Band 4 – Page 64

Calculus, MET1 2008 VCAA 5

The area of the region bounded by the `y`-axis, the `x`-axis, the curve `y = e^(2x)` and the line `x = C`, where `C` is a positive real constant, is `5/2`. Find `C`. (3 marks)

Show Answers Only

`1/2 ln6`

Show Worked Solution

`text(Area)`	`= 5/2`
`int_0^C e^(2x) dx`	`= 5/2`
`[1/2 e^(2x)]_0^C`	`= 5/2`
`e^(2C) – e^0`	`= 5`
`e^(2C)`	`= 6`
`2C`	`= ln6`
`:. C`	`= 1/2 ln6`

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`

Functions, MET1 2008 VCAA 3

Solve the equation `cos((3x)/2) = 1/2` for `x ∈ [−pi/2,pi/2]`. (2 marks)

Show Answers Only

`x = ± (2pi)/9`

Show Worked Solution

`cos((3x)/2) = 1/2`

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

`(3x)/2`	`= (-pi)/3, pi/3, (5pi)/3, …`
`:. x`	`=(-2pi)/9, (2pi)/9, (10pi)/9`
	`= (-2pi)/9, (2pi)/9\ \ \ text(for)\ x ∈ [−pi/2,pi/2]`

Probability, MET1 2009 VCAA 7

The random variable `X` has this probability distribution.

Find

`text(Pr) (X > 1 | X <= 3)` (2 marks)
`text(Var)\ (X),` the variance of `X.` (3 marks)

Show Answers Only

`2/3`
`1.2`

Show Worked Solution

a. `text(Pr) (X > 1 | X <= 3)`

`= (text{Pr} (X = 2) + text{Pr} (X = 3))/(1 – text{Pr} (X = 4))`

`= (0.4 + 0.2)/(1 – 0.1)`

`= 0.6/0.9`

`= 2/3`

b. `text(E) (X)`	`= 0.1 (0) + 1 (0.2) + 2 (0.4) + 3 (0.2) + 4 (0.1)`
	`= 0 + 0.2 + 0.8 + 0.6 + 0.4`
	`= 2`

`text(E) (X^2)`	`= 0^2 (0.1) + 1^2 (0.2) + 2^2 (0.4) + 3^2 (0.2) + 4^2 (0.1)`
	`= 0 + 0.2 + 1.6 + 1.8 + 1.6`
	`= 5.2`

`:.\ text(Var)\ (X)`	`= text(E) (X^2) – [text(E) (X)]^2`
	`= 5.2 – (2)^2`
	`= 1.2`

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`

Algebra, MET1 2009 VCAA 4

Solve the equation `tan (2x) = sqrt 3` for `x in (– pi/4, pi/4) uu (pi/4, (3 pi)/4).` (3 marks)

Show Answers Only

`:. x = pi/6, (2 pi)/3`

Show Worked Solution

`tan (2x) = sqrt 3`

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

`2x`	`= pi/3, (4pi)/3, (-2pi)/3, …`
`:.x`	`= pi/6 + (2pi)/3, (-pi)/3, …`
	`=pi/6 or (2 pi)/3,\ \ \ \ x in (– pi/4, pi/4) uu (pi/4, (3 pi)/4)`

Functions, MET1 2009 VCAA 3

Let `f: R\ text(\)\ {0} -> R` where `f(x) = 3/x - 4.` Find `f^-1`, the inverse function of `f.` (3 marks)

Show Answers Only

`:. f^-1 (x) = 3/(x + 4),\ \ x in R\ text(\)\ {– 4}`

Show Worked Solution

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

MARKER’S COMMENT: “Few students” realised that both the equation and the domain had to be specified.

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

`x`	`= 3/y – 4`
`3/y`	`= x+4`
`y`	`= 3/(x + 4)`

`:. f^-1 (x) = 3/(x + 4),\ \ x in R\ text(\)\ {– 4}`

Calculus, MET1 2009 VCAA 2a

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

Show Answers Only

`-1/2 log_e | 1 – 2x |`

Show Worked Solution

`int (1 – 2x)^-1\ dx = -1/2 log_e | 1 – 2x |`

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`

Functions, MET1 2010 VCAA 3

Let `f: R^+ -> R` where `f(x) = 1/x^2.`

Find `g(x) = f(f(x))`. (1 mark)
Evaluate `g^-1 (16)`, where `g^-1` is the inverse function of `g`. (1 mark)

Show Answers Only

`x^4,\ \ x in R^+`
`g^-1(16) = 2`

Show Worked Solution

a.	`g(x)`	`= f(1/x^2)`
		`= 1/((1/x^2)^2)`
		`=1/(1/x^4)`

	`:. g(x)`	`= x^4,\ \ x in R^+`

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

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

`x`	`= y^4`
`y`	`= +- x^(1/4),`
	`=x^(1/4)\ \ \ text{(Domain of}\ g(x)\ text(is)\ R^+ text{)}`

`:. g^-1 (16)`	`= 16^(1/4)`
	`= 2`

Calculus, MET1 2010 VCAA 1a

Differentiate `x^3 e^(2x)` with respect to `x`. (2 marks)

Show Answers Only

`3x^2 e^(2x) + 2x^3 e^(2x)`

Show Worked Solution

`text(Using Product rule:)`

`(fg)′`	`= f′g + fg′`
`d/(dx) (x^3 e^(2x))`	`= 3x^2 e^(2x) + 2x^3 e^(2x)`

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)`

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`

Functions, MET1 2011 VCAA 3a

State the range and period of the function

`h: R -> R,\ \ h(x) = 4 + 3 cos ((pi x)/2).` (2 marks)

Show Answers Only

`text(Range) = [1, 7];\ \ \ text(Period) = 4`

Show Worked Solution

	`-1`	`<cos ((pi x)/2)<1`
	`-3`	`<3cos ((pi x)/2)<3`
	`1`	`< 4+ 3cos ((pi x)/2)<7`

`:.\ text(Range:)\ [1, 7]`

`text(Period) = (2pi)/n = (2 pi)/(pi/2) = 4`

Calculus, MET1 2011 VCAA 1a

Differentiate `sqrt (4 - x)` with respect to `x.` (1 mark)

Show Answers Only

`(dy)/(dx) = (– 1)/(2 sqrt (4 – x))`

Show Worked Solution

`text(Using the Chain Rule:)`

`text(Let)\ \ y`	`= (4 – x)^(1/2)`
`:. (dy)/(dx)`	`= 1/2 (4 – x)^(-1/2) (- 1)`
	`= (– 1)/(2 sqrt (4 – x))`

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`

Functions, MET1 2012 VCAA 7

Solve the equation `2 log_e (x + 2) - log_e(x) = log_e (2x + 1)`, where `x > 0`, for `x.` (3 marks)

Show Answers Only

`x = 4`

Show Worked Solution

`text(Simplify using log laws:)`

`log_e (x + 2)^2 – log_e (x)`	`= log_e (2x + 1)`
`log_e ((x + 2)^2/x)`	`= log_e (2x + 1)`
`(x + 2)^2/x`	`= 2x + 1`
`x^2 + 4x + 4`	`= 2x^2 + x`
`x^2-3x-4`	`= 0`
`(x-4)(x+1)`	`= 0`
`x`	`= 4 or -1,\ \ text(but)\ x > 0`
`:. x`	`= 4`

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 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)`

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`

Algebra, MET1 2013 VCAA 5a

Solve the equation `2 log_3(5) - log_3 (2) + log_3 (x) = 2` for `x`. (2 marks)

Show Answers Only

`18/25`
`- 6`

Show Worked Solution

	`log_3 (5)^2 – log_3 (2) + log_3 (x)`	`= 2`
	`log_3 (25x) – log_3 (2)`	`=2`
	`log_3 ((25 x)/2)`	`= 2`
	`(25x)/2`	`= 3^2`
	`:. x`	`= 18/25`

Functions, MET1 2013 VCAA 4

Solve the equation `sin (x/2) = -1/2` for `x in [2 pi, 4 pi].` (2 marks)

Show Answers Only

`x = (7 pi)/3, (11 pi)/3`

Show Worked Solution

`x/2`	`=pi/6 + pi, 2pi – pi/6, 2pi + (pi/6 +pi), …`
	`=(7pi)/6, (11pi)/6, (19pi)/6, …`
`:. x`	`=(7pi)/3, (11pi)/3, (19pi)/3, …`

`text(Given)\ \ x in [2 pi, 4 pi],`

`:. x = (7 pi)/3, (11 pi)/3`

Calculus, MET1 2013 VCAA 3

The function with rule `g(x)` has derivative `g prime (x) = sin (2 pi x).`

Given that `g(1) = 1/pi`, find `g(x).` (2 marks)

Show Answers Only

`(3 – cos (2 pi x))/(2 pi)`

Show Worked Solution

`g prime (x)`	`= int sin (2 pi x) dx`
	`= (– cos (2 pi x))/(2 pi) + c`

MARKER’S COMMENT: Mistakes in combining the two fractions for `c` were common.

`text(Substitute)\ \ (1, 1/pi)\ \ text(into)\ g prime(x):`

`1/pi`	`= (– cos (2 pi))/(2 pi) + c`
`1/pi`	`= – 1/(2 pi) + c`
`:. c`	`= 3/(2 pi)`

`:. g(x) = (3 – cos (2 pi x))/(2 pi)`

Calculus, MET1 2013 VCAA 2

Find an anti-derivative of `(4 - 2x)^-5` with respect to `x.` (2 marks)

Show Answers Only

`1/(8(4 – 2x)^4)`

Show Worked Solution

`int(4 – 2x)^-5 dx`	`= – 1/4 xx – 1/2 (4 – 2x)^-4`
	`= 1/(8(4 – 2x)^4)`

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 7

If `f′(x) = 2cos(x) - sin(2x)` and `f(pi/2) = 1/2`, find `f(x)`. (3 marks)

Show Answers Only

`2sinx + 1/2cos(2x) – 1`

Show Worked Solution

`f(x)`	`= int(2cosx – sin2x)dx`
	`= 2sinx + 1/2cos(2x) + c`

`text(Substitute)\ \ f(pi/2) = 1/2:`

`1/2`	`= 2sin(pi/2) + 1/2cos(pi) + c`
`1/2`	`= 2 – 1/2 + c`
`c`	`= −1`
`:. f(x)`	`= 2sinx + 1/2cos(2x) – 1`

Algebra, MET1 2014 VCAA 6

Solve `log_e(x) - 3 = log_e(sqrtx)` for `x`, where `x > 0`. (2 marks)

Show Answers Only

`e^6`

Show Worked Solution

`text(Simplify using log laws:)`

`log_e(x) – log_e(sqrtx)`	`= 3`
`log_e(x/sqrtx)`	`= 3`
`x/sqrtx xx sqrtx/sqrtx`	`=e^3`
`sqrt x`	`= e^3`
`:. x`	`= e^6`

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)`

Functions, MET1 2014 VCAA 3

Solve `2cos(2x) = −sqrt3` for `x`, where `0 <= x <= pi`. (2 marks)

Show Answers Only

`x = (5pi)/12, (7pi)/12`

Show Worked Solution

`cos(2x)`	`= – sqrt3/2`
`2x`	`= (5pi)/6, 2pi – (5pi)/6, 2pi+(5pi)/6`
	`=(5pi)/6, (7pi)/6, (17pi)/6,\ …`
`:. x`	`=(5pi)/12, (7pi)/12\ \ \ text(for)\ \ 0 <= x <= pi`.

Calculus, MET1 2014 VCAA 2

Let `int_4^5 2/(2x - 1) dx = log_e(b)`.

Find the value of `b`. (2 marks)

Show Answers Only

`9/7`

Show Worked Solution

`2 int_4^5(2x – 1)^(−1)dx`	`= 2/2[log_e\|\ 2x – 1\ \|]_4^5`
	`= (log_e(9) – log_e(7))`
	`= log_e(9/7)`

`:. b = 9/7`

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`

Functions, MET1 2015 VCAA 5

On any given day, the depth of water in a river is modelled by the function

`h(t) = 14 + 8sin((pit)/12),\ \ 0 <= t <= 24`

where `h` is the depth of water, in metres, and `t` is the time, in hours, after 6 am.

Find the minimum depth of the water in the river. (1 mark)
Find the values of `t` for which `h(t) = 10`. (2 marks)

Show Answers Only

`6\ text(m)`
`14quadtext(or)quad22`

Show Worked Solution

a. `h_(text(min))\ text(occurs when)\ \ sin((pit)/12)=-1`

MARKER’S COMMENT: Students who used calculus to find the minimum were less successful.

`:. h_(text(min))`	`= 14 – 8`
	`= 6\ text(m)`

b.	`14 + 8sin(pi/12t)`	`= 10`
	`sin(pi/12t)`	`= – 1/2`

`text(Solve in general:)`

`pi/12t`	`=(7pi)/6 + 2pi n\ \ \ \ text(or)\ \ \ `	`pi/12t`	`= (11t)/6 + 2pi n,`
`t`	`= 14 + 24n`	`t`	`=22 + 24n`

`text(Substitute integer values for)\ n,`

`:. t = 14quadtext(or)quad22,\ \ \ (t ∈ [0,24])`

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`

Calculus, MET1 2015 VCAA 3

Evaluate `int_1^4 (1/sqrtx)\ dx`. (2 marks)

Show Answers Only

`2`

Show Worked Solution

MARKER’S COMMENT: This basic integral caused problems with many students answering with a log function.

`int_1^4 x^(−1/2)\ dx`	`= 2[x^(1/2)]_1^4`
	`= 2[4^(1/2) – 1^(1/2)]`
	`= 2(2 – 1)`
	`= 2`

Calculus, MET1 2015 VCAA 2

Let `f′(x) = 1 - 3/x`, where `x != 0`.

Given that `f(e) = −2`, find `f(x)`. (3 marks)

Show Answers Only

`x – 3log_e(x) + 1 – e`

Show Worked Solution

`f(x)`	`= int 1 – 3x^(−1) dx`
`f(x)`	`= x – 3log_e \|\ x\ \| + c`

`text(Substitute)\ \ f(e) = − 2,`

`−2`	`= e – 3log_e(e) + c`
`c`	`= 1- e`

`:. f(x) = x – 3log_e|\ x\ | + 1 – e`

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 2

It is estimated that inflation will average 2% per annum over the next eight years.

If a new machine costs $60 000 now, calculate the cost of a similar new machine in eight years time, adjusted for inflation. Assume no other cost change.

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

Show Answers Only

`$70\ 300\ \ text{(nearest $)}`

Show Worked Solution

`text(Find value)\ (A)\ text(in 8 years.)`

`A`	`= PR^n`
	`= 60\ 000(1.02)^8`
	`= 70\ 299.562…`
	`= $70\ 300\ \ text{(nearest $)}`

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 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 1

Khan wants to buy some office furniture that is valued at $7000.

i. A store requires 25% deposit. Calculate the deposit. (1 mark)

The balance is to be paid in 24 equal monthly instalments. No interest is charged.

ii. Determine the amount of each instalment. Write your answer in dollars and cents. (1 mark)

Another store offers the same $7000 office furniture for $500 deposit and 36 monthly instalments of $220.

i. Determine the total amount paid for the furniture at this store. (1 mark)
ii. Calculate the annual flat rate of interest charged by this store.

Write your answer as a percentage correct to one decimal place. (2 marks)

A third store has the office furniture marked at $7000 but will give 15% discount if payment is made in cash at the time of sale.

Calculate the cash price paid for the furniture after the discount is applied. (1 mark)

Show Answers Only

i. `$1750`
ii.`$218.75`
i. `$8420`
ii. `7.3text{%}`
`$5950`

Show Worked Solution

a.i.	`text(Deposit)`	`= 25text(%) xx 7000`
		`= $1750`

a.ii. `text(Installment amount)`

`= ((7000 – 1750))/24`

`= $218.75`

b.i.	`text(Total paid)`	`= 500 + 36 xx 220`
		`= $8420`

b.ii. `text(Total interest paid)`

`= 8420 – 7000`

`= $1420`

`I`	`= (PrT)/100`
`1420`	`= (6500 xx r xx 3)/100`
`:. r`	`= (1420 xx 100)/(6500 xx 3)`
	`= 7.282…`
	`= 7.3text{% (1 d.p.)}`

c.	`text(Cash price)`	`= 7000 – 15text(%) xx 7000`
		`= 7000 – 1050`
		`= $5950`

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`

GRAPHS, FUR2 2007 VCAA 1

The Goldsmith family are going on a driving holiday in Western Australia.

On the first day, they leave home at 8 am and drive to Watheroo then Geraldton.

The distance–time graph below shows their journey to Geraldton.

At 9.30 am the Goldsmiths arrive at Watheroo.

They stop for a period of time.

For how many minutes did they stop at Watheroo? (1 mark)

After leaving Watheroo, the Goldsmiths continue their journey and arrive in Geraldton at 12 pm.

What distance (in kilometres) do they travel between Watheroo and Geraldton? (1 mark)
Calculate the Goldsmiths' average speed (in km/h) when travelling between Watheroo and Geraldton. (1 mark)

The Goldsmiths leave Geraldton at 1 pm and drive to Hamelin. They travel at a constant speed of 80 km/h for three hours. They do not make any stops.

On the graph above, draw a line segment representing their journey from Geraldton to Hamelin. (1 mark)

Show Answers Only

`text(30 minutes)`
`190 \ text(km)`
`95\ text(km/hr)`

Show Worked Solution

a. `text(30 minutes)`

b. `text(Distance travelled)`

`= 310 – 120`

`= 190\ text(km)`

c. `text(Time taken = 2 hours)`

`:.\ text(Average speed)`	`= 190/2`
	`= 95\ text(km/hr)`

Calculus, MET2 2010 VCAA 1

Part of the graph of the function `g: (-4, oo) -> R,\ g(x) = 2 log_e (x + 4) + 1` is shown on the axes below
1. Find the rule and domain of `g^-1`, the inverse function of `g`. (3 marks)
2. On the set of axes above sketch the graph of `g^-1`. Label the axes intercepts with their exact values. (3 marks)
3. Find the values of `x`, correct to three decimal places, for which `g^-1(x) = g(x)`. (2 marks)
4. Calculate the area enclosed by the graphs of `g` and `g^-1`. Give your answer correct to two decimal places. (2 marks)
The diagram below shows part of the graph of the function with rule

`qquad qquad qquad f (x) = k log_e (x + a) + c`, where `k`, `a` and `c` are real constants.
- The graph has a vertical asymptote with equation `x = –1`.
- The graph has a y-axis intercept at 1.
- The point `P` on the graph has coordinates `(p, 10)`, where `p` is another real constant.

1. State the value of `a`. (1 mark)
2. Find the value of `c`. (1 mark)
3. Show that `k = 9/(log_e (p + 1)`. (2 marks)
4. Show that the gradient of the tangent to the graph of `f` at the point `P` is `9/((p + 1) log_e (p + 1))`. (1 mark)
5. If the point `(– 1, 0)` lies on the tangent referred to in part b.iv., find the exact value of `p`. (2 marks)

Show Answers Only

i. `g^-1(x) = e^((x-1)/2)-4,\ \ x in R`
ii.
iii. `– 3.914 or 5.503`
iv. `52.63\ text(units²)`
i. `1`
ii. `1`
iii. `text(Proof)\ \ text{(See Worked Solutions)}`
iv. `text(Proof)\ \ text{(See Worked Solutions)}`
v. `e^(9/10)-1`

Show Worked Solution

a.i. `text(Let)\ \ y = g(x)`

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

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

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

ii.

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

`text(on the line)\ \ y=x.`

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

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

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

b.i. `text(Vertical Asymptote:)`

`x = – 1`

`:. a = 1`

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

`c = 1`

iii. `f(x)= k log_e (x + 1) + 1`

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

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

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

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

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

`(p, 10),\ \ (– 1, 0)`

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

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

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

CORE*, FUR2 2008 VCAA 4

Michelle intends to keep a car purchased for $17 000 for 15 years. At the end of this time its value will be $3500.

By what amount, in dollars, would the car’s value depreciate annually if Michelle used the flat rate method of depreciation? (1 mark)
Determine the annual flat rate of depreciation correct to one decimal place. (1 mark)

Show Answers Only

`$900`
`5.3text{% (1 d.p.)}`

Show Worked Solution

a.	`text(Depreciation)`	`= (17\ 000 – 3500)/15`
		`= $13\ 500`

`:.\ text(Annual depreciation)`

`= (13\ 500)/15`

`= $900`

b. `:.\ text(Flat rate of depreciation )`

`= 900/(17\ 000) xx 100text(%)`

`= 5.29…`

`= 5.3text{% (1 d.p.)}`

CORE*, FUR2 2008 VCAA 2

Michelle decided to invest some of her money at a higher interest rate. She deposited $3000 in an account paying 8.2% per annum, compounding half yearly.

Write down an expression involving the compound interest formula that can be used to find the value of Michelle’s $3000 investment at the end of two years. Find this value correct to the nearest cent. (2 marks)
How much interest will the $3000 investment earn over a four-year period?

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

Show Answers Only

`$3523.09`
`$1137.40`

Show Worked Solution

a. `text(Compounding periods)\ (n) = 2 xx 2 = 4`

`text(Interest per half year) = 8.2/2 = 4.1text(%)`

`:. A`	`= PR^n`
	`= 3000(1.041)^4`
	`= 3523.093…`
	`= $3523.09\ \ text{(nearest cent)}`

b. `text(After 4 years)\ (n = 8),`

MARKER’S COMMENT: A TVM calculator could also be used to solve this question.

`A`	`= 3000(1.041)^8`
	`= 4137.396…`

`:.\ text(Interest)`	`= 4137.396-3000`
	`= 1137.396…`
	`= $1137.40\ \ text{(nearest cent)}`

CORE*, FUR2 2008 VCAA 1

Michelle has a bank account that pays her simple interest.

The bank statement below shows the transactions on Michelle’s account for the month of July.

What amount, in dollars, was deposited in cash on 11 July? (1 mark)

Interest for this account is calculated on the minimum monthly balance at a rate of 3% per annum.

Calculate the interest for July, correct to the nearest cent. (2 marks)

Show Answers Only

`$620`
`$15.30`

Show Worked Solution

a.	`text(Deposit)`	`= 6870.67-6250.67`
		`= $620`

b. `text(Minimum Balance) = $6120.86`

`:.\ text(Interest)`	`=(PrT)/100`
	`= 6120.86 xx 3/100 xx 1/12`
	`= 15.302…`
	`= $15.30`

GRAPHS, FUR2 2008 VCAA 3

An event involves running for 10 km and cycling for 30 km.

Let `x` be the time taken (in minutes) to run 10 km

`y` be the time taken (in minutes) to cycle 30 km

Event organisers set constraints on the time taken, in minutes, to run and cycle during the event.

Inequalities 1 to 6 below represent all time constraints on the event.

Inequality 1: `x ≥ 0`	Inequality 4: `y <= 150`
Inequality 2: `y ≥ 0`	Inequality 5: `y <= 1.5x`
Inequality 3: `x ≤ 120`	Inequality 6: `y >= 0.8x`

Explain the meaning of Inequality 3 in terms of the context of this problem. (1 mark)

The lines `y = 150` and `y = 0.8x` are drawn on the graph below.

On the graph above
1. draw and label the lines `x = 120` and `y = 1.5x` (2 marks)
2. clearly shade the feasible region represented by Inequalities 1 to 6. (1 mark)

One competitor, Jenny, took 100 minutes to complete the run.

Between what times, in minutes, can she complete the cycling and remain within the constraints set for the event? (1 mark)
Competitors who complete the event in 90 minutes or less qualify for a prize.

Tiffany qualified for a prize.
1. Determine the maximum number of minutes for which Tiffany could have cycled. (1 mark)
2. Determine the maximum number of minutes for which Tiffany could have run. (1 mark)

Show Answers Only

`text(Inequality 3 means that the run must take)`
`text(120 minutes or less for any competitor.)`
i. & ii.
`text(80 – 150 minutes)`
1. `54\ text(minutes)`
2. `50\ text(minutes)`

Show Worked Solution

a. `text(Inequality 3 means that the run must take)`

`text(120 minutes or less for any competitor.)`

b.i. & ii.

c. `text(From the graph, the possible cycling)`

♦♦ Mean mark of parts (c)-(d) (combined) was 19%.

`text(time range is between:)`

`text(80 – 150 minutes)`

d.i. `text(Constraint to win a prize is)`

`x + y <= 90`

`text(Maximum cycling time occurs)`

`text(when)\ y = 1.5x`

`:. x + 1.5x`	`<= 90`
`2.5x`	`<= 90`
`x`	`<= 36`

`:. y_(text(max))`	`= 1.5 xx 36`
	`= 54\ text(minutes)`

d.ii. `text(Maximum run time occurs)`

`text(when)\ \ y = 0.8x`

`:. x + 0.8x`	`<= 90`
`1.8x`	`<= 90`
`x`	`<= 50`

`:. x_(text(max)) = 50\ text(minutes)`

GRAPHS, FUR2 2008 VCAA 2

Tiffany decides to enter a charity event involving running and cycling.

There is a $35 fee to enter.

Write an equation that gives the total amount, `R` dollars, collected from entry fees when there are `x` competitors in the event. (1 mark)

The event costs the organisers $50 625 plus $12.50 per competitor.

Write an equation that gives the total cost, `C`, in dollars, of the event when there are `x` competitors. (1 mark)
1. Determine the number of competitors required for the organisers to break even. (1 mark)

The number of competitors who entered the event was 8670.

Determine the profit made by the organisers. (1 mark)

Show Answers Only

`R = 35x`
`C = 50\ 625 + 12.5x`
1. `2250`
2. `$144\ 450`

Show Worked Solution

a. `R = 35x`

b. `C = 50\ 625 + 12.50x`

c.i. `text(Break even when)\ R = C`

`35x`	`= 50\ 625 + 12.5x`
`22.5x`	`= 50\ 625`
`x`	`= (50\ 625)/22.5`
	`= 2250`

`:. 2250\ text(competitors required to break even.)`

c.ii. `text(When)\ \ x = 8670,`

`text(Profit)`	`= R – C`
	`= 35 xx 8670 – (50\ 625 + 12.5 xx 8670)`
	`= 303\ 450 – 159\ 000`
	`= $144\ 450`

CORE*, FUR2 2009 VCAA 4

The golf club management purchased new lawn mowers for $22 000.

Use the flat rate depreciation method with a depreciation rate of 12% per annum to find the depreciated value of the lawn mowers after four years. (2 marks)
Use the reducing balance depreciation method with a depreciation rate of 16% per annum to calculate the depreciated value of the lawn mowers after four years. Write your answer in dollars correct to the nearest cent. (1 mark)
After 4 years, which method, flat rate depreciation or reducing balance depreciation, will give the greater depreciation? Write down the greater depreciation amount in dollars correct to the nearest cent. (1 mark)

Show Answers Only

`$11\ 440`
`$10\ 953.17`
`$11\ 046.83`

Show Worked Solution

a. `text(When)\ n = 4,`

`text(Depreciation)`	`= 22\ 000 xx 0.12 xx 4`
	`= $10\ 560`

`:.\ text(Depreciated Value)`

`= 22\ 000 – 10\ 560`

`= $11\ 440`

b. `r = 16text(%)`

`text(Value)`	`= 22\ 000(1 – r)^n`
	`= 22\ 000(0.84)^4`
	`= 10\ 953.169…`
	`= $10\ 953.17`

c. `text(Reducing balance gives a greater depreciated)`

`text(amount after 4 years.)`

`text(Greater depreciation amount)`

`= 22\ 000 – 10\ 953.17`

`= $11\ 046.83`

`log_e (x^2) – log_e (x + 3)`	`= log_e (1/2)`
`log_e (x^2/(x + 3))`	`= log_e (1/2)`

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

`text(Sum of probabilities)`	`= 1`
`p^2 + p^2 + p/4 + (4p + 1)/8`	`= 1`
`16p^2 + 2p + 4p + 1`	`= 8`
`16p^2 + 6p – 7`	`= 0`
`(2p – 1) (8p + 7)`	`= 0`

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

a.	`text(Total Area under curve)`	`= 1`
	`a int_0^5 (5x – x^2)\ dx`	`= 1`
	`a [5/2 x^2 – 1/3 x^3]_0^5`	`= 1`
	`a [(125/2 – 125/3) – (0)]`	`= 1`
	`125/6 a`	`= 1`
	`:. a`	`= 6/125`