1 (sqrtx – x^2) dx`
    `= 1/3\ text(u)^2`

 
f.   `text{Case 1:} \ a ≤ 1`

♦♦♦ Mean mark part (f) 18%.

`int_0^a (sqrtx – ax^2) dx = 1/3`

`a =  0.77 \ \ text{or} \ \ a = 1.00`
  

`text{Case 2:} \ a > 1`

`sqrtx = ax^2 \ => \ x = a^{- 3/2}`

`text{Solve for}\ a\ text{(by CAS)}:`

`int_0^{a^{- 3/2}} (sqrtx – ax^2) dx + int_{a^{- 3/2}}^a (sqrtx – ax^2) dx = 1/3`

`:. a = 1.13`

Calculus, MET1-NHT 2018 VCAA 7

Let  `f : [ 0, (pi)/(2)] → R, \ f(x) = 4 cos(x)`  and  `g : [0, (pi)/(2)]  → R, \ g(x) = 3 sin(x)`.

  1. Sketch the graph of `f` and the graph of `g` on the axes provided below.   (2 marks)

     
     
    `qquad qquad `
     

  2. Let `c` be such that  `f(c) = g(c)`,  where  `c∈[0, (pi)/(2)]`

     
    Find the value of  `sin(c)`  and the value of  `cos(c)`.   (3 marks)

     

  3.  Let `A` be the region enclosed by the horizontal axis, the graph of `f` and the graph of `g`.
    1. Shade the region `A` on the axes provided in part a. and also label the position of `c` on the horizontal axis.   (1 mark)

    2. Calculate the area of the region `A`.   (3 marks)

Show Answers Only
  1.  

     

  2. `sin(c) = (4)/(5), \ cos(c) = (3)/(5)`

     

  3. i.

     
    ii. `2 \ u^2`
Show Worked Solution

a.   

 

b.   `text(At intersection:)`

`4cos(c)` `= 3sin(c)`
`tan(c)` `= (4)/(3)`

`sin(c) = (4)/(5)`

`cos(c) = (3)/(5)`

 

c. i.

 

   ii.       `A` `= int_0^c g(x)\ dx + int_c^((pi)/(2)) f(x)\ dx`
  `= int_0^c 3sin x \ dx + int_c^((pi)/(2)) 4cos \ x \ dx`
  `= 3[-cos x]_0^c + 4[sin x]_c^((pi)/(2))`
  `= 3(-cos(c) + cos \ 0) + 4(sin \ (pi)/(2)-sin(c))`
  `= 3(-(3)/(5) + 1) + 4(1-(4)/(5))`
  `= (6)/(5) + (4)/(5)`
  `= 2 \ \ text(u²)`

Calculus, MET2-NHT 2019 VCAA 18 MC

Part of the graph of the function `f`, where  `f(x) = 8 - 2^(x-1)`, is shown below. It intersects the axes at the points `A` and `B`. The line segment joining `A` and `B` is also shown on the graph.
 


 

The area of the shaded region is

  1.  `16 - (15)/(log_e (2))`
  2.  `17 - (15)/(2log_e (2))`
  3.  `(7)/(log_e (2)) - (159)/(16)`
  4.  `16 - (15)/(2log_e (2))`
  5.  `(17)/(2log_e (2)) - 15`
Show Answers Only

`B`

Show Worked Solution

`text(When) \ \ x = 0 \ => \ y = 8 – (1)/(2) = 15/2`

`text(When) \ \ y = 0,\ => \ 8 – 2^(x-1)=0 \ => x=4`

`text(Area)_(AOB)` `= (1)/(2) xx 4 xx (15)/(2)`
  `= 15 \ text(u²)`

 

`text(Shaded Area)` `= int_0^4 8 – 2^(x -1)\ dx – (15)/(2)`
  `= 17 – (15)/(2 log_e 2)`

 
`=> \ B`

Calculus, MET1-NHT 2019 VCAA 7

The shaded region in the diagram below is bounded by the vertical axis, the graph of the function with rule  `f(x) = sin(pix)`  and the horizontal line segment that meets the graph at  `x = a`, where  `1 <= a <= 3/2`.
 


 

Let  `A(a)`  be the area of the shaded region.

  1. Show that  `A(a) = 1/pi-1/pi cos(a pi)-a sin (a pi)`.   (3 marks)

  2. Determine the range of values of `A(a)`.   (2 marks)

    1. Express in terms of  `A(a)`, for a specific value of `a`, the area bounded by the vertical axis, the graph of  `y = 2(sin(pi x) + sqrt 3/2)`  and the horizontal axis.   (2 marks)

    2. Hence, or otherwise, find the area described in part c.i.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `[2/pi, (2 + 3 pi)/(2pi)]`
    1. `2 A(a)`
    2. `(9 + 4 sqrt 3 pi)/(3 pi)`
Show Worked Solution

a.   `text(Lower border of shaded area:)\ y = f(a) = sin(a pi)`

`text(Area)` `= int_0^a sin(pi x)-sin (a pi)\ dx`
  `= [-1/pi cos (pi x)-x sin(a pi)]_0^a`
  `= [-1/pi cos(pi a)-a sin (a pi)-(-1/pi-0)]`
  `= 1/pi-1/pi cos (a pi)-a sin(a pi)`

 

b.   `text(S) text(ince)\ 1 <= a <= 3/2,`

`A(1) = 1/pi-1/pi cos(pi)-1 sin (pi) = 2/pi`

`A(3/2) = 1/pi-1/pi cos ((3 pi)/2)-3/2 sin ((3pi)/2) = 1/pi + 3/2 = (2 + 3pi)/(2 pi)`
 

`:.\ text(Range:)\ \ [2/pi, (2 + 3pi)/(2 pi)]`

 

c.i.  `A(a) = int_0^a sin(pi x)-sin (a pi)\ dx`

`A_1` `=2int _0^a sin(pi x) + sqrt 3/2\ dx`   
  `=2int _0^a sin(pi x)-sin((4pi)/3)\ dx,\ \ \ (a=4/3)`  
  `=2int _0^(4/3) sin(pi x)-sin((4pi)/3)\ dx`  
  `=2 xx A(a)`  

 
`:.\ text(When)\ \ a = 4/3,\ \ text(Area) = 2 xx A(a)`

 

c.ii.  `text(When)\ \ a = 4/3`

`text(Area)` `= 2 xx (1/pi-1/pi cos ((4 pi)/3)-4/3 sin ((4 pi)/3))`
  `= 2 xx (1/pi + 1/(2 pi) + 4/3 xx sqrt 3/2)`
  `= 2(3/(2pi) + (2 sqrt 3)/3)`
  `= 3/pi + (4 sqrt 3)/3`
  `= (9 + 4 sqrt 3 pi)/(3 pi)`

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, MET1 2018 VCAA 9

Consider a part of the graph of  `y = xsin(x)`, as shown below.

  1.  i. Given that  `int(xsin(x))\ dx = sin(x)-xcos(x) + c`, evaluate  `int_(npi)^((n + 1)pi)(xsin(x))\ dx`  when `n` is a positive even integer or 0.
      
    Give your answer in simplest form.   (2 marks)

    ii. Given that  `int(xsin(x))\ dx = sin(x)-xcos(x) + c`, evaluate  `int_(npi)^((n + 1)pi)(xsin(x))\ dx`  when `n` is a positive odd integer.
    Give your answer in simplest form.   (1 mark)

  2.  
  3. Find the equation of the tangent to  `y = xsin(x)`  at the point  `(−(5pi)/2,(5pi)/2)`.   (2 marks)

  4. The translation `T` maps the graph of  `y = xsin(x)`  onto the graph of  `y = (3pi-x)sin(x)`, where
  5. `qquad T: R^2 -> R^2, T([(x),(y)]) = [(x),(y)] + [(a),(0)]`
  6. and `a` is a real constant.
  7. State the value of `a`.   (1 mark)

  8. Let  `f:[0,3pi] -> R, f(x) = (3pi-x)sin(x)`  and  `g:[0,3pi] -> R, g(x) = (x-3pi)sin(x)`.
    The line `l_1` is the tangent to the graph of `f` at the point `(pi/2,(5pi)/2)` and the line `l_2` is the tangent to the graph of `g` at `(pi/2,-(5pi)/2)`, as shown in the diagram below.
     
         
    Find the total area of the shaded regions shown in the diagram above.   (2 marks)

Show Answers Only
  1.  i. `(2n + 1)pi`
    ii. `-(2n + 1)pi`
  2. `:. y =-x`
  3. `-3pi`
  4. `9pi(pi-2)`
Show Worked Solution

a.i.  `text(Given)\ \ n\ \ text(is a positive even integer:)`

♦♦ Mean mark 31%.

`int_(npi)^((n + 1)pi)(xsin(x))dx`

`= [sin(x)-xcos(x)]_(npi)^((n + 1)pi)`

`= [sin((n + 1)pi)-(n + 1)pi · cos((n + 1)pi)]-[sin(npi)-npi · cos(npi)]`

`= [0-(n + 1)pi (−1)]-[0-npi]`

`= (n + 1)pi + npi`

`= (2n + 1)pi`

 

a.ii. `text(Given)\ \ n\ \ text(is a positive odd integer:)`

♦♦♦ Mean mark 19%.

`int_(npi)^((n + 1)pi)(xsin(x))dx`

`= [sin((n + 1)pi)-(n + 1)pi · cos((n + 1)pi)]-[sin(npi)-npi · cos(npi)]`

`= [0-(n + 1)pi(1)]-[0-npi(−1)]`

`= -(n + 1)pi-npi`

`= -(2n + 1)pi`

 

b.    `y` `= xsin(x)`
  `(dy)/(dx)` `= x · cos(x) + sin(x)`

 

♦ Mean mark part (b) 49%.

`(dy)/(dx)` `= -(5pi)/2 · cos(-(5pi)/2) + sin(-(5pi)/2)`
  `= -(5pi)/2 · (0) + (-1)`
  `= -1`

 
`text(T)text(angent has equation)\ \ y =-x + c\ \ text(and passes through)\ \ (-(5pi)/2, (5pi)/2):`

`(5pi)/2` `= +(5pi)/2 + c`
`c` `= 0`

 
`:. y = -x`

 

♦♦ Mean mark part (c) 34%.

c.    `y` `= (3pi-x)sin(x)`
    `=-(x-3pi)sin(x)`

 
`:. a = 3pi`
 

d.   `f(x) = (3pi-x)sin(x)`

♦♦♦ Mean mark 7%.

  `-> l_1\ text(is the tangent)\ \ y =-x\ \ (text{using part (b)})`

`-> g(x)\ text(is)\ \ y=xsin(x)\ \ text(translated 3π to the right.)`

`-> f(x)\ text(is)\ \ g(x)\ \ text(reflected in the)\ \ x text(-axis.)`

 
`text(Area between)\ f(x)\ text(and)\ xtext(-axis)`

`= int_0^pi xsin(x)\ dx + | int_pi^(2pi) xsin(x)\ dx | + int_(2pi)^(3pi) xsin(x)\ dx`

`= (2 xx 0 + 1)pi + |-1(2 xx 1 + 1)pi | + (2 xx 2 + 1)pi\ \ (text{using part (a)})`

`= pi + 3pi + 5pi`

`= 9pi`
 

`:.\ text(Shaded Area)`

`= 2 xx (1/2 xx 3pi xx 3pi)-2 xx 9pi`

`= 9pi^2-18pi`

`= 9pi(pi-2)`

Calculus, MET1 2018 VCAA 8

Let  `f: R -> R, \ f(x) = x^2e^(kx)`, where `k` is a positive real constant.

  1.  Show that  `f^{′}(x) = xe^(kx)(kx + 2)`.   (1 mark)

  2. Find the value of `k` for which the graphs of  `y = f(x)`  and  `y = f^{′}(x)`  have exactly one point of intersection.   (3 marks)

Let  `g(x) = −(2xe^(kx))/k`. The diagram below shows sections of the graphs of `f` and `g` for  `x >= 0`.
 


 

Let `A` be the area of the region bounded by the curves  `y = f(x), \ y = g(x)` and the line  `x = 2`.

  1. Write down a definite integral that gives the value of `A`.   (1 mark)

  2. Using your result from part a., or otherwise, find the value of `k` such that  `A = 16/k`.   (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(One point of intersection)\ x = 0\ text(when)\ k = 1.`
  3. `int_0^2 (x^2e^(kx) + (2xe^(kx))/k)dx`
  4. `1/2 ln 4\ \ text(or)\ \ ln2`
Show Worked Solution

a.   `f(x) = x^2e^(kx)`

`f^{′}(x)` `= 2x · e^(kx) + x^2 · k · e^(kx)`
  `= xe^(kx)(kx + 2)\ \ …text(as required)`

 

b.  `text(Intersection occurs when:)`

♦♦♦ Mean mark 7%.
MARKER’S COMMENT: Most students found the correct quadratic equation but solved for `k`.

`x^2e^(kx)` `= xe^(kx)(kx + 2)`
`x^2-x(kx + 2)` `= 0,\ \ e^(kx) != 0`
`x^2-kx^2-2x` `= 0`
`x^2(1-k)-2x` `= 0`
`x[x(1-k)-2]` `= 0`

 

`:.x = 0\ \ text(or)\ \ x(1-k)-2` `= 0`
`x` `= 2/(1-k)`

 
`text(S)text(ince)\ \ x = 2/(1-k)\ \ text(is undefined when)\ \ k = 1,`

`=>\ text(One point of intersection only at)\ \ x = 0\ \ text(when)\ \ k = 1.`

 

c.    `A` `= int_0^2 f(x)\ dx-int_0^2 g(x)\ dx`
    `= int_0^2 x^2e^(kx)\ dx-int_0^2 −(2xe^(kx))/k \ dx`
    `= int_0^2 (x^2e^(kx) + (2xe^(kx))/k)dx`

♦♦ Mean mark 29%.

 

d.    `int_0^2(x^2e^(kx) + (2xe^(kx))/k)\ dx` `= 16/k`
  `1/k int_0^2(kx^2e^(kx) + 2xe^(kx))\ dx` `= 16/k`
  `1/k int_0^2(xe^(kx)(kx + 2))\ dx` `= 16/k`
  `1/k[x^2e^(kx)]_0^2` `= 16/k\ \ \ text{(using part a)}`
  `[2^2 · e^(2k)-0]` `= 16`
  `e^(2k)` `= 4`
  `2k` `= ln 4`
  `k` `= 1/2 ln 4\ \ text(or)\ \ ln2`

Calculus, MET2 2018 VCAA 19 MC

The graphs  `f: R -> R,\ f(x) = cos ((pi x)/2)`  and  `g: R -> R,\ g(x) = sin (pi x)` are shown in the diagram below.
 


 

An integral expression that gives the total area of the shaded regions is

A.   `int_0^3 (sin(pi x) - cos ((pi x)/2)) dx`

B.   `2 int_(5/3)^3 (sin(pi x) - cos((pi x)/2))dx`

C.   `int_0^(1/3)(cos((pi x)/2) - sin (pi x)) dx - 2 int_(1/3)^1(cos((pi x)/2) - sin (pi x)) dx`

                `- int_(5/3)^3 (cos ((pi x)/2) - sin (pi x)) dx`

D.   `2 int_1^(5/3)(cos((pi x)/2) - sin (pi x)) dx - 2 int_(5/3)^3 (cos ((pi x)/2) - sin (pi x)) dx`

E.   `int_0^(1/3) (cos((pi x)/2) - sin (pi x)) dx + 2 int_(1/3)^1(sin (pi x) - cos ((pi x)/2))dx`

                `+ int_(5/3)^3 (cos((pi x)/2) - sin (pi x)) dx`

Show Answers Only

`C`

Show Worked Solution

`text(The graph contains 2 symmetrical shaded areas:)`

♦ Mean mark 41%.
COMMENT: In each shaded area, it is critical to note which graph is “higher” and which side of the x-axis the shading is.
 

`:.\ text(Area can be represented by:)`

`int_0^(1/3)(cos((pi x)/2) – sin (pi x)) dx – 2 int_(1/3)^1(cos((pi x)/2) – sin (pi x)) dx`

         `- int_(5/3)^3 (cos ((pi x)/2) – sin (pi x)) dx`

 
`=>   C`

Algebra, MET2 2017 VCAA 4

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

  1. The transformation  `T: R^2 -> R^2, \ T([(x),(y)]) = [(x),(y)] + [(c),(d)]`  maps the graph of  `y = 2^x`  onto the graph of  `f`.

     

    State the values of `c` and `d`.   (2 marks)

  2. Find the rule and domain for  `f^(-1)`, the inverse function of  `f`.   (2 marks)

  3. Find the area bounded by the graphs of  `f` and  `f^(-1)`.   (3 marks)

  4. Part of the graphs of  `f` and  `f^(-1)` are shown below.
     

         
     
    Find the gradient of  `f` and the gradient of  `f^(-1)`  at  `x = 0`.   (2 marks)

The functions of  `g_k`, where  `k ∈ R^+`, are defined with domain `R` such that  `g_k(x) = 2e^(kx)-2`.

  1. Find the value of `k` such that  `g_k(x) = f(x)`.  (1 mark)

  2. Find the rule for the inverse functions  `g_k^(-1)` of  `g_k`, where  `k ∈ R^+`.   (1 mark)

  3. i. Describe the transformation that maps the graph of  `g_1` onto the graph of  `g_k`.   (1 mark)

    ii. Describe the transformation that maps the graph of  `g_1^(-1)` onto the graph of  `g_k^(-1)`.   (1 mark)

  4. The lines `L_1` and `L_2` are the tangents at the origin to the graphs of  `g_k`  and  `g_k^(-1)`  respectively.
  5. Find the value(s) of `k` for which the angle between `L_1` and `L_2` is 30°.   (2 marks)

  6. Let `p` be the value of `k` for which  `g_k(x) = g_k^(−1)(x)`  has only one solution.
  7.  i. Find `p`.   (2 marks)

  8. ii. Let  `A(k)`  be the area bounded by the graphs of  `g_k`  and  `g_k^(-1)`  for all  `k > p`.
  9.     State the smallest value of `b` such that  `A(k) < b`.   (1 mark)

Show Answers Only

a.  `c =-1, \ d =-2`

b.  `f^(-1)(x) = log_2 (x + 2)-1, \ x ∈ (-2,∞)`

c.  `3-2/(log_e(2))\ text(units)²`

d. `f^{prime}(0)= 2log_e(2) and f^(-1)^{prime}(0)= 1/(2log_e(2))`

e.  `k = log_e(2)`

f.  `g_k^(-1)(x)= 1/k log_e((x + 2)/2)`

g.i.  `text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

g.ii.  `text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

h.  `k=sqrt3/6\ or\ sqrt3/2`

i.i.  `p = 1/2`

i.ii.  `b=4`

Show Worked Solution

a.   `text(Using the matrix transformation:)`

`x^{prime}` `=x+c\ \ => x=x^{prime}-c`
`y^{prime}` `=y+d\ \ =>y= y^{prime}-d`
   
`y^{prime}-d` `=2^((x)^{prime}-c)`
`y^{prime}` `= 2^((x)^{prime}-c)+d`

 

`:. c = -1, \ d = -2`

 

b.   `text(Let)\ \ y = f(x)`

`text(Inverse : swap)\ x\ ↔ \ y`

`x` `= 2^(y + 1)-2`
`x + 2` `= 2^(y + 1)`
`y + 1` `= log_2(x + 2)`
`y` `= log_2(x + 2)-1`

 

`text(dom)(f^(-1)) = text(ran)(f)`

`:. f^(-1)(x) = log_2 (x + 2)-1, \ x ∈ (-2,∞)`

 

c.   `text(Intersection points occur when)\ \ f(x)=f^(-1)(x)`

MARKER’S COMMENT: Specifically recommends using  `f^(-1)(x)-f(x)`  in this type of integral to avoid errors in stating the integral.

`x` `= -1, 0`

 

`text(Area)` `= int_(-1)^0 (f^(-1)(x)-f(x))\ dx`
  `= 3-2/(log_e(2))\ text(units)²`

 

d.    `f^{prime}(0)` `= 2log_e(2)`
  `f^{(-1)^prime}(0)` `= 1/(2log_e(2))`

 

e.   `g_k(x) = 2e^(kx)-2`

`text(Solve:)\ \ g_k(x) = f(x)quad text(for)\ k ∈ R^+`

`:. k = log_e(2)`

 

f.   `text(Let)\ \ y = g_k(x)`

`text(Inverse : swap)\ x\ text(and)\ y`

`x` `= 2e^(ky)-2`
`e^(ky)` `=(x+2)/2`
`ky` `=log_e((x+2)/2)`
`:. g_k^(-1)(x)` `= 1/k log_e((x + 2)/2)`

 

♦♦ Mean mark part (g)(i) 31%.

g.i.    `g_1(x)` `= 2e^x-2`
  `g_k(x)` `= 2e^(kx)-2`

 
`:. text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

 

♦♦ Mean mark part (g)(ii) 30%.

g.ii.    `g_1^(-1)(x)` `= log_e((x + 2)/2)`
  `g_k^(-1)(x)` `= 1/klog_e((x + 2)/2)`

 
`:. text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

 

h.   `text(When)\ \ x=0,`

♦♦♦ Mean mark part (h) 13%.

`g_k^{prime}(0)=2k\ \ => m_(L_1)=2k`

`g_k^{(-1)^prime}(0)=1/(2k)\ \ => m_(L_2)=1/(2k)`

 

`text(Using)\ \ tan 30^@=|(m_1-m_2)/(1+m_1m_2)|,`

`text(Solve:)\ \ 1/sqrt3` `=+- ((2k-1/(2k)))/2\ \ text(for)\ k`

 
`:. k=sqrt3/6\ or\ sqrt3/2`

 

i.i   `text(By inspection, graphs will touch once if at)\ \ x=0,`

♦♦♦ Mean mark part (i)(i) 4%.

`m_(L_1)` `=m_(L_2)`
`2k` `=1/(2k)`
`k` `=1/2,\ \ (k>0)`

 
`:. p = 1/2`

 

i.ii  `text(As)\ k→oo, text(the graph of)\ g_k\ text(approaches)\ \ x=0`

♦♦♦ Mean mark part (i)(ii) 2%.

`text{(vertically) and}\ \ y=-2\ \ text{(horizontally).}`

`text(Similarly,)\ \ g_k^(-1)\ \ text(approaches)\ \ x=-2`

`text{(vertically) and}\ \ y=0\ \ text{(horizontally).}`

 

`:. lim_(k→oo) A(k) = 4`

`:.b=4`

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, MET2 2017 VCAA 20 MC

The graphs of  `f:[0,pi/2] -> R,  f(x) = cos(x)`  and  `g:[0,pi/2] -> R,  g(x) = sqrt3 sin(x)`  are shown below.

The graph intersect at `B`.

 

The ratio of the area of the shaded region to the area of triangle `OAB` is

  1. `9:8`
  2. `sqrt3 - 1 : (sqrt3 pi)/8`
  3. `8sqrt3 - 3 : 3pi`
  4. `sqrt3 - 1 : (sqrt3 pi)/4`
  5. `1 : (sqrt3 pi)/8`
Show Answers Only

`B`

Show Worked Solution

`text(Find intersection point)\ B:`

`cos(x)` `=sqrt3 sin(x)`
`tan x` `=1/sqrt3`
`x` `=pi/6`

`:. B(pi/6,sqrt3/2)`

 

`A_text(shaded) : A_Δ`

`int_0^(pi/6) sqrt3 sin(x)\ dx + int_(pi/6)^(pi/2) cos(x)\ dx\ :\ 1/2 xx pi/2 xx sqrt3/2`

`sqrt3-1\ :\ (sqrt3 pi)/8`

`=> B`

Calculus, MET1 SM-Bank 6

The diagram shows the function  `f:(2, oo)→R`,  where  `f(x)= log_e(x - 2).`

In the diagram, the shaded region is bounded by  `f(x)`, the `x`-axis and the line  `x = 7`.

Find the exact value of the area of the shaded region.  (4 marks)

Show Answers Only

`5log_e 5 – 4\ \ \ text(u²)`

Show Worked Solution

`text(Shaded Area)\ (A_1)` `= text(Rectangle) – A_2`
`text(Area of Rectangle)` `= 7 xx log_e 5`

 

`text(Finding the Area of)\ A_2`

`y` `= log_e(x – 2)`
`x – 2` `= e^y`
`x` `= e^y + 2`
`:. A_2` `= int_0^(log_e5) x\ dy`
  `= int_0^(log_e5) e^y + 2\ dy`
  `= [e^y + 2y]_0^(log_e5)`
  `= [(e^(log_e 5) + 2log_e5) – (e^0 + 0)]`
  `= (5 + 2log_e 5) – 1`
  `= 4 + 2log_e 5`

 

`:. A_1` `= 7 log_e5 – (4 + 2log_e 5)`
  `= 5log_e 5 – 4\ \ \ text(u²)`

Calculus, MET1 SM-Bank 4

The curves  `y = e^(2x)` and  `y = e^(−x)` intersect at the point `(0,1)` as shown in the diagram.

Find the exact area enclosed by the curves and the line  `x = 2`.  (3 marks)

Show Answers Only

`1/2 e^4 + e^(−2) – 3/2\ \ \ text(u²)`

Show Worked Solution

Note: there’s marker’s comment in this part

`text(Area)` `= int_0^2 e^(2x)\ \ dx – int_0^2 e^(−x)\ \ dx`
  `= int_0^2 (e^(2x) – e^(−x))dx`
  `= [1/2 e^(2x) + e^(−x)]_0^2`
  `= [(1/2 e^4 – e^(−2)) – (1/2 e^0 + e^0)]`
  `= 1/2 e^4 + e^(−2) – 3/2\ \ \ text(u²)`

Calculus, MET2 2009 VCAA 22 MC

Consider the region bounded by the `x`-axis, the `y`-axis, the line with equation  `y = 3`  and the curve with equation  `y = log_e (x - 1).`

The exact value of the area of this region is

A.   `e^-3 - 1`

B.   `16 + 3 log_e (2)`

C.   `3e^3 - e^-3 + 2`

D.   `e^3 + 2`

E.   `3e^2`

Show Answers Only

`D`

Show Worked Solution
♦ Mean mark 45%.

vcaa-2009-22i

`text(Solution 1)`

`text(Area)` `= 3 (e^3 + 1) – int_2^(e^3 + 1) (log_e (x – 1))\ dx`
  `= 3 (e^3 + 1) – (2e^3 + 1)`
  `= e^3 + 2`

`=>   D`

 

`text(Solution 2)`

`text(Inverse of)\ \ y = log_e (x – 1)\ \ text(is)`

`y=e^x+1`

`text(Area)` `=int_0^3 (e^x+1)\ dx`
  `=[e^x+x]_0^3`
  `=[(e^3+3)-e^0]`
  `=e^3 +2\ \ text(u²)`

Calculus, MET2 2016 VCAA 4

  1. Express  `(2x + 1)/(x + 2)`  in the form  `a + b/(x + 2)`,  where `a` and `b` are non-zero integers.   (2 marks)

  2. Let  `f: R text(\){−2} -> R,\ f(x) = (2x + 1)/(x + 2)`.
    1. Find the rule and domain of `f^(-1)`, the inverse function of `f`.   (2 marks)

    2. Part of the graphs of `f` and  `y = x`  are shown in the diagram below.
       

    3. Find the area of the shaded region.   (1 mark)

    4. Part of the graphs of `f` and `f^(-1)` are shown in the diagram below.
       

    5. Find the area of the shaded region.   (1 mark) 
  1. Part of the graph of `f` is shown in the diagram below.
     
       
     

     

    The point `P(c, d)` is on the graph of `f`.

     

    Find the exact values of `c` and `d` such that the distance of this point to the origin is a minimum, and find this minimum distance.   (3 marks)

Let  `g: (−k, oo) -> R, g(x) = (kx + 1)/(x + k)`, where `k > 1`.

  1. Show that  `x_1 < x_2`  implies that  `g(x_1) < g(x_2),` where  `x_1 in (−k, oo) and x_2 in (−k, oo)`.   (2 marks)

  2. Let `X` be the point of intersection of the graphs of  `y = g (x) and y = −x`.
    1. Find the coordinates of `X` in terms of `k`.   (2 marks)

    2. Find the value of `k` for which the coordinates of `X` are  `(-1/2, 1/2)`.   (2 marks)

    3. Let  `Ztext{(− 1, − 1)}, Y(1, 1)`  and `X` be the vertices of the triangle `XYZ`. Let  `s(k)`  be the square of the area of triangle `XYZ`.
       

       

         
       

       

      Find the values of `k` such that  `s(k) >= 1`.   (2 marks)

  3. The graph of `g` and the line  `y = x`  enclose a region of the plane. The region is shown shaded in the diagram below.
     

     

     

    Let  `A(k)`  be the rule of the function `A` that gives the area of this enclosed region. The domain of `A` is  `(1, oo)`.

    1. Give the rule for `A(k)`.   (2 marks)

    2. Show that  `0 < A(k) < 2`  for all  `k > 1`.   (2 marks)

Show Answers Only
  1. `2 + {(−3)}/(x + 2)`
  2.   i. `f^(-1) (x) = (-3)/(x-2)-2,\ \ x in R text(\){2}`
  3.  ii. `4-3 ln(3)\ text(units²)`
  4. iii. `8-6 log_e (3)\ text(units²)`
  5. `c = sqrt 3-2, \ d = 2-sqrt 3`
  6. `text(min distance) = (2 sqrt 2-sqrt 6)`
  7. `text(Proof)\ \ text{(See Worked Solutions)}`
  8.   i. `X (−k + sqrt (k^2-1), k-sqrt (k^2-1))`
  9.  ii. `k = 5/4`
  10. iii. `k in (1, 5/4]`
  11.  i. `A(k) = (k^2-1) log_e ((k-1)/(k + 1)) + 2k`
  12. ii. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.   `text(Solution 1)`

`(2x + 1)/(x + 2)` `= 2-3/(x + 2)`
  `= 2 + {(-3)}/(x + 2) qquad [text(CAS: prop Frac) ((2x + 1)/(x + 2))]`

 

`text(Solution 2)`

`(2x + 1)/(x + 2)` `=(2(x+2)-3)/(x+2)`
  `=2+ (-3)/(x+2)`

 

b.i.  `text(Let)\ \ y = f(x)`

`text(For Inverse: swap)\ \ x ↔ y`

`x` `=2-3/(y + 2)`
`(x-2)(y+2)` `=-3`
`y` `=(-3)/(x-2)-2`

 

`text(Range of)\ f(x):\ \ y in R text(\){2}`

`:. f^(-1) (x) = (-3)/(x-2)-2, \ x in R text(\){2}`

 

b.ii.   `text(Find intersection points:)`

`f(x)` `= x`
`(2x+1)/(x+2)` `=x`
`2x+1` `=x^2+2x`
`:. x` `= +- 1`
`:.\ text(Area)` `= int_(-1)^1 (f(x)-x)\ dx`
  `=int_(-1)^1 (2-3/(x + 2)-x)\ dx`
  `= 4-3 ln(3)\ text(u²)`

 

b.iii.    `text(Area)` `= int_(-1)^1 (f(x)-f^(-1) (x)) dx`
    `=2 xx (4-3 ln(3))\ \ \ text{(twice the area in (b)(ii))}`
    `= 8-6 log_e (3)\ text(u²)`
♦♦♦ Mean mark part (c) 25%.

 

c.   
  `text(Let)\ \ z` `= OP, qquad P(c, -3/(c + 2) + 2)`
  `z` `= sqrt (c^2 + (2-3/(c + 2))^2), \ c > -2`

`text(Stationary point when:)`

`(dz)/(dc) = 0, c > -2`

`:.\ c = sqrt 3-2 overset and (->) d = 2-sqrt 3`

`:. text(Minimum distance) = (2 sqrt 2-sqrt 6)`

 

d.   `text(Given:)\ \ -k < x_1 < x_2`

♦♦♦ Mean mark part (d) 8%.

`text(Must prove:)\ \ g(x_2)-g(x_1) > 0`

`text(LHS:)`

`g(x_2)-g(x_1)`

`= (kx_2 +1)/(x_2+k)-(kx_1 +1)/(x_1+k)`

`=((kx_2 +1)(x_1+k)-(kx_1 +1)(x_2+k))/((x_2+k)(x_1+k))`

`=(k^2(x_2-x_1)-(x_2-x_1))/((x_2+k)(x_1+k))`

`=((k^2-1)(x_2-x_1))/((x_2+k)(x_1+k))`

 

`x_2-x_1 >0,\ and \ k^2-1>0`

`text(S)text(ince)\ \ x_2>x_1> -k,`

`=> -x_2<-x_1<k`

`=>k+x_1 >0, \ and \ k+x_2>0`

 

`:.g(x_2)-g(x_1) >0`

`:. g(x_2) > g(x_1)`

 

♦♦ Mean mark part (e)(i) 32%.
e.i.    `g(x)` `= -x`
  `(kx +1)/(x+k)` `=-x`
  `kx+1` `=-x^2-xk`
  `x^2+2k+1` `=0`
  `:. x` `=(-2k +- sqrt(4k^2-4))/2`
    `= sqrt (k^2-1)-k\ \ text(for)\ \ x > -k`

 

`:. X (-k + sqrt(k^2-1),\ \ k-sqrt(k^2-1))`

 

e.ii.   `text(Equate)\ \ x text(-coordinates:)`

♦ Mean mark part (e)(ii) 44%.
`-k + sqrt(k^2-1)` `= -1/2`
`sqrt(k^2-1)` `=k-1/2`
`k^2-1` `=k^2-k+1/4`
`:. k` `= 5/4`

 

e.iii.   `s(k)` `= (1/2 xx YZ xx XO)^2`
    `= 1/4 xx (YZ)^2 xx (XO)^2`

 

`ZO = sqrt(1^2+1^2) = sqrt2`

♦♦♦ Mean mark (e)(iii) 6%.

`YZ=2 xx ZO = 2sqrt2`

`(YZ)^2 = 8`

`(XO)^2` `=(-k + sqrt(k^2-1))^2-(k-sqrt (k^2-1))^2`
  `=2(-k + sqrt(k^2-1))^2`
   

 `text(Solve)\ \ s(k) >= 1\ \ text(for)\ \ k >= 1,`

`1/4 xx 8 xx 2(-k + sqrt(k^2-1))^2` `>=1`
`(-k + sqrt(k^2-1))^2` `>=1/4`
`k-sqrt(k^2-1)` `>=1/2`
`k-1/2` `>= sqrt(k^2-1)`
`k^2-k+1/4` `>=k^2-1`
`:.k` `<= 5/4`

`:.  1<k<= 5/4`

 

♦♦ Mean mark (f)(i) 28%.
f.i.    `A(k)` `= int_(-1)^1 (g(x)-x)\ dx,\ \ k > 1`
    `= int_(-1)^1 ((kx+1)/(x+k) -x)\ dx`
    `= int_(-1)^1 (k+ (1-k^2)/(x+k) -x)`
    `=[kx + (1-k^2) log_e(x+k)-x^2/2]_(-1)^1`
    `=(k+(1-k^2)log_e(1+k)-1/2)-(-k+(1-k^2)log_e(k-1)-1/2)`
    `=2k+(1-k^2)log_e ((1+k)/(k-1))`
    `= (k^2-1) log_e ((k-1)/(k + 1)) + 2k`

 

♦♦♦ Mean mark part (f)(ii) 4%.
f.ii.  
  `0` `< A(k) < text(Area of)\ Delta ABC`
  `0` `< A(k) < 1/2 xx AC xx BO`
  `0` `< A(k) < 1/2 sqrt(2^2 + 2^2) xx (sqrt(1^2 + 1^2))`
  `0` `< A(k) < 1/2 xx 2 sqrt 2 xx sqrt 2`
  `:. 0` `< A(k) < 2`

Calculus, MET2 2016 VCAA 13 MC

Consider the graphs of the functions `f` and `g` shown below.

The area of the shaded region could be represented by

A.   `int_a^d (f(x) - g(x))\ dx`

B.   `int_0^d (f(x) - g(x))\ dx`

C.   `int_0^b (f(x) - g(x))\ dx + int_b^c (f(x) - g(x))\ dx`

D.   `int_0^a f(x)\ dx + int_a^c (f(x) - g(x))\ dx + int_b^d f(x)\ dx`

E.   `int_0^d f(x)\ dx - int_a^c g(x)\ dx`

Show Answers Only

`E`

Show Worked Solution

`text(Shaded Area:)`

`text(Area under)\ f(x)\ text(between)\ [0, d]\ text(less)`

`text(Area under)\ g(x)\ text(between)\ [a, c]`

`=>   E`

Calculus, MET2 2011 VCAA 14 MC

met2-2011-vcaa-14-mc2 

To find the area of the shaded region in the diagram shown, four different students proposed the following calculations.

  1. `int_0^1 e^(2x)\ dx`
  2. `e^2 - int_0^1 e^(2x)\ dx`
  3. `int_1^(e^2) e^(2y)\ dy`
  4. `int_1^(e^2) (log_e(x))/2 \ dx`

Which of the following is correct?

A.   ii. only

B.   ii. and iii. only

C.   i., ii., iii. and iv.

D.   ii. and iv. only

E.   i. and iv. only 

Show Answers Only

`=> D`

Show Worked Solution
`text(Area)` `=\ text(Area of rect) – text(Area under curve)`
  `= e^2 – int_0^1 e^(2x)\ dx`

`:.\ text(Statement ii is correct.)`

 

`text(Consider the shaded area between the curve and)\ y text(-axis:)`

`y` `=e^(2x)`
`log_e y` `=2x`
`x` `=(log_e(y))/2`

`text(Area)\ = int_1^(e^2) (log_e(x))/2 \ dx`

`:.\ text(Statement iv is correct.)`

`=> D`

Calculus, MET1 2007 VCAA 9

The graph of  `f: R -> R`,  `f(x) = e^(x/2) + 1`  is shown. The normal to the graph of `f` where it crosses the `y`-axis is also shown.
 

MET1 2007 VCAA Q9
 

  1. Find the equation of the normal to the graph of `f` where it crosses the `y`-axis.   (2 marks)

  2. Find the exact area of the shaded region.   (3 marks)

Show Answers Only
  1. `y = -2x + 2`
  2. `(2sqrte-2)\ text(u)²`
Show Worked Solution

a.   `text(Normal is)\ ⊥\ text(to tangent)`

MARKER’S COMMENT: A common error was made in calculating the point of tangency. Be careful!

`text(Point of tangency:)\ (0,2)`

`text(Gradient of normal:)`

`f^{prime}(x)` `= 1/2 e^(x/2)`
`f^{prime}(0)` `= 1/2`

  
`:. m_text(norm) = -2`
  

`text(Equation of normal:)`

`y-y_1` `= m(x-x_1)`
`y-2` `=-2(x-0)`
`y` `=-2x+2`

 

♦ Mean mark 44%.
b.    `:.\ text(Area)` `= int_0^1 (e^(x/2) + 1-(-2x +2))\ dx`
    `= int_0^1 (e^(x/2) + 2x-1)\ dx`
    `= [2e^(x/2) + x^2-x]_0^1`
    `= (2e^(1/2) + 1^2-1)-(2e^0)`
    `= (2sqrte-2)\ text(u)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 2013 VCAA 10

Let  `f: [0, oo) -> R,\ \ f(x) = 2e^(-x/5).`

A right-angled triangle `OQP` has vertex `O` at the origin, vertex `Q` on the `x`-axis and vertex `P` on the graph of `f`, as shown. The coordinates of `P` are `(x, f(x)).`
 

 vcaa-2013-meth-10

  1. Find the area, `A`, of the triangle `OPQ` in terms of `x`.   (1 mark)

  2. Find the maximum area of triangle `OQP` and the value of `x` for which the maximum occurs.   (3 marks)

  3. Let `S` be the point on the graph of `f` on the `y`-axis and let `T` be the point on the graph of `f` with the `y`-coordinate `1/2`.

     

    Find the area of the region bounded by the graph of `f` and the line segment `ST`.   (3 marks)

     

    vcaa-2013-meth-10i

Show Answers Only
  1. `x e^(-x/5)`
  2. `5/e\ text(u²)`
  3. `25/4 log_e (4)-15/2\ text(u²)`
Show Worked Solution
a.   `text(Area)` `= 1/2 xx b xx h`
    `= 1/2x(2e^(-x/5))`
  `:. A` `= xe^(-x/5)`

 

b.   `text(Stationary point when)\ \ (dA)/(dx) = 0,`

♦ Mean mark 35%.
`x(-1/5 e^(-x/5)) + e^(-x/5)` `= 0`
`e^(-x/5)[1-x/5]` `= 0`
`:. x` `= 5\ \ text{(as}\ \ e^(-x/5) >0,\ \ text(for all)\ x text{)}`
`text(When)\ \ x = 5,\ \ A` `= xe^(-x/5)`
  `= 5e^-1`

`:. A_max = 5/e\ text(u²,   when)\ \ x = 5`

 

c.   `text(Find)\ \ S:\ \ F(0) = 2.`

♦♦ Mean mark 32%.

`:. S(0, 2)`
 

`text(Find)\ \ T:\ \ \ ` `2e^(-x/5)` `= 1/2`
  `e^(-x/5)` `= 1/4`
  `-x/5` `= log_e (1/4)`
  `x` `= 5 log_e (4)`

`:. T(5log_e(4), 1/2)`

 

vcaa-2013-meth-10ii

`:.\ text(Area)` `= text(Area)\ SOAT-int_0^(5 log_e(4)) (2e^(-x/5)) dx`
  `=1/2h(a+b) + 10 [e^(-x/5)]_0^(5 log_e (4))`
  `= 5/2 log_e (4) (2 + 1/2) + 10 [e^(-log_e (4))-e^0]`
  `= 25/4 log_e (4) +10 (1/4-1)`
  `= 25/4 log_e (4)-15/2\ text(u²)`

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

 

♦ Mean mark (c) 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 2010 VCAA 1

  1. Part of the graph of the function  `g: (-4, oo) -> R,\ g(x) = 2 log_e (x + 4) + 1`  is shown on the axes below

     

         

    1. Find the rule and domain of  `g^-1`, the inverse function of  `g`.   (3 marks)

    2. On the set of axes above sketch the graph of  `g^-1`. Label the axes intercepts with their exact values.   (3 marks)

    3. Find the values of `x`, correct to three decimal places, for which  `g^-1(x) = g(x)`.   (2 marks)

    4. Calculate the area enclosed by the graphs of  `g`  and  `g^-1`. Give your answer correct to two decimal places.   (2 marks)

  2. The diagram below shows part of the graph of the function with rule
  3.         `f (x) = k log_e (x + a) + c`, where `k`, `a` and `c` are real constants.
     

    • The graph has a vertical asymptote with equation  `x =-1`.
    • The graph has a y-axis intercept at 1.
    • The point `P` on the graph has coordinates  `(p, 10)`, where `p` is another real constant.
       

      VCAA 2010 1b

    1. State the value of `a`.   (1 mark)

    2. Find the value of `c`.   (1 mark)

    3. Show that  `k = 9/(log_e (p + 1)`.   (2 marks)

    4. Show that the gradient of the tangent to the graph of `f` at the point `P` is  `9/((p + 1) log_e (p + 1))`.   (1 mark)

    5. If the point  `(-1, 0)`  lies on the tangent referred to in part b.iv., find the exact value of `p`.   (2 marks)

Show Answers Only
  1.   i. `g^-1(x) = e^((x-1)/2)-4,\ \ x in R`
  2.  ii. 
     
  3. iii. `3.914 or 5.503`
  4. iv. `52.63\ text(units²)`
  5.   i. `1`
  6.  ii. `1`
  7. iii. `text(Proof)\ \ text{(See Worked Solutions)}`
  8.  iv. `text(Proof)\ \ text{(See Worked Solutions)}`
  9.   v. `e^(9/10)-1`
Show Worked Solution

a.i.   `text(Let)\ \ y = g(x)`

`text(Inverse:  swap)\ \ x harr y,\ \ text(Domain)\ (g^-1) = text(Range)\ (g)`

`x = 2 log_e (y + 4) + 1`

`:. g^{-1} (x) = e^((x-1)/2)-4,\ \ x in R`

 

ii.  

 

 iii.  `text(Intercepts of a function and its inverse occur)`

  `text(on the line)\ \ y=x.`

`text(Solve:)\ \ g(x) = g^{-1} (x)\ \ text(for)\ \ x`

`:. x dot = -3.914 or x = 5.503\ \ text{(3 d.p.)}`

 

  iv.   `text(Area)` `= int_(-3.91432…)^(5.50327…) (g(x)-g^-1 (x))\ dx`
    `= 52.63\ text{u²   (2 d.p.)}`

 

b.i.   `text(Vertical Asymptote:)`

`x =-1`

`:. a = 1`

 

  ii.   `text(Solve)\ \ f(0) = 1\ \ text(for)\ \ c,`

`c = 1`

 

iii.  `f(x)= k log_e (x + 1) + 1`

`text(S)text(ince)\ \ f(p)=10,`

`k log_e (p + 1) + 1` `= 10`
`k log_e (p + 1)` `= 9`
`:. k` `= 9/(log_e (p + 1))\ text(… as required)`

 

  iv.   `f^{′}(x)` `= k/(x + 1)`
  `f^{′}(p)` `= k/(p + 1)`
    `= (9/(log_e(p + 1))) xx 1/(p + 1)\ \ \ text{(using part (iii))}`
    `= 9/((p + 1)log_e(p + 1))\ text(… as required)`

 

  v.   `text(Two points on tangent line:)`

♦♦ Mean mark 33%.
MARKER’S COMMENT: Many students worked out the equation of the tangent which was unnecessary and time consuming.

`(p, 10),\ \ (-1, 0)`

`f^{′} (p)` `= (10-0)/(p-(-1))`
  `=10/(p+1)`

`text(Solve:)\ \ 9/((p + 1)log_e(p + 1))=10/(p+1)\ \ \ text(for)\ p,`

`:.p= e^(9/10)-1`

Calculus, MET2 2015 VCAA 4

An electronics company is designing a new logo, based initially on the graphs of the functions

`f(x) = 2 sin (x) and g(x) = 1/2 sin (2x),\ text(for)\ 0 <= x <= 2 pi`

These graphs are shown in the diagram below, in which the measurements in the `x` and `y` directions are in metres.

VCAA 2015 4a

The logo is to be painted onto a large sign, with the area enclosed by the graphs of the two functions (shaded in the diagram) to be painted red.

  1. The total area of the shaded regions, in square metres, can be calculated as  `a int_0^pi sin(x)\ dx`.
  2. What is the value of `a`?   (1 mark)

The electronics company considers changing the circular functions used in the design of the logo.

Its next attempt uses the graphs of the functions  `f(x) = 2 sin(x) and h(x) = 1/3 sin (3x),\ text(for)\ \ 0 <= x <= 2 pi`.

  1. On the axes below, the graph of  `y = f(x)`  has been drawn.
  2. On the same axes, draw the graph of  `y = h(x)`.   (2 marks)

     

          VCAA 2015 4b

  3. State a sequence of two transformations that maps the graph of  `y = f (x)`  to the graph of  `y = h(x)`.   (2 marks)

The electronics company now considers using the graphs of the functions  `k(x) = m sin(x) and q (x) = 1/n sin (nx)`, where `m` and `n` are positive integers with  `m >= 2` and `0<= x <= 2pi`.

    1. Find the area enclosed by the graphs of  `y = k(x)`  and  `y = q(x)`  in terms of `m` and `n` if `n` is even.
    2. Give your answer in the form  `am + b/n^2`, where `a` and `b` are integers.   (2 marks)

    3. Find the area enclosed by the graphs of  `y = k(x)`  and  `y = q(x)`  in terms of `m` and `n` if `n`is odd.
    4. Give your answer in the form  `am + b/n^2`, where `a` and `b` are integers.   (2 marks)

Show Answers Only
  1. `4`

  2.  

    vcaa-2015-4b-answer   

  3. `text(See Worked Solutions)`
    1. `4m + 0/n^2`
    2. `4m + (−4)/(n^2)`
Show Worked Solution
a.   `text(Area)` `= 2 xx int_0^pi (2 sin (x))\ dx`
    `= 4 xx int_0^pi sin (x)\ dx`

`:. a = 4`

♦ Mean mark part (a) 36%.

 

b.   vcaa-2015-4b-answer

 

c.   `text(Find sequence that takes)\ f(x) -> h(x)`

`text(A dilation by factor of)\ 1/6\ text(from the)\ xtext(-axis.)`

`text(A dilation by factor of)\ 1/3\ text(from the)\ ytext(-axis.)`

 

d.i.  `text(Area when)\ n\ text(is even)`

♦♦♦ Mean mark 20%.
MARKER’S COMMENT: Note that `cos(npi)=1` for `n` even.
`= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx`
`=2 [-mcos x + 1/n^2 cos (nx)]_0^pi`
`=2[m+1/n^2 cos (npi) – (-m+1/n^2)]`
`= 2(2m+(cos(npi)-1)/n^2)`
`= 4m + 0/(n^2)`

 

d.ii.  `text(Area when)\ n\ text(is even)`

♦♦♦ Mean mark 14%.
MARKER’S COMMENT: Note that `cos(npi)=-1` for `n` odd.
`= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx`
`=2 [-mcos x + 1/n^2 cos (nx)]_0^pi`
`=2[m+1/n^2 cos (npi) – (-m+1/n^2)]`
`= 2(2m+(cos(npi)-1)/n^2)`
`= 4m + (-4)/(n^2)`

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)

  2.  i. Find the equation of the tangent to the graph of  `f` at the point  `P(1, 4/5)`.   (1 mark)

    ii. 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)`
  2.  i.`y = -9/5x + 13/5`
    ii. `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)`
  `f^{prime}(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 2013 VCAA 4

Part of the graph of a function `g: R -> R, \ g (x) = (16-x^2)/4` is shown below.

VCAA 2013 4a

  1. Points `B` and `C` are the positive `x`-intercept and `y`-intercept of the graph `g`, respectively, as shown in the diagram above. The tangent to the graph of `g` at the point `A` is parallel to the line segment `BC.`
    1. Find the equation of the tangent to the graph of `g` at the point `A.`   (2 marks)

    2. The shaded region shown in the diagram above is bounded by the graph of `g`, the tangent at the point `A`, and the `x`-axis and `y`-axis.
    3. Evaluate the area of this shaded region.   (3 marks)
  2. Let `Q` be a point on the graph of  `y = g(x)`.
  3. Find the positive value of the `x`-coordinate of `Q`, for which the distance `OQ` is a minimum and find the minimum distance.   (3 marks)

The tangent to the graph of `g` at a point `P` has a negative gradient and intersects the `y`-axis at point  `D(0, k)`, where  `5 <= k <= 8.`
 

VCAA 2013 4c
 

  1. Find the gradient of the tangent in terms of `k.`   (2 marks)

  2.   i. Find the rule `A(k)` for the function of `k` that gives the area of the shaded region.   (2 marks)

  3.  ii. Find the maximum area of the shaded region and the value of `k` for which this occurs.   (2 marks)

  4. iii. Find the minimum area of the shaded region and the value of `k` for which this occurs.  (2 marks)

Show Answers Only
    1. `A: y = 5-x`
    2. `text(See Worked Solutions)`
  2. `2sqrt3\ text(units when)\ x = 2sqrt2`
  3. `gprime(2sqrt(k-4)) =-sqrt(k-4)`
    1. `A(k) = (k^2)/(2sqrt(k-4))-32/3, k ∈ [5,8]`
    2. `A_text(max) = 16/3quadtext(when)quadk = 8`
    3. `A_text(min) = (64sqrt3)/9-32/3quadtext(when)quadk = 16/3`
Show Worked Solution

a.i.   `B(4,0), C(0,4), A(a,(16-a^2)/4)`

`m_(BC) = m_text(tan) = (4-0)/(0-4) = -1`

`g(x)` `= (16-x^2)/4`
`g^{prime}(x)` `=-x/2`

 
`text(S)text(ince)\ \ g^{prime}(a) = -1,`

`=>a = 2`

`text(T)text(angent passes through)\ \ (2,3),`

`y-3` `=-(x-2)`
`y` `=-x + 5`

 
`text(Alternatively, using technology:)`

`[text(CAS: tangent Line)]\ (g(x),x,2)`
 

a.ii.   `text(Solution 1)`

♦ Mean mark 37%.
MARKER’S COMMENT: A common incorrect answer:
`int_0^5((-x+5)-g(x))dx=35/12`.  Know why it’s wrong!

  `D(5,0), E(0,5)`

`text(Area)` `= DeltaEOD-int_0^4 g(x)\ dx`
  `= 1/2 xx 5 xx 5-32/3`
  `= 11/6\ text(u²)`

 
`text(Solution 2)`

`text(Area)` `= int_0^5 (-x+5)\ dx-int_0^4 g(x)\ dx`
  `= 11/6\ text(u²)`

 

b.   `Q(x, (16-x^2)/4), O(0,0)`

♦♦♦ Mean mark 20%.
`z` `= OQ`
  `= sqrt(x^2 + ((16-x^2)/4)^2), x > 0`

 
`text(Max or min when)\ \ (dz)/(dx)=0,`

`text(Solve:)\ (dz)/dx = 0quadtext(for)quadx > 0`

`=> x = 2sqrt2`

`=>z(2sqrt2) = 2sqrt3`

`:. text(Min distance of)\ \ 2sqrt3\ \ text(units when)\ \ x = 2sqrt2`
 

c.   `text(Let)\ \ P(p, (16-p^2)/4)`

♦♦♦ Mean mark 8%.

`m_text(tan)\ text(at)\ P =-p/2`

`m_(PD) = ((16-p^2)/4-k)/(p-0)`
 

`text(Equating gradients,)`

`text(Solve:)\ \ ((16-p^2)/4-k)/p=-p/2\ \ text(for)\ p,`

`=> p=2sqrt(k-4)`
 

 `:.\ text(Gradient of tangent:)`

`g^{prime}(2sqrt(k-4)) =-sqrt(k-4)`
 

d.i.   `text(Equation of tangent:)`

♦♦♦ Mean mark 8%.

`y = -sqrt(k-4)x + k`

`text(CAS: tangent Line)\ (g(x),x,2sqrt(k-4))`

`xtext(-int of tangent:)\ x = k/(sqrt(k-4))`
 

`:. A(k)` `= 1/2 xx (k/(sqrt(k-4))) xx k-int_0^4 g(x)\ dx`
  `= (k^2)/(2sqrt(k-4))-32/3, \ \ k ∈ [5,8]`

 

d.ii.   `text(Solve:)\ \ A(k)=0quadtext(for)quadk ∈ [5,8]`

♦♦♦ Mean mark 5%.

`=> k=16/3`

  `text(Sketch the graph of)\ \ A(k)\ \ text(for)\ \ 5<=k<=8`

 

 met2-2013-vcaa-sec4-answer

`:. A_text(max) = 16/3quadtext(when)quadk = 8`
 

d.iii.   `A_text(min)\ text(occurs at the turning point)\ (k=16/3).`

♦♦♦ Mean mark 3%.
 `A_text(min)` `=A(16/3)`
  `=(64sqrt3)/9-32/3`

Calculus, MET2 2012 VCAA 5

The shaded region in the diagram below is the plan of a mine site for the Black Possum mining company.

All distances are in kilometres.

Two of the boundaries of the mine site are in the shape of the graphs of the functions

`f: R -> R,\ f(x) = e^x and g: R^+ -> R,\ g(x) = log_e (x).`

VCAA 2012 5a

    1. Evaluate `int_(−2)^0 f(x)\ dx`.   (1 mark)

    2. Hence, or otherwise, find the area of the region bounded by the graph of `g`, the `x` and `y` axes, and the line `y = –2`.   (1 mark)

    3. Find the total area of the shaded region.   (1 mark)

  2. The mining engineer, Victoria, decides that a better site for the mine is the region bounded by the graph of `g` and that of a new function  `k: (– oo, a) -> R,\ k(x) = – log_e(a-x)`, where `a` is a positive real number.
    1. Find, in terms of `a`, the `x`-coordinates of the points of intersection of the graphs of `g` and `k`.   (2 marks)

    2. Hence, find the set of values of `a`, for which the graphs of `g` and `k` have two distinct points of intersection.   (1 mark)

  3. For the new mine site, the graphs of `g` and `k` intersect at two distinct points, `A` and `B`. It is proposed to start mining operations along the line segment `AB`, which joins the two points of intersection.
  4. Victoria decides that the graph of `k` will be such that the `x`-coordinate of the midpoint of `AB` is `sqrt 2`.
  5. Find the value of `a` in this case.   (2 marks)

Show Answers Only
    1. `1-1/(e^2)`
    2. `1-1/(e^2)\ text(units²)`
    3. `e-1/(e^2)\ text(units²)`
    1. `(a ± sqrt(a^2-4))/2`
    2. `a > 2`
  3. `2sqrt2`
Show Worked Solution
ai.    `int_(−2)^0 f(x)\ dx` `=[e^x]_(-2)^0`
    `= e^0-e^(-2)`
    `=1-1/e^2`

 

a.ii.   `text(S)text(ince)\ \ g(x) = f^(−1)(x),`

♦ Mean mark part (a)(ii) 45%.

`=>\ text(Area is the same as)\ int_(−2)^0f(x)\ dx,`

`:. text(Area) = 1-1/(e^2)\ text(u²)`

 

♦ Mean mark part (a)(iii) 36%.
a.iii.    `text(Area)` `=int_0^1 (e^x)\ dx + text{Area from part (a)(ii)}`
    `= [e^x]_0^1 + (1-1/(e^2))`
    `= e-1/(e^2)\ text(u²)`

 

b.i.    `g(x)` `= k(x)`
  `log_e (x)` `=- log_e(a-x)`
  `log_e (x)+log_e(a-x)` `=0`
  `log_e(x(a-x))` `=0`
  `ax-x^2` `=1`
  `x^2-ax+1` `=0`

 
`:.x= (a ± sqrt(a^2-4))/2`

 

♦♦♦ Mean mark (b.ii.) 11%.

b.ii.   `text(For 2 solutions:)`

`b^2-4ac` `>0`
`a^2-4` `>0`
`:. a` `>2,\  \ (a>0)`

 

c.   `xtext(-coordinate of Midpoint)= sqrt2`

♦♦♦ Mean mark (c) 15%.
`((a + sqrt(a^2-4))/2 +(a-sqrt(a^2-4))/2)/2` `= sqrt2`
`a + sqrt(a^2-4) +a-sqrt(a^2-4)`  `=4 sqrt2`
`a` `=2sqrt2quadtext(for)quada > 2`

Calculus, MET2 2013 VCAA 16 MC

The graph of  `f: [1, 5] -> R,\ f(x) = sqrt (x - 1)`  is shown below.

met2-16

Which one of the following definite integrals could be used to find the area of the shaded region?

A.   `int_1^5 (sqrt (x - 1))\ dx`

B.   `int_0^2 (sqrt (x - 1))\ dx`

C.   `int_0^5 (2 - sqrt (x - 1))\ dx`

D.   `int_0^2 (x^2 + 1)\ dx`

E.   `int_0^2 (x^2)\ dx`

Show Answers Only

`E`

Show Worked Solution

`f(5)=2,\ \ f(1)=0`

♦♦♦ Mean mark 21%.

`f(x)=sqrt(x-1)`

`text(Find inverse:)`

`f^-1(x)=x^2 +1`
 

`:.\ text(Shaded Area)`

`= int_0^2(x^2+1)\ dx – 2 xx 1`

`=int_0^2 (x^2)\ dx`
 

`=>   E`