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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 3

Consider the function \(g:R \to R, g(x)=2^x+5\).

  1. State the value of \(\lim\limits_{x\to -\infty} g(x)\).   (1 mark)
  2. The derivative, \(g'(x)\), can be expressed in the form \(g'(x)=k\times 2^x\).
  3. Find the real number \(k\).   (1 mark)
    1. Let \(a\) be a real number. Find, in terms of \(a\), the equation of the tangent to \(g\) at the point \(\big(a, g(a)\big)\).   (1 mark)
    1. Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places.   (2 marks)

Let \(h:R\to R, h(x)=2^x-x^2\).

  1. Find the coordinates of the point of inflection for \(h\), correct to two decimal places.   (1 mark)
  2. Find the largest interval of \(x\) values for which \(h\) is strictly decreasing.
  3. Give your answer correct to two decimal places.   (1 mark)
  4. Apply Newton's method, with an initial estimate of \(x_0=0\), to find an approximate \(x\)-intercept of \(h\).
  5. Write the estimates \(x_1, x_2,\) and \(x_3\) in the table below, correct to three decimal places.   (2 marks)
      

    \begin{array} {|c|c|}
    \hline
    \rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
    \hline
    \rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \end{array}
  6. For the function \(h\), explain why a solution to the equation \(\log_e(2)\times (2^x)-2x=0\) should not be used as an initial estimate \(x_0\) in Newton's method.   (1 mark)
  7. There is a positive real number \(n\) for which the function \(f(x)=n^x-x^n\) has a local minimum on the \(x\)-axis.
  8. Find this value of \(n\).   (2 marks)
Show Answers Only

a.    \(5\)

b.    \(\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)

ci.  \(y=2^a\ \log_{e}{(2)x}-(a\ \log_{e}{(2)}-1)\times2^a+5\)

\(\text{or}\ \ y=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

cii. \(y=4.255x\)

d. \((2.06 , -0.07)\)

e. \([0.49, 3.21]\)

f. 

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  -1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.773 \\
\hline
\end{array}

g. \(\text{See worked solution.}\)

h. \(n=e\)

Show Worked Solution

a.    \(\text{As }x\to -\infty,\ \ 2^x\to 0\)

\(\therefore\ 2^x+5\to 5\)
  

b.     \(g(x)\) \(=2^x+5\)
    \(=\Big(e^{\log_{e}{2}}\Big)^x\)
    \(=e\ ^{x\log_{e}{2}}+5\)
  \(g'(x)\) \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\)
     \(=\log_{e}{2}\times 2^x\)
  \(\therefore\ k\) \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)
   
ci.  \(\text{Tangent at}\ (a, g(a)):\)

\(y-(2^a+5)\) \(=\log_{e}{2}\times2^a(x-a)\)
\(\therefore\ y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

♦♦ Mean mark (c)(i) 50%.

cii.  \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)

\( y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

  
\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)

\(\text{Equation of tangent when }\ a\approx 2.6178\)

\( y\) \(=2^{2.6178..}\ \log_{e}{(2)x}+0\)
\(\therefore\  y\) \(=4.255x\)

♦♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students did not substitute (0, 0) into the correct equation or did not find the value of \(a\) and used \(a=0\).
d.     \(h(x)\) \(=2^x-x^2\)
  \(h'(x)\) \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\)
  \(h”(x)\) \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\)

\(\text{Solving }h”(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)

\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)

\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)

e.    \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)

\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)


♦♦ Mean mark (e) 40%.
MARKER’S COMMENT: Round brackets were often used which were incorrect as endpoints were included. Some responses showed the interval where the function was strictly increasing.

f.    \(\text{Newton’s Method }\Rightarrow\  x_a-\dfrac{h(x_a)}{h'(x_a)}\) 

\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)

\(h(x)=2x-x^2\ \text{and }h'(x)=\ln{2}\times 2^x-2x\)

\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  0-\dfrac{2^0-2\times 0}{\ln2\times 2^0\times 0}=-1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -1.433-\dfrac{2^{-1.433}-2\times -1.433}{\ln2\times 2^{-1.433}\times -1.433}=-0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.897-\dfrac{2^{-0.897}-2\times -0.897}{\ln2\times 2^{-0.897}\times -0.897}=-0.773 \\
\hline
\end{array}

g.    \(\text{The denominator in Newton’s Method is}\ h'(x)=\log_{e}{(2)}\cdot 2^x-2x\)

\(\text{and the calculation will be undefined if }h'(x)=0\ \text{as the tangent lines are horizontal}.\)

\(\therefore\ \text{The solution to }h'(x)=0\ \text{cannot be used for }x_0.\)

♦♦♦ Mean mark (g) 20%.

h.    \(\text{For a local minimum }f(x)=0\)

\(\rightarrow\ n^x-x^n=0\)

\(\rightarrow\ n^x=x^n\ \ \ (1)\)

\(\text{Also for a local minimum }f'(x)=0\)

\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)

\(\text{Substitute (1) into (2)}\)

\(\ln(n)\cdot x^n-nx^{n-1}=0\) 

\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)

\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)

\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)

\(\therefore\ n=e\)


♦♦♦ Mean mark (h) 10%.
MARKER’S COMMENT: Many students showed that \(f'(x)=0\) but failed to couple it with \(f(x)=0\). Ensure exact values are given where indicated not approximations.

Calculus, MET2 2023 VCAA 14 MC

A polynomial has the equation  \(y=x(3x-1)(x+3)(x+1)\).

The number of tangents to this curve that pass through the positive \(x\)-intercept is

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Positive}\ x\text{-intercept occurs at}\ \Big(\dfrac{1}{3}, 0\Big) \)

\(\text{Find tangent line at}\ \ x=a\ \text{(by CAS):}\)

\(y_{\text{tang}}=\Big(12a^3+33a^2+10a-3\Big)x-a^2\Big(9a^2+22a+5\Big)\)

\(\text{Solve}\ \ \ y_{\text{tang}}\Bigg(\dfrac{1}{3}\Bigg)=0\ \text{for }a:\)

\(a=\dfrac{-\sqrt{7}-4}{3},\ a=\dfrac{\sqrt{7}-4}{3}\text{ or}\ a=\dfrac{1}{3}\)

\(\therefore\ \text{3 solutions exist.}\)

\(\Rightarrow D\)

\(\text{NOTE: Graphical methods could also be used}\)


♦♦♦ Mean mark 29%.

Calculus, MET2 2020 VCAA 5

Let  `f: R to R, \ f(x)=x^{3}-x`.

Let  `g_{a}: R to R`  be the function representing the tangent to the graph of `f` at  `x=a`, where  `a in R`.

Let `(b, 0)` be the `x`-intercept of the graph of `g_{a}`.

  1. Show that  `b= {2a^{3}}/{3 a^{2}-1}`.   (3 marks)
  2. State the values of `a` for which `b` does not exist.   (1 mark)
  3. State the nature of the graph of `g_a` when `b` does not exist.   (1 mark)
  4. i.  State all values of `a` for which  `b=1.1`. Give your answer correct to four decimal places.   (1 mark)
  5. ii. The graph of `f` has an `x`-intercept at (1, 0).
  6.      State the values of  `a`  for which  `1 <= b <= 1.1`.
  7.      Give your answers correct to three decimal places.   (1 mark)

The coordinate `(b, 0)` is the horizontal axis intercept of `g_a`.

Let `g_b` be the function representing the tangent to the graph of `f` at  `x=b`, as shown in the graph below.
 
 
     
 

  1. Find the values of `a` for which the graphs of `g_a` and `g_b`, where `b` exists, are parallel and where  `b!=a`.   (3 marks)

Let  `p:R rarr R, \ p(x)=x^(3)+wx`, where  `w in R`.

  1. Show that  `p(-x)=-p(x)`  for all  `w in R`.   (1 mark)

A property of the graphs of `p` is that two distinct parallel tangents will always occur at `(t, p(t))` and `(-t,p(-t))` for all  `t!=0`.

  1. Find all values of `w` such that a tangent to the graph of `p` at `(t, p(t))`, for some  `t > 0`, will have an `x`-intercept at `(-t, 0)`.   (1 mark)
  2. Let  `T:R^(2)rarrR^(2),T([[x],[y]])=[[m,0],[0,n]][[x],[y]]+[[h],[k]]`, where  `m,n in R text(\{0})`  and  `h,k in R`.
     
    State any restrictions on the values of `m`, `n`, `h`, and `k`, given that the image of `p` under the transformation `T` always has the property that parallel tangents occur at  `x = -t`  and  `x = t`  for all  `t!=0`.   (1 mark)
Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `a=+-sqrt3/3`
  3. `text(Horizontal line.)`
  4.  i. `a=-0.5052, 0.8084, 1.3468`
  5. ii. `a in (-0.505,-0.500]uu(0.808,1.347)`
  6. `a=+- sqrt5/5`
  7. `text(See Worked Solutions.)`
  8. `w=-5t^2`
  9. `h=0`
Show Worked Solution

a.   `f^(‘)(a) = 3a^2-1`

`g_a(x)\ \ text(has gradient)\ \ 3a^2-1\ \ text(and passes through)\ \ (a, a^3-a)`

`g_a(x)-(a^3-a)` `=(3a^2- 1)(x-a)`  
`g_a(x)` `=(3a^2-1)(x-a)+a^3-a`  

  
`x”-intercept occurs at”\ (b,0):`

`0=(3a^2-1)(b-a) + a^3-a`

`(3a^2-1)(b-a)` `=a-a^3`  
`3a^2b-3a^3-b+a` `=a-a^3`  
`b(3a^2-1)` `=a-a^3+3a^3-a`  
`:.b` `=(2a^3)/(3a^2-1)`  

 
b.   `b\ text{does not exist when:}`

♦ Mean mark part (b) 46%.

`(3a^2-1)=0`

`a=+-sqrt3/3`

♦♦ Mean mark part (c) 23%.
 

c.   `text{If}\ \ a=+-sqrt3/3,\ \ g_a^(‘)(x) = 0`

`=>\ text{the graph is a horizontal line (does not cross the}\ xtext{-axis).}`
 

d.i.  `”Solve “{2a^{3}}/{3 a^{2}-1}=1.1” for “a:`

`a=-0.5052″ or “a=0.8084” or “a=1.3468\ \ text{(to 4 d.p.)}`
  

d.ii.  `” Solve “1 <= (2a^(3))/(3a^(2)-1) < 1.1” for “a:`

♦♦♦ Mean mark part (d)(ii) 13%.

`a in (-0.505,-0.500]uu(0.808,1.347)\ \ text{(to 3 d.p.)}`
 

e.   `f^(‘)(b) = 3b^2-1`

`g_b(x)\ \ text(has gradient)\ \ 3b^2-1\ \ text(and passes through)\ \ (b, b^3-b)`

`g_b(x)-(b^3-b)` `=(3b^2- 1)(x-b)`  
`g_b(x)` `=(3b^2-1)(x-b)+b^3-b`  

 
`g_a(x)\ text{||}\ g_b(x)\ \ text{when}`

♦♦♦ Mean mark part (e) 13%.
`3a^2-1` `=3b^2-1`  
  `=3(2a^3)/(3a^2-1)-1`  

 
`=> a=+-1, +- sqrt5/5, 0`

`text(Test each solution so that)\ \ b!=a :`

`text(When)\ \ a=+-1, 0 \ => \ b=a`

`:. a=+- sqrt5/5`
 

f.    `p(-x)` `=(-x)^3-wx`
    `=-x^3-wx`
    `=-(x^3+wx)`
    `=-p(x)`

 
g.  `p^(‘)(t) = 3t^2+w`

♦♦♦ Mean mark part (g) 3%.

`p(t)\ \ text(has gradient)\ \ 3t^2+w\ \ text(and passes through)\ \ (t, t^3+wt)`

`p(t)-(t^3+wt)` `=(3t^2+w)(x-t)`  
`p(t)` `=(3t^2+w)(x-t) + t^3+wt`  

 
`text{If}\ p(t)\ text{passes through}\ \ (-t, 0):`

`0=(3t^2+w)(-2t) + t^3+wt`

`=>w=-5t^2\ \ (t>0)`
 

h.   `text{Property of parallel tangents is retained under transformation}`

♦♦♦ Mean mark part (h) 2%.

`text{if rotational symmetry remains (odd function).}`

`=>h=0`

`text(No further restrictions apply to)\ m, n\ \ text{or}\ \ k.`

Calculus, MET2 2020 VCAA 4

The graph of the function  `f(x)=2xe^((1-x^(2)))`, where  `0 <= x <= 3`, is shown below.
 

  1. Find the slope of the tangent to `f` at  `x=1`.   (1 mark)
  2. Find the obtuse angle that the tangent to `f` at  `x = 1`  makes with the positive direction of the horizontal axis. Give your answer correct to the nearest degree.   (1 mark)
  3. Find the slope of the tangent to `f` at a point  `x =p`. Give your answer in terms of  `p`.   (1 mark)
  4.  i. Find the value of `p` for which the tangent to `f` at  `x=1` and the tangent to `f` at  `x=p`  are perpendicular to each other. Give your answer correct to three decimal places.   (2 marks)
  5. ii. Hence, find the coordinates of the point where the tangents to the graph of `f` at  `x=1`  and  `x=p`  intersect when they are perpendicular. Give your answer correct to two decimal places.   (3 marks)

Two line segments connect the points `(0, f(0))`  and  `(3, f(3))`  to a single point  `Q(n, f(n))`, where  `1 < n < 3`, as shown in the graph below.
 
         
 

  1.   i. The first line segment connects the point `(0, f(0))` and the point `Q(n, f(n))`, where `1 < n < 3`.
  2.      Find the equation of this line segment in terms of  `n`.   (1 mark)
  3.  ii. The second line segment connects the point `Q(n, f(n))` and the point  `(3, f(3))`, where  `1 < n < 3`.
  4.      Find the equation of this line segment in terms of `n`.   (1 mark)
  5. iii. Find the value of `n`, where  `1 < n < 3`, if there are equal areas between the function `f` and each line segment.
  6.      Give your answer correct to three decimal places.   (3 marks)
Show Answers Only
  1. `-2`
  2. `117^@`
  3. `2(1-2p^(2))e^(1-p^(2))” or “(2e-4p^(2)e)e^(-p^(2))`
  4.  i. `0.655`
  5. ii. `(0.80, 2.39)`
  6.   i. `y_1=2e^((1-n^2))x`
  7.  ii. `y_2=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3) + 6e^(-8)`
  8. iii. `n= 1.088`
Show Worked Solution

a.   `f(x)=2xe^((1-x^(2)))`

`f^{′}(1)=-2`

♦ Mean mark part (b) 37%.

 

b.   `text{Solve:}\ tan theta = -2\ \ text{for}\ \ theta in (pi/2, pi)`

`theta = 117^@`
 

c.   `text{Slope of tangent}\ = f^{′}(p)`

`f^{′}(p)=2(1-2p^(2))e^(1-p^(2))\ \ text{or}\ \ (2e-4p^(2)e)e^(-p^(2))`
 

d.i.   `text{If tangents are perpendicular:}`

`f^{′}(p) xx -2 =-1\ \ =>\ \ f^^{′}(p)=1/2`

`text{Solve}\ \ 2(1-2p^(2))e^(1-p^(2))=1/2\ \ text{for}\ p:`

`p=0.655\ \ text{(to 3 d.p.)}`
 

d.ii.  `text{Equation of tangent at}\ \ x=1: \ y=4-2x`

♦ Mean mark part (d)(ii) 41%.

`text{Equation of tangent at}\ \ x=p: \ y= x/2 + 1.991…`

`text{Solve}\ \ 4-2x = x/2 + 1.991…\ \ text{for}\ x:`

`=> x = 0.8035…`

`=> y=4-2(0.8035…) = 2.392…`

`:.\ text{T}text{angents intersect at (0.80, 2.39)}`

♦ Mean mark part (e)(i) 44%.

 
e.i.  `Q (n, 2n e^(1-n^2))`

`m_(OQ) = (2n e^((1-n^2)) – 0)/(n-0) = 2e^((1-n^2))`

`:.\ text{Equation of segment:}\ \ y_1=2e^((1-n^2))x`

♦♦ Mean mark part (e)(ii) 28%.
 

e.ii.  `P(3, f(3)) = (3, 6e^(-8))`

`m_(PQ) = (2n e^((1-n^2)) – 6e^(-8))/(n-3)`

`text{Equation of line segment:}`

`y_2-6e^(-8)` `=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3)`  
`y_2` `=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3) + 6e^(-8)`  
♦♦ Mean mark part (e)(iii) 28%.

 

e.iii.  `text{Find}\ n\ text{where shaded areas are equal.}`

`text{Solve}\ int_(0)^(n)(f(x)-y_(1))\ dx=int_(n)^(3)(y_(2)-f(x))\ dx\ \ text{for} n:`

`=> n= 1.088\ \ text{(to 3 d.p.)}`

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′(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 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. 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)

    2. 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. Find the equation of the tangent to  `y = xsin(x)`  at the point  `(−(5pi)/2,(5pi)/2)`.  (2 marks)
  3. The translation `T` maps the graph of  `y = xsin(x)`  onto the graph of  `y = (3pi - x)sin(x)`, where
     
    `qquad T: R^2 -> R^2, T([(x),(y)]) = [(x),(y)] + [(a),(0)]`
     
    and `a` is a real constant.

     

    State the value of `a`.  (1 mark)

  4. 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.   
    1. `(2n + 1)pi`
    2. `−(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, MET2 2018 VCAA 9 MC

A tangent to the graph of  `y = log_e(2x)`  has a gradient of 2.

This tangent will cross the `y`-axis at

A.   `0`

B.   `-0.5`

C.   `-1`

D.   `-1 - log_e(2)`

E.   `-2 log_e(2)`

Show Answers Only

`C`

Show Worked Solution
`y` `= ln 2x`
`(dy)/(dx)` `= 1/x`

 
`text(When)\ \ (dy)/(dx) = 2:`

`1/x` `= 2`
`x` `= 1/2`

 
`text(When)\ \ x = 1/2,\ \ y = ln 1 = 0`

`text(Find equation)\ \ m = 2,\ \ text(through)\ \ (1/2, 0):`

`y – 0` `= 2 (x – 1/2)`
`y` `= 2x – 1`

 
`=>   C`

Calculus, MET1 SM-Bank 21

Find the equation of the tangent to the curve  `y = cos 2x`  at the point whose `x`-coordinate is  `pi/6`.  (3 marks)

Show Answers Only

`y = – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

Show Worked Solution

`y = cos 2x`

`dy/dx = -2 sin 2x`

`text(When)\ \ x = pi/6,`

`y` `= cos (2 xx pi/6)`
  `= cos (pi/3)`
  `= 1/2`

 

`dy/dx` `= -2 sin (pi/3)`
  `= -2 xx sqrt 3 / 2`
  `= – sqrt 3`

 

`:. text(Equation of tangent,)\ \ m = – sqrt 3, text(through)\ \ (pi/6, 1/2) :`

`y – y_1` `= m(x – x_1)`
`y – 1/2` `= – sqrt 3 ( x – pi/6)`
`y – 1/2` `= – sqrt 3 x + (sqrt 3 pi)/6`
`y` `= – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

Calculus, MET2 2016 VCAA 2

Consider the function  `f(x) = -1/3 (x + 2) (x-1)^2.`

  1.  i. Given that  `g^{′}(x) = f (x) and g (0) = 1`, show that  `g(x) = -x^4/12 + x^2/2-(2x)/3 + 1`.  (1 mark)
  2. ii. Find the values of `x` for which the graph of  `y = g(x)`  has a stationary point.  (1 mark)

The diagram below shows part of the graph of  `y = g(x)`, the tangent to the graph at  `x = 2`  and a straight line drawn perpendicular to the tangent to the graph at  `x = 2`. The equation of the tangent at the point `A` with coordinates  `(2, g(2))`  is  `y = 3-(4x)/3`.

The tangent cuts the `y`-axis at `B`. The line perpendicular to the tangent cuts the `y`-axis at `C`.
 


 

  1.   i. Find the coordinates of `B`.  (1 mark)
  2.  ii. Find the equation of the line that passes through `A` and `C` and, hence, find the coordinates of `C`.  (2 marks)
  3. iii. Find the area of triangle `ABC`.  (2 marks)
  4. The tangent at `D` is parallel to the tangent at `A`. It intersects the line passing through `A` and `C` at `E`.

     


     
     i. Find the coordinates of `D`.  (2 marks)

  5. ii. Find the length of `AE`.  (3 marks)
Show Answers Only
  1.  i. `text(Proof)\ \ text{(See worked solutions)}`
  2. ii. `x = −2, 1`
  3.   i. `(0, 3)`
  4. ii. `y = 3/4 x-7/6`
  5.      `(0, −7/6)`
  6. iii. `25/6\ text(u²)`
  7. `(−1, 25/12)`
  8. `27/20\ text(u)`
Show Worked Solution
a.i.    `g(x)` `= int f(x)\ dx`
    `=-1/3 int (x + 2) (x-1)^2\ dx`
    `=-1/3int(x^3-3x+2)\ dx`
  `:.g(x)` `= -x^4/12 + x^2/2-(2x)/3 + c`

 

`text(S)text(ince)\ \ g(0) = 1,`

`1` `= 0 + 0-0 + c`
`:. c` `= 1`

 
`:. g(x) = -x^4/12 + x^2/2-(2x)/3 + 1\ \ …\ text(as required)`

 

a.ii.   `text(Stationary point when:)`

`g^{′}(x) = f(x) = 0`

`-1/3(x + 2) (x-1)^2=0`

`:. x = −2, 1`

 

b.i.   

`B\ text(is the)\ y text(-intercept of)\ \ y = 3-4/3 x`

`:. B (0, 3)`

 

b.ii.   `m_text(norm) = 3/4, \ text(passes through)\ \ A(2, 1/3)`

   `text(Equation of normal:)`

`y-1/3` `=3/4(x-2)`
`y` `=3/4 x -7/6`
   

`:. C (0, −7/6)`

 

b.iii.   `text(Area)` `= 1/2 xx text(base) xx text(height)`
    `= 1/2 xx (3 + 7/6) xx 2`
    `= 25/6\ text(u²)`

 

♦ Mean mark part (c)(i) 43%.
MARKER’S COMMENT: Many students gave an incorrect `y`-value here. Be careful!
c.i.    `text(Solve)\ \ \ g^{′}(x)` `= -4/3\ \ text(for)\ \ x < 0`
  `=> x` `= −1`
  `g(-1)` `=-1/12+1/2+2/3+1`
    `=25/12`

 
`:. D (−1, 25/12)`

 

c.ii.   `text(T) text(angent line at)\ \ D:`

`y-25/12` `=- 4/3(x+1)`
`y` `=- 4/3x + 3/4`

 

`DE\ \ text(intersects)\ \ AE\ text(at)\ E:`

♦♦ Mean mark 32%.
`-4/3 x + 3/4` `= 3/4 x-7/6`
`25/12 x` `= 23/12`
`x` `=23/25`

 
`:. E (23/25, -143/300)`

 

`:. AE` `= sqrt((2-23/25)^2 + (1/3-(-143/300))^2)`
  `= 27/20\ text(units)`

Calculus, MET2 2016 VCAA 1

Let  `f: [0, 8 pi] -> R, \ f(x) = 2 cos (x/2) + pi`.

  1. Find the period and range of  `f`.  (2 marks)
  2. State the rule for the derivative function  `f prime`.  (1 mark)
  3. Find the equation of the tangent to the graph of  `f` at  `x = pi`.  (1 mark)
  4. Find the equations of the tangents to the graph of  `f: [0, 8 pi] -> R,\ \ f(x) = 2 cos (x/2) + pi`  that have a gradient of 1.  (2 marks)
  5. The rule of  `f prime` can be obtained from the rule of  `f` under a transformation `T`, such that
      
    `qquad T: R^2 -> R^2,\ T([(x), (y)]) = [(1, 0), (0, a)] [(x), (y)] + [(−pi), (b)]`
     

     

    Find the value of `a` and the value of `b`.  (3 marks)

  6. Find the values of  `x, \ 0 <= x <= 8 pi`, such that  `f(x) = 2 f prime (x) + pi`.  (2 marks)
Show Answers Only
  1. `text(Period:)\ 4 pi; qquad text(Range:)\ [pi – 2, pi + 2]`
  2. `f prime (x) = −sin (x/2)`
  3. `y = −x + 2 pi`
  4. `y = x – 2 pi and y = x – 6 pi`
  5. `a = 1/2 and b = -pi/2`
  6. `x = (3 pi)/2, (7 pi)/2, (11 pi)/2, (15 pi)/2`
Show Worked Solution

a.   `text(Period)= (2pi)/n = (2 pi)/(1/2) = 4pi`

MARKER’S COMMENT: Including round brackets rather than square ones was a common mistake.

`text(Range:)\ [pi – 2, pi + 2]`
 

b.   `f prime (x) = text(−sin) (x/2)`

 

c.   `[text(CAS: tangentLine)\ (f(x), x, pi)]`

`y = -x + 2 pi`

 

d.   `text(Solve)\ \ f prime (x) = 1\ \ text(for)\ x in [0, 8 pi]`

♦ Mean mark part (d) 50%.

`-> x = 3 pi or 7 pi`

`:. y = x – 2 pi and y = x – 6 pi\ \ [text(CAS)]`

 

e.   `text(Using the transition matrix,)`

♦♦ Mean mark part (e) 27%.
`x_T` `=x-pi`
`x` `=x_T+pi`
`y_T` `=ay+b`
`y` `=(y_T-b)/a`
   

`f(x)= cos (x/2) + pi/2\ \ ->\ \ f′(x) = -sin(x/2)`

`(y_T-b)/a` `=2cos((x_T+pi)/2)+pi`
`y_T` `=2a cos((x_T+pi)/2)+a pi +b`
  `=-2a sin(x_T/2)+a pi + bqquad [text(Complementary Angles)]`
   
`-2a` `=-1`
`:. a` `=1/2`
`1/2 pi +b` `=0`
`:.b` `=-pi/2`

 

f.   `text(Solve)\ \ f(x) = 2 f prime (x) + pi\ \ text(for)\ \ x in [0, 8 pi]`

♦ Mean mark part (f) 50%.
`2 cos (x/2) + pi` `= -2 sin(x/2)+pi`
`tan(x/2)` `=-1`
`x/2` `=(3pi)/4, (7pi)/4, (11pi)/4, (15pi)/4`
`:.x` `= (3 pi)/2, (7 pi)/2, (11 pi)/2, (15 pi)/2`

Calculus, MET1 2009 VCAA 8

Let  `f: R -> R,\ f(x) = e^x + k`,  where `k` is a real number. The tangent to the graph of  `f` at the point where  `x = a`  passes through the point `(0, 0).` Find the value of `k` in terms of `a.`  (3 marks)

Show Answers Only

`:. k = ae^a – e^a`

Show Worked Solution

`text(Point of tangency:)\ \ P (a, e^a + k)`

♦ Mean mark 45%.

`text(Find gradient of tangent at)\ \ x = a:`

`f(x)` `=e^x+k`
`f′ (x)` `= e^x`
`:. m_tan` `= e^a`

 

`text(Find equation of tangent:)`

`y – y_1` `= m (x – x_1)`
`y – (e^a + k)` `= e^a (x – a)`

 

`text(Substitute)\ \ (0, 0):`

`0 – e^a – k` `= – ae^a`
`:. k` `= ae^a – e^a`
  `=e^a(a-1)`

Calculus, MET1 2010 VCAA 10

The line  `y = ax - 1`  is a tangent to the curve  `y = x^(1/2) + d`  at the point  `(9, c)`  where `a, c` and `d` are real constants.

Find the values of `a, c` and `d.`  (4 marks)

Show Answers Only

`a = 1/6,\ \ c = 1/2,\ \ d = – 5/2`

Show Worked Solution

`text(Find)\ \ m_tan\ \ text(at)\ \ (9,c),`

`(dy)/(dx)` `= 1/(2 sqrt x)`
`m_tan` `= 1/(2 sqrt 9)=1/6`
`:. a` `= 1/6`

 

`text(Find equation of tangent:)`

`y – y_1` `= m (x – x_1)`
`y – c` `= 1/6 (x – 9)`
`y` `= 1/6 x – 3/2 + c`
   

`text(S)text(ince)\ y text(-intercept = – 1),`

`-1` `= – 3/2 +c`
`:.c` `=1/2`

 

`text(S)text(ince)\ \ (9,1/2)\ \ text(lies on)\ \ y = x^(1/2) + d`

`1/2` `= 3 + d`
`:. d` `= – 5/2`

Calculus, MET1 2012 VCAA 10

Let  `f: R -> R,\ f(x) = e^(– mx) + 3x`,  where `m` is a positive rational number.

  1.  i. Find, in terms of `m`, the `x`-coordinate of the stationary point of the graph of  `y = f(x)`.  (2 marks)
  2. ii. State the values of `m` such that the `x`-coordinate of this stationary point is a positive number.  (1 mark)
  3. For a particular value of `m`, the tangent to the graph of  `y = f(x)`  at  `x = – 6`  passes through the origin.
  4. Find this value of `m`.  (3 marks)
Show Answers Only
  1.  i. `1/m log_e(m/3)`
  2. ii. `m > 3`
  3. `1/6`
Show Worked Solution

a.i.   `text(SP’s occur when)\ \ f prime(x)=0`

`– me^(- mx) + 3` `= 0`
`me^(-mx)` `3`
`– mx` `= log_e (3/m)`
`:. x` `= – 1/m log_e (3/m)`
  `= 1/m log_e (m/3), \ m>0`

 

♦♦♦ Parrt (a)(ii) mean mark 18%.
  ii.    `1/m log_e (m/3)` `> 0`
  `log_e (m/3)` `> 0`
  `m/3` `> 1`
  `:. m` `> 3`

 

b.   `text(Point of tangency:)\ \ (– 6, e^(– 6m) – 18)`

♦♦ Mean mark 33%.
MARKER’S COMMENT: Many confused `m` with the more common use of `m` for gradient in  `y=mx+c`.

`text(At)\ \ x=-6,`

`m_tan` `= f prime (– 6)`
  `= – me^(– 6m) + 3`

 

`:.\ text(Equation of tangent:)`

`y – y_1` `= m (x – x_1)`
`y – (e^(– 6m) – 18)` `= (– me^(– 6m) + 3) (x – (– 6))`

 

`text(S)text{ince tangent passes through (0, 0):}`

`– e^(– 6m) + 18` `= (–me^(– 6m) + 3)(6)`
`– e^(– 6m) + 18` `= – 6 me^(– 6m) + 18`
`e^(– 6m) – 6me^(– 6m)` `= 0`
`e^(– 6m) (1 – 6m)` `= 0`
`1-6m` `=0`
`:.m` `=1/6`

Calculus, MET2 2014 VCAA 5

Let  `f: R -> R, \ \ f (x) = (x - 3)(x - 1)(x^2 + 3)  and  g: R-> R, \ \ g (x) = x^4 − 8x.`

  1. Express  `x^4 - 8x`  in the form  `x(x - a) ((x + b)^2 + c)`.  (2 marks)
  2. Describe the translation that maps the graph of  `y = f (x)`  onto the graph of  `y = g (x)`.  (1 mark)
  3. Find the values of `d` such that the graph of  `y = f (x + d)` has
  4.  i. one positive `x`-axis intercept  (1 mark)
  5. ii. two positive `x`-axis intercepts.  (1 mark)
  6. Find the value of `n` for which the equation  `g (x) = n`  has one solution.  (1 mark)
  7. At the point  `(u, g(u))`, the gradient of  `y = g(x)`  is `m` and at the point `(v, g(v))`, the gradient is  `-m`, where `m` is a positive real number.
  8.  i. Find the value of  `u^3 + v^3`.  (2 marks)
  9. ii. Find `u` and `v` if  `u + v = 1`.  (1 mark)
  10.  i. Find the equation of the tangent to the graph of  `y = g(x)`  at the point  `(p, g(p))`.  (1 mark)
  11. ii. Find the equations of the tangents to the graph of  `y = g(x)`  that pass through the point with coordinates  `(3/2, text(−12))`.  (3 marks)
Show Answers Only
  1. `x(x- 2)((x + 1)^2 + 3)`
  2. `text(See Worked Solutions)`
  3.  i. `[1,3)`
  4. ii. `d < 1`
  5. `−6 xx 2^(1/3)`
  6.  i. `4`
  7. ii. `u = (sqrt5 + 1)/2,quad v = (−sqrt5 + 1)/2`
  8. i. `y = 4(p^3 – 2)x – 3p^4`
  9. ii. `y = −8x;quady = 24x – 48`
Show Worked Solution

a.  `text(Solution 1)`

`g(x) = x^4 – 8x`

`= x(x – 2)(x^2 + 2x + 4)\ \ \ text([by CAS])`

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

 

`text(Solution 2)`

`x^4 – 8x` `=x(x^3-2^3)`
  `=x(x-2)(x^2 +2x+4)`
  `=x(x-2)(x^2 +2x+1+3)`
  `=x(x-2)((x+1)^2+3)`

 

b.    `f(x + 1)` `= ((x + 1) – 3)((x + 1) – 1)((x + 1)^2 + 3)`
    `= x(x – 2)((x + 1)^2 + 3)`
    `= g(x)`

 

♦ Mean mark part (b) 37%.

`:.\ text(Horizontal translation of 1 unit to the left.)`

 

c.i.  `text(Consider part of the)\ \ f(x)\ text(graph below:)`

♦♦♦ Mean mark 7%.

 
met2-2014-vcaa-sec5-answer
 

`text(For one positive)\ xtext(-axis intercept, translate at least)`

`text(1 unit left, but not more than 3 units left.)`

`:. d ∈ [1,3)`

 

c.ii.   `text(Translate less than 1 unit left, or translate)`

♦♦♦ Mean mark part (c)(ii) 19%.

`text(right.)`

`:. d < 1`

 

d.   `text(If)\ \ g(x)=n\ \ text(has one solution, then it)`

`text(will occur when)\ \ g′(x)=0  and  x>0.`

♦♦♦ Mean mark 17%.
`g′(x)` `=4x^3-8`
`4x^3` `=8`
`x` `=2^(1/3)`

 

met2-2014-vcaa-sec5-answer1
 

`n` `=g(2^(1/3))`
  `=2^(4/3)-8xx2^(1/3)`
  `=-6 xx 2^(1/3)`

 

e.i.   `gprime(u) = mqquadgprime(v) = −m`

♦♦ Mean mark 28%.
`gprime(u)` `= −gprime(v)`
`4u^3 – 8` `= −(4v^3 – 8)`
`4u^3 + 4v^3` `= 16`
`:. u^3 + v^3` `= 4`

 

e.ii.   `u^3 + v^3 = 4\ …\ (1)`

♦♦♦ Mean mark 10%.

`u + v = 1\ …\ (2)`
 

`text(Solve simultaneous equation for)\ \ u > 0:`

`:. u = (sqrt5 + 1)/2,quad v = (−sqrt5 + 1)/2`

 

f.i.   `text(Solution 1)`

`text(Using the point-gradient formula,)`

`y-g(p)` `=g′(p)(x-p)`
`y-(p^4-8p)` `=(4p^3-8)(x-p)`
`y` `=(4p^3-8)x -3p^4`

 

`text(Solution 2)`

♦♦ Mean mark (f)(i) 22%.

`y = 4(p^3 – 2)x – 3p^4`

`text([CAS: tangentLine)\ (g(u),x,p)]`

 

f.ii.   `text(Sub)\ (3/2,−12)\ text(into tangent equation,)`

`text(Solve:)\ \ −12 = 4(p^3 – 2)(3/2) – 3p^4\ \ text(for)\ p,`

♦♦♦ Mean mark 14%.

`p = 0\ \ text(or)\ \ p = 2`

`text(When)\ \ p = 0:`    `y` `= 4(−2)x`
    `= −8x`
`text(When)\ \ p = 2:`    `y` `= 4(2^3 – 2)x – 3(2)^4`
    `= 24x – 48`

 

`:. text(Equations are:)\ \ y = −8x, \ y = 24x – 48`

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 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.`
  2.  i. Find the equation of the tangent to the graph of `g` at the point `A.`  (2 marks)
  3. ii. 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.
  4.     Evaluate the area of this shaded region.  (3 marks)
  5. Let `Q` be a point on the graph of  `y = g(x)`.
  6. 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′(x)` `=- x/2`

 

`text(S)text(ince)\ \ g′(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:)`

`gprime(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`

Graphs, MET2 2013 VCAA 1

Trigg the gardener is working in a temperature-controlled greenhouse. During a particular 24-hour time interval, the temperature  `(Ttext{°C})` is given by  `T(t) = 25 + 2 cos ((pi t)/8), \ 0 <= t <= 24`, where `t` is the time in hours from the beginning of the 24-hour time interval.

  1. State the maximum temperature in the greenhouse and the values of `t` when this occurs.  (2 marks)
  2. State the period of the function `T.`  (1 mark)
  3. Find the smallest value of `t` for which  `T = 26.`  (2 marks)
  4. For how many hours during the 24-hour time interval is  `T >= 26?`  (2 marks)

Trigg is designing a garden that is to be built on flat ground. In his initial plans, he draws the graph of  `y = sin(x)`  for  `0 <= x <= 2 pi`  and decides that the garden beds will have the shape of the shaded regions shown in the diagram below. He includes a garden path, which is shown as line segment `PC.`

The line through points  `P((2 pi)/3, sqrt 3/2)`  and  `C (c, 0)`  is a tangent to the graph of  `y = sin (x)`  at point `P.`

    1. Find  `(dy)/(dx)`  when  `x = (2 pi)/3.`  (1 mark)
    2. Show that the value of `c` is  `sqrt 3 + (2 pi)/3.`  (1 mark)

In further planning for the garden, Trigg uses a transformation of the plane defined as a dilation of factor `k` from the `x`-axis and a dilation of factor `m` from the `y`-axis, where `k` and `m` are positive real numbers.

  1. Let `X prime, P prime` and `C prime` be the image, under this transformation, of the points `X, P` and `C` respectively. 

     

    1. Find the values of `k` and `m`  if  `X prime P prime = 10`  and  `X prime C prime = 30.`  (2 marks)
    2. Find the coordinates of the point `P prime.`  (1 mark)
Show Answers Only
  1. `27^@Cquadtext(when)\ t = 0, 16`
  2. `text(16 hours)`
  3. `8/3`
  4. `text(8 hours)`
    1. `−1/2`
    2. `text(See Worked Solutions)`
    1. `k = 20/sqrt3, quadm = 30/sqrt3`
    2. `Pprime((20pi sqrt3)/3,10)`
Show Worked Solution

a.   `T_text(max)\ text(occurs when)\ \ cos((pit)/8) = 1,`

`T_text(max)= 25 + 2 = 27^@C`

`text(Max occurs when)\ \ t = 0, or 16\ text(h)`

 

b.    `text(Period)` `= (2pi)/(pi/8)`
    `= 16\ text(hours)`

 

c.   `text(Solve:)\ \ 25 + 2 cos ((pi t)/8)=26\ \ text(for)\ t,`

`t`  `= 8/3,40/3,56/3\ \ text(for)\ t ∈ [0,24]`
`t_text(min)` `= 8/3`

 

d.   `text(Consider the graph:)`

met2-2013-vcaa-sec1-answer1

`text(Time above)\ 26 text(°C)` `= 8/3 + (56/3 – 40/3)`
  `= 8\ text(hours)`

 

e.i.   `(dy)/(dx) = cos(x)`

`text(At)\ x = (2pi)/3,`

`(dy)/(dx)` `= cos((2pi)/3)= −1/2`

 

e.ii.  `text(Solution 1)`

`text(Equation of)\ \ PC,`

`y-sqrt3/2` `=-1/2(x-(2pi)/3)`
`y` `=-1/2 x +pi/3 +sqrt3/2`

 

`PC\ \ text(passes through)\ \ (c,0),`

`0` `=-1/2 c +pi/3 + sqrt3/2`
`c` `=sqrt3 + (2 pi)/3\ …\ text(as required)`

 

`text(Solution 2)`

`text(Equating gradients:)`

`- 1/2` `= (sqrt3/2 – 0)/((2pi)/3 – c)`
`-1` `= sqrt3/((2pi – 3c)/3)`
`3c – 2pi` `= 3sqrt3`
`3c` `= 3 sqrt3 + 2pi`
`:. c` `= sqrt3 + (2pi)/3\ …\ text(as required)`

 

f.i.   `Xprime ((2pi)/3 m,0)qquadPprime((2pi)/3 m, sqrt3/2 k)qquadCprime ((sqrt3 + (2pi)/3)m, 0)`

`XprimePprime` `= 10`
`sqrt3/2 k` `= 10`
`:. k` `= 20/sqrt3`
  `=(20sqrt3)/3`
♦♦♦ Mean mark part (f)(i) 14%.

 

`XprimeCprime=30`

`((sqrt3 + (2pi)/3)m) – (2pi)/3 m` `= 30`
`:. m` `= 30/sqrt3`
  `=10sqrt3`
♦♦♦ Mean mark part (f)(ii) 12%.

 

f.ii.    `Pprime((2pi)/3 m, sqrt3/2 k)` `= Pprime((2pi)/3 xx 10sqrt3, sqrt3/2 xx 20/sqrt3)`
    `= Pprime((20pisqrt3)/3,10)`

Calculus, MET2 2012 VCAA 2

Let  `f: R text(\{2}) -> R,\ f(x) = 1/(2x-4) + 3.`

  1. Sketch the graph of  `y = f(x)` on the set of axes below. Label the axes intercepts with their coordinates and label each of the asymptotes with its equation.  (3 marks)
     

     

          VCAA 2012 2a
     

  2.   i. Find  `f^{′}(x)`.  (1 mark)
  3.  ii. State the range of  `f ^{′}`.  (1 mark)
  4. iii. Using the result of part ii. explain why `f` has no stationary points.  (1 mark)
  5. If  `(p, q)`  is any point on the graph of  `y = f(x)`, show that the equation of the tangent to  `y = f(x)`  at this point can be written as  `(2p-4)^2 (y-3) = -2x + 4p-4.`  (2 marks)
  6. Find the coordinates of the points on the graph of  `y = f(x)`  such that the tangents to the graph at these points intersect at  `(-1, 7/2).`  (4 marks)
  7. A transformation  `T: R^2 -> R^2`  that maps the graph of  `f` to the graph of the function  `g: R text(\{0}) -> R,\ g(x) = 1/x`  has rule
  8.      `T([(x), (y)]) = [(a, 0), (0, 1)] [(x), (y)] + [(c), (d)]`, where `a`, `c` and `d` are non-zero real numbers.
  9. Find the values of `a, c` and `d`.  (2 marks)
Show Answers Only
  1.  
    met2-2012-vcaa-sec2-answer
    1. `f^{′}(x) = (−2)/((2x-4)^3)`
    2. `text(Range) = (−∞,0)`
    3. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(Coordinates:) (1,5/2)\ text(or)\ (5,19/6)`
  5. `a = 2, c = −4, d = −3`
Show Worked Solution

a.   `text(Asymptotes:)`

`x = 2`

`y = 3`

met2-2012-vcaa-sec2-answer

 

b.i.   `f^{′}(x) = (−2)/((2x-4)^2)`

 

b.ii.   `text(Range) = (−∞,0), or  R^-`

MARKER’S COMMENT: Incorrect notation in part (b)(ii) was common, including `{-oo,0}, -R, (0,-oo)`.

 

b.iii.    `text(As)\ \ ` `f^{′}(x) < 0quadtext(for)quadx ∈ R text(\{2})`
    `f^{′}(x) != 0`

 

`:. f\ text(has no stationary points.)`

 

c.   `text(Point of tangency) = P(p,1/(2p-4) + 3)`

♦♦ Mean mark 29%.
`m_text(tang)` `= f^{′}(p)`
  `= (−2)/((2p-4)^2)`

 

`text(Equation of tangent using:)`

`y-y_1` `= m(x-x_1)`
`y-(1/(2p-4) + 3)` `= (−2)/((2p-4)^2)(x-p)`
`y-3` `= (−2(x-p))/((2p-4)^2) + (2p-4)/((2p-4)^2)`
`(2p-4)^2(y-3)` `= −2x + 2p + 2p-4`
`:. (2p-4)^2(y-3)` `= −2x + 4p-4\ \ text(… as required)`

 

d.   `text(Substitute)\ \ (−1,7/2)\ text{into tangent (part c),}`

♦♦♦ Mean mark 19%.

`text(Solve)\ \ (2p-4)^2(7/2-3) = −2(−1) + 4p-4\ \ text(for)\ p:`

`:. p = 1,\ text(or)\ 5`

`text(Substitute)\ \ p = 1\ text(and)\ p = 5\ text(into)\ \ P(p,1/(2p-4) + 3)`

`:. text(Coordinates:)\ (1,5/2)\ text(or)\ (5,19/6)`

 

e.   `text(Determine transformations that that take)\ f -> g:`

`text(Dilate the graph of)\ \ f(x) = 1/(2x-4) + 3\ \ text(by a)`

`text(factor of 2 from the)\ \ ytext(-axis).`

`y = 1/(2(x/2)-4) + 3= 1/(x-4) + 3`

`text(Translate the graph 4 units to the left and 3)`

`text(units down to obtain)\ \ g(x).`

 

`text(Using the transformation matrix,)`

`x^{′}` `=ax+c`
`y^{′}` `=y+d`

 

`f -> g:\ \ 1/(2x-4) -> 1/(x^{′})`

`x^{′}=2x-4`

`=> a=2,\ \ c=-4`

 

`f -> g:\ \ y -> y^{′} + 3`

`y^{′}=y -3`

`=>\ \ d=-3`

Calculus, MET2 2015 VCAA 4 MC

Consider the tangent to the graph of  `y = x^2`  at the point  `(2, 4).`

Which of the following points lies on this tangent?

A.   `text{(1, −4)}`

B.   `(3, 8)`

C.   `text{(−2, 6)}`

D.   `(1, 8)`

E.   `text{(4, −4)}`

Show Answers Only

`B`

Show Worked Solution

`text(Find equation of tangent line)`

`text(at)\ x = 2:`

`y = 4x – 4qquadtext([CAS: tangentLine) (x^2,x,2)]`

`(3,8)\ \ text(is on)\ y = 4x – 4`

`=>   B`

Calculus, MET2 2012 VCAA 18 MC

The tangent to the graph of  `y = log_e(x)`  at the point  `(a, log_e(a))`  crosses the `x`-axis at the  point  `(b, 0)`, where  `b < 0.`

Which of the following is false?

A.   `1 < a < e`

B.   The gradient of the tangent is positive

C.   `a > e`

D.   The gradient of the tangent is `1/a`

E.   `a > 0`

Show Answers Only

`A`

Show Worked Solution

`text(Consider Option)\ A:`

♦♦ Mean mark 30%.

`text(T)text{angent at}\ x=1\ text{(from graph),}\ \ b>0.`

`text(T)text{angent at}\ \ x=e\ \ text{is}\ \ y=1/e x,\ \ b=0.`

 `text(Graphing the tangents:)`

`text(For negative)\ xtext{-intercept  (i.e.}\ \ b<0text{)}`

`a > e`

`:. A\ text(is false.)`

`=>   A`

Calculus, MET2 2013 VCAA 11 MC

If the tangent to the graph of  ` y = e^(ax), a != 0,\ \ text(at)\ \ x = c`  passes through the origin, then  `c`  is equal to

A.   `0`

B.   `1/a`

C.   `1`

D.   `a`

E.   `-1/a`

Show Answers Only

`B`

Show Worked Solution

`y = e^(ax)\ \ \ => dy/dx = ae^(ax)`

♦ Mean mark 47%.

`text(At)\ \ x = c,`

`y = e^(ac)\ \ \ => (dy)/(dx) = ae^(ac)`

`text(T)text(angent passes through)\ \ (0,0) and (c,e^(ac))`

`:.\ text(Gradient of tangent) = (e^(ac)-0)/(c-0) = e^(ac)/c`

`text(Equating the gradients,)`

`ae^(ac)` `=e^(ac)/c`
`ac` `=1`
`c` `=1/a`

 
`=>   B`