Band 5 – Page 45

Functions, 2ADV F1 2016 HSC 4 MC

Which diagram shows the graph of an odd function?

Show Answers Only

`A`

Show Worked Solution

`text(Odd functions occur when:)`

♦ Mean mark 38%.

`f(x) = – f(x)`

`text(Graphically, this occurs when a function has)`

`text(symmetry when rotated 180° about the origin.)`

`=> A`

Calculus, MET1 2006 VCAA 9

A rectangle `XYZW` has two vertices, `X` and `W`, on the `x`-axis and the other two vertices, `Y` and `Z`, on the graph of `y = 9 - 3x^2`, as shown in the diagram below. The coordinates of `Z` are `(a, b)` where `a` and `b` are positive real numbers.

Find the area, `A`, of rectangle `XYZW` in terms of `a`. (1 mark)
Find the maximum value of `A` and the value of `a` for which this occurs. (3 marks)

Show Answers Only

`A = 18a – 6a^3`
`A_max = 12\ text(u² when)\ \ a = 1`

Show Worked Solution

♦ Mean mark 37%.

a.	`A`	`=\ text(length × height)`
		`=2a (9 – 3a^2)`
		`= 18a – 6a^3`

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

♦♦ Mean mark 29%.

`18 – 18 a^2`	`=0`
`a^2`	`=1`
`a`	`= 1,\ \ \ \ \ a > 0`

`text(Find)\ \ A\ \ text(when)\ \ \ a = 1:`

`A = 18 (1) – 6 (1)^3 = 12`

`:. A_max = 12\ text(u² when)\ \ a = 1`

Calculus, MET1 2006 VCAA 8

A normal to the graph of `y = sqrt x` has equation `y = -4x + a`, where `a` is a real constant. Find the value of `a.` (4 marks)

Show Answers Only

`a = 18`

Show Worked Solution

`text(Normal) = text(line) _|_ text(to tangent)`

♦ Mean mark 40%.

`m_{text(norm)}=-4\ \ => \ m_tan=1/4`

`y=sqrt x\ \ => \ dy/dx=1/2 x^(-1/2)`

`text(Find)\ x\ text(when:)`

`1/(2 sqrt x)`	`=1/4`
`sqrt x`	`=2`
`x`	`=4`

`:.\ text(Point of tangency is)\ \ (4, 2)`

`text(Find equation of normal:)`

`y-y_1`	`= m (x-x_1)`
`y-2`	`= -4 (x-4)`
`:. y`	`= -4x + 18`
`:. a`	`= 18`

Probability, MET1 2006 VCAA 6

The probability density function of a continuous random variable `X` is given by

`f(x) = {(x/12,\ \ 1 <= x <= 5), (\ 0,\ \ text(otherwise)):}`

Find `text(Pr) (X < 3)` (2 marks)
If `text(Pr) (X >= a) = 5/8`, find the value of `a`. (2 marks)

Show Answers Only

`1/3`
`sqrt 10`

Show Worked Solution

a. `text(Pr) (X < 3)`

`= int_1^3 1/12x\ dx`

`= 1/12 [1/2 x^2]_1^3`

`= 1/24 [3^2 – 1^2]`

`= 1/3`

b. `text(Pr) (X >= a) = 5/8`

♦ Mean mark 47%.

`int_a^5 x/12\ dx`	`= 5/8`
`1/24 [x^2]_a^5`	`= 5/8`
`(5^2 – a^2)`	`= 15`
`a^2`	`= 10`
`:. a`	`= sqrt 10,\ \ \ a in (1, 5)`

Probability, MET1 2006 VCAA 5

Let `X` be a normally distributed random variable with a mean of 72 and a standard deviation of 8. Let `Z` be the standard normal random variable. Use the result that `text(Pr) (Z < 1) = 0.84`, correct to two decimal places, to find

the probability that `X` is greater than `80` (1 mark)
the probability that `64 < X < 72` (1 mark)
the probability that `X < 64` given that `X < 72` (2 marks)

Show Answers Only

`0.16`
`0.34`
`8/25`

Show Worked Solution

`text(Pr) (X > 80)`

`= text(Pr) (Z > (80 – 72)/8)`

`= text(Pr) (Z > 1)`

`= 0.16`

b. `text(Pr) (64 < X < 72)`

♦ Mean mark 45%.

`= text(Pr) (72 < X < 80)\ \ \ text(due to symmetry)`

`= 0.5 – 0.16`

`= 0.34`

♦ Mean mark 40%.
MARKER’S COMMENT: Notation was poor, showing a lack of understanding in this area.

c. `text(Conditional probability)`

`text(Pr) (X < 64 \| X < 72)`	`= (text{Pr} (X < 64))/(text{Pr} (X < 72))`
	`= 0.16/0.50`
	`= 8/25 or 0.32`

Functions, MET1 2006 VCAA 4

For the function `f: [– pi, pi] -> R, f(x) = 5 cos (2 (x + pi/3))`

write down the amplitude and period of the function (2 marks)
sketch the graph of the function `f` on the set of axes below. Label axes intercepts with their coordinates.

Label endpoints of the graph with their coordinates. (3 marks)

Show Answers Only

`text(Amplitude) = 5;\ \ \ text(Period) = pi`

Show Worked Solution

a. `text(Amplitude) = 5`

`text(Period) = (2 pi)/2 = pi`

`text(Shift)\ \ y = 5 cos (2x)\ \ text(left)\ \ pi/3\ \ text(units).`

`text(Period) = pi`

`text(Endpoints are)\ \ (-pi, -5/2) and (pi,-5/2)`

Functions, MET1 2006 VCAA 2

For the function `f: R -> R, f(x) = 3e^(2x)-1,`

find the rule for the inverse function `f^-1` (2 marks)
find the domain of the inverse function `f^-1` (1 mark)

Show Answers Only

`f^-1 (x) = 1/2 log_e ((x + 1)/3)`
`x in (– 1, oo)`

Show Worked Solution

a. `text(Let)\ \ y = f(x)`

`text(For inverse, swap)\ \ x harr y`

`x`	`= 3e^(2y)-1`
`(x + 1)/3`	`= e^(2y)`
`2y`	`= log_e ((x + 1)/3)`
`y`	`= 1/2 log_e ((x + 1)/3)`

`:. f^-1 (x) = 1/2 log_e ((x + 1)/3)`

♦ Mean mark 45%.

b. `text(Domain)\ (f^-1) = text(Range)\ f(x),`

`:.x in (– 1, oo)`

Calculus, MET1 2007 VCAA 12

`P` is the point on the line `2x + y-10 = 0` such that the length of `OP`, the line segment from the origin `O` to `P`, is a minimum. Find the coordinates of `P` and this minimum length. (4 marks)

Show Answers Only

`P(4,2)\ text(for minimum length)\ 2sqrt5\ text(units)`

Show Worked Solution

`P(x, − 2x + 10)`

♦ Mean mark 26%.

`text(Let)\ z`	`=\ text(length of)\ OP`
`z`	`= sqrt((x – 0)^2 + (−2x + 10)^2)`
`z^2`	`= x^2 + 4x^2 – 40x + 100`
`z^2`	`= 5x^2 – 40x + 100`

`text(Minimum)\ z\ text(occurs when minimum)\ z^2\ text(occurs.)`

`text(Stationary point when:)`

`d/(dx) (z^2)`	`= 0`
`10x – 40`	`= 0`
`x`	`= 4`

`text(Find)\ ytext(-coordinate:)`

`2(4) + y – 10`	`= 0`
`y`	`= 2`

`text(Find length)\ OP:`

`z`	`= sqrt(4^2 + (−2 xx 4 + 10)^2)`
	`= sqrt20`
	`=2 sqrt5\ \ text(units)`

`:. P(4,2)\ text(for minimum length)\ 2sqrt5\ text(units)`

Probability, MET1 2007 VCAA 11

There is a daily flight from Paradise Island to Melbourne. The probability of the flight departing on time, given that there is fine weather on the island, is 0.8, and the probability of the flight departing on time, given that the weather on the island is not fine, is 0.6.

In March the probability of a day being fine is 0.4.

Find the probability that on a particular day in March

the flight from Paradise Island departs on time (2 marks)
the weather is fine on Paradise Island, given that the flight departs on time. (2 marks)

Show Answers Only

`0.68`
`8/17`

Show Worked Solution

`text{Pr(FT)}\ +\ text{Pr(F′T)}`

`= 0.4 xx 0.8 + 0.6 xx 0.6`

`= 0.32 + 0.36`

`= 0.68`

b. `text(Conditional probability:)`

♦♦ Mean mark 29%.
MARKER’S COMMENT: Students continue to struggle with conditional probability. Attention required here.

`text(Pr)(F\ text(\|)\ T)`	`= (text(Pr)(F ∩ T))/(text(Pr)(T))`
	`= 0.32/0.68`

`:. text(Pr)(F\ text(|)\ T) = 8/17`

Calculus, MET1 2007 VCAA 9

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

Find the equation of the normal to the graph of `f` where it crosses the `y`-axis. (2 marks)
Find the exact area of the shaded region. (3 marks)

Show Answers Only

`y = – 2x + 2`
`(2sqrte – 2)\ text(u)²`

Show Worked Solution

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

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

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

`text(Gradient of normal:)`

`f′(x)`	`= 1/2 e^(x/2)`
`f′(0)`	`= 1/2`

`:. m_text(norm) = -2`

`text(Equation of normal:)`

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

♦ Mean mark 44%.

b.	`:.\ text(Area)`	`= int_0^1 (e^(x/2) + 1 – (-2x +2))\ dx`
		`= int_0^1 (e^(x/2) + 2x – 1)\ dx`
		`= [2e^(x/2) + x^2 – x]_0^1`
		`= (2e^(1/2) + 1^2 – 1) – (2e^0)`
		`= (2sqrte – 2)\ text(u)²`

Functions, MET1 2007 VCAA 8

Let `f: R -> R`, `f(x) = sin((2pix)/3)`.

Solve the equation `sin((2pix)/3) = -sqrt3/2` for ` x ∈ [0,3]`. (2 marks)
Let `g: R -> R`, `g(x) = 3f(x - 1) + 2`.
Find the smallest positive value of `x` for which `g(x)` is a maximum. (2 marks)

Show Answers Only

`x = 2, 5/2`
`7/4`

Show Worked Solution

a. `sin((2pix)/3) = -sqrt3/2`

`=>\ text(Base angle)\ = pi/3`

`(2 pi x)/3`	`=(4pi)/3, (5pi)/3, (10pi)/3, …`
`:.x`	`=2 or 5/2, \ \ \ text(for)\ x ∈ [0,3]`

b. `g(x) = 3sin[(2pi)/3 (x – 1)] + 2`

♦♦ Mean mark 29%.
STRATEGY: Max/min questions involving trig functions can often use the powerful identity, `-1 <= sin theta“<=1` to solve efficiently and reduce errors.

`text(Maximum occurs when:)`

`sin[(2pi)/3 (x – 1)]`	`= 1`
`(2pi)/3 (x – 1)`	`= pi/2`
`x-1`	`= pi/2 xx 3/(2pi)`
`:. x`	`=7/4`

Calculus, MET1 2007 VCAA 7

If `f(x) = xcos(3x)`, then `f′(x) = cos(3x) - 3xsin(3x)`.

Use this fact to find an antiderivative of `xsin(3x)`. (3 marks)

Show Answers Only

`- 1/3xcos(3x) + 1/9sin(3x)`

Show Worked Solution

`text(If)\ \ f′(x) = g(x) \ -> \ int g(x)\ dx = f(x)`

♦ Mean mark 36%.
MARKER’S COMMENT: This standard and familiar calculus problem was surprisingly poorly answered.

`int (cos(3x) -3xsin(3x))\ dx`	`= xcos(3x)`
`1/3 sin(3x) – 3 int xsin(3x)\ dx`	`= xcos(3x)`
`-3 int xsin(3x)\ dx`	`= xcos(3x) – 1/3sin(3x)`
`:. int xsin(3x)\ dx`	`= 1/9sin(3x) – 1/3xcos(3x)`

Probability, MET1 2007 VCAA 6

Two events, `A` and `B`, from a given event space, are such that `text(Pr)(A) = 1/5` and `text(Pr)(B) = 1/3`.

Calculate `text(Pr)(A′ ∩ B)` when `text(Pr)(A ∩ B) = 1/8`. (1 mark)
Calculate `text(Pr)(A′ ∩ B)` when `A` and `B` are mutually exclusive events. (1 mark)

Show Answers Only

`5/24`
`1/3`

Show Worked Solution

a. `text(Sketch Venn diagram:)`

♦♦ Mean mark 31%.
MARKER’S COMMENTS: Students who drew a Venn diagram or Karnaugh map were the most successful.

`:. text(Pr)(A′ ∩ B)`	`=text(Pr)(B) – text(Pr)(A ∩B)`
	`=1/3 – 1/8`
	`=5/24`

♦♦ Mean mark 23%.

`text(Mutually exclusive means)\ \ text(Pr)(A ∩ B)=0,`

`:. text(Pr)(A′ ∩ B) = 1/3`

Probability, MET1 2007 VCAA 5

It is known that 50% of the customers who enter a restaurant order a cup of coffee. If four customers enter the restaurant, what is the probability that more than two of these customers order coffee? (Assume that what any customer orders is independent of what any other customer orders.) (2 marks)

Show Answers Only

`text(Pr)(X > 2) = 5/16`

Show Worked Solution

`text(Pr)(X > 2)`	`= text(Pr)(X = 3) + text(Pr)(X = 4)`
	`= ((4),(3))(1/2)^3(1/2) + ((4),(4))(1/2)^4(1/2)^0`
	`= 4 xx 1/16 + 1 xx 1/16`

♦ Mean mark 41%.
MARKER’S COMMENT: Notation was often poor and confused. Common error: evaluating co-efficients.

`:. text(Pr)(X > 2) = 5/16`

Calculus, MET1 2007 VCAA 3

The diagram shows the graph of a function with domain `R`.

For the graph shown above, sketch on the same set of axes the graph of the derivative function. (3 marks)
Write down the domain of the derivative function. (1 mark)

Show Answers Only

`R\ text(\)\ {0,3}`

Show Worked Solution

♦ Part (a) mean mark 42%, and part (b) mean mark 45%.

b. `text(Derivative does not exist at either sharp)`

`text(points or discontinuous points.)`

`:. R\ text(\)\ {0,3}`

Functions, MET1 2008 VCAA 10

Let `f: R -> R,\ \ f(x) = e^(2x)-1`.

Find the rule and domain of the inverse function `f^(−1)`. (2 marks)
On the axes provided, sketch the graph of `y = f(f^(−1)(x))` for its maximal domain. (1 mark)
Find `f(−f^(−1)(2x))` in the form `(ax)/(bx + c)` where `a`, `b` and `c` are real constants. (2 marks)

Show Answers Only

`f^(−1)(x) = 1/2log_e(x + 1), x ∈ (−1,∞)`
`f(−f^(−1)(2x)) = (−2x)/(2x + 1)`

Show Worked Solution

a. `text(Let)\ \ y = f(x)`

`text(For Inverse, swap)\ x ↔ y`

`x`	`= e^(2y)-1`
`x + 1`	`= e^(2y)`
`2y`	`= log_e(x + 1)`
`y`	`= 1/2 log_e(x + 1)`

`text(Domain)(f^(−1))`	`=\ text(Range)\ (f)`
	`= (−1,∞)`

`:. f^(−1)(x) = 1/2log_e(x + 1),quadx ∈ (−1,∞)`

b. `f(f^(−1)(x)) = x`

♦ Mean mark part (b) 19%.
MARKER’S COMMENT: Few students were aware that a function of its own inverse function is the line `y=x` over the appropriate domain.

`text(Domain is)\ \ (-1, oo)`

♦ Mean mark 34%.

c.	`−f^(−1)(2x)`	`= −1/2 ln(2x + 1)`
	`:. f(−f^(−1)(2x))`	`= e^(-log_e(2x + 1))-1`
		`=(2x+1)^-1 -1`
		`= 1/(2x + 1)-1`
		`= (−2x)/(2x + 1)`

Calculus, MET1 2008 VCAA 9

A plastic brick is made in the shape of a right triangular prism. The triangular end is an equilateral triangle with side length `x` cm and the length of the brick is `y` cm.

The volume of the brick is 1000 cm³.

Find an expression for `y` in terms of `x`. (2 marks)
Show that the total surface area, `A` cm², of the brick is given by
`A = (4000sqrt3)/x + (sqrt3 x^2)/2` (2 marks)
Find the value of `x` for which the brick has minimum total surface area. (You do not have to find this minimum.) (3 marks)

Show Answers Only

`y = (4000 sqrt3)/(3 x^2)`
`text(See Worked Solutions)`
`10 root(3)(4)`

Show Worked Solution

a. `text(Cross-section is an equilateral triangle and)`

`text(angles of)\ Delta\ text(are)\ 60^@.`

`sin 60^@`	`= h/x`
`:. h`	`= sqrt3/2 x`

`text(Volume)`	`= 1000`
`(1/2 x)(sqrt3/2 x)y`	`= 1000`
`sqrt3/4 x^2y`	`= 1000`
`:. y`	`= 4000/(sqrt3 x^2)`
	`=(4000 sqrt3)/(3 x^2)`

♦ Mean mark 45%.

b.	`A`	`= 2 xx (sqrt3/4 x^2) + 3 xx (xy)`
		`= sqrt3/2 x^2 + 3x(4000/(sqrt3 x^2))`
		`= (12\ 000)/(sqrt3 x) xx sqrt3/sqrt3 + sqrt3/2 x^2`
		`= (4000sqrt3)/x + (x^2 sqrt3)/2`

♦ Mean mark 42%.

c. `A = 4000 sqrt3 x^(−1) + sqrt3/2 x^2,quadx > 0`

`text(Stationary Point when)\ \ (dA)/(dx)=0,`

`-4000sqrt3 x^(−2) + sqrt3x`	`=0`
`(4000sqrt3)/(x^2)`	`= sqrt3x`
`x^3`	`= 4000`
`:. x`	`= root(3)(4000)`
	`= 10 root(3)(4)`

Probability, MET1 2008 VCAA 4

The function

`f(x) = {{:(k),(0):}{:(sin(pix)qquadtext(if)qquadx ∈ [0,1]),(qquadqquadqquadqquadquadtext(otherwise)):}`

is a probability density function for the continuous random variable `X`.

Show that `k = pi/2`. (2 marks)
Find `text(Pr)(X <= 1/4 | X <= 1/2)`. (3 marks)

Show Answers Only

`text(See Worked Solutions)`
`1 – (sqrt2)/2`

Show Worked Solution

a. `text(Total Area under curve) = 1\ text(u²)`

`int_0^1 k sin(pix)dx`	`= 1`
`- k/pi [cos(pix)]_0^1`	`= 1`
`- k/pi[cos(pi) – cos(0)]`	`= 1`
`- k/pi[(−1) – (1)]`	`= 1`
`2k`	`= pi`
`:.k`	`= pi/2`

b. `text(Conditional Probability:)`

♦ Mean mark 47%.
MARKER’S COMMENT: Few students used the symmetry of the probability density function to calculate the denominator.

`text(Pr)(X <= 1/4 | X <= 1/2)`

`= (text(Pr)(X <= 1/4))/(text(Pr)(X <= 1/2))`

`= (pi/2 int_0^(1/4) sin(pix)dx)/(1/2)`

`= -pi/pi [cos(pix)]_0^(1/4)`

`= -1 [cos(pi/4) – cos(0)]`

`= -1 [1/(sqrt2) – 1]`

`= 1 – (sqrt2)/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)`

Probability, MET1 2009 VCAA 5

Four identical balls are numbered 1, 2, 3 and 4 and put into a box. A ball is randomly drawn from the box, and not returned to the box. A second ball is then randomly drawn from the box.

What is the probability that the first ball drawn is numbered 4 and the second ball drawn is numbered 1? (1 mark)
What is the probability that the sum of the numbers on the two balls is 5? (1 mark)
Given that the sum of the numbers on the two balls is 5, what is the probability that the second ball drawn is numbered 1? (2 marks)

Show Answers Only

`1/12`
`1/3`
`1/4`

Show Worked Solution

a. `text(Pr) (4, 1)`

`= 1/4 xx 1/3`

`= 1/12`

b. `text(Pr) (text(Sum) = 5)`

`= text(Pr) (1, 4) + text(Pr) (2, 3) + text(Pr) (3, 2) + text(Pr) (4, 1)`

`= 4 xx (1/4 xx 1/3)`

`= 1/3`

c. `text(Conditional Probability)`

♦ Mean mark 46%.

`text(Pr) (2^{text(nd)} = 1\ |\ text(Sum) = 5)`

`= (text{Pr} (4, 1))/(text{Pr} (text(Sum) = 5))`

`= (1/12)/(1/3)`

`= 1/4`

Calculus, MET1 2010 VCAA 11

A cylinder fits exactly in a right circular cone so that the base of the cone and one end of the cylinder are in the same plane as shown in the diagram below. The height of the cone is 5 cm and the radius of the cone is 2 cm.

The radius of the cylinder is `r` cm and the height of the cylinder is `h` cm.

For the cylinder inscribed in the cone as shown above

find `h` in terms of `r` (2 marks)

The total surface area, `S` cm², of a cylinder of height `h` cm and radius `r` cm is given by the formula

`S = 2 pi r h + 2 pi r^2`.

find `S` in terms of `r` (1 mark)
find the value of `r` for which `S` is a maximum. (2 marks)

Show Answers Only

`(10 -5r)/2`
`10 pi r – 3 pi r^2,\ \ 0 < r < 2`
`5/3\ text(cm)`

Show Worked Solution

a. `text(Sketch cross-section:)`

`text(Using ratios of similar triangles,)`

♦ Mean mark 43%.

`Delta ABC\ text(|||)\ Delta PQC`

`5/h`	`= 2/(2 – r)`
`2h`	`= 10 – 5r`
`:. h`	`= (10 – 5r)/2`

♦ Mean mark 47%.

b.	`S`	`= 2 pi r ((10 – 5 r)/2) + 2 pi r^2`
		`= 10 pi r – 5 pi r^2 + 2 pi r^2`
		`= 10 pi r – 3 pi r^2,\ \ 0 < r < 2`

c. `text(SP’s occur when)\ \ (dS)/(dr) = 0,`

♦ Mean mark 46%.

`10 pi – 6 pi r`	`= 0`
`3r`	`= 5`
`:. r`	`= 5/3\ text(cm)`

Calculus, MET1 2010 VCAA 9

Part of the graph of `f: R^+ -> R, \ f(x) = x log_e (x)` is shown below.

Find the derivative of `x^2 log_e (x)`. (1 mark)
Use your answer to part a. to find the area of the shaded region in the form `a log_e (b) + c` where `a, b` and `c` are non-zero real constants. (3 marks)

Show Answers Only

`x + 2x log_e (x)`
`(9/2 log_e (3) – 2)\ text(u²)`

Show Worked Solution

a. `text(Using Product Rule:)`

`(fg)′`	`= f g′ + f′ g`
`d/(dx) (x^2 log_e (x))`	`= x^2 (1/x) + 2x log_e (x)`
	`= x + 2x log_e (x)`

b. `text{Integrating the answer from part (a):}`

`int (x + 2x log_e (x))\ dx`	`= x^2 log_e (x)`
`1/2 x^2 + 2 int x log_e (x)\ dx`	`= x^2 log_e (x)`
`2 int x log_e (x)\ dx`	`= x^2 log_e (x) – 1/2 x^2`
`:. int x log_e (x)\ dx`	`= 1/2 x^2 log_e (x) – 1/4 x^2`

♦♦ Mean mark 33%.
MARKER’S COMMENT: The most common error was not dividing everything through by 2. Be careful!

`:.\ text(Area)`	`= int_1^3 (x log_e (x)) dx`
	`= [1/2 x^2 log_e (x) – 1/4 x^2]_1^3`
	`= (9/2 log_e (3) – 9/4) – (0 – 1/4)`
	`= (9/2 log_e (3) – 2)\ \ text(u²)`

Functions, MET1 2010 VCAA 6

The transformation `T: R^2 -> R^2` is defined by

`T([(x), (y)]) = [(3, 0), (0, 2)] [(x), (y)] + [(– 1), (4)].`

The image of the curve `y = 2x^2 + 1` under the transformation `T` has equation `y = ax^2 + bx + c.`

Find the values of `a, b,` and `c.` (3 marks)

Show Answers Only

`a = 4/9,\ \ \ b = 8/9,\ \ \ c = 58/9`

Show Worked Solution

`text(Expanding the matrix equation:)`

`x′`	`= 3x – 1`	`qquad qquad y′`	`= 2y + 4`
`x`	`= (x′+1)/3`	`qquad qquad y`	`= (y′ -4)/2`

`text(Substitute)\ \ x, y\ \ text(into)\ \ y = 2x^2 + 1`

♦ Mean mark 40%.

`(y′ – 4)/2`	`= 2 ((x′ + 1)/3)^2 + 1`
`y′ – 4`	`= 4/9 ((x prime)^2 + 2x′ + 1) + 2`
`y′`	`= 4/9 (x prime)^2 + 8/9 x′ + (4/9 + 6)`
`:. y′`	`= 4/9 (x prime)^2 + 8/9 x′ + 58/9`

`:. a = 4/9,\ \ \ b = 8/9,\ \ \ c = 58/9`

Statistics, MET1 2010 VCAA 5

Let `X` be a normally distributed random variable with mean 5 and variance 9 and let `Z` be the random variable with the standard normal distribution.

Find `text(Pr) (X > 5)`. (1 mark)
Find `b` such that `text(Pr) (X > 7) = text(Pr) (Z < b)`. (2 marks)

Show Answers Only

`0.5`
`-2/3`

Show Worked Solution

a. `text(Pr) (X > 5) = 0.5`

♦ Part (b) mean mark 43%.

b.	`text(Pr) (X > 7)`	`= text(Pr) (Z > (7 – 5)/3)`
		`= text(Pr) (Z > 2/3)`
		`= text(Pr) (Z < -2/3)`

`:. b=-2/3`

Calculus, MET1 2011 VCAA 10

The figure shown represents a wire frame where `ABCE` is a convex quadrilateral. The point `D` is on line segment `EC` with `AB = ED = 2\ text(cm)` and `BC = a\ text(cm)`, where `a` is a positive constant.

`/_ BAE = /_ CEA = pi/2`

Let `/_ CBD = theta` where `0 < theta < pi/2.`

Find `BD` and `CD` in terms of `a` and `theta`. (2 marks)
Find the length, `L` cm, of the wire in the frame, including length `BD`, in terms of `a` and `theta`. (1 mark)
Find `(dL)/(d theta)`, and hence show that `(dL)/(d theta) = 0` when `BD = 2CD`. (2 marks)
Find the maximum value of `L` if `a = 3 sqrt 5`. (1 mark)

Show Answers Only

`BD = a cos theta,\ \ \ CD = a sin theta`
`L = 4 + a + 2 a cos theta + a sin theta`
`text(Proof)\ text{(See Worked Solutions)}`
`L_max = 19 + 3 sqrt 5`

Show Worked Solution

a. `text(In)\ \ Delta BCD,`

`cos theta`	`= (BD)/a`
`:. BD`	`= a cos theta`
`sin theta`	`= (CD)/a`
`:. CD`	`= a sin theta`

b.	`L`	`= 4 + 2 BD + CD + a`
		`= 4 + 2a cos theta + a sin theta + a`
		`= 4 + a + 2a cos theta + a sin theta`

c. `text(Noting that)\ a\ text(is a constant:)`

♦ Mean mark 35%.

`(dL)/(d theta)= – 2 a sin theta + a cos theta`

`text(When)\ \ (dL)/(d theta) = 0`,

`- 2 a sin theta+ a cos theta`	`= 0`
`a cos theta`	`= 2 a sin theta`
`:. BD`	`= 2CD\ \ text{(using part (a))}`

d. `text(SP’s when)\ \ (dL)/(d theta)=0,`

♦♦♦ Mean mark 5%.

`- 2 a sin theta+ a cos theta`	`= 0`
`sin theta`	`=1/2 cos theta`
`tan theta`	`=1/2`

`text(If)\ \ tan theta=1/2,\ \ cos theta = 2/sqrt5,\ \ sin theta = 1/sqrt5`

`L_(max)`	`= 4 + a + 2a cos theta + a sin theta`
	`= 4 + (3 sqrt 5) + 2 (3 sqrt 5) (2/sqrt 5) + (3 sqrt 5) (1/sqrt 5)`
	`= 4 + 3 sqrt 5 + 12 + 3`
	`= 19 + 3 sqrt 5\ text(cm)`

Calculus, MET1 2011 VCAA 9

Parts of the graphs of the functions

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

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

The area of the shaded region is 64.

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

Show Answers Only

`a = 8,\ \ m = 4`

Show Worked Solution

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

♦ Mean mark 41%.
MARKER’S COMMENT: Few students correctly used the intersection to achieve the final answer.

`f(x)`	`= g(x)`
`x^3 – ax`	`= ax`
`x^3 – 2ax`	`= 0`
`x (x^2 – 2a)`	`= 0`

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

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

`text{Shaded Area = 64 (given)},`

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

`:. m`

`= sqrt (2 xx 8)=4`

`:. a = 8,\ \ m = 4`

Probability, MET1 2011 VCAA 8

Two events, `A` and `B`, are such that `text(Pr) (A) = 3/5` and `text(Pr) (B) = 1/4.`

If `A prime` denotes the compliment of `A`, calculate `text(Pr) (A prime nn B)` when

`text(Pr) (A uu B) = 3/4` (2 marks)
`A` and `B` are mutually exclusive. (1 mark)

Show Answers Only

`text(Pr) (A prime nn B) = 3/20`
`text(Pr) (A prime nn B) = 1/4`

Show Worked Solution

a. `text(Sketch Venn Diagram)`

`text(Pr) (A uu B)`	`= text(Pr) (A) + text(Pr) (B) – text(Pr) (A nn B)`
`3/4`	`= 3/5 + 1/4 – text(Pr) (A nn B)`
`text(Pr) (A nn B)`	`= 1/10`

`:.\ text(Pr) (A prime nn B) = 1/4 – 1/10 = 3/20`

`text(Pr) (A∩ B)=0\ \ text{(mutually exclusive)},`

`:.\ text(Pr) (A prime nn B) = text(Pr) (B) = 1/4`

Probability, MET1 2011 VCAA 7

A biased coin tossed three times. The probability of a head from a toss of this coin is `p.`

Find, in terms of `p`, the probability of obtaining
1. three heads from the three tosses (1 mark)
2. two heads and a tail from the three tosses. (1 mark)
If the probability of obtaining three heads equals the probability of obtaining two heads and a tail, find `p`. (2 marks)

Show Answers Only

1. `p^3`
2. `3p^2 (1 – p)`
`0 or 3/4`

Show Worked Solution

a.i.	`text(Pr) (HHH)`	`= p xx p xx p`
		`= p^3`

ii. `text(Pr) text{(2 Heads and 1 Tail from 3 tosses)}`

♦ Part (a)(ii) mean mark 41%.

`= ((3), (2)) xx p xx p xx (1 – p)`

`= 3 p^2 (1 – p)`

b. `text(If probabilities are equal:)`

♦ Mean mark 48%.
MARKER’S COMMENT: Many students incorrectly assumed `p` could not be zero.

`p^3`	`= 3p^2 – 3p^3`
`4p^3 – 3p^2`	`= 0`
`p^2 (4p – 3)`	`= 0`

`:. p = 0 or p = 3/4`

Algebra, MET1 2011 VCAA 6

Consider the simultaneous linear equations

`kx - 3y`	`= k + 3`
`4x + (k + 7) y`	`= 1`

where `k` is a real constant.

Find the value of `k` for which there are infinitely many solutions. (3 marks)
Find the values of `k` for which there is a unique solution. (1 mark)

Show Answers Only

`– 4`
`k in R\ text(\)\ {– 4, – 3}`

Show Worked Solution

a. `text(Infinite solutions if gradients and)`

♦ Mean mark 39%.
MARKER’S COMMENT: A number of solutions are possible here, including using the determinant.

`y text(-intercepts are the same.)`

`kx-3y`	`=k+3`
`3y`	`=kx-k-3`
`y`	`=k/3 x – (k+3)/3`
`:.m_1=k/3 and c_1= -((k+3)/3)`

`4x + (k + 7) y`	`=1`
`y`	`=((-4)/(k+7)) x + 1/(k+7)`
`:. m_2=(-4)/(k+7) and c_2=1/(k+7)`

`text(Equating gradients and)\ y text(-intercepts:)`

`m_1`	`= m_2`	`\ \ \ and`	`\ \ \ c_1`	`=c_2`
`k/3`	`= (– 4)/(k + 7)`		`-((k+3)/3)`	`= 1/(k+7)`
`k^2 + 7k`	`= – 12`		`-(k+3)(k+7)`	`=3`
`k^2 + 7k + 12`	`= 0`		`k^2+10k+24`	`=0`
`(k + 3) (k + 4)`	`= 0`		`(k+6)(k+4)`	`= 0`
`k`	`= – 3, – 4`		`k`	`= – 6, – 4`

`:. k = – 4\ \ \ text{(satisfies both)}`

b. `text(Unique solution if:)`

♦♦ Mean mark 33%.

`m_1 != m_2`

`:. k in R\ text(\)\ {– 4, – 3}`

Calculus, MET1 2011 VCAA 2a

Find an antiderivative of `1/(3x - 4)` with respect to `x.` (1 mark)

Show Answers Only

`1/3 log_e | 3x – 4 |`

Show Worked Solution

`int (3x – 4)^-1\ dx`

♦ Mean mark 43%.
MARKER’S COMMENT: The constant `c` is required for the general derivative `int f(x)\ dx` but is not necessary for an antiderivative.

`= 1/3 log_e (| 3x – 4 |)+c`

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, MET1 2012 VCAA 9

Let `f: R -> R,\ \ f(x) = x sin (x)`.
Find `f prime (x)`. (1 mark)
Use the result of part a. to find the value of `int_(pi/6)^(pi/2) x cos (x)\ dx` in the form `a pi + b`. (3 marks)

Show Answers Only

`x cos (x) + sin (x)`
`(5 pi)/12 – sqrt 3/2`

Show Worked Solution

a. `f(x)=x sin(x)`

`text(Using Product Rule:)`

`(gh)′`	`= g′ h + g h′`
`:. f prime (x)`	`= x cos (x) + sin (x)`

b. `text(Integrating)\ \ f′(x)\ \ text{from part (a),}`

`int (x cos (x) + sin (x))\ dx`	`= x sin (x)`
`int (x cos x)\ dx – cos (x)`	`= x sin (x)`
`:. int (x cos x)\ dx`	`= x sin (x) + cos x`

`text(Evaluate definite integral:)`

`int_(pi/6)^(pi/2) (x cos x) dx`

`= [x sin (x) + cos (x)]_(pi/6)^(pi/2)`

`= (pi/2 sin (pi/2) + cos (pi/2)) – (pi/6 sin (pi/6) + cos (pi/6))`

`= (pi/2 + 0) – (pi/6 xx 1/2 + sqrt 3/2)`

`= (5 pi)/12 – sqrt 3/2`

Probability, MET1 2012 VCAA 8a

The random variable `X` is normally distributed with mean 100 and standard deviation 4.

If `text(Pr) (X < 106) = q`, find `text(Pr) (94 < X < 100)` in terms of `q`. (2 marks)

Show Answers Only

`q – 1/2`

Show Worked Solution

`text(Given)\ \ text(Pr) (X < 106) = q,`

♦ Mean mark 40%.
MARKER’S COMMENT: Those who drew a diagram to visualise the symmetry were the most successful.

`=> text(Pr) (X < 94) = 1-q`

`text(By symmetry of normal curve:)`

`:.\ text(Pr) (94 < X < 100)`

`= 1/2 – (1 – q)`

`= q – 1/2`

Functions, MET1 2012 VCAA 6

The graphs of `y = cos (x) and y = a sin (x)`, where `a` is a real constant, have a point of intersection at `x = pi/3.`

Find the value of `a`. (2 marks)
If `x in [0, 2 pi]`, find the `x`-coordinate of the other point of intersection of the two graphs. (1 mark)

Show Answers Only

`1/sqrt 3`
`(4 pi)/3`

Show Worked Solution

a. `text(Intersection occurs when)\ \ x=pi/3,`

`a sin(pi/3)`	`= cos (pi/3)`
`tan(pi/3)`	`= 1/a`
`sqrt 3`	`=1/a`
`:. a`	`=1/sqrt3`

b.	`tan (x)`	`= sqrt 3`
	`x`	`= pi/3, (4 pi)/3, 2pi+ pi/3, …\ text(but)\ x in [0, 2 pi]`
	`:. x`	`= (4 pi)/3`

Probability, MET1 2012 VCAA 4

On any given day, the number `X` of telephone calls that Daniel receives is a random variable with probability distribution given by

Find the mean of `X`.(2 marks)
What is the probability that Daniel receives only one telephone call on each of three consecutive days? (1 mark)
Daniel receives telephone calls on both Monday and Tuesday.

What is the probability that Daniel receives a total of four calls over these two days? (3 marks)

Show Answers Only

`1.5`
`0.008`
`29/64`

Show Worked Solution

a.	`text(E) (X)`	`= 0 xx 0.2 + 1 xx 0.2 + 2 xx 0.5 + 3 xx 0.1`
		`= 0 + .2 + 1 + 0.3`
		`= 1.5`

b. `text(Pr) (1, 1, 1)`

MARKER’S COMMENT: Many students understood that the calculation of 0.2³ was required but were not able evaluate correctly.

`= 0.2 xx 0.2 xx 0.2`

`= 0.008`

c. `text(Conditional Probability:)`

♦ Mean mark 36%.

`text(Pr) (x = 4 | x >= 1\ text{both days})`

`= (text{Pr} (1, 3) + text{Pr} (2, 2) + text{Pr} (3, 1))/(text{Pr}(x>=1\ text{both days}))`

`= (0.2 xx 0.1 + 0.5 xx 0.5 + 0.1 xx 0.2)/(0.8 xx 0.8)`

`= (0.02 + 0.25 + 0.02)/0.64`

`= 0.29/0.64`

`= 29/64`

Calculus, MET1 2012 VCAA 2

Find an anti-derivative of `1/(2x - 1)^3` with respect to `x.` (2 marks)

Show Answers Only

`(-1)/(4(2x – 1)^2) + c`

Show Worked Solution

♦ Mean mark 37%.
MARKER’S COMMENT: The most common error was answering with a log equation. The low mean mark was a surprise here. Worth attention!

`int (2x – 1)^-3 dx`	`= -1/(2 xx 2) (2x – 1)^-2 +c`
	`= (-1)/(4(2x – 1)^2) + c`

Calculus, MET1 2013 VCAA 10

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

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

Find the area, `A`, of the triangle `OPQ` in terms of `x`. (1 mark)
Find the maximum area of triangle `OQP` and the value of `x` for which the maximum occurs. (3 marks)
Let `S` be the point on the graph of `f` on the `y`-axis and let `T` be the point on the graph of `f` with the `y`-coordinate `1/2`.

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

Show Answers Only

`x e^(-x/5)`
`5/e\ text(u²)`
`25/4 log_e (4) – 15/2\ text(u²)`

Show Worked Solution

a.	`text(Area)`	`= 1/2 xx b xx h`
		`= 1/2x(2e^(-x/5))`
	`:. A`	`= xe^(-x/5)`

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

♦ Mean mark 35%.

`x(-1/5 e^(-x/5)) + e^(-x/5)`	`= 0`
`e^(-x/5)[1 – x/5]`	`= 0`
`:. x`	`= 5\ \ text{(as}\ \ e^(-x/5) >0,\ \ text(for all)\ x text{)}`

`text(When)\ \ x = 5,\ \ A`	`= xe^(-x/5)`
	`= 5e^-1`

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

c. `text(Find)\ \ S:\ \ F(0) = 2.`

♦♦ Mean mark 32%.

`:. S(0, 2)`

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

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

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

Functions, MET1 2013 VCAA 9

The graph of `f(x) = (x - 1)^2 - 2, x in [– 2, 2]`, is shown below. The graph intersects the `x`-axis where `x = a.`

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

Note: other parts of this question are out of the syllabus and not included.

Show Answers Only

`1 – sqrt 2`

Show Worked Solution

`text(Find)\ \ x text(-intercept:)`

♦ Mean mark 50%.

`(x – 1)^2 – 2`	`= 0`
`(x – 1)^2`	`= 2`
`x – 1`	`= +- sqrt 2`
`x`	`= 1 +- sqrt 2,\ \ a < 0`
`:. a`	`= 1 – sqrt 2`

Probability, MET1 2013 VCAA 8

A continuous random variable, `X`, has a probability density function

`f(x) = { (pi/4 cos ((pi x)/4),\ \ \ text(if)\ x in [0, 2]), (\ \ \ 0,\ \ \ text(otherwise)):}`

Given that `d/(dx) (x sin ((pi x)/4)) = (pi x)/4 cos ((pi x)/4) + sin ((pi x)/4)`, find `text(E)(X).` (3 marks)

Show Answers Only

`2 – 4/pi`

Show Worked Solution

`text(Find)\ \ int (pix)/4 cos ((pi x)/4)\ dx`

`text(Integrating the given equation:)`

`int (pi x)/4 cos ((pi x)/4) + sin ((pi x)/4)\ dx`	`= x sin ((pi x)/4)`
`int (pi x)/4 cos ((pi x)/4)\ dx – 4/pi cos ((pi x)/4)`	`= x sin ((pi x)/4)`
`:. int (pi x)/4 cos ((pi x)/4)\ dx`	`= x sin ((pi x)/4) + 4/pi cos ((pi x)/4)`

♦♦ Mean mark 32%.

`:. text(E)(X)`	`= int_0^2 x (pi/4 cos (pi/4 x)) dx`
	`= [x sin ((pi x)/4) + 4/pi cos ((pi x)/4)]_0^2`
	`= (2 sin (pi/2) + 4/pi cos (pi/2)) – (0 + 4/pi cos (0))`
	`= (2 + 0) – (0 +4/pi)`
	`= 2 – 4/pi`

Probability, MET1 2013 VCAA 7

The probability distribution of a discrete random variable, `X`, is given by the table below.

Show that `p = 2/3 or p = 1`. (3 marks)
Let `p = 2/3`.
1. Calculate `text(E) (X)`. (2 marks)
2. Find `text(Pr) (X >= text(E) (X))`. (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
1. `28/15`
2. `8/15`

Show Worked Solution

a. `text(S)text(ince probabilities must sum to 1:)`

`0.2 + 0.6p^2 + 0.1 + 1 – p + 0.1`	`= 1`
`0.6p^2 – p + 0.4`	`= 0`
`6p^2 – 10p + 4`	`= 0`
`3p^2 – 5p + 2`	`= 0`
`(p – 1) (3p – 2)`	`= 0`

`:. p = 1 or p = 2/3`

b.i.	`text(E) (X)`	`= sum x text(Pr) (X = x)`
		`= 1 xx (3/5 xx 2^2/3^2) + 2 (1/10) + 3 (1-2/3) + 4 (1/10)`
		`= 4/15 + 1/5 + 1 + 2/5`
		`= 28/15`

♦♦ Part (b)(ii) mean mark 32%.

ii.	`text(Pr) (X >= 28/15)`	`= text(Pr) (X = 2) + text(Pr) (X = 3) + text(Pr) (X = 4)`
		`= 1/10 + 1/3 + 1/10`
		`= 8/15`

Calculus, MET1 2013 VCAA 6

Let `g: R -> R,\ \ g(x) = (a - x)^2`, where `a` is a real constant.

The average value of `g` on the interval `[– 1, 1]` is `31/12.`

Find all possible values of `a.` (3 marks)

Show Answers Only

`+- 3/2`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 39%.
MARKER’S COMMENT: “Far too many” students misunderstood “average value” in this context.

`1/(1 – (– 1)) int_-1^1 (a – x)^2 dx`	`= 31/12`
`[ax^2 – ax^2 + x^3/3]_-1^1`	`= 31/6`
`[(a^2 – a+1/3) – (-a^2-a-1/3)]`	`=31/6`
`2a^2+2/3`	`=31/6`
`a^2`	`=27/4`
`:. a`	`=+- 3/2`

`text(Solution 2)`

`1/(1 – (– 1)) int_-1^1 (a – x)^2 dx`	`= 31/12`
`1/2 [(a – x)^3/-3]_-1^1`	`= 31/12`
`[(a – x)^3]_-1^1`	`= – 31/2`
`(a – 1)^3 – (a + 1)^3`	`= – 31/2`

`(a^3 – 3a^2 + 3a – 1) – (a^3 + 3a^2 + 3a + 1)=- 31/2`

`-6a^2 – 2`	`= – 31/2`
`6a^2`	`= 27/2`
`a^2`	`= 9/4`
`:. a`	`= +- 3/2`

Calculus, MET1 2014 VCAA 10

A line intersects the coordinate axes at the points `U` and `V` with coordinates `(u, 0)` and `(0, v)`, respectively, where `u` and `v` are positive real numbers and `5/2 <= u <= 6`.

When `u = 6`, the line is a tangent to the graph of `y = ax^2 + bx` at the point `Q` with coordinates `(2, 4)`, as shown.

If `a` and `b` are non-zero real numbers, find the values of `a` and `b`. (3 marks)
The rectangle `OPQR` has a vertex at `Q` on the line. The coordinates of `Q` are `(2, 4)`, as shown.
1. Find an expression for `v` in terms of `u`. (1 mark)
2. Find the minimum total shaded area and the value of `u` for which the area is a minimum. (2 marks)
3. Find the maximum total shaded area and the value of `u` for which the area is a maximum. (1 mark)

Show Answers Only

`a = −3/2, b = 5`
1. `v=(4u)/(u-2)`
2. `8\ text{u² (when}\ u=4)`
3. `17\ text{u²}\ \ (text{when}\ u=5/2)`

Show Worked Solution

a. `text(S)text{ince (2, 4) lies on parabola:}`

♦ Mean mark 48%.
MARKER’S COMMENT: The most common error incorrectly stated that (6,0) lay on the parabola.

`4`	`= a(2)^2 + b(2)`
`4`	`= 4a + 2b`
`2`	`= 2a + b\ …\ (1)`

`y`	`=ax^2+bx`
`dy/dx`	`=2ax+b`

`text(At)\ \ x=2,`

`m_(QU)`	`=m_(text(tang))`
`(4 – 0)/(2 – 6)`	`= 2a(2) + b`
`−1`	`= 4a + b\ …\ (2)`

`text(Subtract)\ \ (2) – (1)`

`−3 = 2a`

`:. a = −3/2, b = 5`

b.i.

`text(Solution 1)`

`text(Using similar triangles:)`

`Delta VOU\ text(|||)\ Delta QPU`

♦♦ Mean mark 32%.

`v/4`	`= u/(u -2)`
`:. v`	`= (4u)/(u -2)`

`text(Solution 2)`

`m_(VQ)`	`=m_(UQ)`
`(v-4)/(0-2)`	`=(0-4)/(u-2)`
`v`	`=8/(u-2) + 4`
	`=(8+4(u-2))/(u-2)`
	`=(4u)/(u-2)`

♦♦♦ Part (b)(ii) mean mark 18%.

b.ii.	`text(Area)`	`= 1/2uv – 2 xx 4`
		`= 1/2u((4u)/(u -2)) – 8`
		`= (2u^2)/(u -2) – 8`

`text(SP occurs when)\ \ (dA)/(du)=0,`

`(4u(u – 2) – 2u^2)/((u – 2)^2)`	`= 0`
`2u^2 – 8u`	`= 0`
`2u(u – 4)`	`= 0`

`u = 0quadtext(or)quadu = 4`

`:. u = 4, \ \ u ∈ [5/2,6]`

`text(When)\ u=4,\ \ A`	`=(2 xx 4^2)/(4-2) – 8`
	`= 8\ text(u²)`

`text(Test areas at end points,)`

`text(When)\ u=5/2,\ A=17`

`text(When)\ u=6,\ A=10`

`:. A_(text(min))=8\ text{u² (when}\ u=4)`

b.iii. `text{As only one (local minimum) stationary point}`

♦♦♦ Mean mark 8%.

`text(exists over)\ 5/2 <= u <= 6, text(the maximum area)`

`text(must occur at an endpoint.)`

`:. A_(text(max)) = 17\ text{u²}\ \ (text{when}\ u=5/2)`

Probability, MET1 2014 VCAA 9

Sally aims to walk her dog, Mack, most mornings. If the weather is pleasant, the probability that she will walk Mack is `3/4`, and if the weather is unpleasant, the probability that she will walk Mack is `1/3`.

Assume that pleasant weather on any morning is independent of pleasant weather on any other morning.

In a particular week, the weather was pleasant on Monday morning and unpleasant on Tuesday morning.
Find the probability that Sally walked Mack on at least one of these two mornings. (2 marks)
In the month of April, the probability of pleasant weather in the morning was `5/8`.
i. Find the probability that on a particular morning in April, Sally walked Mack. (2 marks)
ii. Using your answer from part b.i., or otherwise, find the probability that on a particular morning in April, the weather was pleasant, given that Sally walked Mack that morning. (2 marks)

Show Answers Only

`5/6`
1. `19/32`
2. `15/19`

Show Worked Solution

a.	`text{Pr(at least 1 walk)}`	`= 1 – text{Pr(no walk)}`
		`= 1 – 1/4 xx 2/3`
		`= 5/6`

b.i. `text(Construct tree diagram:)`

`text(Pr)(PW) + text(Pr)(P′W)`	`= 5/8 xx 3/4 + 3/8 xx 1/3`
	`= 19/32`

♦ Part (b)(ii) mean mark 38%.

b.ii.	`text(Pr)(P \| W)`	`= (text(Pr)(P ∩ W))/(text(Pr)(W))`
		`= (5/8 xx 3/4)/(19/32)`
		`= 15/32 xx 32/19`
		`= 15/19`

Probability, MET1 2014 VCAA 8

A continuous random variable, `X`, has a probability density function given by

`f(x) = {{:(1/5e^(−x/5),x >= 0),(0, x < 0):}`

The median of `X` is `m`.

Determine the value of `m`. (2 marks)
The value of `m` is a number greater than 1.

Find `text(Pr)(X < 1 | X <= m)`. (2 marks)

Show Answers Only

`−5log_e(1/2)\ \ text(or)\ \ 5log_e(2)\ \ text(or)\ \ log_e 32`
`2(1 – e^(−1/5))`

Show Worked Solution

a.	`1/5 int_0^m e^(−x/5)dx`	`= 1/2`
	`1/5 xx (−5)[e^(−x/5)]_0^m`	`= 1/2`
	`[-e^(- x/5)]_0^m`	`= 1/2`
	`-e^(−m/5) + 1`	`= 1/2`
	`e^(−m/5)`	`= 1/2`
	`- m/5`	`= log_e(1/2)`

`:. m = −5log_e(1/2)\ \ \ (text(or)\ \ 5log_e(2),\ text(or)\ \ log_e 32)`

b. `text(Using Conditional Probability:)`

♦ Part (b) mean mark 38%.
MARKER’S COMMENT: A common error was assuming `m` obtained in part (a) was equivalent to `text(Pr)(X<=m)`.

`text(Pr)(X < 1 \| X <= m)`	`= (text(Pr)(X < 1))/(text(Pr)(X <= m))`
	`= (1/5 int_0^1 e^(−x/5)dx)/(1/2)`
	`= (1/5(−5)[e^(−x/5)]_0^1)/(1/2)`
	`= −2[(e^(−1/5)) – e^0]`
	`= 2(1 – e^(−1/5))`

Calculus, MET1 2014 VCAA 5

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

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

Label any end points with their coordinates. (2 marks)
Find the area enclosed by the graph of the function and the horizontal line given by `y = 4`. (3 marks)

Show Answers Only

`(0, 0) and (2, 4)`
`27/4\ text(u²)`

Show Worked Solution

`text(SP’s occur when)\ \ f′(x)`

`= 0`

`6x – 3x^2`	`= 0`
`3x(2 – x)`	`=0`

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

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

♦ Part (c) mean mark 48%.

`text(Solution 1)`

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

`:.\ text(Area)`	`= 27/4 text(units²)`

`text(Solution 2)`

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

Calculus, MET1 2015 VCAA 10

The diagram below shows a point, `T`, on a circle. The circle has radius 2 and centre at the point `C` with coordinates `(2, 0)`. The angle `ECT` is `theta`, where `0 < theta <= pi/2`.

The diagram also shows the tangent to the circle at `T`. This tangent is perpendicular to `CT` and intersects the `x`-axis at point `X` and the `y`-axis at point `Y`.

Find the coordinates of `T` in terms of `theta`. (1 mark)
Find the gradient of the tangent to the circle at `T` in terms of `theta`. (1 mark)
The equation of the tangent to the circle at `T` can be expressed as
`qquad cos(theta)x + sin(theta)y = 2 + 2cos(theta)`
i. Point `B`, with coordinates `(2, b)`, is on the line segment `XY`.
Find `b` in terms of `theta`. (1 mark)
ii. Point `D`, with coordinates `(4, d)`, is on the line segment `XY`.
Find `d` in terms of `theta`. (1 mark)
Consider the trapezium `CEDB` with parallel sides of length `b` and `d`.
Find the value of `theta` for which the area of the trapezium `CEDB` is a minimum. Also find the minimum value of the area. (3 marks)

Show Answers Only

`T(2 + 2costheta, 2 sintheta)`
`(−1)/(tan(theta))`
i. `2/(sintheta)`
ii. `(2 – 2 costheta)/(sintheta)`
`theta = pi/3`
`A_text(min) = 2sqrt3\ text(u²)`

Show Worked Solution

a.	`cos theta`	`= (CM)/(CT)`
		`=(CM)/2`
	`CM`	`= 2costheta`

♦♦♦ Part (a) mean mark 20%.
MARKER’S COMMENT: Many students did not include the “+2” in the `x`-coordinate.

`sintheta`	`= (TM)/2`
`TM`	`= 2sintheta`

`:. T\ text(has coordinates)\ \ (2 + 2costheta, 2 sintheta)`

♦♦♦ Part (b) mean mark 16%.

b.	`m_(CT)`	`=(TM)/(CM)`
		`=(2 sin theta)/(2 cos theta)`
		`=tan theta`

	`:.m_(XY)`	`=- 1/tan theta,\ \ \ (CT ⊥ XY)`

c.i. `text(Substitute)\ \ (2,b)\ \ text(into equation:)`

`2costheta + bsintheta`	`= 2 + 2costheta`
`:. b`	`= 2/(sintheta)`

c.ii. `text(Substitute)\ \ (4,d)\ \ text(into equation:)`

♦ Part (c)(ii) mean mark 47%.

`4costheta + dsintheta`	`= 2 + 2costheta`
`d sin theta`	`=2 – 2cos theta`
`:.d`	`= (2 – 2 costheta)/(sintheta)`

♦♦♦ Part (d) mean mark 19%.

d.	`text(A)_text(trap)`	`= 1/2 xx 2 xx (b + d)`
		`= 2/(sintheta) + (2 – 2costheta)/(sintheta)`
		`= (4 – 2costheta)/(sintheta)`

`text(Stationary point when)\ \ (dA)/(d theta)=0`

`(2sin^2theta – costheta(4 – 2costheta))/(sin^2theta)`	`= 0`
`2sin^2theta – 4costheta + 2cos^2theta`	`= 0`
`2[sin^2theta + cos^2theta] – 4costheta`	`= 0`
`2 – 4costheta`	`= 0`
`costheta`	`= 1/2`
`theta`	`= pi/3,\ \ \ \ theta ∈ (0, pi/2)`

`A(pi/3)`	`= (4 – 2(1/2))/(sqrt3/2)`
	`=3 xx 2/sqrt3`
	`= 2sqrt3`

`:. A_text(min) = 2sqrt3\ text(u²)`

Calculus, MET2 2010 VCAA 4

Consider the function `f: R -> R,\ f(x) = 1/27 (2x - 1)^3 (6 - 3x) + 1.`

Find the `x`-coordinate of each of the stationary points of `f` and state the nature of each of these stationary points. (4 marks)

In the following, `f` is the function `f: R -> R,\ f(x) = 1/27 (ax - 1)^3 (b - 3x) + 1` where `a` and `b` are real constants.

Write down, in terms of `a` and `b`, the possible values of `x` for which `(x, f (x))` is a stationary point of `f`. (3 marks)
For what value of `a` does `f` have no stationary points? (1 mark)
Find `a` in terms of `b` if `f` has one stationary point. (2 marks)
What is the maximum number of stationary points that `f` can have? (1 mark)
Assume that there is a stationary point at `(1, 1)` and another stationary point `(p, p)` where `p != 1`.

Find the value of `p`. (3 marks)

Show Answers Only

`text(Point of inflection at)\ \ x = 1/2`
`text(Local max at)\ \ x = 13/8`
`x = (ab + 1)/(4a) or x = 1/a`
`0`
`3/b`
`2`
`4`

Show Worked Solution

a. `text(S.P. occurs when)\ \ f prime (x) = 0`

`f(x)`	`=1/27 (2x – 1)^3 (6 – 3x) + 1`
`f′(x)`	`=- 1/9 (2x – 1)^2 (8x – 13) `

`text(Solve:)\ \ f′(x)=0\ \ text(for)\ x,`

`:. x = 1/2 or x = 13/8`

`text(Sketch the graph:)`

`=>\ text(Point of inflection at)\ \ x = 1/2`

`=>\ text(Local max at)\ \ x = 13/8`

b. `text(S.P. occurs when)\ \ f prime (x) = 0`

♦ Mean mark part (b) 46%.
MARKER’S COMMENT: Issues with use of CAS caused significant difficulties in this question.

`f(x)`	`=1/27 (ax – 1)^3 (b – 3x) + 1`
`f′(x)`	`=1/9 (ax – 1)^2 (ab+1-4ax)`

`text(Solve:)\ \ f prime (x) = 0\ \ text(for)\ \ x,`

`:. x = (ab + 1)/(4a) or x = 1/a`

c. `text(For)\ \ x = (ab + 1)/(4a) or x = 1/a\ \ text(to exist,)`

♦ Mean mark part (c) 47%.

`a != 0`

`:.\ text(No stationary points when)\ \ a = 0`

d. `text(If there is 1 S.P.,)`

♦♦♦ Mean mark part (d) 16%.

`(ab + 1)/(4a)`	`= 1/a`
`:. a`	`= 3/b`

♦ Mean mark part (e) 37%.

e. `text(The maximum number of SP’s for a quartic)`

`text(polynomial is 3. In the function given, one of)`

`text(the SP’s is a point of inflection.)`

`:.f(x)\ \ text(has a maximum of 2 SP’s.)`

f. `text(Solution 1)`

`text(SP’s occur at)\ \ (1,1) and (p,p),\ \ text(where,)`

♦♦♦ Mean mark part (f) 9%.

`x = (ab + 1)/(4a) or x = 1/a`

`text(Consider)\ \ p=1/a,`

`f(p)`	`=f(1/a)`
	`=1/27 (a1/a-1)^3(b-31/a)+1=1`

`f(p)=1,\ \ text(SP at)\ (1,1) and p!=1`

`=> p!=1/a`

`text(Consider)\ \ 1=1/a,`

`=> a=1 and b=4p-1`

`f(1)=1`

`f(p)=p`

`1/27 (p-1)^3(4p-1-3p)+1`	`=p`
`1/27(p-1)^4-(p-1)`	`=0`
`(p-1)(1/27(p-1)^3-1)`	`=0`
`(p-1)^3`	`=27`
`p`	`=4`

`text{Solution 2 [by CAS]}`

`text(Define)\ \ f(x) = 1/27 (x – 1)^3 (b – 3x) + 1`

`text(Solve:)\ \ f(p) = p, f prime (1) = 1 and f prime (p) = p\ \ text(for)\ \ p,`

`:. p = 4,\ \ \ (p != 1)`

Calculus, MET2 2010 VCAA 3

An ancient civilisation buried its kings and queens in tombs in the shape of a square-based pyramid, `WABCD.`

The kings and queens were each buried in a pyramid with `WA = WB = WC = WD = 10\ text(m).`

Each of the isosceles triangle faces is congruent to each of the other triangular faces.

The base angle of each of these triangles is `x`, where `pi/4 < x < pi/2.`

Pyramid `WABCD` and a face of the pyramid, `WAB`, are shown here.

`Z` is the midpoint of `AB.`

i. Find `AB` in terms of `x`. (1 mark)
ii. Find `WZ` in terms of `x`. (1 mark)
Show that the total surface area (including the base), `S\ text(m)^2`, of the pyramid, `WABCD`, is given by
`S = 400(cos^2 (x) + cos (x) sin (x))`. (2 marks)
Find `WY`, the height of the pyramid `WABCD`, in terms of `x`. (2 marks)
The volume of any pyramid is given by the formula `text(Volume) = 1/3 xx text(area of base) xx text(vertical height)`.
Show that the volume, `T\ text(m³)`, of the pyramid `WABCD` is `4000/3 sqrt(cos^4 x - 2 cos^6 x)`. (1 mark)

Queen Hepzabah’s pyramid was designed so that it had the maximum possible volume.

Find `(dT)/(dx)` and hence find the exact volume of Queen Hepzabah’s pyramid and the corresponding value of `x`. (4 marks)

Queen Hepzabah’s daughter, Queen Jepzibah, was also buried in a pyramid. It also had

`WA = WB = WC = WD = 10\ text(m.)`

The volume of Jepzibah’s pyramid is exactly one half of the volume of Queen Hepzabah’s pyramid. The volume of Queen Jepzibah’s pyramid is also given by the formula for `T` obtained in part d.

Find the possible values of `x`, for Jepzibah’s pyramid, correct to two decimal places. (2 marks)

Show Answers Only

1. `20 cos (x)`
2. `10 sin (x)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`10 sqrt (sin^2(x) – cos^2 (x))`
`text(Proof)\ \ text{(See Worked Solutions)}`
`x = cos^-1 (sqrt 3/3) -> T_max = (4000 sqrt 3\ m^3)/27`
`x dot = 0.81 or x dot = 1.23`

Show Worked Solution

a.i.	`cos x`	`= (1/2 AB)/10`
	`:. AB`	`= 20 cos(x)`

ii.	`sin (x)`	`= (wz)/10`
	`:. wz`	`= 10 sin (x)`

b.	`text(Area base)`	`= (20 cos (x))^2`
		`= 400 cos^2(x)`
	`4 xx text(Area)_Delta`	`= 4 xx (1/2 xx 20 cos (x) xx 10 sin (x))`
		`= 400 cos (x) sin (x)`

♦ Mean mark 47%.

`:. S`	`= 400 cos^2 (x) + 400 cos (x) sin (x)`
	`= 400 (cos^2 (x) + cos (x) sin (x))\ \ text(… as required.)`

c. `text(Using)\ \ Delta WYZ,`

`text(Using Pythagoras,)`

`WY`	`= sqrt (10^2 sin^2 (x) – 10^2 cos^2 (x))`
	`= 10 sqrt (sin^2(x) – cos^2 (x))`

♦♦♦ Mean mark part (d) 22%.

d.	`T`	`= 1/3 xx text(base) xx text(height)`
		`= 1/3 xx (400 cos^2 (x)) xx (10 sqrt(sin^2 (x) – cos^2 (x)))`
		`= 4000/3 sqrt (cos^4 (x) (sin^2 (x) – cos^2 (x))`

`text(Using)\ \ sin^2 (x) = 1 – cos^2 (x),`

`T`	`= 4000/3 sqrt (cos^4 (x) (1 – cos^2 (x) – cos^2 (x))`
	`= 4000/3 sqrt (cos^4 (x) – 2 cos^6 (x))`

e. `(dT)/(dx) = (8000 cos (x) sin (x) (3 cos^2 (x) – 1))/(3 sqrt(1 – 2 cos^2 (x)))`

♦ Mean mark part (e) 45%.

`text(Stationary point when,)`

`(dT)/(dx) = 0\ \ text(for)\ \ x in (pi/4, pi/2)`

`:. x = cos^-1 (sqrt 3/3)`

`:.T_max`	`=T(cos^-1 (sqrt 3/3))`
	`= (4000 sqrt 3)/27\ \ text(m³)`

♦♦♦ Mean mark part (f) 16%.

f. `text(Solve)\ \ T(x) = (2000 sqrt 3)/27\ \ text(for)\ \ x in (pi/4, pi/2)`

`:. x = 0.81 or x = 1.23\ \ text{(2 d.p.)}`

Probability, MET1 2015 VCAA 9

An egg marketing company buys its eggs from farm A and farm B. Let `p` be the proportion of eggs that the company buys from farm A. The rest of the company’s eggs come from farm B. Each day, the eggs from both farms are taken to the company’s warehouse.

Assume that `3/5` of all eggs from farm A have white eggshells and `1/5` of all eggs from farm B have white eggshells.

An egg is selected at random from the set of all eggs at the warehouse.

Find, in terms of `p`, the probability that the egg has a white eggshell. (1 mark)
Another egg is selected at random from the set of all eggs at the warehouse.
1. Given that the egg has a white eggshell, find, in terms of `p`, the probability that it came from farm B. (2 marks)
2. If the probability that this egg came from farm B is 0.3, find the value of `p`. (1 mark)

Show Answers Only

`(2p+1)/5`
1. `(1 – p)/(2p + 1)`
2. `7/16`

Show Worked Solution

`text(Pr)(AW) + text(Pr)(BW)`	`= p xx 3/5 + (1-p) xx 1/5`
	`=(3p)/5+1/5-p/5`
	`= (2p+1)/5`

♦ Part (b) mean mark 41%.
MARKER’S COMMENT: Algebraic fractions “were not handled well”!

b.i.	`text(Pr)(B \| W)`	`= (text(Pr)(B ∩ W))/(text(Pr)(W))`
		`= ((1 – p)/5)/((2p + 1)/5)`
		`=(1-p)/5 xx 5/(2p+1)`
		`= (1 – p)/(2p + 1)`

♦♦♦ Part (c) mean mark 19%.
STRATEGY: Previous parts of a question are gold dust for directing your strategy in many harder questions.

b.ii.	`text(Pr)(B \| W)`	`= 3/10`
	`(1 – p)/(2p + 1)`	`= 3/10`
	`10 – 10p`	`= 6p + 3`
	`7`	`= 16p`
	`:. p`	`= 7/16`

Probability, MET1 2015 VCAA 8

For events `A` and `B` from a sample space, `text(Pr)(A | B) = 3/4` and `text(Pr)(B) = 1/3`.

Calculate `text(Pr)(A ∩ B)`. (1 mark)
Calculate `text(Pr)(A′ ∩ B)`, where `A′` denotes the complement of `A`. (1 mark)
If events `A` and `B` are independent, calculate `text(Pr)(A ∪ B)`. (1 mark)

Show Answers Only

`1/4`
`1/12`
`5/6`

Show Worked Solution

a. `text(Using Conditional Probability:)`

`text(Pr)(A \| B)`	`= (text(Pr)(A ∩ B))/(text(Pr)(B))`
`3/4`	`= (text(Pr)(A ∩ B))/(1/3)`
`:. text(Pr)(A ∩ B)`	`= 1/4`

`text(Pr)(A′ ∩ B)`	`= text(Pr)(B) – text(Pr)(A ∩B)`
	`= 1/3 – 1/4`
	`= 1/12`

c. `text(If)\ A, B\ text(independent)`

♦♦ Mean mark 28%.
MARKER’S COMMENT: A lack of understanding of independent events was clearly evident.

`text(Pr)(A ∩ B)`	`= text(Pr)(A) xx Pr(B)`
`1/4`	`= text(Pr)(A) xx 1/3`
`:. text(Pr)(A)`	`= 3/4`

`text(Pr)(A ∪ B)`	`= text(Pr)(A) + text(Pr)(B) – text(Pr)(A ∩ B)`
	`= 3/4 + 1/3 – 1/4`
`:. text(Pr)(A ∪ B)`	`= 5/6`

Probability, MET1 2015 VCAA 6

Let the random variable `X` be normally distributed with mean 2.5 and standard deviation 0.3

Let `Z` be the standard normal random variable, such that `Z ∼\ text(N)(0, 1)`.

Find `b` such that `text(Pr)(X > 3.1) = text(Pr)(Z < b)`. (1 mark)
Using the fact that, correct to two decimal places, `text(Pr)(Z < –1) = 0.16`, find `text(Pr)(X < 2.8 | X > 2.5)`.

Write the answer correct to two decimal places. (2 marks)

Show Answers Only

`−2`
`0.68`

Show Worked Solution

♦ Part (a) mean mark 50%.

a.	`text(Pr)(X > 3.1)`	`= text(Pr)(Z > (3.1 – 2.5)/0.3)`
		`= text(Pr)(Z > 2)`
		`= text(Pr)(Z < − 2)`

`:. b = −2`

♦ Part (b) mean mark 48%.
MARKER’S COMMENT: Students who drew a diagram of a “normal” curve with relevent shaded areas were more successful.

`text(Pr)(X < 2.8 | X > 2.5)`

`= (text(Pr)(2.5 < X < 2.8))/(text(Pr)(X > 2.5))`

`= 0.34/0.5`

`= 34/50`

`= 68/100`

`= 0.68`

Calculus, MET1 2015 VCAA 4

Consider the function `f:[−3,2] -> R, \ \ f(x) = 1/2(x^3 + 3x^2 - 4)`.

Find the coordinates of the stationary points of the function. (2 marks)

The rule for `f` can also be expressed as `f(x) = 1/2(x - 1)(x + 2)^2`.

On the axes below, sketch the graph of `f`, clearly indicating axis intercepts and turning points.

Label the end points with their coordinates. (2 marks)
Find the average value of `f` over the interval `0 <= x <=2`. (2 marks)

Show Answers Only

`(0, − 2), ( − 2,0)`
`1`

Show Worked Solution

a. `text(Stationary points when)\ \ f´(x)=0,`

`1/2(3x^2 + 6x)`	`= 0`
`3x(x + 2)`	`= 0`

`:. x = 0, − 2`

`:.\ text(Coordinates of stationary points:)`

`(0, − 2), (− 2,0)`

♦ Part (c) mean mark 50%.
MARKER’S COMMENT: Most students recalled the average value definition but then did not integrate correctly.

c.	`text(Avg value)`	`= 1/(2 – 0) int_0^2 f(x) dx`
		`= 1/2 int_0^2 1/2(x^3 + 3x^2 – 4)dx`
		`= 1/4[1/4x^4 + x^3 – 4x]_0^2`
		`= 1/4[(16/4 + 2^3 – 4(2))-0]`
		`= 1`

CORE*, FUR2 2006 VCAA 4

A company anticipates that it will need to borrow $20 000 to pay for a new machine.

It expects to take out a reducing balance loan with interest calculated monthly at a rate of 10% per annum.

The loan will be fully repaid with 24 equal monthly instalments.

Determine the total amount of interest that will be paid on this loan.

Write your answer to the nearest dollar. (2 marks)

Show Answers Only

`$2150\ \ text{(nearest $)}`

Show Worked Solution

`text(Find monthly payment by TVM Solver,)`

♦♦ Mean mark 21%.

`N`	`= 2 xx 12 = 24`
`I(text(%))`	`= 10`
`PV`	`= 20\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=>PMT = −922.898…`

`text(Total repayments)`

`= 24 xx 922.898…`

`= $22\ 149.65…`

`:.\ text(Total interest paid)`

`= 22\ 149.56… – 20\ 000`

`= 2149.56…`

`= $2150\ \ text{(nearest $)}`

CORE*, FUR2 2006 VCAA 3

The company prepares for this expenditure by establishing three different investments.

$7000 is invested at a simple interest rate of 6.25% per annum for eight years.

Determine the total value of this investment at the end of eight years. (2 marks)
$10 000 is invested at an interest rate of 6% per annum compounding quarterly for eight years.

Determine the total value of this investment at the end of eight years.

Write your answer correct to the nearest dollar. (1 mark)
$500 is deposited into an account with an interest rate of 6.5% per annum compounding monthly.

Deposits of $200 are made to this account on the last day of each month after interest has been paid.

Determine the total value of this investment at the end of eight years.

Write your answer correct to the nearest dollar. (1 mark)

Show Answers Only

`$10\ 500`
`$16\ 103`
`$25\ 935`

Show Worked Solution

a.	`I`	`= (PrT)/100`
		`= (7000 xx 6.25 xx 8)/100`
		`= $3500`

`:.\ text(Total value of investment)`

`= 7000 + 3500`

`= $10\ 500`

b. `text(Compounding periods) = 8 xx 4 = 32`

`text(Interest rate)`	`= (text(6%))/4`
	`= 1.5text(% per quarter)`

`:.\ text(Total value of investment)`

`= PR^n`

`= 10\ 000(1.015)^32`

`= 16\ 103.24…`

`= $16\ 103\ \ text{(nearest $)}`

c. `text(By TVM Solver,)`

`N`	`= 8 xx 12 = 96`
`I(text(%))`	`= 6.5`
`PV`	`= 500`
`PMT`	`= 200`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = −25\ 935.30…`

`:.\ text(Total value of investment is $25 935.)`

CORE*, FUR2 2006 VCAA 1

A company purchased a machine for $60 000.

For taxation purposes the machine is depreciated over time.

Two methods of depreciation are considered.

Flat rate depreciation

The machine is depreciated at a flat rate of 10% of the purchase price each year.
1. By how many dollars will the machine depreciate annually? (1 mark)
2. Calculate the value of the machine after three years. (1 mark)
3. After how many years will the machine be $12 000 in value? (1 mark)
Reducing balance depreciation

The value, `V`, of the machine after `n` years is given by the formula `V = 60\ 000 xx (0.85)^n`
1. By what percentage will the machine depreciate annually? (1 mark)
2. Calculate the value of the machine after three years. (1 mark)
3. At the end of which year will the machine's value first fall below $12 000? (1 mark)
At the end of which year will the value of the machine first be less using flat rate depreciation than it will be using reducing balance depreciation? (2 marks)

Show Answers Only

1. `$6000`
2. `$42\ 000`
3. `8\ text(years)`
1. `text(15%)`
2. `$36\ 847.50`
3. `10\ text(years)`
`text(7th year)`

Show Worked Solution

a.i.	`text(Annual depreciation)`	`= 10text(%) xx 60\ 000`
		`= $6000`

a.ii. `text(After 3 years,)`

`text(Value)`	`= 60\ 000 – (3 xx 6000)`
	`= $42\ 000`

a.iii. `text(Find)\ n\ text(when value = $12 000)`

`12\ 000`	`= 60\ 000 – 6000 xx n`
`6000n`	`= 48\ 000`
`:.n`	`=(48\ 000)/6000`
	`= 8\ text(years)`

b.i.	`1 – r`	`= 0.85`
	`r`	`= 0.15`

`:.\ text(Annual depreciation is 15%.)`

b.ii. `text(After 3 years,)`

`text(Value)`	`= 60\ 000 xx (0.85)^3`
	`= $36\ 847.50`

b.iii. `text(Find)\ n\ text(when)\ \ V = $12\ 000`

`12\ 000`	`= 60\ 000 xx (0.85)^n`
`(0.85)^n`	`= 0.2`
`:. n`	`= 9.90…\ \ text(years)`

`:.\ text(Machine value falls below $12 000)`

`text(after 10 years.)`

c. `text(Sketching both graphs,)`

`text(From the graph, at the end of the 7th year the)`

`text(value using flat rate drops below reducing)`

`text(balance for the 1st time.)`

CORE, FUR2 2007 VCAA 4

The books in Khan's office are valued at $10 000.

Calculate the value of these books after five years if they are depreciated by 12% per annum using the reducing balance method. Write your answer correct to the nearest dollar. (1 mark)

Khan believes his books should be valued at $4000 after five years.

Determine the annual reducing balance depreciation rate that will produce this value. Write your answer as a percentage correct to one decimal place. (2 marks)

Show Answers Only

`$5277\ \ text{(nearest $)}`
`text(16.7%)`

Show Worked Solution

a. `text(Value after 5 years)`

`= 10\ 000(1 – r)^n`

`= 10\ 000(1 – 0.12)^5`

`= 10\ 000(0.88)^5`

`= 5277.319…`

`= $5277\ \ text{(nearest $)}`

b. `text(Value after 5 years = $4000)`

`:. 4000`	`= 10\ 000 xx (1 – r)^5`
`(1 – r)^5`	`= 0.4`
`1 – r`	`= root(5)(0.4)`
	`= 0.832…`
`:. r`	`= 0.1674…`
	`= 16.7text{% (1 d.p.)}`

`text(Alternative solution)`

`text(By TVM Solver:)`

`N`	`= 5`
`I(text(%))`	`= ?`
`PV`	`= 10\ 000`
`PMT`	`= 0`
`FV`	`= −4000`
`text(P/Y)`	`= 1`
`text(C/Y)`	`= 1`

`Itext(%) = −16.74`

`:.\ text(Depreciation rate is 16.7%.)`

CORE*, FUR2 2007 VCAA 3

Khan paid $900 for a fax machine.

This price includes 10% GST (goods and services tax).

Determine the price of the fax machine before GST was added. Write your answer correct to the nearest cent. (1 mark)
Khan will depreciate his $900 fax machine for taxation purposes.

He considers two methods of depreciation.

Flat rate depreciation

Under flat rate depreciation the fax machine will be valued at $300 after five years.
i. Calculate the annual depreciation in dollars. (1 mark)

Unit cost depreciation

Suppose Khan sends 250 faxes a year. The $900 fax machine is depreciated by 46 cents for each fax it sends.
ii. Determine the value of the fax machine after five years. (1 mark)

Show Answers Only

`$818.18`
1. `$120`
2. `$325`

Show Worked Solution

a. `text(Let $)P = text(price ex-GST)`

MARKER’S COMMENT: Reverse GST questions continue to cause problems for many students.

`:. P + 10text(%)P`	`= 900`
`1.1P`	`= 900`
`P`	`= 900/1.1`
	`= 818.181…`
	`= $818.18\ \ text(nearest cent)`

b.i. `text(Annual depreciation)`

`= ((900 – 300))/5`

`= $120`

b.ii. `text(Value after 5 years)`

`= 900 – (250 xx 0.46 xx 5)`

`= $325`

CORE*, FUR2 2007 VCAA 2

Khan decides to extend his home office and borrows $30 000 for building costs. Interest is charged on the loan at a rate of 9% per annum compounding monthly.

Assume Khan will pay only the interest on the loan at the end of each month.

Calculate the amount of interest he will pay each month. (1 mark)

Suppose the interest rate remains at 9% per annum compounding monthly and Khan pays $400 each month for five years.

Determine the amount of the loan that is outstanding at the end of five years.

Write your answer correct to the nearest dollar. (1 mark)

Khan decides to repay the $30 000 loan fully in equal monthly instalments over five years.

The interest rate is 9% per annum compounding monthly.

Determine the amount of each monthly instalment. Write your answer correct to the nearest cent. (1 mark)

Show Answers Only

`$225`
`$16\ 801`
`$622.75`

Show Worked Solution

a. `text(Interest paid each month)`

♦ Mean mark of all parts (combined) was 37%.
MARKER’S COMMENT: Many students were confused by the fact that part (a) did not have a given loan period.

`= 1/12 xx 0.09 xx 30\ 000`

`= $225`

b. `text(Find principal left after 5 years.)`

`text(By TVM Solver:)`

`N`	`= 5 xx 12 = 60`
`I(text(%))`	`= 9`
`PV`	`= 30\ 000`
`PMT`	`= −400`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = −16\ 800.77…`

`:. $16\ 801\ text(is outstanding after 5 years.)`

c. `text(By TVM Solver,)`

`N`	`= 5 xx 12 = 60`
`I(text(%))`	`= 9`
`PV`	`= 30\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=>PMT = −622.750…`

`:.\ text(Monthly installment is $622.75)`

GRAPHS, FUR2 2007 VCAA 3

Gas is generally cheaper than petrol.

A car must run on petrol for some of the driving time.

Let `x` be the number of hours driving using gas

`y` be the number of hours driving using petrol

Inequalities 1 to 5 below represent the constraints on driving a car over a 24-hour period.

Explanations are given for Inequalities 3 and 4.

Inequality 1: `x ≥ 0`

Inequality 2: `y ≥ 0`

Inequality 3: `y ≤ 1/2x`	The number of hours driving using petrol must not exceed half the number of hours driving using gas.
Inequality 4: `y ≥ 1/3x`	The number of hours driving using petrol must be at least one third the number of hours driving using gas.

Inequality 5: `x + y ≤ 24`

Explain the meaning of Inequality 5 in terms of the context of this problem. (1 mark)

The lines `x + y = 24` and `y = 1/2x` are drawn on the graph below.

On the graph above
1. draw the line `y = 1/3x` (1 mark)
2. clearly shade the feasible region represented by Inequalities 1 to 5. (1 mark)

On a particular day, the Goldsmiths plan to drive for 15 hours. They will use gas for 10 of these hours.

Will the Goldsmiths comply with all constraints? Justify your answer. (1 mark)

On another day, the Goldsmiths plan to drive for 24 hours.

Their car carries enough fuel to drive for 20 hours using gas and 7 hours using petrol.

Determine the maximum and minimum number of hours they can drive using gas while satisfying all constraints. (2 marks)

Maximum = ___________ hours

Minimum = ___________ hours

Show Answers Only

`text(Inequality 5 means that the total hours driving)`
`text(with gas PLUS the total hours driving with petrol)`
`text(must be less than or equal to 24 hours.)`
i. & ii.
`text(If they drive for 10 hours on gas,)`
`text(5 hours is driven on petrol,)`
`text{(10, 5) is in the feasible region.}`
`:.\ text(They comply with all constraints.)`
`text(Maximum = 18 hours)`
`text(Minimum = 17 hours)`

Show Worked Solution

a. `text(Inequality 5 means that the total hours driving)`

`text(with gas PLUS the total hours driving with petrol)`

`text(must be less than or equal to 24 hours.)`

♦♦ Mean mark of parts (b)-(d) (combined) was 33%.

b.i. & ii.

c. `text(If they drive for 10 hours on gas, 5 hours)`

MARKER’S COMMENT: A mark was only awarded if a reference was made to the 5 hours driving.

`text(is driven on petrol.)`

`=>\ text{(10, 5) is in the feasible region.}`

`:.\ text(They comply with all constraints.)`

d. `text(Maximum = 18 hours)`

♦♦♦ “Very few” obtained both answers here.
MARKER’S COMMENT: Many ignored that the Goldsmiths planned to drive for 24 hours.

`text{(6 hours of petrol available)}`

`text(Minimum = 17 hours)`

`text{(7 hours of petrol is the highest available)}`

GRAPHS, FUR2 2007 VCAA 2

The Goldsmiths car can use either petrol or gas.

The following equation models the fuel usage of petrol, `P`, in litres per 100 km (L/100 km) when the car is travelling at an average speed of `s` km/h.

`P = 12 - 0.02s`

The line `P = 12 - 0.02s` is drawn on the graph below for average speeds up to 110 km/h.

Determine how many litres of petrol the car will use to travel 100 km at an average speed of 60 km/h.

Write your answer correct to one decimal place. (1 mark)

The following equation models the fuel usage of gas, `G`, in litres per 100 km (L/100 km) when the car is travelling at an average speed of `s` km/h.

`G = 15 - 0.06s`

On the axes above, draw the line `G = 15 - 0.06s` for average speeds up to 110 km/h. (1 mark)
Determine the average speeds for which fuel usage of gas will be less than fuel usage of petrol. (1 mark)

The Goldsmiths' car travels at an average speed of 85 km/h. It is using gas.

Gas costs 80 cents per litre.

Determine the cost of the gas used to travel 100 km.

Write your answer in dollars and cents. (2 marks)

Show Answers Only

`10.8\ text(litres)`
`text(75 km/hr) < text(speed) <= 110\ text(km/hr)`
`$7.92`

Show Worked Solution

a. `text(When)\ S = 60\ text(km/hr)`

`P`	`= 12 – 0.02 xx 60`
	`= 10.8\ text(litres)`

b. `text(When)\ s = 110,`

`G = 15 – 0.06 xx 100 = 8.4`

`:.\ text{(0, 15) and (110, 8.4) are on the line}`

c. `text(Intersection of graphs occur when)`

MARKER’S COMMENT: Since fuel usage is less for gas, a speed of 75 km/hr was incorrect, as it was equal.

`12 – 0.02s`	`= 15 – 0.06s`
`0.04s`	`= 3`
`s`	`= 75`

`:.\ text(Gas usage is less than fuel for)`

`text(average speeds over 75 km/hr.)`

d. `text(When)\ x = 85,`

`text(Gas usage)`	`= 15 – 0.06 xx 85`
	`= 9.9\ text(L/100 km.)`

`:.\ text(C)text(ost of gas for 100km journey)`

`= 9.9 xx 0.80`

`= $7.92`