Band 4 – Page 50

Calculus, SPEC1-NHT 2017 VCAA 3

Find the gradient of the curve with equation `x = sin (y/15)` when `x = 1/4`. Give your answer in the form `a sqrt b`, where `a, b \ in Z^+`. (3 marks)

Show Answers Only

`m = 4 sqrt 15`

Show Worked Solution

`x`	`= sin (y/15)`
`sin^(-1)x`	`= y/15`
`y`	`= 15 sin^(-1)x`
`dy/dx`	`= 15/(sqrt(1-x^2))`

`text(When)\ \ x=1/4:`

`dy/dx`	`= 15/(sqrt(1-(1/4)^2))`
	`= 15/(sqrt(15/16))`
	`= 60/sqrt15`
	`=4sqrt15`

`:. m = 4 sqrt 15`

Calculus, SPEC1 2017 VCAA 10

Show that `d/dx(x arccos(x/a)) = arccos(x/a)−x/(sqrt(a^2 - x^2))`, where `a > 0`. (1 mark)
State the maximum domain and the range of `f(x) = sqrt(arccos(x/2))`. (2 marks)
Find the volume of the solid of revolution generated when the region bounded by the graph of `y = f(x)`, and the lines `x = −2` and `y = 0`, is rotated about the `x`-axis. (4 marks)

Show Answers Only

`text(See Worked Solutions)`
`x ∈ [−2, 2], \ y ∈ [0, sqrtpi]`
`2pi^2`

Show Worked Solution

a.	`u`	`= x,`	`v`	`= cos^(−1)(x/a)`
	`uprime`	`= 1,`	`vprime`	`= (−1)/sqrt(a^2 – x^2)`

`d/(dx)(xcos^(−1)(x/a))`	`= uprimev + vprimeu`
	`= cos^(−1)(x/a) + (x(−1))/sqrt(a^2 – x^2)`
	`= arccos(x/a) – x/sqrt(a^2 – x^2)`

b. `arccos(x/2)>=0`

`text(Maximal domain:)\ x ∈ [−2, 2]`

`f(x) = (arccos(x/2))^(1/2)`

`text(Range:)\ \ y ∈ [0, sqrtpi]`

c. `V`	`= pi int_(−2)^2 y^2\ dx`
	`= pi int_(−2)^2 cos^(-1)(x/2)\ dx`
	`= pi int_(−2)^2 cos^(-1) (x/2) – x/sqrt(4 – x^2) + x/sqrt(4 – x^2)\ dx`
	`= pi [x cos^(-1)(x/2)]_(−2)^2 + pi int_(−2)^2 x/sqrt(4 – x^2)\ dx`

`text(Let)\ \ u = 4 – x^2`

♦ Mean mark part (c) 35%.

`(du)/(dx) = -2x\ \ =>\ \ -1/2 (du) = x\ dx`

`text(When)\ \ x=2\ \ => \ u=0`

`text(When)\ \ x=-2\ \ =>\ \ u=0`

`:. V`	`= pi [2cos^(−1)(1) – (-2)cos^(−1)(−1)] – pi/2 int_0^0 1/sqrtu du`
	`= pi(2 xx 0 + 2 xx pi)`
	`= 2pi^2`

Mechanics, SPEC1-NHT 2017 VCAA 1

A 5 kg mass on a smooth plane inclined at 30° is held in equilibrium by a horizontal force of magnitude `P` newtons, as shown in the diagram below.

On the diagram above, show all other forces acting on the mass and label them. (1 mark)
Find `P`. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`P = (5g sqrt 3)/3`

Show Worked Solution

b.	`P cos 30^@`	`= 5g sin 30^@`
	`(P sqrt 3)/2`	`= (5g)/2`
	`P sqrt 3`	`= 5g`
	`P`	`= (5g)/sqrt 3`
	`P`	`= (5g sqrt 3)/3`

Complex Numbers, SPEC2 2014 VCAA 9 MC

The circle `| z - 3 - 2i | = 2` is intersected exactly twice by the line given by

A. `| z - i | = | z + 1 |`

B. `| z - 3 - 2i | = | z - 5 |`

C. `| z - 3 - 2i | = | z - 10i |`

D. `text(Im)(z) = 0`

E. `text(Re)(z) = 5`

Show Answers Only

`B`

Show Worked Solution

`text{Circle has centre (3,2) and radius 2.}`

`text{Consider each option (using CAS):}`

`text(Options A and C do not intersect.)`

`text(Options D and E are tangents.)`

`text(Only option B intersects twice).`

`=> B`

Complex Numbers, SPEC2 2015 VCAA 7 MC

If `z = sqrt3 + 3i`, then `z^63` is

real and negative
equal to a negative real multiple of `i`
real and positive
equal to a positive real multiple of `i`
a positive real multiple of `1 + isqrt3`

Show Answers Only

`A`

Show Worked Solution

`z`	`= sqrt3 + 3i`
	`= 2sqrt3 text(cis)\ (pi/3)`

`:. z^63`	`= (2sqrt3)^63 text(cis)\ ((63pi)/12)`
	`= (2sqrt3)^63 text(cis)\ (21pi)`
	`= (2sqrt3)^63 text(cis)\ (pi)`
	`= -(2sqrt3)^63`

`=> A`

Graphs, SPEC2 2015 VCAA 2 MC

The range of the function with rule `f(x) = (2 - x)arcsin(x/2 - 1)` is

A. `[-pi,0]`

B. `[-pi/2,pi/2]`

C. `[-((2 - x)pi)/2,((2 - x)pi)/2]`

D. `[0,4]`

E. `[0,pi]`

Show Answers Only

`A`

Show Worked Solution

`text(By CAS, create a graph of)\ \ f(x) = (2 – x)sin^(-1)(x/2 – 1).`

`=> A`

Calculus, SPEC1 2015 VCAA 9

Consider the curve represented by `x^2 - xy + 3/2 y^2 = 9.`

Find the gradient of the curve at any point `(x, y).` (2 marks)
Find the equation of the tangent to the curve at the point `(3, 0)` and find the equation of the tangent to the curve at the point `(0, sqrt 6).`

Write each equation in the form `y = ax + b.` (2 marks)
Find the acute angle between the tangent to the curve at the point `(3, 0)` and the tangent to the curve at the point `(0, sqrt 6).`

Give your answer in the form `k pi`, where `k` is a real constant (2 marks)

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`(dy)/(dx) = (2x – y)/(x – 3y)`
`y = 2(x – 3);\ \ \ y = 1/3 x + sqrt 6`
`theta = pi/4`

Show Worked Solution

a.	`d/(dx)(x^2) – d/(dx)(xy) + 3/2* d/(dx) (y^2)`	`= 0`
	`2x – x(dy)/(dx) – y + 3/2(2y)(dy)/(dx)`	`= 0`
	`(dy)/(dx)(−x + 3y)`	`= y – 2x`
	`:. (dy)/(dx)`	`= (y – 2x)/(3y – x)`

b. `m_{(3,0)} = (0 – 6)/(0 – 3) = 2`

`:.\ text{Equation of tangent at (3, 0):}`

`y = 2(x – 3)`

`= 2x – 6`

`m_{(0,sqrt6)} = (sqrt6 – 0)/(3sqrt6 – 0) = 1/3`

`:.\ text{Equation of tangent at}\ (0,sqrt6):`

`y -sqrt6`	`= 1/3(x – 0)`
`y`	`=1/3 x + sqrt6`

c. `m_1 = 2 = tan(theta_1), \ \ m_2 = 1/3 = tan(theta_2)`

`alpha`	`= theta_1 – theta_2`
	`= tan^(−1)(2) – tan^(−1)(1/3)`

`tan(alpha)`	`= tan(tan^(−1)(2) – tan^(−1)(1/3))`
	`= (tan(tan^(−1)(2)) – tan(tan^(−1)(1/3)))/(1 + tan(tan^(−1)(2)tan(tan^(−1)(1/3))))`
	`= (2 – 1/3)/(1 + 2/3)`
	`= 1`

`:. alpha = pi/4\ \ \ (alpha ∈ (0, pi/2))`

Calculus, SPEC1 2015 VCAA 8

Show that `int tan (2x)\ dx = 1/2 log_e |\ sec (2x)\ | + c.` (2 marks)
The graph of `f(x) = 1/2 arctan (x)` is shown below
i. Write down the equations of the asymptotes. (1 mark)
ii. On the axes above, sketch the graph of `f^-1`, labelling any asymptotes with their equations. (1 mark)
Find `f(sqrt 3).` (1 mark)
Find the area enclosed by the graph of `f`, the `x`-axis and the line `x = sqrt 3.` (2 marks)

Show Answers Only

a. `text(Proof)\ \ text{(See Worked Solutions)}`

b.	(i)	`y = -pi/4,\ \ y = pi/4`
	(ii)

c. `pi/6`

d. `(sqrt 3 pi)/6 – log_e sqrt 2`

Show Worked Solution

a. `text(Let)\ \ u = cos(2x)`

`(du)/(dx) = -2 sin (2x)\ \ =>\ \ -1/2 xx du = sin (2x)\ dx`

`int tan (2x)\ dx`	`= int (sin(2x))/(cos(2x))\ dx`
	`= -1/2 int 1/u\ du`
	`= -1/2 log_e \|\ u\ \| + c`
	`= -1/2 log_e \|\ cos (2x)\ \| + c`
	`= 1/2 log_e \|\ sec (2x)\ \| + c`

b.i. `text(Range:)\ \ tan^(-1)(x) in (- pi/2,pi/2)`

`=>1/2 tan^(-1)(x) in (- pi/4, pi/4)`

`:.\ text(Asymptotes:)\ \ y = -pi/4,\ \ y = pi/4`

b.ii.

c.	`f(sqrt 3)`	`= 1/2 tan^(-1) (sqrt 3)`
		`= 1/2 xx pi/3`
		`= pi/6`

d.	`y`	`= 1/2tan^(−1)(x)`
	`2y`	`= tan^(−1)(x)`
	`x`	`= tan(2y)`

♦ Mean mark part (d) 49%.

`text(Area)`	`=\ text(Area of rectangle – Area between graph and y-axis)`
	`= sqrt3 xx pi/6 – int_0^(pi/6) tan(2y)\ dy`
	`= (sqrt3 pi)/6 -1/2[ln\ \| sec(2y) \|]_0^(pi/6)`
	`= (sqrt3 pi)/6 – 1/2[ ln(sec(pi/3)) – ln(sec 0)]`
	`= (sqrt3 pi)/6 – 1/2 (ln2 -ln1)`
	`= (sqrt3 pi)/6 – 1/2 ln2`

Trigonometry, SPEC1 2015 VCAA 7

Solve `sin(2x) = sin(x), \ x in [0, 2 pi].` (3 marks)
Find `{x:\ text(cosec)(2x) <\ text(cosec)(x),\ x in (0, pi/2) uu (pi/2, pi)}.` (2 marks)

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`x = 0, pi, 2 pi or x = pi/3, (5 pi)/3`
`x ∈ (0, pi/3) ∪ (pi/2, pi)`

Show Worked Solution

a. `2sin(x)cos(x) = sin(x)`

MARKER’S COMMENT: Many students expanded sin(2x) and then cancelled sin(x) on both sides which lost a set of solutions!

`sin(x)(2cos(x) – 1) = 0`

`sin(x) = 0, cos(x) = 1/2`

`x = 0, pi/3, pi, 2pi – pi/3, 2pi`

`x = 0, pi/3, pi, (5pi)/3, 2pi`

b. `text(Solving cosec)(2x) = text(cosec)(x), \ x ∈ (0, pi) \\ {pi/2}:`

♦♦ Mean mark 26%.

`text{Using part (a):}`

`sin(2x) = sin(x)\ \ text(when)\ \ x = pi/3`

`:. text(cosec)(x) = text(cosec)(2x)\ \ text(at)\ \ x=pi/3.`

`text(Sketch graphs:)`

`text(When cosec)(2x) < text(cosec)(x):`

`x ∈ (0, pi/3) ∪ (pi/2, pi)`

Calculus, SPEC1 2015 VCAA 5

Find the volume generated when the region bounded by the graph of `y = 2x^2 - 3`, the line `y = 5` and the `y`-axis is rotated about the `y`-axis. (3 marks)

Show Answers Only

`16 pi`

Show Worked Solution

`V = pi int_(−3)^5 x^2\ dy`

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

`:. V`	`= pi/2 int_(−3)^5 (y + 3)\ dy`
	`= pi/2[(y^2)/2 + 3y]_(−3)^5`
	`= pi/2(25/2 + 15 – (9/2 – 9))`
	`= pi/2(16/2 + 24)`
	`= 16pi`

Complex Numbers, SPEC1 2015 VCAA 4

Find all solutions of `z^3 = 8i, \ z in C` in cartesian form. (3 marks)
Find all solutions of `(z − 2i)^3 = 8i, \ z in C` in cartesian form. (1 mark)

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`-2i, +- sqrt 3 + i`
`z = 0 or z = +- sqrt 3 + 3i`

Show Worked Solution

a. `z^3 = 8text(cis)(pi/2)`

`z_1`	`= 8^(1/3)text(cis)(pi/6)`
	`= 2text(cis)(pi/6)`
	`= sqrt3 + i`

`z_2`	`= 2text(cis)(pi/6 + (2pi)/3)`
	`= 2text(cis)((5pi)/6)`
	`= −sqrt3 + i`

`z_3`	`= 2text(cis)(pi/6 – (2pi)/3)`
	`= 2text(cis)(−pi/2)`
	`= −2i`

`:. z = sqrt3 + i, \ −sqrt3 + i, \ −2i`

♦ Mean mark part (b) 44%.

b.	`z_1 – 2i`	`= sqrt3 + i`
	`z_1`	`= sqrt3 + 3i`

`z_2 – 2i`	`= −sqrt3 + i`
`z_2`	`= −sqrt3 + 3i`

`z_3 – 2i`	`= −2i`
`z_3`	`= 0`

`:. z = sqrt3 + 3i, \ −sqrt3 + 3i, \ 0`

Mechanics, SPEC1 2015 VCAA 2

A 20 kg parcel sits on the floor of a lift.

The lift is accelerating upwards at 1.2 ms¯².

Find the reaction force of the lift floor on the parcel in newtons. (2 marks)
Find the acceleration of the lift downwards in ms¯² so that the reaction of the lift floor on the parcel is 166 N. (2 marks)

Show Answers Only

`220`
`1.5\ text(ms)^-2`

Show Worked Solution

`text(Acceleration:)\ \ ↑ 1.2\ text(ms)^(−2)`

`∑F`	`= 20 xx 1.2`
	`= R – 20text(g)`

`R – 20 xx 9.8`	`=20 xx 1.2`
`:. R`	`= 20(1.2 + 9.8)`
	`= 20 xx 11`
	`= 220\ text(N upwards)`

b. `↓a\ text(ms)^(−2)`

`∑F`	`= 20a`
	`= 20text(g) – R_2`

`20a`	`= 20text(g) – 166`
	`= 20text(g) – 2 xx 83`
	`= 20text(g) – 20 xx 8.3`
`:. a`	`= text(g) – 8.3`
	`= 9.8 – 8.3`
	`= 1.5\ text(ms)^(−2)`

A taut rope of length `1 2/3` m suspends a mass of 20 kg from a fixed point `O`. A horizontal force of `P` newtons displaces the mass by 1 m horizontally so that the taut rope is then at an angle of `theta` to the vertical.

Show all the forces acting on the mass on the diagram below. (1 mark)

Show that `sin (theta) = 3/5`. (1 mark)
Find the magnitude of the tension force in the rope in newtons. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`text(Proof)\ text{(See Worked Solutions)}`
`T = 245\ text(N)`

Show Worked Solution

b.	`sin(theta)`	`= 1/(1 2/3)`
		`= 1/((5/3)`
		`= 3/5`

`cos theta = (20g)/T`

`sin theta = 3/5\ \ text{(using part b)}`

`underbrace{(400g^2)/T^2 + 9/25}_(cos^2 theta + sin^2 theta = 1)`	`= 1`
`(400g^2)/T^2`	`= 16/25`
`(20g)/T`	`=4/5`
`T/(20g)`	`= 5/4`
`T`	`= 25g`
	`=245\ text(N)`

Complex Numbers, SPEC2 2014 VCAA 8 MC

The principal argument of `(−3sqrt2 - isqrt6)/(2 + 2i)` is

A. `(−13pi)/12`

B. `(7pi)/12`

C. `(11pi)/12`

D. `(13pi)/12`

E. `(−11pi)/12`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ z_1=−3sqrt2 – isqrt6`

`=>\ text(Arg)\ z_1 = pi + tan^(-1) (sqrt6/(3 sqrt2)) = (7pi)/6`

`text(Let)\ \ z_2 = 2 + 2i`

`=>\ text(Arg)\ z_2 = pi/4`

`:.\ text(Arg)\ (z_1/z_2)`	`= text(Arg)\ ((−3sqrt2 – isqrt6)/(2 + 2i))`
	`= (7pi)/6 – pi/4`
	`= (11pi)/12`

`=> C`

Complex Numbers, SPEC2 2014 VCAA 7 MC

The sum of the roots of `z^3 - 5z^2 + 11z - 7 = 0`, where `z ∈ C`, is

`1 + 2sqrt3i`
`5i`
`4 - 2sqrt3i`
`2sqrt3i`
`5`

Show Answers Only

`E`

Show Worked Solution

`text(Coefficients are real)`

`=>\ text(Conjugate root theorem applies.)`

`z_1=alpha + beta i,\ \ z_2=alpha – beta i, \ \ z_3=gamma`

`z_1+z_2+z_3`	`= alpha + beta i + alpha – beta i + gamma`
	`= 2 alpha + gamma\ \ \ (∈R)`

`=> E`

Complex Numbers, SPEC2 2014 VCAA 6 MC

Given that `i^n = p` and `i^2 = −1`, then `i^(2n + 3)` in terms of `p` is equal to

`p^2 - i`
`p^2 + i`
`−p^2`
`−ip^2`
`ip^2`

Show Answers Only

`D`

Show Worked Solution

`i^(2n + 3)`	`= i^(2n)i^2i`
	`= (i^n)^2i^2i`
	`= −ip^2`

`=> D`

Graphs, SPEC2 2014 VCAA 3 MC

The features of the graph of the function with rule `f(x) = (x^2 - 4x + 3)/(x^2 - x - 6)` include

A. asymptotes at `x = 1` and `x = −2`

B. asymptotes at `x = 3` and `x = −2`

C. an asymptote at `x = 1` and a point of discontinuity at `x = 3`

D. an asymptote at `x = −2` and a point of discontinuity at `x = 3`

E. an asymptote at `x = 3` and a point of discontinuity at `x = −2`

Show Answers Only

`D`

Show Worked Solution

`f(x)`	`= (x^2 – 4x + 3)/(x^2 – x – 6)`
	`=(x^2 -x-6 -3x+9)/(x^2 – x – 6)`
	`= 1 – (3(x – 3))/((x + 2)(x – 3))`
	`=1 – 3/(x + 2)\ text(where)\ x != 3`

`=> D`

Mechanics, SPEC1 2014 VCAA 8

A body of mass 5 kg is held in equilibrium by two light inextensible strings. One string is attached to a ceiling at `A` and the other to a wall at `B`. The string attached to the ceiling is at an angle `theta` to the vertical and has tension `T_1` newtons, and the other string is horizontal and has tension `T_2` newtons. Both strings are made of the same material.

i. Resolve the forces on the body vertically and horizontally, and express `T_1` in terms of `theta`. (2 marks)
ii. Express `T_2` in terms of `theta`. (1 mark)
Show that `tan (theta) < sec (theta)` for `0 < theta < pi/2`. (1 mark)
The type of string used will break if it is subjected to a tension of more than 98 N.

Find the maximum allowable value of `theta` so that neither string will break. (3 marks)

Show Answers Only

i. `T_1 = 5g sec theta`
ii. `T_2 = 5g tan theta`
`text(Proof)\ text{(See Worked Solutions)}`
`theta_max = pi/3`

Show Worked Solution

a.i.

`text(Resolve vertically): \ 5g = T_1 cos theta`

`text(Resolve horizontally): \ T_1 sin theta = T_2`

`:. T_1 = 5g sec theta`

a.ii.	`T_2`	`= 5g sec theta sin theta`
		`= 5g tan theta`

b.	`cos theta in (0, 1), \ theta in (0, pi/2)`
	`sin theta in (0, 1), \ theta in (0, pi/2)`

`:. sin theta < 1, \ theta in (0, pi/2)`

♦ Mean mark part (b) 40%.

`text(If)\ sin theta < 1 and cos theta > 0:`

`(sin theta)/(cos theta) < 1/(cos theta)`

`:. tan theta < sec theta, quad 0 < theta < pi/2`

c. `tan theta < sec theta \ \ =>\ \ T_2\ text(will be smaller) => \ T_1\ text(will break first)`

♦ Mean mark part (c) 46%.

`5text(g)\ sec theta_max`	`= 98`
`sec theta_max`	`= 98/(5 xx 9.8)`
`sec theta_max`	`= 10/5`
`sec theta_max`	`= 2`
`:. theta_max`	`= pi/3`

Calculus, SPEC1 2014 VCAA 7

Consider `f(x) = 3x arctan (2x)`.

Write down the range of `f`. (1 mark)
Show that `f prime(x) = 3 arctan (2x) + (6x)/(1 + 4x^2)`. (1 mark)
Hence evaluate the area enclosed by the graph of `g(x) = arctan (2x)`, the `x`-axis and the lines `x = 1/2` and `x = sqrt 3/2`. (3 marks)

Show Answers Only

`f(x) in [0, oo)`
`text(Proof)\ text{(See Worked Solutions)}`
`(pi sqrt 3)/6 – pi/8 – 1/4 ln2`

Show Worked Solution

a. `f(x) = 3x arctan (2x)`

`f(0) = 0`

♦♦♦ Mean mark 4%.

`text(When)\ \ x < 0, \ \ 3x<0,\ \ text(and)\ \ y = tan^(-1) (2x) < 0`

`text(When)\ \ x>0,\ \ 3x>0,\ \ text(and)\ \ y = tan^(-1) (2x) > 0`

`:. f(x) in [0, oo)`

b. `u = 3x, qquad v = tan^(-1)(2x)`

`u prime = 3, qquad v prime = 2/(1 + 4x^2)`

`:. f prime (x)`	`= 3(arctan (2x)) + (2/(1 + 4x^2))(3x)`
	`= 3 arctan (2x) + (6x)/(1 + 4x^2)`

c.	`A`	`= int_(1/2)^(sqrt 3/2) arctan (2x)\ dx`
		`= 1/3 int_(1/2)^(sqrt 3/2) 3arctan (2x)\ dx`
		`=1/3 int_(1/2)^(sqrt 3/2) (3 arctan (2x) + (6x)/(1 + 4x^2) – (6x)/(1 + 4x^2))\ dx`
		`= 1/3 int_(1/2)^(sqrt 3/2) 3 arctan (2x) + (6x)/(1 + 4x^2) dx – int_(1/2)^(sqrt 3/2) (2x)/(1 + 4x^2)\ dx`
		`= 1/3 [3x arctan (2x)]_(1/2)^(sqrt 3/2) – 1/4 int_(1/2)^(sqrt 3/2) (8x)/(1 + 4x^2)\ dx`
		`= [sqrt 3/2 arctan (sqrt 3) – 1/2 arctan (1)] – 1/4[ln (1 + 4x^2)]_(1/2)^(sqrt 3/2)`
		`= sqrt 3/2 (pi/3) – 1/2 (pi/4) – 1/4 [ln (1 + 4(3/4)) – ln(1 + 4 (1/4)]`
		`= (pi sqrt 3)/6 – pi/8 – 1/4 (ln4 – ln 2)`
		`= (pi sqrt 3)/6 – pi/8 – 1/4 ln2`

Calculus, SPEC1 2014 VCAA 6

Verify that `a/(a - 4) = 1 + 4/(a - 4)`. (1 mark)

Part of the graph of `y = x/sqrt(x^2 - 4)` is shown below.

The region enclosed by the graph of `y = x/sqrt(x^2 - 4)` and the lines `y = 0, \ x = 3` and `x = 4` is rotated about the `x`-axis

Find the volume of the resulting solid of revolution. (4 marks)

Show Answers Only

`text(Proof)\ text{(See Worked Solutions)}`
`pi(1 + l n (5/3))`

Show Worked Solution

a.	`a/(a – 4)`	`= (a – 4 + 4)/(a – 4)`
		`= (a – 4)/(a – 4) + 4/(a – 4)`
		`= 1 + 4/(a – 4), quad a != 4`

b. `r = y`

`V`	`= pi int_3^4 y^2\ dx`
	`= pi int_3^4 underbrace{x^2/(x^2 – 4)}_text(using part a.)\ dx`
	`= pi int_3^4 1 + 4/(x^2 – 4) dx`
	`= pi int_3^4 1 + 4/((x – 2)(x + 2)) dx`

`text(Using partial fractions:)`

`4/((x – 2)(x + 2)) = A/(x – 2) + B/(x + 2)`

`A(x + 2) + B(x – 2) = 4`

`text(When)\ \ x = 2: \ 4A = 4\ \ =>\ \ A = 1`

`text(When)\ \ x = -2: -4B = 4\ \ =>\ \ B = -1`

`:. V`	`= pi int_3^4 1 + 1/(x – 2) – 1/(x + 2)\ dx`
	`= pi [x + ln \|x – 2\| – ln \|x + 2\|]_3^4`
	`= pi [4 + ln2 – ln6 – (3 + ln 1 – ln5)]`
	`= pi (1 + ln2 – ln6 + ln5)`
	`= pi (1 + ln ((2 xx 5)/6))`
	`= pi (1 + ln (5/3))`

Calculus, SPEC1 2014 VCAA 5

For the function with rule `f(x) = 96 cos (3x) sin (3x)`, Find the value of `a` such that `f(x) = a sin (6x)`. (1 mark)
Use an appropriate substitution in the form `u = g(x)` to find an equivalent definite integral for

`int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx` in terms of `u` only. (3 marks)
Hence evaluate `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`, giving your answer in the form `sqrt k, \ k in Z`. (1 mark)

Show Answers Only

`48`
`int_(sqrt 3/2)^0 – 8u^2 du`
`sqrt 3`

Show Worked Solution

a. `96 cos (3x) sin(3x)`

`= 48 (2 cos(3x) sin(3x))`

`= 48 sin (6x)`

`:. a = 48`

b. `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`

`= int_(pi/36)^(pi/12) 48 sin (6x) cos^2 (6x)\ dx`

`u`	`= cos (6x)`
`(du)/(dx)`	`= -6 sin (6x)\ \ =>\ \ du = -6 sin(6x)\ dx`

`u(pi/12)`	`= cos (pi/2) = 0`
`u(pi/36)`	`= cos (pi/6) = sqrt 3/2`

`:. int_(pi/36)^(pi/12) 48 sin (6x) cos^2 (6x)\ dx`

`= int_(sqrt 3/2)^0 – 8u^2\ du`

c. `:. int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`

`= -8 [u^3/3]_(sqrt 3/2)^0`

`= -8 (0 – (sqrt 3/2)^3/3)`

`= 8/3 xx (3sqrt3)/8`

`= sqrt 3`

Calculus, SPEC1 2014 VCAA 4

Find the gradient of the line perpendicular to the tangent to the curve defined by `y = -3e^(3x) e^y` at the point `(1, -3)`. (3 marks)

Show Answers Only

`4/9`

Show Worked Solution

`(dy)/(dx)`	`= -3 d/(dx) (e^(3x) e^y)`
`(dy)/(dx)`	`= -9e^(3x) e^y + -3 e^(3x) e^y (dy)/(dx)`
` -9e^(3x) e^y`	`=(dy)/(dx) (1 + 3e^(3x) e^y)`
`(dy)/(dx)`	`= (-9e^(3x) e^y)/(1 + 3e^(3x) e^(-3))`

`text{At (1, –3):}`
`:. m_text(norm)`	`= (1 + 3e^(3 xx 1) e^(-3))/(9e^(3 xx 1) e^(-3))`
	`= (1 + 3e^3 e^(-3))/(9e^3 e^(-3))`
	`= 4/9`

Complex Numbers, SPEC1 2014 VCAA 3

Let `f` be a function of a complex variable, defined by the rule `f(z) = z^4 - 4z^3 + 7z^2 - 4z + 6`.

Given that `z = i` is a solution of `f(z) = 0`, write down a quadratic factor of `f(z)`. (2 marks)
Given that the other quadratic factor of `f(z)` has the form `z^2 + bz + c`, find all solutions of `z^4 - 4z^3 + 7z^2 - 4z + 6 = 0` in a cartesian form. (3 marks)

Show Answers Only

`z^2 + 1`
` z = i, quad z = -i, quad z = 2 + i sqrt 2, quad z = 2 – i sqrt 2`

Show Worked Solution

a. `text(Real coefficients) \ => \ z = -i\ \ text(is also a solution)`

`:. (z – i)(z + i)= z^2 + 1\ \ text(is a quadratic factor)`

b.	`(z^2 + 1)(z^2 + bz + c)`	`= z^4 – 4z^3 + 7z^2 – 4z + 6`
	`z^4 + bz^3 + (c + 1)z^2 + bz + c`	`= z^4 – 4z^3 + 7z^2 – 4z + 6`

`text(Equating co-efficients:)`

`bz^3 = -4z^3\ \ =>\ \ b=-4`

`(c + 1)z^2 = 7z^2\ \ =>\ \ c = 6`

`(z^2 + 1)(z^2 + 4z + 6)`	`= 0`
`(z^2 + 1)(z^2 – 4z + 2^2 – 4 + 6)`	`= 0`
`(z^2 + 1)((z – 2)^2 + 2)`	`= 0`
`(z^2 + 1)((z – 2)^2 – 2i^2)`	`= 0`
`(z^2 + 1)(z – 2 – i sqrt 2)(z – 2 + i sqrt 2)`	`= 0`

`:.z = i, quad z = -i, quad z = 2 + i sqrt 2, quad z = 2 – i sqrt 2`

Vectors, SPEC1 2014 VCAA 2

The position vector of a particle at time `t >= 0` is given by

`underset ~r (t) = (t - 2) underset ~ i + (t^2 - 4t + 1) underset ~j`

Show that the cartesian equation of the path followed by the particle is `y = x^2 - 3`. (1 mark)
Sketch the path followed by the particle on the axes below, labelling all important features. (2 marks)

Find the speed of the particle when `t = 1`. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(See Worked Solutions)`
`sqrt 5`

Show Worked Solution

`x`

`= t – 2\ \ =>\ \ t=x+2`

`y`	`= t^2 – 4t + 1`
	`= (x + 2)^2 – 4 (x + 2) + 1`

`y`	`= x^2 + 4x + 4 – 4x – 8 + 1`
	`= x^2 – 3`

b. `t >= 0`

`x = t – 2\ \ =>\ \ x >= -2`

`y(-2)`	`= (-2)^2 – 3`
	`= 1`
`y(0)`	`= -3`

`0`	`= x^2 – 3`
`x^2`	`= 3`
`x`	`= +- sqrt 3`

c.	`underset ~ dot r (t)`	`= underset ~ dot i + (2t – 4) underset ~j`
	`underset ~ dot r (1)`	`= underset ~ dot i + (2 – 4) underset ~j`
		`= underset ~ dot i – 2 underset ~j`
	`\|underset ~ dot r(1)\|`	`= sqrt (1^2 + (-2)^2)`
		`= sqrt 5`

Vectors, SPEC1 2014 VCAA 1

Consider the vector `underset ~a = sqrt 3 underset ~i - underset ~j - sqrt 2 underset ~k`, where `underset ~i, underset ~j` and `underset ~k` are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

Find the unit vector in the direction of `underset ~a`. (1 mark)
Find the acute angle that `underset ~a` makes with the positive direction of the `x`-axis. (2 marks)
The vector `underset ~b = 2 sqrt 3 underset ~i + m underset ~j - 5 underset ~k`.

Given that `underset ~b` is perpendicular to `underset ~a,` find the value of `m`. (2 marks)

Show Answers Only

`1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`
`theta = 45^@`
`m = 6 + 5 sqrt 2`

Show Worked Solution

a.	`\|underset ~a\|`	`= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)`
		`= sqrt 6`

`:. hat underset ~a`	`= underset ~a/\|underset ~a\|`
	`= 1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`

Mean mark part (b) 51%.

b.	`underset ~a ⋅ underset ~i`	`= sqrt 3 xx 1 = sqrt 3`
	`underset ~a ⋅ underset ~i`	`= \|underset ~a\|\|underset ~i\| cos theta`
		`= sqrt 6 cos theta`
	`sqrt 3`	`= sqrt 6 cos theta`

`cos theta`	`= 1/sqrt 2`
`:. theta`	`= pi/4 = 45^@`

c. `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6 – m + 5 sqrt 2`	`=0`
`:. m`	`=6 + 5 sqrt 2`

Vectors, SPEC1 2015 VCAA 1

Consider the rhombus `OABC` shown below, where `vec (OA) = a underset ~i` and `vec (OC) = underset ~i + underset ~j + underset ~k`, and `a` is a positive real constant.

Find `a.` (1 mark)
Show that the diagonals of the rhombus `OABC` are perpendicular. (2 marks)

Show Answers Only

`sqrt 3`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

a. `|\ vec(OA)\ | = |\ vec(OC)\ |,`

`:. a`	`= sqrt (1^2 + 1^2 + 1^2)`
	`= sqrt 3`

b. `overset(->)(CB) = overset(->)(OA), \ \ overset(->)(AB) = overset(->)(OC)`

MARKER’S COMMENT: Vector notation was poor in many answers.

`overset(->)(OB) = overset(->)(OC) + overset(->)(CB)`

`= underset~i + underset~j + underset~k + sqrt3 underset~i`

`= (sqrt3 + 1)underset~i + underset~j + underset~k`

`overset(->)(AC) = overset(->)(AO) + overset(->)(OC)`

`= −sqrt3underset~i + underset~i + underset~j + underset~k`

`= (1 – sqrt3)underset~i + underset~j + underset~k`

`overset(->)(AC) · overset(->)(OB)`	`= (1 + sqrt3)(1 – sqrt3) + 1 + 1`
	`= 1 – 3 + 1 + 1`
	`= 0`

`:. overset(->)(AC) ⊥ overset(->)(OB)`

Mechanics, SPEC2 2017 VCAA 16 MC

An object of mass 20 kg, initially at rest, is pulled along a rough horizontal surface by a force of 80 N acting at an angle of 40° upwards from the horizontal. A friction force of 20 N opposes the motion.

After the pulling force has acted for 5 seconds, the magnitude of the momentum, in kg ms⁻¹, of the object is closest to

10
40
160
210
4100

Show Answers Only

`D`

Show Worked Solution

`∑underset~F = 80cos(40°) – 20`

`ma`	`= 80cos(40°) – 20`
`a`	`=4 cos(40°) – 1`

`Deltav`	`= 0+at`
	`= 5(4cos(40°) – 1)`

`m Deltav`	`= 100(4cos(40°) – 1)`
	`~~ 206.4`

`=> D`

Vectors, SPEC2 2017 VCAA 15 MC

A body has displacement of `3underset~i + underset~j` metres at a particular time. The body moves with constant velocity and two seconds later its displacement is `−underset~i + 5underset~j` metres.

The velocity, in ms⁻¹, of the body is

`2underset~i + 6underset~j`
`−2underset~i + 2underset~j`
`−4underset~i + 4underset~j`
`4underset~i - 4underset~j`
`underset~i + 3underset~j`

Show Answers Only

`B`

Show Worked Solution

`Deltaunderset~s`	`= (−1 – 3)underset~i + (5 – 1)underset~j`
	`= −4underset~i + 4underset~j`

`underset~v`	`= (Deltaunderset~s)/t`
	`= (−4underset~i + 4underset~j)/2`
	`= −2underset~i + 2underset~j`

`=> B`

Mechanics, SPEC2 2017 VCAA 14 MC

Two particles with mass `m_1` kilograms and `m_2` kilograms are connected by a taut light string that passes over a smooth pulley. The particles sit on smooth inclined planes, as shown in the diagram below.

If the system is in equilibrium, then `m_1/m_2` is equal to

`(sec(theta))/2`
`2sec(theta)`
`2cos(theta)`
`1/2`
`1`

Show Answers Only

`A`

Show Worked Solution

Mean mark 51%.

`∑F`	`=m_1gsin(2theta) – m_2gsin(theta) = 0`

`m_1gsin(2theta)`	`= m_2gsin(theta)`
`m_1/m_2`	`= (sin(theta))/(sin(2theta))`
	`= (sin(theta))/(2sin(theta)cos(theta))`
	`= 1/(2 cos(theta))`
	`= 1/2 sec(theta)`

`=> A`

Vectors, SPEC1 2017 VCAA 9

A particle of mass 2 kg with initial velocity `3underset~i + 2underset~j` ms⁻¹ experiences a constant force for 10 seconds.

The particle's velocity at the end of the 10-second period `43underset~i - 18underset~j` ms⁻¹ .

Find the magnitude of the constant force in newtons. (2 marks)
Find the displacement of the particle from its initial position after 10 seconds. (3 marks)

Show Answers Only

`4sqrt5 N`
`230underset~i – 80underset~j`

Show Worked Solution

a.	`u_i`	`= 3`	`v_i`	`= 43`
	`u_j`	`= 2`	`v_j`	`= −18`

`43`	`= 3 + 10a`
`40`	`= 10a`
`a_i`	`= 4`

♦ Mean mark 38%.

`−18`	`= 2 + 10a_j`
`−20`	`= 10a_j`
`a_j`	`= −2`

`underset~a`	`= 4underset~i – 2underset~j`
`\|underset~a\|`	`= sqrt(4^2 + (−2)^2)`
	`= sqrt(20)`
	`= sqrt(4 xx 5)`
	`= 2sqrt5`

`:.\|F\|`	`= 2 xx 2sqrt5`
	`= 4sqrt5 N`

b. `underset~a = 4underset~i – 2underset~j`

`underset~v`	`= int4underset~i – 2underset~j\ dt`
	`= (4t + C_i)underset~i + (−2t + C_j)underset~j\ \ \ (C_i, C_j ∈ R)`

`underset~v(0) = C_i underset~i + C_j underset~j = 3underset~i + 2underset~j`

`=> C_i = 3, C_j = 2`

`underset~v(t) = (4t + 3)underset~i + (2 – 2t)underset~j`

`underset~x(t)`	`= int(4t + 3)underset~i + (2 – 2t)underset~j\ dt`
	`= (((4t^2)/2 +3t) + b_i)underset~i + ((2t – (2t^2)/2) + b_j)underset~j \ \ \ (b_i, b_j ∈ R)`
	`= (2t^2 + 3t + b_i)underset~i + (2t – t^2 + b_j)underset~j`

`underset~x(0) = b_iunderset~i + b_junderset~j`

`underset~0 => b_i = b_j = 0`

`underset~x(t) = (2t^2 + 3t)underset~i + (2t – t^2)underset~j`

`underset~x(10)`	`= (2(10)^2 + 3(10))underset~i + (2(10) – 10^2)underset~j`
	`= 230underset~i – 80underset~j`

Calculus, SPEC1 2017 VCAA 8

A slope field representing the differential equation `dy/dx = −x/(1 + y^2)` is shown below.

Sketch the solution curve of the differential equation corresponding to the condition `y(−1) = 1` on the slope field above and, hence, estimate the positive value of `x` when `y = 0`. Give your answer correct to one decimal place. (2 marks)
Solve the differential equation `(dy)/(dx) = (−x)/(1 + y^2)` with the condition `y(−1) = 1`. Express your answer in the form `ay^3 + by + cx^2 + d = 0`, where `a`, `b`, `c` and `d` are integers. (2 marks)

Show Answers Only

`2y^3 + 6y + 3x^2 – 11 = 0`

Show Worked Solution

♦♦ Mean mark part (a) 32%.
MARKER’S COMMENT: Solution curve should follow slope ticks and not cross them.

b.	`(1 + y^2)(dy)/(dx)`	`= −x`
	`int 1 + y^2 dy`	`= −int x\ dx`
	`y + (y^3)/3`	`= −(x^2)/2 + C, C ∈ R`

`text(Substituting)\ (-1,1):`

`1 + (1^3)/3`	`= −((−1)^2)/2 + C`
`1 + 1/3`	`= −1/2 + C`
`:. C`	`= 1/2 + 4/3`
	`= 11/6`

`y + 1/3y^3`	`= −1/2x^2 + 11/6`
`6y + 2y^3`	`= −3x^2 + 11`

`:. 2y^3 + 6y + 3x^2 – 11 = 0`

Complex Numbers, SPEC1 VCAA 2017 3

Let `z^3 + az^2 + 6z + a = 0, \ z ∈ C`, where `a` is a real constant.

Given that `z = 1 - i` is a solution to the equation, find all other solutions. (3 marks)

Show Answers Only

`z_2`	`= 1 + i`
`z_3`	`= 2`

Show Worked Solution

`z_1 = 1 – i`

`=>\ z_2 = 1 + i\ \ text{(conjugate pair)}`

`=>\ z_3 ∈ R`

`z^3 + az^a + 6z + a`

	`= (z – z_1)(z – z_2)(z – z_3)`
	`= (z – (1 – i))(z – (1 + i))(z – z_3)`
	`= ((z – 1) + i)((z – 1) – i)(z – z_3)`
	`= ((z – 1)^2 – i^2)(z – z_3)`
	`= (z^2 – 2z + 1 + 1)(z – z_3)`
	`= (z^2 – 2z + 2)(z – z_3)`
	`= z^3 – (z + z_3)z^2 + (2z_3 + 2)z – 2z_3`

`text(Equating the co-efficients of)\ \ z:`

`6`	`= 2z_3 + 2`
`4`	`= 2z_3`
`z_3`	`= 2`

`:.\ text(Other solutions are:)`

`z_2`	`= 1 + i`
`z_3`	`= 2`

Calculus, SPEC1 2017 VCAA 1

Find the equation of the tangent to the curve given by `3xy^2 - 2y = x` at the point (1, –1). (3 marks)

Show Answers Only

`y = 1/2x – 3/2`

Show Worked Solution

`d/dx(3xy^2) + d/dx(2y)`	`= d/dx(x)`
`d/(dx)(3x) · y^2 + d/(dx)(y^2)(3x) + 2(dy)/(dx)`	`= 1`
`3y^2 + 6xy*(dy)/(dx) + 2(dy)/(dx)`	`= 1`
`text(At)\ \ (1, -1):`
`3(−1)^2 + 6(1)(−1)m + 2m`	`= 1`
`3 – 6m + 2m`	`= 1`
`−4m`	`= −2`
`m`	`= 1/2`

`y+1`	`=1/2(x-1)`
`:. y`	`=1/2x -3/2`

Statistics, SPEC2-NHT 2018 VCAA 19 MC

A local supermarket sells apples in bags that have negligible mass. The stated mass of a bag of apples is 1 kg.

The mass of this particular type of apple is known to be normally distributed with a mean of 115 grams and a standard deviation of 7 grams. A particular bag contains nine randomly selected apples.

The probability that the nine apples in this bag have a total mass of less than 1 kg is

0.0478
0.1132
0.4265
0.5373
0.9522

Show Answers Only

`A`

Show Worked Solution

`M~N (115, 7^2)`

`T = M + M + M + M + M + M + M + M + M`

`T~N (9 xx 115, 9 xx 7^2)~N(1035, 21^2)`

`\text{Pr} (T < 1000) ~~ 0.0478`

`=> A`

Statistics, SPEC2-NHT 2018 VCAA 18 MC

The heights of all six-year-old children in a given population are normally distributed. The mean height of a random sample of 144 six-year-old children from this population is found to be 115 cm.

If a 95% confidence interval for the mean height of all six-year-old children is calculated to be (113.8, 116.2) cm, the standard deviation used in this calculation is closest to

A. 1.20

B. 7.35

C. 15.09

D. 54.02

E. 88.13

Show Answers Only

`B`

Show Worked Solution

`M`	`=(116.2 – 113.8)/2`
`(1.96 xx sigma)/sqrt 144`	`= 1.2`
`sigma`	`= (1.2 xx 12)/1.96`
	`= 7.35`

`=> B`

Calculus, SPEC2-NHT 2018 VCAA 17 MC

An object travels in a straight line relative to an origin `O`.

At time `t` seconds its velocity, `v` metres per second, is given by

`v(t) = {(sqrt(4 - (t - 2)^2), text(,) quad 0 <= t <= 4), (-sqrt(9 - (t - 7)^2), text(,) quad 4 < t <= 10):}`

The graph of `v(t)` is shown below.

The object will be back at its initial position when `t` is closest to

A. 4.0

B. 6.5

C. 6.7

D. 6.9

E. 7.0

Show Answers Only

`C`

Show Worked Solution

`text(Solve for)\ k:`

`int_0^4 sqrt(4 – (t – 2)^2)\ dt= int_4^k sqrt(9 -(t – 7)^2)\ dt`

`k~~ 6.7`

`=> C`

Mechanics, SPEC2-NHT 2018 VCAA 16 MC

A body of mass 2 kg is moving in a straight line with constant velocity when an external force of `8 N` is applied in the direction of motion for `t` seconds.

If the body experiences a change in momentum of 40 kg ms¯¹, then `t` is

A. 3

B. 4

C. 5

D. 6

E. 7

Show Answers Only

`C`

Show Worked Solution

`sum F = 8 = 2a\ \ =>\ \ a = 4`

`m(v – u)`	`= 40`
`2(v – u)`	`= 40`
`v – u`	`= 20`

`v`	`= u + at`
`v – u`	`= at`
`20`	`= 4t`
`:. t`	`= 5`

`=> C`

Mechanics, SPEC2-NHT 2018 VCAA 15 MC

An 80 kg person stands in an elevator that is accelerating downwards at `1.2\ text(ms)^(-2)`.

The reaction force of the elevator floor on the person, in newtons, is

A. 688

B. 704

C. 784

D. 880

E. 896

Show Answers Only

`A`

Show Worked Solution

`sum F`	`=80 text(g) – R`
`80 text(g) – R`	`= 80 xx 1.2`
`:. R`	`= 80xx9.8 – 80 xx 1.2`
	`= 688`

`=> A`

Mechanics, SPEC2-NHT 2018 VCAA 14 MC

The diagram above shows a particle at `O` in equilibrium in a plane under the action of three forces of magnitudes `P, Q` and `R`.

Which one of the following statements is false?

A. `R = Q sin (60^@)`

B. `Q = R sin (60^@)`

C. `P = R sin(30^@)`

D. `Q cos (60^@) = P cos (30^@)`

E. `P cos (60^@) + Q cos (30^@) = R`

Show Answers Only

`A`

Show Worked Solution

`text(By elimination:)`

`sin 60^@`	`= Q/R`
`Q`	`= R\ sin 60^@ \ \ text{(B correct)}`
`sin30^@`	`= P/R`
`P`	`= R\ sin30^@\ \ text{(C correct)}`

`text(Equating horizontal forces:)`

`Q\ cos 60^@ = P\ cos 30^@\ \ text{(D correct)}`

`text(Equating vertical forces:)`

`P\ cos 60^@ + Q\ cos 30^@ = R\ \ text{(E correct)}`

`=> A`

Vectors, SPEC2-NHT 2018 VCAA 12 MC

In the diagram above, `LOM` is a diameter of the circle with centre `O`.

`N` is a point on the circumference of the circle.

If `underset~r = vec(ON)` and `underset ~s = vec(MN)`, then `vec(LN)` is equal to

`2 underset ~r - 2 underset ~s`
`underset ~r - 2 underset ~s`
`underset ~r + 2 underset ~s`
`2 underset ~r + underset ~s`
`2 underset ~r - underset ~s`

Show Answers Only

`E`

Show Worked Solution

`vec(LN)`	`= vec(LM) + vec(MN)`
`vec(LN)`	`= vec(LO) + vec(ON)`
	`= 1/2\ vec(LM) + vec(ON)`

`:. vec(LM) + underset~s`	`= 1/2\ vec(LM) + underset~r`
`1/2\ vec(LM)`	`= underset ~r – underset~s`
`vec(LM)`	`= 2 underset~r – 2 underset~s`

`:. vec(LN)`	`= 2 underset~r – 2 underset~s + underset~s`
	`= 2 underset~r – underset~s`

`=> E`

Calculus, SPEC2-NHT 2018 VCAA 10 MC

The graph of an antiderivative of a function `g` is shown below.

Which one of the following could best represent the graph of `g`?

Show Answers Only

`E`

Show Worked Solution

`=> E`

Calculus, SPEC2-NHT 2018 VCAA 8 MC

Using a suitable substitution, `int_1^2(3/(2 + (4x + 1)^2))\ dx` can be expressed as

A. `3/4 int_1^2 (1/(2 + u^2))\ du`

B. `3/4 int_5^9 (1/(2 + u^2))\ du`

C. `3 int_5^9 (1/(2 + u^2))\ du`

D. `3 int_1^2 (1/(2 + u^2))\ du`

E. `-12 int_9^5 (1/(2 + u^2))\ du`

Show Answers Only

`B`

Show Worked Solution

`u = 4x + 1 qquad qquad qquad u(1) = 5`

`(du)/(dx) = 4 qquad qquad qquad qquad \ \ u(2) = 9`

`3/4 \ du =3\ dx`

`int_1^2(3/(2 + (4x + 1)^2))\ dx`

`=int_5^9 ((3/4)/(2 + u^2))\ du`

`= 3/4 int_5^9 (1/(2 + u^2))\ du`

`=> B`

Complex Numbers, SPEC2-NHT 2018 VCAA 6 MC

Given that `(z - 3i)` is a factor of `P(z) = z^3 + 2z^2 + 9z + 18`, which one of the following statements is false?

`P(3i) = 0`
`P(-3i) = 0`
`P(z)` has three linear factors over `C`
`P(z)` has no real roots
`P(z)` has two complex conjugate roots

Show Answers Only

`D`

Show Worked Solution

`z – 3i\ \ text(is a factor) \ => \ z + 3i\ text(is a factor)`

`text(deg) (P) = 3`

`text(Complex roots occur in conjugate pairs.)`

`text(S)text(ince 3 roots → 3rd root must be real.)`

`=> D`

Graphs, SPEC2-NHT 2018 VCAA 3 MC

The implied domain of the function with rule `f(x) = (3x)/(pi/2 - arccos (2 - x))` is

A. `[1, 3]`

B. `[-1, 1]`

C. `[0, 1) uu (1, 2]`

D. `[-1, 0) uu (0, 1]`

E. `[1, 2) uu (2, 3]`

Show Answers Only

`E`

Show Worked Solution

`-1 <= 2 -x <= 1`

`-3 <= – x <= – 1`

`3 >= x >= 1\ \ …\ (1)`

`(2):`

`pi/2 != cos^(-1) (2 – x)`

`0 != 2 – x`

`x != 2\ \ …\ (2)`

`(1) nn (2)`

`1 <= x <= 3 nn x != 2`

`:. x in [1, 2) uu (2, 3]`

`=> E`

Algebra, SPEC2-NHT 2018 VCAA 2 MC

Let `f(x) = (sqrt(x + 1))/x` and `g(x) = tan^2 (x)`, where `0 < x < pi/2`.

`f(g(x))` is equal to

`sin (x) sec^2 (x)`
`sec (x) tan^2 (x)`
`cos (x) cot^2 (x)`
`cos (x) text(cosec)^2 (x)`
`text(cosec) (x) cos^2 (x)`

Show Answers Only

`D`

Show Worked Solution

`f(tan^2 (x))`	`= (sqrt (tan^2(x) + 1))/(tan^2(x))`
	`= (sqrt(sec^2 (x)))/(tan^2(x))`

`0 < x < pi/2 => sqrt (sec^2(x)) = sec(x)`

`f(g(x))`	`= (sec (x))/(tan^2(x))`
	`= (cos^2(x))/(sin^2(x) cos(x))`
	`= (cos(x))/(sin^2(x))`
	`= cos(x) text(cosec)^2 (x)`

`=> D`

Graphs, SPEC2-NHT 2018 VCAA 1 MC

Let `f(x) = text(cosec) (x)`. The graph of `f` is transformed by:

a dilation by a factor of 3 from the `x`-axis, followed by
translation of 1 unit horizontally to the right, followed by
a dilation by a factor of `1/2` from the `y`-axis

The rule of the transformed graph is

`g(x) = 2\ text(cosec) (3x + 1)`
`g(x) = 3\ text(cosec) (2x - 1)`
`g(x) = 3\ text(cosec) (2 (x - 1))`
`g(x) = 2\ text(cosec) (x/3 - 1)`
`g(x) = 3\ text(cosec) ((x - 1)/2)`

Show Answers Only

`B`

Show Worked Solution

`f_1 (x) = 3\ text(cosec) (x)`

`f_2 (x) = 3\ text(cosec) (x – 1)`

`f_3 (x) = g(x) = 3\ text(cosec) (2x – 1)`

`=> B`

Calculus, SPEC1-NHT 2018 VCAA 9

i. Given that `cot(2 theta) = a`, show that `tan^2(theta) + 2a tan(theta) - 1 = 0`. (2 marks)
ii. Show that `tan(theta) = -a +- sqrt(a^2 + 1)`. (1 mark)
iii. Hence, show that `tan(pi/12) = 2 - sqrt 3`, given that `cot(2 theta) = sqrt 3`, where `theta in (0, pi)`. (1 mark)
Find the gradient of the tangent to the curve `y = tan (theta)` at `theta = pi/12`. (2 marks)
A solid of revolution is formed by rotating the region between the graph of `y = tan(theta)`, the horizontal axis, and the lines `theta = pi/12` and `theta = pi/3` about the horizontal axis.

Find the volume of the solid of revolution. (3 marks)

Show Answers Only

1. `text(Proof)\ \ text{(See Worked Solutions)}`
2. `text(Proof)\ \ text{(See Worked Solutions)}`
3. `text(Proof)\ \ text{(See Worked Solutions)}`
`8 – 4 sqrt 3`
`pi (2 sqrt 3 – 2 – pi/4)`

Show Worked Solution

a.i.	`1/(tan 2 theta)`	`= a`
	`1`	`= a tan (2 theta)`
	`1`	`= a ((2 tan (theta))/(1 – tan^2(theta)))`

	`1/a (1 – tan^2 (theta))`	`= 2 tan (theta)`
	`1 – tan^2 (theta)`	`= 2a tan (theta)`
	`:. tan^2 (theta) + 2 a tan (theta) – 1`	`=0\ \ text(… as required)`

a.ii.	`[tan^2 (theta) + 2a tan (theta) + a^2] – a^2 – 1`	`= 0`
	`(tan (theta) + a)^2`	`= a^2 + 1`
	`tan (theta) + a`	`= +- sqrt(a^2 + 1)`
	`:. tan (theta)`	`= -a +- sqrt(a^2 + 1)`

a.iii. `theta in (0, pi) \ => \ 2 theta in (0, 2 pi)`

`text(S)text(ince)\ \ cot(2theta)=sqrt3\ \ \ =>\ \ \ tan(2theta)=1/sqrt3`

	`:. 2 theta`	`= pi/6, pi + pi/6`
	`theta`	`= pi/12, (7 pi)/12`
	`tan (theta)`	`= – sqrt 3 +- sqrt((sqrt 3)^2 + 1)`
		`= – sqrt 3 +- sqrt(3 + 1)`
		`= – sqrt 3 +- 2`

`:. tan (pi/12) = 2 – sqrt 3,\ \ \ \ (tan (pi/12) > 0)`

b.	`y`	`=tan(theta)`
	`y prime`	`= sec^2 (theta)`
		`= 1 + tan^2 (theta)`

`y prime (pi/12)`	`= 1 + tan^2 (pi/12)`
	`= 1 + (2 – sqrt 3)^2`
	`= 1 + 4 – 4 sqrt 3 + 3`
	`= 8 – 4 sqrt 3`

c.	`V`	`= pi int_(pi/12)^(pi/3) y^2\ d theta`
		`= pi int_(pi/12)^(pi/3) tan^2 (theta)\ d theta`
		`= pi int_(pi/12)^(pi/3) (1 + tan^2 (theta) – 1)\ d theta`
		`= pi int_(pi/12)^(pi/3) (sec^2 (theta) – 1)\ d theta`
		`= pi [tan (theta) – theta]_(pi/12)^(pi/3)`
		`= pi (tan (pi/3) – pi/3 – (tan (pi/12) – pi/12))`
		`= pi (sqrt 3 – (4 pi)/12 – (2 – sqrt 3) + pi/12)`
		`= pi (2 sqrt 3 – 2 – pi/4)\ \ text(u³)`

Complex Numbers, SPEC1-NHT 2018 VCAA 8

A circle in the complex plane is given by the relation `|z - 1 - i| = 2, \ z in C`.

Sketch the circle on the Argand diagram below. (1 mark)

i. Write the equation of the circle in the form `(x - a)^2 + (y - b)^2 = c` and show that the gradient of a tangent to the circle can be expressed as `(dy)/(dx) = (1 - x)/(y - 1)`. (2 marks)
ii. Find the gradient of the tangent to the circle where `x = 2` in the first quadrant of the complex plane. (1 mark)
Find the equations of all rays that are perpendicular to the circle in the form `text(Arg) (z) = alpha`. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
1. `text(Proof)\ \ text{(See Worked Solutions)}`
2. `(-1)/sqrt 3`
`text(Arg) (z) = pi/4; quad text(Arg) (z) = (-3 pi)/4`

Show Worked Solution

b.i.	`(x – 1)^2 + (y – 1)^2`	`= 4`
	`2(x – 1) + d/(dx) ((y – 1)^2)`	`= 0`
	`2(x – 1) + 2(y-1)*(dy)/(dx)`	`= 0`
	`2 (y-1)*(dy)/(dx)`	`= -2(x – 1)`
	`(dy)/(dx)`	`= (-(x – 1))/(y-1)`
		`= (1 – x)/(y – 1)`

b.ii.	`(2 – 1)^2 + (y – 1)^2`	`= 4`
	`1 + (y – 1)^2`	`= 4`
	`(y – 1)^2`	`= 3`
	`y`	`= 1 + sqrt 3\ \ \ (y > 0)`

	`(dy)/(dx)\|_{(2, 1 + sqrt 3)}`	`= (1 – 2)/(1 + sqrt 3 – 1`
		`= (-1)/sqrt 3`

c.	`P (0, 0) quad C(1, 1)`
	`-> y = x`
	`:. alpha = pi/4, quad (-3 pi)/4`

`text(Arg) (z) = pi/4`

`text(Arg) (z) = (-3 pi)/4`

Calculus, SPEC1-NHT 2018 VCAA 7

Find `d/(dx) ((1 - x^2)^(1/2))`. (2 marks)
Hence, find the length of the curve specified by `y = sqrt (1 - x^2)` from `x = 1/2` to `x = sqrt 3/2`.

Give your answer in the form `k pi, k in R`. (2 marks)

Show Answers Only

`(-x)/sqrt(1 – x^2)`
`pi/6`

Show Worked Solution

a. `d/(dx) ((1 – x^2)^(1/2))`

`= (1/2 (-2x) (1 – x^2)^(-1/2))`

`= -x(1 – x^2)^(-1/2)`

`= (-x)/sqrt (1 – x^2)`

b.	`l`	`= int_(1/2)^(sqrt 3/2) sqrt(1 + ((-x)/sqrt(1 – x^2))^2)\ dx`
		`= int_(1/2)^(sqrt 3/2) sqrt((1 – x^2 + x^2)/(1 – x^2))\ dx`
		`= int_(1/2)^(sqrt 3/2) sqrt(1/(1 – x^2))\ dx`
		`= int_(1/2)^(sqrt 3/2) 1/sqrt(1 – x^2)\ dx`
		`= [sin^(-1) (x)]_(1/2)^(sqrt 3/2)`
		`= sin^(-1) (sqrt 3/2) – sin^(-1) (1/2)`
		`= pi/3 – pi/6`
		`= pi/6`

Calculus, SPEC1-NHT 2018 VCAA 6

Given that `y = (x-1)e^(2x)` is a solution to the differential equation `a(d^2y)/(dx^2) + b (dy)/(dx) = y`, find the values of `a` and `b`, where `a` and `b` are real constants. (4 marks)

--- 9 WORK AREA LINES (style=lined) ---

Show Answers Only

`a = -1/4, b = 1`

Show Worked Solution

`y`	`= (x-1)e^(2x)`
`y^{′}`	`= e^(2x) + 2e^(2x) (x-1)`
	`= e^(2x) + 2y`
`y^{″}`	`= 2e^(2x) + 2y^{′}`
	`= 2e^(2x) + 2e^(2x) + 4e^(2x) (x-1)`
	`= 4e^(2x) + 4e^(2x) (x-1)`
	`= 4e^(2x) (1 + x-1)`
	`= 4xe^(2x)`

`ay^{″} + by^{′} = y`

`4axe^(2x) + be^(2x) (1 + 2(x-1))`	`= (x-1)e^(2x)`
`e^(2x)(4ax + b (2x-1)-(x-1))`	`= 0`
`4ax + b + 2bx-2b-x + 1`	`= 0,\ \ \ (e^(2x) != 0)`
`(4a + 2b-1) x + 1-b`	`= 0x + 0`

`:.b = 1`

`4a + 2-1 = 0`

`:. a = -1/4`

Calculus, SPEC1-NHT 2018 VCAA 5

Evaluate `int_1^(2 sqrt(3) - 1) (1/(x^2 + 2x + 5))\ dx`. (4 marks)

Show Answers Only

`pi/24`

Show Worked Solution

` int_1^(2 sqrt(3) – 1) (1/(x^2 + 2x + 5))\ dx`

`= int_1^(2 sqrt(3) – 1) 1/((x + 1)^2 + 4)\ dx`

`text(Let)\ \ u = x + 1, quad (du)/(dx) = 1\ \ =>\ \ du=dx`

`text(Limits:)\ \ u(1) = 2, quad u(2 sqrt 3 – 1) = 2 sqrt 3`

`:. int_1^(2 sqrt(3) – 1) 1/(x^2 + 2x + 5)\ dx`

`= int_2^(2 sqrt 3) (1/(u^2 + 4))\ du`

`= 1/2 int_2^(2 sqrt 3) 2/(u^2 + 4)\ du`

`= 1/2 [tan^(-1) (u/2)]_2^(2 sqrt 3)`

`= 1/2(tan^(-1) (sqrt 3) – tan^(-1)(1))`

`= 1/2 (pi/3 – pi/4)`

`= 1/2 ((4 pi – 3 pi)/12)`

`= pi/24`

Statistics, SPEC1-NHT 2018 VCAA 4

Throughout this question, use an integer multiple of standard deviations in calculations.

The standard deviation of all scores on a particular test is 21.0

From the results of a random sample of `n` students, a 95% confidence interval for the mean score for all students was calculated to be `(44.7, 51.7)`.

Calculate the mean score and the size of this random sample. (2 marks)
Determine the size of another random sample for which the endpoints of the 95% confidence interval for the population mean of the particular test would be 1.0 either side of the sample mean. (2 marks)

Show Answers Only

`bar x = 48.2; quad n = 144`
`1764`

Show Worked Solution

a. `bar x = (51.7 + 44.7)/2 = 96.4/2 = 48.2`

`text{Difference: z-score –2 to 0 = 3.5}`

`2 xx 21/sqrt n`	`= 7/2`
`7 sqrt n`	`= 2 xx 2 xx 21`
`sqrt n`	`= (2 xx 2 xx 7 xx 3)/7`
	`=12`
`:.n`	`=144`

b.	`2 xx 21/sqrtn`	`= 1`
	`sqrt n`	`= 42`
	`n`	`= 42^2`
		`= (40 + 2)^2`
		`= 1600 + 160 + 4`
		`= 1764`

Algebra, SPEC1-NHT 2018 VCAA 3

Find `sin(t)` given that `t = arccos (12/13) + arctan (3/4)`. (3 marks)

Show Answers Only

`56/65`

Show Worked Solution

`sin (t)`	`= sin (cos^(-1) (12/13) + tan^(-1)(3/4))`
	`= sin (cos^(-1) (12/13)) cos (tan^(-1) (3/4)) + sin (tan^(-1)(3/4)) cos (cos^(-1)(12/13))`

`cos^(-1) (12/13) in (0, pi/2), quad tan^(-1)(3/4) in (0, pi/2)`

`:. sin (t)`	`= 5/13 xx 4/5 + 3/5 xx 12/13`
	`= 20/65 + 36/65`
	`= 56/65`

Vectors, SPEC1-NHT 2018 VCAA 2

Let `underset ~a = 3 underset ~i - 2 underset ~j + m underset ~k` and `underset ~b = 2 underset ~i - underset ~j + 3 underset ~k`, where `m in R`.

Find the value(s) of `m` such that the magnitude of the vector resolute of `underset ~a` parallel to `underset ~b` is equal to `sqrt 14`. (3 marks)

Show Answers Only

`m = (-22)/3, 2`

Show Worked Solution

`underset ~a ⋅ underset ~ hat b = (3 xx 2 +(-2 xx – 1) + m xx 3)/sqrt(2^2 + (-1)^2 + 3^2) = +- sqrt 14`

`(8 + 3m)/sqrt 14`	`= +- sqrt 14`
`8 + 3m`	`= +- 14`
`3m`	`= -8 +- 14`
`m`	`= (-8 +- 14)/3`
`:. m`	`= – 22/3, 2`

Mechanics, SPEC1-NHT 2018 VCAA 1

A light inextensible string hangs over a frictionless pulley connecting masses of 3 kg and 7 kg, as shown below.

Draw all of the forces acting on the two masses on the diagram above. (1 mark)
Calculate the tension in the string. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`(21g)/5 N`

Show Worked Solution

b.	`sum F = 7g – 3g = 4g`	` = (7 + 3)a`
	`a = (4g)/10= (2g)/5`
	`sum F_3 = T – 3g`	`= 3a`
	`T – 3g`	`= (6g)/5`
	`T`	`= 3g + (6g)/5`
		`= (21g)/5\ \ N`

Statistics, SPEC2 2018 VCAA 20 MC

The scores on the Mathematics and Statistics tests, expressed as percentages, in a particular year were both normally distributed. The mean and the standard deviation of the Mathematics test scores were 71 and 10 respectively, while the mean and the standard deviation of the statistics test scores were 75 and 7 respectively.

Assuming the sets of tests scores were independent of each other, the probability, correct to four decimal places, that a randomly chosen Mathematics score is higher than a randomly chosen Statistics score is

0.2877
0.3716
0.4070
0.7123
0.9088

Show Answers Only

`B`

Show Worked Solution

`M~N (71 , 10^2)`

`S~N (75, 7^2)`

`text(Pr)(M > S) = text(Pr)(M-S > 0)`

`M-S = X~N(71-75, 1^2 xx 10^2 + (-1)^2 xx 7^2)`

`X~N(-4, 149)`

`text(Pr)(M > S) = text(Pr)(X > 0) ~~ 0.3716`

`=> B`

Statistics, SPEC2 2018 VCAA 19 MC

The gestation period of cats is normally distributed with mean `mu = 66` days and variance `sigma^2 = 16/9`.

The probability that a sample of five cats chosen at random has an average gestation period greater than 65 days is closest to

0.5000
0.7131
0.7734
0.8958
0.9532

Show Answers Only

`E`

Show Worked Solution

`G~N (66 , 16/9)`

`bar G~N (66, 16/(9 xx 5))`

`text(Pr)(bar G > 65) ~~ 0.9532`

`=> E`

Mechanics, SPEC2 2018 VCAA 16 MC

The diagram below shows a mass being acted on by a number of forces whose magnitudes are labelled. All forces are measured in newtons and the system is in equilibrium.

The value of `F_2` is

A. `sqrt 2/2 (8 + 3 sqrt 3)`

B. `(11 sqrt 2)/2`

C. `(3 sqrt 2)/2`

D. `7.78`

E. `7.0`

Show Answers Only

`B`

Show Worked Solution

`text(vertical:)\ \ sum F_y = 4 + 3 sin 30^@ +\ ^(−)F_2 sin 45^@ = 0`

`4 + 3/2 – F_2/sqrt 2`	`= 0`
`F_2/sqrt 2`	`= 4 + 3/2`
`:. F_2`	`= sqrt 2 (4 + 3/2)`
	`= (11 sqrt 2)/2`

`=> B`

Statistics, SPEC2 2018 VCAA 18 MC

A 95% confidence interval for the mean height `mu`, in centimetres, of a random sample of 36 Irish setter dogs is `58.42 < mu < 67.31`

The standard deviation of the height of the population of Irish setter dogs, in centimetres, correct to two decimal places, is

A. 2.26

B. 2.27

C. 13.60

D. 13.61

E. 62.87

Show Answers Only

`D`

Show Worked Solution

`M`	`= (67.31 – 58.42)/2`
	`= 8.89/2`
	`= 4.445`

`M`	`= (1.96 sigma)/sqrt 36`
`4.445`	`= (1.96 sigma)/6`
`:. sigma`	`=(6 xx 4.445)/1.96`
	`= 13.61`

`=> D`

Mechanics, SPEC2 2018 VCAA 15 MC

A constant force of magnitude `P` newtons accelerates a particle of mass 8 kg in a straight line from a speed of 4 ms¯¹ to a speed of 20 ms¯¹ over a distance of 15 m.

The magnitude of `P` is

A. 9.8

B. 12.5

C. 12.8

D. 100

E. 102.4

Show Answers Only

`E`

Show Worked Solution

`u = 4, quad v = 20, quad s = 15`

`v^2`	`= u^2 + 2as`
`20^2`	`= 4^2 + 30a`
`400`	`= 16 + 30a`
`384`	`= 30a`
`a`	`= 64/5`

`:. P`	`= 8a`
	`= (8 xx 64)/5`
	`= 102.4`

`=> E`