Band 5 – Page 72

Harder Ext1 Topics, EXT2 2006 HSC 8b

For `x > 0`, let `f(x) = x^n e^-x`, where `n` is an integer and `n >= 2.`

The two points of inflection of `f(x)` occur at `x = a` and `x = b`, where `0 < a < b`. Find `a` and `b` in terms of `n.` (4 marks)
Show that
1. `(f(b))/(f(a)) = ((1 + 1/sqrt n)/(1 - 1/sqrt n))^n e^(-2 sqrt n).` (2 marks)
Using the following
If `0<=x<=1/sqrt2` then `1 <= ((1 + x)/(1 - x)) e^(-2x) <= e^((4x^3)/3),` (DO NOT prove this)
2. show that `1 <= (f(b))/(f(a)) <= e^(4/(3 sqrt n)).` (2 marks)
What can be said about the ratio `(f(b))/(f(a))` as `n -> oo?` (1 mark)

Show Answers Only

`a = n – sqrt n,\ \ \ \ \ b = n + sqrt n`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`lim_(n -> oo) (f(b))/(f(a)) = 1`

Show Worked Solution

(i)	`f(x)`	`= x^n e^-x,\ \ \ \ n >= 2,\ \ \ x > 0`
	`f′(x)`	`= nx^(n – 1) e^-x – x^n e^-x`
	`f″(x)`	`= n[(n – 1) x^(n -2) e^-x – x^(n – 1) e^-x] – (nx^(n – 1) e^-x – x^n e^-x)`
		`= x^(n – 2) e^-x [n(n – 1) – nx – nx + x^2]`
		`= x^(n – 2) e^-x (x^2-2nx +n^2 – n)`

`text(P.I.’s occur when) \ \ f″(x)=0`

`text(i.e. when)\ \ \ x^2-2nx +n^2 – n = 0,\ \ \ \ (e^-x>0 and x>0)`

`:.x=`	`(2n+-sqrt(4n^2- 41(n^2-n)))/2`
`=`	`\ \ n+- sqrtn`

`text(S)text(ince)\ \ a<b`

`:. a = n – sqrt n\ and\ b = n + sqrt n.`

(ii) `(f(b))/(f(a)) =`	`((n + sqrt n)^n e^-(n + sqrt n))/((n – sqrt n)^n e^-(n – sqrt n))`
`=`	`(n^n(1 + 1/sqrt n)^n)/(n^n(1 – 1/sqrt n)^n) xx e^(-n – sqrt n + n – sqrt n)`
`=`	`((1 + 1/sqrt n)/(1 – 1/sqrt n))^n e^(-2 sqrt n)`

(iii) `text(S)text(ince)\ \ n>=2\ \ \ => 1/sqrt n <=1/sqrt2`

`text(Substituting)\ \ x=1/sqrt n\ \ text(into the inequality)`

`1`	`<= ((1 + 1/sqrt n)/(1 – 1/sqrt n)) e^((-2)/sqrt n) <= e^(4/(3(sqrt n)^3))`
`1^n`	`<= ((1 + 1/sqrt n)/(1 – 1/sqrt n))^n e^((-2)/sqrt n xx n) <= e^(4/(3(sqrt n)^3) xx n)`
`:.1`	`<= (f(b))/(f(a)) <= e^(4/(3 sqrt n))`

(iv)	`lim_(n -> oo) e^(4/(3 sqrt n))`	`=1`
	`:.lim_(n -> oo) (f(b))/(f(a))`	`=1`

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`

Calculus, EXT2 C1 2006 HSC 7b

Let `I_n = int_0^x sec^n t\ dt`, where `0 <= x <= pi/2`.

Show that `I_n = (sec^(n - 2) x tan x)/(n - 1) + (n - 2)/(n - 1) I_(n - 2).` (3 marks)

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Hence find the exact value of

`int_0^(pi/3) sec^4 t\ dt.` (2 marks)

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Show Answers Only

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

Show Worked Solution

STRATEGY: The choice of `u=sec^(n-2)t` and `v=sec^2 t` was extremely important in being able to answer this question. Take note!

i. `I_n`	`=int_0^x sec^n t\ dt,\ \ \ 0 <= x < pi/2`
	`=int_0^x sec^(n – 2)t sec^2 t\ dt`

`text(Integrating by parts)`

`u`	`=sec^(n-2)t,`	`u′`	`=(n-2) sec^(n-2)t tan\ t`
`v`	`=tan\ t,`	`v′`	`=sec^2 t`

`I_n`	`=[tan t sec^(n – 2)t]_0^x – int_0^x tan t (n – 2) sec^(n – 3) t sec t tan t\ dt`
	`=tan x sec^(n – 2) x – (n – 2) int_0^x sec^(n – 2) t tan^2 t\ dt`
	`=tan x sec^(n – 2) x – (n – 2) int_0^x sec^(n – 2) t (1+sec^2 t)\ dt`
	`=tan x sec^(n – 2) x – (n – 2) int_0^x sec^n t\ dt + (n – 2) int_0^x sec^(n – 2)t\ dt`
	`=tan x sec^(n – 2) x-(n-2)I_n+(n-2)I_(n-2)`

`I_n + (n – 2) I_n`	`= tan x sec^(n – 2) x + (n – 2) I_(n – 2)`
`(n – 1)I_n`	`= tan x sec^(n – 2) x + (n – 2) I_(n – 2)`
`:.I_n `	`= (tan x sec^(n – 2) x)/(n – 1) + ((n – 2)/(n – 1)) I_(n – 2)`

ii. `int_0^(pi/3) sec^4 t\ dt =`	`(tan\ pi/3 sec^2\ pi/3)/3 + 2/3 int_0^(pi/3) sec^2 t\ dt`
`=`	`(sqrt 3 xx 4)/3 + 2/3 [tan t]_0^(pi/3)`
`=`	`(4 sqrt 3)/3 + 2/3 (sqrt 3 – 0)`
`=`	`2 sqrt 3`

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)

--- 5 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)}`
`(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`

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 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`

Harder Ext1 Topics, EXT2 2007 HSC 8c

The diagram shows a regular `n`-sided polygon with vertices `X_1, X_2, …, X_n`. Each side has unit length. The length `d_k` of the ‘diagonal’ `X_n X_k` where `k = 1, 2, …, n - 1` is given by

`d_k = (sin\ (k pi)/n)/(sin\ pi/n).` (Do NOT prove this.)

Show, using the identity,
`sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n - 1) pi)/n = cot\ pi/(2n),` (Do NOT prove this.)
1. that `d_1 + … + d_(n - 1) = 1/(2 sin^2\ pi/(2n)).` (2 marks)
Let `p` be the perimeter of the polygon and
`q = 1/n (d_1 + … + d_(n - 1))`.
Show that `p/q = 2 (n sin\ pi/(2n))^2.` (2 marks)
Hence calculate the limiting value of `p/q` as `n -> oo.` (1 mark)

Show Answers Only

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

Show Worked Solution

(i) `d_k = (sin\ (k pi)/n)/(sin\ pi/n)`

`d_1 + d_2 + … + d_(n – 1)`

`=(sin\ pi/n)/(sin\ pi/n) + (sin\ (2 pi)/n)/(sin\ pi/n) + … + (sin\ ((n – 1) pi)/n)/(sin\ pi/n)`

`=1/(sin\ pi/n) xx cot\ pi/(2n)`

`=(cos\ pi/(2n))/(2 sin\ pi/(2n) cos\ pi/(2n) xx sin\ pi/(2n))`

`=1/(2 sin^2\ pi/(2n))`

(ii)	`p/q`	`=n/(1/n (d_1 + d_2 + … + d_(n – 1))`
		`=n^2 xx (2 sin^2\ pi/(2n))/1`
		`=2 (n sin\ pi/(2n))^2`

MARKER’S COMMENT: Applying the result `lim_(theta->0)\ sin theta/theta = 1` was poorly done.

(iii)	`lim_(n -> oo)\ p/q`	`= lim_(n -> oo)\ 2(n sin\ pi/(2n))^2`
		`=2n^2 xx pi^2/(4n^2) xx lim_(n -> oo)\ ((sin\ pi/(2n))^2)/(pi^2/(4n^2))`
		`=pi^2/2 xx lim_(n -> oo)\ (sin\ pi/(2n))/(pi/(2n)) xx lim_(n -> oo)\ (sin\ pi/(2n))/(pi/(2n))`
		`=pi^2/2 xx 1 xx 1\ \ \ \ \ \ text{(noting that as}\ \ n->oo, pi/(2n)->0text{)}`
		`=pi^2/2`

Complex Numbers, EXT2 N2 2007 HSC 8b

Let `n` be a positive integer. Show that if `z^2 != 1` then

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

--- 5 WORK AREA LINES (style=lined) ---
By substituting `z = cos theta + i sin theta` where `sin theta != 0`, into part (i), show that

`1 + cos 2 theta + … + cos (2n - 2) theta + i[sin 2 theta + … + sin (2n - 2) theta]`

`= (sin n theta)/(sin theta) [cos (n - 1) theta + i sin (n - 1) theta].` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Suppose `theta = pi/(2n)`. Using part (ii), show that

`sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n - 1) pi)/n = cot\ pi/(2n).` (2 marks)

--- 8 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 + z^2 + z^4 + … + z^(2n – 2),\ z^2 != 1`

`text(GP where)\ a = 1,\ \ r = z^2,\ \ n\ text(terms):`

`S_n`	`=(1((z^2)^n – 1))/(z^2 – 1)`
	`=(z^(2n) – 1)/(z^2 – 1)`
	`=((z^n – z^-n))/(z – z^-1) xx z^n/z`
	`=((z^n – z^-n)/(z – z^-1))z^(n – 1)`

ii.	`z`	`= cos theta + i sin theta`
	`z^n`	`= cos n theta + i sin n theta\ \ …\ text(etc)\ \ \ \ text{(De Moivre)}`
	`z^-n`	`= cos( -n theta) + i sin (-n theta)`
		`= cos n theta – i sin n theta`

`text(LHS)`	`= 1 + (cos 2 theta + i sin 2 theta) + (cos 4 theta + i sin 4 theta) + `
	`… + (cos(2n – 2) theta + i sin (2n – 2) theta)`
	`= 1 + cos 2 theta + cos 4 theta + … + cos (2n – 2) theta + `
	`i (sin 2 theta + sin 4 theta + … + sin (2n – 2) theta)`

`text{Using part (i):}`

`text(LHS)`	`=((cos n theta + i sin n theta – cos n theta + i sin n theta))/(cos theta + i sin theta – cos theta + i sin theta) xx`
	`[cos (n – 1) theta + i sin (n – 1) theta]`
	`=(2 i sin n theta)/(2 i sin theta) [cos (n – 1) theta + i sin (n – 1) theta]`
	`=(sin n theta)/(sin theta) [cos (n – 1) theta + i sin (n – 1) theta]\ \ text(… as required.)`

iii. `text{Equating the imaginary parts in part (ii):}`

`sin 2 theta + sin 4 theta + … + sin 2 (n – 1) theta = (sin n theta sin (n – 1) theta)/(sin theta)`

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

`sin\ (2 pi)/(2n) + sin\ (4 pi)/(2n) + … + sin\ (2(n – 1) pi)/(2 n) = (sin\ (n pi)/(2n) sin\ ((n – 1) pi)/(2n))/(sin\ pi/(2n))`

`:. sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n – 1) pi)/n`

`=(sin\ pi/2)/(sin\ pi/(2n)) xx sin\ ((n – 1) pi)/(2n)`

`=1/(sin\ pi/(2n)) sin (pi/2 – pi/(2n))`

`=(cos\ pi/(2n))/(sin\ pi/(2n))`

`=cot\ pi/(2n)`

Calculus, EXT2 C1 2007 HSC 8a

Using a suitable substitution, show that

`int_0^a f(x)\ dx = int_0^a f(a - x)\ dx.` (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
A function `f(x)` has the property that `f(x) + f(a - x) = f(a).`

Using part (i), or otherwise, show that

`int_0^a f(x)\ dx = a/2\ f(a).` (2 marks)

--- 8 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(Show)\ \ int_0^a f(x)\ dx = int_0^a f(a – x)\ dx.`

`text(Let)\ \ x = a – u,\ \ dx = -du`

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

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

`:. int_0^a f(x)\ dx`	`=int_a^0 f(a – u) (-du)`
	`=int_0^a f(a – u)\ du`
	`=int_0^a f(a – x)\ dx\ \ text{.. as required}`

MARKER’S COMMENT: Integrating `f(a)` was poorly done in part (ii). Pay careful attention to this in the Worked Solution.

ii. `f(x) = f(a) – f(a – x)`

`int_0^a f(x)\ dx`	`=int_0^a [f(a) – f(a – x)]\ dx`
	`=int_0^a f(a)\ dx – int_0^a f(a – x)\ dx`
	`=int_0^a f(a)\ dx – int_0^a f(x)\ dx`
`2 int_0^a f(x)\ dx`	`=int_0^a f(a)\ dx`
	`=[f(a) xx x]_0^a`
`int_0^a f(a)\ dx`	`=(f(a))/2 (a – 0)`
	`=a/2\ f(a)`

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)

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

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

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)

--- 2 WORK AREA LINES (style=lined) ---
Hence show that, for `n >= 2`,

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

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

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

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.`

Combinatorics, EXT1′ A1 2015 HSC 7 MC

The numbers `1, 2, . . . n`, for `n >= 4`, are randomly arranged in a row.

What is the probability that the number 1 is somewhere to the left of the number 2?

`1/2`
`1/n`
`1/(2(n - 2)!)`
`1/(2(n - 1)!)`

Show Answers Only

`A`

Show Worked Solution

`text(The number 1 is either to the left or to the right)`

♦ Mean mark 35%.

`text(of the number 2, so the probability is just half.)`

`=> A`

Harder Ext1 Topics, EXT2 2014 HSC 9 MC

A particle is moving along a straight line so that initially its displacement is `x = 1`, its velocity is `v = 2`, and its acceleration is `a = 4`.

Which is a possible equation describing the motion of the particle?

`v = 2sin(x − 1) + 2`
`v = 2 + 4log_e x`
`v^2= 4(x^2 − 2)`
`v = x^2 + 2x + 4`

Show Answers Only

`A`

Show Worked Solution

`text(By elimination)`

♦ Mean mark 40%.

`text(Consider when)\ \ x=1`

`=>\ text{Not (C) as}\ \ v\ \ text(is undefined)`

`=>\ text{Not (D) where}\ \ v=7`

`text{For (A), find}\ \ a\ \ text(when)\ \ x=1`

`a`	`= v (dv)/(dx)`
	`= [2 sin(x − 1) + 2] × 2cos(x − 1).1`
	`=[2sin(1 − 1) + 2] xx 2cos(1 − 1)`
	`= 4`

`text{For (B), find}\ \ a\ \ text(when)\ \ x=1`

`a`	`= (2 + 4log_e x) × 4/x`
	`= (2 + 4log_e 1) × 4/1`
	`= 8`

`=> A`

Integration, EXT2 2014 HSC 7 MC

Which expression is equal to `int 1/(1 − sin x)\ dx`?

`tan x − sec x + c`
`tan x + sec x + c`
`log_e(1 − sin x)+ c`
`(log_e(1 − sin x))/(− cos x) + c`

Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 44%.

`int(dx)/(1 − sin x)`	`= int ((1 + sin x))/((1 − sin x)(1 + sin x))\ dx`
	`= int ((1 + sinx))/(cos^2 x)\ dx`
	`=int 1/cos^2x + sinx/cosx * 1/cos x\ dx`
	`= int sec^2 x + tan x sec x\ dx`
	`= tan x + sec x + c`

`=> B`

Volumes, EXT2 2014 HSC 6 MC

The region bounded by the curve `y^2 = 8x` and the line `x = 2` is rotated about the line `x = 2` to form a solid.

Which expression represents the volume of the solid?

`pi int_0^4 2^2 − ((y^2)/8)^2 dy`
`2pi int_0^4 2^2 − ((y^2)/8)^2 dy`
`pi int_0^4 (2 − (y^2)/8)^2 dy`
`2pi int_0^4 (2 − (y^2)/8)^2 dy`

Show Answers Only

`D`

Show Worked Solution

`text(Radius)\ (r)`	`=2-x`
	`=2-y^2/8`

♦ Mean mark 45%.

`δV`	`= pi r^2\ δy`
`:.V`	`= int_(−4)^4 pi(2 − y^2/8)^2 dy`
	`= 2pi int_0^4 (2 − y^2/8)^2 dy`

`=> D`

Harder Ext1 Topics, EXT2 2009 HSC 8c

A game is being played by `n` people, `A_1, A_2, ..., A_n`, sitting around a table. Each person has a card with their own name on it and all the cards are placed in a box in the middle of the table. Each person in turn, starting with `A_1`, draws a card at random from the box. If the person draws their own card, they win the game and the game ends. Otherwise, the card is returned to the box and the next person draws a card at random. The game continues until someone wins.

Let `W` be the probability that `A_1` wins the game.

Let `p = 1/n and q = 1 - 1/n`.

Show that `W = p + q^n W.` (1 mark)
Let `m` be a fixed positive integer and let `W_m` be the probability that `A_1` wins in no more than `m` attempts.
Use `e^(-n/(n - 1)) < (1 - 1/n)^n < e^-1,`
to show that, if `n` is large, `W_m/W` is approximately equal to `1 - e^-m.` (3 marks)

Show Answers Only

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

Show Worked Solution

(i) `text(Chance of drawing own card is)\ \ p = 1/n`

`text(Chance of not drawing own card is)\ \ q = 1 – p = 1 – 1/n`

`P(A_1\ text{wins in round 1)}`	`= p`
`P(A_1\ text{wins in round 2)}`	`= q^n*p`
`P(A_1\ text{wins in round 3)}`	`= q^nq^np`
`:.P(A_1\ \ text{wins)}`	`= p + pq^n + pq^(2n) + …`

`=> text(G.P. where)\ \ a=p,\ \ r=q^n,\ \ and\ \ 0<q^n<1`

`W`	`=p/(1-q^n)`
`W-W q^n`	`=p`
`:.W`	`=p+W q^n`

(ii) `W_m`	`=p + q^n p + q^(2n) p + … + q^((m – 1)n) p`
	`=p ((1 – q^(nm)))/(1 – q^n)\ \ \ \ \ \ text{(G.P. with}\ m\ text{terms)}`
`W_m/W`	`=p ((1 – q^(nm)))/(1 – q^n) xx (1 – q^n)/p`
	`=1 – q^(nm)`
	`= 1 – (1 – 1/n)^(nm)`

`text(Considering)\ \ \ e^(-n/(n – 1)) < (1 – 1/n)^n < e^-1`

`text{As}\ \ n -> oo, \ \ n/(n-1)->1, \ \ e^(-n/(n – 1))->e^-1, and`

`:.(1-1/n)^n`	`~~e^-1`
`(1-1/n)^(mn)`	`~~e^-m`
`:.\ W_m/W`	`~~ 1 – e^-m`

Harder Ext1 Topics, EXT2 2009 HSC 8b

Let `n` be a positive integer greater than `1`.

The area of the region under the curve `y = 1/x` from `x = n - 1` to `x = n` is between the areas of two rectangles, as shown in the diagram.

Show that

`e^(-n/(n - 1)) < (1 - 1/n)^n < e^-1.` (2 marks)

Show Answers Only

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

Show Worked Solution

`text(Area of small rectangle)=1*1/n=1/n`

`text(Area of large rectangle)=1/(n-1)`

`1 xx 1/n <int_(n – 1)^n (dx)/x< 1 xx 1/(n – 1)`
`1/n <[log_e x]_(n – 1)^n< 1/(n – 1)`
`1/n <log_e n – log_e (n – 1)< 1/(n – 1)`
`1/n <log_e n/(n – 1)< 1/(n – 1)`
`e^(1/n) <n/(n – 1)<e^(1/(n – 1))`
`e<(n/(n – 1))^n<e^(n/(n – 1))`	`text{(using if}\ \ a>b,\ text(then)\ 1/a<1/btext{)}`
`e^(-n/(n – 1)) <((n – 1)/n)^n<e^-1`
`:.e^(-n/(n – 1))<(1-1/n)^n<e^-1`

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`

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`

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)

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)	`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^@)`

Integration, EXT2 2010 HSC 8

Let

`A_n = int_0^(pi/2) cos^(2n) x\ dx` and `B_n = int_0^(pi/2) x^2cos^(2n)x\ dx`,

where `n` is an integer, `n ≥ 0`. (Note that `A_n > 0`, `B_n > 0`.)

Show that
`nA_n = (2n − 1)/2 A_(n − 1)` for `n ≥ 1`. (2 marks)
Using integration by parts on `A_n`, or otherwise, show that
1. `A_n = 2n int_0^(pi/2) x sin x cos^(2n − 1) x\ dx` for `n ≥ 1`. (1 mark)
Use integration by parts on the integral in part (ii) to show that
1. `(A_n)/(n^2) = ((2n − 1))/n B_(n − 1) − 2B_n` for `n ≥ 1`. (3 marks)
Use parts (i) and (iii) to show that
1. `1/(n^2) = 2((B_(n − 1))/(A_(n − 1)) − (B_n)/(A_n))` for `n ≥ 1`. (1 mark)
Show that
1. `sum_(k = 1)^n 1/(k^2) = (pi^2)/6 − 2 (B_n)/(A_n)`. (2 marks)
Use the fact that
`sin x ≥ 2/pi x` for `0 ≤ x ≤ pi/2` to show that
1. `B_n ≤ int_0^(pi/2) x^2(1 − (4x^2)/(pi^2))^n dx`. (1 mark)
Show that
`int_0^(pi/2) x^2(1 − (4x^2)/(pi^2))^n dx = (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − (4x^2)/(pi^2))^(n + 1) dx`. (1 mark)
From parts (vi) and (vii) it follows that
1. `B_n ≤ (pi^2)/(8(n + 1)) int_0^(pi/2)(1 − (4x^2)/(pi^2))^(n + 1) dx`.
Use the substitution `x = pi/2 sin t` in this inequality to show that
1. `B_n ≤ (pi^3)/(16(n + 1)) int_0^(pi/2) cos^(2n + 3)t\ dt ≤ (pi^3)/(16(n + 1)) A_n`. (2 marks)
Use part (v) to deduce that
1. `(pi^2)/6 − (pi^3)/(8(n + 1)) ≤ sum_(k = 1)^n 1/(k^2) < (pi^2)/6`. (1 mark)
What is
`lim_(n → ∞) sum_(k = 1)^n 1/(k^2)`? (1 mark)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`(pi^2)/6`

Show Worked Solution

♦♦ Mean mark part (i) 21%.

(i)	`u`	`=cos^(2n − 1)x`
	`du`	`=-(2n-1) sinx cos^(2n − 2) x\ dx`
	`v`	`=sin x`
	`dv`	`=cos x\ dx`

`A_n`	`= int_0^(pi/2) cos x cos^(2n − 1)x\ dx`
	`= [sin x cos^(2n − 1)x]_0^(pi/2) − int_0^(pi/2) sin x*-(2n-1) sinx cos^(2n − 2) x\ dx`
	`= 0 + (2n − 1) int_0^(pi/2) sin^2 x cos^(2n − 2) x\ dx`
	`= (2n − 1) int_0^(pi/2) (1 − cos^2x)cos^(2n − 2)x\ dx`
	`= (2n − 1)(int_0^(pi/2) cos^(2n − 2)x\ dx − int_0^(pi/2) cos^(2n)x\ dx)`
	`= (2n − 1) A_(n − 1) − (2n − 1) A_n`

`A_n + (2n − 1) A_n`	`= (2n − 1) A_(n − 1)`
`2nA_n`	`= (2n − 1) A_(n − 1)`
`:.nA_n`	`= (2n − 1)/2 A_(n − 1)\ \ text(for)\ \ n ≥ 1.`

♦♦ Mean mark part (ii) 32%.

(ii)	`u`	`=cos^(2n) x`
	`du`	`=-2n\ sinx cos^(2n − 1) x\ dx`
	`v`	`= x`
	`dv`	`= dx`

`A_n`	`= int_0^(pi/2) 1 xx cos^(2n)x\ dx`
	`= [x cos^(2n) x]_0^(pi/2) − int_0^(pi/2) x * -2n\ sinx cos^(2n − 1) x\ dx`
	`= 0 + 2n int_0^(pi/2) x sin x cos^(2n − 1) x\ dx`
	`= 2n int_0^(pi/2) x sin x cos^(2n − 1)x\ dx\ \ \ text(for)\ \ n ≥ 1`

♦♦♦ Mean mark part (iii) 8%.

(iii)	`u`	`=sin x cos^(2n-1) x`
	`du`	`=cosx cos^(2n-1)x -(2n-1) cos^(2n-2)x *sin^2x\ dx`
		`=cos^(2n)x – (2n-1)cos^(2n-2) x* (1-cos^2 x)\ dx`
		`=cos^(2n)x – (2n-1)cos^(2n-2)x +(2n-1)cos^(2n)\ dx`
		`=2n cos^(2n)x-(2n-1)cos^(2n-2)x`
	`v`	`= x^2/2`
	`dv`	`= x\ dx`

`A_n`	`= 2n int_0^(pi/2) x sin x cos^(2n − 1)x\ dx`
	`= 2n[x^2/2 sin x cos^(2n − 1) x]_0^(pi/2)`
	`− 2n int_0^(pi/2) x^2/2 (2n cos^(2n)x-(2n-1)cos^(2n-2)x)\ dx`
	`= 0 −2n[n int_0^(pi/2) x^2 cos^(2n) x\ dx − (2n − 1)/2 int_0^(pi/2) x^2 cos^(2n − 2) x \ dx]`
	`= -2n(nB_n-(2n-1)/2 B_(n-1))`
	`= -2n^2B_n+n(2n-1)B_(n-1)`
`:.(A_n)/(n^2)`	`= (2n − 1)/n B_(n − 1) − 2B_n\ \ \ text(for)\ \ n ≥ 1`

♦♦♦ Mean mark part (iv) 13%.

(iv)	`(A_n)/(n^2)`	`= ((2n −1)B_(n − 1))/n − 2B_n\ \ \ text(for)\ \ n ≥ 1`
	`:.1/(n^2)`	`= ((2n − 1)B_(n − 1))/(nA_n) − (2B_n)/(A_n)`
		`= ((2n − 1)B_(n − 1))/((2n − 1)/2 A_(n − 1)) − (2B_n)/(A_n)\ \ \ \ \ text{(using part (i))}`
		`= (2B_(n − 1))/(A_(n − 1)) − (2B_n)/(A_n)`
		`= 2((B_(n − 1))/(A_(n − 1)) − (B_n)/(A_n))\ \ \ text(for)\ \ n ≥ 1`

♦♦♦ Mean mark part (v) 9%.

(v)	`sum_(k = 1)^n 1/(k^2)`	`= 1/(1^2) + 1/(2^2) + 1/(3^2) + … + 1/(n^2)`
	`sum_(k = 1)^n 1/(k^2)`	`= 2((B_0)/(A_0) − (B_1)/(A_1)) + 2((B_1)/(A_1) − (B_2)/(A_2)) + … + 2((B_(n − 1))/(A_(n − 1)) − (B_n)/(A_n))`
		`= 2(B_0)/(A_0) − 2(B_n)/(A_n)`

`A_0 = int_0^(pi/2) dx = [x]_0^(pi/2) = pi/2`

`B_0 = int_0^(pi/2)x^2\ dx = [(x^3)/3]_0^(pi/2) = (pi^3)/24`

`sum_(k = 1)^n 1/(k^2)`	`= 2 xx ((pi^3)/24)/(pi/2) − 2(B_n)/(A_n)`
	`= (pi^2)/6 − 2(B_n)/(A_n)`

♦♦♦ Mean mark part (vi) 8%.

(vi)	`B_n`	`= int_0^(pi/2) x^2 cos^(2n)x\ dx`
	`B_n`	`= int_0^(pi/2) x^2 (1 − sin^2 x)^n\ dx`

`text(S)text(ince)\ \ sin x ≥ 2/pi x,\ \ 0 ≤ x ≤ pi/2`

`:.B_n`	`≤ int_0^(pi/2) x^2 (1 − (2/pi x)^2)^n\ dx`
	`≤ int_0^(pi/2) x^2 (1 − (4x^2)/(pi^2))^n\ dx`

(vii)	`u`	`=x,\ \ \ du=dx`
	`dv`	`=x(1-(4x^2)/(pi^2))^n\ dx`
	`v`	`=(- pi^2)/(8(n+1)) (1-(4x^2)/(pi^2))^(n+1)`

♦♦♦ Mean mark part (vii) 2%.

`int_0^(pi/2) x * x(1 − (4x^2)/(pi^2))^n dx`

`= [(-pi^2)/(8(n + 1))(1 − (4x^2)/(pi^2))^(n + 1) * x]_0^(pi/2) – int_0^(pi/2) (-pi^2)/(8(n + 1))(1 − (4x^2)/(pi^2))^(n + 1)\ dx`

`= (-pi^2)/(8(n + 1)) (0 − 0) + (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − (4x^2)/(pi^2))^(n + 1) dx`

`= (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − (4x^2)/(pi^2))^(n + 1) dx`

♦♦♦ Mean mark part (viii) 4%.

(viii)	`text(Using)\ \ x`	`=pi/2 sin t\ \ \ \ dx=pi/2 cos t\ dt`
	`text(When)\ \ x`	`=0,\ \ \ t=0`
	`text(When)\ \ x`	`=pi/2,\ \ \ t=pi/2`

`B_n`	`≤ (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − (4x^2)/(pi^2))^(n + 1)\ dx`
	`≤ (pi^2)/(8(n + 1)) int_0^(pi/2) (1 − sin^2 t)^(n + 1) *pi/2 cos t\ dt`
	`≤ (pi^3)/(16(n + 1)) int_0^(pi/2) (cos^2 t)^(n + 1) cos t\ dt`
	`≤ (pi^3)/(16(n + 1)) int_0^(pi/2) cos^(2n + 3) t\ dt`

`text(Consider the RHS of the inequality)`

`text(RHS)`	`=(pi^3)/(16(n + 1)) int_0^(pi/2) cos^(2n + 3) t\ dt`
	`≤(pi^3)/(16(n + 1)) int_0^(pi/2) cos^(2n) t\ dt,\ \ \ \ text{(as cos t ≤ 1)}`
	`≤(pi^3)/(16(n + 1)) A_n\ \ \ \ text(… as required)`

♦♦♦ Mean mark part (ix) 3%.

(ix) `sum_(k = 1)^n 1/(k^2) = (pi^2)/6 − 2(B_n)/(A_n)\ \ \ \ text{(from part (v))}`

`:.sum_(k = 1)^n 1/(k^2) < (pi^2)/6\ \ \ \ (A_n > 0`, `B_n > 0)`

`text{Using part (viii)}`

`B_n`	`≤(pi^3)/(16(n + 1)) A_n`
`(2B_n)/A_n`	`≤(pi^3)/(8(n + 1)) `

`:.sum_(k = 1)^n 1/(k^2) ≥ (pi^2)/6 − (pi^3)/(8(n + 1))`

`:. (pi^2)/6 − (pi^3)/(8(n + 1)) ≤ sum_(k = 1)^n 1/(k^2) < (pi^2)/6`

♦♦♦ Mean mark part (x) 19%.

(x) `text(S)text(ince)\ (pi^3)/(8(n + 1)) → 0\ text(as)\ n → ∞`

`lim_(n → ∞) sum_(k = 1)^n 1/(k^2) = (pi^2)/6`

Polynomials, EXT2 2010 HSC 7c

Let `P(x) = (n − 1)x^n − nx^(n − 1) + 1`, where `n` is an odd integer, `n ≥ 3`.

Show that `P(x)` has exactly two stationary points. (1 mark)
Show that `P(x)` has a double zero at `x = 1`. (1 mark)
Use the graph `y = P(x)` to explain why `P(x)` has exactly one real zero other than `1`. (2 marks)
Let `α` be the real zero of `P(x)` other than `1`.
Given that `2^x>=3x-1` for `x>=3`, or otherwise, show that `-1 < α ≤ -1/2`. (2 marks)
Deduce that each of the zeros of `4x^5 − 5x^4 + 1` has modulus less than or equal to `1`. (2 marks)

Show Answers Only

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

Show Worked Solution

(i)	`P′(x)`	`= n(n − 1)x^(n − 1) − n(n − 1)x^(n − 2)`
		`= n(n − 1)x^(n − 2)(x − 1)`

`P′(x) = 0\ \ text(when)\ \ x = 0\ \ text(or)\ \ x = 1.`

`:.P(x)\ text(has exactly two stationary points.)`

(ii)	`P(1)`	`= (n − 1) xx 1 − n xx 1 + 1 = 0, and`
	`P′(1) `	`= 0`

`:. P(x)\ text(has a double zero at)\ \ x=1`

(iii) `P(x)\ text{has exactly two stationary points at}\ \ x=0 and 1`

♦♦ Mean mark part (iii) 26%.

`text(S)text(ince)\ \ P(0) = 1 and P(1)=0`

`=>\ text{A local maximum occurs at (0,1)}`

`text(Noting that)\ \ P(x)`	`-> oo\ \ text(as)\ \ x-> oo, and`
`P(x)`	`-> -oo\ \ text(as)\ \ x-> -oo`

`:.\ text(The double zero at)\ \ x=1,\ text{means that}\ \ P(x)`

`text(has exactly one real zero other than)\ \ x=1.`

♦♦♦ Mean mark part (iv) 3%.

(iv)	`P(-1)`	`=(n-1)(-1)^n-n xx (-1)^(n-1) + 1`
		`=(n-1)(-1) – nxx1 +1\ \ \ \ text{(given}\ n\ text{is odd)}`
		`=-2n+2`
		`<0\ \ \ text{(for odd}\ n>=3text{)}`

`P(-1/2)`	`= (n − 1)(-1/2)^n − n(-1/2)^(n − 1)+1`
	`= -(n − 1) xx 1/(2^n) − n/(2^(n − 1)) + 1\ \ \ \ text{(given}\ n\ text{is odd)}`
	`= (-n + 1 − 2n + 2^n)/(2^n)`
	`= (2^n − (3n − 1))/(2^n)`
	`>=0\ \ \ \ \ text{(given}\ \ 2^n>=3n-1\ \ text{for}\ \ n>=3, and 2^n > 0)`

`text(S)text(ince)\ \ P(x)\ \ text(is continuous, when it changes sign, it cuts the)\ x text(-axis)`

`:. -1 < α ≤ -1/2`

(v) `P(x) = 4x^5 − 5x^4 + 1\ \ text(is of the form)`

♦♦♦ Mean mark part (v) 2%.

`P(x) = (n − 1)x^n − nx^(n − 1) + 1`

`:.P(x)\ \ text(has 5 zeros)\ \=> 1, 1, alpha, beta, bar beta\ \ \ \ (text(where)\ beta\ text{is not real})`

`text(The zeros 1 and α have a modulus ≤ 1.)`

`text(Consider the product of the roots)`

`11alphabetabar beta`	`=- 1/4`
`alphabetabar beta`	`=- 1/4`
`\|\ alphabetabar beta\ \|`	`=\|\ – 1/4\ \|`
`\|\ alpha\ \|*\|\ beta\ \|^2`	`= 1/4`

`text(S)text(ince)\ \ |\ α\ | > 1/2,\ \ |\ beta\ |<1`

`:.text(Each of the zeros of)\ x^5 − 5x^4 + 1\ text(has a modulus)\ <=1.`

Harder Ext1 Topics, EXT2 2010 HSC 7a

In the diagram `ABCD` is a cyclic quadrilateral. The point `K` is on `AC` such that `∠ADK = ∠CDB`, and hence `ΔADK` is similar to `ΔBDC`.

Copy or trace the diagram into your writing booklet.

Show that `ΔADB` is similar to `ΔKDC`. (2 marks)
Using the fact that `AC = AK + KC`,
show that `BD xx AC = AD xx BC + AB xx DC`. (2 marks)
A regular pentagon of side length `1` is inscribed in a circle, as shown in the diagram.

Let `x` be the length of a chord in the pentagon.
Use the result in part (ii) to show that `x = (1 + sqrt5)/2`. (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

Harder Ext1 Topics, EXT2 2010 HSC 7a Answer

`text(S)text(ince)\ \ ∠ADB`	`=∠ADK +∠KDB, and`
`∠KDC`	`=∠CDB +∠KDB`
`=>∠ADB`	`= ∠KDC`
`∠ABD`	`= ∠ACD\ \ \ text{(angles in the same segment on arc}\ ADtext{)}`

`∠ADK`	`= ∠CDB\ \ \ `	`text{(corresponding angles of similar}`
		`text{triangles,}\ ΔADK\ text(\|\|\|)\ ΔBDC text{)}`

`:.ΔADB\ text(|||)\ ΔKDC\ \ text{(equiangular)}`

(ii) `text(S)text(ince)\ ΔADB\ text(|||)\ ΔKDC`

`(BD)/(DC) = (AD)/(DK) = (AB)/(KC)\ \ text{(corresponding sides of similar triangles)}`

`:.AB xx DC = KC xx BD\ \ \ \ …\ (1)`

♦♦ Mean mark part (ii) 31%.

`text(Similarly, using)\ \ ΔADK\ text(|||)\ ΔBDC\ \ text{(given)}`

`(AD)/(BD)=(AK)/(BC)`

`:. BC xx AD = AK xx BD\ \ \ \ …\ (2)`

`text{Add (1) + (2)}`

`AB xx DC + AD xx BC`	`=KC xx BD+AK xx BD`
	`=BD(AK+KC)`
`:.BD xx AC`	`=AD xx BC+AB xx DC`

(iii) `text(Each diagonal of the pentagon is)\ x.`

♦♦ Mean mark part (iii) 24%.

`text{Hence in applying the result in (i) you have}`

`BD = AC = x, AD = DC = CB = 1, BA = x`

`x xx x`	`= 1 xx 1 + x xx 1`
`x^2 − x − 1`	`= 0`
`x`	`= (1 ± sqrt(1 + 4))/2`
	`= (1 ± sqrt(5))/2`
`:.x`	`= (1 + sqrt(5))/2,\ \ \ (x>0)`

Polynomials, EXT2 2010 HSC 6c

Expand `(cos theta + i sin theta)^5` using the binomial theorem. (1 mark)
Expand `(cos theta + i sin theta)^5` using de Moivre’s theorem, and hence show that
1. `sin 5theta = 16 sin^5 theta − 20sin^3 theta + 5 sin theta`. (3 marks)
Deduce that
`x = sin (pi/10)` is one of the solutions to
1. `16x^5 − 20x^3 + 5x − 1 = 0`. (1 mark)
Find the polynomial `p(x)` such that `(x − 1) p(x) = 16x^5 − 20x^3 + 5x − 1`. (1 mark)
Find the value of `a` such that `p(x) = (4x^2 + ax − 1)^2`. (1 mark)
Hence find an exact value for
1. `sin (pi/10)`. (1 mark)

Show Answers Only

`cos^5 theta + 5i cos^4 theta sin theta − 10 cos^3 theta sin^2 theta − 10i cos^2 theta sin^3 theta`
`+ 5 cos theta sin^4 theta + i sin^5 theta`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`16x^4 + 16x^3 − 4x^2 − 4x + 1`
`a = 2`
`(-1 + sqrt5)/4`

Show Worked Solution

(i) `(cos theta + i sin theta)^5`

`=cos^5 theta + 5cos^4 theta (i sin theta) + 10 cos^3 theta (i sin theta)^2 + `

`10 cos^2 theta (isin theta)^3 + 5 cos theta (i sin theta)^4 + (i sin theta)^5`

`= cos^5 theta + 5i cos^4 theta sin theta − 10 cos^3 theta sin^2 theta − 10i cos^2 theta sin^3 theta +`

`5 cos theta sin^4 theta + i sin^5 theta`

(ii) `text(Using De Moivre)`

`(cos theta + i sin theta)^5 = cos 5theta + i sin 5theta`

`text(Equating imaginary parts)`

`sin 5theta`	`= 5 cos^4theta sin theta − 10cos^2 theta sin^3 theta + sin^5 theta`
	`= 5 sin theta (1 − sin^2 theta)^2 − 10 sin^3 theta (1 − sin^2 theta) + sin^5 theta`
	`= 5 sin theta (1 − 2 sin^2 theta + sin^4 theta) − 10 sin^3 theta + 10 sin^5 theta + sin^5 theta`
	`= 5 sin theta − 10 sin^3 theta + 5 sin^5 theta − 10 sin^3 theta + 11 sin^5 theta`
	`= 16 sin^5 theta − 20 sin^3 theta + 5 sin theta`

(iii) `text(If)\ x = sin (pi/10)`

`16 sin^5 (pi/10) − 20 sin^3 (pi/10) + 5 sin (pi/10)`	`=sin (5 xx pi/10)`
`16x^5 − 20x^3 + 5x`	`= sin\ pi/2`
`16x^5 − 20x^3 + 5x`	`=1`

`:. sin\ pi/10\ \ text(is one solution to)\ \ 16x^5 − 20x^3 + 5x − 1=0`

(iv) `16x^5 − 20x^3 + 5x − 1`

`=(x-1)(16x^4 + 16x^3 − 4x^2 − 4x + 1)`

`:.p(x) = 16x^4 + 16x^3 − 4x^2 − 4x + 1`

(v) `(4x^2 + ax − 1)^2`

`= 16x^4 + 4ax^3 − 4x^2 + 4ax^3 + a^2x^2 − ax − 4x^2 − ax + 1`

`= 16x^4 + 8ax^3 − 8x^2 + a^2x^2 − 2ax +1`

`text(By equating coefficients of)\ \ x^3`

`:.a = 2`

(vi) `4 x^2 + 2 x − 1 = 0`

♦♦ Mean mark part (vi) 27%.

`x`	`=(-2 ± sqrt(4 + 16))/8`
	`=(-1 ± sqrt5)/4`

`:. sin\ pi/10 = (-1 + sqrt5)/4\ \ \ \ (sin\ pi/10\ > 0)`

Proof, EXT2 P2 2010 HSC 6b

A sequence `a_n` is defined by

`a_n = 2a_(n − 1) + a_(n − 2)`,

for `n ≥ 2`, with `a_0 = a_1 = 2`.

Use mathematical induction to prove that

`a_n = (1 + sqrt2)^n + (1 − sqrt2)^n` for all `n ≥ 0`. (3 marks)

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

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Prove)\ \ a_n = (1 + sqrt2)^n + (1 − sqrt2)^n\ text(for all)\ n ≥ 0.`

`text(where)\ a_0 = a_1 = 2, and \ a_n = 2a_(n − 1) + a_(n − 2)\ \ text(for)\ \ n ≥ 2,\ `

`text(When)\ \ n=0,`	`a_0`	`= (1 + sqrt2)^0 + (1 − sqrt2)^0`
		`= 1 + 1=2`
`text(When)\ \ n=1,`	`a_1`	`= (1 + sqrt2)^1 + (1 − sqrt2)^1`
		`= 1 + sqrt2 + 1 − sqrt2=2`

`:.\ text(True for)\ n=0\ \ and\ n=1`

♦ Mean mark 36%.

`text(Assume that)`

`a_k = (1 + sqrt2)^k + (1 − sqrt2)^k\ text(and)`

`a_(k − 1) = (1 + sqrt2)^(k − 1) + (1 − sqrt2)^(k − 1)\ text(for all)\ k ≥ 1.`

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

`text(i.e.)\ \ a_(k + 1) = (1 + sqrt2)^(k + 1) + (1 − sqrt2)^(k + 1).`

`a_(k + 1)`	`= 2a_k + a_(k − 1).`
	`= 2(1 + sqrt2)^k + 2(1 − sqrt2)^k + (1 + sqrt2)^(k − 1) + (1 − sqrt2)^(k − 1)`
	`= (1 + sqrt2)^(k − 1)(2 + 2sqrt2 + 1) + (1 − sqrt2)^(k − 1)(2 − 2sqrt2 + 1)`
	`= (1 + sqrt2)^(k − 1)(1 + sqrt2)^2 + (1 − sqrt2)^(k − 1)(1− sqrt2)^2`
	`= (1 + sqrt2)^(k + 1) + (1 − sqrt2)^(k + 1)`

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

`:.\ text(S)text(ince true for)\ n = 0 and n=1,\ text(by PMI, it is)`

`text(true for integral)\ n ≥ 0.`

Volumes, EXT2 2010 HSC 6a

The diagram shows the frustum of a right square pyramid. (A frustum of a pyramid is a pyramid with its top cut off.)

The height of the frustum is `h` m. Its base is a square of side `a` m, and its top is a square of side `b` m (with `a > b > 0`).

A horizontal cross-section of the frustum, taken at height `x` m, is a square of side `s` m, shown shaded in the diagram.

Show that
`s = a − ((a − b))/h x` (2 marks)
Find the volume of the frustum. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`h/3(a^2 + ab + b^2)\ \ text(m³)`

Show Worked Solution

(i)

`text(Using similar triangles)`

♦ Mean mark part (i) 46%.

`((s − b)/2)/((a − b)/2)`	`= (h − x)/h`	`text{(corresponding sides of}`
		`text{similar triangles)}`
`(s − b)/(a − b)`	`= 1 − x/h`
`s − b`	`= (a − b)(1 − x/h)`
`s − b`	`= a − b − (a − b) x/h`
`:.s`	`= a − (a − b) x/h`

(ii) `text(Volume of one slice)\ = s^2\ δx`

♦ Mean mark part (ii) 46%.

`:.V`	`= int_0^h s^2\ dx`
	` = int_0^h(a − ((a − b))/h x)^2 dx`
	`= int_0^h(a^2 − (2a(a − b))/h x + ((a − b)^2)/(h^2) x^2)dx`
	`= [a^2x − (a(a − b))/h x^2 + ((a − b)^2)/(3h^2) x^3]_0^h`
	`= a^2h − (a(a − b))/h h^2 + ((a − b)^2)/(3h^2) h^3`
	`= a^2h − (a^2 − ab)h + ((a^2 − 2ab + b^2))/3 h`
	`= h(a^2 − a^2 + ab + (a^2)/3 − (2ab)/3 + (b^2)/3)`
	`= h/3(a^2 + ab + b^2)\ \ text(m³)`

Harder Ext1 Topics, EXT2 2010 HSC 5c

A TV channel has estimated that if it spends `$x` on advertising a particular program it will attract a proportion `y(x)` of the potential audience for the program, where

`(dy)/(dx) = ay(1 − y)`

and `a > 0` is a given constant.

Explain why
`(dy)/(dx)` has its maximum value when `y = 1/2`. (1 mark)
Using
`int (dy)/(y(1 − y)) = ln (y/(1 − y)) + c`, or otherwise, deduce that
`y(x) = 1/(ke^(-ax) + 1)` for some constant `k > 0`. (3 marks)
The TV channel knows that if it spends no money on advertising the program then the audience will be one-tenth of the potential audience.
Find the value of the constant `k` referred to in part (c)(ii). (1 mark)
What feature of the graph
`y = 1/(ke^(-ax) + 1)` is determined by the result in part (c)(i)? (1 mark)
Sketch the graph
`y(x) = 1/(ke^(-ax) + 1)` (1 mark)

Show Answers Only

`text(See Worked Solutions.)`
`text{Proof (See Worked Solutions)}`
`9`
`y = 1/2`
`text(See Worked Solutions.)`

Show Worked Solution

(i) `(dy)/(dx) = ay(1 − y)`

♦♦ Mean mark part (i) 32%.

`=>text(Given)\ \ a>0,\ \ ay(1 − y)\ text(is an inverted parabola)`

`text(with zeros at)\ \ y = 0\ \ text(and)\ \ y = 1`

`=>\ text(Parabola symmetry means that a maximum occurs when)`

`y = (0+1)/2=1/2`

`:.(dy)/(dx)\ \ text(has its maximum value when)\ y = 1/2.`

♦ Mean mark part (ii) 41%.

(ii) `dy/dx=ay(1 − y)`

`int (dy)/(y(1 − y))`	`=int a\ dx`
`ax`	`= ln (y/(1 − y))+c`
`e^(ax − c)`	`= y/(1 − y)`
`e^(ax − c) − ye^(ax − c)`	`= y`
`y(1 + e^(ax − c))`	`= e^(ax − c)`
`y`	`= (e^(ax − c))/(1 + e^(ax − c))`
`y`	`= 1/(e^c e^(− ax) + 1)`

`text(Let)\ k = e^c`

`:.y(x) = 1/(ke^(-ax) + 1)\ text(for some constant)\ \ k > 0.`

(iii) `text(When)\ \ x = 0, y = 0.1`

`0.1`	`= 1/(ke^0 + 1)`
`k + 1`	`= 10`
`k`	`= 9`

♦♦ Mean mark part (iv) 25%.

(iv) `text(The gradient the curve is greatest at)\ y = 1/2\ \ \ text{from part (i)}`

`:. text(A point of inflection occurs at)\ y =1/2.`

(v) `text(As)\ \ x->oo,\ \ y->1/(0+1)=1`

♦♦ Mean mark part (v) 12%.

Harder Ext1 Topics, EXT2 2010 HSC 4d

A group of `12` people is to be divided into discussion groups.

In how many ways can the discussion groups be formed if there are `8` people in one group, and `4` people in another? (1 mark)
In how many ways can the discussion groups be formed if there are `3` groups containing `4` people each? (2 marks)

Show Answers Only

`495`
`5775`

Show Worked Solution

(i)	`text(Number of combinations)`	`=\ ^(12)C_4`
		`= 495`

(ii) `text(Number of discussion groups)`

♦ Mean mark part (ii) 35%.

`=(\ ^12C_4 xx\ ^8C_4 xx\ ^4C_4) / (3!)`

`= (34\ 650) / (3!)`

`= 5775`

Graphs, EXT2 2010 HSC 4a

A curve is defined implicitly by `sqrtx + sqrty = 1`.
Use implicit differentiation to find `(dy)/(dx)`. (2 marks)
Sketch the curve `sqrtx + sqrty = 1`. (2 marks)
Sketch the curve `sqrt(|\ x\ |) + sqrt(|\ y\ |) = 1` (1 mark)

Show Answers Only

`- sqrty/sqrtx`
`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

Show Worked Solution

(i)	`sqrtx + sqrt y`	`= 1`
	`1/(2sqrtx) + 1/(2sqrty)*(dy)/(dx)`	`= 0`
	`:.(dy)/(dx)`	`= – sqrty/sqrtx`

(ii)

♦♦ Mean mark part (iii) 46%.

(iii)

Complex Numbers, EXT2 N2 2010 HSC 2d

Let `z = cos theta + i sin theta` where `0 < theta < pi/2`.

On the Argand diagram the point `A` represents `z`, the point `B` represents `z^2` and the point `C` represents `z + z^2`.

Copy or trace the diagram into your writing booklet.

Explain why the parallelogram `OACB` is a rhombus. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Show that `text(arg)\ (z + z^2) = (3theta)/2`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Show that `| z + z^2 | = 2 cos theta/2`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
By considering the real part of `z + z^2`, or otherwise deduce that
`cos theta + cos 2theta = 2 cos theta/2 cos (3theta)/2`. (1 mark)

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

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

Show Worked Solution

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

`|\ OA\ |=|\ z\ |=1`

`|\ OB\ | =|\ z^2\ |=|\ z\ |^2 = 1`

`:.\ OACB\ text(is a parallelogram with a pair of adjacent sides equal.)`

`:.\ OACB\ text(is a rhombus.)`

ii. `text(arg)\ z^2 = 2\ text(arg)\ z = 2 theta`

`∠BOA = 2 theta − theta = theta`

`text(S)text(ince)\ OACB\ text(is a rhombus then)\ CO\ text(bisects)\ ∠BOA`

`:.∠COA = theta/2`

`:.\ text(arg)(z + z^2) = theta + theta/2 = (3theta)/2`

iii. `OC = |\ z + z^2\ |`

`text(Join)\ \ AB\ \ text(so that it meets)\ \ OC\ \ text(at)\ \ M`

`AB ⊥ OC, and OM=OC\ \ \ text{(diagonals of a rhombus)}`

♦♦ Mean mark part (iii) 26%.

`text(In)\ \ Delta OAM:`

`cos\ theta/2`	`=(OM)/(OA)`
	`=OM`
`:.OC`	`=2 xx OM`
	`=2 cos\ theta/2`

iv.	`z + z^2`	`= cos theta + i sin theta + (cos theta + i sin theta)^2`
		`= cos theta + cos 2theta + i(sin theta + sin 2theta)\ \ \ \ text{(De Moivre)}`
`:.\ text(Re)(z+z^2)=cos theta + cos 2theta`

♦♦ Mean mark part (iv) 44%.

`text{Using parts (ii) and (iii),}`

`z + z^2=2 cos\ theta/2(cos (3theta)/2+ i sin (3theta)/2)`

`text(Equating real parts:)`

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

Harder Ext1 Topics, EXT2 2011 HSC 8b

A bag contains seven balls numbered from `1` to `7`. A ball is chosen at random and its number is noted. The ball is then returned to the bag. This is done a total of seven times.

What is the probability that each ball is selected exactly once? (1 mark)
What is the probability that at least one ball is not selected? (1 mark)
What is the probability that exactly one of the balls is not selected? (2 marks)

Show Answers Only

`(6!)/7^6=720/(117\ 649)`
`1 – (6!)/7^6=(116\ 929)/(117\ 649)`
`(3 xx 6!)/7^5=2160/(16\ 807)`

Show Worked Solution

(i) `P text{(each ball is selected once)}`

♦♦ Mean mark part (i) 32%.

`=7/7 xx 6/7 xx 5/7 xx 4/7 xx 3/7 xx 2/7 xx 1/7`

`=(6!)/7^6`

`=720/(117\ 649)`

(ii) `P text{(at least one ball is not selected)}`

`=1 – P text{(each ball once)}`

`=1 – (6!)/7^6`

`=(116\ 929)/(117\ 649)`

(iii) `text{Consider when 1 of the balls is not selected (say #1).}`

♦♦♦ Mean mark part (iii) 2%.

`=>\ text(One of the other 6 balls is selected twice.)`

`text(e.g. 2, 2, 3, 4, 5, 6, 7).`

`text(This can be done in)\ \ (7!)/(2!)\ \ text(ways.)`

`text(With 6 differently numbered pairs possible,)`

`text(This can be done in)\ \ 6 xx (7!)/(2!)\ \ text(ways)`

`text(S)text(ince there are 7 numbers that can be left out,)`

`text(Total number of ways to leave 1 number out)`

`=7 xx 6 xx (7!)/(2!)`

`=21 xx 7!`

`:.P text{(exactly one ball not selected)}`	`=(21 xx 7!)/7^7`
	`=(3 xx 6!)/7^5`
	`=2160/(16\ 807)`

Conics, EXT2 2011 HSC 7c

The diagram shows the ellipse `x^2/a^2 + y^2/b^2 = 1`, where `a > b.` Let `e` be the eccentricity of the ellipse.

The line `l` is the tangent to the ellipse at the point `P`. The line `l` has equation `y = mx + c`, where `m` is the slope and `c` is the `y`-intercept.

The point `S` and `S prime` are the focal points of the ellipse, where `S` is on the positve `x`-axis. The perpendiculars to `l` through `S` and `S prime` intersect `l` at `Q` and `Q prime` respectively.

By substituting the equation for `l` into the equation for the ellipse, show that
1. `a^2 m^2 + b^2 = c^2.` (3 marks)
Show that the perpendicular distance from `S` to `l` is given by
1. `QS = (|\ mae + c\ |)/sqrt (1 + m^2).` (1 mark)
It is given that
`Q prime S prime = (|\ mae - c\ |)/sqrt (1 + m^2)`
Hence, prove that `QS xx Q prime S prime = b^2.` (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) `text(Substitute)\ \ y = mx + c\ \ text(into)\ \ x^2/a^2 + y^2/b^2 = 1`

` x^2/a^2 + ((mx + c)^2)/b^2`	`= 1`
`b^2 x^2 + a^2 (m^2 x^2 + 2mcx + c^2)`	`= a^2 b^2`
`(a^2 m^2 + b^2)x^2 + 2a^2 mcx + a^2 c^2 – a^2 b^2`	`= 0`

♦ Mean mark part (i) 35%.

`text(At)\ P,\ \ Delta = 0`

`(2a^2 mc)^2 – 4(a^2 m^2 + b^2) (a^2 c^2 – a^2 b^2) `	`= 0`
`4a^2[a^2 m^2 c^2 – (a^2 m^2 + b^2)(c^2 – b^2)]`	`=0`
`a^2 m^2 c^2 – (a^2 m^2 c^2 – a^2 m^2 b^2 + b^2 c^2 – b^4)`	`=0`
`a^2 m^2 b^2 – b^2 c^2 + b^4`	`=0`
`:.a^2 m^2 + b^2`	`=c^2`

(ii) `S(ae, 0),\ \ \ mx – y + c = 0`

♦ Mean mark part (ii) 49%.

`SQ`	`= \|\ (ax + by + c)/(sqrt (a^2+b^2))\ \|`
	`= \|\ (mae – 0 + c)/(sqrt (m^2 + (-1)^2))\ \|`
	`= (\|\ mae + c\ \|)/sqrt (1 + m^2)`

♦ Mean mark part (iii) 49%.

(iii) `Q prime S prime = (|\ mae – c\ |)/sqrt (1 + m^2)`

`QS xx Q prime S prime`	`= (\|\ mae + c\ \|)/sqrt (1 + m^2) xx (\|\ mae – c\ \|)/sqrt (1 + m^2)`
	`= 1/(1 + m^2) xx \|\ m^2 a^2 e^2 – c^2\ \|`
	`= 1/(1 + m^2) xx \|\ m^2 a^2 e^2 – (a^2 m^2 + b^2)\ \|`
	`= 1/(1 + m^2) xx \|\ m^2 a^2 e^2 – a^2 m^2 – b^2\ \|`
	`= 1/(1 + m^2) xx \|\ m^2 a^2 (e^2 – 1) – b^2\ \|`
	`= 1/(1 + m^2) xx \|\ -m^2 b^2 – b^2\ \|,\ \ \ text{(using}\ \ b^2=a^2(e^2-1)text{)}`
	`= 1/(1 + m^2) xx \|\ -b^2\ \| \|\ m^2 + 1\ \|`
	`= b^2/(1 + m^2) xx \|\ m^2 + 1\ \|`
	`= b^2`

Calculus, EXT2 C1 2011 HSC 7b

Let `I = int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx.`

Use the substitution `u = 4-x` to show that

`I = int_1^3 (sin^2(pi/8 u))/(u(4-u))\ du.` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, find the value of `I`. (3 marks)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`1/4 log_e 3`

Show Worked Solution

i. `text(Let)\ \ u = 4-x,\ \ du = -dx,\ \ x = 4-u`

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

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

`I`	`=int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx`
	`=int_3^1 (cos^2 (pi/8(4-u)))/((4-u)u) (-du)`
	`=-int_3^1 (cos^2(pi/2-pi/8 u))/((4-u)u)\ du`
	`=int_1^3 (sin^2(pi/8 u))/(u (4-u))\ du`

ii. `text{S}text{ince}\ \ int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx = int_1^3 (sin^2(pi/8 x))/(x(4-x))\ dx`

`text(We can add the integrals such that)`

`2I`	`=int_1^3 (cos^2(pi/8 x))/(x(4-x)) dx + int_1^3 (sin^2(pi/8 x))/(x(4-x)) dx`
	`=int_1^3 1/(x(4-x)) dx`

`text(Using partial fractions:)`

♦ Mean mark 36%.

`1/(x(4-x))`	`=A/x+B/(4-x)`
`1`	`=A(4-x)+Bx`

`text(When)\ \ x=0,\ \ A=1/4`

`text(When)\ \ x=0,\ \ B=1/4`

`2I`	`=1/4 int_1^3 (1/x + 1/(4-x))\ dx`
	`=1/4 [log_e x-log_e (4-x)]_1^3`
	`=1/4 [log_e 3-log_e 1-(log_e 1-log_e 3)]`
	`=1/2 log_e 3`
`:.I`	`=1/4 log_e 3`

Mechanics, EXT2 M1 2011 HSC 6a

Jac jumps out of an aeroplane and falls vertically. His velocity at time `t` after his parachute is opened is given by `v(t)`, where `v(0) = v_0` and `v(t)` is positive in the downwards direction. The magnitude of the resistive force provided by the parachute is `kv^2`, where `k` is a positive constant. Let `m` be Jac’s mass and `g` the acceleration due to gravity. Jac’s terminal velocity with the parachute open is `v_T.`

Jac’s equation of motion with the parachute open is

`m (dv)/(dt) = mg - kv^2.` (Do NOT prove this.)

Explain why Jac’s terminal velocity `v_T` is given by

`sqrt ((mg)/k).` (1 mark)

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By integrating the equation of motion, show that `t` and `v` are related by the equation

`t = (v_T)/(2g) ln[((v_T + v)(v_T - v_0))/((v_T - v)(v_T + v_0))].` (3 marks)

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Jac’s friend Gil also jumps out of the aeroplane and falls vertically. Jac and Gil have the same mass and identical parachutes.

Jac opens his parachute when his speed is `1/3 v_T.` Gil opens her parachute when her speed is `3v_T.` Jac’s speed increases and Gil’s speed decreases, both towards `v_T.`

Show that in the time taken for Jac's speed to double, Gil's speed has halved. (3 marks)

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Show Answers Only

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

Show Worked Solution

i. `m (dv)/(dt) = mg – kv^2`

`text(As)\ \ v ->text(terminal velocity,)\ \ (dv)/(dt) -> 0`

`mg – kv_T^2`	`= 0`
`v_T^2`	`= (mg)/k`
`v_T`	`= sqrt ((mg)/k)`

ii. `m (dv)/(dt) = mg – kv^2`

`int_0^t dt`	`=int_(v_0)^v m/(mg – kv^2)\ dv`
`t`	`= m/k int_(v_0)^v (dv)/((mg)/k – v^2)\ \ \ \ text{(using}\ \ v_T^2= (mg)/k text{)}`
	`= (v_T^2)/g int_(v_0)^v (dv)/(v_T^2 – v^2)`

♦ Mean mark part (ii) 39%.

`text(If)\ \ v_0 < v_T,\ \ text(then)\ \ v < v_T\ \ text(at all times.)`

`:. t`	`=(v_T^2)/g int_(v_0)^v (dv)/(v_T^2 – v^2)`
	`=(v_T^2)/g int_(v_0)^v (dv)/((v_T – v)(v_T + v))`
	`=(v_T^2)/(2 g v_T) int_(v_0)^v (1/((v_T – v)) + 1/((v_T + v))) dv`
	`=v_T/(2g)[-ln (v_T – v) + ln (v_T + v)]_(v_0)^v`
	`=v_T/(2g)[ln(v_T + v)-ln(v_T – v)-ln(v_T + v_0)+ln(v_T – v_0)]`
	`=v_T/(2g) ln[((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

`text(If)\ \ v_0 > v_T,\ \ text(then)\ \ v > v_T\ \ text(at all times.)`

`text(Replace)\ \ v_T – v\ \ text(by) – (v – v_T)\ \ text(in the preceeding)`

`text(calculation, leading to the same result.)`

iii. `text{From (ii) we have}:\ \ t = v_T/(2g) ln [((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

`text(For Jac,)\ \ v_0 = V_T/3\ \ text(and we have to find the time)`

`text(taken for his speed to be)\ \ v = (2v_T)/3.`

`t`	`=v_T/(2g) ln[((v_T + (2v_T)/3)(v_T – v_T/3))/((v_T – (2v_T)/3)(v_T + v_T/3))]`
	`=v_T/(2g) ln [(((5v_T)/3)((2v_T)/3))/((v_T/3)((4v_T)/3))]`
	`=v_T/(2g) ln[(10/9)/(4/9)]`
	`=v_T/(2g) ln (5/2)`

`text(For Gil),\ \ v_0 = 3v_T\ \ text(and we have to find)`

♦♦♦ Mean mark part (iii) 16%.

`text(the time taken for her speed to be)\ \ v = (3v_T)/2.`

`t`	`=v_T/(2g) ln [((v_T + (3v_T)/2)(v_T – 3v_T))/((v_T – (3v_T)/2)(v_T + 3v_T))]`
	`=v_T/(2g) ln [(((5v_T)/2)(-2v_T))/((-v_T/2)(4v_T))]`
	`=v_T/(2g) ln [(-5)/-2]`
	`=v_T/(2g) ln (5/2)`

`:.\ text(The time taken for Jac’s speed to double is)`

`text(the same as it takes for Gil’s speed to halve.)`

Conics, EXT2 2011 HSC 5c

The diagram shows the ellipse `x^2/a^2 + y^2/b^2 = 1`, where `a > b`. The line `l` is the tangent to the ellipse at the point `P`. The foci of the ellipse are `S` and `S prime`. The perpendicular to `l` through `S` meets `l` at the point `Q`. The lines `SQ` and `S prime P` meet at the point `R`.

Copy or trace the diagram into your writing booklet.

Use the reflection property of the ellipse at `P` to prove that `SQ = RQ.` (2 marks)
Explain why `S prime R = 2a.` (1 mark)
Hence, or otherwise, prove that `Q` lies on the circle `x^2 + y^2 = a^2.` (3 marks)

Show Answers Only

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

Show Worked Solution

(i)

`text(Let)\ \ l\ \ text(cut the)\ \ y text(-axis at)\ \ M`

`/_ MPS prime`	`= /_ QPS\ \ \ \ \ text{(reflection property of an ellipse)}`
`/_ RPQ`	`=/_ MPS prime\ \ \ \ \ text{(vertically opposite angles)}`

♦♦♦ Mean mark 21%.

`text(In)\ \ Delta SPQ and Delta RPQ`

`/_ SPQ = /_ RPQ\ \ \ text{(both equal}\ \ /_ MPS prime text{)}`

`/_ SQP = /_ RQP=90^@\ \ \ text{(given that}\ \ PQ⊥SRtext{)}`

`PQ\ \ text(is a common side)`

`:.\ Delta SPQ -= Delta RPQ\ \ \ \ text{(AAS)}`

`:. SQ = RQ\ \ \ text{(corresponding sides of congruent triangles)}`

♦ Mean mark part (ii) 46%.

(ii) `SP = PR\ \ \ text{(corresponding sides of congruent triangles)}`

`S prime P+ PS`	`= 2a\ \ \ \ text{(locus property of an ellipse)`
`:.S prime P+PR`	`= 2a`
`:.S prime R`	`= 2a`

(iii) `text(Join)\ \ QO`

♦♦♦ Mean mark 5%.

`text(Consider)\ \ Delta SS prime R and Delta SOQ`

`(SO)/(SS prime) = 1/2\ \ \ \ text{(}O\ \ text(is the midpoint of)\ \ SS prime text{)}`

`(SQ)/(SR) = 1/2\ \ \ \ text{(}Q\ \ text(is the midpoint of)\ \ SR prime text{)}`

`/_ S prime SR = /_ OSQ\ \ text{(common angle)}`

`:.\ Delta SS prime R\ text(|||)\ Delta SOQ`

`\ \ \ text{(AAS)}`

`(OQ)/(S prime R)`	`= 1/2`	`\ \ \ text{(corresponding sides of}`
		`\ \ \ text{similar triangles)}`
`(OQ)/(2a)`	`=1/2`
`:.\ OQ`	`=a`

`:.\ Q\ \ text(lies on the circle)\ \ x^2 + y^2 = a^2`

Proof, EXT2 P1 2011 HSC 5b

If `p, q` and `r` are positive real numbers and `p + q >= r`, prove that

`p/(1 + p) + q/(1 + q) - r/(1 + r) >= 0.` (3 marks)

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

Show Worked Solution

`text(If)\ \ p + q >= r,\ \ text(then)\ \ p + q – r >= 0`

♦ Mean mark 41%.

`text(LHS)`	`= p/(1 + p) + q/(1 + q) – r/(1 + r)`
	`= (p(1 + q)(1 + r) + q(1 + p)(1 + r) – r(1 + p)(1 + q))/((1 + p)(1 + q)(1 + r))`
	`= (p(1 + q + r + qr) + q(1 + p + r + pr) – r(1 + p + q + pq))/((1 + p)(1 + q)(1 + r))`
	`= (p + pq + pr + pqr + q + pq + qr + pqr – r – pr – qr – pqr)/((1 + p)(1 + q)(1 + r))`
	`= ((p + q – r) + pq(2 + r))/((1 + p)(1 + q)(1 + r))`
	`>=(pq(2 + r))/((1 + p)(1 + q)(1 + r))\ \ \ \ text{(S}text{ince}\ \ p + q – r >= 0 text{)}`
	`>= 0\ \ \ \ \ \ \ text{(S}text{ince}\ \ p > 0,\ q > 0,\ r > 0text{)}`

Harder Ext1 Topics, EXT2 2011 HSC 4b

In the diagram, `ABCD` is a cyclic quadrilateral. The point `E` lies on the circle through the points `A, B, C` and `D` such that `AE\ text(||)\ BC`. The line `ED` meets the line `BA` at the point `F`. The point `G` lies on the line `CD` such that `FG\ text(||)\ BC.`

Copy or trace the diagram into your writing booklet.

Prove that `FADG` is a cyclic quadrilateral. (2 marks)
Explain why `/_ GFD =/_ AED.` (1 mark)
Prove that `GA` is a tangent to the circle through the points `A, B, C` and `D.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`/_ GFD = /_ AED\ \ text{(alternate angles,}\ \ FG\ text(||)\ AE text{)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`text(Let)\ /_ BCD`	`= alpha`
`/_ FAD`	`= alpha\ \ text{(exterior angle of a cyclic quadrilateral}\ ABCD text{)}`
`/_ FGC`	`= pi – alpha\ \ text{(cointerior angles,}\ \ FG\ text(\|\|)\ BC text{)}`

`:.\ \ /_ FAD + /_ FGD = pi`

`:.\ FADG\ \ text{is a cyclic quadrilateral (opposite angles are supplementary)}`

(ii) `/_ GFD = /_ AED\ \ text{(alternate angles,}\ \ FG\ text(||)\ AE text{)}`

(iii) `text(Join)\ GA`

`/_ GAD = /_ GFD\ \ text{(angles in the same segment on arc}\ GDtext{)}`

`text(S)text(ince)\ /_ GFD`	`= /_ AED\ \ \ text{(part (ii))}`
`/_ GAD`	`= /_ AED`

`:.GA\ \ text(is a tangent to the circle through)\ \ A, B, C and D.`

`text{(angle in the alternate segment equals the angle}`

`text(between)\ GA\ text(and chord)\ AD text{).}`

Complex Numbers, EXT2 N2 2011 HSC 4a

Let `a` and `b` be real numbers with `a != b`. Let `z = x + iy` be a complex number such that

`|\ z - a\ |^2 - |\ z - b\ |^2 = 1.`

Prove that `x = (a + b)/2 + 1/(2 (b - a)).` (2 marks)

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Hence, describe the locus of all complex numbers `z` such that `|\ z - a\ |^2 - |\ z - b\ |^2 = 1.` (1 mark)

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`text(Proof)\ \ text{(See Worked Solutions)`
`x = (a + b)/2 + 1/(2(b – a))`

Show Worked Solution

i.	`\|\ z – a\ \|^2 – \|\ z – b\ \|^2`	`= 1`
	`\|\ (x-a)+iy\ \|^2-\|\ (x-b)+iy\ \|^2`	`=1`
	`(x – a)^2 + y^2 – ((x – b)^2 + y^2)`	`=1`
	`(x – a)^2 – (x – b)^2`	`=1`
	`(x – a – (x – b)) (x – a + x – b)`	`=1`
	`(b – a) (2x – a – b)`	`=1`

`2x – a – b`	`= 1/(b – a)`
`2x`	`= a + b + 1/(b – a)`
`:. x`	`= (a + b)/2 + 1/(2(b – a))`

♦ Mean mark part (ii) 44%.

ii. `text(The locus is the vertical line)`

`x = (a + b)/2 + 1/(2(b – a)).`

Proof, EXT2 P1 2012 HSC 16c

Let `n` be an integer where `n > 1`. Integers from `1` to `n` inclusive are selected randomly one by one with repetition being possible. Let `P(k)` be the probability that exactly `k` different integers are selected before one of them is selected for the second time, where `1 ≤ k ≤ n`.

Explain why `P(k) = ((n − 1)!k)/(n^k(n − k)!)`. (2 marks)

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Suppose `P(k) ≥ P(k − 1)`. Show that `k^2- k- n ≤ 0`. (2 marks)

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Show that if `sqrt(n + 1/4) > k − 1/2` then the integers `n` and `k` satisfy `sqrtn > k − 1/2`. (2 marks)

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Hence show that if `4n + 1` is not a perfect square, then `P(k)` is greatest when `k` is the closest integer to `sqrtn`.

You may use part (iii) and also that `k^2 − k − n >0` if `P(k)< P(k − 1)`. (2 marks)

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

Show Worked Solution

i. `text(Let)\ P(1)=text{(After 1st number chosen, the second draw matches)}`

`P(1) =n/n xx 1/n=1/n`

`text(Let)\ P(2)=text{(After 1st two numbers chosen, the third draw matches)}`

`P(2) = n/n xx (n − 1)/n xx 2/n`

♦♦♦ Mean mark part (i) 7%.

`P(3)=n/n xx (n-1)/n xx (n-2)/n xx 3/n`

`vdots`

`=>text{On the}\ (k+1)text(th draw)`

`P(k)`	`= n/n xx (n − 1)/n xx (n − 2)/n xx …\ xx (n − k+1)/n xx k/n`
	`=((n-1)!)/n^k xx k/(1 xx 2 xx … xx (n-k))`
	`=((n-1)!\ k)/(n^k (n-k)!)`

♦ Mean mark part (ii) 38%.

ii.	`P(k)`	`≥ P(k − 1)`
	`((n − 1)!\ k)/(n^k(n − k)!)`	`≥ ((n − 1)! (k − 1))/(n^(k − 1)(n − k + 1)!)`
	`k(n − k + 1)`	`≥ n(k − 1)`
	`kn − k^2 + k`	`≥ nk − n`
	`:.k^2 − k − n`	`≤ 0`

♦♦♦ Mean mark part (iii) 10%.

iii.	`sqrt(n + 1/4)`	`> k − 1/2`
	`n + 1/4`	`> (k − 1/2)^2`
	`n + 1/4`	`> k^2 − k + 1/4`
	`n`	`>k^2 − k`
	`n`	`> k(k − 1)`

`text(S)text(ince)\ n\ text(and)\ k\ text(are integers such that)\ 1 ≤ k ≤ n.`

`=>n\ text(is at least the next integer after)\ k.`

`=>(k −1)\ text(is the integer before)\ k.`

`:.n`	`>k^2 − k +1/4`
`n`	`>(k-1/2)^2`
`:.sqrt n`	`>k-1/2`

iv. `text(Find the largest integer)\ \ k\ \ text(for which)\ \ P(k) ≥ P(k − 1)`

`P(k) ≥ P(k − 1)\ \ text(when)\ \ k^2 − k − n ≤ 0\ \ \ text{(part (ii))}`

`text(Solving)\ \ k^2 − k − n ≤ 0`

♦♦♦ Mean mark part (iii) 1%.

`(1 – sqrt(4n + 1))/2 ≤ k ≤ (1 + sqrt(4n+1))/2`

`text(Given)\ \ n>0,\ \ (1 – sqrt(4n + 1))/2<0`

`:.1 ≤ k ≤ (1 + sqrt(4n+1))/2`

`text(S)text(ince)\ \ 4n+1\ \ text(is not a perfect square and)\ \ k\ \ text(is an integer)`

`k<`	` (1 + sqrt(1 + 4n))/2`
`k<`	` 1/2 + sqrt(n + 1/4)`
`k-1/2<`	`sqrt(n + 1/4)`
`k-1/2<`	`sqrt n\ \ \ \ \ text{(from part (iii))}`
`k<`	`1/2+sqrt n`

`:. P(k)\ text(is greatest when)\ \ k\ \ text(is the closest integer to)\ n.`

Harder Ext1 Topics, EXT2 2012 HSC 16a

In how many ways can `m` identical yellow discs and `n` identical black discs be arranged in a row? (1 mark)
In how many ways can 10 identical coins be allocated to 4 different boxes? (1 mark)

Show Answers Only

`((m + n)!)/(m!n!)`
`(13!)/(10!3!)`

Show Worked Solution

(i) `((m + n)!)/(m!n!)\ text{ways (By definition)}`

♦ Mean mark part (i) 43%.

(ii) `text{Consider this as being “arrange 10 coins in 4 boxes}`

♦♦♦ Mean mark part (ii) 1%!

`text{with 3 separators between the boxes, making a total}`

`text{of 13 items” (as per the diagram).}`

`:.\ text(10 identical coins and 3 identical separators to arrange.)`

`:.\ text(Number of ways) = (13!)/(10!3!)\ \ \ \ text{(from part (i))}`

Proof, EXT2 P1 2012 HSC 15a

Prove that `sqrt(ab) ≤ (a + b)/2`, where `a ≥ 0` and `b ≥ 0`. (1 mark)

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If `1 ≤ x ≤ y`, show that `x(y − x + 1) ≥ y`. (2 marks)

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Let `n` and `j` be positive integers with `1 ≤ j ≤ n`.

Prove that `sqrtn ≤ sqrt(j(n − j + 1)) ≤ (n + 1)/2.` (2 marks)

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For integers `n ≥ 1`, prove that

`(sqrtn)^n ≤ n! ≤ ((n + 1)/2)^n`. (1 mark)

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Show Answers Only

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

Show Worked Solution

i.	`(sqrta − sqrtb)^2`	`≥ 0`
	`a − 2sqrt(ab) + b`	`≥ 0`
	`a + b`	`≥ 2sqrt(ab)`
	`sqrt(ab)`	`≤ (a + b)/2`

ii. `text(Solution 1)`

♦♦ Mean mark part (ii) 28%.

`text(S)text(ince)\ \ 1 ≤ x ≤ y`

`y-x`	`>=0`
`y(x-1)-x(x-1)`	`>=0,\ \ \ \ (x-1>=0)`
`xy-x^2+x-y`	`>=0`
`:.xy-x^2+x`	`>=y`

`text(Solution 2)`

`x( y − x + 1)`	`= xy − x^2 + x`
	`= -y + xy − x^2 + x + y`
	`= y(x − 1) − x(x − 1)+ y`
	`= (x − 1)( y − x) + y`

`text(S)text(ince)\ \ x − 1 ≥ 0\ \ text(and)\ \ y − x ≥ 0`

`=>(x − 1)( y − x) + y`	`>=y`
`:.x(y − x + 1)`	`>=y`

iii. `text(Let)\ c = n − j + 1, \ c > 0\ text(as)\ n ≥ j.`

♦ Mean mark part (iii) 39%.

`=> sqrt(j(n − j + 1))\ \ text(can be written as)\ \ sqrt(jc).`

`sqrt(jc)`	`≤ (j + c)/2\ \ \ \ text{(part (i))}`
`sqrt(j(n − j + 1))`	`≤ (j + n-j+1)/2`
`=>sqrt(j(n − j + 1))`	`≤ (n+1)/2`
`j(n − j +1)`	`≥ n\ \ \ \ text{(part (ii))}`
`=>sqrt(j(n − j + 1))`	`≥ sqrt n`

`:.sqrtn ≤ sqrt(j(n − j + 1)) ≤ (n + 1)/2`

iv. `text{From (iii)}\ n ≤ j(n− j +1) ≤ (n + 1)/2`

♦♦♦ Mean mark part (iv) just 2%!

`text(Let)\ j\ text(take on the values from 1 to)\ n.`

`j = 1:`	`sqrtn ≤ sqrt(1(n)) ≤ (n + 1)/2`
`j = 2:`	`sqrtn ≤ sqrt(2(n−1)) ≤ (n + 1)/2`
	`vdots`
`j = n:`	`sqrtn ≤ sqrt(n(1)) ≤ (n + 1)/2`

`text{Multiply the corresponding parts of each line}`

`(sqrtn)^n ≤ sqrt(n! xx n!) ≤ ((n + 1)/2)^n`

`(sqrtn)^n ≤ n! ≤ ((n + 1)/2)^n`

Harder Ext1 Topics, EXT2 2012 HSC 14d

The diagram shows points `A` and `B` on a circle. The tangents to the circle at `A` and `B` meet at the point `C`. The point `P` is on the circle inside `ΔABC`. The point `E` lies on `AB` so that `AB ⊥ EP`. The points `F` and `G` lie on `BC` and `AC` respectively so that `FP ⊥ BC` and `GP ⊥ AC`.

Copy or trace the diagram into your writing booklet.

Show that `ΔAPG` and `ΔBPE` are similar. (2 marks)
Show that `EP^2 = FP xx GP`. (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

`text(Join)\ AP\ text(and)\ BP.`

`text(In)\ ΔAPG\ text(and)\ ΔBPE`

`∠GAP`	`= ∠ABP\ \ text{(angle in alternate segment)}`
`∠AGP`	`= ∠BEP = 90^@\ \ text{(given)}`
`:.ΔAPG\ text(\|\|\|)\ ΔBPE\ \ text{(equiagular)}`

(ii) `text(Similarly, ΔBPF ||| ΔAPE)`

`:. (FP)/(EP)`	`= (BP)/(AP)\ \ `	`text{(corresponding sides in}`
		`Delta BPF and Delta APEtext{)}`
`(AP)/(BP) `	`= (GP)/(EP)`	`text{(corresponding sides in}`
		`Delta APG and Delta BPEtext{)}`
`=>(EP)/(GP)`	`=(FP)/(EP)`
`:.EP^2`	`=FP xx GP`

Conics, EXT2 2012 HSC 13c

Let `P` be a point on the hyperbola given parametrically by `x = a\ sec\ theta` and `y = b\ tan\ theta`, where `a` and `b` are positive. The foci of the hyperbola are `S(ae,0)` and `S′(–ae,0)` where `e` is the eccentricity. The point `Q` is on the `x`-axis so that `PQ` bisects `∠SPS′`.

Show that `SP = a(e\ sec\ theta\ – 1)`. (1 mark)
It is given that
`S′P = a(e\ sec\ theta + 1),\ \ and\ \ (PS)/(QS) = (PS′)/(QS′)`.
Using this, or otherwise, show that the `x`-coordinate of `Q` is
1. `a/(sec\ theta)`. (2 marks)
The slope of the tangent to the hyperbola at `P` is
1. `(b\ sec\ theta)/(a\ tan\ theta)`. (Do NOT prove this.)
Show that the tangent at `P` is the line `PQ`. (1 mark)

Show Answers Only

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

Show Worked Solution

(i) `text(Solution 1)`

♦ Mean mark 50%.

`text(Let)\ M\ text(be on the directrix at)\ x = a/e`

`M (a/e, b\ tan theta)`

`PM`	`= a\ sec\ theta − a/e`
	`= a/e(e\ sec\ theta − 1)`

`text(S)text(ince)\ PS = ePM\ \ \ text{(by definition)}`

`:.PS`	`= e xx a/e(e\ sec\ theta − 1) `
	`=a(e\ sec\ theta − 1)`

`text(Solution 2)`

`PS`	`= sqrt((a\ sec\ theta − ae)^2 + b^2\ tan^2\ theta)`
	`= sqrt(a^2\ sec^2\ theta − 2a^2e\ sec\ theta + a^2e^2 + b^2\ tan^2\ theta)`
	`= sqrt(a^2\ sec^2\ theta − 2a^2e\ sec\ theta + a^2e^2 + (a^2e^2 − a^2)tan^2\ theta)`
	`= sqrt(a^2\ sec^2\ theta − a^2\ tan^2\ theta − 2a^2e\ sec\ theta + a^2e^2 + a^2e^2\ tan^2\ theta)`
	`= sqrt(a^2 (sec^2\ theta − tan^2\ theta) − 2a^2e\ sec\ theta + a^2e^2 + a^2e^2\ tan^2\ theta)`
`text{(Using}\ \ sec^2 theta – tan^2 theta = 1 text{)}`
	`= sqrt(a^2 − 2a^2e\ sec\ theta + a^2e^2 + a^2e^2 (sec^2\ theta −1))`
	`= a sqrt(1− 2e\ sec\ theta + e^2 + e^2\ sec^2\ theta − e^2)`
	`= a sqrt(e^2\ sec^2\ theta − 2e\ sec\ theta +1)`
	`= a sqrt((e\ sec\ theta −1)^2)`
	`= a(e\ sec\ theta −1)`

(ii) `text(Let)\ Q\ text(be)\ (x, 0)`

`text{Using}\ \ (PS)/(QS)“= (PS′)/(QS′)`

`(a(e\ sec theta − 1))/(ae-x)`	`=(a(e\ sec\ theta + 1))/(x+ae)`
`a(x + ae)(e\ sec\ theta − 1)`	`= a(ae − x)(e\ sec\ theta + 1)`
`xe\ sec\ theta − x + ae^2\ sec\ theta − ae`	`= ae^2\ sec\ theta + ae − xe\ sec\ theta − x`
`2xe\ sec\ theta`	`= 2ae`
`:.x`	`= a/(sec\ theta)\ \ \ text(… as required)`

(iii) `text(Solution 1)`

`m_text(tan)=(b\ sec theta)/(a\ tan theta)`

`P(a\ sec theta, b\ tan\ theta),\ \ Q(a/(sec\ theta), 0)`

`m_(PQ)`	`= (b tan theta)/(a sec theta – a/sec theta)`
	`=(b tan theta sec theta)/(a(sec^2 theta-1))`
	`=(b tan theta sec theta)/(a tan^2 theta)`
	`=(b sec theta)/(a tan theta)`

`:. PQ\ text(and the tangent at)\ P\ text(are the same line.)`

`text(Alternative Solution)`

`text(The equation of the tangent at)\ P\ text(is)`

`y − b\ tan\ theta = (b\ sec\ theta)/(a\ tan\ theta)(x − a\ sec\ theta)`

`text(Check if)\ Q(a/(sec\ theta),0)\ text(satisfies)`

`-b\ tan\ theta`	`= (b\ sec\ theta)/(a\ tan\ theta)(x − a\ sec\ theta)`
`-ab\ tan^2\ theta`	`= bx\ sec\ theta − ab\ sec^2\ theta`
`x\ sec\ theta`	`= a(sec^2\ theta − tan^2\ theta)`
`x\ sec\ theta`	`= a`
`x`	`=a/ sec theta\ \ \ \ text{(proven true in part (ii))}`

`:.\ text(This line passes through)\ Q\ text(so the tangent at)\ P\ text(is the line)\ PQ.`

Harder Ext1 Topics, EXT2 2012 HSC 13b

The diagram shows `ΔS′SP`. The point `Q` is on `S′S` so that `PQ` bisects `∠S′PS`. The point `R` is on `S′P` produced so that `PQ\ text(||)\ RS`.

Show that `PS = PR`. (1 mark)
Show that `(PS)/(QS) = (PS′)/(QS′)`. (2 marks)

Show Answers Only

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

Show Worked Solution

(i)	`∠S′PQ`	`= ∠PRS = α`	`text{(corresponding angles,}\ QP\ text(\|\|)\ SR)`
	`∠QPS`	`= ∠PSR = α`	`text{(alternate angles,}\ QP\ text(\|\|)\ SR)`
	`:. ∠PSR`	`= ∠PRS = α`

`:. ΔPSR\ text{is isosceles (two angle equal)}`

`:. PS = PR\ \ \ text{(opposite equal angles in}\ Delta PSR text{)}`

(ii) `text(Draw a line through)\ S′\ text(parallel to)\ QP`

`:.(PS′)/(PR)`	`= (QS′)/(QS)\ \ \ text{(parallel lines preserve ratios)}`
`(PS′)/(QS′)`	`= (PR)/(QS)`
`text(S)text(ince)\ \ PS = PR\ \ \ text{(from part (i))}`
`(PS′)/(QS′)`	`= (PS)/(QS)\ \ \ text(… as required)`

Complex Numbers, EXT2 N2 2012 HSC 12d

On the Argand diagram the points `A_1` and `A_2` correspond to the distinct complex numbers `u_1` and `u_2` respectively. Let `P` be a point corresponding to a third complex number `z`.

Points `B_1` and `B_2` are positioned so that `ΔA_1PB_1` and `ΔA_2B_2P`, labelled in an anti-clockwise direction, are right-angled and isosceles with right angles at `A_1` and `A_2`, respectively. The complex numbers `w_1` and `w_2` correspond to `B_1` and `B_2`, respectively.

Explain why `w_1 = u_1 + i(z − u_1)`. (1 mark)

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Find the locus of the midpoint of `B_1B_2` as `P` varies. (2 marks)

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

Show Answers Only

`text(See Worked Solutions.)`
`(u_1 + u_2)/2 + (u_2 − u_1)/2 i`

Show Worked Solution

i.	`vec (A_1P)`	`= z − u_1`
	`vec (A_1B_1)`	`= w_1 − u_1`

`B_1A_1 ⊥ A_1P\ text(and)\ |vec (A_1P)| = |vec (A_1B_1)|`

`vec (A_1B_1)\ text(is an anticlockwise rotation of)\ vec (A_1P)\ text(through)\ 90^@`

`:.w_1 − u_1 = i(z −u_1)`

`:.w_1 = u_1+ i(z −u_1)`

ii.	`vec (A_2B_2)`	`= w_2 − u_2`
	`vec (A_2P)`	`= z − u_2`

`A_2B_2 ⊥ A_2P\ text(and)\ |vec (A_2B_2)| = |vec (A_2P)|`

`vec (A_2P)\ text(is an anticlockwise rotation of)\ vec (A_2B_2)\ text(through)\ 90^@`

`z − u_2`	`= i(w_2 −u_2)`
`iw_2`	`= z − u_2 + iu_2`
`−w_2`	`= iz − iu_2 − u_2`
`:. w_2`	`= u_2 + i(u_2 − z)`

`:.\ text(The midpoint of)\ B_1B_2\ text(is)\ (w_1 + w_2)/2`

`= 1/2[u_1 + i(z − u_1) + u_2 + i(u_2 − z)]`

`= 1/2[u_1 + u_2 + i(u_2 − u_1)]`

`= (u_1 + u_2)/2 + (u_2 − u_1)/2 i\ \ \ \ text{(which is a fixed point)}`

Mechanics, EXT2 2013 HSC 16b

A small bead `P` of mass `m` can freely move along a string. The ends of the string are attached to fixed points `S` and `S′`, where `S′` lies vertically above `S`. The bead undergoes uniform circular motion with radius `r` and constant angular velocity `omega` in a horizontal plane.

The forces acting on the bead are the gravitational force and the tension forces along the string. The tension forces along `PS` and `PS′` have the same magnitude `T`.

The length of the string is `2a` and `SS′= 2ae`, where `0 < e < 1`. The horizontal plane through `P` meets `SS′` at `Q`. The midpoint of `SS′` is `O` and `beta = /_S′PQ`. The parameter `theta` is chosen so that `OQ =a cos theta.`

What information indicates that `P` lies on an ellipse with foci `S` and `S′`, and with eccentricity `e`? (1 mark)
Using the focus–directrix definition of an ellipse, or otherwise, show that
`SP = a(1 − e cos theta ).` (1 mark)
Show that
`sin beta = (e + cos theta)/(1 + e cos theta).` (2 marks)
By considering the forces acting on `P` in the vertical direction, show that
1. `mg = (2T(1 - e^2) cos theta)/(1 - e^2 cos^2 theta).` (2 marks)
Show that the force acting on `P` in the horizontal direction is
1. `mr omega^2 = (2T sqrt(1 - e^2) sin theta)/(1 - e^2 cos^2 theta).` (3 marks)
Show that
1. `tan theta = (r omega^2)/g sqrt(1 - e^2).` (1 mark)

Show Answers Only

`text(See Worked Solutions)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`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) `SP + S′P = 2a and SS prime = 2ae,\ \ 0 < e < 1`

♦♦ Mean mark 22%. 

`text(This is the condition for an ellipse with foci)`

`text(at)\ \S and S′ text(and eccentricity)\ e.`

♦ Mean mark 40%.

(ii)	`SP`	`= ePM\ \ \ \ \ text{(where}\ M\ text{is on the directrix)}`
		`=e(OM-OQ)`
		`= e (a/e – a cos theta)`
		`= a – a e cos theta`
		`= a(1 – e cos theta)`

♦ Mean mark 39%. 

(iii) `S prime P`	`= 2a – (a – a e cos theta)`
	`= a + a e cos theta`

`sin beta`	`= (QS prime)/(S prime P)`
	`= (ae + a cos theta)/(a + a e cos theta)`
	`= (e + cos theta)/(1 + e cos theta)`

♦♦ Mean mark 43%. 

(iv)

`T sin beta = T ((e + cos theta)/(1 + e cos theta))`

`sin /_ QPS`	`= (QS)/(SP)`
	`= (ae – a cos theta)/(a (1 – e cos theta)`
	`= (e – cos theta)/(1 – e cos theta)`

`text(Resolving the forces vertically)`

`mg`	`= T sin beta – T sin /_ QPS`
`mg`	`= T ((e + cos theta)/(1 + e cos theta) – (e – cos theta)/(1 – e cos theta))`
`mg`	`= T ((e-e^2 cos theta + cos theta – e cos^2 theta-(e + e^2 cos theta – cos theta – e cos^2 theta))/(1 – e^2 cos^2 theta))`
`mg`	`= T((-2e^2 cos theta + 2 cos theta)/(1 – e^2 cos^2 theta))`
`mg`	`= (2T(1 – e^2) cos theta)/(1 – e^2 cos^2 theta)`

(v) `text(Resolving the forces horizontally)`

♦♦♦ Mean mark 8%. 

`mr omega^2= T cos beta + T cos /_ QPS`

`cos beta`	`= r/(S prime P)`
	`= r/(2a – SP)`
	`= r/(2a – (a – a e cos theta))`
	`= r/(a (1 + e cos theta))`

`cos /_ QPS`	`= r/(SP)`
	`= r/(a (1 – e cos theta))`

`r`	`= sqrt (SP^2 – SQ^2)`
	`= sqrt (a^2 (1 – e cos theta)^2 – a^2 (e – cos theta)^2)`
	`= a sqrt (1 – 2e cos theta + e^2 cos ^2 theta – (e^2 – 2 e cos theta + cos^2 theta))`
	`= a sqrt (1 – e^2 – (1 – e^2) cos^2 theta)`
	`= a sqrt ((1 – e^2) (1 – cos^2 theta))`
	`= a sin theta sqrt (1 – e^2)`

`:. mr omega^2`	`= T ((a sin theta sqrt(1 – e^2))/(a(1 + e cos theta)) + (a sin theta sqrt(1 – e^2))/(a(1 – e cos theta)))`
	`= T sin theta sqrt (1 – e^2) ((1 – e cos theta + 1 + e cos theta)/(1 – e^2 cos^2 theta))`
	`= (2T sqrt (1 – e^2) sin theta)/(1 – e^2 cos^2 theta)`

♦ Mean mark 38%.

(vi) `cos theta`	`= (mg (1 – e^2 cos^2 theta))/(2T (1 – e^2))`
`sin theta`	`= (mr omega^2(1 – e^2 cos^2 theta))/(2T sqrt (1 – e^2))`
`tan theta`	`= (mr omega^2(1 – e^2 cos^2 theta))/(2T sqrt(1 – e^2)) xx (2T(1 – e^2))/(mg(1 – e^2 cos^2 theta))`
	`= (r omega^2 sqrt (1 – e^2))/g`

Proof, EXT2 P1 2013 HSC 16a

Find the minimum value of `P(x) = 2x^3 - 15x^2 + 24x + 16`, for `x >= 0.` (2 marks)

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Hence, or otherwise, show that for `x >= 0`,

`(x + 1) (x^2 + (x + 4)^2) >= 25x^2.` (1 mark)

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Hence, or otherwise, show that for `m >= 0` and `n >= 0`,

`(m + n)^2 + (m + n + 4)^2 >= (100mn)/(m + n + 1).` (2 marks)

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

Show Answers Only

`0`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

♦ Mean mark 43%. 
STRATEGY: The best responses carefully considered the domain restrictions, notably `P(0).`

i.	`P(x)`	`= 2x^3 – 15x^2 + 24x + 16,\ \ \ x >= 0`
	`P prime(x)`	`= 6x^2 – 30x +24`
		`= 6 (x^2 – 5x + 4)`
		`= 6 (x – 1) (x – 4)`
	`P″(x)`	` = 12x – 30`

`text(MAX or MIN when)\ \ P prime (x)=0`

`text(i.e. when)\ \ x=1 or 4`

`P″(1) = -6 < 0\ \ \ \ P″(4) = 18 > 0`

`:.text(Minimum turning point of)\ \ P(x)\ \ text(at)\ \ x = 4,`

`P(4) = 128 – 240 + 96 + 16 = 0`

`text(Checking limits:)`

`P(0) = 16`

`text(As)\ \ x->oo,\ \ y->oo`

`:.text(Minimum value of)\ \ P(x)\ \ text(is)\ \ 0\ \ text(when)\ \ x = 4.`

♦ Mean mark 47%. 

ii. `text(LHS)`	`= (x + 1) (x^2 + (x + 4)^2)`
	`= (x + 1) (2x^2 + 8x + 16)`
	`= 2x^3 + 8x^2 + 16x + 2x^2 + 8x + 16`
	`= 2x^3 + 10x^2 + 24x + 16`
	`= (2x^3 – 15x^2 + 24x + 16) + 25x^2`
	`= P(x) + 25x^2`

`text(The minimum value of)\ \ P(x)\ \ text(for)\ \ x>= 0\ \ text(is)\ \ 0,`

`=>P(x) + 25x^2`	`>= 25x^2`
`:.(x + 1) (x^2 + (x + 4)^2)`	`>= 25x^2,\ \ \ text(for)\ x >= 0`

♦♦ Mean mark 19%. 
STRATEGY: The best responses carefully considered the domain restrictions, notably `P(0).`

iii.	`(m + n)^2 + (m + n + 4)^2`
	`= (m + n)^2 + (m + n)^2 + 8(m + n) + 16`
	`= 2 (m + n)^2 + 8 (m + n) + 16`

`text(Let)\ \ x = m + n`

`(m + n)^2 + (m + n + 4)^2 = 2x^2 + 8x + 16`

`text{Using part (ii)}`

`(x + 1) (2x^2 + 8x + 16)`	`>= 25x^2`
`2x^2 + 8x + 16`	`>= (25x^2)/(x + 1),\ \ \ x >= 0`
`(m + n)^2 + (m + n + 4)^2`	`>= (25 (m + n)^2)/(m + n + 1)`

`text(Consider)\ \ (m – n)^2`

`text(S)text(ince)\ \ (m – n)^2`	`>= 0`
`m^2 + n^2`	`>= 2mn`
`(m+n)^2-2mn`	`>= 2mn`
`(m + n)^2`	`>= 4mn`
`:.(m + n)^2 + (m + n + 4)^2`	`>= (100mn)/(m + n + 1)`