Band 4 – Page 100

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)

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Hence show that for `0 <= x <= 1/sqrt 2`

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

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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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`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)

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`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)

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`(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)

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`y = 1/(f(x))` (2 marks)

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

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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) ---

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`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)

--- 6 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.	`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)

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

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

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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)

--- 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)}`
`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)

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

Show Answers Only

`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)

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)	`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)

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

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)

Show Answers Only

`2e – e^2/2 – 1/2\ \ text(u³)`

Show Worked Solution

`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)

Show Answers Only

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

Show Worked Solution

(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`

Show Answers Only

`C`

Show Worked Solution

`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`?

Show Answers Only

`A`

Show Worked Solution

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

`=>A`

Harder Ext1 Topics, EXT2 2007 HSC 4a

Two circles intersect at `A` and `B`.

The lines `LM` and `PQ` pass through `B`, with `L` and `P` on one circle and `M` and `Q` on the other circle, as shown in the diagram.

Copy or trace this diagram into your writing booklet.

Show that `/_ LAM = /_ PAQ.` (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

`text(Join)\ \ LA, PA, MA, QA`

`text(Let)\ \ /_ ALM = x^@,\ \ /_AML = y^@`

`/_ APB`	`= /_ ALB = x^@\ \ text{(angles in the same segment on arc}\ AB text{)}`
`text(Similarly, but in the circle through)\ \ ABMQ`
`/_ AQB`	`= /_ AMB = y^@\ \ text{(angles in the same segment on arc}\ AB text{)}`

`/_ LAM`	`= 180^@ – (x + y)^@\ \ \ text{(angle sum of}\ \ Delta LAMtext{)}`
`/_ PAQ`	`= 180^@ – (x + y)^@\ \ text{(angle sum of}\ \ Delta PAQtext{)}`
`:.\ /_ LAM = /_ PAQ`

Mechanics, EXT2 2007 HSC 3d

A particle `P` of mass `m` undergoes uniform circular motion with angular velocity `omega` in a horizontal circle of radius `r` about `O`. It is acted on by the force due to gravity, `mg`, a force `F` directed at an angle `theta` above the horizontal and a force `N` which is perpendicular to `F`, as shown in the diagram.

By resolving forces horizontally and vertically, show that
1. `N = mg cos theta - m r omega^2 sin theta.` (3 marks)
For what values of `omega` is `N > 0?` (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`g/r cot theta`

Show Worked Solution

(i)

`text(Resolving forces vertically)`

`F sin theta + N cos theta`	`= mg`
`N cos theta`	`=mg-F sin theta\ \ \ …\ (1)`
`text(Resolving forces horizontally)`
`F cos theta – N sin theta`	`= m r omega^2`
`N sin theta`	`=F cos theta-m r omega^2\ \ \ …\ (2)`

`text{Multiply (1)}\ xx cos theta and text{(2)}\ xx sin theta`

`N cos^2 theta`	`=mg cos theta -F sin theta cos theta\ \ \ …\ (3)`
`N sin^2 theta`	`=F cos theta sin theta-m r omega^2 sin theta\ \ \ …\ (4)`

`text{Add (3) + (4)}`

`:.N`	`=mg cos theta-F sin theta cos theta + F cos theta sin theta-m r omega^2 sin theta`
	`=mg cos theta – m r omega^2 sin theta`

(ii) `text(When)\ \ N > 0,`

`mg cos theta`	`>m r omega^2 sin theta`
`omega^2`	`<(g cos theta)/(r sin theta)`
	`<g/r cot theta`

Volumes, EXT2 2007 HSC 3c

Use the method of cylindrical shells to find the volume of the solid formed when the shaded region bounded by

`y = 0,\ \ y = (log_e x)/x,\ \ x = 1 and x = e`

is rotated about the `y`-axis. (4 marks)

Show Answers Only

`2 pi\ \ text(u³)`

Show Worked Solution

(i) `y = 0,\ \ y = (log_e x)/x,\ \ x = 1 and x = e`

MARKER’S COMMENT: A “significant number” of students were confused about the radius of the cylindrical shell.

`δV`	`=2 pi x y\ δx`
`V`	`=2 pi int_1^e x*(log_e x)/x\ dx`
	`= 2pi int_1^e log_e x\ dx`

`text(Integrating by parts,)`

`u`	`=log_e x,`	`u′`	`=1/x`
`v′`	`=1,`	`v`	`=x`

`:.V`	`=2pi [[x log_e x]_1^e – int_1^e x * 1/x\ dx]`
	`=2pi [(e – log_e 1) – [x]_1^e]`
	`=2pi (e – (e – 1))`
	`=2pi\ \ text(u³)`

Functions, EXT1′ F2 2007 HSC 3b

The zeros of `x^3 - 5x + 3` are `alpha, beta` and `gamma.`

Find a cubic polynomial with integer coefficients whose zeros are `2 alpha, 2 beta` and `2 gamma.` (2 marks)

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

Show Answers Only

`x^3 – 20x + 24`

Show Worked Solution

`text(Solution 1)`

`α+β+γ`	`=0`
`:.2α+2β+2γ`	`=2(α+β+γ)`
	`=0`
`αβ+βγ+γα`	`=-5`
`:.2α2β+2β2γ+2γ2α`	`=4(αβ+βγ+γα)`
	`=-20`
`αβγ`	`=-3`
`:.2α2β2γ`	`=8αβγ`
	`=-24`

`:.\ text(Polynomial is)\ \ x^3 – 20x + 24`

`text(Solution 2)`

`P(x) = x^3 – 5x + 3`

`text(New zeros are)\ \ 2 alpha, 2 beta and 2 gamma`

`:.H(x)`	`=(x/2)^3 – 5(x/2) + 3`
	`=x^3/8-(5x)/2+3`

`:.\ text(New Polynomial with integer coefficients is)`

`x^3 – 20x + 24`

Functions, EXT1′ F1 2007 HSC 3a

The diagram shows the graph of `y = f(x)`. The line `y = x` is an asymptote.

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

`f(-x).` (1 mark)

--- 8 WORK AREA LINES (style=lined) ---
`f(|\ x\ |).` (2 marks)

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`f(x) - x.` (2 marks)

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

Show Answers Only

Show Worked Solution

MARKER’S COMMENT: In part (ii), a significant number of students graphed `y=|f(x)|`.

ii.

iii.

Complex Numbers, EXT2 N2 2007 HSC 2d

The points `P,Q` and `R` on the Argand diagram represent the complex numbers `z_1, z_2` and `a` respectively.

The triangles `OPR` and `OQR` are equilateral with unit sides, so `|\ z_1\ | = |\ z_2\ | = |\ a\ | = 1.`

Let `omega = cos pi/3 + i sin pi/3.`

Explain why `z_2 = omega a.` (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Show that `z_1 z_2 = a^2.` (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Show that `z_1` and `z_2` are the roots of `z^2 - az + a^2 = 0.` (2 marks)

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

Show Answers Only

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

Show Worked Solution

i. `text(S) text(ince)\ \ Delta ORQ\ \ text(is equilateral, each angle is)\ \ pi/3\ \ text(radians.)`

`text(From)\ \ R(a):`

`Q(z_2)\ \ text(is an anticlockwise rotation through)\ \ pi/3.`

`:. z_2 = a(cos\ pi/3+i sin\ pi/3)=omega a.`

ii. `text(Solution 1)`

`text(Similarly,)\ \ a`	`=z_1 omega`
`z_1`	`=a/omega`
`:z_1z_2`	`=a/omega xx omega a`
	`=a^2`

`text(Solution 2)`

`P(z_1)\ \ text(is a clockwise rotation of)\ \ R(a)\ \ text(through)\ \ pi/3.`

`:.z_1`	`= bar omega a.`
`:. z_1 z_2`	`= bar omega a xx omega a`
	`=a^2(cos pi/3 – i sin pi/3) xx (cos pi/3 + i sin pi/3)`
	`=a^2(cos^2 pi/3 + sin^2 pi/3)`
	`= a^2`

iii. `z^2-az + a^2 = 0`

`text(Let the roots be)\ \ alpha and beta.`

`alpha + beta`	`=-b/a=a`
`alpha beta`	`=c/a=a^2`

`z_1 z_2`	`= a^2\ \ \ \ \ text{(part (ii))}`
`z_1 + z_2`	`=bar omega a + omega a`
	`=(cos pi/3 + i sin pi/3 + cos pi/3-i sin pi/3) a`
	`=2 cos pi/3 xx a`
	`=2 xx 1/2 xx a`
	`=a`

`:.\ z_1 and z_2\ \ text(are the roots of)\ \ z^2-az + a^2 = 0.`

Complex Numbers, EXT2 N2 2007 HSC 2c

The point `P` on the Argand diagram represents the complex number `z`, where `z` satisfies

`1/z + 1/bar z = 1.`

Give a geometrical description of the locus of `P` as `z` varies. (3 marks)

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

Show Answers Only

`text(Locus is a circle, centre)\ (1, 0), text(radius 1,)`

`text(excluding the point)\ (0, 0).`

Show Worked Solution

`1/z + 1/bar z`	`=1/(x + iy) +1/(x – iy)`
	`=(x – iy + x+iy)/(x^2+y^2)`
	`=(2x)/(x^2+y^2)`

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

`:.\ text(Locus is a circle, centre)\ (1, 0), text(radius 1,)`

`text(excluding the point)\ (0, 0).`

Calculus, EXT2 C1 2007 HSC 1e

It can be shown that

`2/(x^3 + x^2 + x + 1) = 1/(x + 1) - x/(x^2 + 1) + 1/(x^2 + 1).` (Do NOT prove this.)

Use this result to evaluate `int_(1/2)^2 2/(x^3 + x^2 + x + 1)\ dx.` (4 marks)

Show Answers Only

`tan^-1 2 – tan^-1 1/2`

Show Worked Solution

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

`=int_(1/2)^2 (1/(x + 1) – x/(x^2 + 1) + 1/(x^2 + 1)) dx`

`=[log_e(x + 1) – 1/2 log_e (x^2 + 1) + tan^-1 x]_(1/2)^2`

`=[(log_e 3 – 1/2 log_e 5 + tan^-1 2) – (log_e 3/2 – 1/2 log_e 5/4 + tan^-1 1/2)]`

`=log_e\ 3/sqrt5 -log_e (3/2 xx 2/sqrt5) + tan^-1 2 – tan^-1 1/2`

`=log_e (3/sqrt(5) xx sqrt (5)/3) + tan^-1 2 – tan^-1 1/2`

`=tan^-1 2 – tan^-1 1/2`

Calculus, EXT2 C1 2007 HSC 1d

Evaluate `int_0^(3/4) x/sqrt (1 - x)\ dx.` (4 marks)

Show Answers Only

`5/12`

Show Worked Solution

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

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

`text(When)\ \ x = 3/4,\ \ u = 1/4`

`int_0^(3/4) (x\ dx)/sqrt (1 – x)`	`=int_1^(1/4) ((1 – u)(-du))/sqrt u`
	`=int_(1/4)^1 (u^(-1/2) – u^(1/2)) du`
	`=[2 u^(1/2) – 2/3 u^(3/2)]_(1/4)^1`
	`=[(2 – 2/3) – (1 – 1/12)]`
	`=5/12`

Harder Ext1 Topics, EXT2 2015 HSC 10 MC

Consider the expansion of

`(1 + x + x^2 + … + x^n) (1 + 2x + 3x^2 + … + (n + 1) x^n).`

What is the coefficient of `x^n` when `n = 100?`

`4950`
`5050`
`5151`
`5253`

Show Answers Only

`C`

Show Worked Solution

`text(By expanding, the co-efficients of)\ \ x^n\ \ text(will be,)`

`1+2+3+…+100+101`

`=> text(A.P. with)\ \ a=1,\ \ d=1,\ \ n=101`

`:.S_n`	`=n/2(a + l)`
	`=101/2(1+101)`
	`=5151`

`=> C`

Complex Numbers, EXT2 N2 2015 HSC 9 MC

The complex number `z` satisfies `| z - 1 | = 1.`

What is the greatest distance that `z` can be from the point `i` on the Argand diagram?

`1`
`sqrt 5`
`2 sqrt 2`
`sqrt 2 + 1`

Show Answers Only

`D`

Show Worked Solution

`| z – 1 | = 1\ \ text{is a circle, centre (1,0), radius 1.}`

`text(Consider the graph:)`

`text(Distance from)\ \(0, i)\ \ text(to the centre)=sqrt2`

`:.\ text(Greatest distance)=sqrt2 + text(radius)=sqrt2+1`

`=> D`

Complex Numbers, EXT2 N1 2014 HSC 8 MC

The Argand diagram shows the complex numbers `w`, `z` and `u`, where `w` lies in the first quadrant, `z` lies in the second quadrant and `u` lies on the negative real axis.

Which statement could be true?

`u = zw` and `u = z + w`
`u = zw` and `u = z − w`
`z = uw` and `u = z + w`
`z = uw` and `u = z − w`

Show Answers Only

`B`

Show Worked Solution

`text(Using the parallelogram method, it could be true that)`

`u = z − w`

`=>\ text{Eliminate (A) and (C)}`

`text{arg}(u)`	`=\ text{arg}(w) + text{arg}(z)`
`u`	` = zw`

`=> B`

Functions, EXT1′ F1 2013 HSC 9 MC

Which diagram best represents the graph `y = (sin x)/x?`

Show Answers Only

`B`

Show Worked Solution

`f(x)`	`=(sin x)/x`
`f(−x)`	`=(sin (−x))/-x`
	`=(sin (x))/x`
	`=f(x)`

`=>f(x)\ \ text(is even)`

`lim_(x -> 0) (sin(x))/x =1`

`=> B`

Mechanics, EXT2 2013 HSC 7 MC

The angular speed of a disc of radius `5` cm is `10` revolutions per minute.

What is the speed of a mark on the circumference of the disc?

`50\ \ text(cm min)^-1`
`1/2\ \ text(cm min)^-1`
`100 pi\ \ text(cm min)^-1`
`1/(4 pi)\ \ text(cm min)^-1`

Show Answers Only

`C`

Show Worked Solution

`omega`	`= 2 pi f`
	`= 2 pi xx 10`
	`= 20 pi`

` v`	`=r omega`
	`= 5 xx 20 pi`
	`= 100 pi\ \ text(cm min)^-1`

`=> C`

Complex Numbers, EXT2 N2 2013 HSC 5 MC

Which region on the Argand diagram is defined by `pi/4 <= | z - 1 | <= pi/3?`

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`B`

Show Worked Solution

`| z-1 |=r\ \ text{is a circle with centre (1, 0), and}`

`text(radius)\ \ r.`

`=> B`

Complex Numbers, EXT2 N1 2013 HSC 3 MC

The Argand diagram below shows the complex number `z.`

Which diagram best represents `z^2?`

Show Answers Only

`D`

Show Worked Solution

`text(Consider)\ \ z\ \ text(in polar form:)`

`\|\ z\ \|`	`< 1`
`:.\|\ z^2\ \|`	`< \|\ z\ \|`

`text(arg) (z^2) = 2text(arg) (z)`

`=> D`

Proof, EXT2 P2 2009 HSC 8a

Using the substitution `t = tan\ theta/2`, or otherwise, show that

`qquad cot theta + 1/2 tan\ theta/2 = 1/2 cot\ theta/2.` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Use mathematical induction to prove that, for integers `n >= 1`,

`qquad sum_(r = 1)^n 1/2^(r - 1) tan x/2^r = 1/2^(n - 1) cot\ x/2^n - 2 cot x.` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Show that `lim_(n -> oo) sum_(r = 1)^n 1/2^(r - 1) tan\ x/2^r = 2/x - 2 cot x.` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence find the exact value of
`qquad tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + ….` (2 marks)

--- 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)}`
`4/pi`

Show Worked Solution

i. `t=tan\ theta/2, \ \ \ sin theta=(2t)/(1+t^2),\ \ \ cos theta=(1-t^2)/(1+t^2)`

`text(Prove)\ \ cot theta + 1/2 tan\ theta/2 = 1/2 cot\ theta/2`

`text(LHS)`	`=cot theta + 1/2 tan\ theta/2`
	`=cos theta/sin theta+ 1/2 tan\ theta/2`
	`=(1-t^2)/(2t)+t/2`
	`=(1-t^2+t^2)/(2t)`
	`=1/(2t)`
	`=1/2 cot\ theta/2`
	`=\ text(RHS)`

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

`text(LHS)`	`=1/2^0 tan\ x/2^1=tan\ x/2`
`text(RHS)`	`=1/2^0 cot\ x/2 – 2 cot x`
	`=cot\ x/2 – 2 cot x`

`text{Using part (i)},\ \ 1/2 tan\ theta/2`	`= 1/2 cot\ theta/2 – cot theta,`
`:.tan\ theta/2`	`= cot\ theta/2 – 2 cot theta`

`text(RHS)`	`=tan\ x/2`
	`=\ text(LHS)`
`:.\ text(True for)\ \ n=1`

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

`text(i.e.)\ \ sum_(r = 1)^k 1/2^(r – 1) tan\ x/2^r = 1/2^(k – 1) cot\ x/2^k – 2 cot x.`

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

`text(i.e.)\ \sum_(r = 1)^(k + 1) 1/2^(r – 1) tan x/2^r = 1/2^k cot x/2^(k + 1) – 2 cot x`

`text(LHS)`	`=sum_(r = 1)^(k) 1/2^(r – 1) tan x/2^r + 1/2^k tan x/2^(k + 1)`
	`=1/2^(k – 1) cot x/2^k – 2 cot x + 1/2^k tan x/2^(k + 1)`
	`=1/2^(k – 1) (cot x/2^k + 1/2 tan x/2^(k +1)) – 2 cot x,\ \ \ \ text{(Let}\ \ theta=x/2^ktext{)}`
	`=1/2^(k – 1)(cot\ theta + 1/2 tan\ theta/2) – 2 cot x`
	`=1/2^(k – 1)(1/2 cot\ theta/2) – 2 cot x,\ \ \ \ text{(from part (i))}`
	`=1/2^k cot\ theta/2 – 2 cot x`
	`=1/2^k cot x/2^(k + 1) – 2 cot x`
	`=\ 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.`

iii. `lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan x/2^r `

`=lim_(n -> oo) (1/2^(n-1) cot x/2^n – 2 cot x)`

`=lim_(n -> oo) (2/x * x/2^n * 1/(tan x/2^n) – 2 cot x)`

`=2/x xx lim_(n -> oo) ((x/2^n)/(tan x/2^n)) – 2 cot x`

`=>text(As)\ \ n -> oo,\ \ x/2^n=theta->0, and`

`=>lim_(theta-> 0) (theta)/(tan theta) =1`

`:.lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan x/2^r =2/x – 2 cot x`

iv. `tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + …`

`=lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan (pi/2)/2^r`

`=(2/(pi/2)) – 2 cot pi/2`

`=4/pi`

Complex Numbers, EXT2 N2 2009 HSC 7b

Let `z = cos theta + i sin theta.`

Show that `z^n + z^-n = 2 cos n theta`, where `n` is a positive integer. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Let `m` be a positive integer. Show that

`(2 cos theta)^(2m) = 2 [cos 2 m theta + ((2m), (1)) cos (2m - 2) theta + ((2m), (2)) cos (2m - 4) theta`

`+ … + ((2m), (m - 1)) cos 2 theta] + ((2m), (m)).` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Hence, or otherwise, prove that

`int_0^(pi/2) cos^(2m) theta\ d theta = pi/(2^(2m + 1)) ((2m), (m))`

where `m` is a positive integer. (2 marks)

--- 6 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.	`z`	`= cos theta + i sin theta`
	`z^n`	`= cos n theta + i sin n theta\ \ \ \ text{(De Moivre)}`
	`z^-n`	`= cos (-n theta) + i sin (-n theta)\ \ \ \ text{(De Moivre)}`
		`= cos n theta – i sin n theta`
	`z^n + z^-n`	`= cos n theta + i sin n theta + cos n theta – i sin n theta`
		`= 2 cos n theta,\ \ \ \ n > 0`

ii. `z + z^-1 = 2 cos theta`

`:.(2 cos theta)^(2m)`

`=(z + z^-1)^(2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 1) z^-1 + ((2m), (2)) z^(2m – 2) z^-2+`

` … + ((2m), (2m – 1)) z^1 z^-(2m – 1) + z^-(2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 2) + ((2m), (2)) z^(2m – 4)+`

` … + ((2m), (2m – 1)) z^-(2m – 2) + z^(-2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 2) + ((2m), (2)) z^(2m – 4) + … + ((2m), (m)) z^(2m-2m) …`

`+ ((2m), (2)) z^-(2m – 4) + ((2m), (1)) z^-(2m – 2) + z^(-2m)`

`=(z^(2m) + z^(-2m)) + ((2m), (1)) (z^(2m – 2) + z^-(2m – 2)) + ((2m), (2))`

`(z^(2m – 4) + z^-(2m – 4)) + … + ((2m), (m – 1)) (z + z^-1) + ((2m), (m))`

`=2 [cos 2 m theta + ((2m), (1)) cos (2m – 2) theta + ((2m), (2)) cos (2m – 4) theta`

`+ … + ((2m), (m – 1)) cos 2 theta] + ((2m), (m))`

iii. `int_0^(pi/2) cos^(2m) d theta`

`=1/(2^(2m)) int_0^(pi/2) (2 cos theta)^(2m)`

`=1/(2^(2m)) int_0^(pi/2)[2(cos 2 m theta + ((2m), (1)) cos (2m – 2) theta + ((2m), (2))`

`cos (2m – 4) theta + … + ((2m), (m – 1)) cos 2 theta) + ((2m), (m))] d theta`

`=1/(2^(2m)) [2((sin 2 m theta)/(2m) + ((2m), (1)) (sin (2m – 2) theta)/(2m – 2)`

`+ … + ((2m), (m – 1)) (sin 2 theta)/2) + ((2m), (m)) theta]_0^(pi/2)`

`=1/(2^(2m)) [2(0 + 0 + … + 0) + ((2m), (m)) pi/2 – (0)]`

`=pi/(2^(2m + 1)) ((2m), (m))`

Mechanics, EXT2 M1 2009 HSC 7a

A bungee jumper of height 2 m falls from a bridge which is 125 m above the surface of the water, as shown in the diagram. The jumper’s feet are tied to an elastic cord of length `L` m. The displacement of the jumper’s feet, measured downwards from the bridge, is `x` m.

The jumper’s fall can be examined in two stages. In the first stage of the fall, where `0 <= x <= L`, the jumper falls a distance of `L` m subject to air resistance, and the cord does not provide resistance to the motion. In the second stage of the fall, where `x > L`, the cord stretches and provides additional resistance to the downward motion.

The equation of motion for the jumper in the first stage of the fall is

`ddot x = g - rv`

where `g` is the acceleration due to gravity, `r` is a positive constant, and `v` is the velocity of the jumper.

(1) Given that `x = 0` and `v = 0` initially, show that

`qquad x = g/r^2 ln (g/(g - rv)) - v/r.` (3 marks)

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

(2) Given that `g = 9.8\ text(ms)^-2` and `r = 0.2\ text(s)^-1`, find the length, `L`, of the cord such that the jumper’s velocity is `30\ text(ms)^-1` when `x = L`. Give your answer to two significant figures. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
In the second stage of the fall, where `x > L`, the displacement `x` is given by

`x = e^(-t/10)(29 sin t - 10 cos t) + 92`

where `t` is the time in seconds after the jumper’s feet pass `x = L`.

Determine whether or not the jumper’s head stays out of the water. (4 marks)

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

Show Answers Only

`(1)\ \ text{Proof}\ \ text{(See Worked Solutions)}`
`(2)\ \ 82\ \ text(m)`
`text(The jumper’s head will stay out of the water)`

Show Worked Solution

i. (1)	`ddot x`	`=g-rv`
	`v (dv)/(dx)`	`= g – rv`
	`(dx)/(dv)`	`=v/(g-rv)`

`:. int dx`	`= int v/(g-rv)\ dv`
`x`	`=-1/r int ((g-rv-g)/(g-rv))\ dv`
	`=-1/r int (1-g/(g-rv))\ dv`
	`=-1/r (v + g/r ln(g-rv)) +c`

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

`:.c=1/r(g/r lng)=g/r^2 ln g`

`:.x`	`=-1/r (v + g/r ln(g-rv))+g/r^2 ln g`
	`=-v/r- g/r^2 ln(g-rv) +g/r^2 ln g`
	`=g/r^2 ln (g/(g – rv)) – v/r`

i. (2) `g = 9.8\ \ text(ms)^-1,\ \ r = 0.2\ \ text(s)^-1`

`text(If)\ \ x = L,\ \ v = 30\ \ text(ms)^-1`

`:.L`	`=9.8/0.2^2 log_e (9.8/(9.8 – 0.2 xx 30)) – 30/0.2`
	`=82\ \ text(m)\ \ \ \ text{(2 sig.)`

ii.	`x`	`= e^(-t/10) (29 sin t – 10 cos t) + 92`
	`dx/dt`	`=e^(-t/10) (29cos t + 10 sin t)`
		`+(-1/10 e^(-t/10) )(29 sin t – 10 cos t)`
		`=e^(-t/10)(30 cos t+7.1 sin t)`

`text(When)\ \ dx/dt=0\ \ => text(maximum occurs)`

`30 cos t+7.1 sin t`	`=0`
`tan t`	`=-30/7.1`
`:.t`	`=tan^-1 (-30/7.1)`
	`=pi-1.338…`
	`~~1.8\ \ text(s)`

`text(When)\ \ t=1.8`

`x`	`=e^-0.18 (29 sin 1.8 – 10 cos 1.8) + 92`
	`~~117.5\ \ text(m)`

`:.\ text(Distance from the bridge to the jumper’s head) = 119.5\ \ text(m)`

`:.\ text(The jumper’s head will not enter the water.)`

Harder Ext1 Topics, EXT2 2009 HSC 6c

The diagram shows a circle of radius `r`, centred at the origin, `O`. The line `PQ` is tangent to the circle at `Q`, the line `PR` is horizontal, and `R` lies on the line `x = c`.

Find the length of `PQ` in terms of `x, y and r.` (1 mark)
The point `P` moves such that `PQ = PR`.
Show that the equation of the locus of `P` is
1. `y^2 = r^2 + c^2 - 2cx.` (2 marks)
Find the focus, `S`, of the parabola in part (ii). (2 marks)
Show that the difference between the length `PS` and the length `PQ` is independent of `x.` (2 marks)

Show Answers Only

`PQ = sqrt (x^2 + y^2 – r^2)`
`text(Proof)\ \text{(See Worked Solutions)}`
`(r^2/(2c), 0)`
`text(Proof)\ \text{(See Worked Solutions)}`

Show Worked Solution

(i)	`OP`	`= sqrt (x^2 + y^2)`
	`PQ^2`	`= OP^2 – OQ^2`
	`PQ^2`	`= x^2 + y^2 – r^2`
	`:.\ PQ`	`= sqrt (x^2 + y^2 – r^2)`

(ii) `text(When)\ \ PQ = PR`

`sqrt (x^2 + y^2 – r^2)`	`=c-x`
`x^2 + y^2 – r^2`	`= (c – x)^2`
`x^2 + y^2 – r^2`	`= c^2 – 2cx + x^2`
`y^2 – r^2`	`= c^2 – 2cx`

`:.\ y^2 = r^2 + c^2 – 2cx\ \ text(is the locus of)\ \ P`

(iii) `text(Rearranging the locus of)\ \ P\ \ text(in the form)`

♦♦♦ “Few” students answered part (iii) correctly (exact data unavailable).

`(y-y_0)^2`	`=4a(x-x_0)`
`y^2`	`= r^2 + c^2 – 2cx`
	`= -2c(x – (r^2 + c^2)/(2c))`

`=>\ text(The parabola is lying on its side and opening to the left.)`

`text(Vertex) = ((r^2 + c^2)/(2c),0)`

`text(Focal length) \ \ \ \ 4a`	`=-2c`
`a`	`=- c/2`

`:.S\ text(has coordinates)\ \ ((r^2 + c^2)/(2c) – c/2, 0) -=(r^2/(2c), 0)`

(iv) `text(The directrix of the parabola has the equation)`

♦♦♦ “Few” students answered part (iv) correctly (exact data unavailable).
MARKER’S COMMENT: Very few used the efficient focus-directrix definition here.

`x`	`=(r^2+c^2)/(2c) + c/2`
	`=(r^2+2c^2)/(2c)`

`text(The definition of a parabola requires that)`

`PS`	`=PM`
	`=(r^2+2c^2)/(2c)-x`

`text(S)text(ince)\ \ \ PQ`	`=PR\ \ \ \ \ \ text{(from part (ii))}`
	`=c-x`

`=> PS-PQ`	`=(r^2+2c^2)/(2c)-x-(c-x)`
	`=r^2/(2c)`

`:. PS-PQ\ \ text(is independent of)\ x.`

Polynomials, EXT2 2009 HSC 6b

Let `P(x) = x^3 + qx^2 + qx + 1`, where `q` is real. One zero of `P(x)` is `-1`.

Show that if `alpha` is a zero of `P(x)` then `1/alpha` is a zero of `P(x).` (1 mark)
Suppose that `alpha` is a zero of `P(x)` and `alpha` is not real.
(1) Show that `|\ alpha\ | = 1.` (2 marks)
(2) Show that `text(Re)(alpha) = (1 - q)/2.` (2 marks)

Show Answers Only

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

Show Worked Solution

(i) `P(x) = x^3 + qx^2 + qx + 1`

`text(Let roots be)\ \ -1, alpha, beta`

`text(Product of roots)`	`=d/a`
` -1 xx alpha xx beta`	`=-1`
`:. beta`	`=1/alpha`

(ii)(1) `text(If)\ \ alpha\ \ text(is a zero of)\ \ P(x)`

`=>1/alpha\ \ text{is also a zero}\ \ \ \ text{(part (i))}`

`text(S)text(ince all coefficients are real)`

`=>bar alpha\ \ text(is also a zero)`

`:.bar alpha`	`=1/alpha`
`alpha bar alpha`	`=1`
`\|\ alpha\ \|^2`	`=1`
`:.\|\ alpha\ \|`	`=1`

(ii)(2) `text{Sum of roots} = -b/a`

`:.\ -1 + alpha + bar alpha =`	`-q`
`alpha + bar alpha =`	`1 – q`
`2 text(Re)(alpha) =`	`1 – q`
`:.text(Re)(alpha) =`	`(1 – q)/2`

Volumes, EXT2 2009 HSC 6a

The base of a solid is the region enclosed by the parabola `x =4 – y^2` and the `y`-axis. The top of the solid is formed by a plane inclined at `45^@` to the `xy`-plane. Each vertical cross-section of the solid parallel to the `y`-axis is a rectangle. A typical cross-section is shown shaded in the diagram.

Find the volume of the solid. (3 marks)

Show Answers Only

`128/5\ \ text(u³)`

Show Worked Solution

`text(Shaded cross section)`

`text(Height)`	`= 4 – x,\ \ \ \ 0 <= x <= 4`
`text(Length)`	`=2y\ \ \ \ \ \ (text{using}\ \ x=4-y^2)`
	`=2 sqrt(4-x)`
`:. text(Area)\ \ `	`=2 sqrt(4-x) (4-x)`
	`=2 (4-x)^(3/2),\ \ \ \ 0 <= x <= 4`

`delta V`	` = 2 (4 – x)^(3/2)\ delta x,\ \ 0 <= x <= 4`
`:.V `	`=2 int_0^4 (4 – x)^(3/2)\ dx`
	`=2[-2/5 (4 – x)^(5/2)]_0^4`
	`=- 4/5[(4 – x)^(5/2)]_0^4`
	`=- 4/5 (0 – 2^5)`
	`=128/5\ \ text(u³)`

Harder Ext1 Topics, EXT2 2009 HSC 5c

Let `f(x) = (e^x - e^-x)/2 - x.`

Show that `f″(x) > 0` for all `x > 0.` (2 marks)
Hence, or otherwise, show that
1. `f prime (x) > 0` for all `x > 0.` (2 marks)
Hence, or otherwise, show that
1. `(e^x - e^-x)/2 > x` for all `x > 0.` (1 mark)

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)	`f(x) =`	`(e^x – e^-x)/2 – x`
	`f prime (x) =`	`(e^x + e^-x)/2 – 1`
	`f″(x) =`	`(e^x – e^-x)/2`

`text(When)\ \ x > 0, \ \ e^x > e^-x`

`=>e^x – e^-x > 0,`

`:.f ″(x) > 0\ \ text(for)\ \ x > 0`

(ii) `text(Solution 1)`

`f prime (0) = (e^0 – e^0)/2 – 1 = 0`

`text(S)text(ince)\ \ f″(x) > 0\ \ text(for)\ \ x > 0,`

`=>f prime(x)\ \ text(is increasing.)`

`:.f prime (x) > 0\ \ \ \ (text{for}\ \ x > 0)`

`text(Solution 2)`

`f prime (x)`	`=(e^x + e^-x)/2 – 1`
	`=(e^x + e^-x -2)/2`
	`=(e^(2x) – 2e^x + 1)/(2e^x)`
	`=(e^x – 1)^2/(2e^x),\ \ \ \ (e^x > 0)`

`:.f prime (x) > 0\ \ \ \ (text{for}\ \ x > 0)`

(iii) `f(0) = (e^0 – e^0)/2 – 0 = 0`

`text(S)text(ince)\ \ f′(x) > 0,\ \ \ \ (x > 0)`

`=>f(x)\ \ text(is increasing.)`

`:.f (x)> 0`

`(e^x – e^-x)/2 – x`	`> 0`
`:.(e^x – e^-x)/2`	`>x\ \ \ \ (text{for}\ \ x > 0)`

Calculus, EXT2 C1 2009 HSC 5b

For each integer `n >= 0`, let

`I_n = int_0^1 x^(2n + 1) e^(x^2)\ dx.`

Show that for `n >= 1,`

`I_n = e/2 - nI_(n-1).` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, calculate `I_2.` (2 marks)

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

Show Answers Only

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

Show Worked Solution

i. `text(Integrating by parts:)`

`u`	`=x^(2n),`	`u′`	`=2nx^(2n-1)`
`v′`	`=xe^(x^2),`	`v`	`=1/2 e^(x^2)`

`:.I_n`	`= int_0^1 x^(2n + 1) e^(x^2)\ dx`
	`= int_0^1 x^(2n) x e^(x^2)\ dx`
	`= [(x^(2n) e^(x^2))/2]_0^1 – int_0^1 (2n x^(2n) e^(x^2))/2 \ dx`
	`= e/2 – n int_0^1 x^(2n) e^(x^2)\ dx`
	`= e/2 – nI_(n-1)`

ii. `I_2 = e/2 – 2I_1`

`I_1 = e/2 – I_0`

`I_0`	`= int_0^1 xe^(x^2)\ dx`
	`= [(e^(x^2))/2]_0^1`
	`= (e-1)/2`

`I_1`	`= e/2 – (e – 1)/2=1/2`
`:.I_2`	`= e/2 – 2 xx 1/2`
	`= (e – 2)/2`

Harder Ext1 Topics, EXT2 2009 HSC 5a

In the diagram `AB` is the diameter of the circle. The chords `AC` and `BD` intersect at `X`. The point `Y` lies on `AB` such that `XY` is perpendicular to `AB`. The point `K` is the intersection of `AD` produced and `YX` produced.

Copy or trace the diagram into your writing booklet.

Show that `/_ AKY = /_ ABD.` (2 marks)
Show that `CKDX` is a cyclic quadrilateral. (2 marks)
Show that `B, C and K` are collinear. (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)

`/_ ADB = 90^@\ \ \ text{(angle in a semicircle on diameter}\ \ AB text{)}`

`text(In)\ \ Delta BDA and Delta KYA`

`/_ BAD`	`= /_ KAY\ \ text{(common angle)}`
`/_ ADB`	`= /_ KYA = 90^@`
`:./_ AKY`	`= /_ ABD\ \ text{(angle sum of triangle)}`

(ii) `text(Join)\ \ DC`

`/_ DCA`	`= /_ DBA\ \ text{(angles in the same segment on chord}\ DA text{)}`
`/_ DCA`	`= /_ XKD\ \ text{(both equal to}\ /_ DBA text{)}`

`=>text(S)text(ince)\ \ /_ DKX = /_ DCX\ \ text(are a pair of equal angles)`

`text{standing on arc}\ DX\ \ text{(in the same segment)}.`

`:.CKDX\ \ text(is a cyclic quadrilateral.)`

(iii) `/_ KDX`

`= 90^@\ \ text{(} /_ ADK\ text{is a straight angle)}`

`/_ KCX`

`= 90^@\ \ text{(opposite angles of a cyclic}`

`text{quadrilateral are supplementary)}`

`/_ ACB`

`= 90^@\ \ text{(angle in a semicircle)}`

`/_ KCX + /_ ACB = 180^@`

`:./_ BCK\ \ text(is a straight angle.)`

`:.B, C and K\ \ text(are collinear.)`

Mechanics, EXT2 2009 HSC 4b

A light string is attached to the vertex of a smooth vertical cone. A particle `P` of mass `m` is attached to the string as shown in the diagram. The particle remains in contact with the cone and rotates with constant angular velocity `omega` on a circle of radius `r`. The string and the surface of the cone make an angle of `alpha` with the vertical, as shown.

The forces acting on the particle are the tension, `T`, in the string, the normal reaction, `N`, to the cone and the gravitational force `mg`.

Resolve the forces on `P` in the horizontal and vertical directions. (2 marks)
Show that `T = m (g cos alpha + r omega^2 sin alpha)` and find a similar expression for `N.` (2 marks)
Show that if `T = N` then
1. `omega^2 = g/r ((tan alpha - 1)/(tan alpha + 1)).` (2 marks)
For which values of `alpha` can the particle rotate so that `T = N`? (1 mark)

Show Answers Only

`text(Horizontal:)\ \ \ \ \ T sin alpha – N cos alpha = mr omega^2`
`text(Vertical:)\ \ \ \ \ \ T cos alpha + N sin alpha = mg`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`pi/4 < alpha < pi/2`

Show Worked Solution

(i)

`text(Resolving the forces at)\ \ P`

`text(Horizontal)`

`T sin alpha – N cos alpha = mrω^2\ \ \ \ …\ (1)`

`text(Vertical)`

`T cos alpha + N sin alpha = mg\ \ \ \ …\ (2)`

(ii) `text{Multiply}\ \ (1) xx sin alpha\ \ and\ \ (2) xx cos alpha`

`T sin^2 alpha – N cos alpha sin alpha`	`= mr omega^2 sin alpha\ \ \ \ …\ (3)`
`T cos^2 alpha + N sin alpha cos alpha`	`= mg cos alpha\ \ \ \ …\ (4)`
`text{Add (3) + (4)}`
`T (sin^2 alpha + cos^2 alpha)`	`= mr omega^2 sin alpha + mg cos alpha`
`:.T`	`= m(g cos alpha + r omega^2 sin alpha)`

`text{Multiply}\ \ (1) xx cos\ α\ \ and\ \ (2) xx sin\ α`

`T sin alpha cos alpha – N cos^2 alpha`	`= mr omega^2 cos alpha\ \ \ \ …\ (5)`
`T cos alpha sin alpha + N sin^2 alpha`	`= mg sin alpha\ \ \ \ …\ (6)`
`text{Subtract (6) – (5)}`
`N (sin^2 alpha + cos^2 alpha)`	`= mg sin alpha – mr omega^2 cos alpha`
`:.N`	`= m(g sin alpha – r omega^2 cos alpha)`

(iii) `text(When) \ \ T = N`

`m(g cos alpha + r omega^2 sin alpha)`	`= m(g sin alpha – r omega^2 cos alpha)`
`g cos alpha + r omega^2 sin alpha`	`= g sin alpha – r omega^2 cos alpha`
`r omega^2 sin alpha + r omega^2 cos alpha`	`= g sin alpha – g cos alpha`
`r omega^2(sin alpha + cos alpha)`	`= g(sin alpha – cos alpha)`

`:.omega^2`	`=g/r((sin alpha – cos alpha)/(sin alpha + cos alpha))`
	`=g/r((tan alpha – 1)/(tan alpha + 1))`

♦ Part (iv) proved challenging (exact data not available).

MARKER’S COMMENT: Most students did not realise `ω² >0`.

(iv) `text(S) text(ince)\ \ omega^2 > 0`

`=>(tan alpha-1) > 0\ \ \ \ text{(using part (iii))}`

`:.\ \ pi/4 < alpha < pi/2.`

Conics, EXT2 2009 HSC 4a

The ellipse `x^2/a^2 + y^2/b^2 = 1` has foci `S(ae, 0)` and `S prime (– ae, 0)` where `e` is the eccentricity, with corresponding directrices `x = a/e` and `x = -a/e`. The point `P(x_0, y_0)` is on the ellipse. The points where the horizontal line through `P` meets the directrices are `M` and `M prime`, as shown in the diagram.

Show that the equation of the normal to the ellipse at the point `P` is
1. `y - y_0 = (a^2y_0)/(b^2x_0) (x - x_0).` (2 marks)
The normal at `P` meets the `x`-axis at `N`. Show that `N` has coordinates `(e^2x_0, 0).` (2 marks)
Using the focus-directrix definition of an ellipse, or otherwise, show that
1. `(PS)/(PS prime) = (NS)/(NS prime).` (2 marks)
Let `alpha = /_ S prime PN` and `beta = /_ NPS.`
By applying the sine rule to `Delta S prime PN` and to `Delta NPS`, show that `alpha = beta.` (2 marks)

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`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)	`x^2/a^2 + y^2/b^2`	`= 1`
	`(2x)/a^2 + (2y)/b^2 * (dy)/(dx)`	`= 0`
	`(dy)/(dx)`	`= (-b^2x)/(a^2y)`

`text(At)\ \ P(x_0, y_0)`

`(dy)/(dx) = (-b^2 x_0)/(a^2 y_0)`

`:.m\ text{of normal at}\ \ P = (a^2 y_0)/(b^2 x_0)`

`:.\ text(Equation of normal at)\ \ P\ \ text(is)`

`y – y_0 = (a^2 y_0)/(b^2 x_0) (x – x_0)`

(ii) `N\ \ text(occurs when)\ \ y = 0`

`0-y_0`	`= (a^2 y_0)/(b^2 x_0) (x – x_0)`
`-b^2 x_0`	`= a^2 x – a^2 x_0`
`a^2x`	`=x_0(a^2-b^2)`
`x`	`= (a^2 – b^2)/a^2 x_0`
	`= (a^2-a^2(1-e^2))/a^2 x_0`
	`=e^2 x_0`

`:.N\ \ text(has coordinates)\ \ (e^2 x_0, 0)`

(iii) `(PS)/(PM)`	`= (P S′)/(PM′) = e`
`:.(PS)/(PS prime)`	`= (PM)/(PM prime)`

`PM`	`=a/e-x_0,`	`\ \ PM′`	`=a/e +x_0`
`text(S)text(ince)\ \ N(e^2 x_0, 0)`
`NS`	`=ae-e^2x_0,`	`NS′`	`=ae+e^2 x_0`

`(PM)/(PM prime)`	`= (a/e – x_0)/(a/e+x_0) xx e^2/e^2`
	`= (ae – e^2x_0)/(ae+e^2 x_0)`
`:.(PS)/(PS prime)`	`= (NS)/(NS prime)`

(iv) `text(In)\ \ Delta S prime PN,`

`(P S′)/(sin /_ S′NP)`	`= (NS′)/(sin alpha)`
`=>P S′`	`= (NS′sin /_ S′NP)/(sin alpha)`

`text(In)\ \ Delta NPS,`

`(PS)/(sin /_ SNP)`	`= (NS)/(sin beta)`
`=>PS`	`= (NS sin /_ SNP)/(sin beta)`

`(PS)/(PS prime)`	`=(NS sin /_ SNP)/(sin beta) xx (sin alpha)/(NS′sin /_ S′NP)`
	`=(NS)/(NS prime) xx (sin /_ SNP sin alpha)/(sin beta sin /_ S′NP)`
`1`	`=sin alpha/sin beta xx (sin /_ SNP)/(sin /_ S′NP)\ \ \ \ text{(using part (iii))}`

`text(S)text(ince)\ \ /_ SNP + /_ S′NP = 180^@`

`=> sin/_ SNP`	`= sin /_ S′NP`
`:. sin alpha`	`= sin beta`
`:. alpha`	`=beta\ \ \ \ \ (text{since}\ \ alpha + beta<180^@)`

Volumes, EXT2 2009 HSC 3d

The diagram shows the region enclosed by the curves `y = x + 1` and `y = (x - 1)^2.`

The region is rotated about the `y`-axis.

Use the method of cylindrical shells to find the volume of the solid formed. (3 marks)

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`(27 pi)/2\ \ text(u³)`

Show Worked Solution

`text(Intersection occurs when)`

`x + 1`	`= x^2 – 2x + 1`
`x^2 – 3x`	`= 0`
`:.x = 0, 3`

`delta V`	`=2pi xy\ δx`
	`=2pi x[x+1 -(x-1)^2]\ δx`
	`=2pi x(3x-x^2)\ δx`
`:.V`	`=2 pi int_0^3 x (3x – x^2)\ dx`
	`=2 pi int_0^3 (3x^2 – x^3)\ dx`
	`=2 pi [x^3 – x^4/4]_0^3`
	`=2 pi [(27 – 81/4)-0]`
	`=(27 pi)/2\ \ text(u³)`

Functions, EXT1′ F2 2009 HSC 3c

Let `P(x) = x^3 + ax^2 + bx + 5`, where `a` and `b` are real numbers.

Find the values of `a` and `b` given that `(x - 1)^2` is a factor of `P(x).` (3 marks)

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

Show Answers Only

`a = 3,\ \ \ b = -9`

Show Worked Solution

`P(x)`	`= x^3 + ax^2 + bx + 5`
`P(1)`	`= 1 + a + b + 5=0`
`:.b`	`= -a -6\ \ \ \ …\ (1)`

`P prime (x)`	`= 3x^2 + 2ax + b`
`P prime (1)`	`= 3 + 2a + b=0`
`2a+b`	`= -3\ \ \ \ …\ (2)`

`text(Substitute)\ \ b=-a-6\ \ text{from (1) into (2)}`

`2a+(-a-6)=-3`

`:.a = 3,\ \ \ b = -9`