Band 4 – Page 71

Probability, MET2 2012 VCAA 13 MC

`A` and `B` are events of a sample space `S.`

`text(Pr)(A nn B) = 2/5` and `text(Pr)(A nn B prime) = 3/7`

`text(Pr)(B prime | A)` is equal to

`6/35`
`15/29`
`14/35`
`29/35`
`2/3`

Show Answers Only

`B`

Show Worked Solution

`text(Pr)(B prime \| A)`	`= (text(Pr)(B prime nn A))/(text(Pr)(A))`
	`= (text(Pr)(B prime nn A))/(text(Pr)(B prime nn A)+ text(Pr)(A nn B))`
	`= (3/7)/(3/7 + 2/5)`
	`= 15/29`

`=> B`

Statistics, MET2 2012 VCAA 11 MC

The weights of bags of flour are normally distributed with mean 252 g and standard deviation 12 g. The manufacturer says that 40% of bags weigh more than `x` g.

The maximum possible value of `x` is closest to

`249.0`
`251.5`
`253.5`
`254.5`
`255.0`

Show Answers Only

`E`

Show Worked Solution

`X = text(weight), X ∼ N(252, 12^2)`

`Pr(X < x)`	`= 0.6`
`:. x`	`= 255.04`

`[text(CAS: invNorm) (0.6,252,12)]`

`=> E`

Calculus, MET2 2012 VCAA 10 MC

The average value of the function `f: [0, 2 pi] -> R,\ f(x) = sin^2(x)` over the interval `[0, a]` is 0.4.

The value of `a`, to three decimal places, is

`0.850`
`1.164`
`1.298`
`1.339`
`4.046`

Show Answers Only

`C`

Show Worked Solution

`1/(a – 0) int_0^a(sin^2x)\ dx`	`= 0.4`
`[x- (sin 2x)/2]_0^a`	`=0.4a`
`a-(sin 2a)/2`	`=0.4a`
`:. a`	`= 1.298, quad a∈ (0, 2pi)`

`=> C`

CORE*, FUR2 2014 VCAA 1

The adult membership fee for a cricket club is $150.

Junior members are offered a discount of $30 off the adult membership fee.

Write down the discount for junior members as a percentage of the adult membership fee. (1 mark)

Adult members of the cricket club pay $15 per match in addition to the membership fee of $150.

If an adult member played 12 matches, what is the total this member would pay to the cricket club? (1 mark)

If a member does not pay the membership fee by the due date, the club will charge simple interest at the rate of 5% per month until the fee is paid.

Michael paid the $150 membership fee exactly two months after the due date.

Calculate, in dollars, the interest that Michael will be charged. (1 mark)

The cricket club received a statement of the transactions in its savings account for the month of January 2014.

The statement is shown below.

1. Calculate the amount of the withdrawal on 17 January 2014. (1 mark)
2. Interest for this account is calculated on the minimum balance for the month and added to the account on the last day of the month.
  
  What is the annual rate of interest for this account?
  
  Write your answer, correct to one decimal place. (1 mark)

Show Answers Only

`text(20%)`
`$330`
`$15`
1. `$17\ 000`
2. `3.5`

Show Worked Solution

a. `text(Discount for junior members)`

`= 30/150 xx 100text(%)`

`= 20text(%)`

b.	`text(Match Payments)`	`= 12 xx 15=$180`

`:.\ text(Total paid to the club)`

`= 150 + 180`

`= $330`

c.	`I`	`= (PrT)/100`
		`= (150 xx 5 xx 2)/100`
		`= $15`

d.i. `text(Withdrawal on 17-Jan)`

`= 59\ 700 – 42\ 700`

`= $17\ 000`

d.ii. `text(Minimum Jan balance) = $42\ 700`

`47\ 200 xx r xx 1/12`	`= 125.12`
`:. r`	`= (125.12 xx 12)/(42\ 700)`
	`= 0.0351…`

`:.\ text(Annual interest rate) = 3.5text(%)`

Calculus, MET2 2012 VCAA 7 MC

The temperature, `T^@C`, inside a building `t` hours after midnight is given by the function

`f: [0, 24] -> R,\ T(t) = 22 - 10\ cos (pi/12 (t - 2))`

The average temperature inside the building between 2 am and 2 pm is

`10°text(C)`
`12°text(C)`
`20°text(C)`
`22°text(C)`
`32°text(C)`

Show Answers Only

`D`

Show Worked Solution

`text(Period) = (2pi)/n = (2pi)/(pi/12) = 24`

`text(At 2 am,)\ \ t=2,`

`T(2) = 22 – 10\ cos (0) = 12`

`text(At 2 pm,)\ \ t=14,`

`T(14) = 22 – 10\ cos (pi) = 32`

`text(Symmetry of graph means the average)`

`text(temperature occurs at)\ \ t=8:`

`T(8) = 22 – 10\ cos ((pi)/2) = 22`

`=> D`

Graphs, MET2 2012 VCAA 3 MC

The range of the function `f: text{[−2, 3)} -> R,\ f(x) = x^2 - 2x - 8` is

A. `R`

B. `text{(−9, −5]}`

C. `text{(−5, 0)}`

D. `text{[−9, 0]}`

E. `text{[−9, −5)}`

Show Answers Only

`D`

Show Worked Solution

`text(Sketching the quadratic:)`

`:.\ text(Range) = [−9,0]`

`=> D`

Graphs, MET2 2012 VCAA 5 MC

Let the rule for a function `g` be `g (x) = log_e ((x - 2)^2).` For the function `g`, the

maximal domain `= R^+` and range `= R`
maximal domain `= R text(\{2})` and range `= R`
maximal domain `= R text(\{2})` and range `=\ text{(−2, ∞)`
maximal domain `= [2, oo)` and range `= (0, oo)`
maximal domain `= [2, oo)` and range `= [0, oo)`

Show Answers Only

`B`

Show Worked Solution

`text(Domain:)\ (x – 2)^2 > 0`

`:. x != 2`

`text(Range: sketch graph)`

`:. text(Range) = R`

`=> B`

Algebra, STD2 A1 SM-Bank 9

The volume of a sphere is given by `V = 4/3 pi r^3` where `r` is the radius of the sphere.

If the volume of a sphere is `220\ text(cm)^3`, find the radius, to 1 decimal place. (3 marks)

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

Show Answers Only

`3.7\ \ text{cm (to 1 d.p.)}`

Show Worked Solution

`V`	`= 4/3 pi r^3`
`3V`	`= 4 pi r^3`
`r^3`	`= (3V)/(4 pi)`

`text(When)\ \ V = 220`

`r^3`	`= (3 xx 220)/(4 pi)`
	`= 52.521…`
`:. r`	`=root3 (52.521…)`
	`= 3.744…\ \ \ text{(by calc)}`
	`= 3.7\ \ text{cm (to 1 d.p.)}`

Algebra, MET2 2013 VCAA 13 MC

If the equation `f(2x) - 2f(x) = 0` is true for all real values of `x`, then the rule for `f` could be

`x^2/2`
`sqrt (2x)`
`2x`
`log_e (x/2)`
`x - 2`

Show Answers Only

`C`

Show Worked Solution

`text(We need)\ \ f(2x)=2\ f(x),`

`text(Consider)\ C,`

`f(x)`	`=2x,`
`f(2x)`	`= 2(2x)`
	`= 2\ f(x)`

`text(CAS can be used to quickly prove other answers)`

`text(do not satisfy equation.)`

`=> C`

Probability, MET2 2013 VCAA 9 MC

Harry is a soccer player who practises penalty kicks many times each day.

Each time Harry takes a penalty kick, the probability that he scores a goal is 0.7, independent of any other penalty kick.

One day Harry took 20 penalty kicks.

Given that he scored at least 12 goals, the probability that Harry scored exactly 15 goals is closest to

A. `0.1789`

B. `0.8867`

C. `0.8`

D. `0.6396`

E. `0.2017`

Show Answers Only

`E`

Show Worked Solution

`text(Let)\ \ X =\ text(number of goals,)`

`X ∼\ text(Bi)(20, 0.7)`

`text(Pr) (X = 15 | X >= 12)`

`= (text(Pr) (X = 15))/(text(Pr) (X >= 12))`

`~~ 0.2017`

`=> E`

Calculus, MET2 2013 VCAA 6 MC

For the function `f(x) = sin (2 pi x) + 2x,` the average rate of change for `f(x)` with respect to `x` over the interval `[1/4, 5]` is

A. `0`

B. `34/19`

C. `7/2`

D. `(2 pi + 10)/4`

E. `23/4`

Show Answers Only

`B`

Show Worked Solution

`text(Average rate of change for)\ \f(x)\ \ text(over)\ \ [1/4, 5]`

`= (f(5) – f(1/4))/(5 – 1/4)`

`= (sin (10 pi) + 10 – (sin\ pi/2 +1/2))/(19/4)`

`= 34/19`

`=> B`

Polynomials, EXT2 2015 HSC 16b

Let `n` be a positive integer.

By considering `(cos alpha + i sin alpha)^(2n)`, show that
`cos(2n alpha) = cos^(2n) alpha - ((2n), (2)) cos^(2n - 2) alpha sin^2 alpha + ((2n), (4)) cos^(2n - 4) alpha sin^4 alpha - …`
1. `+ … + (-1)^(n - 1) ((2n), (2n - 2)) cos^2 alpha sin^(2n - 2) alpha + (-1)^n sin^(2n) alpha.`
Let `T_(2n) (x) = cos(2n cos^-1 x)`, for `-1 <= x <= 1.` (2 marks)
Show that
`T_(2n)(x) = x^(2n) - ((2n), (2)) x^(2n - 2)(1 - x^2) + ((2n), (4)) x^(2n - 4) (1 - x^2)^2 +`
1. `… + (-1)^n (1 - x^2)^n.` (2 marks)
By considering the roots of `T_(2n) (x)`, find the value of
1. `cos(pi/(4n)) cos((3 pi)/(4n)) …\ cos (((4n - 1) pi)/(4n)).` (3 marks)
Prove that
1. `1 - ((2n), (2)) + ((2n), (4)) - ((2n), (6)) + … + (-1)^n ((2n), (2n)) = 2^n cos ((n pi)/2).` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(-1)^n/2^(2n – 1)`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `text(By De Moivre)`

`(cos alpha + i sin alpha)^(2n)=cos (2n alpha) + i sin (2n alpha)`

`text(By the Binomial Theorem)`

`(cos alpha + i sin alpha)^(2n)`

`=cos^(2n) alpha + ((2n), (1)) cos^(2n – 1) alpha (i sin alpha) + ((2n), (2)) cos^(2n – 2) alpha (isin alpha)^2 +`

`… ((2n), (2n – 2)) cos^2 alpha (isin alpha)^(2n-2) + ((2n), (2n – 1)) cos alpha (i sin alpha)^(2n-1) + (i sin alpha)^(2n)`

`text(Equating real parts,)`

`cos(2n alpha) = cos^(2n) alpha – ((2n), (2)) cos^(2n – 2) alpha sin^2 alpha + … `

`+ (-1)^(n – 1) ((2n), (2n – 2)) cos^2 alpha sin^(2n – 2) alpha + (-1)^n sin^(2n) alpha`

(ii) `T_(2n) (x) = cos (2n cos^-1 x)\ \ text(for)\ \ -1 <= x <= 1`

`text(Given)\ \ alpha = cos^-1 x,\ \ =>x = cos alpha`

`text(Substituting)\ \ x = cos alpha\ \ text{into part (i),}`

`cos (2n cos^-1 x)`

`= x^(2n) – ((2n), (2)) x^(2n – 2) (1 – cos^2 alpha) + ((2n), (4)) x^(2n – 4) (1 – cos^2 alpha)^2 +`

`… + (-1)^n (1 – cos^2 alpha)^n`

`:. T_(2n) (x) = x^(2n) – ((2n), (2)) x^(2n – 2) (1 – x^2) + ((2n), (4)) x^(2n – 4)(1 – x^2)^2 +`

`… + (-1)^n (1 – x^2)^n`

♦♦♦ Mean mark 8%!
STRATEGY: Note that finding the correct roots gained 2 full marks in this part.

(iii) `T_(2n) (x) = 0\ \ text(when)`

`cos (2n cos^-1 x)`	`=0`
`2n cos^-1 x`	`=pi/2, (3 pi)/2, (5 pi)/2, … , ((2k + 1) pi)/2,\ \ k = 0, 1, 2 … , (2n – 1)`
`cos^-1 x`	`= ((2k + 1) pi)/(4n),\ \ k = 0, 1, 2 …, (2n – 1)`
`:. x`	`= cos(pi/(4n)), cos((3 pi)/(4n)), … , cos((4n – 1) pi)/(4n)`

`=>T_(2n) (x) = 0\ \ text(has degree)\ \ 2n`

`=>T_(2n) (x) = 0\ \ text(has)\ \ 2n\ \ text(distinct roots)`

`:.\ text(Product of roots of)\ \ T_(2n) (x) = 0\ \ text(is the constant term)`

`text(divided by the coefficient of)\ \ x^(2n).`

`text(Constant term is)\ \ T_(2n) (0)=(-1)^n`

`text(Coefficient of)\ \ x^(2n)\ \ text(is)\ \ (1 + ((2n), (2)) + ((2n), (4)) + … + 1)`

`:.cos(pi/(4n)) cos((3 pi)/(4n)) cos ((5 pi)/(4n)) …\ cos(((4n – 1) pi)/(4n))`

`=((-1)^n)/((1 + ((2n), (2)) + ((2n), (4)) + … + 1))`

`text(*The denominator can be further simplified to)\ \ 2^(2n-1)\ \ text(by)`

`text(using the binomial expansion of)\ \ (1+x)^(2n)`

♦♦ Mean mark 14%.

(iv) `cos(2n cos^-1 x)= cos ((n pi)/2)\ \ text(when)`

`cos^-1 x=pi/4\ \ \ =>x=1/sqrt2`

`text{Using part (ii)}`

`cos((n pi)/2)`	`=(1/sqrt 2)^(2n) – ((2n), (2)) (1/sqrt 2)^(2n – 2) (1/2) +`
	`((2n), (4)) (1/sqrt 2)^(2n – 4) (1/2)^2 – … + (-1)^n (1/2)^n`
	` = 1/2^n – ((2n), (2)) 1/2^(n-2) 1/2^2 + ((2n), (4)) 1/2^(n-4) 1/2^4 – … + (-1)^n 1/2^n`
	`=1/2^n – ((2n), (2)) 1/2^n + ((2n), (4)) 1/2^n – … + (-1)^n 1/2^n`
`:.2^n cos ((n pi)/2)`	`=1 – ((2n), (2)) + ((2n), (4)) – ((2n), (6)) + … + (-1)^n`

Harder Ext 1 Topics, EXT2 2015 HSC 16a

A table has `3` rows and `5` columns, creating `15` cells as shown.
Counters are to be placed randomly on the table so that there is one counter in each cell. There are `5` identical black counters and `10` identical white counters.
Show that the probability that there is exactly one black counter in each column is `81/1001.` (2 marks)
The table is extended to have `n` rows and `q` columns. There are `nq` counters, where `q` are identical black counters and the remainder are identical white counters. The counters are placed randomly on the table with one counter in each cell.
Let `P_n` be the probability that each column contains exactly one black counter.
Show that `P_n = n^q/(((nq), (q))).` (2 marks)
Find `lim_(n -> oo) P_n.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(q!)/q^q`

Show Worked Solution

(i) `text(Arrangements of 5 black counters in 15 cells)`

`=\ ^15C_5`

`text(In each column, there are 3 cells to place a black counter.)`

`:.\ text(Ways to place a black counter in each column)`

`\ =3^5=243`

`:.P text{(1 black counter in each column)`

`=243/(\ ^15C_5`

`=81/1001`

(ii) `text{Similarly to part (i),}`

♦♦ Mean mark 16%.

`text(Arrangements of)\ \ q\ \ text(black counters in)\ \ nq\ \ text(cells)`

`=\ ^(nq)C_q`

`text(In each column, there are)\ \ n\ \ text(cells to place a black counter.)`

`:.\ text(Ways to place a black counter in each of)\ \ q\ \ text(columns)`

`\ =n^q`

`:.P text{(1 black counter in each column)`

`=n^q/(\ ^(nq)C_q)`

♦♦♦ Mean mark 5%.
COMMENT: Lowest State mean mark on the 2015 exam.

(iii)	`lim_(n -> oo) P_n`	`= lim_(n -> oo) n^q/(\ ^(nq)C_q)`
		`= lim_(n -> oo)[n^q xx (q!)/((nq)(nq – 1)…(nq – q + 1))]`
		`= lim_(n -> oo)[(n^q q!)/(n^q q(q – 1/n)(q – 2/n) … (q – (q – 1)/n))]`
		`= lim_(n -> oo)[(q!)/(q(q – 1/n)(q – 2/n) … (q – (q – 1)/n))]`
		`= (q!)/q^q`

Proof, EXT2 P1 2015 HSC 15c

For positive real numbers `x` and `y`, `sqrt (xy) <= (x + y)/2`. (Do NOT prove this.)

Prove `sqrt (xy) <= sqrt ((x^2 + y^2)/2)`, for positive real numbers `x` and `y.` (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Prove `root4(abcd) <= sqrt ((a^2 + b^2 + c^2 + d^2)/4)`, for positive real numbers `a, b, c` and `d.` (2 marks)

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

Show Answers Only

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

Show Worked Solution

i. `text(S)text(ince)\ \ (x – y)^2 = x^2 + y^2 – 2xy`

`and \ (x – y)^2 >= 0`

`0`	`≤x^2 + y^2 – 2xy`
`2xy`	`≤x^2 + y^2`
`:.sqrt (xy)`	`≤sqrt ((x^2 + y^2)/2)`

ii. `sqrt(ab) <= sqrt((a^2 + b^2)/2),\ \ \ \ text{(part (i))}`

♦♦ Mean mark 29%.

`sqrt(cd) <= sqrt((c^2 + d^2)/2),\ \ \ \ text{(part (i))}`

`sqrt(ab) sqrt(cd)`	`<=sqrt((a^2 + b^2)/2)*sqrt((c^2 + d^2)/2)`
	`<=sqrt(((a^2 + b^2)/2) * ((c^2 + d^2)/2))`

`sqrt (xy) <= (x + y)/2\ \ \ \ text{(given):}`
`sqrt(ab) sqrt(cd)`	`<=1/2((a^2 + b^2+c^2+d^2)/2)`
`sqrt(abcd)`	`<=(a^2 + b^2+c^2+d^2)/4`
`:.root4(abcd)`	`<=sqrt((a^2 + b^2+c^2+d^2)/4)`

Proof, EXT2 P1 2015 HSC 15b

Suppose that `x >= 0` and `n` is a positive integer.

Show that `1 - x <= 1/(1 + x) <= 1.` (2 marks)

--- 2 WORK AREA LINES (style=lined) ---
Hence, or otherwise, show that

`1 - 1/(2n) <= n ln (1 + 1/n) <= 1.` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Hence, explain why

`lim_(n -> oo) (1 + 1/n)^n = e.` (1 mark)

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

Show Answers Only

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

Show Worked Solution

i.	`1-x^2`	`<=1\ \ \ text(for)\ x>=0`
	`(1-x)(1+x)`	`<=1`
	`(1-x)`	`<=1/(1+x)`

`text(S)text(ince)\ \ 1 + x >= 1\ \ text(when)\ \ x >= 0`

`=>1/(1 + x) <= 1`

`:. 1 – x <= 1/(1 + x) <= 1.`

♦♦ Mean mark 21%.
STRATEGY: The conversion of the middle term from a fraction into a logarithm should flag the need for integration of each term.

ii.	`int_0^(1/n) (1 – x)\ dx`	`<= int_0^(1/n) (dx)/(1 + x) <= int_0^(1/n) 1\ dx`
	`[x – x^2/2]_0^(1/n)`	`<= [ln (1 + x)]_0^(1/n) <= [x]_0^(1/n)`
	`1/n-1/(2n^2)`	`<= ln (1 + 1/n) <= 1/n`
	`1 – 1/(2n) `	`<= n ln (1 + 1/n) <= 1,\ \ \ \ \ (n>=1)`

iii.	`lim_(n -> oo) (1 – 1/(2n))`	`<= lim_(n -> oo){n ln (1 + 1/n)} <= lim_(n -> oo) (1)`
	`1`	`<= lim_(n -> oo) {l ln (1 + 1/n)} <= 1`

♦♦ Mean mark 29%.

`:. lim_(n -> oo) (n ln (1 + 1/n))`	`=1`
`lim_(n -> oo) ln (1 + 1/n)^n`	`=1`
`:.lim_(n -> oo) (1 + 1/n)^n`	`=e`

Mechanics, EXT2 M1 2015 HSC 15a

A particle `A` of unit mass travels horizontally through a viscous medium. When `t = 0`, the particle is at point `O` with initial speed `u`. The resistance on particle `A` due to the medium is `kv^2`, where `v` is the velocity of the particle at time `t` and `k` is a positive constant.

When `t = 0`, a second particle `B` of equal mass is projected vertically upwards from `O` with the same initial speed `u` through the same medium. It experiences both a gravitational force and a resistance due to the medium. The resistance on particle `B` is `kw^2`, where `w` is the velocity of the particle `B` at time `t`. The acceleration due to gravity is `g`.

Show that the velocity `v` of particle `A` is given by

`1/v = kt + 1/u.` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
By considering the velocity `w` of particle `B`, show that

`t = 1/sqrt(gk) (tan^-1(u sqrt(k/g)) - tan^-1 (w sqrt(k/g))).` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Show that the velocity `V` of particle `A` when particle `B` is at rest is given by

`1/V = 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g)).` (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Hence, if `u` is very large, explain why

`V ~~ 2/pi sqrt(g/k).` (1 mark)

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

Show Answers Only

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

Show Worked Solution

i. `text(Particle)\ A:`

`ddot x`	`= -kv^2`
`(dv)/(dt)`	`= -kv^2`
`(dt)/(dv)`	`=- 1/(kv^2)`
`t`	`= -1/k int 1/v^2\ dv`
`-kt`	`= -1/v + c`
`text(When)\ \ t=0,\ \ v=u\ \ \ \ =>c=1/u`
`-kt`	`= -1/v + 1/u`
`:.1/v`	`= kt + 1/u`

ii. `text(Particle)\ B:`

`ddot x`	`= -g – kw^2`
`(dw)/(dt)`	`= -g – kw^2`
`(dt)/(dw)`	`=-1/(g + kw^2)`
`t`	`= – int (dw)/(g + kw^2)`
	`= -1/k int (dw)/(g/k + w^2)`
	`= -1/k xx 1/sqrt (g/k) tan^-1(w/sqrt (g/k)) + c`
	`= -1/sqrt (gk)\ tan^-1 ((sqrt k w)/sqrt g) + c`

`text(When)\ \ t = 0,\ \ w = u`

`=>c= 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`

`:. t`	`= -1/sqrt (gk) tan^-1 ((sqrt k w)/sqrt g) + 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`
	`=1/sqrt (gk) (tan^-1 (u sqrt(k/g)) – 1/sqrt (gk) tan^-1 (w sqrt (k/g)))`

iii. `B\ \ text(at rest when)\ \ w = 0`

`t = 1/sqrt (gk) (tan^-1 (u sqrt (k/g)))`

`:.1/V`	`= k xx 1/sqrt(gk) tan^-1 (u sqrt (k/g)) + 1/u,\ \ \ \ \ text{(part (i))}`
	`= 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g))`

iv. `1/V = 1/u + sqrt (k/g) tan^-1 (u sqrt (k/g))`

`text(As)\ \ u -> oo,\ \ tan^-1 (u sqrt (k/g)) -> pi/2`

`:.\ text(If)\ \ u\ \ text(is very large,)`

`1/V`	`~~ 0 + sqrt (k/g) xx pi/2`
`:.V`	`~~ 2/pi sqrt (g/k)`

Mechanics, EXT2 2015 HSC 14c

A car of mass `m` is driven at speed `v` around a circular track of radius `r`. The track is banked at a constant angle `theta` to the horizontal, where `0 < theta < pi/2`. At the speed `v` there is a tendency for the car to slide up the track. This is opposed by a frictional force `mu N`, where `N` is the normal reaction between the car and the track, and `mu > 0`. The acceleration due to gravity is `g`.

Show that
`v^2 = rg((tan theta + mu)/(1 - mu tan theta)).` (3 mark)
At the particular speed `V`, where `V^2 = rg`, there is still a tendency for the car to slide up the track.
Using the result from part (i), or otherwise, show that `mu < 1.` (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

`text(Resolving the forces vertically)`

`N cos theta`	`= mg + mu N sin theta`
`N (cos theta – mu sin theta)`	`= mg\ \ \ …\ (1)`

`text(Resolving the forces horizontally)`

`N sin theta + mu N cos theta`	`=(m v^2)/r`
`N (sin theta + mu cos theta)`	`=(m v^2)/r\ \ \ …\ (2)`

`text(Divide)\ \ (2)÷(1)`

`((m v^2)/r)/(mg)`	`=(sin theta + mu cos theta)/(cos theta – mu sin theta)`
`v^2/(rg)`	`=(sin theta + mu cos theta)/(cos theta – mu sin theta)`
`v^2`	`=rg ((sin theta + mu cos theta)/(cos theta – mu sin theta))`
	`= rg ((tan theta + mu)/(1 – mu tan theta))`

(ii) `text(Given that)\ \ V^2=rg`

♦ Mean mark 46%.

`=> (tan theta + mu)/(1 – mu tan theta)=1`

`(tan theta + mu)`	`=(1 – mu tan theta)`
`mu(1+ tan theta)`	`=1-tan theta`
`mu`	`=(1-tan theta)/(1+ tan theta)`
	`=1- (2tan theta)/(1+ tan theta)`

`text(S)text(ince)\ \ tan theta>0\ \ text(for)\ \ 0<theta<pi/2`

`=> (2tan theta)/(1+ tan theta) >0`

`:. mu<1`

Functions, EXT1′ F2 2015 HSC 14b

The cubic equation `x^3 – px + q = 0` has roots `alpha, beta` and `gamma`.

It is given that `alpha^2 + beta^2 + gamma^2 = 16` and `a^3 + beta^3 + gamma^3 = -9`.

Show that `p = 8.` (1 mark)
Find the value of `q.` (2 marks)
Find the value of `alpha^4 + beta^4 + gamma^4.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`3`
`128`

Show Worked Solution

(i) `alpha + beta + gamma = – b/a=0`

`alpha beta + beta gamma + gamma alpha = c/a= -p`

`(alpha + beta + gamma)^2`	`= alpha^2 + beta^2 + gamma^2 + 2(alpha beta + beta gamma + gamma alpha)`
`0`	`= 16 – 2p`
`:. p`	`= 8`

(ii)	`alpha^3 – 8 alpha + q`	`=0\ \ \ …\ (1)`
	`beta^3 – 8 beta + q`	`=0\ \ \ …\ (2)`
	`gamma^3 – 8 gamma + q`	`=0\ \ \ …\ (3)`

`(1)+(2)+(3)`

`(alpha^3 + beta^3 + gamma^3) – 8(alpha + beta + gamma) + 3q = 0`

`-9 – 0 + 3q`	`=0`
`q`	`=3`

COMMENT: Part (iii) proved challenging for many students with the State mean mark just over 50%.

(iii)	`alpha(alpha^3 – 8 alpha + 3)`	`=0\ \ \ …\ (1)`
	`beta(beta^3 – 8 beta + 3)`	`=0\ \ \ …\ (2)`
	`gamma(gamma^3 – 8 gamma + 3)`	`=0\ \ \ …\ (3)`

`(1)+(2)+(3)`

`alpha^4 + beta^4 + gamma^4 – 8(alpha^2 + beta^2 + gamma^2) + 3(alpha + beta + gamma) = 0`

`alpha^4 + beta^4 + gamma^4`	`= 8 xx 16 – 0`
	`= 128`

Calculus, EXT2 C1 2015 HSC 14a

Differentiate `sin^(n - 1) theta cos theta`, expressing the result in terms of `sin theta` only. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, deduce that

`int_0^(pi/2) sin^n theta\ d theta = ((n-1))/n int_0^(pi/2) sin^(n - 2) theta\ d theta`, for `n>1.` (2 marks)

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Find `int_0^(pi/2) sin^4 theta\ d theta.` (1 mark)

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

Show Answers Only

`text(See Worked Solutions)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(3 pi)/16`

Show Worked Solution

i. `d/(d theta) (sin^(n – 1) theta cos theta)`

`=(n – 1) sin^(n – 2) theta cos theta cos theta + sin^(n – 1) theta xx (-sin theta)`

`=(n – 1) sin^(n – 2) theta cos^2 theta – sin^n theta`

`=(n – 1) sin^(n – 2) theta (1 – sin^2 theta) – sin^n theta`

`=(n – 1) sin^(n – 2) theta – (n – 1) sin^n theta – sin^n theta`

`=(n – 1) sin^(n – 2) theta – n sin^n theta`

ii. `text{From part (i)}`

`n sin^n theta = (n – 1) sin^(n – 2) theta – d/(d theta) (sin^(n – 1) theta cos theta)`

`:. int_0^(pi/2) sin^n theta\ d theta`

`=1/n int_0^(pi/2) ((n – 1) sin^(n – 2) theta – d/(d theta) (sin^(n – 1) theta cos theta)) d theta`

`=1/n int_0^(pi/2) (n – 1) sin^(n – 2) theta\ d theta – 1/n int_0^(pi/2) d/(d theta) (sin^(n – 1) theta cos theta)\ d theta`

`= 1/n int_0^(pi/2) (n – 1) sin^(n – 2) theta\ d theta – 1/n int_0^(pi/2) d/(d theta) (sin^(n – 1) theta cos theta)\ d theta`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta – 1/n [sin^(n – 1) theta cos theta]_0^(pi/2)`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta – 1/n (0 – 0)`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta,\ \ \ \ (n>1)`

iii.	`int_0^(pi/2) sin^4 theta\ d theta`	`= 3/4 int_0^(pi/2) sin^2 theta\ d theta`
		`= 3/4 xx [(2-1)/2 int_0^(pi/2) d theta]`
		`= 3/8 xx [theta]_0^(pi/2)`
		`= (3 pi)/16`

Harder Ext1 Topics, EXT2 2015 HSC 13c

A small spherical balloon is released and rises into the air. At time `t` seconds, it has radius `r` cm, surface area `S = 4 pi r^2` and volume `V = 4/3 pi r^3`.

As the balloon rises it expands, causing its surface area to increase at a rate of `((4 pi)/3)^(1/3)\ \text(cm)^2 text(s)^-1`. As the balloon expands it maintains a spherical shape.

By considering the surface area, show that
1. `(dr)/(dt) = 1/(8 pi r) (4/3 pi)^(1/3).` (2 marks)
Show that
1. `(dV)/(dt) = 1/2 V^(1/3).` (2 marks)
When the balloon is released its volume is `8000\ text(cm³)`. When the volume of the balloon reaches `64000\ text(cm³)` it will burst.
How long after it is released will the balloon burst? (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`1\ \ text(hour)`

Show Worked Solution

(i)	`(dS)/(dt)`	`= (dS)/(dr) xx (dr)/(dt)`
	`((4 pi)/3)^(1/3)`	`=8 pi r xx (dr)/(dt)`
	`(dr)/(dt)`	`=((4 pi)/3)^(1/3) xx 1/(8 pi r)`
		`=1/(8 pi r) (4/3 pi)^(1/3)`

(ii) `(dV)/(dt) = (dV)/(dr) xx (dr)/(dt)`

`(dV)/(dt) =`	`4 pi r^2 xx ((4 pi)/3)^(1/3) xx 1/(8 pi r)`
`=`	`r/2 xx ((4 pi)/3)^(1/3)`
`=`	`1/2 (4/3 pi r^3)^(1/3)`
`=`	`1/2 V^(1/3)`

(iii) `text(Find)\ \ t\ \ text(when)\ \ V=64\ 000`

`text(When)\ \ t = 0,\ \ V = 8000`

`(dV)/(dt)`	`=1/2 V^(1/3)`
`(dt)/(dV)`	`=2/V^(1/3)`
`int_0^t\ dt`	`= int_8000^64000 2/V^(1/3)\ dV`
`:.t`	`= [3V^(2/3)]_8000^64000`
	`= 3(1600 – 400)`
	`= 3600\ \ text(seconds)`
	`= 1\ \ text(hour)`

Volumes, EXT2 2015 HSC 13b

Two quarter cylinders, each of radius `a`, intersect at right angles to form the shaded solid.

A horizontal slice `ABCD` of the solid is taken at height `h` from the base. You may assume that `ABCD` is a square, and is parallel to the base.

Show that `AB = sqrt(a^2 - h^2).` (1 mark)
Find the volume of the solid. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`(2a^3)/3\ \ text(u³)`

Show Worked Solution

(i)

`text(The cylinder has equation)\ \ x^2 + y^2 = a^2`

`text(When)\ \ y = h,`

`x`	`= sqrt (a^2 – h^2)`
`:. AB`	`= sqrt (a^2 – h^2)`

(ii) `BC = sqrt (a^2 – h^2)`

`text(Area of)\ \ ABCD = (a^2 – h^2)`

`:.text(Volume of solid) =`	`int_0^a (a^2 – h^2) dh`
`=`	`[a^2 h – h^3/3]_0^a`
`=`	`[(a^3 – a^3/3) – 0]`
`=`	`(2a^3)/3\ \ text(u³)`

Volumes, EXT2 2015 HSC 12d

The diagram shows the graph `y = sqrt (x + 1)` for `0 <= x <= 3`. The shaded region is rotated about the line `x = 3` to form a solid.

Use the method of cylindrical shells to find the volume of the solid. (4 marks)

Show Answers Only

`(188 pi)/15\ \ text(u³)`

Show Worked Solution

`text(Solution 1)`

`δV=2pi r h\ δx,\ \ \ text(where)`

`r=3-x,\ \ and\ \ h=y=sqrt (x+1)`

`:.V`	`=2 pi int_0^3 (3 – x) sqrt (x + 1)\ dx`
	`=2 pi int_0^3 (3 sqrt (x + 1) – x sqrt (x + 1))\ dx`
	`=2 pi [2 (x + 1)^(3/2)]_0^3- 2 pi int_0^3 x sqrt (x + 1))\ dx`

`text(Integrating by parts)`

`u`	`=x`	`u′`	`=1`
`v′`	`=(x+1)^(1/2)`	`v`	`=2/3 (x+1)^(3/2)`

`:.V`	`=2 pi [2 (x + 1)^(3/2)]_0^3 – 2 pi ([2/3 x (x + 1)^(3/2)]_0^3 – int_0^3 2/3 (x + 1)^(3/2)\ dx)`
	`=4 pi (8 – 1) – (4 pi)/3 (24 – 0 – [2/5 (x + 1)^(5/2)]_0^3)`
	`=28 pi – (4 pi)/3 (24 – [2/5 xx 32 – 2/5])`
	`=28 pi – (4 pi)/3 xx 58/5`
	`=(188 pi)/15\ \ text(u³)`

`text{Solution 2 (Substitution)}`

`int_0^3 x sqrt(x + 1)\ dx`

`text(Let)\ \ u = x + 1,\ \ du = dx,\ \ x = u – 1`

`text(When)\ \ x = 0,\ \ u = 1`

`text(When)\ \ x = 3,\ \ u = 4`

`int_0^3 x sqrt (x + 1)\ dx =`	`int_1^4 (u – 1) sqrt u\ du`
`=`	`int_1^4 (u^(3/2) – u^(1/2)) du`
`=`	`[2/5 u^(5/2) – 2/3 u^(3/2)]_1^4`
`=`	`(64/5 – 16/3) – (2/5 – 2/3)`
`=`	`62/5 – 14/3`
`=`	`116/15`

`:.V =`	`2 pi [2 (x + 1)^(3/2)]_0^3 – 2 pi xx 116/15`
`=`	`28 pi – (232 pi)/15`
`=`	`(188 pi)/15\ \ text(u³)`

Polynomials, EXT2 2015 HSC 12b

The polynomial `P(x) = x^4 - 4x^3 + 11x^2 - 14x + 10` has roots `a + ib` and `a + 2ib` where `a` and `b` are real and `b != 0.`

By evaluating `a` and `b`, find all the roots of `P(x).` (3 marks)
Hence, or otherwise, find one quadratic polynomial with real coefficients that is a factor of `P(x).` (1 mark)

Show Answers Only

`1 +- i,\ \ 1 +- 2i`
`x^2 − 2x + 2\ \ or\ \ x^2 − 2x + 5`

Show Worked Solution

(i) `text(S)text(ince coefficients of)\ \ P(x)\ \ text(are real,)`

`=>\ text(Complex roots occur in conjugate pairs)`

`=>\ text(Roots are)\ \ a +- ib\ \ and\ \ a +- 2ib`

`text(Sum of roots) = -b/a=4`

`4`	`=a + ib + a – ib + a + 2ib + a – 2ib`
`4a`	`=4`
`:.a`	`=1`

`text(Products of roots)`

`(a + ib) (a – ib) (a – 2ib) (a – 2ib)`	`= 10`
`(a^2 + b^2) (a^2 + 4b^2)`	`= 10`
`(1 + b^2) (1 + 4b^2)`	`= 10`
`4b^4 + 5b^2 + 1`	`= 10`
`4b^4 + 5b^2 – 9`	`= 0`
`(4b^2 + 9) (b^2 – 1)`	`= 0`

`:.b^2 = 1,\ \ \ \ (b\ \ text{is real})`

`:.b = +- 1`

`:.P(x)\ text(has roots)\ \ \ 1 +- i,\ 1 +- 2i.`

(ii)	`P(x)`	`=(x – 1 – i) (x – 1 + i)(x-1-2i)(x-1+2i)`
		`=(x^2 – 2x + 2)(x^2 – 2x + 5)`

Complex Numbers, EXT2 N1 2015 HSC 12a

The complex number `z` is such that `|\ z\ |=2` and `text(arg)(z) = pi/4.`

Plot each of the following complex numbers on the same half-page Argand diagram.

`z` (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
`u = z^2` (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
`v = z^2 - bar z` (1 mark)

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

Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

i. `z =`	`2 text(cis) pi/4`
`=`	`sqrt 2 (1 + i)`

ii. `u =`	`z^2`
`=`	`4 text(cis)\ pi/2`
`=`	`4i`

COMMENT: 12a(iii) had the lowest mean mark (55%) of any part within Q12 and deserves attention.

iii. `v =`	`z^2 – bar z`
`=`	`4i – sqrt 2 (1 – i)`
`=`	`- sqrt 2 + (4 + sqrt 2) i`

Calculus, EXT2 C1 2015 HSC 11f

Show that

`cot theta + text(cosec)\ theta = cot(theta/2).` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find

`int (cot theta + text(cosec)\ theta)\ d theta.` (1 mark)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`2 ln\ |sin\ theta/2| + c`

Show Worked Solution

i. `cot theta + text(cosec)\ theta =`	`(cos theta)/(sin theta) + 1/(sin theta)`
`=`	`(1 + cos theta)/(sin theta)`
`=`	`(1 + 2 cos^2 (theta/2) – 1)/(2 sin (theta/2) cos (theta/2))`
`=`	`(2 cos^2(theta/2))/(2 sin (theta/2) cos (theta/2))`
`=`	`(cos (theta/2))/(sin (theta/2))`
`=`	`cot (theta/2)`

COMMENT: Part (ii) mean mark 51%.

ii. `int (cot theta + text(cosec)\ theta)\ d theta =`	`int cot (theta/2)\ d theta`
`=`	`int (cos (theta/2))/(sin (theta/2))\ d theta`
`=`	`2 ln\ \|sin\ theta/2\| + c`

Complex Numbers, EXT2 N1 2015 HSC 11b

Consider the complex numbers `z = -sqrt 3 + i` and `w = 3 (cos\ pi/7 + i sin\ pi/7).`

Evaluate `|\ z\ |.` (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Evaluate `text(arg)(z).` (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the argument of `z/w.` (1 mark)

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

Show Answers Only

`2`
`(5 pi)/6`
`(29 pi)/42`

Show Worked Solution

i. `\|\ z\ \|`	`= sqrt ((-sqrt3)^2 + 1^2)`
	`= 2`

ii. `text(arg)\ (z) =`	`tan^-1 (1- sqrt 3)`
`=`	`pi – pi/6`
`=`	`(5 pi)/6`

iii. `text(arg) (z/w) =`	`text(arg)\ z – text(arg)\ w`
`=`	`(5 pi)/6 – pi/7`
`=`	`(29 pi)/42`

Proof, EXT2 P1 2006 HSC 8a

Suppose `0 <= t <= 1/sqrt 2.`

Show that `0 <= (2t^2)/(1 - t^2) <= 4t^2.` (2 marks)

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Hence show that `0 <= 1/(1 + t) + 1/(1 - t) - 2 <= 4t^2.` (1 mark)

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By integrating the expressions in the inequality in part (ii) with respect to `t` from `t = 0` to `t = x\ \ text{(where}\ \ 0 <= x <= 1/sqrt2\ \ text{)}`, show that

`0 <= log_e ((1 + x)/(1 - x)) - 2x <= (4x^3)/3.` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Hence show that for `0 <= x <= 1/sqrt 2`

`1 <= ((1 + x)/(1 - x)) e^(-2x) <= e^((4x^3)/3).` (1 mark)

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

Show Answers Only

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

Show Worked Solution

i.	`0`	`<= t <= 1/sqrt 2`
	`0`	`<= t^2 <= 1/2`
	`0`	`>= -t^2 >= -1/2`
	`1`	`>= 1 – t^2 >= 1/2`
	`1`	`<= 1/(1 – t^2) <= 2`
	`2t^2`	`<= (2t^2)/(1 – t^2) <= 4t^2,\ \ \ \ (2t^2>0)`
	`:. 0`	`<= (2t^2)/(1 – t^2) <= 4t^2`

ii. `1/(1 + t) + 1/(1 – t) – 2`

`=((1-t)+(1+t)-2(1-t^2))/(1-t^2)`

`=(2t^2)/(1-t^2)`

`text{Substituting into part (i)}`

`:. 0 <= 1/(1 + t) + 1/(1 – t) – 2 <= 4t^2`

iii.	`int_0^x 0\ dt`	`<= int_0^x (1/(1 + t) + 1/(1 – t) – 2)\ dt <= int_0^x 4t^2\ dt`
	`0`	`<= [log_e (1 + t) – log_e (1 – t) – 2t]_0^x <= [(4t^3)/3]_0^x`
	`0`	`<= [log_e (1 + x) – log_e (1 – x) – 2x]<= [(4x^3)/3]`
	`0`	`<= log_e ((1 + x)/(1 – x)) – 2x <= (4x^3)/3`

iv. `text(S)text(ince)\ \ e^a>e^b\ \ text(when)\ \ a>b`

`e^0`	`<= e^([ln ((1 + x)/(1 – x)) – 2x]) <= e^((4x^3)/3)`
`1`	`<= e^(ln ((1 + x)/(1 – x))) xx e^(-2x) <= e^((4x^3)/3)`
`1`	`<= ((1 + x)/(1 – x)) e^(-2x) <= e^((4x^3)/3)`

Proof, EXT2 P2 2006 HSC 7c

The sequence `{x_n}` is given by

`x_1 = 1` and `x_(n + 1) = (4 + x_n)/(1 + x_n)` for `n >= 1.`

Prove by induction that for `n >= 1`, `x_n = 2 ((1 + alpha^n)/(1 - alpha^n))`, where `alpha = -1/3.` (4 marks)

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Hence find the limiting value of `x_n` as `n -> oo.` (1 mark)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`2`

Show Worked Solution

i. `text(If)\ \ n = 1`

`x_1=2 ((1 – 1/3)/(1 + 1/3))=(2 xx 2/3)/(4/3)=1`

`:.text(True for)\ \ n = 1`

`text(Assume true for)\ \ n=k,`

`text(i.e.)\ \ \ \ x_k = 2 ((1 + alpha^k)/(1 – alpha^k))`

`text(Prove true for)\ \ n=k+1`

`text(i.e.)\ \ \ \ x_(k + 1) = 2 ((1 + alpha^(k + 1))/(1 – alpha^(k+1)))`

`x_(k + 1) =`	`(4 + x_k)/(1 + x_k)`
`=`	`(4 + 2 ((1 + alpha^k)/(1 – alpha^k)))/(1 + 2((1 + alpha^k)/(1 – alpha^k)))`
`=`	`2 [(2(1 – alpha^k) + (1 + alpha^k))/(1 – alpha^k + 2 (1 + alpha^k))]`
`=`	`2 [(2 – 2 alpha^k + 1 + alpha^k)/(1 – alpha^k + 2 + 2 alpha^k)]`
`=`	`2 [(3 – alpha^k)/(3 + alpha^k)]`
`=`	`2 [(1 – alpha^k xx 1/3)/(1 + alpha^k xx 1/3)]`
`=`	`2 [(1 + alpha^k xx alpha)/(1 – alpha^k xx alpha)],\ \ \ \ (alpha=-1/3)`
`=`	`2 [(1 + alpha^(k + 1))/(1 – alpha^(k + 1))]`
`=`	`text(RHS)`

`:.text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k.`

`:.text(S)text(ince true for)\ \ n=1, text(by PMI, true for all integral)\ \ n >= 1.`

ii. `text(S)text(ince)\ \ lim_(n -> oo) (-1/3)^n=0`

`:.lim_(n -> oo) x_n`	`=2 ((1 + (-1/3)^n)/(1 – (-1/3)^n))`
	`=2`

Harder Ext1 Topics, EXT2 2006 HSC 7a

The curves `y = cos x` and `y = tan x` intersect at a point `P` whose `x`-coordinate is `alpha.`

Show that the curves intersect at right angles at `P.` (3 marks)
Show that
`sec^2 alpha = (1 + sqrt 5)/2.` (2 marks)

Show Answers Only

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

Show Worked Solution

(i) `y = cos x,\ \ y prime = -sin x`

`text(When)\ \ x`	`=alpha,`
`y`	`= cos\ alpha`
`m_1`	`=- sin\ alpha`

`y = tan x,\ \ y prime = sec^2 x`

`text(When)\ \ x`	`=alpha,`
`y`	`=tan\ alpha`
`m_2`	`=sec^2 alpha`

`text(S)text(ince intersection occurs when)\ \ x=alpha`

`=> cos\ alpha = tan\ alpha`

`m_1 xx m_2`	`=-sin\ alpha xx sec^2 alpha`
	`=(-sin\ alpha)/(cos^2 alpha)`
	`=(-tan\ alpha)/(cos\ alpha)`
	`=-1`

`:.text(The curves intersect at right-angles at)\ \ P.`

(ii)	`text(At)\ \ P,\ \ \ cos\ α`	`= tan\ α`
	`cos^2 α`	`=tan^2 α`
		`=sec^2 α-1`
	`1`	`=sec^4 α-sec^2 α`
	`0`	`=sec^4 α-sec^2 α-1`

`:. sec^2 α`	`=(1+-sqrt(1+4))/2`
	`=(1+sqrt5)/2\ \ \ \ \ (sec^2 alpha >0)`

Mechanics, EXT2 M1 2006 HSC 6b

In an alien universe, the gravitational attraction between two bodies is proportional to `x^(–3)`, where `x` is the distance between their centres.

A particle is projected upward from the surface of a planet with velocity `u` at time `t = 0`. Its distance `x` from the centre of the planet satisfies the equation

`ddot x = - k/x^3.`

Show that `k =gR^3`, where `g` is the magnitude of the acceleration due to gravity at the surface of the planet and `R` is the radius of the planet. (1 mark)

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Show that `v`, the velocity of the particle, is given by

`v^2 = (gR^3)/x^2 - (gR - u^2).` (3 marks)

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It can be shown that `x = sqrt (R^2 + 2uRt - (gR - u^2) t^2).` (Do NOT prove this.)

Show that if `u >= sqrt (gR)` the particle will not return to the planet. (2 marks)

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If `u < sqrt (gR)` the particle reaches a point whose distance from the centre of the planet is `D`, and then falls back.
(1) Use the formula in part (ii) to find `D` in terms of `u, R` and `g.` (1 mark)

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(2) Use the formula in part (iii) to find the time taken for the particle to return to the surface of the planet in terms of `u, R` and `g.` (1 mark)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(1)\ \ D = sqrt ((gR^3)/(gR – u^2))`

`(2)\ \ (2uR)/(gR – u^2)`

Show Worked Solution

`text(When)\ x=R,\ \ ddot x = – k/(R^3)`

`text(Given)\ \ ddot x = -g\ \ text(on the surface)`

`-g =`	`-k/R^3`
`:.k =`	`gR^3`

ii.	`ddot x`	`=- (gR^3)/x^3`
	`1/2 v^2`	`= int – (gR^3)/x^3\ dx`
		`= (gR^3)/(2x^2)+c_1`
	`:.v^2`	`=(gR^3)/x^2 +c_2`

`text(When)\ t=0, x=R and v=u`

`u^2`	`=(gR^3)/R^2 +c_2`
`c_2`	`=u^2-gR`
`:.v^2`	`=(gR^3)/x^2 +u^2-gR`
	`=(gR^3)/x^2 -(gR-u^2)`

iii. `text(Solution 1)`

`text(If)\ \ u >= sqrt (gR)\ \ text(then)\ \ u^2 >= gR`

`x`	`= sqrt (R^2 + 2uRt – (gR – u^2) t^2)`
	`≥sqrt (R^2 + 2sqrt(gR)Rt – (gR – gR) t^2)`
	`≥sqrt (R^2 + 2sqrt(gR)Rt)`
	`>sqrt (R^2)\ \ \ \ (t>0)`
	`>R`

`:. x>R\ \ text(when)\ \ t>0,\ text(and the particle does not)`

`text(return to the surface of the planet.)`

`text(Solution 2)`

`v^2`	`=(gR^3)/x^2 -(gR-u^2)`
	`>=(gR^3)/x^2\ \ \ \ text{(since}\ u^2 >= gR text{)}`
`v`	`>=0`

`:.\ text(S)text(ince)\ \ v>=0,\ \ text(the particle is never moving back)`

`text(towards the planet and will never return.)`

iv. (1) `v^2 = (gR^3)/D^2 – (gR – u^2)`

`x = D\ \ text(occurs when)\ \ v = 0`

`:.0 =`	`(gR^3)/D^2 – (gR – u^2)`
`D^2 =`	`(gR^3)/(gR – u^2)`
`:.D =`	`sqrt ((gR^3)/(gR – u^2))`

iv. (2) `text(Find)\ \ t\ \ text(when)\ \ x = R`

`text{Using part (iii)}`

`R`	`=sqrt (R^2 + 2uRt – (gR – u^2) t^2)`
`R^2`	`= R^2 + 2uRt – (gR – u^2) t^2`
`0`	`= t(2uR – (gR – u^2)t)`
`:.t`	` = (2uR)/(gR – u^2)\ \ \ \ (t>0)`

`:.\ text(It takes)\ \ (2uR)/(gR – u^2)\ \ text(seconds to return to the planet.)`

Harder Ext1 Topics, EXT2 2006 HSC 6a

In `Delta ABC, /_ CAB = alpha, /_ ABC = beta` and `/_ BCA = gamma`. The point `O` is chosen inside `Delta ABC` so that `/_ OAB = /_ OBC = /_ OCA = theta`, as shown in the diagram.

Show that
1. `(OA)/(OB) = (sin (beta - theta))/(sin theta).` (1 mark)
Hence show that
1. `sin^3 theta = sin (alpha - theta)\ sin (beta - theta)\ sin (gamma - theta).` (2 marks)
Prove the identity
1. `cot x - cot y = (sin (y - x))/(sin x sin y).` (1 mark)
Hence show that
`(cot theta - cot alpha) (cot theta - cot beta) (cot theta - cot gamma) = text(cosec)\ alpha\ text(cosec)\ beta\ text(cosec)\ gamma.` (1 mark)
Hence find the value of `theta` when `Delta ABC` is an isosceles right triangle. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`tan^-1\ 1/2`

Show Worked Solution

(i)

`/_ CAB = alpha,\ \ \ /_ ABC = beta,\ \ \ /_ BCA = gamma`

`text(Using the sine rule in)\ \ Delta OAB:`

`(OA)/(sin (beta – theta))`	`= (OB)/(sin theta)`
`(OA)/(OB)`	`= (sin (beta – theta))/(sin theta)`

(ii)	`text(Similarly,)`	`\ \ \ (OA)/(OC)`	`= (sin theta)/(sin (alpha – theta))`
		`\ \ \ (OB)/(OC)`	`= (sin (gamma – theta))/(sin theta)`

`(OA)/(OB)(OB)/(OC) (OC)/(OA)`	`= (sin (beta – theta))/(sin theta) * (sin (gamma – theta))/(sin theta) * (sin (alpha – theta))/(sin theta)`
`1`	`= (sin (beta – theta) sin (gamma – theta) sin (alpha – theta))/(sin^3 theta)`
`:. sin^3 theta`	`=sin (alpha – theta) sin (beta – theta) sin (gamma – theta)`

(iii) `text(RHS) =`	`(sin (y – x))/(sin x sin y)`
`=`	`(sin y cos x – cos y sin x)/(sin x sin y)`
`=`	`(cos x)/(sin x) – (cos y)/(sin y)`
`=`	`cot x – cot y`
`=`	`text(LHS)`

(iv) `(cot theta – cot alpha) (cot theta – cot beta) (cot theta – cot gamma) `

`=(sin (alpha – theta))/(sin alpha sin theta) * (sin (beta – theta))/(sin beta sin theta) * (sin (gamma – theta))/(sin gamma sin theta)\ \ \ \ text{(using part (iii))}`

`=sin^3theta/(sin alpha sin beta sin gamma sin^3 theta)\ \ \ \ text{(using part (ii))}`

`=text(cosec)\ alpha\ text(cosec)\ beta\ text(cosec)\ gamma`

♦♦ Mean mark sub 30% (exact % not available).
MARKER’S COMMENT: Very few students correctly calculated the values of `cot\ pi/2, cot\ pi/4,“ text(cosec)\ pi/2 and text(cosec)\ pi/4!`

(v) `text(Let)\ \ gamma = 90^@,\ \ =>alpha = beta = 45^@`

`text{Substituting into part (iv)}`

`(cot theta – 1) (cot theta – 1) (cot theta – 0)`	`= sqrt 2 xx sqrt2 xx 1`
`(cot^2 theta – 2 cot theta + 1) cot theta`	`= 2`
`cot^3 theta – 2 cot^2 theta + cot theta – 2`	`= 0`
`cot^2 theta (cot theta – 2) + 1 (cot theta – 2)`	`= 0`
`(cot theta – 2) (cot^2 theta + 1)`	`= 0`

`cot theta`	`=2,\ \ \ \ \ (cot^2 theta ≠ -1)`
`tan theta`	`=1/2`
`:. theta`	`=tan^-1\ 1/2`
	`=0.46\ \ text(radians)`

Harder Ext1 Topics, EXT2 2006 HSC 5d

In a chess match between the Home team and the Away team, a game is played on each of board 1, board 2, board 3 and board 4.

On each board, the probability that the Home team wins is `0.2`, the probability of a draw is `0.6` and the probability that the Home team loses is `0.2`.

The results are recorded by listing the outcomes of the games for the Home team in board order. For example, if the Home team wins on board 1, draws on board 2, loses on board 3 and draws on board 4, the result is recorded as `WDLD`.

How many different recordings are possible? (1 mark)
Calculate the probability of the result which is recorded as `WDLD.` (1 mark)
Teams score 1 point for each game won, ½ a point for each game drawn and 0 points for each game lost.
What is the probability that the Home team scores more points than the Away team? (3 marks)

Show Answers Only

`81`
`0.0144`
`0.344`

Show Worked Solution

(i) `text(3 possibilities on each board)`

`:.\ text(Number of different recordings)`

`=3^4`

`= 81`

(ii)	`P (WDLD)`	`= 0.2 xx 0.6 xx 0.2 xx 0.6`
		`= 0.0144`

(iii) `text(Solution 1)`

`text(Home team scores more points)`

`WWWW`	`=0.2^4`
`WWWD`	`=\ ^4C_3 xx 0.2^3 xx 0.6`
`WWDD`	`=(4!)/(2!2!) xx 0.2^2 xx 0.6^2`
`WDDD`	`=\ ^4C_3 xx 0.2 xx 0.6^3`
`WWWL`	`=\ ^4C_3 xx 0.2^4`
`WWDL`	`= (4!)/(2!) xx 0.2^2 xx 0.6 xx 0.2`

`:. P text{(Home team scores more points)}`

`= 0.2^4 + 4 xx 0.2^3 xx 0.6 + 6 xx 0.2^2 xx 0.6^2 + 4 xx 0.2 xx 0.6^3`

`+4 xx 0.2^4 + 12 xx 0.2^2 xx 0.6 xx 0.2`

`= 0.344`

STRATEGY: Recognising symmetry in probability outcomes can often provide a very efficient solution involving less calculation as shown in part (iii) here.

`text(Solution 2)`

`text(S)text(ince probabilities are symmetric,)`

`Ptext{(one team winning)}=(1-Ptext{(draw)})/2`

`Ptext{(draw)}`	`=Ptext{(4D)} + P text{(1W,1L,2D)} + P text{(2W,2L)}`
	`=0.6^4+(4!)/(2!) xx 0.2^2 xx 0.6^2 + (4!)/(2!2!) xx 0.2^4`
	`=0.1296+0.1728+0.0096`
	`=0.312`

`:.Ptext{(Home team winning)}`	`=(1-0.312)/2`
	`=0.344`

Mechanics, EXT2 2006 HSC 5c

A particle, `P`, of mass `m` is attached by two strings, each of length `l`, to two fixed points, `A` and `B`, which lie on a vertical line as shown in the diagram.

The system revolves with constant angular velocity `omega` about `AB`. The string `AP` makes an angle `alpha` with the vertical. The tension in the string `AP` is `T_1` and the tension in the string `BP` is `T_2` where `T_1 >= 0` and `T_2 >= 0`. The particle is also subject to a downward force, `mg`, due to gravity.

Resolve the forces on `P` in the horizontal and vertical directions. (2 marks)
If `T_2 = 0`, find the value of `omega` in terms of `l, g` and `alpha.` (1 mark)

Show Answers Only

`text(Vertical):\ (T_1 – T_2) cos alpha = mg\ \ \ \ \ text(Horizontal):\ T_1 + T_2 = ml omega^2`
`omega = sqrt (g/(l cos alpha))`

Show Worked Solution

(i)

`text(Resolving the forces vertically)`

`T_1 cos alpha = mg + T_2 cos alpha\ \ \ …\ (1)`

`text(Resolving the forces horizontally)`

` T_1 sin alpha+ T_2 sin alpha = mr omega^2\ \ \ …\ (2)`

`r = l sin alpha`

`:. (T_1 + T_2) sin alpha = m l sin alpha omega^2`

`T_1 + T_2 = m l omega^2`

(ii) `text(Substitute)\ \ T_2 = 0\ \ text{into (1) and (2)}`

`T_1`	`= (mg)/cos alpha\ \ \ …\ (1)`
`T_1 sin alpha`	`= m r omega^2`
	`=ml sin alpha omega^2\ \ \ \ text{(since}\ \ r=l sin alpha text{)}`
`T_1`	`=ml omega^2\ \ \ …\ (2)`

`:. m l omega^2`	`= (mg)/(cos alpha)`
`omega^2`	`= g/(l cos alpha)`
`:.omega`	`= sqrt(g/(l cos alpha))`

Harder Ext1 Topics, EXT2 2006 HSC 5b

Show that `cos (alpha + beta) + cos (alpha - beta) = 2 cos alpha cos beta.` (1 mark)
Hence, or otherwise, solve the equation
1. `cos theta + cos 2 theta + cos 3 theta + cos 4 theta = 0`
for `0 <= theta <= 2 pi.` (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`theta = pi/5,\ pi/2,\ (3 pi)/5,\ pi,\ (7 pi)/5,\ (3 pi)/2,\ (9 pi)/5`

Show Worked Solution

(i) `text(LHS)`	`= cos (alpha + beta) + cos (alpha – beta)`
	`= cos alpha cos beta – sin alpha sin beta + cos alpha cos beta + sin alpha sin beta`
	`= 2 cos alpha cos beta`

STRATEGY: Students who applied the rule in part (i) 3 times solved this problem in the most efficient way. Many however, only applied it twice and lost valuable time.

(ii) `cos theta + cos 2 theta + cos 3 theta + cos 4 theta`

`=(cos 3 theta + cos theta)+ (cos 4 theta + cos 2 theta)`

`=2cos 2 theta cos theta + 2 cos 3 theta cos theta\ \ \ \ \ text{(using part (i))}`

`=2 cos theta (cos 2 theta+cos 3 theta)`

`=2cos theta(2 cos\ (5 theta)/2 cos\ theta/2)\ \ \ \ \ text{(using part (i))}`

`=4cos theta cos\ (5 theta)/2 cos\ theta/2`

`4cos theta cos\ (5 theta)/2 cos\ theta/2=0\ \ \ text(when)`

`cos theta=0\ \ => theta=pi/2, (3pi)/2`

`cos\ theta/2=0\ \ => theta=pi`

`cos\ (5 theta)/2=0\ \ => theta=pi/5, (3pi)/5, (7pi)/5, (9pi)/5`

`:. theta = pi/5,\ pi/2,\ (3 pi)/5,\ pi,\ (7 pi)/5,\ (3 pi)/2,\ (9 pi)/5`

Volumes, EXT2 2006 HSC 5a

A solid is formed by rotating the region bounded by the curve `y = x (x - 1)^2` and the line `y = 0` about the `y`-axis. Use the method of cylindrical shells to find the volume of this solid. (3 marks)

Show Answers Only

`pi/15\ \ text(u³)`

Show Worked Solution

`delta V`	`= 2 pi xy\ delta x`
	`= 2 pi x * x (x – 1)^2\ delta x`
`:.V`	`=int_0^1 2 pi x^2 (x – 1)^2\ dx`
	`=2 pi int_0^1 (x^4 – 2x^3 + x^2)\ dx`
	`=2 pi [x^5/5 – x^4/2 + x^3/3]_0^1`
	`=2 pi [(1/5 – 1/2 + 1/3)-0]`
	`=pi/15\ \ text(u³)`

Harder Ext1 Topics, EXT2 2006 HSC 4d

In the acute-angled triangle `ABC, \ K` is the midpoint of `AB, \ L` is the midpoint of `BC` and `M` is the midpoint of `CA`. The circle through `K, L` and `M` also cuts `BC` at `P` as shown in the diagram.

Copy or trace the diagram into your writing booklet.

Prove that `KMLB` is a parallelogram. (1 mark)
Prove that `/_ KPB = /_ KML.` (1 mark)
Prove that `AP _|_ BC.` (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

`(AK)/(KB) = (AM)/(MC) = 1/1`

`:. KM\ text(\|\|)\ BC`	`\ \ \ text{(parallel lines cut in the}` `\ \ \ \ text{same proportion)`

`text(Similarly,)\ \ (CL)/(LB) = (CM)/(MA) = 1/1`

`:. ML\ text(||)\ AB`

`:. KMLB\ \ text(is a parallelogram)`

(ii)

`/_ BPK`

`= /_ KML`

`text{(exterior angle of a cyclic}`
`text{quadrilateral}\ \ KMLP text{)}`

(iii)	`/_ KBP`	`= /_ KML\ \ \ text{(opposite angles of a parallelogram)}`
	`:. /_ KBP`	`= /_ KPB\ \ \ text{(both equal}\ \ /_ KML text{)}`

`:. Delta BKP\ \ text(is isosceles)`

`text(S)text(ince)\ \ BK = KP=KA\ \ \ text{(given}\ K\ \ text(is the midpoint of)\ \ ABtext{)}`

`=>K\ \ text(is the centre of a circle, diameter)\ \ AB,`

`text(that passes through)\ \ A, B and P`

`/_ APB = 90^@\ \ \ \ text{(angle in semi-circle)}`

`:. AP _|_ BC.`

Conics, EXT2 2006 HSC 4c

Let `P(p, 1/p), Q(q, 1/q)` and `R(r, 1/r)` be three distinct points on the hyperbola `xy = 1.`

Show that the equation of the line, `l`, through `R`, perpendicular to `PQ`, is `y = pqx - pqr + 1/r.` (2 marks)
Write down the equation of the line, `m`, through `P`, perpendicular to `QR.` (1 mark)
The lines `l` and `m` intersect at `T.`
Show that `T` lies on the hyperbola. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`y = qrx – pqr + 1/p`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `text(Gradient)\ \ PQ =`	`(1/p – 1/q)/(p – q)`
`=`	`((q – p)/(pq))/(p – q)`
`=`	`-1/(pq)`

`=>text(Gradient of perpendicular) = pq`

`:.\ text(Equation of)\ \ l,\ \ text(through)\ \ R(r, 1/r)`

`y – 1/r`	`= pq (x – r)`
`y – 1/r`	`= pqx – pqr`
`:.y`	`= pqx – pqr + 1/r`

(ii) `text(Equation of)\ \ m,\ \ text(through)\ \ P(p, 1/p),\ \ text(is)`

`y = qrx – pqr + 1/p`

(iii) `T\ \ text(occurs at the intersection of)`

MARKER’S COMMENT: Many students had difficulty simplifying `1/p-1/r -: (p-r).` Be clear on how to do this.

`y = pqx – pqr + 1/r\ \ \ …\ (1)`

`y = qrx – pqr + 1/p\ \ \ …\ (2)`

`text{Subtract (1) – (2)}`

`(pq – qr)x + 1/r – 1/p`	`=0`
`q(p – r)x`	`= 1/p – 1/r`
`q(p – r)x`	`= (r – p)/(pr)`
`x`	`= -1/(pqr)`

`text{Substitute into (1)}`

`y = (-pq)/(pqr) – pqr + 1/r = -pqr`

`:.T ((-1)/(pqr), -pqr)`

`text(Substituting into)\ \ xy = 1`

`text(LHS)`	`=(-1)/(pqr) xx (-pqr)`
	`=1`
	`=\ text(RHS)`

`:. T\ \ text(lies on the hyperbola)`

Volumes, EXT2 2006 HSC 4b

The base of a solid is the parabolic region `x^2 <= y <= 1` shaded in the diagram.

Vertical cross-sections of the solid perpendicular to the `y`-axis are squares.

Find the volume of the solid. (3 marks)

Show Answers Only

`2\ \ text(u³)`

Show Worked Solution

`text(Length of square)=2x`

MARKER’S COMMENT: A substantial number of students incorrectly treated this part as a solid of revolution question. Make sure you understand why it is not!

`:.\ text(Area of cross-section)`

`= (2x)^2`

`= 4x^2`

`delta V`	`= 4x^2\ δy\ \ \ text(where)\ \ y = x^2`
`:.V`	`=int_0^1 4x^2\ dy`
	`=int_0^1 4y\ dy`
	`=[2y^2]_0^1`
	`=2\ \ text(u³)`

Polynomials, EXT2 2006 HSC 3c

Two of the zeros of `P(x) = x^4 - 12x^3 + 59x^2 - 138x + 130` are `a + ib` and `a + 2ib`, where `a` and `b` are real and `b > 0.`

Find the values of `a` and `b`. (3 marks)
Hence, or otherwise, express `P(x)` as the product of quadratic factors with real coefficients. (1 mark)

Show Answers Only

`a = 3,\ \ \ b = 1`
`(x^2 – 6x + 10) (x^2 – 6x + 13)`

Show Worked Solution

(i) `text(S) text(ince the coefficients are real)`

STRATEGY: The observation that complex zeros occur in conjugate pairs for any `P(x)` with real coefficients was critical to making any progress here and is a regularly examined concept.

`=>\ text(Roots occur in conjugate pairs.)`

`:.\ text(Roots are)\ \ \ a + ib,\ a – ib,\ a + 2ib,\ a – 2ib`

`text(Sum of roots) = 4a=-b/a=12`

`:. a=3`

`text(Product of the roots)=130`

`(3 + ib) (3 – ib) (3 + 2ib) (3 – 2ib)`	`= 130`
`(9 + b^2 ) (9 + 4b^2 )`	`=130`
`81 + 45b^2 + 4b^4`	`=130`
`4b^4 + 45b^2 – 49`	`=0`
`(4b^2 + 49) (b^2 − 1)`	`=0`

`:.b = 1,\ \ \ \ \ (b>0)`

`:.a = 3,\ \ \ b = 1`

(ii) `text(Zeros are)\ \ 3 + i,\ 3 – i,\ 3 + 2i,\ 3 – 2i`

`P(x)`	`= (x – 3 – i) (x − 3+ i) (x − 3 – 2i) (x − 3 + 2i)`
	`=((x-3)^2+1)((x-3)^2+4)`
	`= (x^2 − 6x +10) (x^2 − 6x + 13)`

Conics, EXT2 2006 HSC 3b

The diagram shows the graph of the hyperbola

`x^2/144 - y^2/25 = 1.`

Find the coordinates of the points where the hyperbola intersects the `x`-axis. (1 mark)
Find the coordinates of the foci of the hyperbola. (2 marks)
Find the equations of the directrices and the asymptotes of the hyperbola. (2 marks)

Show Answers Only

`(12, 0),\ (– 12, 0)`
`(13, 0),\ (– 13, 0)`
`text(Directrices:)\ \ x = +- 144/13;\ \ text(Asymptotes:)\ \ y = +- 5/12 x`

Show Worked Solution

(i) `text(Intersection when)\ \ y=0`

`x^2/144 + 0`	`=1`
`x`	`= +-12`

`:. xtext(-axis intersection at)\ \ (12, 0),\ (– 12, 0)`

(ii) `a=12,\ \ b = 5`

`text(Using)\ \ \ b^2`	`=a^2(e^2-1)`
`25`	`= 144 (e^2 – 1)`
`25/144`	`= e^2 – 1`
`e^2`	`= 169/144`
`e`	`= 13/12`

`:.S(ae,0)-=(13,0)`

`:. S′(– ae,0)-=(– 13,0)`

(iii)	`text(Directrices when)\ \ \ x`	`=+-a/e`
		`=+-144/13`
	`text(Asymptotes when)\ \ \ y`	`=+-b/a x`
		`=+- 5/12 x`

Functions, EXT1′ F1 2006 HSC 3a

The diagram shows the graph of `y =f(x)`. The graph has a horizontal asymptote at `y =2`.

Draw separate one-third page sketches of the graphs of the following:

`y = (f(x))^2` (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
`y = 1/(f(x))` (2 marks)

--- 10 WORK AREA LINES (style=lined) ---
`y = x\ f(x)` (2 marks)

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

Show Answers Only

ii.

iii.

Show Worked Solution

ii.

iii.

Conics, EXT2 2006 HSC 2d

The equation `|\ z - 1 - 3i\ | + |\ z - 9 - 3i\ | = 10` corresponds to an ellipse in the Argand diagram.

Write down the complex number corresponding to the centre of the ellipse. (1 mark)
Sketch the ellipse, and state the lengths of the major and minor axes. (3 marks)
Write down the range of values of arg`(z)` for complex numbers `z` corresponding to points on the ellipse. (1 mark)

Show Answers Only

`5 + 3i`
`text(Major) = 10;\ \ text(Minor) = 6`
`0 <= text(arg)(z) <= pi/2`

Show Worked Solution

(i) `|\ z – 1 – 3i\ | + |\ z – 9 – 3i\ | = 10`

`text(This equation is Argand equivalent of)\ \ \ PS+PS′=2a,`

`text(where)\ \ z=P,\ \ text(and the foci are the points)`

♦♦ A “significant portion” of students could not do this question at all (exact data not available).

`1+3i\ \ and\ \ 9+3i.`

`:.\ text(Centre of the ellipse)`	`= ((1+9)/2,(3+3)/2)`
	`=(5,3)`
	`=5+3i`

(ii) `text(Major axis length)\ =2a=10`

`:.a=5`

`text(Distance from centre to)\ S=ae=4`

`text(Minor axis length)\ =2b`

`text(Using Pythagoras,)`

`b^2`	`=5^2-4^2=9`
`b`	`=3`

`:.\ text(Length of the minor axis)\ = 6`

(iii) `text(Looking at the graph, we can see that)`

`text(arg)\ (5+0i)=0`

`text(arg)\ (0+3i)=pi/2`

`:.\ text(Range is)\ \ 0 <= text(arg)(z) <= pi/2`

Complex Numbers, EXT2 N2 2006 HSC 2c

Find, in modulus-argument form, all solutions of `z^3 = -1.` (2 marks)

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

Show Answers Only

`text(cis)\ pi/3,\ \ \ -1,\ \ \ text(cis)\ -pi/3`

Show Worked Solution

`z^3`	`=-1`
`-1`	`=\ text(cis)\ (pi+2k pi)`
`:.z`	`=\ text(cis)\ (pi+2k pi)/3\ \ \ text{(De Moivre)}`

MARKER’S COMMENT: Another successful solution strategy was graphical, using the unit circle and three equal angles generated from pi.

`text(When)\ k=0`

`z=\ text(cis)\ pi/3`

`text(When)\ k=1`

`z=\ text(cis)\ pi = -1`

`text(When)\ k=-1`

`z=\ text(cis)\ -pi/3`

`:.\ text(The 3 solutions to)\ \ z^3=-1\ \ text(are)`

`z=\ text(cis)\ pi/3,\ \ \ -1,\ \ \ text(cis)\ -pi/3`

Conics, EXT2 2007 HSC 7c

The diagram shows an ellipse with eccentricity `e` and foci `S` and `S prime`.

The tangent at `P` on the ellipse meets the directrices at `R` and `W`. The perpendicular to the directrices through `P` meets the directrices at `N` and `M` as shown. Both `/_ PSR` and `/_ PS prime W` are right angles.

Let `/_ MPW = /_ NPR = beta.`

Show that
1. `(PS)/(PR) = e cos beta`
where `e` is the eccentricity of the ellipse. (2 marks)
By also considering `(PS prime)/(PW)` show that `/_ RPS = /_ WPS prime.` (2 marks)

Show Answers Only

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

Show Worked Solution

(i)	`(PS)/(PN)`	`= (PS prime)/(PM)=e`
	`:.\ PS`	`= ePN`

`text(In)\ \ Delta PNR,`

`(PN)/(PR)`	`=cos beta`
`PN`	`=PR cos beta`
`:.\ PS`	`= e PR cos beta`
`:.\ (PS)/(PR)`	`= e cos beta`

(ii) `text(Similarly)\ \ PS prime = ePM`

`(PM)/(PW)`	`= cos beta`
`PM`	`= PW cos beta`
`PS prime`	`= ePW cos beta`
`:.\ (PS prime)/(PW)`	`= e cos beta`

`=>(PS prime)/(PW) = cos /_ WPS prime = e cos beta`

`=>(PS)/(PR) = cos /_ RPS = e cos beta`

`:.\ cos /_ RPS = cos /_ WPS prime`

`:.\ /_ RPS = /_ WPS prime`

Conics, EXT2 2007 HSC 7b

In the diagram the secant `PQ` of the ellipse `x^2/a^2 + y^2/b^2 = 1` meets the directrix at `R`. Perpendiculars from `P` and `Q` to the directrix meet the directrix at `U` and `V` respectively. The focus of the ellipse which is nearer to `R` is at `S`.

Copy or trace this diagram into your writing booklet.

Prove that
1. `(PR)/(QR) = (PU)/(QV).` (1 mark)
Prove that
1. `(PU)/(QV) = (PS)/(QS).` (1 mark)
Let `/_ PSQ = phi,\ \ /_ RSQ = theta and /_ PRS = alpha.`
By considering the sine rule in `Delta PRS and Delta QRS`, and applying the results of part (i) and part (ii),
show that `phi = pi - 2 theta.` (2 marks)
Let `Q` approach `P` along the circumference of the ellipse, so that `phi -> 0.`
What is the limiting value of `theta?` (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`theta -> pi/2`

Show Worked Solution

(i) `text(In)\ \ Delta PUR and Delta QVR`

`/_ PUR`	`= /_ QVR=90^@`
`/_ PRU`	`= /_ QRV\ \ \ \ \ text{(common angle)}`
`:.\ Delta PUR\ text(\|\|\|)\ Delta QVR\ \ \ \ \ text{(equiangular)}`

`:.\ (PR)/(QR)`

`= (PU)/(QV)\ \ \ \ `

`text{(corresponding sides in similar}`
`text{triangles are proportional)}`

(ii) `(SP)/(PU) = e and (SQ)/(QV) = e`

`(SP)/(PU)`	`= (SQ)/(QV)`
`:.\ (PU)/(QV)`	`= (PS)/(QS)`

(iii) `text(In)\ \ Delta PRS`

`(PS)/(sin alpha)`	`= (PR)/(sin(phi + theta))`
`(PS)/(PR) `	`= (sin alpha)/(sin (phi + theta))`

`text(In)\ \ Delta QRS`

`(QS)/(sin alpha)`	`= (QR)/(sin theta)`
`(QS)/(QR)`	`= (sin alpha)/(sin theta)`

`text(Using)\ \ (PR)/(QR) = (PU)/(QV) = (PS)/(QS)\ \ \ text{(from parts (i) and (ii))}`

`=>(PS)/(PR)`	`= (QS)/(QR)`
`(sin alpha)/(sin (phi + theta))`	`= (sin alpha)/(sin theta)`

`:.\ sin (phi + theta) = sin theta`

`:.\ phi + theta = pi – theta\ \ \ or\ \ phi + theta = theta`

`:.\ phi = pi – 2 theta,\ \ \ \ \ (phi ≠ 0)`

(iv)	`text(As)\ \ phi`	`-> 0`
	`pi – 2 theta`	`-> 0`
	`:.theta`	`-> pi/2`

Proof, EXT2 P1 2007 HSC 7a

Show that `sin x < x` for `x > 0.` (2 marks)

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Let `f(x) = sin x - x + x^3/6`.

Show that the graph of `y = f(x)` is concave up for `x > 0.` (2 marks)

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By considering the first two derivatives of `f(x)`,show that `sin x > x - x^3/6` for `x > 0.` (2 marks)

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

Show Worked Solution

i.	`text(Let)\ \ g(x)`	`=sin x-x`
	`g′(x)`	`=cosx-1<=1\ \ \ text(for all)\ x>0`

MARKER’S COMMENT: A geometric proof using arc length and a right angled triangle caused problems as few students could deal with the case `x>pi`.

`=>g(x)\ \ text(is a decreasing function)`

`text(When)\ \ x=0,\ \ g(0)=0`

`text(Considering)\ \ g(x)\ \ text(when)\ \ x>0,`

`g(x)`	`<0`
`sinx -x`	`<0`
`sin x`	`<x\ \ \ text(for all)\ x>0`

ii.	`f(x)`	`=sin x – x + x^3/6`
	`f prime (x)`	`=cos x – 1 + x^2/2`
	`f ″ (x)`	`=x – sin x`
	`:.\ f″ (x)`	`> 0\ \ \ \ text{(using part (i))}`

`:. f(x)\ \ text(is concave up for)\ \ x > 0.`

iii. `f″(x)>0\ \ \ \ text{(part (ii))}`

`=>f′(x)\ \ text(is an increasing function)`

`text(When)\ \ x=0,\ \ f′(0)=0`

`=>f′(x)>0\ \ \ text(for)\ \ x>0`

`:. f(x)\ \ text(has a positive gradient that steepens)`

`text(for)\ \ x>0, and f(0)=0`

`f(x)`	`>0`
`sin x – x + x^3/6`	`>0`

`:.sin x > x – x^3/6\ \ \ text(for)\ \ x>0`

Mechanics, EXT2 M1 2007 HSC 6b

A raindrop falls vertically from a high cloud. The distance it has fallen is given by

`x = 5 log_e ((e^(1.4 t) + e^(-1.4 t))/2)`

where `x` is in metres and `t` is the time elapsed in seconds.

Show that the velocity of the raindrop, `v` metres per second, is given by

`v = 7 ((e^(1.4 t) - e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))` (2 marks)

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Hence show that `v^2 = 49 (1 - e^(-(2x)/5)).` (2 marks)

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Hence, or otherwise, show that `ddot x = 9.8 - 0.2v^2.` (2 marks)

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The physical significance of the 9.8 in part (iii) is that it represents the acceleration due to gravity.

What is the physical significance of the term `–0.2 v^2?` (1 mark)

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Estimate the velocity at which the raindrop hits the ground. (1 mark)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`-0.2v^2\ \ text(is the air resistance)`
`7\ \ text(ms)^-1`

Show Worked Solution

i.	`x`	`= 5 log_e ((e^(1.4t) + e^(-1.4t))/2)`
	`v=dx/dt`	`= (5(1.4e^(1.4t) – 1.4e^(-1.4t)))/(e^(1.4t) + e^(-1.4t))`
		`= 7 ((e^(1.4 t) – e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))`

ii.	`v^2`	`=49 ((e^(1.4 t) – e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))^2`
		`=49 ((e^(2.8 t) + e^(-2.8 t)-2)/(e^(2.8 t) + e^(-2.8 t)+2))`
		`=49 ((e^(2.8 t) + e^(-2.8 t)+2-4)/(e^(2.8 t) + e^(-2.8 t)+2))`
		`=49 (((e^(1.4t) + e^(-1.4t))^2-2^2)/((e^(1.4t) + e^(-1.4t))^2))`
		`=49 (1 – (2/(e^(1.4t) + e^(-1.4t)))^2)`

`text(S)text(ince)\ \ x`	`= 5 log_e ((e^(1.4 t) + e^(-1.4 t))/2)`
`e^(x/5)`	`=(e^(1.4 t) + e^(-1.4 t))/2`

`:. v^2`	`=49 (1 – (e^(- x/5))^2)`
	`=49(1-e^(- (2x)/5))`

iii.	`ddotx`	`=d/(dx) (1/2 v^2)`
		`=49/2 xx 2/5 xx e^(-(2x)/5)`
		`=49/5 e^(-(2x)/5)`
		`=49/5 (1- v^2/49),\ \ \ \ text{(from part (ii))}`
		`=9.8 – 0.2v^2`

iv. `-0.2v^2\ \ text(is the wind resistance acting on the rain drop.)`

v. `text(Terminal velocity occurs when)\ \ ddot x=0`

`9.8 – 0.2v^2`	`=0`
`v^2`	`=49`
`v`	`=7,\ \ \ \ (v > 0)`

`:.\ text(The rain drop hits the ground travelling at)\ \ 7\ \ text(ms)^-1`

Proof, EXT2 P2 2007 HSC 6a

Use the binomial theorem `(a + b)^n = a^n + ((n), (1)) a^(n - 1) b + … + b^n`

to show that, for `n >= 2`,

`2^n > ((n), (2)).` (1 mark)

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Hence show that, for `n >= 2`,

`(n + 2)/2^(n - 1) < (4n + 8)/(n(n - 1)).` (2 marks)

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Prove by induction that, for integers `n >= 1`,

`1 + 2 (1/2) + 3 (1/2)^2 + … + n (1/2)^(n - 1) = 4 - (n + 2)/2^(n - 1).` (3 marks)

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Hence determine the limiting sum of the series

`1 + 2 (1/2) + 3 (1/2)^2 + ….` (1 mark)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`4`

Show Worked Solution

i. `text(Let)\ \ a = 1,\ \ b = 1`

`2^n`	`= 1 + ((n), (1)) + ((n), (2)) + … + ((n), (n))`
`:. 2^n`	`> ((n), (2))\ \ \ \ text(for)\ n>=2`

ii.	`((n), (2))`	`= (n(n – 1))/(2 xx 1)`
	`:.2^n`	`>(n(n – 1))/(2 xx 1),\ \ \ \ \ text{(part (i))}`
	`1/2^n`	`<2/(n (n – 1))`
	`2/2^n`	`<4/(n (n – 1))`
	`1/2^(n – 1) `	`<4/(n (n – 1))`
	`:.(n + 2)/2^(n – 1)`	`< (4n + 8)/(n (n – 1)),\ \ \ \ \ (n+2>0)`

iii. `text(If)\ n = 1,`

`text(LHS) = 1`

`text(RHS) = 4 – (1+2)/2^0 = 1 = text(LHS)`

`:.\ text(True for)\ \ n = 1`

`text(Assume result is true for)\ \ n = k,\ \ text(i.e.)`

`1 + 2 (1/2) + 3 (1/2)^2 + … + k (1/2)^(k – 1) = 4 – (k + 2)/2^(k – 1).`

`text(Prove result is true for)\ \ n = k + 1,\ \ text(i.e.)`

`1 + 2 (1/2) + … + k (1/2)^(k – 1) + (k + 1) (1/2)^k = 4 – (k + 3)/2^k`

`text(LHS)`	`=1 + 2 (1/2) + … + k (1/2)^(k – 1) + (k + 1) (1/2)^k`
	`=4 – (k + 2)/2^(k – 1) + (k + 1)/2^k`
	`=4 – (2k + 4 – k – 1)/2^k`
	`=4 – (k + 3)/2^k`
	`=\ text(RHS)`

`=>text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k.`

`:.\ text(S)text(ince true for)\ \ n = 1, text(by PMI, true for integral)\ \ n>=1`

iv. `lim_(n -> oo) (4 – (n + 2)/2^(n – 1))=4 – lim_(n -> oo) ((n + 2)/2^(n – 1))`

`text{Using part (ii):}`

`(n + 2)/2^(n – 1)`	`< (4n + 8)/(n (n – 1))`
`lim_(n -> oo) ((4n + 8)/(n (n – 1)))`	`= 0`
`:. lim_(n -> oo) ((n + 2)/2^(n – 1)) `	`= 0`

`:.lim_(n -> oo) (4 – (n + 2)/2^(n – 1)) = 4 – 0 = 4`

Polynomials, EXT2 2007 HSC 5d

In the diagram, `ABCDE` is a regular pentagon with sides of length `1`. The perpendicular to `AC` through `B` meets `AC` at `P.`

Copy or trace this diagram into your writing booklet.

Let `u = cos\ pi/5`.
Use the cosine rule in `Delta ACD` to show that `8u^3 - 8u^2 + 1 = 0.` (2 marks)
One root of `8x^3 - 8x^2 + 1 = 0` is `1/2`.
Find the other roots of `8x^3 - 8x^2 + 1 = 0` and hence find the exact value of `cos\ pi/5.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`(1 + sqrt 5)/4`

Show Worked Solution

(i)	`u`	`= cos\ pi/5`
	`AP`	`=u`
	`AC`	`=AD=2u`

`text(By symmetry,)\ \ /_ DAE = pi/5`

`text(Angle sum of a pentagon)=(n-2) pi=3 pi`

`:.text(Each angle of a regular pentagon)=(3 pi)/5`

`:.\ /_ CAD = pi/5`

`text(Using the cosine rule in)\ \ Delta ACD`

`1^2`	`= (2u)^2 + (2u)^2 – 2 xx 2u xx 2u cos\ pi/5`
`1`	`= 8u^2 – 8u^2 xx u`
`:.8u^3 – 8u^2 + 1 = 0`

(ii) `(2x – 1)\ \ text(is a factor)`

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

`text(Other roots occur when)`

`4x^2 – 2x – 1 = 0`

`x`	`=(2 +- sqrt (4 + 16))/8`
	`=(2 +- sqrt 20)/8`
	`=(1 +- sqrt 5)/4`

`=>u\ \ text(is a root of)\ \ 8u^3 – 8u^2 + 1 = 0,\ \ \ \ (u>0)`

`:.cos\ pi/5 = (1 + sqrt 5)/4`

Calculus, EXT2 C1 2007 HSC 5c

Write `(x - 1) (5 - x)` in the form `b^2 - (x - a)^2`, where `a` and `b` are real numbers. (1 mark)

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Using the values of `a` and `b` found in part (i) and making the substitution `x - a = b sin theta`, or otherwise, evaluate

` int_1^5 sqrt ((x - 1) (5 - x))\ dx.` (2 marks)

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`2^2 – (x – 3)^2`
`2 pi`

Show Worked Solution

i. `(x – 1) (5 – x)`	`= 5x – x^2 -5+ x`
	`= -(x^2 – 6x + 5)`
	`= 4 – (x – 3)^2`
	`= 2^2 – (x – 3)^2`

ii.	`text(Let)\ \ x – 3`	`= 2 sin theta`
	`dx`	`= 2 cos theta\ d theta`

`text(When)\ \ x = 1,\ \ theta = -pi/2`

`text(When)\ \ x = 5,\ \ theta = pi/2`

`:.int_1^5 sqrt (4 – (x – 3)^2)\ dx`

`=int_((-pi)/2)^(pi/2) sqrt ((4 – 4 sin^2 theta))*2 cos theta\ d theta`

`=4 int_((-pi)/2)^(pi/2) sqrt(cos^2 theta) * cos theta\ d theta`

`=4 int_((-pi)/2)^(pi/2) cos^2 theta\ d theta`

`=2 int_((-pi)/2)^(pi/2) (1 + cos 2 theta)\ d theta`

`=2[theta + (sin 2 theta)/2]_((-pi)/2)^(pi/2)`

`=2[(pi/2 + 0) – ((-pi)/2 – 0)]`

`=2 pi`

Conics, EXT2 2007 HSC 5b

The points at `P(x_1, y_1)` and `Q(x_2, y_2)` lie on the same branch of the hyperbola

`x^2/a^2 - y^2/b^2 = 1.`

The tangents at `P` and `Q` meet at `T(x_0, y_0).`

Show that the equation of the tangent at `P` is
1. `(x x_1)/a^2 - (y y_1)/b^2 = 1.` (2 marks)
Hence show that the chord of contact, `PQ`, has equation
1. `(x x_0)/a^2 - (y y_0)/b^2 = 1.` (2 marks)
The chord `PQ` passes through the focus `S(ae, 0)`, where `e` is the eccentricity of the hyperbola. Prove that `T` lies on the directrix of the hyperbola. (1 mark)

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

Show Worked Solution

(i)	`x^2/a^2 – y^2/b^2`	`= 1`
`text(S)text(ince)\ \ P(x_1, y_1)\ \ text(lies on the hyperbola)`
	`x_1^2/a^2 – y_1^2/b^2`	`=1\ \ \ …\ text{(*)}`
	`(2x)/a^2 – (2y)/b^2 * (dy)/(dx)`	`=0`
	`:.\ (dy)/(dx)`	`= (b^2 x)/(a^2 y)`

`text(At)\ \ P(x_1, y_1),\ \ (dy)/(dx) = (b^2 x_1)/(a^2 y_1)`

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

`y – y_1`	`=(b^2 x_1)/(a^2 y_1) (x – x_1)`
`(y y_1)/b^2 – (y_1^2)/b^2`	`=(x x_1)/a^2 – (x_1^2)/a^2`
`(x x_1)/a^2 – (y y_1)/b^2`	`=(x_1^2)/a^2 – (y_1^2)/b^2\ \ \ \ text{(using (*) above)}`
`(x x_1)/a^2 – (y y_1)/b^2`	`=1`

(ii) `text(Equation of tangent at)\ \ Q\ \ text(is)`

`(x x_2)/a^2 – (y y_2)/b^2 = 1`

`T (x_0, y_0)\ \ text(sits on both tangents)`

`:.\ (x_0 x_1)/a^2 – (y_0 y_1)/b^2 = 1`

`:.\ (x_0 x_2)/a^2 – (y_0 y_2)/b^2 = 1`

`=>P(x_1,y_1) and Q(x_2,y_2)\ \ text(both lie on the line)`

`(x_0 x)/a^2 – (y_0 y)/b^2 = 1`

`:.\ text(The equation of)\ \ PQ\ \ text(is)\ \ (x_0 x)/a^2 – (y_0 y)/b^2 = 1.`

(iii) `text(Substitute)\ \ S(ae, 0)\ \ text(into)\ \ PQ,`

`(aex_0)/a^2 – 0/b^2`	`=1`
`aex_0`	`=a^2`
`x_0`	`= a/e`

`:.\ T\ \ text(lies on the directrix of the hyperbola.)`

Combinatorics, EXT1′ A1 2007 HSC 5a

A bag contains 12 red marbles and 12 yellow marbles. Six marbles are selected at random without replacement.

Calculate the probability that exactly three of the selected marbles are red. Give your answer correct to two decimal places. (1 mark)

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Hence, or otherwise, calculate the probability that more than three of the selected marbles are red. Give your answer correct to two decimal places. (2 marks)

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Show Answers Only

`0.36`
`0.32`

Show Worked Solution

i. `\ ^12C_3= text(# Ways of selecting 3 R or Y from 12.)`

`\ ^24C_6=text(# Ways of selecting 6 from 24.)`

`P text{(exactly 3R)}`	`=(\ ^12C_3 xx \ ^12C_3)/(\ ^24C_6)`
	`=(220 xx 220)/(134\ 596)`
	`=0.36\ \ text{(to 2 d.p.)}`

ii. `text(Solution 1)`

`text(S)text(ince)\ \ P text{(> 3 Red)`	`=Ptext{(< 3 Red)}`
`P text{(> 3 Red)`	`=1/2[1-Ptext{(exactly 3R)}]`
	`=1/2(1-0.36)`
	`=0.32`

`text(Solution 2)`

`P (> 3R)`	`=P (4R) + P (5R) + P (6R)`
	`=(\ ^12C_4 \ ^12C_2 + \ ^12C_5 \ ^12C_1 + \ ^12C_6 xx \ ^12C_0)/(\ ^24C_6)`
	`=(43\ 098)/(134\ 596)`
	`=0.32\ \ text{(to 2 d.p.)}`

Polynomials, EXT2 2007 HSC 4d

The polynomial `P(x) = x^3 + qx^2 + rx + s` has real coefficients. It has three distinct zeros, `alpha, - alpha and beta.`

Prove that `qr = s.` (3 marks)
The polynomial does not have three real zeros. Show that two of the zeros are purely imaginary. (A number is purely imaginary if it is of the form `iy`, with `y` real and `y != 0.`) (2 marks)

Show Answers Only

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

Show Worked Solution

(i) `text(Solution 1)`

`P(x) = x^3 + qx^2 + rx + s\ \ \ =>\ text(Roots)\ \ alpha, -alpha, beta.`

`alpha – alpha + beta`	`= -b/a=-q`
`=>beta`	`=-q\ \ \ …\ (1)`
`alpha * (- alpha) * beta`	`=-d/a= -s`
`=>alpha^2 beta`	`=s\ \ \ …\ (2)`
`-alpha^2 + alpha beta – alpha beta`	`= c/a=r`
`=>alpha^2`	`=-r\ \ \ …\ (3)`
`text{Substitute (1) and (3) into (2)`
`(-r) xx (-q)`	`= s`
`:.\ qr`	`= s`

`text(Solution 2)`

`text(S)text(ince)\ \ beta`	`= -q,`
`P(-q)`	`=0`

`(-q)^3 + q(-q)^2 + r(-q) + s`	`= 0`
`:. qr`	`=s`

(ii) `text(S) text(ince)\ \ q, r, s\ \ text(are real,)`

`=>beta = -q\ \ text(is real and)\ \ (x+q)\ \ text(is a factor)`

`=>alpha and – alpha\ \ text(are both complex.)`

`:.\ P(x) = (x + q) (x^2 + r)`

`text(If)\ \ x^2 + r`	`=0`
`x`	`= +- sqrt (-r)`

`text(S)text(ince)\ \ α^2=-r\ \ \ \ \ text{(see (3) in part (i))}`

`=> r>0\ \ text(for α to be complex)`

`:. x= +- i sqrt r\ \ \ \ text{(which are both purely imaginary)`

Volumes, EXT2 2007 HSC 4c

The base of a solid is the region bounded by the curve `y = log_e x`, the `x`-axis and the lines `x = 1` and `x = e`, as shown in the diagram.

Vertical cross-sections taken through this solid in a direction parallel to the `x`-axis are squares. A typical cross-section, `PQRS`, is shown.

Find the volume of the solid. (3 marks)

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`2e – e^2/2 – 1/2\ \ text(u³)`

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`text(Length of)\ PQ`	`=e-x`
`text(Area of)\ PQRS`	`= (e-x)^2`

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

`text(When)\ x = e,\ \ y = 1`

`δV`	`=(e-x)^2\ δy`
`:.V`	`=int_0^1 (e – x)^2\ dy`
	`=int_0^1 (e – e^y)^2\ dy`
	`=int_0^1 (e^2 – 2e^(y + 1) + e^(2y))\ dy`
	`=[e^2 y – 2e^(y + 1) + e^(2y)/2]_0^1`
	`=[(e^2 – 2e^2 + e^2/2) – (0 – 2e + 1/2)]`
	`=2e – e^2/2 – 1/2\ \ text(u³)`

Harder Ext1 Topics, EXT2 2007 HSC 4b

Show that
`sin 3 theta = 3 sin theta cos^2 theta - sin^3 theta.` (2 marks)
Show that
`4 sin theta sin (theta + pi/3) sin (theta + (2 pi)/3) = sin 3 theta.` (2 marks)
Write down the maximum value of
`sin theta sin (theta + pi/3) sin (theta + (2 pi)/3).` (1 mark)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`1/4`

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(i) `sin (2 theta + theta)`	`= sin 2 theta cos theta + cos 2 theta sin theta`
	`= 2 sin theta cos theta cos theta + (cos^2 theta – sin^2 theta) sin theta`
	`= 2 sin theta cos^2 theta + cos^2 theta sin theta – sin^3 theta`
	`= 3 sin theta cos^2 theta – sin^3 theta`

(ii)	`text(LHS)`	`=4 sin theta (sin theta cos\ pi/3 + cos theta sin\ pi/3) (sin theta cos\ (2 pi)/3 + cos theta sin\ (2 pi)/3)`
		`=4 sin theta ((sin theta)/2 + (sqrt 3 cos theta)/2)(- (sin theta)/2 + (sqrt 3 cos theta)/2)`
		`=sin theta (sqrt 3 cos theta + sin theta)(sqrt 3 cos theta – sin theta)`
		`=sin theta (3 cos^2 theta – sin^2 theta)`
		`=3 sin theta cos^2 theta – sin^3 theta`
		`=sin 3 theta`

(iii) `sin theta sin (theta + pi/3) sin (theta + (2 pi)/3)=1/4 sin\ 3 theta\ \ \ text{(part (ii))}`

`:.\ text(Maximum value)=1/4`

Volumes, EXT2 2012 HSC 9 MC

The diagram shows the graph `y = x(2 − x)` for `0 ≤ x ≤ 2`. The region bounded by the graph and the `x`-axis is rotated about the line `x = –2` to form a solid.

Which integral represents the volume of the solid?

`2pi int_0^2 x(2 − x)^2\ dx`
`2pi int_0^2 x^2(2 − x)\ dx`
`2pi int_0^2 x(2 − x)(2 + x)\ dx`
`2pi int_0^2 x(2 − x)(x − 2)\ dx`

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`C`

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`δV`	`= 2pi r\ y\ δx`
	`= 2pi(x + 2)x(2 − x)δx`
`:.V`	`= 2pi int_0^2 x(2 − x)(x + 2) dx.`

`=>C`

Complex Numbers, EXT2 N1 2012 HSC 3 MC

The complex number `z` is shown on the Argand diagram below.

Which of the following best represents `i barz`?

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`A`

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`ibarz\ text(is)\ bar z\ text(rotated)\ 90^@\ text(anticlockwise.)`

`=>A`

`y`	`=sqrt (b – x^2)`
`dy/dx`	`=(-2x)/(2 sqrt(b-x^2))`

`(-1)/sqrt(b-1)`	`=- 1/3`
`sqrt(b-1)`	`=3`
`b-1`	`=9`
`b`	`=10`

`-7a^3 + 14a^2`	`=0`
`-7a^2 (a – 2)`	`=0`
`:. a = 2,`	`\ \ \ a!=0`

`dy/dx`	`=(-1)/sqrt(b-1)`

`text(Midpoint)`	`= ((d + 1)/2, (2 + (-5))/2)`
	`= ((d + 1)/2, -3/2)`