smc-634-50-Find tangent given curve

Calculus, MET2 2024 VCAA 16 MC

Suppose that a differentiable function \( f: R \rightarrow R \) and its derivative \(f^{\prime}: R \rightarrow R\) satisfy \(f(4)=25\) and \(f^{\prime}(4)=15\).

Determine the gradient of the tangent line to the graph of \( {\displaystyle y=\sqrt{f(x)} } \) at \( x=4 \).

\(\sqrt{15}\)
\(\dfrac{1}{10}\)
\(\dfrac{15}{2}\)
\(\dfrac{3}{2}\)

Show Answers Only

\(D\)

Show Worked Solution

\(y\)	\(=\sqrt{f(x)}={f(x)}^{\frac{1}{2}}\)
\(y^{\prime}\)	\(=\dfrac{1}{2}\left({f(x)}^{-\frac{1}{2}}\right)\times f^{\prime}(x)\)
	\(=\dfrac{f^{\prime}(x)}{2\sqrt{f(x)}}\)
\(\text{Given}\ \ f(4)=25,\ f^{\prime}(4)=15\)
\(\therefore\ y^{\prime}\)	\(=\dfrac{15}{2\sqrt{25}}\)
	\(=\dfrac{3}{2}\)

\(\Rightarrow D\)

Calculus, MET2 2022 VCAA 1

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

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

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

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

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

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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`.

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

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

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

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

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

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

Find the coordinates of the point of inflection for \(h\), correct to two decimal places. (1 mark)
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Find the largest interval of \(x\) values for which \(h\) is strictly decreasing.
Give your answer correct to two decimal places. (1 mark)
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Apply Newton's method, with an initial estimate of \(x_0=0\), to find an approximate \(x\)-intercept of \(h\).
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}

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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)
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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.
Find this value of \(n\). (2 marks)
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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. \([0.49, 3.21]\)

\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

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\(D\)

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}`.

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

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`.

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`.

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)
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

`text(See Worked Solutions.)`
`a=+-sqrt3/3`
`text(Horizontal line.)`
i. `a=-0.5052, 0.8084, 1.3468`
ii. `a in (-0.505,-0.500]uu(0.808,1.347)`
`a=+- sqrt5/5`
`text(See Worked Solutions.)`
`w=-5t^2`
`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.

Find the slope of the tangent to `f` at `x=1`. (1 mark)
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)
Find the slope of the tangent to `f` at a point `x =p`. Give your answer in terms of `p`. (1 mark)
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)
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.

i. The first line segment connects the point `(0, f(0))` and the point `Q(n, f(n))`, where `1 < n < 3`.
Find the equation of this line segment in terms of `n`. (1 mark)
ii. The second line segment connects the point `Q(n, f(n))` and the point `(3, f(3))`, where `1 < n < 3`.
Find the equation of this line segment in terms of `n`. (1 mark)
iii. Find the value of `n`, where `1 < n < 3`, if there are equal areas between the function `f` and each line segment.
Give your answer correct to three decimal places. (3 marks)

Show Answers Only

`-2`
`117^@`
`2(1-2p^(2))e^(1-p^(2))” or “(2e-4p^(2)e)e^(-p^(2))`
i. `0.655`
ii. `(0.80, 2.39)`
i. `y_1=2e^((1-n^2))x`
ii. `y_2=(2n e^((1-n^2)) – 6e^(-8))/(n-3) (x-3) + 6e^(-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)`.

State the maximal domain and the range of `q`. (2 marks)
i. Find the equation of the tangent to the graph of `q` when `x = –2`. (1 mark)
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.`

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

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

`text{Domain:} \ x ∈ (-∞, -1)`
`text{Range:} \ y ∈ R`
i. `-x – 2`
ii. `x + 2`
`text{Not a one-to-one function as it fails the horizontal line test.}`
`-2e^{-2a} + 2e^{-a}`
`-0.67`
`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}`

`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.

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

Find the rule of `A`, in terms of `a`. (3 marks)
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.

Find the value of `b` for which the total area of these regions is a minimum. (2 marks)
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

`y = 2a^3 – 3a^2x + 1`
`(2a^3 + 1)/(3a^2)`
`-2a`
`(80a^6 + 8a^3 – 9a^2 + 2)/(12a^2)`
`10^(-1/3)`
`9/10`
`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)
Find the equation of the tangent to `y = xsin(x)` at the point `(−(5pi)/2,(5pi)/2)`. (2 marks)
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)
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. `(2n + 1)pi`
2. `−(2n + 1)pi`
`:. y = −x`
`−3pi`
`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, MET1 SM-Bank 2

Let `f: (0,oo) → R,` where `f(x) = log_e (x).`

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

Show Answers Only

`y = x/e`

Show Worked Solution

`y = log_ex`

`dy/dx = 1/x`

`text(At)\ (e, 1),\ \ m = 1/e`

`text(Equation of tangent,)\ \ m = 1/e,\ text(through)\ (e, 1):`

`y − 1`	`= 1/e(x −e)`
`:. y`	`= x/e`

Calculus, MET2 2016 VCAA 2

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

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)
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`.

i. Find the coordinates of `B`. (1 mark)
ii. Find the equation of the line that passes through `A` and `C` and, hence, find the coordinates of `C`. (2 marks)
iii. Find the area of triangle `ABC`. (2 marks)
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)
ii. Find the length of `AE`. (3 marks)

Show Answers Only

i. `text(Proof)\ \ text{(See worked solutions)}`
ii. `x = −2, 1`
i. `(0, 3)`
ii. `y = 3/4 x-7/6`
`(0, −7/6)`
iii. `25/6\ text(u²)`
`(−1, 25/12)`
`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`.

Find the period and range of `f`. (2 marks)
State the rule for the derivative function `f prime`. (1 mark)
Find the equation of the tangent to the graph of `f` at `x = pi`. (1 mark)
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)
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)
Find the values of `x, \ 0 <= x <= 8 pi`, such that `f(x) = 2 f prime (x) + pi`. (2 marks)

Show Answers Only

`text(Period:)\ 4 pi; qquad text(Range:)\ [pi – 2, pi + 2]`
`f prime (x) = −sin (x/2)`
`y = −x + 2 pi`
`y = x – 2 pi and y = x – 6 pi`
`a = 1/2 and b = -pi/2`
`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.

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

Show Answers Only

i. `1/m log_e(m/3)`
ii. `m > 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.`

Express `x^4 - 8x` in the form `x(x - a) ((x + b)^2 + c)`. (2 marks)
Describe the translation that maps the graph of `y = f (x)` onto the graph of `y = g (x)`. (1 mark)
Find the values of `d` such that the graph of `y = f (x + d)` has
i. one positive `x`-axis intercept (1 mark)
ii. two positive `x`-axis intercepts. (1 mark)
Find the value of `n` for which the equation `g (x) = n` has one solution. (1 mark)
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.
i. Find the value of `u^3 + v^3`. (2 marks)
ii. Find `u` and `v` if `u + v = 1`. (1 mark)
i. Find the equation of the tangent to the graph of `y = g(x)` at the point `(p, g(p))`. (1 mark)
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

`x(x- 2)((x + 1)^2 + 3)`
`text(See Worked Solutions)`
i. `[1,3)`
ii. `d < 1`
`−6 xx 2^(1/3)`
i. `4`
ii. `u = (sqrt5 + 1)/2,quad v = (−sqrt5 + 1)/2`
i. `y = 4(p^3 – 2)x – 3p^4`
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%.

`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)`

`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.`

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)
Find the distance `PS` and express it in the form `sqrt b/c`, where `b` and `c` are positive integers. (2 marks)

Find the area of the shaded region in the graph above. (3 marks)

Show Answers Only

`−3/5(x – 4)(x – 2)`
1. `y = – 9/5x + 13/5`
2. `S (0, 13/5)`
  
  `Q (13/9, 0)`
`sqrt 106/5`
`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.

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

Find the gradient of the tangent in terms of `k.` (2 marks)
i. Find the rule `A(k)` for the function of `k` that gives the area of the shaded region. (2 marks)
ii. Find the maximum area of the shaded region and the value of `k` for which this occurs. (2 marks)
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)`
`2sqrt3\ text(units when)\ x = 2sqrt2`
`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`

`:. 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.

State the maximum temperature in the greenhouse and the values of `t` when this occurs. (2 marks)
State the period of the function `T.` (1 mark)
Find the smallest value of `t` for which `T = 26.` (2 marks)
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.

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

`27^@Cquadtext(when)\ t = 0, 16`
`text(16 hours)`
`8/3`
`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:)`

`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.`

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)
i. Find `f^{′}(x)`. (1 mark)
ii. State the range of `f ^{′}`. (1 mark)
iii. Using the result of part ii. explain why `f` has no stationary points. (1 mark)
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)
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)
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
`T([(x), (y)]) = [(a, 0), (0, 1)] [(x), (y)] + [(c), (d)]`, where `a`, `c` and `d` are non-zero real numbers.
Find the values of `a, c` and `d`. (2 marks)

Show Answers Only

1. `f^{′}(x) = (−2)/((2x-4)^3)`
2. `text(Range) = (−∞,0)`
3. `text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(Coordinates:) (1,5/2)\ text(or)\ (5,19/6)`
`a = 2, c = −4, d = −3`

Show Worked Solution

a. `text(Asymptotes:)`

`x = 2`

`y = 3`

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`