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Calculus, EXT2 C1 2011 HSC 8a

For every integer `m >= 0`  let

`I_m = int_0^1 x^m (x^2 - 1)^5\ dx.`
 

Prove that for  `m >= 2`

`I_m = (m - 1)/(m + 11) \ I_(m - 2).`   (3 marks)

Show Answers Only

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

Show Worked Solution

`I_m = int_0^1 x^m(x^2 – 1)^5 dx`

`u` `=x^(m – 1)\ \ \ \ \ \ ` `\ \ \ \ u′` `=(m – 1)x^(m – 2)`
`v′` `=x(x^2 – 1)^5` `v` `=1/12 (x^2 – 1)^6`
♦♦♦ Mean mark 24%.

 

`I_m` `=[x^(m-1) xx 1/12(x^2 – 1)^6]_0^1 – int_0^1 1/12(x^2 – 1)^6 xx (m – 1)x^(m – 2) dx`
  `=[0 – 0] – (m – 1)/12 int_0^1 x^(m – 2) (x^2 – 1)^6 dx`
  `=-(m – 1)/12 int_0^1 x^(m – 2) (x^2 – 1)^5 (x^2 – 1) dx`
  `=-(m – 1)/12 int_0^1 x^m (x^2 – 1)^5 dx + (m – 1)/12 int _0^1 x^(m – 2) (x^2 – 1)^5 dx`
  `=-(m – 1)/12 I_m + (m – 1)/12 I_(m – 2)`

 

`12I_m + mI_m – I_m` `= (m – 1)I_(m – 2)`
`(m + 11)I_m` ` = (m – 1)I_(m-2)`
`:.I_m` `= (m – 1)/(m + 11) I_(m – 2)`

Complex Numbers, EXT2 N2 2011 HSC 6c

On an Argand diagram, sketch the region described by the inequality

    `|\ 1 + 1/z\ | <= 1.`  (2 marks)

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Show Worked Solution

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: Substituting `z=x+iy` immediately was common and caused major algebraic problems.
`|\ 1 + 1/z\ |<= ` `1`
`|\ (z + 1)/z\ |<= ` `1`
`|\ z + 1\ |/|\ z\ |<= ` `1`
`|\ z + 1\ |<= ` `|\ z\ |`
`sqrt ((x + 1)^2 + y^2) <=` `sqrt (x^2 + y^2)`
`(x + 1)^2 + y^2 <=` `x^2 + y^2`
`x^2 + 2x + 1 <=` ` x^2`
`:.x <=` `-1/2`

Graphs, EXT2 2011 HSC 6b

Let `f (x)` be a function with a continuous derivative.

  1. Prove that  `y = (f(x))^3`  has a stationary point at  `x = a`  if  `f(a) = 0`  or  `f prime(a) = 0.`  (2 marks)
  2.  
  3. Without finding  `f″(x)`, explain why  `y = (f(x))^3`  has a horizontal point of inflection at  `x = a`  if  `f(a) = 0`  and  `f prime (a) != 0.`  (1 mark)
  4.  
  5. The diagram shows the graph  `y = f(x).` 
     

    1. HSC 2011 6bi
       
  6. Copy or trace the diagram into your writing booklet. 
  7.  
    On the diagram in your writing booklet, sketch the graph  `y = (f(x))^3`, clearly distinguishing it from the graph  `y = f(x).`  (3 marks)

 

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(See Worked Solutions)`

 

 

 

 

 

 

Show Worked Solution

(i)   `text(If)\ \ f(x)\ \ text(is a function with a continuous derivative,)`

`=> f prime (x)\ \ text(exists for all)\ \  x.`

`y` `= (f(x))^3`
`(dy)/(dx)` `= 3 xx (f(x))^2 xx f prime(x)`

 

`:.\ (dy)/(dx) = 0\ \ text(when)\ \ f(x) = 0 or f prime (x) = 0`

`:.\ x = a\ \ text(is a stationary point if)\ \ f(a) = 0 or f prime(a) = 0`

 

 

♦♦♦ Mean mark part (ii) 0%!
A BEAST!

(ii)    `text(If)\ \ f prime (a) != 0\ \ text(then either)\ \ f prime (a) > 0 or f prime (a) < 0,`

`and f prime (x)\ \ text(keeps that sign either side of)\ \ x = a.`

 

`:. text(If)\ \ f(a) = 0 and f prime(a) != 0, text(there is a stationary point at)\ \ x = a`

`text(where the slope of the curve does not change either side of)\ \ x = a.`

`text(i.e. a horizontal point of inflection occurs at)\ \ x=a`

 

(iii)  `y = (f(x))^3`

`text(When)\ \ f(x) = 0,\ \ (f(x))^3 = 0\ \ text{(horizontal P.I. from part (ii))}`

`text(When)\ \ f(x) = 1,\ \ (f(x))^3 = 1`
 

`text(If)\ \ 0 < f(x) < 1,\ \ 0 < (f(x))^3 < f(x)`

`text(If)\ \ f(x) > 1,\ \ (f(x))^3 > f(x)`
 

`text(When)\ f prime (a) = 0,\ \ (f(x))^3\ \ text(has a maximum turning point)`

`f(x) < 0,\ \ (f(x))^3 < 0`
 

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.)

  1. Explain why Jac’s terminal velocity  `v_T`  is given by  
     
         `sqrt ((mg)/k).`  (1 mark)

  2. 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)

  3. 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)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `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.

  1. Use the reflection property of the ellipse at  `P`  to prove that  `SQ = RQ.`  (2 marks)
  2. Explain why  `S prime R = 2a.`  (1 mark)
  3. Hence, or otherwise, prove that  `Q`  lies on the circle  `x^2 + y^2 = a^2.`  (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(See Worked Solutions)`
  3. `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`

Harder Ext1 Topics, EXT2 2011 HSC 4c

A mass is attached to a spring and moves in a resistive medium. The motion of the mass satisfies the differential equation

`(d^2y)/(dt^2) + 3 (dy)/(dt) + 2y = 0,`

where  `y`  is the displacement of the mass at time  `t.`

  1. Show that, if  `y = f(t)`  and  `y = g(t)`  are both solutions to the differential equation and  `A`  and  `B`  are constants, then
    1. `y = A f (t) + Bg (t)`
  2. is also a solution.  (2 marks)
  5. A solution of the differential equation is given by  `y = e^(kt)`  for some values of  `k`, where  `k`  is a constant.
  6. Show that the only possible values of  `k`  are  `k = -1`  and  `k = -2.`  (2 marks)
  9. A solution of the differential equation is
    1. `y = Ae^(-2t) + Be^-t.`
  10. When  `t = 0`, it is given that  `y = 0`  and  `(dy)/(dt) = 1`.
  11. Find the values of  `A`  and  `B.`  (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `A = -1,\ \ B = 1`
Show Worked Solution

(i)   `(d^2y)/(dt^2) + 3(dy)/(dt) + 2y = 0`

♦♦♦ Mean mark 18%.
`text(Given)\ \ y` `=Af(t) + Bg (t)\ \ text(then)`
`y′` `=Af′(t)+ Bg′(t)`
`y″` `=Af″(t)+Bg″(t)`

 

`text(LHS)` `=d^2/(dt^2) (Af (t) + Bg (t)) + 3 d/(dt) (Af (t) + Bg (t)) + 2 (Af (t) + Bg(t))`
  `=Af″ (t) + Bg″ (t) + 3 Af prime (t) + 3 Bg prime (t) + 2 Af (t) + 2 Bf (t)`
  `=A (f″(t) + 3 f prime (t) + 2 f (t)) + B(g″ (t) + 3g prime (t) + 2 f(t))`
  `=A(0) + B(0)`
  `=0`

`:.y = Af(t) + Bg(t)\ \ text(is a solution of)\ \ (d^2y)/(dt^2) + 3 (dy)/(dt) + 2y = 0`

 

(ii)   `y = e^(kt),\ \ (dy)/(dt) = ke^(kt),\ \ (d^2y)/(dt^2) = k^2 e^(kt)`

`text(Substituting into)`

`(d^2y)/(dt^2) + 3 (dy)/(dt) + 2y` `= 0`
`k^2 e^(kt) + 3 ke^(kt) + 2e^(kt)` `=0`
`e^(kt) (k^2 + 3k + 2)` `=0`
`k^2 + 3k + 2` `=0\ \ \ \ (e^(kt) != 0)`
`(k + 1) (k + 2)` `=0`

 

`:.k = -1 or -2`

 

(iii)   `y` `= Ae^(-2t) + Be^-t`
  `(dy)/(dt)` `= -2 Ae^(-2t) – Be^-t`

 

`text(When)\ \ t = 0,\ \ y = 0`

`=>A + B = 0\ \ \ …\ text{(i)}`

`text(When)\ \ t = 0,\ \ (dy)/(dt) = 1`

`=>-2A – B = 1\ \ \ …\ text{(ii)}`

`text{Add (i) + (ii)}`

`-A = 1 `

`:.A = -1 and B=1`

Conics, EXT2 2011 HSC 3d

The equation  `x^2/16 - y^2/9 = 1`  represents a hyperbola.

  1. Find the eccentricity  `e.`  (1 mark)
  2. Find the coordinates of the foci.  (1 mark)
  3. State the equations of the asymptotes.  (1 mark)
  4. Sketch the hyperbola.  (1 mark)
  5. For the general hyperbola  
  6. `x^2/a^2 - y^2/b^2 = 1`,
  8. describe the effect on the hyperbola as  `e -> oo.`  (1 mark)

 

Show Answers Only
  1. `e = 5/4`
  2. `text(Foci) (+-5, 0)`
  3. `y = +- 3/4 x`
  5. `text(The hyperbola approaches the)\ \ y text(-axis from both sides.)`
Show Worked Solution

(i)   `x^2/16 – y^2/9 = 1,\ \ \ =>a^2 = 16,\ \ \ b^2 = 9`

`b^2` `= a^2 (e^2 – 1)`
`9` `= 16 (e^2 – 1)`
`e^2` `= 1 + 9/16 `
  `= 25/16`
`:. e` `= 5/4`

 

 

(ii)   `S (ae, 0),\ \ S prime (-ae, 0)`

`S(5, 0),\ \ S prime (–5, 0)`

 

(iii)  `y = +- b/a x`

`y = +- 3/4 x`

 

(iv)   HSC 2011 3di

 

♦♦♦ Mean mark part (v) 24%.

 

(v)   `text(As)\ \ e -> oo,\ \ b -> oo\ \ text(and the asymptotes approach the)`

`y text{-axis from both sides (i.e. the gradients of the asymptotes →∞)}`

` text(Thus the hyperbola approaches the)\ \ y text(-axis from both sides.)`

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`.

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

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

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

  4. 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)

Show Answers Only
  1. `text{Proof (See Worked Solutions.)}`
  2. `text{Proof (See Worked Solutions.)}`
  3. `text{Proof (See Worked Solutions.)}`
  4. `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

  1. In how many ways can  `m`  identical yellow discs and  `n`  identical black discs be arranged in a row?  (1 mark)

  2. In how many ways can 10 identical coins be allocated to 4 different boxes?  (1 mark)
Show Answers Only
  1. `((m + n)!)/(m!n!)`

  2. `(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).}`

 

Harder Ext1 Topics, EXT2 2012 HSC 16a Answer1

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

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

Polynomials, EXT2 2012 HSC 15b

Let `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`, where `k` is real.

Let  `α = x + iy`, where  `x`  and  `y`  are real.

Suppose that  `α`  and  `iα`  are zeros of  `P(z)`, where  `bar α ≠ iα`.

  1. Explain why  `bar α`  and  `-i bar α`  are zeros of  `P(z)`.   (1 mark)

  2. Show that  `P(z) = z^2(z − k)^2 + (kz − 1)^2`.   (1 mark)

  3. Hence show that if  `P(z)`  has a real zero then
    1. `P(z) = (z^2 + 1)(z+ 1)^2` or  `P(z) = (z^2 + 1)(z − 1)^2.`   (2 marks)

  4. Show that all zeros of  `P(z)`  have modulus `1`.   (2 marks)
  5. Show that  `k = x − y`.   (1 mark)
  6. Hence show that  `-sqrt2 ≤ k ≤ sqrt2`.   (2 marks) 
Show Answers Only
  1. `text{Proof (See Worked Solutions.)}`
  2. `text{Proof (See Worked Solutions.)}`
  3. `text{Proof (See Worked Solutions.)}`
  4. `text{Proof (See Worked Solutions.)}`
  5. `text{Proof (See Worked Solutions.)}`
  6. `text{Proof (See Worked Solutions.)}`
Show Worked Solution

(i)   `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz +1`

`P(α) = P(iα) = 0, \ \ bar α ≠ iα.`

`text(S)text(ince)\ \ P(z)\ text(has real coefficients,)`

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

`text(i.e.)\ \ P(α) = 0\ text(and)\ P(bar α) = 0`

`bar α` `= x − iy`
`bar(i α)` `=bar (i(x+iy))`
  `=bar (ix-y)`
  `=-y-ix`
  `=-i(x-iy)`
  `=-i barα`

 

`:. -i bar α\ text(is a zero of)\ P(z)\ text(as it is the conjugate of)\ iα.`

 

(ii)   `H(z)` `= z^2(z − k)^2 + (kz − 1)^2`
    `= z^2 (z^2 − 2kz + k^2) + (k^2z^2 − 2kz + 1)`
    `= z^4 − 2kz^3 + k^2z^2 + k^2z^2 − 2kz +1`
    `= z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`
    `= P(z)` 

 

(iii)  `text(If)\ z\ text(is real then)\ z^2(z − k)^2\ text(and)\ (kz −1)^2\ text(are real and)`

`text(either positive or zero.)`

♦♦♦ Mean mark part (iii) 11%.

`:.text(If)\ \ P(z) = z^2(z − k)^2 + (kz − 1)^2 = 0`

`=>(z − k)^2 = 0\ text(and)\ (kz − 1)^2 = 0`

`text(When)\ \ z = k,\ \ \ k^2 − 1 = 0\ \ text(or)\ k = ±1.`

`text(If)\ k = 1`

`P(z)` `= z^2(z − 1)^2 + (z − 1)^2`
  `= (z^2 + 1)(z − 1)^2.`

`text(If)\ k = -1`

`P(z)` `= z^2(z + 1)^2 + (-z − 1)^2`
  `= (z^2 + 1)(z + 1)^2`

  

(iv)  `text(Product of the roots) = e/a=1`

♦♦♦ Mean mark part (iv) 20%.
`:. α *bar α *iα *(-i barα)`  `=1`
`(α bar α)^2` `=1`
`(|α|)^4` `=1`
`|α|` `=1`

`:.\ text(All zeros have modulus 1.)`

 

(v)  `text(Sum of zeroes)\ = -b/a=-(-2k)/1 = 2k`

♦♦♦ Mean mark part (v) 20%.
`:. 2k` `=α + bar α + iα + (-i bar α )`
  `=x + iy + x − iy + (-y + ix) − i(x − iy)`
  `=2x − y + ix − ix − y`
  `=2x − 2y`
`:. k` `=x-y`

 

 

(vi)  `text(S)text(ince)\ \ |α| = 1\ \ \ text{(part (iv))}`

♦♦♦ Mean mark part (vi) 2%.
`=>x^2 + y^2` `=1`
`text(Substitute)\ \ y=x-k\ \ \ text{(part (v))}`
`x^2 + (x − k)^2` `=1`
`2x^2 − 2kx + k^2 − 1` `=0`

 

`text(For a real solution to exist), Δ ≥ 0`

`4k^2 − 8(k^2 − 1) ` `≥ 0`
`-4k^2 + 8` `≥ 0`
`k^2` `≥ 2`

 

`:. -sqrt2 ≤ k ≤ sqrt2`

Proof, EXT2 P1 2012 HSC 15a

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

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

  3. Let  `n`  and  `j`  be positive integers with  `1 ≤ j ≤ n`.
     
    Prove that  `sqrtn ≤ sqrt(j(n − j + 1)) ≤ (n + 1)/2.`  (2 marks)

  4. For integers  `n ≥ 1`, prove that
     
        `(sqrtn)^n ≤ n! ≤ ((n + 1)/2)^n`.   (1 mark)

     
Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
  4. `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`

Volumes, EXT2 2012 HSC 14c

The solid  `ABCD`  is cut from a quarter cylinder of radius  `r`  as shown. Its base is an isosceles triangle  `ABC`  with  `AB = AC`. The length of  `BC`  is  `a`  and the midpoint of  `BC`  is  `X`.

The cross-sections perpendicular to  `AX`  are rectangles. A typical cross-section is shown shaded in the diagram.

Volumes, EXT2 2012 HSC 14c

Find the volume of the solid  `ABCD`.  (4 marks)

Show Answers Only

`(ar^2)/3\ \ text(u³)`

Show Worked Solution

`text(Need to find the width of the rectangular cross-section)`

`text(Using similar triangles)`

Volumes, EXT2 2012 HSC 14c Answer1

`(AF)/(AX)` `= (DF)/(BX)`
`h/r` `= (DF)/(a/2)`
`DF` `= (ah)/(2r)`
`:.DE` `= (ah)/r`

 

`text(Need to find the height of the cross-section)`

Volumes, EXT2 2012 HSC 14c Answer2

`JF` `=sqrt((AJ)^2-(AF)^2)`
  ` = sqrt(r^2 − h^2)`

 

`=>text(Rectangle has base length)\ (ah)/r\ text(and height)\ sqrt(r^2 − h^2)`

`text(Volume of solid)` `=lim_(δh→0) ∑_(h=0)^r (ah)/r  sqrt(r^2 − h^2)\ δh`
  `= int_0^r (ah)/r  sqrt(r^2 − h^2)\ dh`
  `= a/r int_0^r h sqrt(r^2 − h^2)\ dh` 
`text(Let)\ \ u^2` `=r^2-h^2`
`2u\ du` `=-2h\ dh`
`-u\ du` `=h\ dh`

`text(If)\ \ h=r,\ \ u^2=0,\ \ u=0`

`text(If)\ \ h=0,\ \ u^2=r^2,\ \ u=r\ \ (r>0)`

`:.\ text(Volume of solid)` `=a/r int_r^0 u xx -u\ du`
  `=-a/r [u^3/3]_r^0`
  `=-a/r((-r^3)/3)`
  `=(ar^2)/3\ \ text(u³)`

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.`

 

  1. What information indicates that  `P`  lies on an ellipse with foci  `S`  and  `S′`, and with eccentricity  `e`?  (1 mark)

  2. Using the focus–directrix definition of an ellipse, or otherwise, show that
  3. `SP = a(1 − e cos theta ).`  (1 mark)
  5. Show that
  6. `sin beta = (e + cos theta)/(1 + e cos theta).`  (2 marks)
  8. 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)
  9. 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)
  10. Show that
    1. `tan theta = (r omega^2)/g sqrt(1 - e^2).`  (1 mark)

 

 

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
  6. `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

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

  2. Hence, or otherwise, show that for  `x >= 0`,
     
        `(x + 1) (x^2 + (x + 4)^2) >= 25x^2.`  (1 mark)

  3. Hence, or otherwise, show that for  `m >= 0`  and  `n >= 0`,
     
        `(m + n)^2 + (m + n + 4)^2 >= (100mn)/(m + n + 1).`  (2 marks)

Show Answers Only
  1. `0`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `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)`

Mechanics, EXT2 M1 2013 HSC 15d

A ball of mass  `m`  is projected vertically into the air from the ground with initial velocity  `u`. After reaching the maximum height  `H`  it falls back to the ground. While in the air, the ball experiences a resistive force  `kv^2`, where  `v`  is the velocity of the ball and  `k`  is a constant.

The equation of motion when the ball falls can be written as

`m dot v = mg - kv^2.`      (Do NOT prove this.)

  1. Show that the terminal velocity  `v_T` of the ball when it falls is
     
         `sqrt ((mg)/k).`  (1 mark)

     

  2. Show that when the ball goes up, the maximum height  `H`  is
     
         `H = (v_T^2)/(2g) ln (1 + u^2/(v_T^2)).`  (3 marks)

     

  3. When the ball falls from height  `H`  it hits the ground with velocity  `w`.
     
    Show that  `1/w^2 = 1/u^2 + 1/(v_T^2).`  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `m dot v = mg – kv^2`

`t = 0,\ \ \ v = 0,\ \ \ x = 0\ \ \ text(Ball falling)`

`text{For terminal velocity}\ \(v_T),\ \ \  dot v = 0`

`v_T^2` `= (mg)/k`
`:.v_T` `= sqrt ((mg)/k)`

 

ii.  `text(When the ball rises),\ \ m dot v = -mg – kv^2`

♦ Mean mark 43%.

MARKER’S COMMENT: More than half of students incorrectly wrote the equation to solve as `m dot v=mg-kv^2!`
`text(Using)\ \ dot v` `= v (dv)/(dx)`
`mv (dv)/(dx)` `= -mg – kv^2`
`dx` `=(-mv)/(mg + kv^2) dv` 
`int_0^H dx` `= -int_u^0 (mv)/(mg + kv^2) dv`
`[x]_0^H` `= -m/(2k) [log_e (mg + kv^2)]_u^0`
`H` `= -m/(2k) (log_e (mg) – log_e (mg + ku^2))`
  `= m/(2k) log_e ((mg + ku^2)/(mg))`
  `= m/(2k) log_e (1 + (ku^2)/(mg))\ \ \ \ text{(using}\ \ k = (mg)/v_T^2 text{)}`
`:.H` `=(v_T^2)/(2g) log_e (1 + u^2/(v_T^2))`

 

 

iii.  `text(When the ball falls),\ \ m dot v = mg – kv^2`

♦♦ Mean mark 14%.
`mv (dv)/(dx)` `= mg – kv^2`
`dx` `=(mv)/(mg – kv^2) dv` 

 

`int_0^H dx` `= int_0^w (mv)/(mg – kv^2)dv`
`[x]_0^H` ` = – m/(2k)[log_e (mg – kv^2)]_0^w`
`H` `= -m/(2k) (log_e (mg – kw^2) – log_e (mg))`
  `= -m/(2k) log_e ((mg – kw^2)/(mg))`
  `= -m/(2k) log_e (1-(kw^2)/(mg))\ \ \ \ text{(using}\ \ k = (mg)/v_T^2 text{)}`
`H` `=-(v_T^2)/(2g) log_e (1 – w^2/(v_T^2))`
  `=-(v_T^2)/(2g) log_e ((v_T^2 – w^2)/(v_T^2))`
  `=(v_T^2)/(2g) log_e ((v_T^2)/(v_T^2 – w^2))`

 

`text{Using part (ii):}`

`(v_T^2)/(2g) log_e ((v_T^2)/(v_T^2 – w^2))` `=(v_T^2)/(2g) log_e (1 + u^2/(v_T^2))`
`(v_T^2)/(v_T^2 – w^2)` `=(v_T^2 + u^2)/(v_T^2)`
`(v_T^2 + u^2)(v_T^2 – w^2)` `=v_T^4`
`v_T^4-v_T^2 w^2+v_T^2 u^2-w^2 u^2` `=v_T^4`
`v_T^2 w^2+w^2 u^2` `=v_T^2 u^2`
`(v_T^2 w^2)/(v_T^2w^2u^2)+(w^2 u^2)/(v_T^2w^2u^2)` `=(v_T^2 u^2)/(v_T^2w^2u^2)`
`:.1/u^2 + 1/(v_T^2)` `=1/w^2`

Harder Ext1 Topics, EXT2 2013 HSC 14d

A triangle has vertices `A, B` and `C`. The point `D` lies on the interval `AB` such that `AD = 3` and `DB = 5`. The point `E` lies on the interval `AC` such that `AE = 4`, `DE = 3` and `EC = 2`.

  1. Prove that `Delta ABC` and `Delta AED` are similar.  (1 mark)
  2. Prove that `BCED` is a cyclic quadrilateral.  (1 mark)
  3. Show that `CD = sqrt 21`.  (2 marks)
  4. Find the exact value of the radius of the circle passing through the points `B, C, E and D`.  (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `(3 sqrt 105)/10`
Show Worked Solution

(i)  `text(In)\ \ Delta ABC and Delta AED`

`/_ BAC = /_ EAD\ \ \ \ text{(common angle)}`

`(AB)/(AE)` `= 8/4 = 2/1`
`(AC)/(AD)` `= 6/3 = 2/1`
`(AB)/(AE)` `= (AC)/(AD) = 2/1`

 

`:.\ Delta ABC\ \ text(|||)\ \ Delta AED` `\ \ \ \ \ text{(sides about equal angles}`
  `\ \ \ \ \ text{are in the same ratio)}`

 

 

(ii)   `/_ ABC = /_ AED\ \ \ \ \ text{(corresponding angles of similar triangles)}`

`/_ AED\ \ text(is an exterior angle of quadrilateral)\ \ BCED`

`:.\ BCED\ \ text(is a cyclic quadrilateral as an exterior)`

`text(angle equals the interior opposite angle.)`

 

(iii)  `cos /_ AED = (3^2 + 4^2 – 3^2)/(2 xx 3 xx 4) = 2/3`

♦♦ Mean mark 33%.

`cos /_ CED` `= cos(pi – /_AED)`
  `=-cos/_AED`
  `=-2/3`

`text(In)\ \ Delta CDE`

`CD^2` `= 2^2 + 3^2 – 2 xx 2 xx 3 xx (-2/3)`
  `= 13 + 8`
  `=21`
`:.CD` `= sqrt 21`

  

(iv)

`text(Mark the centre)\ \ O,\ \ text(draw the radii)\ \ OC, OD`

`text(Let)\ \ /_ CBD` `= alpha`  
`:. /_ COD` `= 2 alpha` ` \ \ \ \ text{(angles at circumference and}`
    `\ \ \ \ text{centre on arc}\ \ CD text{)}`
`/_ DEC` `= pi – alpha` ` \ \ \ \ text{(opposite angles of cyclic}`
    `\ \ \ \ text{quadrilateral}\ \ BCED text{)}`

 

♦♦♦ Mean mark 7%.

COMMENT: This part had the lowest mean mark in the entire 2013 exam.

`cos alpha = cos /_ AED = 2/3\ \ \ \ text{(part (iii))}`

`text(Let)\ \ r= text(circle radius)`

`text(In)\ \ Delta DOC`

`CD^2` `=r^2 + r^2 – 2r xx r xx cos 2 alpha`
`21` `=2r^2 – 2r^2 (2 cos^2 alpha – 1)`
  `=2r^2 – 2r^2 (8/9 – 1)`
  `=2r^2 + (2r^2)/9`
  `=(20r^2)/9`
`r^2` `=(9 xx 21)/20`
`:.r` `=(3 sqrt 21)/(2 sqrt 5)`
  `=(3 sqrt 105)/10`

Proof, EXT2 P1 2014 HSC 16b

Suppose `n` is a positive integer. 

  1. Show that 
     
        `-x^(2n) ≤ 1/(1 + x^2) − (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2)) ≤ x^(2n)`.  (3 marks)

  2. Use integration to deduce that
     
    `-1/(2n + 1) ≤ pi/4 − (1 − 1/3 + 1/5 − … + (-1)^(n − 1) 1/(2n − 1)) ≤ 1/(2n + 1)`.  (2 marks)

  3. Explain why   `pi/4 = 1 − 1/3 + 1/5 − 1/7 + …`.  (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.   `text(Let)\ \ S_n=1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2)`

`=>text(GP where)\ \ a=1, r=-x^2`

♦♦ Mean mark 12%.
STRATEGY: Applying the GP formula and simplifying the middle term is worth 2 full marks in this question.
`:.S_n` `=(1(1-(-x^2)^n))/(1-(-x^2))`
  `= (1 − (-x^2)^n)/(1 + x^2)`

 

`:.1/(1 + x^2) − (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2))​`

`=1/(1 + x^2)-(1 − (-x^2)^n)/(1 + x^2)`

`=((-x^2)^n)/(1 + x^2)`

`=((-1)^nx^(2n))/(1 + x^2)`

 

`text(S)text(ince)\ \ (1 + x^2)>=1,\ \ \ -x^(2n) ≤ ((-1)^nx^(2n))/(1 + x^2) ≤ x^(2n)`

`:.\ text(We can conclude)`

`-x^(2n) ≤ 1/(1 + x^2) − (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2))\ ≤ x^(2n)` 

 

ii.   `text{Integrating part (i) between 0 and 1}`

♦ Mean mark 43%.
`int_0^1 -x^(2n)\ dx` `=(-1)/(2n + 1)[x^(2n + 1)]_0^1`
  `=(-1)/(2n + 1)`
`int_0^1 1/(1 + x^2)` `=[tan^(-1) x]_0^1`
  `=pi/4`
`int_0^1 (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2))\ dx` 
`=[x − (x^3)/3 + (x^5)/5 −… + ((-1)^(n − 1)x^(2n − 1))/(2n − 1)]_0^1`
`=1 − 1/3 + 1/5 − … + ((-1)^(n − 1))/(2n − 1)`
`int_0^1 x^(2n)\ dx` `=1/(2n + 1)[x^(2n + 1)]_0^1`
  `=1/(2n + 1)`

 
`:.\ text(We can conclude)`

`(-1)/(2n + 1) ≤ pi/4 − (1 − 1/3 + 1/5 − … + (-1)^(n − 1) 1/(2n − 1)) ≤ 1/(2n + 1)`

 

iii.  `(-1)/(2n + 1) ≤ pi/4 − (1 − 1/3 + 1/5 − … + ((-1)^(n − 1))/(2n − 1)) ≤ 1/(2n + 1)`

`text(As)\ n → ∞,`

`=>(-1)/(2n + 1) → 0^-\ \ text(and)\ \ 1/(2n + 1) → 0^+`

`=> pi/4 − (1 − 1/3 + 1/5 − … + 1/(2n − 1)) → 0`

♦♦ Mean mark 28%.

`:. pi/4 = 1 -1/3 + 1/5 − 1/7 + …`

Mechanics, EXT2 2014 HSC 15c

A toy aeroplane `P` of mass `m` is attached to a fixed point `O` by a string of length `l`. The string makes an angle  `ø`  with the horizontal. The aeroplane moves in uniform circular motion with velocity `v` in a circle of radius `r` in a horizontal plane.

Mechanics, EXT2 2014 HSC 15c

The forces acting on the aeroplane are the gravitational force  `mg`, the tension force  `T`  in the string and a vertical lifting force  `kv^2`, where `k` is a positive constant.

  1. By resolving the forces on the aeroplane in the horizontal and the vertical directions, show that
    `(sin\ ø)/(cos^2\ ø) = (lk)/m − (lg)/(v^2)`.  (3 marks)

  3. Part (i) implies that
    `(sin\ ø)/(cos^2\ ø) < (lk)/m`.   (Do NOT prove this.)
  5. Use this to show that
    1. `sin\ ø < (sqrt(m^2 + 4l^2k^2) − m)/(2lk).`  (2 marks)

  6. Show that
    `(sin\ ø)/(cos^2\ ø)` is an increasing function of  `ø`  for  `-pi/2 < ø < pi/2`.  (2 marks)

  8. Explain why `ø` increases as `v` increases.  (1 mark)
Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
  4. `text{Proof (See Worked Solutions)}`
Show Worked Solution

(i)    `text(Vertically)`

 `kv^2 = mg + T\ sin\ ø\ \ \ \ \ …\ (1)`

`text(Horizontally)`

`T cos\ ø` `=(mv^2)/r`
  `=(mv^2)/(l cos\ ø)\ \ \ \ \ \ (r=lcos\ ø)`
`T` `=(mv^2)/(l cos^2\ ø)\ \ \ \ …\ (2)`

 

`text{Substitute (2) into (1)}`

`mg + (mv^2)/(l cos^2\ ø)\ sin\ ø` `=kv^2`
`lmg + (mv^2)/(cos^2\ ø)\ sin\ ø` `=lkv^2`
`(mv^2)/(cos^2\ ø)\ sin\ ø` `=lkv^2 – lmg`
`(sin\ ø)/(cos^2\ ø)` `=(lk)/m − (lg)/(v^2)\ \ \ text(… as required)`

 

♦♦♦ Mean mark 14%.
STRATEGY: The form of the required proof strongly hints that the quadratic formula is relevant – a clue that was ignored by many students. 
(ii)   `(sin\ ø)/(cos^2\ ø)` `< (lk)/m`
   `sin\ ø` `< (lk)/m(1 − sin^2\ ø)`
  `m sin\ ø` `< lk – lksin^2\ ø`
    `lk\ sin^2\ ø + m\ sin\ ø − lk< 0` 

 

`text(Using the quadratic formula)`

`sin\ ø = (-m ± sqrt(m^2 + 4l^2k^2))/(2lk)`

`text(S)text(ince)\ \ 0<ø<pi/2\ \ \ \ =>\ 0 < sin\ ø < 1`

`:. sin\ ø = (sqrt(m^2 + 4l^2k^2)-m)/(2lk)`

 

`=> lk\ sin^2\ ø + m\ sin\ ø − lk< 0\ \ text(is true)` 

`text(when)\ \ sin\ ø = 0`

`=>sqrt(m^2 + 4l^2k^2)>m`

`:.sin\ ø < (sqrt(m^2 + 4l^2k^2) − m)/(2lk)\ \ \ \ text(… as required)`

 

(iii)  `f(x)` `= (sin\ ø)/(cos^2\ ø)`
  `f′(x)` `= (cos\ ø\ cos^2\ ø − sin\ ø xx 2\ cos\ ø(-sin\ ø))/(cos^4\ ø)`
    `= (cos^2\ ø + 2sin^2\ ø)/(cos^3\ ø)`

 

♦♦ Mean mark 25%.

`text(Consider)\ \ -pi/2 < ø < pi/2`

`cos\ ø>0, \ \ cos^3\ ø > 0, \ \ 2sin^2\ ø>=0`

`=> f′(x) > 0`

 

`:. f(x)=(sin\ ø)/(cos^2\ ø)\ text(is an increasing function for)\ \ \ -pi/2 < ø < pi/2.`

 

(iv)  `text(Consider)\ \ (sin\ ø)/(cos^2\ ø) = (lk)/m − (lg)/(v^2)`

♦♦♦ Mean mark 8%.
COMMENT: This question had the lowest mean mark in the entire 2014 exam.
`text(As)\ v\ text(increases)` `=>(lg)/(v^2)\ text(decreases)`
  `=>(lk)/m – (lg)/(v^2)\ text(increases)`
  `=>(sin\ ø)/(cos^2\ ø)\ text(increases.)`

 

`text{From (iii)},\ (sin\ ø)/(cos^2\ ø)\ text(is an increasing function as)\ ø\ text(increases.)`

`:.ø\  text(increases as)\ \ v\ \ text(increases.)`

Complex Numbers, EXT2 N2 2014 HSC 15b

  1. Using de Moivre’s theorem, or otherwise, show that for every positive integer `n`,
     
        `(1 + i)^n + (1 − i)^n = 2(sqrt2)^n\ cos\ (npi)/4`.  (2 marks)

  2. Hence, or otherwise, show that for every positive integer `n` divisible by 4,
     
        `((n),(0)) − ((n),(2)) + ((n),(4)) − ((n),(6)) + … + ((n),(n)) = (-1)^(n/4)(sqrt2)^n`  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution
i. `1+i` `=sqrt2(1/sqrt2 + 1/sqrt2 i)`
     `=sqrt2(cos\ pi/4 + i sin\ pi/4)`
  `(1+i)^n` `=(sqrt2)^n (cos\ (n pi)/4 + i sin\ (n pi)/4)`
     
  `text(Similarly,)`
  `1-i` `=sqrt2(cos (-pi/4) + i sin (-pi/4))`
    `=sqrt2(cos\ pi/4 – i sin\ pi/4)`
  `(1-i)^n` `=(sqrt2)^n (cos\ (n pi)/4 – i sin\ (n pi)/4)`

 

`:.(1 + i)^n + (1 − i)^n`

`= (sqrt2)^n (cos\ (n pi)/4 + i sin\ (n pi)/4) +(sqrt2)^n (cos\ (n pi)/4 – i sin\ (n pi)/4)`
`= (sqrt2)^n(cos\ (npi)/4 + i\ sin\ (npi)/4 + cos\ (npi)/4 − i\ sin\ (npi)/4) `
`= 2(sqrt2)^n\ cos\ (npi)/4\ \ \ text(… as required)`

 

ii.   `(1 + i)^n= ((n), (0)) + i((n),(1)) − ((n),(2)) − i((n), (3)) + ((n),(4)) + … + i^n((n),(n))`

 `(1 − i)^n=((n),(0)) − i((n),(1)) − ((n),(2)) + i((n),(3)) + … + (-1)^ni^n((n),(n))`

`(1 + i)^n + (1 − i)^n`

♦♦ Mean mark 22%.
MARKER’S COMMENT: Few students could link part (i) to solving part (ii).

`= 2((n),(0)) − 2((n),(2)) + 2((n),(4)) + … + 2((n),(n))`

`text(Given)\ n\ text(is divisible by 4, the last term)=+((n),(n))`
 

`text{Using part (i):}`

`2((n),(0)) − 2((n),(2)) + 2((n),(4)) + … + 2((n),(n))` `=2(sqrt2)^n cos\ (npi)/4`
`((n),(0)) − ((n),(2)) + ((n),(4)) + … + ((n),(n))` `=(sqrt2)^n cos\ (npi)/4`

 
`text(If)\ n\ text(is a positive integer divisible by 4 then)\ cos\ (npi)/4 = ±1,`

`text(depending on whether)\ n\ text(is an even or odd multiple of 4.)`
 

`:. ((n),(0)) − ((n),(2)) + ((n),(4)) + … + ((n),(n)) = (-1)^(n/4)(sqrt2)^n`

Proof, EXT2 P1 2014 HSC 15a

Three positive real numbers `a`, `b` and `c` are such that  `a + b + c = 1`  and  `a ≤ b ≤ c`.

By considering the expansion of  `(a + b + c)^2`, or otherwise, show that 

`qquad 5a^2 + 3b^2 +c^2 ≤ 1`.  (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

 

Show Worked Solution

`a + b+ c = 1,\ \ a ≤ b ≤ c`

♦♦ Mean mark 13%.

`(a + b+ c)^2 = 1`

`a^2 + b^2 + c^2 + 2ab+ 2bc + 2ac = 1`

`text(S)text(ince)\ a ≤ b\ \ =>a^2 ≤ ab`

`text(S)text(ince)\ b ≤ c\ \ =>b^2 ≤ bc`

`text(S)text(ince)\ a ≤ c\ \ =>a^2 ≤ ac`
 

`=> a^2 + b^2 + c^2 + 2a^2 + 2b^2 + 2a^2` `≤ 1`
`:.5a^2 + 3b^2 + c^2` `≤ 1\ \ \ text(… as required)`

Mechanics, EXT2* M1 2004 HSC 7a

The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours. A ship is to sail from a wharf to the harbour entrance and then out to sea. On the morning the ship is to sail, high tide at the wharf occurs at 2 am. The water depths at the wharf at high tide and low tide are 10 metres and 4 metres respectively.

  1. Show that the water depth, `y` metres, at the wharf is given by
     
         `y = 7 + 3 cos\ ((4pit)/(25))`, where  `t`  is the number of hours after high tide.  (2 marks)  

  2. An overhead power cable obstructs the ship’s exit from the wharf. The ship can only leave if the water depth at the wharf is 8.5 metres or less.

     

    Show that the earliest possible time that the ship can leave the wharf is 4:05 am.  (2 marks)

  3. At the harbour entrance, the difference between the water level at high tide and low tide is also 6 metres. However, tides at the harbour entrance occur 1 hour earlier than at the wharf. In order for the ship to be able to sail through the shallow harbour entrance, the water level must be at least 2 metres above the low tide level.

     

    The ship takes 20 minutes to sail from the wharf to the harbour entrance and it must be out to sea by 7 am. What is the latest time the ship can leave the wharf?  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `4:28\ text(am)`
Show Worked Solution

i.   `text(Period)`

`(2pi)/n =` ` 12.5`
`12.5n =` ` 2pi`
`25n =` ` 4pi`
`n =` ` (4pi)/25`

 
`text(Amplitude)`

`a` `= 1/2(10 − 4)`
  `= 3`

 
`=>\ text(Motion centres around)\ \ x = 7\ \ text(with)`

`y = 10\ \ text(when)\ \ t= 0`

 

`:.\ text(Water depth is given by)`

`y` `= 7+a cos nt`
  `= 7 + 3 cos\ ((4pit)/(25))\ \ …\ text(as required.)`

  

ii.  

 Calculus in the Physical, EXT1 2004 HSC 7a Answer

`text(Find)\ \ t\ \ text(when)\ \ y = 8.5:`

`8.5` `= 7 + 3\ cos\ ((4pit)/25)`
`3\ cos\ ((4pit)/25)` `= 1.5`
`cos\ ((4pit)/25)` `= 1/2`
`(4pit)/25` `=pi/3`
`t` `=(25pi)/(3 xx 4pi)`
  `=2 1/12`
  `= 2\ text(hrs 5 mins)`

 

`:.\ text(The earliest time the ship can leave)`

`text(is 4:05 am  … as required.)`

 

iii.  `text(2 metres above low tide = 6 m)`

`text(Find)\ t\ text(when)\ y = 6:`

`6 = 7 + 3\ cos\ ((4pit)/(25))`

`3\ cos\ ((4pit)/(25))` `= -1`
`cos\ ((4pit)/(25))` `= -1/3`
`(4pit)/25` `= 1.9106…`
`:.t` `= (25 xx 1.9106…)/(4pi)`
  `= 3.801…`
  `= 3\ text{hr 48 min  (nearest min)}`

 

`:.\ text(At the harbour entrance, water depth)`

`text(of 6 m occurs when)\ \ t = 2\ text(hr 48 min.)`

 

`:.\ text(Given 20 mins sailing time, the latest the ship can)`

`text(leave the wharf is at)\ \ t = 2\ text(hr 28 min, or 4:28 am.)`

Mechanics, EXT2* M1 2007 HSC 7b

A small paintball is fired from the origin with initial velocity `14` metres per second towards an eight-metre high barrier. The origin is at ground level, `10` metres from the base of the barrier.

The equations of motion are

`x = 14t\ cos\ theta`

`y = 14t\ sin\ theta – 4.9t^2`

where  `theta`  is the angle to the horizontal at which the paintball is fired and  `t`  is the time in seconds. (Do NOT prove these equations of motion)
 

  1. Show that the equation of trajectory of the paintball is
     
         `y = mx − ((1 + m^2)/40)x^2`, where  `m = tan\ theta`.  (2 mark)

  2. Show that the paintball hits the barrier at height  `h`  metres when
     
         `m = 2 ± sqrt(3 − 0.4h)`.

     

    Hence determine the maximum value of  `h`.  (2 marks)

  3. There is a large hole in the barrier. The bottom of the hole is `3.9` metres above the ground and the top of the hole is `5.9` metres above the ground. The paintball passes through the hole if  `m`  is in one of two intervals. One interval is  `2.8 ≤ m ≤ 3.2`.

     

    Find the other interval.  (2 marks)

  4. Show that, if the paintball passes through the hole, the range is 
     
         `(40m)/(1 + m^2)\ \ text(metres.)`
     
    Hence find the widths of the two intervals in which the paintball can land at ground level on the other side of the barrier.  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `0.8 ≤ m ≤ 1.2`
  4. `text{1.3 m  (to 1 d.p.)  and  0.5 m  (to 1 d.p.)}`
Show Worked Solution
i.    `x` `= 14t\ cos\ theta` `\ \ …\ (1)`
  `y` `= 14t\ sin\ theta-4.9t^2` `\ \ …\ (2)`

 

`text(Substitute)\ \ t = x/(14\ cos\ theta)\ \ text{from (1) into (2)}`

`y` `= 14(x/(14\ cos\ theta))\ sin\ theta − 4.9(x/(14\ cos\ theta))^2`
  `= x\ tan\ theta − (4.9/(14^2))((x^2)/(cos^2\ theta))`
  `= x\ tan\ theta − (x^2)/40\ sec^2\ theta`
  `= x\ tan\ theta − (x^2)/40(1 + tan^2\ theta)`
  `= mx − ((1 + m^2)/40)x^2\ \ \ \ \ (text(Given)\ \ m = tan\ theta)`

 

ii.  `text(Show paintball hits at)\ \ h\ \ text(when)`

`m = 2 ± sqrt(3 − 0.4h)`

`text(i.e.)\ \ y = h\ \ text(when)\ \ x = 10`

`10m − ((1 + m^2)/40) · 10^2` `= h`
`10m − 5/2(1 + m^2)` `= h`
`20m − 5 − 5m^2` `= 2h`
`5m^2 − 20m + 2h + 5` `= 0`

 

`text(Using the quadratic formula)`

`m` `=(20 ± sqrt((-20)^2 − 4 · 5 · (2h + 5)))/(2 · 5)`
  `= (20 ± sqrt(400 − 40h −100))/10`
  `= (20 ± sqrt(300 − 40h))/10`
  `= (20 ± 10sqrt(3 − 0.4h))/10`
  `= 2 ± sqrt(3 − 0.4h)\ \ \ text(… as required)`

 

`text(Find maximum)\ \ h`

`sqrt(3 − 0.4h)` `≥ 0`
`3 − 0.4h` `≥ 0`
`0.4h` `≤ 3`
`h` `≤ 7.5`

 

`:.\ text(Maximum)\ \ h = 7.5\ text(m)`

 

iii.   EXT1 2007 7bi

`text{Using part (ii)}`

`text(When)\ \ h = 3.9`

`m` `= 2 ± sqrt(3 − 0.4(3.9))`
  `= 2 ± sqrt(1.44)`
  `= 2 ± 1.2`
  `= 3.2\ \ text(or)\ \ 0.8`

 

`text(When)\ \ h = 5.9`

`m` `= 2 ± sqrt(3 − 0.4(5.9))`
  `= 2 ± sqrt(0.64)`
  `= 2 ± 0.8`
  `= 2.8\ \ text(or)\ \ 1.2`

 

`:.\ text(The other interval is)\ \ \ 0.8 ≤ m ≤ 1.2`

 

iv.  `text(Find)\ \ x\ \ text(when)\ \ y = 0`

`mx − ((1 + m^2)/40)x^2` `= 0`
`x[m − ((1 + m^2)/40)x]` `= 0`
`((1 + m^2)/40)x` `= m,\ \  \ \ x ≠ 0`
`:. x` `= (40m)/(1 + m^2)\ \ …\ text(as required)`

 

`text(Consider the interval)\ \ \ 2.8 ≤ m ≤ 3.2`

`text(When)\ \ m = 2.8`

`=> x = (40(2.8))/(1 + 2.8^2) = 12.669…\ text(m)`

`text(When)\ \ m = 3.2`

`=>x = (40(3.2))/(1 + 3.2^2) = 11.387…\ text(m)`

 

`:.\ text(Landing width interval)`

`= 12.669… − 11.387…`

`= 1.281…`

`= 1.3\ text(m)\ \ text{to 1 d.p.}`

 

`text(Consider the interval)\ \ \ 0.8 ≤ m ≤ 1.2`

`text(When)\ \ m = 0.8`

`=>x = (40(0.8))/(1 + 0.8^2) = 19.512…\ text(m)`

`text(When)\ \ m = 1.2`

`=>x = (40(1.2))/(1 + 1.2^2) = 19.672…`

 

`text(S)text(ince interval includes)\ \ m = 1\ \ text(where the)`

`text(paintball has maximum range.)`

`x_(text(max)) = (40(1))/(1 + 1^2) = 20\ text(m)`

 

`:.\ text(Landing width interval)`

`= 20 − 19.512…`

`= 0.487…`

`= 0.5\ text(m)\ \ \ text{(to 1 d.p.)}`

L&E, EXT1 2007 HSC 7a

2007 7a

The graphs of the functions  `y = kx^n`  and  `y = log_e x`  have a common tangent at  `x = a`, as shown in the diagram.

  1. By considering gradients, show that 
    1. `a^n = 1/(nk)`. (1 mark)
  2.  
  3. Express  `k`  as a function of  `n`  by eliminating `a`.   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `1/(en)`
Show Worked Solution
(i)    `y_1` `= kx^n`
  `(dy_1)/(dx)` `= nkx^(n − 1)`

 

`text(At)\ \ x = a,\ \ (dy_1)/(dx) = nka^(n − 1)`

 

`y_2` `= log_e x`
`(dy_2)/(dx)` `= 1/x`

 

`text(At)\ \ x = a,\ \ (dy_2)/dx = 1/a`

 

`text(S)text(ince tangents have the same gradient at)\ \ x=a,` 

`:. nka^(n − 1)` `= 1/a`
`a^n nk` `= 1`
`a^n` `= 1/(nk)\ \ …\ text(as required)`

 

(ii)  `text(At)\ \ x = a`

`y_1` `= ka^n`
`y_2` `= log_e a`

 

`text(Given a common tangent)`

♦♦♦ “Vast majority” could not eliminate `a` in part (ii).
`ka^n` `= log_e a`
`k(1/(nk))` `= log_e a\ \ \ text{(from (i))}`
`1/n` `= log_e a`
`:.a` `= e^(1/n)`

 

`text(Substitute)\ \ a = e^(1/n)\ \ text(into)\ \ a^n = 1/(nk)`

`(e^(1/n))^n` `= 1/(nk)`
`e` `= 1/(nk)`
`:.k` `= 1/(en)`

Functions, EXT1 F1 2007 HSC 6b

Consider the function  `f(x) = e^x − e^(-x)`.

  1. Show that  `f(x)`  is increasing for all values of `x`.  (1 mark)

  2. Show that the inverse function is given by
     
    `qquad qquad f^(-1)(x) = log_e((x + sqrt(x^2 + 4))/2)` (3 marks)

  3. Hence, or otherwise, solve  `e^x - e^(-x) = 5`. Give your answer correct to two decimal places.  (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions.)}`
  2. `text{Proof (See Worked Solutions.)}`
  3. `1.65\ \ text{(to 2 d.p.)}`
Show Worked Solution
i.    `f(x)` `= e^x − e^(-x)`
  `f′(x)` `= e^x + e^(-x)`

 

`text(S)text(ince)\ \ ` `e^x` `> 0\ \ text(for all)\ x`
  `e^(-x)` `> 0\ \ text(for all)\ x`
  `f′(x)` `> 0\ \ text(for all)\ x`

 

`:.f(x)\ \ text(is an increasing function for all)\ x.`

 

ii.  `y = e^x − e^(-x)`

`text(Inverse function)`

`x` `= e^y − 1/(e^y)`
`xe^y` `= e^(2y) − 1`
`e^(2y) − xe^y − 1` `= 0`

 

`text(Let)\ \ A = e^y`

`:.A^2 − xA − 1 = 0`

 

`text(Using the quadratic formula)`

`A` `=(x ± sqrt((-x)^2 − 4 · 1 · (-1)))/(2 · 1)`
  `=(x ± sqrt(x^2 + 4))/2`

 

`text(S)text(ince)\ \ (x – sqrt(x^2 + 4))/2<0\ \ text(and)\ \ e^y>0,`

`:.e^y` `= (x + sqrt(x^2 + 4))/2`
`log_e e^y` ` = log_e((x + sqrt(x^2 + 4))/2)`
`y` `= log_e((x + sqrt(x^2 + 4))/2)`
`:.f^(-1)(x)` `= log_e((x + sqrt(x^2 + 4))/2)\ \  …\ text(as required)`

  

iii.   `e^x − e^(-x)` `= 5`
  `f(x)` `= 5`
  `f^(-1)(5)` `= x`

 

`f^(-1)(5)` `= log_e((5 + sqrt(5^2 + 4))/2)`
  `= log_e((5 + sqrt29)/2)`
  `= 1.647…`
  `= 1.65\ \ text{(to 2 d.p.)}`

Trig Calculus, EXT1 2006 HSC 7

A gutter is to be formed by bending a long rectangular metal strip of width `w` so that the cross-section is an arc of a circle.

Let `r` be the radius of the arc and `2 theta` the angle at the centre, `O`, so that the cross-sectional area, `A`, of the gutter is the area of the shaded region in the diagram on the right.

  1. Show that, when  `0 < theta <= pi/2`, the cross-sectional area is
    2. `A = r^2 (theta - sin theta cos theta).`  (2 marks)

  2. The formula in part (i) for  `A`  is true for  `0 < theta < pi.`      (Do NOT prove this.)
  3. By first expressing  `r`  in terms of  `w`  and  `theta`, and then differentiating, show that
    2. `(dA)/(d theta) = (w^2 cos theta (sin theta - theta cos theta))/(2 theta^3).`
  4. for  `0 < theta < pi.`  (3 marks)

  5. Let  `g(theta) = sin theta - theta cos theta.`
  6. By considering  `g prime(theta)`, show that  `g(theta) > 0`  for   `0 < theta < pi.`  (3 marks)

  8. Show that there is exactly one value of  `theta`  in the interval  `0 < theta < pi` for which
    1. `(dA)/(d theta) = 0.`  (2 marks)

  9. Show that the value of  `theta`  for which  `(dA)/(d theta) = 0`  gives the maximum cross-sectional area. Find this area in terms of `w.`  (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `w^2/(2 pi)\ \ text(u²)`
Show Worked Solution

(i)   `text(Show)\ \ A = r^2(theta – sin theta cos theta)`

`text(Area of segment)\ \ OBC`

`= (2 theta)/(2 pi) xx pi r^2`

`= r^2 theta`

`text(Area of)\ \ Delta OBC` `= 1/2 ab sin C`
  `= 1/2 r * r * sin 2 theta`
  `= 1/2 r^2 * 2 sin theta cos theta`
  `= r^2 sin theta cos theta`

 

`:.\ text(Shaded Area (A))`

`= text(Area of segment)\ \ OBC – text(Area of)\ \ Delta OBC`

`= r^2 theta – r^2 sin theta cos theta`

`= r^2 (theta – sin theta cos theta)\ \ text(…  as required.)`

 

(ii)   `text(Consider Arc length)\ \ BC`

`w` `= (2 theta)/(2 pi) xx 2 pi r`
  `= 2 theta r`
`:.\ r` `= w/(2 theta)`

 

`:. A` `= (w/(2 theta))^2 (theta – sin theta cos theta)`
  `= w^2/(4 theta) – (w^2 sin theta cos theta)/(4 theta^2)`

 

`:. (dA)/(d theta)` `= (-w^2)/(4 theta^2) – (w^2/4) [((sin theta xx -sin theta + cos theta * cos theta) theta^2 – 2 theta sin theta cos theta)/theta^4]`
  `= (-w^2)/(4 theta^2) – (w^2/(4 theta^4))[(cos^2 theta – sin^2 theta) theta^2 – 2 theta sin theta cos theta]`
  `= (-w^2)/(4 theta^2) – (w^2/(4 theta^3))[(2 cos^2 theta – 1) theta – 2 sin theta cos theta]`
  `= (-w^2)/(4 theta^2) – (w^2/(4 theta^3))[(2 cos^2 theta – 1) theta – 2 sin theta cos theta]`
  `= w^2/(4 theta^3) [-theta – 2 theta cos^2 theta + theta + 2 sin theta cos theta]`
  `= w^2/(4 theta^3) [2 sin theta cos theta – 2 theta cos^2 theta]`
  `= w^2/(2 theta^3) (sin theta cos theta – theta cos^2 theta)`
  `= (w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3)\ \ text(… as required)`

 

(iii)   `g(theta)`  `= sin theta- theta cos theta`
  `g prime (theta)` `= cos theta – (theta xx -sin theta + cos theta * 1)`
    `= cos theta + theta sin theta – cos theta`
    `= theta sin theta`

 

`text(S)text(ince)\ \ 0 < theta < pi,`

`=> sin theta > 0`

`:.\ g prime (theta) > 0`

`g(0) = sin 0 – 0 * cos 0 = 0`

 

`:.\ text(S)text(ince)\ \ g(0) = 0 and g(theta)\ \ text(is an increasing function)`

`(g prime (theta) > 0)\ \ text(for)\ \ 0 < theta < pi , text(then)\ \ g(theta) > 0.`

 

(iv)   `text(If)\ \ (dA)/(d theta) = 0\ ,\ \ \ 0 < theta < pi`

`(w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3) = 0`

`text(Consider)`

`(w^2 cos theta)/(2 theta^3) = 0\ ,\ \ theta != 0`

`=>theta = pi/2`

`text(Consider)`

`sin theta – theta cos theta=` ` 0`
`text(i.e.)\ \ g(theta)=` ` 0`

 

`=>\ text(No solution for)\ \ 0 < theta < pi\ \ text{(using part (iii))}`

 `:.\ text(There is only one value of)\ \ theta.`

 

(v)   `(dA)/(d theta)` `= (w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3)`
  `= (w^2 cos theta * g(theta))/(2 theta^3)`

 

`text(If)\ \ theta = pi/4\ ,\ \ g(pi/4) > 0\ ,\ \ cos (pi/4) > 0`

`:.\ (dA)/(d theta) > 0`

`text(If)\ \ theta = (3 pi)/4\ ,\ \ g((3 pi)/4) > 0\ ,\ \ cos ((3 pi)/4) < 0`

`:.\ (dA)/(d theta) < 0`

`:.\ text(Maximum when)\ \ theta = pi/2`

 

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

`A` `= r^2 (theta – sin theta cos theta)`
  `= (w/(2 theta))^2 (theta – sin theta cos theta)`
  `= w^2/(2^2 xx (pi/2)^2) (pi/2 – sin\ pi/2 cos\ pi/2)`
  `= w^2/pi^2 (pi/2)`
  `= w^2/(2 pi)\ \ text(u²)`

Mechanics, EXT2* M1 2006 HSC 6a

Two particles are fired simultaneously from the ground at time  `t = 0.`

Particle 1 is projected from the origin at an angle  `theta, \ \ 0 < theta < pi/2`, with an initial velocity  `V.`

Particle 2 is projected vertically upward from the point  `A`, at a distance  `a`  to the right of the origin, also with an initial velocity of  `V.`
 


 

It can be shown that while both particles are in flight, Particle 1 has equations of motion:

`x = Vt cos theta`

`y = Vt sin theta -1/2 g t^2,`

and Particle `2` has equations of motion:

`x = a`

`y = Vt -1/2 g t^2.`   Do NOT prove these equations of motion.

Let  `L`  be the distance between the particles at time  `t.`

  1. Show that, while both particles are in flight,
     
         `L^2 = 2V^2t^2 (1 - sin theta) - 2aVt cos theta + a^2.`  (2 marks)

  2. An observer notices that the distance between the particles in flight first decreases, then increases.

     

    Show that the distance between the particles in flight is smallest when
     
         `t = (a cos theta)/(2V(1 - sin theta))`  and that this smallest distance is  `a sqrt ((1 - sin theta)/2).`  (3 marks)

  3. Show that the smallest distance between the two particles in flight occurs while Particle 1 is ascending if  
     
         `V > sqrt((a g cos theta)/(2 sin theta \ (1 - sin theta))).`  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.   

`text(Show)\ \ L^2 = 2V^2t^2 (1 – sin theta) – 2aVt cos theta + a^2`

`text(Consider)\ \ P_1`

`x_1 = Vt cos theta`

`y_1 = Vt sin theta – 1/2 g t^2`

`text(Consider)\ \ P_2`

`x_2 = a`

`y_2 = Vt -1/2 g t^2`

`text(Let)\ \ d=\ text(Vertical distance between particles)`

`d= Vt -1/2 g t^2 – (Vt sin theta – 1/2 g t^2)`

`d= Vt (1 – sin theta)`

 

`text(Using Pythagoras:)`

`L^2` `= (a – x_1)^2 + d^2`
  `= (a – Vt cos theta)^2 + V^2t^2 (1 – sin theta)^2`
  `= a^2 – 2aVt cos theta + V^2t^2 cos ^2 theta + V^2t^2 (1 – 2 sin theta + sin^2 theta)`
  `= a^2 – 2aVt cos theta + V^2 t^2 (cos^2 theta + sin^2 theta + 1 – 2 sin theta)`
  `= a^2 – 2aVt cos theta + V^2 t^2 (2 – 2 sin theta)`
  `= 2 V^2 t^2 (1 – sin theta) – 2aVt cos theta + a^2\ \ text(…  as required.)`

 

ii.   `L^2 = 2V^2 t^2 (1 – sin theta) – 2a Vt cos theta + a^2`

`(d(L^2))/(dt) = 4 V^2 t\ (1 – sin theta) – 2aV cos theta`

`text(Max or min when)\ \ (d(L^2))/(dt) = 0`

`4V^2t\ (1 – sin theta)` `= 2aV cos theta`
 `t` `= (2a V cos theta)/(4V^2 (1 – sin theta))`
  `= (a cos theta)/(2V(1 – sin theta)`

 

`(d^2(L^2))/(dt^2) = 4V^2(1 – sin theta) > 0\ \ text(for)\ \ V > 0,\ \ 0 < theta < pi/2`

`:.\ L^2\ \ text(is a minimum)`

`:.\ L\ \ text(is a minimum when)\ \ t = (a cos theta)/(2V (1 – sin theta)`

 

`text(Show minimum distance is)\ \ a sqrt {(1 – sin theta)/2}`

`text(When)\ \ t = (a cos theta)/(2V(1 – sin theta))`

`L^2` `= 2V^2 ((a^2 cos ^2 theta)/(4V^2 (1 – sin theta)^2)) (1 – sin theta)`
  `\ \ – 2aV ((a cos theta)/(2V (1 – sin theta))) cos theta + a^2`
  `= (a^2 cos^2 theta)/(2 (1 – sin theta)) – (a^2 cos^2 theta)/((1 – sin theta)) + a^2`
  `= a^2[(cos^2 theta – 2 cos^2 theta + 2 (1 – sin theta))/(2(1 – sin theta))]`
  `= a^2[(-cos^2 theta + 2 – 2 sin theta)/(2(1 – sin theta))]`
  `= a^2[(-(1 – sin^2 theta) + 2 – 2 sin theta)/(2 (1 – sin theta))]`
  `= a^2 [(sin^2 theta – 2 sin theta + 1)/(2(1 – sin theta))]`
  `= a^2 [(1 – sin theta)^2/(2 (1 – sin theta))]`
  `= a^2 [((1 – sin theta))/2]`
`:.\ L` `= sqrt ((a^2(1 – sin theta))/2)`
  `= a sqrt ((1 – sin theta)/2)\ \ text(…  as required.)`

 

iii.   `text(Smallest distance occurs when)`

`t = (a cos theta)/(2V (1 – sin theta)`

`text(If)\ \ P_1\ \ text(is ascending,)\ \ dot y_1 > 0`

`y_1 = Vt sin theta – 1/2 g t^2`

`dot y_1 = V sin theta – g t`

`:.\ V sin theta – g ((a cos theta)/(2V (1 – sin theta)))` `> 0`
`2V^2 sin theta\ (1 – sin theta) – a g cos theta` `> 0`
`2V^2 sin theta\ (1 – sin theta)` `> ag cos theta`
`V^2` `> (a g cos theta)/(2 sin theta\ (1 – sin theta))`
`V` `> sqrt ((a g cos theta)/(2 sin theta\ (1 – sin theta)))\ \ text(…  as required.)`

Mechanics, EXT2* M1 2004 HSC 6b

A fire hose is at ground level on a horizontal plane. Water is projected from the hose. The angle of projection, `theta`, is allowed to vary. The speed of the water as it leaves the hose, `v` metres per second, remains constant. You may assume that if the origin is taken to be the point of projection, the path of the water is given by the parametric equations

`x = vt\ cos\ theta`

`y = vt\ sin\ theta − 1/2 g t^2`

where  `g\ text(ms)^(−2)`  is the acceleration due to gravity.  (Do NOT prove this.)

  1. Show that the water returns to ground level at a distance`(v^2\ sin\ 2theta)/g`  metres from the point of projection.   (2 marks)

This fire hose is now aimed at a 20 metre high thin wall from a point of projection at ground level 40 metres from the base of the wall. It is known that when the angle  `theta`  is 15°, the water just reaches the base of the wall.  
 

Calculus in the Physical World, EXT1 2004 HSC 6b
 

  1. Show that  `v^2 = 80g`.  (1 mark)

  2. Show that the cartesian equation of the path of the water is given by
     
         `y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`.  (2 marks)

  3. Show that the water just clears the top of the wall if
     
         `tan^2\ theta − 4\ tan\ theta + 3 = 0`.  (2 marks)

  4. Find all values of  `theta`  for which the water hits the front of the wall.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(See Worked Solutions)`
  5. `15^@ ≤ theta ≤ 45^@ \ \ text(and)\ \ 71.6^@ ≤ theta ≤ 75^@`
Show Worked Solution
i.    `x` `= vt\ cos\ theta`
  `y` `= vt\ sin\ theta − 1/2 g t^2`

 

`text(Find)\ \ t\ \ text(when)\ \ y = 0`

`vt\ sin\ theta − 1/2 g t^2` `= 0`
`t(v\ sin\ theta − 1/2 g t)` `= 0`
`v\ sin\ theta − 1/2 g t` `= 0, \ \ t ≠ 0`
`1/2 g t` `= v\ sin\ theta`
`t` `= (2v\ sin\ theta)/g`

 

`text(Find)\ \ x\ \ text(when)\ \ t = (2v\ sin\ theta)/g`

`x` `= v · (2v\ sin\ theta)/g\ cos\ theta`
  `= (v^2 · \ 2\ sin\ theta\ cos\ theta)/g`
  `= (v^2\ sin\ 2theta)/g`

 

`:.\ text(When)\ \ x = (v^2\ sin\ 2theta)/g,\ \ \text(the water returns)`

`text(to ground level  … as required.)`

 

ii.   `text(Show)\ \ v^2 = 80\ text(g)`

`text(When)\ \ theta = 15^@, \ x = 40`

`text{From part (i)}`

`40` `= (v^2\ sin\ 30^@)/g`
`v^2 xx 1/2` `= 40g`
`v^2` `= 80g\ \ \ …text(as required)`

 

iii.  `text(Show)\ \ y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`

`x` `= vt\ cos\ theta`
`:.t` `= x/(v\ cos\ theta)`

 

`text(Substitute into)`

`y` `= vt\ sin\ theta − 1/2 g t^2`
  `= v · x/(v\ cos\ theta) · sin\ theta −1/2 g  (x/(v\ cos\ theta))^2`
  `= x\ tan\ theta − 1/2  g (x^2/(v^2\ cos^2\ theta))`
  `= x\ tan\ theta − 1/2  g · x^2/(80g\ cos^2\ theta)`
  `= x\ tan\ theta − (x^2\ sec^2\ theta)/160\ \ \ …text(as required.)`

 

iv.   `text(Water clears the top if)\ \ y = 20\ \ text(when)\ \ x = 40`

`text{Substitute into equation from (iii)}`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160` `= 20`
`40\ tan\ theta − 10\ sec^2\ theta` `= 20`
`40\ tan\ theta − 10(1 + tan^2\ theta)` `= 20`
`40\ tan\ theta − 10 − 10\ tan^2\ theta` `= 20`
`10\ tan^2\ theta − 40\ tan\ theta\ + 30` `= 0`
`tan^2\ theta − 4\ tan\ theta + 3` `= 0\ \ \ text(… as required)`

 

v.   

Calculus in the Physical World, EXT1 2004 HSC 6b Answer

`text(Water hits the bottom of the wall when)`

`x = 40\ \ text(and)\ \ y = 0`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160` `= 0`
`40\ tan\ theta − 10\ sec^2\ theta` `= 0`
`4\ tan\ theta − (1 + tan^2\ theta)` `= 0`
`tan^2\ theta − 4\ tan\ theta + 1` `= 0`

 

`text(Using the quadratic formula)`

`tan\ theta` `= (+4 ± sqrt(16 − 4 · 1 · 1))/ (2 xx 1)`
  `= (4 ± sqrt12)/2`
  `= 2 ± sqrt3`
`theta` `= 15^@\ \ text(or)\ \ 75^@`

 

`text(Water hits the top of the wall when)`

`x = 40\ text(and)\ \ y = 20`

`tan^2\ theta − 4\ tan\ theta + 3` `= 0\ \ \ text{from (iv)}`
`(tan\ theta − 1)(tan\ theta − 3)` `= 0`
`tan\ theta` `= 1` `text(or)` `tan\ theta` `= 3`
`theta` `= 45^@`   `theta` `= tan^(−1)\ 3`
        `= 71.565…`
        `= 71.6^@\ \ \text{(to 1 d.p.)}`

 

`:.\ text(Following the motion of the water as)\ \ theta\ \ text(increases,)`

`text(water hits the wall when)`

`15^@ ≤ theta ≤ 45^@` `\ text(and)`
`71.6^@ ≤ theta ≤ 75^@`  

Mechanics, EXT2* M1 2006 HSC 4b

A particle is undergoing simple harmonic motion on the `x`-axis about the origin. It is initially at its extreme positive position. The amplitude of the motion is 18 and the particle returns to its initial position every 5 seconds.

  1. Write down an equation for the position of the particle at time  `t`  seconds.  (2 marks)

  2. How long does the particle take to move from a rest position to the point halfway between that rest position and the equilibrium position?  (3 marks)

Show Answers Only
  1. `x = 18 cos ((2 pi)/5 t)`
  2. `5/6\ \ text(seconds)`
Show Worked Solution

i.   `text{Amplitude (A)} = 18`

♦♦♦ “Very poorly” answered (exact data unavailable).

`text(Period) = (2 pi)/n = 5`

`5n` `= 2 pi`
`n` `= (2 pi)/5`

 

`text(Using)\ \ x` `= A cos n t`
`x` `= 18 cos ((2 pi)/5 t)`

 

ii.   `text(When)\ \ t= 0,\ \ \ x = 18`

`text(Find)\ \ t\ \ text(when)\ \ x = 9`

`9` `= 18 cos ((2 pi)/5 t)`
`cos ((2 pi)/5 t)` `= 1/2`
`(2 pi)/5 t` `= pi/3`
`t` `= (5 pi)/(3 xx 2 pi)`
  `= 5/6\ \ text(seconds)`

 

`:.\ text(It takes the particle)\ \  5/6\ \ text(seconds to move from)`

`text(rest position and half way to equilibrium.)`

Plane Geometry, EXT1 2006 HSC 3d

The points  `P, Q`  and  `T`  lie on a circle. The line  `MN`  is tangent to the circle at  `T`  with  `M`  chosen so that  `QM`  is perpendicular to  `MN`. The point  `K`  on  `PQ`  is chosen so that  `TK`  is perpendicular to  `PQ`  as shown in the diagram.

  1. Show that  `QKTM`  is a cyclic quadrilateral.  (1 mark)
  2. Show that  `/_KMT = /_KQT.`  (1 mark)
  3. Hence, or otherwise, show that  `MK`  is parallel to  `TP.`  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   

`/_ QMT = 90^@\ \ \ (QM _|_ MN)`

`/_ QKT = 90^@\ \ \ (PQ _|_ TK)`

`:.\ /_ QMT + /_ QKT = 180^@`

 

`:.\ QKTM\ \ text(is cyclic)\ \ text{(opposite angles are supplementary)}`

`text(…  as required.)`

 

(ii)  `text(Show)\ \ /_ KMT = /_ KQT`

`/_ KMQ = /_ KTQ = theta`

`text{(angles in the same segment on arc}\ \ KQ text{)}`

 

`/_ KQT` `= 90 – theta\ \ text{(angle sum of}\ \ Delta KQT text{)}`
`/_ KMT` `= /_ QMT – /_ KMQ`
  `= 90 – theta`

 

`:.\ /_ KMT = /_ KQT\ \ text(…  as required.)`

 

(iii)   `text(Show)\ \ MK\ text(||)\ TP`

`/_ NTP` `= /_ KQT\ \ text{(angle in alternate segment)`
  `= 90 – theta`

 

`:.\ /_ NTP = /_ KMT\ \ text{(from part (ii))}`

`:.\ MK\ text(||)\ TP\ \ text{(corresponding angles are equal)}`

Mechanics, EXT2* M1 2005 6b

An experimental rocket is at a height of  5000 m, ascending with a velocity of  ` 200 sqrt 2\ text(m s)^-1`  at an angle of  45°  to the horizontal, when its engine stops.
 

 
After this time, the equations of motion of the rocket are:

`x = 200t`

`y = -4.9t^2 + 200t + 5000,`

where `t` is measured in seconds after the engine stops. (Do NOT show this.)

  1. What is the maximum height the rocket will reach, and when will it reach this height?  (2 marks)

  2. The pilot can only operate the ejection seat when the rocket is descending at an angle between 45° and 60° to the horizontal. What are the earliest and latest times that the pilot can operate the ejection seat?  (3 marks)

  3. For the parachute to open safely, the pilot must eject when the speed of the rocket is no more than  `350\ text(m s)^-1`. What is the latest time at which the pilot can eject safely?  (2 marks)

Show Answers Only
  1. `7041\ text(metres)`

     

    `20.4\ text(seconds.)`

  2. `40.8\ text(seconds)\ ;\ 55.8\ text(seconds)`
  3. `49.7\ text(seconds)`
Show Worked Solution

i.

    

`y = -4.9t^2 + 200t + 5000`

`dot y = -9.8t + 200`

`text(Max height occurs when)\ \ dot y = 0`

`9.8t` `= 200`
`t` `= 20.408…`
  `= 20.4\ text{seconds  (to 1 d.p.)}`

 

`text(When)\ t = 20.408…`

`y` `= -4.9 (20.408…)^2 + 200 (20.408…) + 5000`
  `= 7040.816…`
  `= 7041\ text{m  (to nearest metre)}`

 

`:.\ text(The rocket will reach a maximum height of)`

`text(7041 metres when)\ \ t = 20.4\ text(seconds.)`

 

ii.  `text(The rocket is descending at 45° at point)\ A\ text(on the graph.)`

`text(The symmetry of the parabolic motion means that)\ \ A`

`text(occurs when)\ \ t = 2 xx text(time to reach max height, or)`

`40.8\ text(seconds.)`

 

`text(Point)\ B\ text(occurs when the rocket is descending at 60°)`

`text(to the horizontal.)`

 

`text(At)\ \ B,`

 

`-tan 60° = (dot y)/(dot x)`

`text{(The gradient of the projectile becomes}`

`text{negative after reaching its max height.)}`

`dot y = -9.8t + 200`

`dot x = d/(dt) (200t) = 200`

`:.\ – sqrt 3` `= (-9.8t + 200)/200`
`-200 sqrt 3` `= -9.8t + 200`
`9.8t` `= 200 + 200 sqrt 3`
`t` `= (200 + 200 sqrt 3)/9.8`
  `= (546.410…)/9.8`
  `= 55.756…`
  `= 55.8\ text{seconds  (nearest second)}`

 

`:.\ text(The pilot can operate the ejection seat)`

`text(between)\ \ t = 40.8 and t = 55.8\ text(seconds.)`

 

iii.

`v^2 = (dot x)^2 + (dot y)^2`

`text(When)\ \ v = 350`

`350^2` `= 200^2 + (-9.8t + 200)^2`
`122\ 500` `= 40\ 000 + (-9.8t + 200)^2`
`(-9.8t + 200)^2` `= 82\ 500`
`-9.8t + 200` `= +- sqrt (82\ 500)`
`9.8t` `= 200 +- sqrt (82\ 500)`
`t` `= (200 +- sqrt (82\ 500))/9.8`
  `= 49.717…\ \ \ (t > 0)`
  `= 49.7\ text{seconds  (to 1 d.p.)}`

 

`:.\ text(The latest time the pilot can eject safely)`

`text(is when)\ \ t = 49.7\ text(seconds.)`

Polynomials, EXT1 2005 HSC 3a

  1. Show that the function  `g(x) = x^2 - log_e (x + 1)`  has a zero between  `0.7`  and  `0.9.`  (1 mark)
  2. Use the method of halving the interval to find an approximation to this zero of  `g(x)`, correct to one decimal place.  (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `0.7\ \ text{(to 1 d.p.)}`
Show Worked Solution
(i)    `g(x)` `= x^2 – log_e (x + 1)`
`g(0.7)` `= 0.7^2 – log_e (0.7 + 1)`
  `= 0.49 – log_e 1.7`
  `= -0.04…`
`g(0.9)` `= 0.9^2 – log_e (0.9 + 1)`
  `= 0.81 – log_e 1.9`
  `= 0.168…`

 

`:.\ text(S)text(ince the sign changes, a zero exists)`

`text(between 0.7 and 0.9.)`

 

(ii)  `text(Halving the interval)`

`g(0.8)` `= 0.8^2 – log_e (0.8 + 1)`
  `= 0.64 – log_e 1.8`
  `= 0.0522…`

 

`:.\ text(S)text(ince)\ \ g(0.8) > 0,\ \ text(the zero exists)`

`text(between 0.7 and 0.8.)`

♦♦♦ Part (ii) was “very poorly done”.

 

`g(0.75)` `= 0.75^2 – log_e 1.75`
  `= 0.00288…`

 

`=>text(S)text(ince)\ \ g(0.75) > 0,\ \ text(the zero exists)`

`text(between 0.7 and 0.75.)`

`:.\ text{The zero will be  0.7  (to 1 d.p.)}`

Calculus, 2ADV C3 2007 HSC 10b

The noise level, `N`, at a distance `d` metres from a single sound source of loudness `L` is given by the formula

`N = L/d^2.`

Two sound sources, of loudness `L_1` and `L_2` are placed `m` metres apart.
 

The point `P` lies on the line between the sound sources and is `x` metres from the sound source with loudness `L_1.`

  1. Write down a formula for the sum of the noise levels at `P` in terms of `x`.  (1 mark)

  2. There is a point on the line between the sound sources where the sum of the noise levels is a minimum.

     

    Find an expression for `x` in terms of `m`, `L_1` and `L_2` if `P` is chosen to be this point.  (4 marks)

Show Answers Only
  1. `N = L_1/x^2 + L_2/(m-x)^2`
  2. `x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`
Show Worked Solution
i.  

`N = L/d^2`

`text(Noise from)\ L_1` `= L_1/x^2`
`text(Noise from)\ L_2` `= L_2/(m-x)^2`
`:. N` `= L_1/x^2 + L_2/(m-x)^2`

 

ii.  `N = L_1\ x^-2 + L_2 (m – x)^-2`

`(dN)/(dx)` `= -2 L_1 x^-3 + -2 L_2 (m – x)^-3 xx d/(dx) (m – x)`
  `= (-2 L_1)/x^3 + (2 L_2)/(m – x)^3`

 

`text(Max or min when)\ (dN)/(dx) = 0`

`(2 L_1)/x^3` `= (2 L_2)/(m – x)^3`
`2 L_1 (m – x)^3` `= 2 L_2\ x^3`
`L_1 (m – x)^3` `= L_2\ x^3`
`root 3 L_1 (m – x)` `= root 3 L_2\ x`
`root 3 L_1\ m – root 3 L_1\ x` `= root 3 L_2\ x`
`root 3 L_2\ x + root 3 L_1\ x` `= root 3 L_1\ m`
`x (root 3 L_2 + root 3 L_1)` `= root 3 L_1\ m`
`x` `= (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`

 

`(dN)/(dx)` `= -2 L_1\ x^-3 + 2 L_2 (m – x)^-3`
`(d^2N)/(dx^2)` `= 6 L_1\ x^-4 – 6 L_2 (m – x)^-4 xx -1`
  `= (6 L_1)/x^4 + (6 L_2)/(m – x)^4 > 0`

 

`:.\ text(A minimum occurs when)`

`x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`

Calculus in the Physical World, 2UA 2007 HSC 10a

An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are  `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for  `t >= 6`.
 


 

  1. Using Simpson’s rule, estimate the distance travelled between  `t = 0`  and  `t = 4`.  (2 marks)
  2. The object is initially at the origin. During which time(s) is the displacement of the object decreasing?  (1 mark)
  3. Estimate the time at which the object returns to the origin. Justify your answer.  (2 marks)
  4. Sketch the displacement, `x`, as a function of time.  (2 marks)
Show Answers Only
  1. `~~ 6\ \ text(units)`
  2. `t > 5\ \ text(seconds)`
  3. `7.2\ \ text(seconds)`

 

 

 

 

 

 

 

 

Show Worked Solution

(i)

`text(Distance travelled)`

`= int_0^4 (dx)/(dt)\ dt`

`~~  h/3 [y_0 + 4y_1 + y_2]`

`~~ 2/3 [0 + 4 (1) + 5]`

`~~ 2/3 [9]`

`~~ 6\ \ text(units)`

 

(ii)  `text(Displacement is reducing when the velocity is negative.)`

`:. t > 5\ \ text(seconds)`

 

(iii)  `text(At)\ B,\ text(the displacement) = 6\ text(units)`

`text(Considering displacement from)\ B\ text(to)\ D.`

`text(S)text(ince the area below the graph from)`

`B\ text(to)\ C\ text(equals the area above the)`

`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`

`text(in displacement from)\ B\ text(to)\ D.`

 

`text(Considering)\ t >= 6`

`text(Time required to return to origin)`

`t` `= d/v`
  `= 6/5`
  `= 1.2\ \ text(seconds)`

 

`:.\ text(The particle returns to the origin after 7.2 seconds.)`

(iv)

 2UA HSC 2007 10ai

Financial Maths, 2ADV M1 2007 HSC 9c

Mr and Mrs Caine each decide to invest some money each year to help pay for their son’s university education. The parents choose different investment strategies.

Mr Caine makes 18 yearly contributions of $1000 into an investment fund. He makes his first contribution on the day his son is born, and his final contribution on his son’s seventeenth birthday. His investment earns 6% compound interest per annum.

  1. Find the total value of Mr Caine’s investment on his son’s eighteenth birthday.  (3 marks)

Mrs Caine makes her contributions into another fund. She contributes $1000 on the day of her son’s birth, and increases her annual contribution by 6% each year. Her investment also earns 6% compound interest per annum.

  1. Find the total value of Mrs Caine’s investment on her son’s third birthday (just before she makes her fourth contribution).  (2 marks)

  2. Mrs Caine also makes her final contribution on her son’s seventeenth birthday. Find the total value of Mrs Caine’s investment on her son’s eighteenth birthday.  (1 mark)

Show Answers Only
  1. `$32\ 760\ \ \ text{(nearest dollar)}`
  2. `$3573\ \ \ text{(nearest dollar)}`
  3. `$51\ 378\ \ \ text{(nearest dollar)}`
Show Worked Solution

i.  `text(Let)\ A_n = text(value of the investment after)\ n\ text(years)`

`A_1` `= 1000 (1.06)`
`A_2` `= A_1 (1.06) + 1000 (1.06)`
  `= [1000 (1.06)] (1.06) + 1000 (1.06)`
  `= 1000 (1.06)^2 + 1000 (1.06)`
`A_3` `= A_2 (1.06) + 1000 (1.06)`
  `= 1000 (1.06)^3 + 1000 (1.06)^2 + 1000 (1.06)`
  `= 1000 [1.06 + 1.06^2 + 1.06^3]`

`vdots`

`A_n =` `1000 [1.06 + 1.06^2 + … + 1.06^n]`
  `text(Note that)  (1.06 + 1.06^2 + … + 1.06^n)\ text(is a)`
  `text(GP where)\ a = 1.06 and r = 1.06`
`:. A_n =` `1000 [(a(r^n – 1))/(r – 1)]`

 

`text(The son’s 18th birthday occurs when)\ n = 18`

`:. A_18` `= 1000 [(1.06 (1.06^18 -1))/(1.06 – 1)]`
  `= 1000 xx 32.7599…`
  `= 32\ 759.991…`
  `= $32\ 760\ \ \ text{(nearest $)}`

 

`:.\ text(The value of Mr Caine’s investment is $32 760.)`

 

ii.  `text(Let)\ V_n = text(value of investment after)\ n\ text(years)`

`V_1` `= 1000 (1.06)`
`V_2` `= V_1 (1.06) + 1000 (1.06) (1.06)`
  `= 1000 (1.06)^2 + 1000 (1.06)^2`
  `= 2000 (1.06)^2`
`V_3` `= V_2 (1.06) + 1000 (1.06)^3`
  `= 2000 (1.06)^3 + 1000 (1.06)^3`
  `= 3000 (1.06)^3`
  `= 3573.048…`
  `= $3573\ \ \ text{(nearest dollar)}`

 

iii.  `text(Continuing the pattern)`

`V_4 = 4000 (1.06)^4`

`vdots`

`V_n = n xx 1000 (1.06)^n`

`:. V_18` `= 18 xx 1000 xx (1.06)^18`
  `= 51\ 378.104…`
  `= $51\ 378\ \ \ text{(nearest dollar)}`

 

`:.\ text(The value of Mrs Caine’s investment on her)`

`text(son’s 18th birthday will be $51 378.)`

Geometry and Calculus, EXT1 2005 HSC 7b

Let  `f(x) = Ax^3 - Ax + 1`,  where  `A > 0`.

  1. Show that  `f(x)`  has stationary points at
    1. `x = +- (sqrt 3)/3.`  (1 mark)

  2. Show that  `f(x)`  has exactly one zero when
    1. `A < (3 sqrt 3)/2.`  (2 marks)

  3. By observing that  `f(-1) = 1`, deduce that  `f(x)`  does not have a zero in the interval  `-1 <= x <= 1`  when
    1. `0 < A < (3 sqrt 3)/2.`  (1 mark)

  4. Let  `g(theta) = 2 cos theta + tan theta`,  where  `-pi/2 < theta < pi/2.`
  6. By calculating  `g prime (theta)`  and applying the result in part (iii), or otherwise, show that  `g(theta)`  does not have any stationary points.  (3 marks)
  7. Hence, or otherwise, deduce that  `g(theta)`  has an inverse function.  (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)   `f(x) = Ax^3 – Ax + 1\ \ ,\ \ \ A > 0`

`f prime (x) = 3Ax^2 – A`

`text(S.P.’s when)\ \ f prime (x) = 0`

`3Ax^2 – A` `= 0`
`A (3x^2 – 1)` `= 0`
`3x^2` `= 1`
`x^2` `= 1/3`
`x` `= +- 1/sqrt3 xx (sqrt 3)/(sqrt 3)`
  `= +- (sqrt 3)/3\ \ text(…  as required)`

 

(ii)   `text(Consider)\ \ f(x)\ \ text(with only 1 zero.)`

`f(0) = 1`

`:.\ f(x)\ \ text(passes through)\ \ (0, 1)\ \ text(and has)`

`text(turning points either side at)\ \ x = +- (sqrt 3)/3.`

 

`text(Given 1 zero)`

`=> f((sqrt 3)/3) > 0`

`A ((sqrt 3)/3)^3 – A ((sqrt 3)/3) + 1` `> 0`
`A((3 sqrt 3)/27) – A ((9 sqrt 3)/27)` `> -1`
`A ((-6 sqrt 3)/27)` `> -1`
`A` `< 27/(6 sqrt 3)`
`A` `< 9/(2 sqrt 3) xx (sqrt 3)/(sqrt 3)`
`A` `< (9 sqrt 3)/6`
 `A` `< (3 sqrt 3)/2\ \ text(…  as required.)`

 

(iii)  `text(S)text(ince 1 zero occurs when)\ \ A < (3 sqrt 3)/2\ \ text{(part (ii))}`

`=> text(Minimum TP when)\ \ x = (sqrt3)/3`

`=> text(Maximum TP when)\ \ x = -(sqrt 3)/3\ \ text{(see graph)}`

 

`text(S)text(ince)\ \ f(-1) = 1`

`:. f(x)\ \ text(cannot have a zero for)\ \ \ -1 <= x <= 1`

 

(iv)   `g(theta) = 2 cos theta + tan theta,\ \ \ \ \ \ -pi/2 < theta < pi/2`

`g prime (theta) = -2 sin theta + sec^2 theta`

 

`text(S.P.’s when)\ \ g prime (theta) = 0`

`-2 sin theta + sec^2 theta` `= 0`
`-2 sin theta + 1/(cos^2 theta)` `= 0`
`-2 sin theta + 1/((1 – sin^2 theta))` `= 0`
`(-2 sin theta (1 – sin^2 theta) + 1)/((1 – sin^2 theta))` `= 0`
`-2 sin theta (1 – sin^2 theta) + 1` `= 0`
 `-2 sin theta + 2 sin^3 theta + 1` `= 0`
 `2 sin^3 theta – 2 sin theta + 1` `= 0\ \ text(…  (1))`

 

`text(Let)\ \ x = sin theta and A = 2`

`text{Equation (1) becomes}`

`Ax^3 – 2x + 1 = 0`

`text(S)text(ince)\ \ A = 2`

`0 < A < (3 sqrt 3)/2`

`text(S)text(ince)\ -pi/2` `< \ \ \ theta` `< pi/2`
`-1` `< sin theta` `< 1`
`-1` `< \ \ \ x` `< 1`

 

`:.\ text{Using Part (iii)} => g prime (theta)\ \ text(has no zeros and)`

`text(therefore)\ \ g (theta)\ \ text(has no S.P.’s for)\ \ -pi/2 < theta < pi/2.`

 

(v)   `text(Given)\ \ g(theta)\ \ text(has no stationary points, it will be)`

`text(either an increasing or decreasing function in the)`

`text(range)\ \ (-pi)/2 < theta < pi/2\ \ text(and therefore it has an inverse)`

`text(function.)`

Calculus, EXT1 C1 2005 HSC 7a

An oil tanker at  `T`  is leaking oil which forms a circular oil slick. An observer is measuring the oil slick from a position  `P`, 450 metres above sea level and 2 kilometres horizontally from the centre of the oil slick.
 


 

  1. At a certain time the observer measures the angle, `alpha`, subtended by the diameter of the oil slick, to be 0.1 radians. What is the radius, `r`, at this time?  (2 marks)

  2. At this time, `(d alpha)/(dt) = 0.02`  radians per hour. Find the rate at which the radius of the oil slick is growing.  (2 marks)

Show Answers Only
  1. `102.6\ text{m  (to 1 d.p.)}`
  2. `20.6\ text{metres  (to 1 d.p.)}`
Show Worked Solution

i.

`text(Let)\ \ Q\ \ text(be the point on the sea such)\ \ /_PQT`

`text(is a right angle.)`

 
`text(Consider)\ \ Delta PQT,`

`text(Using Pythagoras,)`

`PT^2` `= PQ^2 + QT^2`
  `= 450^2 + 2000^2`
`PT` `= sqrt(450^2 + 2000^2)`
  `= 2050\ text(m)`

 
`text(Consider)\ \ Delta PMT`

`tan\  /_ MPT` `= r/(PT)`
`tan 0.05` `= r/2050`
`:. r` `= 2050 xx tan 0.05`
  `= 102.585…`
  `= 102.6\ text{m  (to 1 d.p.)}`

 

ii.  `text(Find)\ \ (dr)/(dt)\ \ text(when)\ \ alpha = 0.1`

`(dr)/(dt)` `= (dr)/(d alpha) * (d alpha)/(dt)`
   
`r` `= 2050 xx tan\ alpha/2`
`:.\ (dr)/(d alpha)` `= 2050 xx 1/2 xx sec^2\ alpha/2`
  `= 1025 sec^2\ alpha/2`

  
`text(When)\ \ (d alpha)/(dt)= 0.02\ \ text(radians per hour),`

`alpha = 0.1\ \ \ text{(given)}.`

`:. (dr)/(dt)` `= 1025 sec^2 0.05 xx 0.02`
  `= 20.5513…`
  `= 20.6\ text{metres per hour  (to 1 d.p.)}`

Linear Functions, 2UA 2005 HSC 10b

Xuan and Yvette would like to meet at a cafe on Monday. They each agree to come to the cafe sometime between 12 noon and 1 pm, wait for 15 minutes, and then leave if they have not seen the other person.

Their arrival times can be represented by the point  `(x, y)`  in the Cartesian plane, where  `x`  represents the fraction of an hour after 12 noon that Xuan arrives, and  `y`  represents the fraction of an hour after 12 noon that Yvette arrives.

Thus  `(1/3, 2/5)`  represents Xuan arriving at 12:20 pm and Yvette arriving at 12:24 pm. Note that the point  `(x, y)`  lies somewhere in the unit square  `0 ≤ x ≤ 1`  and   `0 ≤ y ≤ 1`  as shown in the diagram.
 

Probability, 2UA 2005 HSC 10b
 

  1. Explain why Xuan and Yvette will meet if
  2. `x - y ≤ 1/4`  or  `y - x ≤1/4`.  (1 mark)
  3.  
  4. The probability that they will meet is equal to the area of the part of the region given by the inequalities in part (i) that lies within the unit square  `0 ≤ x ≤1`  and  `0≤ y ≤ 1`.
     
  5. Find the probability that they will meet.  (2 marks)
  6.  
  7. Xuan and Yvette agree to try to meet again on Tuesday. They agree to arrive between 12 noon and 1 pm, but on this occasion they agree to wait for `t` minutes before leaving.
  8.  
  9. For what value of `t` do they have a 50% chance of meeting?  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `7/16`
  3.  
  4. `text(17.6 minutes)`
Show Worked Solution

(i)   `text(Xuan and Yvette must arrive within)\ 1/4\ text(of an hour)`

`text(of each in order to meet.)`

`text(If Xuan arrives first,)\ \ y − x ≤ 1/4.`

`text(If Yvette arrives first,)\ \ x − y ≤ 1/4.`
 

(ii)  `text{From part (i), the inequalities are}`

`y ≤ x + 1/4`

`y ≥ x − 1/4`
 

`text(The shaded area satisfies the inequalities)`

`text(and the conditions)`

`0 ≤ x ≤ 1`

`0 ≤ y≤ 1`
 

Probability, 2UA 2005 HSC 10b Answer
 

`text(Shaded Area)`

`=\ text(Area of square − 2 triangles)`

`= 1^2 −2 xx (1/2 xx b xx h)`

`= 1 − 2 xx (1/2 xx 3/4 xx 3/4)`

`= 1 − 9/16`

`= 7/16\ \ text(u²)`
 

`:.\ text{P(Xuan and Yvette meet)}`

`=\ text(Shaded Area)/text(Area of square)`

`= 7/16`
 

(iii)  `text(We need the shaded area to be)\ 1/2\ text(u²)`

`text(for a 50% chance.)`

`text(If they wait)\ t\ text{minutes, the inequalities from part (i) are:}`

`y ≤ x + t/(60)`

`y ≥ x − t/(60)`

Probability-2UA-2005-HSC-10b-Answer2 

`text(Shaded Area)\ ` `= 1 − 2 xx (1/2 xx b xx h)`
 `1/2` `= 1 − 2 xx [1/2 xx (1 − t/(60))(1 − t/(60))]`
 `1/2` `= 1 − (1 − t/(60))^2`
`(1 − t/(60))^2` `= 1/2`
`1 − t/(60)` `= 1/sqrt2`
`t/(60)` `= 1 − 1/sqrt2`
`t` `= 60(1 − 1/sqrt2)`
  `= 17.593…`
  `= text(17.6 minutes  (to 1 d.p.))`

 

`:.\ text(They have 50% chance of meeting if they)`

`text(wait 17.6 minutes.)`

Quadratic, 2UA 2005 HSC 10a

2005 10a

The parabola  `y = x^2`  and the line  `y = mx + b`  intersect at the points  `A(α,α^2)`  and  `B(β, β^2)`  as shown in the diagram.

  1. Explain why  `α + β = m`  and  `αβ = –b`.  (1 mark)
  2. Given that
  3. `(α − β)^2 + (α^2 − β^2)^2 = (α − β)^2[1 + (α + β)^2]`, show that the distance  `AB = sqrt((m^2 + 4b)(1 + m^2)).`  (2 marks)
  4. The point  `P(x, x^2)`  lies on the parabola between  `A`  and  `B`. Show that the area of the triangle  `ABP`  is given by  `1/2(mx − x^2 + b)sqrt(m^2 + 4b).`  (2 marks)
  5.  
  6. The point  `P`  in part (iii) is chosen so that the area of the triangle  `ABP`  is a maximum.
  7. Find the coordinates of  `P`  in terms of  `m`.  (2 marks)
  8.  
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `P(m/2, m^2/4)`
Show Worked Solution

(i)   `text(Instersection)`

`y = x^2` `\ \ …\ (1)`
`y = mx + b` `\ \ …\ (2)`

`text(Substitute)\ y = x^2\ text(into)\ \ (2)`

`x^2 = mx + b`

`x^2 − mx − b = 0`

`text(We know the intersection occurs when)`

`x = α\ \ text(and)\ \ x = β`

`:.α\ \ text(and)\ \ β\ \ text(are roots of)\ \ x^2 − mx − b = 0`

`:.α + β` `= (−b)/a` `= m`  
`αβ` `= c/a` `= −b` `\ \ …text(as required)`

 

 

(ii)   `A(α, α^2),\ \ B(β, β^2)`

`AB` `= sqrt((α − β)^2 + (α^2 − β^2)^2)`
  `= sqrt((α − β)^2[1 + (α + β)^2])`
  `= sqrt([(α + β)^2 − 4αβ][1 + (α + β)^2])`
  `= sqrt([m^2 − 4(−b)][1 + m^2])`
  `= sqrt((m^2 + 4b)(1 + m^2))\ \ …text(as required)`

 

(iii) `⊥\ text(distance of)\ \ (x, x^2)\ \ text(from)\ \ \ y=mx+b`

`text(i.e.)\ \ mx – y+b=0`

`= |(ax_1 + by_1 + c)/(sqrt(a^2 + b^2))|`

`= |(mx − 1(x^2) + b)/(sqrt(m^2 + (−1)^2))|`

`= |(mx − x^2 + b)/(sqrt(m^2 + 1))|`

 

`text(Area of)\ ΔABP`

`= 1/2 xx AB xx h`

`= 1/2 xx sqrt((m^2 + 4b)(1 + m^2)) xx (mx − x^2 + b)/(sqrt(m^2 + 1))`

`= 1/2 (mx − x^2 + b)sqrt(m^2 + 4b)`

 

(iv) `A` `= 1/2 sqrt(m^2 + 4b)(mx − x^2 + b)`
  `(dA)/(dx)` `= 1/2 sqrt(m^2 + 4b)(m − 2x)` 
  `(d^2A)/(dx^2)` `= 1/2 sqrt(m^2 + 4b)(−2)`
    `= −sqrt(m^2 + 4b)` 

 

`text(Max or min when)\ (dA)/(dx) = 0`

`1/2sqrt(m^2 + 4b)(m − 2x)` `= 0`
`m-2x` `=0`
`2x` `= m`
`x` ` = m/2`

`text(When)\ x = 2`

`(d^2A)/(dx^2) = −sqrt(m^2 + 4b) < 0\ \ \ text{(constant)}`

`:.\ text(Maximum when)\ x = m/2`

`:.P\ text(is)\ (m/2, m^2/4)` 

Combinatorics, EXT1 A1 2015 HSC 14c

Two players `A` and `B` play a series of games against each other to get a prize. In any game, either of the players is equally likely to win.

To begin with, the first player who wins a total of 5 games gets the prize.

  1. Explain why the probability of player `A` getting the prize in exactly 7 games is  `((6),(4))(1/2)^7`.  (1 mark)

  2. Write an expression for the probability of player `A` getting the prize in at most 7 games.  (1 mark)

  3. Suppose now that the prize is given to the first player to win a total of `(n + 1)` games, where `n` is a positive integer.

     

    By considering the probability that `A` gets the prize, prove that
     

  4. `((n),(n))2^n + ((n + 1),(n))2^(n − 1) + ((n + 2),(n))2^(n − 2) + … + ((2n),(n)) = 2^(2n)`.   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `\ ^4C_4(1/2)^5 +\ ^5C_4(1/2)^6 +\ ^6C_4(1/2)^7`
  3. `text{Proof (See Worked Solutions.)}`
Show Worked Solution

i.   `text(To win exactly 7 games, player)\ A`

♦♦♦ Mean mark 19%.

`text(must win the 7th game.)`
 

`:.P(A\ text{wins in 7 games)}`

`=\ ^6C_4 · (1/2)^4(1/2)^2 xx 1/2`

`=\ ^6C_4(1/2)^7`

 

ii.  `Ptext{(wins in at most 7 games)}`

♦♦♦ Mean mark 23%.

`=Ptext{(wins in 5, 6 or 7 games)}`

`=\ ^4C_4(1/2)^4 xx 1/2 +\ ^5C_4(1/2)^4(1/2) xx 1/2 +\ ^6C_4(1/2)^7`

`=\ ^4C_4(1/2)^5 +\ ^5C_4(1/2)^6 +\ ^6C_4(1/2)^7`

♦♦♦ Mean mark part (iii) 9%.

 

iii. `text(Prove that)`

`\ ^nC_n · 2^n +\ ^(n + 1)C_n · 2^(n − 1) + … +\ ^(2n)C_n = 2^(2n)`

`P(A\ text(wins in)\ (n + 1)\ text{games)}`

`=\ ^nC_n(1/2)^(n + 1) +\ ^(n + 1)C_n(1/2)^(n + 2) + … +\ ^(2n)C_n(1/2)^(2n + 1)`

 

`text{One player must have won after (2n + 1) games are played.}`

`text(S)text(ince each player has an equal chance,)`
 

`\ ^nC_n(1/2)^(n + 1) +\ ^(n + 1)C_n(1/2)^(n + 2) + … +\ ^(2n)C_n(1/2)^(2n + 1) = 1/2`

 

`text(Multiply both sides by)\ 2^(2n + 1)`
 

`\ ^nC_n2^(-(n + 1)) · 2^(2n + 1) +\ ^(n + 1)C_n · 2^(-(n + 2)) · 2^(2n + 1) + …`

`… +\ ^(2n)C_n · 2^(-(2n + 1)) · 2^(2n + 1) = 2^(-1) · 2^(2n + 1)`
 

`:. \ ^nC_n · 2^n +\ ^(n + 1)C_n · 2^(n − 1) + … +\ ^(2n)C_n = 2^(2n)`

Measurement, STD2 M6 2015 HSC 30e

From point `S`, which is 1.8 m above the ground, a pulley at `P` is used to lift a flat object `F`. The lengths `SP` and `PF` are 5.4 m and 2.1 m respectively. The angle `PSC` is 108°.
 

 

  1. Show that the length  `PC`  is 6.197 m, correct to 3 decimal places.  (1 mark)

  2. Calculate `h`, the height of the object above the ground.  (4 marks)

Show Answers Only
  1. `6.197\ text{m  (to 3 d.p.)  … as required}`
  2. `1.37\ text{m  (to 2 d.p.)}`
Show Worked Solution
i.   

2UG 2015 30e Answer

`text(Show)\ PC = 6.197\ text(m)`

♦ Mean mark 41%.

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

`PC^2` `= PS^2 + SC^2-2 xx PS xx SC xx cos\ 108^@`
  `= 5.4^2 + 1.8^2-2 xx 5.4 xx 1.8 xx cos\ 108^@`
  `= 38.4072…`
`:.PC` `= 6.19736…`
  `= 6.197\ text{m  (to 3 d.p.)  …as required}`

 

ii.   `text(Let)\ \ SD⊥PE`

♦♦ Mean mark below 19%.
STRATEGY: Finding `PC` in part (i) and needing `PE` to find `h` should flag the strategy of finding `EC` and using Pythagoras.

`∠DSC\ text(is a right angle)`

`:.∠DSP = 108^@-90^@ = 18^@`

 

`text(In)\ ΔPDS`

`cos\ 18^@` `= (DS)/5.4`
 `DS` `= 5.4 xx cos\ 18^@`
  `= 5.1357…\ text(m)`

 
`EC = DS = 5.1357…\ text{m  (opposite sides of rectangle}\ DECS text{)}`
 

`text(Using Pythagoras in)\ Delta PEC:`

`PE^2 + EC^2 = PC^2`

`PE^2 + 5.1357^2` `= 6.197^2`
`PE^2` `= 12.027…`
`PE` `= 3.468…\ text(m)`

 
`text(From the diagram,)`

`h` `= PE-PF`
  `= 3.468…-2.1`
  `= 1.368…`
  `= 1.37\ text{m  (to 2 d.p.)}`

Algebra, STD2 A4 2015 HSC 29e

A diver springs upwards from a diving board, then plunges into the water. The diver’s height above the water as it varies with time is modelled by a quadratic function. Graphing software is used to produce the graph of this function.
 

Explain how the graph could be used to determine how high above the height of the diving board the diver was when he reached the maximum height.  (2 marks)

Show Answers Only

`1.2\ text(m)`

Show Worked Solution

`text(We can calculate the height of the board by)`

♦♦ Mean mark 19%.

`text(finding the)\ ytext(-value at)\ t = 0,\ text(which is 8 m.)`

`text(The diver’s maximum height occurs at)\ t=0.5,`

`text(which is approximately 9.2 m.)`
 

`:.\ text(Maximum height above the board)`

`= 9.2-8`

`= 1.2\ text(m)`

Statistics, STD2 S1 2015 HSC 29d

Data from 200 recent house sales are grouped into class intervals and a cumulative frequency histogram is drawn.
 

2UG 2015 29d1 
 

  1. Use the graph to estimate the median house price.   (1 mark)

  2. By completing the table, calculate the mean house price.   (3 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Class Centre} \rule[-1ex]{0pt}{0pt} & \text{Frequency} \\ \text{(\$'000)} &  \\
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `text(See Worked Solutions.)`
Show Worked Solution
i.   

2UG 2015 29d Answer

`text(From the graph, the estimated median)`

`text(house price = $392 500)`

♦♦♦ Mean marks of 9% for part (i) and 34% for part (ii)!

 
ii.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Class Centre} \rule[-1ex]{0pt}{0pt} & \text{Frequency} \\ \text{(\$’000)} &  \\
\hline
\rule{0pt}{2.5ex}  375 \rule[-1ex]{0pt}{0pt} & 30\\
\hline
\rule{0pt}{2.5ex}  385 \rule[-1ex]{0pt}{0pt} &  50 \\
\hline
\rule{0pt}{2.5ex}  395 \rule[-1ex]{0pt}{0pt} & 70 \\
\hline
\rule{0pt}{2.5ex}  405 \rule[-1ex]{0pt}{0pt} & 50 \\
\hline
\end{array}

 
`text(Mean house price ($’000))`

`= (375xx30 + 385xx50 + 395xx70 + 405xx50)/200`

`=$392`

`:. text(Mean house price is)\ $392\ 000`

Measurement, STD2 M7 2015 HSC 29c

The image shows a rectangular farm shed with a flat roof.
 

Measurement, 2UG 2015 HSC 29c

 
The real width of the shed indicated by the dotted line was measured using an online ruler tool, and found to be approximately 12 metres.

  1. By measurement and calculation, show that the area of the roof of the shed is approximately 216 m².  (2 marks)

  2. All the rain that falls onto this roof is diverted into a cylindrical water tank which has a diameter of 3.6 m. During a storm, 5 mm of rain falls onto the roof.

     

    Calculate the increase in the depth of water in the tank due to the rain that falls onto the roof during the storm.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `10.6\ text{cm  (to 1 d.p.)}`
Show Worked Solution

i.    `text(By measurement,)`

♦ Mean mark 35%.

`text(Roof length = 1.5 times the roof width)`

`text{Width = 12 m  (given)}`

`text(Length)` `= 1.5 xx 12`
  `= 18\ text(m)`

 

`:.\ text(Area of roof)` `= 12 xx 18`
  `= 216\ text(m²)\ \ text(…as required)`

 

ii.  `text(Volume of water)`

♦♦♦ Mean mark 14%.

`= Ah`

`= 216 xx 0.005\ \ \ (5\ text(mm) = 0.005\ text(m))`

`= 1.08\ text(m³)`
 

`text(Volume of cylinder =)\ pir^2h`

`r = 3.6/2 = 1.8\ text(m)`
 

`text(Find)\ h\ text(when)\ V= 1.08\ text(m³),`

`pi xx 1.8^2 xx h` `= 1.08`
 `:. h` `= 1.08/(pi xx 1.8^2)`
  `= 0.1061…\ text(m)`
  `= 10.6\ text{cm  (to 1 d.p.)}`

Calculus, 2ADV C3 2015 HSC 16c

The diagram shows a cylinder of radius `x` and height `y` inscribed in a cone of radius `R` and height `H`, where `R` and `H` are constants.
  

The volume of a cone of radius `r` and height `h` is  `1/3 pi r^2 h.`

The volume of a cylinder of radius `r` and height `h` is  `pi r^2 h.`

  1. Show that the volume, `V`, of the cylinder can be written as
     
         `V = H/R pi x^2 (R-x).`   (3 marks)

  2. By considering the inscribed cylinder of maximum volume, show that the volume of any inscribed cylinder does not exceed `4/9` of the volume of the cone.   (4 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `text(Show)\ \ V = H/R pi x^2 (R-x)`
 

♦♦ Mean mark (i) 16%.

 

`text(Consider)\ \ Delta ABC and Delta DEC`

`/_ ABC = /_DEC = 90°`

`/_ BCA\ \ text(is common)`

`:. Delta ABC\ \ text(|||)\ \ Delta DEC\ \ text{(equiangular)}`

`:.\ (DE)/(EC)`

`= (AB)/(BC)`

`\ text{(corresponding sides of}`

`\ text{similar triangles)}`
`y/(R-x)` `= H/R`  
`y` `= H/R (R-x)`  

 

`text(Volume of cylinder)`

`= pi r^2 h`

`= pi x^2 xx H/R (R-x)`

`= H/R pi x^2 (R-x)\ \ text(…  as required.)`
 

♦♦ Mean mark (ii) 21%.

ii.   `V= H/R pi x^2 (R-x)`

`(dV)/(dx)` `= H/R pi [(x^2 xx -1) + 2x (R-x)]`
  `= H/R pi [-x^2 + 2xR-2x^2]`
  `= H/R pi [2xR-3x^2]`
`(dV^2)/(dx^2)` `= H/R pi [2R-6x]`

 

`text(Max or min when)\ \ (dV)/(dx) = 0`

`H/R pi [2xR-3x^2]` `= 0`
`2xR-3x^2` `= 0`
`x (2R-3x)` `= 0`
`x = 0\ \ or\ \ 3x` `= 2R`
`x` `= (2R)/3`

 

`text(When)\ \ x = 0:`

`(d^2V)/(dx^2) = H/R pi [2R-0] > 0\ \ =>\  text(MIN)`
 

`text(When)\ \ x = (2R)/3:`

`(d^2V)/(dx^2)` `= H/R pi [2R-6 xx (2R)/3]`
  `= H/R pi [-2R] < 0\ \ =>\ \text{MAX}`

 
`text(Maximum Volume of cylinder)`

`= H/R pi ((2R)/3)^2 (R-(2R)/3)`

`= H/R pi xx (4R^2)/9 xx R/3`

`= (4 pi H R^2)/27`

 

`text(Volume of cone)\ = 1/3 pi R^2 H`

`:.\ text(Max Volume of Cylinder)/text(Volume of cone)`

`= ((4 pi H R^2)/27)/(1/3 pi R^2 H)`

`= (4 pi H R^2)/27 xx 3/(pi R^2 H)`

`= 4/9`

 

`:.\ text(The volume of the inscribed cylinder does)`

`text(not exceed)\ 4/9\ text(of the cone volume.)`

Plane Geometry, 2UA 2015 HSC 15b

The diagram shows `Delta ABC` which has a right angle at `C`. The point `D` is the midpoint of the side `AC`. The point `E` is chosen on `AB` such that `AE = ED`. The line segment `ED` is produced to meet the line `BC` at `F`.

Copy or trace the diagram into your writing booklet.

  1. Prove that `Delta ACB` is similar to `Delta DCF.`  (2 marks)
  2. Explain why `Delta EFB` is isosceles.  (1 mark)
  3. Show that `EB = 3AE.`  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)  `text(Prove)\ \ Delta ACB\ \ text(|||)\ \ Delta DCF`

`/_ EAD = /_ ADE = theta\ \ text{(base angles of isosceles}\ \ Delta AED text{)}`

`/_ CDF = /_ ADE = theta\ \ text{(vertically opposite angles)}`

`/_ DCF = /_ ACB = 90°\ \ text{(}/_ FCB\ text{is a straight angle)}`

`:.\ Delta ACB\ \ text(|||)\ \ Delta DCF\ \ text{(equiangular)}`

 

♦ Mean mark 45%.

(ii)  `/_ DFC = 90 – theta\ \ text{(angle sum of}\ \ Delta DCF text{)}`

`/_ ABC = 90 – theta\ \ text{(angle sum of}\ \ Delta ACB text{)}`

`:.\ Delta EFB\ \ text{is isosceles  (base angles are equal)}`

 

(iii)  `text(Show)\ \ EB = 3AE`

♦♦ Mean mark 23%.
`(DC)/(AC)` `= (DF)/(AB)`

`\ \ \ \ \ text{(corresponding sides of}`

`\ \ \ \ \ text{similar triangles)}`
`1/2` `= (DF)/(AB)`  
`2DF` `= AB`  
`2(EF – ED)` `= AE + EB`  
`2(EB – AE)` `= AE + EB` `\ \ \ \ \ text{(given}\ EF = EB, ED = AE text{)}`
`2EB – 2AE` `= AE + EB`  

 

`:. EB = 3AE\ \ text(…  as required)`

Calculus, 2ADV C4 2015 HSC 9 MC

A particle is moving along the `x`‑axis. The graph shows its velocity `v` metres per second at time `t` seconds.
 

2012 2ua 9 mc
 

When `t = 0` the displacement `x` is equal to `2` metres.

What is the maximum value of the displacement `x`?

  1. `text(8 m)`
  2. `text(14 m)`
  3. `text(16 m)`
  4. `text(18 m)`
Show Answers Only

`D`

Show Worked Solution

`text(Distance travelled)`

♦♦ Mean mark 31%.

`= int_0^4 v\ dt`

`= text(Area under the velocity curve)`

`= 1/2 xx b xx h`

`= 1/2 xx 4 xx 8`

`= 16\ \ text(metres.)`

 

`text(S)text(ince velocity is always positive between)\ t=0`

`text(and)\  t = 4,\ text(and the original displacement = 2,)`

`text(the maximum displacement) = 16 + 2 = 18\ \ text(metres)`

`=> D`

Plane Geometry, 2UA 2006 HSC 10b

A rectangular piece of paper `PQRS` has sides `PQ = 12` cm and `PS = 13` cm. The point `O` is the midpoint of `PQ`. The points `K` and `M` are to be chosen on `OQ` and `PS` respectively, so that when the paper is folded along `KM`, the corner that was at `P` lands on the edge `QR` at `L`. Let `OK = x` cm and `LM = y` cm.

Copy or trace the diagram into your writing booklet.

  1. Show that `QL^2 = 24x`.  (1 mark)
  2. Let `N` be the point on `QR` for which `MN` is perpendicular to `QR`.
  3. By showing that `Delta QKL\ text(|||)\ Delta NLM`, deduce that `y = {sqrt 6 (6 + x)}/sqrt x`.  (3 marks)
  4. Show that the area, `A`, of `Delta KLM` is given by
    1. `A = {sqrt 6 (6 + x)^2}/(2 sqrt x)`  (1 mark)
    2.  
  5. Use the fact that `12 <= y <= 13` to find the possible values of `x`.  (2 marks)
  6. Find the minimum possible area of `Delta KLM`.  (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `2 2/3 <= x <= 6`
  5.  
  6. `53.33…\ text(cm²)`
Show Worked Solution
(i)     

 `text(Show)\ QL^2 = 24x`

`KP = KL = 6 + x`

`KQ = 6 – x`

`text(Using Pythagoras)`

`QL^2` `= KL^2 – KQ^2`
  `= (6 + x)^2 – (6 – x)^2`
  `= 36 + 12x + x^2 – 36 + 12x – x^2`
  `= 24x\ …\ text(as required)`

 

(ii)  `text(Show)\ y = {sqrt 6 (6 + x)}/sqrt x`

`text(In)\ Delta QKL`

`/_LQK=90°\ \ text{(given)}`

`text(Let)\ /_QKL = theta`

`:. /_QLK = 90 – theta\ \ text{(angle sum of}\ Delta QKL text{)}`

`text(In)\ Delta NLM`

`/_LNM=90°\ \ text{(given)}`

`/_NLM` `= 180 – (90 + 90 – theta)\ \ (/_QLN\ text(is a straight angle))`
  `= theta`

`:. Delta QKL\ text(|||)\ Delta NLM\ \ \ text{(equiangular)}`

`:. y/(MN)` `= (KL)/(QL)\ \ text{(corresponding sides of}`
  `text{similar triangles)}` 
`y/12` `= (6 + x)/sqrt(24x)`
`y` `= (12(6 + x))/(2 sqrt (6x))`
  `= (6 (6 + x))/(sqrt x xx sqrt 6) xx sqrt 6/sqrt 6`
  `= (sqrt 6 (6 + x))/sqrt x\ …\ text(as required)`

 

(iii)  `text(Area)\ DeltaKLM` `= 1/2 xx y xx KL`
  `= 1/2 xx (sqrt 6 (6 + x))/sqrt x xx (6 + x)`
  `= (sqrt 6 (6 + x)^2)/(2 sqrt x)\ …\ text(as required.)`

 

(iv)  `text(Given that)\ \ \ \ \ \ \ \ 12 <= y <= 13`

`12 <= (sqrt 6 (6 + x))/sqrt x <= 13`

`text(Consider)\ (sqrt 6 ( 6 + x))/sqrt x` `>= 12`
`(6 (6 + x)^2)/x` `>= 144`
`(6 + x)^2` `>= 24x`
`36 + 12x + x^2` `>= 24x`
`x^2 – 12x + 36` `>= 0`
`(x – 6)^2` `>= 0`
`:. x` `>= 6`

`text(However, we know)\ OP=6,\ text(and)\ x <= 6`.

`:. x = 6\ text(satisfies both conditions)`

`text(Consider)\ (sqrt 6 (6 + x))/sqrt x` `<= 13`
`(6 (6 + x)^2)/x` `<= 169`
`6 (6 + x)^2` `<= 169x`
`6 (36 + 12x + x^2)` `<= 169x`
`216 + 72x + 6x^2` `<= 169x`
`6x^2 – 97x + 216` `<= 0`

 

`text(Using the quadratic formula)`

`x` `= (-b +- sqrt (b^2 -4ac))/(2a)`
  `= (97 +- sqrt ((-97)^2 -4 xx 6 xx 216))/(2 xx 6)`
  `= (97 +- sqrt 4225)/12`
  `= (97 +- 65)/12`
  `= 13 1/2 or 2 2/3`

`:. 2 2/3 <= x <= 13 1/2`

`text(However, we know)\ x <= 6,\ text(so)`

`2 2/3 <= x <= 6\ text(satisfies both conditions)`

`:.\ text(All possible values of)\ x\ text(are)`

`2 2/3 <= x <= 6`.

 

(v)  `A = (sqrt 6 (6 + x)^2)/(2 sqrt x)`

`text(Using the quotient rule)`

`(dA)/(dx)` `= (u prime v – uv prime)/v^2`
  `= {2 sqrt 6 (6 + x) xx 2 sqrt x – sqrt 6 (6 + x)^2 xx 1/2 xx 2 xx x^(-1/2)}/(2 sqrt x)^2`
  `= {4 sqrt x sqrt 6 (6 + x) – sqrt 6 (6 + x)^2 * 1/sqrt x}/(4x)`
  `= {4 sqrt 6 x (6 + x) – sqrt 6 (6 + x)^2}/(4x sqrt x)`
  `= (sqrt 6 (6 + x) (4x – 6 – x))/(4x sqrt x)`
  `= (sqrt 6 (6 + x) (3x – 6))/(4x sqrt x)`

`text(Max or min when)\ (dA)/(dx) = 0`

`sqrt 6 (6 + x) (3x – 6) = 0`

`:. 3x = 6\ \ ,\ \ x ≠ -6`

`x = 2`

`text(However,)\ x = 2\ text(lies outside the range)`

`text(of possible values)\ \ 2 2/3 <= x <= 6`

`:.\ text(Check limits)`

`text(At)\ \ x = 2 2/3`

`A` `= (sqrt 6 (6 + 2 2/3)^2)/(2 sqrt (2 2/3))`
  `= 56.33…\ text(cm²)`

 

`text(At)\ \ x = 6`

`A` `= (sqrt 6 (6 + 6)^2)/(2 sqrt 6)`
  `= 72.0\ text(cm²)`

 

`:.\ text(Minimum area of)\ Delta KLM\ text(is 56.33… cm²)`

Calculus, 2ADV C3 2006 HSC 9c

A cone is inscribed in a sphere of radius `a`, centred at `O`. The height of the cone is `x` and the radius of the base is `r`, as shown in the diagram.

  1. Show that the volume, `V`, of the cone is given by
     
         `V = 1/3 pi(2ax^2 - x^3)`.  (2 marks)

  2. Find the value of `x` for which the volume of the cone is a maximum. You must give reasons why your value of `x` gives the maximum volume.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x = 4/3 a`
Show Worked Solution

i.  `text(Show)\ V = 1/3 pi (2ax^2 – x^3)`

`V = 1/3 pi r^2 h`

`text(Using Pythagoras)`

`(x – a)^2 + r^2` `= a^2`
`r^2` `= a^2 – (x – a)^2`
  `= a^2 – x^2 + 2ax – a^2`
  `= 2ax – x^2`
`:. V` `= 1/3 xx pi xx (2ax – x^2) xx x`
  `= 1/3 pi (2ax^2 – x^3)\ …\ text(as required)`

 

ii.  `(dV)/(dx) = 1/3 pi (4ax – 3x^2)`

`(d^2V)/(dx^2) = 1/3 pi (4a – 6x)`

`text(Max or min when)\ (dV)/(dx) = 0`

`1/3 pi (4ax – 3x^2)` `= 0`
`4ax – 3x^2` `= 0`
`x(4a – 3x)` `= 0`
`3x` `= 4a,` ` \ \ \ \ x ≠ 0`
`x` ` =4/3 a`  

 

`text(When)\ \ x = 4/3 a`

`(d^2V)/(dx^2)` `= 1/3 pi (4a – 6 xx 4/3 a)`
  `= 1/3 pi (-4a) < 0`
`=>\ text(MAX)`

 

`:.\ text(Cone volume is a maximum when)\ \ x = 4/3 a.`

Calculus, 2ADV C4 2006 HSC 9b

During a storm, water flows into a 7000-litre tank at a rate of `(dV)/(dt)` litres per minute, where `(dV)/(dt) = 120 + 26t-t^2` and `t` is the time in minutes since the storm began.

  1. At what times is the tank filling at twice the initial rate?  (2 marks)

  2. Find the volume of water that has flowed into the tank since the start of the storm as a function of `t`.  (1 mark)

  3. Initially, the tank contains 1500 litres of water. When the storm finishes, 30 minutes after it began, the tank is overflowing.

     

    How many litres of water have been lost?  (2 marks)

Show Answers Only
  1. `t = 6 or 20\ text(minutes)`
  2. `V = 120t + 13t^2 – 1/3t^3`
  3. `800\ text(litres)`
Show Worked Solution

i.  `(dV)/(dt) = 120 + 26t-t^2`

`text(When)\ t = 0, (dV)/(dt) = 120`

`text(Find)\ t\ text(when)\ (dV)/(dt) = 240`

`240 = 120 + 26t-t^2`

`t^2-26t + 120 = 0`

`(t-6)(t-20) = 0`

`t = 6 or 20`

`:.\ text(The tank is filling at twice the initial rate)`

`text(when)\ t = 6 and t = 20\ text(minutes)`

 

ii.  `V` `= int (dV)/(dt)\ dt`
  `= int 120 + 26t-t^2\ dt`
  `= 120t + 13t^2-1/3t^3 + c`

 
`text(When)\ t = 0, V = 0`

`=>  c = 0`

`:. V= 120t + 13t^2-1/3t^3`

 

iii.  `text(Storm water volume into the tank when)\ t = 30`

`= 120(30) + 13(30^2)-1/3 xx 30^3`

`= 3600 + 11\ 700-9000`

`= 6300\ text(litres)`
 

`text(Total volume)` `= 6300 + 1500`
  `= 7800\ text(litres)`

 

`:.\ text(Overflow)` `= 7800-7000`
  `= 800\ text(litres)`

Trigonometry, 2ADV T1 2005 HSC 9b

Trig Ratios, 2UA 2005 HSC 9b
 

The triangle  `ABC`  has a right angle at `B, \ ∠BAC = theta` and  `AB = 6`. The line `BD` is drawn perpendicular to `AC`. The line `DE` is then drawn perpendicular to `BC`. This process continues indefinitely as shown in the diagram.

  1. Find the length of the interval `BD`, and hence show that the length of the interval  `EF`  is  `6 sin^3\ theta`.  (2 marks)

  2. Show that the limiting sum 
     
    `qquad BD + EF + GH + ···`
     
    is given by  `6 sec\ theta tan\ theta`.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution
i.   Trig Ratios, 2UA 2005 HSC 9b Answer

`text(Show)\ EF = 6\ sin^3\ theta`

`text(In)\ ΔADB`

`sin\ theta` `= (DB)/6`
`DB` `= 6\ sin\ theta`
`∠ABD` `= 90 − theta\ \ \ text{(angle sum of}\ ΔADB)`
`:.∠DBE` `= theta\ \ \ (∠ABE\ text{is a right angle)}`

 

`text(In)\ ΔBDE:`

`sin\ theta` `= (DE)/(DB)`
  `= (DE)/(6\ sin\ theta)`
`DE` `= 6\ sin^2\ theta`
`∠BDE` `= 90 − theta\ \ \ text{(angle sum of}\ ΔDBE)`
`∠EDF` `= theta\ \ \ (∠FDB\ text{is a right angle)}`

 

`text(In)\ ΔDEF:`

`sin\ theta` `= (EF)/(DE)`
  `= (EF)/(6\ sin^2\ theta)`
`:.EF` `= 6\ sin^3\ theta\ \ …text(as required)`

 

ii. `text(Show)\ \ BD + EF + GH\ …`

`text(has limiting sum)\ =6 sec theta tan theta`

`underbrace{6\ sin\ theta + 6\ sin^3\ theta +\ …}_{text(GP where)\ \ a = 6 sin theta, \ \ r = sin^2 theta}`
 

`text(S)text(ince)\ \ 0 < theta < 90^@`

`−1` `< sin\ theta` `< 1`
`0` `< sin^2\ theta` `< 1`

 
`:. |\ r\ | < 1`
 

`:.S_∞` `= a/(1 − r)`
  `= (6\ sin\ theta)/(1 − sin^2\ theta)`
  `= (6\ sin\ theta)/(cos^2\ theta)`
  `= 6 xx 1/(cos\ theta) xx (sin\ theta)/(cos\ theta)`
  `= 6 sec\ theta\ tan\ theta\ \ …text(as required.)`

Calculus, EXT1* C1 2005 HSC 7b

Calculus in the Physical World, 2UA 2005 HSC 7b
 

The graph shows the velocity, `(dx)/(dt)`, of a particle as a function of time. Initially the particle is at the origin. 

  1. At what time is the displacement, `x`, from the origin a maximum?  (1 mark)

  2. At what time does the particle return to the origin? Justify your answer.  (2 marks)

  3. Draw a sketch of the acceleration,  `(d^2x)/(dt^2)`, as a function of time for  `0 ≤ t ≤ 6`.  (2 marks)

Show Answers Only
  1. `2`
  2. `4`
  3. `text(See Worked Solutions.)`
Show Worked Solution

i.   `text(Maximum displacement in graph when)`

`(dx)/(dt)` `= 0`
`:.t` `= 2`

 

ii.  `text(Particle returns to the origin when)\ t = 4.`

`text(The displacement can be calculated by the)`

`text(net area below the curve and since the)`

`text(area above the curve between)\ t = 0\ text(and)\ t = 2`

`text(is equal to the area below the curve between)`

`t = 2\ text(and)\ t = 4,\ text(the displacement returns to)`

`text{the initial displacement (i.e. the origin).}`

 

iii.   Calculus in the Physical World, 2UA 2005 HSC 7b Answer

Algebra, STD2 A4 2006 HSC 28b

A new tunnel is built. When there is no toll to use the tunnel, 6000 vehicles use it each day. For each dollar increase in the toll, 500 fewer vehicles use the tunnel.

  1. Find the lowest toll for which no vehicles will use the tunnel.  (1 mark)

  2. For a toll of $5.00, how many vehicles use the tunnel each day and what is the total daily income from tolls?  (2 marks)

  3. If `d` (dollars) represents the value of the toll, find an equation for the number of vehicles `(v)` using the tunnel each day in terms of `d`.  (2 marks)

  4. Anne says ‘A higher toll always means a higher total daily income’.

     

    Show that Anne is incorrect and find the maximum daily income from tolls. (Use a table of values, or a graph, or suitable calculations.)  (3 marks)

Show Answers Only
  1. `text($12 toll is the lowest for which)`

     

    `text(no vehicles will use the tunnel.)`

  2. `$17\ 500`
  3. `v = 6000 – 500d`
  4. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.  `text(500 less vehicles per $1 toll)`

`12 xx 500 = 6000`

`:. $12\ text(toll is the lowest for which no)`

`text(vehicles will use the tunnel.)`

 

ii.  `text(If the toll is $5)`

`5 xx 500 = 2500\ text(less vehicles)`

`:.\ text(Vehicles using the tunnel)`

`= 6000 – 2500`

`= 3500`

`:.\ text(Daily toll income)` `= 3500 xx $5`
  `= $17\ 500`

 

iii.    `d` `=\ text(toll)`
`v` `=\ text(Number of vehicles using the tunnel)`
    `:. v` `= 6000 – 500d`

 

iv.  `text(Income from tolls)`

`=\ text(Number of vehicles) xx text(toll)`

`= (6000 – 500d) xx d`

`= 6000d – 500d^2`

`= 500d (12 – d)`
 

 

`text(From the graph, the maximum income from tolls)`

`text(occurs when the toll is $6.)`

`:.\ text(Anne is incorrect.)`

 

`text(Alternate Solution)`

`text{The table of values shows that income (I) increases}`

`text(and peaks when the toll hits $6 before decreasing)`

`text(again as the toll gets more expensive.)`

`:.\ text(Anne is incorrect.)`

Statistics, STD2 S4 2006 HSC 27b

Each member of a group of males had his height and foot length measured and recorded. The results were graphed and a line of fit drawn.
 

  1. Why does the value of the `y`-intercept have no meaning in this situation?  (1 mark)

  2. George is 10 cm taller than his brother Harry. Use the line of fit to estimate the difference in their foot lengths.  (1 mark)

  3. Sam calculated a correlation coefficient of  −1.2  for the data. Give TWO reasons why Sam must be incorrect.  (2 marks)

Show Answers Only
  1. `text(The y-intercept occurs when)\ x = 0.\ text(It has`
    `text(no meaning to have a height of 0 cm.)`
  2. `text(A 10 cm height difference means George should)`
    `text(have a 3 cm longer foot.)`
  3. `text(A correlation co-efficient must be between –1 and 1.)`
    `text(Foot length is positively correlated to a person’s)`
    `text(height and therefore can’t be a negative value.)`
Show Worked Solution

i.  `text(The y-intercept occurs when)\ x = 0.\ text(It has)`

`text(no meaning to have a height of 0 cm.)`

 

ii.  `text(A 20 cm height difference results in a foot length)`

`text(difference of 6 cm.)`
 

`:.\ text(A 10 cm height difference means George should)`

`text(have a 3 cm longer foot.)`

 

iii.  `text(A correlation co-efficient must be between –1 and 1.)`

`text(Foot length is positively correlated to a person’s)`

`text(height and therefore isn’t a negative value.)`

Statistics, STD2 S1 2007 HSC 22 MC

A set of examination results is displayed in a cumulative frequency histogram and polygon (ogive).
 

2007 22 mc
 
Sanath knows that his examination mark is in the 4th decile.

Which of the following could have been Sanath’s examination mark?

  1.    37
  2.    57
  3.    67
  4.    77
Show Answers Only

`B`

Show Worked Solution

Data, 2UG 2007 HSC 22 MC Answer

`text(4th decile occurs when cumulative frequency)`

`text(is between 15 and 20.)`

`:.\ text(Examination mark must be between 55 and 60.)`

`=>  B`

Probability, STD2 S2 2006 HSC 28a

On a bridge, the toll of $2.50 is paid in coins collected by a machine. The machine only accepts two-dollar coins, one-dollar coins and fifty-cent coins.

  1. List the different combinations of coins that could be used to pay the $2.50 toll.  (1 mark)

  2. Jill has three two-dollar coins, six one-dollar coins and two fifty-cent coins. She selects two coins at random.

     

    What is the probability that she selects exactly $2.50?  (3 marks)

  3. At the end of a day, the machine contains `x` two-dollar coins, `y` one-dollar coins and `w` fifty-cent coins.

     

    Write an expression for the total value of coins in dollars in the machine.  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `6/55`
  3. `$(2x + y + 0.5w)`
Show Worked Solution

i.  `text(Combinations for $2.50)`

`$2, 50c`

`$1, $1, 50c`

`$1, 50c, 50c, 50c`

`50c, 50c, 50c, 50c, 50c`

 

ii.  `3 xx $2, 6 xx $1, 2 xx 50c`

`P($2.50)` `= (3/11 xx 2/10) + (2/11 xx 3/10)`
  `= 6/110 + 6/110`
  `= 6/55`

 

iii.  `text(Total value) = $ (2x + y + 0.5w)`

Measurement, STD2 M1 2005 HSC 28a

The Mitchell family has moved to a new house which has an empty swimming pool. The base of the pool is in the shape of a rectangle, with a semicircle on each end.

2005 28a1

  1. Explain why the expression for the area of the base of the pool is  `2xy + πy^2`.   (1 mark) 

      
            2005 28a2
      

The pool is 1.1 metres deep.

  1. The sides and base of the pool are covered in tiles. If  `x =6`  and  `y = 2.5`, find the total area covered by tiles. (Give your answer correct to the nearest square metre.)   (4 marks)

Before filling the pool, the Mitchells need to install a new shower head, which saves 6 litres of water per minute.

The shower is used 5 times every day, for 3 minutes each time.

  1. If the charge for water is $1.013 per kilolitre, how much money would be saved in one year by using this shower head? (Assume there are 365 days in a year.)   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(80 m)^2\ text{(nearest m}^2text{)}`
  3. `$33.28\ \ text{(nearest cent)}`
Show Worked Solution
i.    `text(Area of base)` `=\ text(Area of rectangle +)`
    `\ \ \ \ \ \2 xx text(Area of semi-circle)`
    `= (x xx 2y) + 2 xx (1/2 xx pi xx y^2)`
    `= 2xy + piy^2`

 

ii.   `text(Area of base)` `= (2 xx 6 xx 2.5) + (pi xx 2.5^2)`
    `= 49.634…\ text(m²)`

 

`text(Area of walls) = text(Length) xx text(Height)`

`text(Length)` `= 2x + 2 xx text(semi-circle perimeter)`
  `= (2 xx 6) + 2 xx (1/2 xx 2 xx pi xx 2.5)`
  `= 12 + 15.707…`
  `= 27.707…\ text(m)`

 

`:.\ text(Area of walls)` `= 27.707 xx 1.1`
  `= 30.478…\ text(m)^2`

 

`:.\ text(Total Area covered by tiles)`

`= 49.634… + 30.478…`

`= 80.11…`

`= 80\ text{m²  (nearest m²)}`

 

iii.  `text(Water saved)` `= 5 xx 3 xx 6`
    `= 90\ text(L per day.)`

 

`text(Water saved per year)`

`= 90 xx 365`

`= 32\ 850\ text(L)`

`= 32.85\ text(kL)`

`:.\ text(Money saved)` `= 32.85 xx $1.013`
  `= $33.277…`
  `= $33.28\ text{(nearest cent)}`

Statistics, STD2 S1 2005 HSC 27d

Nine students were selected at random from a school, and their ages were recorded.
 

2UG-2005-27d 

  1. What is the sample standard deviation, correct to two decimal places?   (2 marks)

  2. Briefly explain what is meant by the term standard deviation.   (1 mark)

Show Answers Only
  1. `text{1.69  (to 2 d.p.)}`
  2. `text(Standard deviation is a measure of how much)`

     

    `text(members of a data group differ from the mean)`

     

    `text(value of the group)`

Show Worked Solution

i.  `text(Sample standard deviation)`

`= 1.6914…\ text{(by calculator)}`

`= 1.69\ \ \ text{(to 2 d.p.)}`

 

ii.  `text(Standard deviation is a measure of how much)`

`text(members of a data group differ from the mean)`

`text(value of the group.)`

Algebra, 2UG 2004 HSC 28b

2004 28b

A set of garden gnomes is made so that the cost (`$C`) varies directly with the cube of the base length (`b` centimetres). A gnome with a base length of `text(10 cm)` has a cost of `$50`.

  1. Write an equation relating the variables `C` and `b`, and a constant `k`.  (1 mark)
  2. Find the value of `k`.  (1 mark)
  3. Felicity says, ‘If you double the base length, you double the cost.’ Is she correct? Justify your answer with mathematical calculations.  (2 marks)

 

Show Answers Only
  1. `C = kb^3`
  2. `0.05`
  3. `text(Felicity is incorrect, because doubling)`
    `text(the base from 10 cm to 20 cm causes the)`
    `text(cost to increase 8 times.)`

 

Show Worked Solution
(i)  `C` `∝ b^3`
 `:.C` `= kb^3`

 

(ii)  `C = 50\ \ text(when)\ \ b = 10`

`50` `= k xx 10^3`
`:.k` `= 50/10^3`
  `= 0.05`

 

(iii)  `text(If)\ b = 20`

`C` `= 0.05 xx 20^3`
  `= $400`

 

`:.\ text(Felicity is incorrect, because doubling)`

`text(the base from 10 cm to 20 cm causes the)`

`text(cost to increase 8 times.)`

Algebra, STD2 A4 2004 HSC 28a

A health rating, `R`, is calculated by dividing a person’s weight, `w`, in kilograms by the square of the person’s height, `h`, in metres.

  1. Fred is 150 cm and weighs 72 kg. Calculate Fred’s health rating.  (1 mark)

  2. Over several years, Fred expects to grow 10 cm taller. By this time he wants his health rating to be 25. How much weight should he gain or lose to achieve his aim? Justify your answer with mathematical calculations.  (2 marks)

Show Answers Only
  1. `32`
  2. `text(8 kg)`
Show Worked Solution

i.  `R = w/h^2`

`text(When)\ \ w = 72\ \ and\ \ h = 1.5\ text(m)`

`R` `= 72/1.5^2`
  `= 32`

 

ii.  `text(Find)\ \ w\ \ text(if)\ \ R = 25\ \ and\ \ h = 1.6`

`25` `= w/1.6^2`
`w` `= 25 xx 1.6^2`
  `= 64\ text(kg)`

 

`:.\ text(Weight Fred should lose)`

`= 72 – 64`

`= 8\ text(kg)`

Financial Maths, STD2 F1 2006 HSC 22 MC

This income tax table is used to calculate Evelyn’s tax payable.
 

Evelyn’s taxable income increases from $50 000 to $80 000.

What percentage of her increase will she pay in additional tax?

  1. `text(15.25%)`
  2. `text(40.7%)`
  3. `text(43.5%)`
  4. `text(52%)`
Show Answers Only

`B`

Show Worked Solution
`text(Tax on $50 000)` `= 2500 + 0.35 xx (50\ 000-45\ 000)`
  `= 2500 + 1750`
  `= $4250`
`text(Tax on $80 000)` `= 11\ 250 + 0.52 xx (80\ 000-70\ 000)`
  `= 11\ 250 + 5200`
  `= $16\ 450`
`:.\ text(Extra tax)` `= 16\ 450-4250`
  `= $12\ 200`

 

`:.\ text(% Increase paid in tax)`

`= (12\ 200) / (30\ 000) xx 100`

`=\ text(40.66… %)`
 

`=>  B`