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Binomial, EXT1 2008 HSC 6c

Let  `p`  and  `q`  be positive integers with  `p ≤ q`.

  1. Use the binomial theorem to expand  `(1 + x) ^(p+ q)`, and hence write down the term of
     
  2. `((1 + x)^(p + q))/(x^q)`  which is independent of  `x`.  (2 marks)
     

  3. Given that
     
  4. `((1 + x)^(p + q))/(x^q) = (1 + x)^p(1 + 1/x)^q`,
     
    apply the binomial theorem and the result of part(i) to find a simpler expression for
     
  5. `1 + ((p),(1))((q),(1)) + ((p),(2))((q),(2)) + … + ((p),(p))((q),(p))`.  (3 marks)

 

 

Show Answers Only
  1. `\ ^(p + q)C_q`
  2. `\ ^(p + q)C_q`
Show Worked Solution

(i)   `(1 + x)^(p + q)`

`=\ ^(p + q)C_0 +\ ^(p + q)C_1 x + … +\ ^(p + q)C_q x^q + … +\ ^(p + q)C_(p + q) · x^(p + q)`

`:.\ text(Independent term of)\ ((1 + x)^(p + q))/(x^q)`

`= (\ ^(p + q)C_q·x^q)/(x^q)`

`=\ ^(p + q)C_q`

 

(ii)   `(1 + x)^p(1 + 1/x)^q`

`= (\ ^pC_0 +\ ^pC_1 x + … +\ ^pC_p x^p)`

`xx (\ ^qC_0 +\ ^qC_1 · 1/x + … +\ ^qC_p · 1/(x^p) + … +\ ^qC_q · 1/(x^q))`
 

`text(The independent term in this expansion)`

`=\ ^pC_0 ·\ ^qC_0 +\ ^pC_1 ·\ ^qC_1 + … +\ ^pC_p ·\ ^qC_p\ text{(since}\ p ≤ q)`

`= 1 +\ ^pC_1 ·\ ^qC_1 + … +\ ^pC_p ·\ ^qC_p`
 

`text(S)text(ince)\ ((1 + x)^(p + q))/(x^q) = (1 + x)^p(1 + 1/x)^q,\ text(the independent)`

`text(terms are equal.)`
 

`:.\ ^(p + q)C_q\ text(is a simpler expression for)`

`1 +\ ^pC_1 ·\ ^qC_1 +\ ^pC_2 ·\ ^qC_2 + … +\ ^pC_p ·\ ^qC_p`

Polynomials, EXT2 2015 HSC 12b

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

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

 

Show Answers Only
  1. `1 +- i,\ \ 1 +- 2i`
  2. `x^2 − 2x + 2\ \  or\ \  x^2 − 2x + 5`
Show Worked Solution

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

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

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

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

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

 

`text(Products of roots)`

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

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

`:.b = +- 1`

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

 

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

 

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`

Mechanics, EXT2 2011 HSC 5a

A small bead of mass  `m`  is attached to one end of a light string of length  `R`. The other end of the string is fixed at height  `2h`  above the centre of a sphere of radius  `R`, as shown in the diagram. The bead moves in a circle of radius  `r`  on the surface of the sphere and has constant angular velocity  `omega > 0`. The string makes an angle of  `theta`  with the vertical.

Three forces act on the bead: the tension force  `F`  of the string, the normal reaction force  `N`  to the surface of the sphere, and the gravitational force  `mg`.

  1. By resolving the forces horizontally and vertically on a diagram, show that
    1. `F sin theta - N sin theta = m omega^2 r`
  2. and
    1. `F cos theta + N cos theta = mg.`  (2 marks)
  3. Show that
    1. `N = 1/2 mg sec theta - 1/2 m omega^2 r\ text(cosec)\ theta.`  (2 marks)
  4. Show that the bead remains in contact with the sphere if  
    1. `omega <= sqrt (g/h).`  (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(Resolving forces horizontally)`

`text(Net force)` `= m r omega^2\ \ \ text{(towards the centre of the circle)}`
`F sin theta – N sin theta` `= mr omega^2\ \ \ \ text{… (1)}`
♦ Mean mark part (i) 50%.

 

`text(Resolving forces vertically)`

`text(Net force)` `= mg\ \ \ \ text{(gravitational force)}`
`F cos theta + N cos theta` `= mg\ \ \ \ text{… (2)}`

 

(ii)      `text{Divide (1) by}\ \ sin theta`

`F – N = mr omega^2\ text(cosec)\ theta\ \ \ \ text{… (3)}`

`text{Divide (2) by}\ \ cos theta`

`F+N = mg sec theta\ \ \ \ text{… (4)}`

`text{Subtract (4) – (3)}`

`2N` `= mg sec theta – mr omega^2\ text(cosec)\ theta`
`:.N` `= 1/2 mg sec theta – 1/2 mr omega^2\ text(cosec)\ theta`

 

(iii)  `text(When in contact with the sphere,)\ \ N >= 0.`

`1/2 mg sec theta – 1/2 mr omega^2` `>= 0`
`1/2 mg sec theta` `>= 1/2 mr omega^2\ text(cosec)\ theta`
`g sec theta` `>= r omega^2\ text(cosec)\ theta`
`omega^2` `<= (g sec theta)/(r\ text(cosec)\ theta)`
  `<= g/r tan theta,\ \ \ \ (text{since}\ \ tan theta = r/h)`
  `<= g/r xx r/h`
  `<=g/h`
`:. omega` `<= sqrt (g/h)`

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))}`

Harder Ext1 Topics, EXT2 2014 HSC 16a

The diagram shows two circles `C_1` and `C_2` . The point `P` is one of their points of intersection. The tangent to `C_2` at `P` meets `C_1` at `Q`, and the tangent to `C_1` at `P` meets `C_2` at `R`.

The points `A` and `D` are chosen on `C_1` so that `AD` is a diameter of `C_1` and parallel to `PQ`. Likewise, points `B` and `C` are chosen on `C_2` so that `BC` is a diameter of `C_2` and parallel to `PR`.

 

The points `X` and `Y` lie on the tangents `PR` and `PQ`, respectively, as shown in the diagram.

Harder Ext1 Topics, EXT2 2014 HSC 16a 

 

Copy or trace the diagram into your writing booklet.

  1. Show that `∠APX = ∠DPQ`.  (2 marks)
  2. Show that `A`, `P` and `C` are collinear.  (3 marks)
  3. Show that `ABCD` is a cyclic quadrilateral. (1 mark)
Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `text(See Worked Solutions.)`
  3. `text(See Worked Solutions.)`
Show Worked Solution
(i)

Harder Ext1 Topics, EXT2 2014 HSC 16a Answer

 
 `∠APX` `= ∠ADP\ \ \ text{(angle in alternate segment)`βαγ
 `∠ADP` `= ∠DPQ\ \ \ text{(alternate angles,}\ AD\ text(||)\ PQ)`
`:.∠APX` `= ∠DPQ`

 

(ii)  `text(Join)\ PC\ text(and)\ PB.`

♦♦ Mean mark 31%.
MARKER’S COMMENT: Be vigilant that your reasoning does not implicitly assume `APC` or `DPB` are straight lines.

`text{Let}\ \ ∠APX = ∠DPQ=α\ \ \ text{(from part (i))}`

`text{Similarly,}\ \ ∠YPB = ∠CPR=β`

`∠XPY = ∠RPQ=γ\ \ \ text{(vertically opposite angles.)}`

`∠APD` `= 90^@\ \ \ text{(angle in a semicircle in}\ C_1)`
`∠CPB` `= 90^@\ \ \ text{(angle in a semicircle in}\ C_2)`

 

`90^@ + 90^@ + 2(α+β+γ)` `= 360^@ \ \ text{(sum of angles at a point}\ Ptext{)}`
`:. (α+β+γ)` `= 90^@`
`∠APC` `=90+(α+β+γ)=180^@`

 

`:.\ A, P\ text(and)\ C\ text(are collinear.)`

 

♦ Mean mark 44%.

(iii)  `:.∠APX = ∠CPR\ text{(vertically opposite angles,}\ APC\ text{is a straight line)}`

`:.α=β\ \ =>∠BCA = ∠BDA`

`text(S)text(ince chord)\ AB\ text(subtends a pair of equal angles at)\ C and D`

`=>ABCD\ text(is a cyclic quadrilateral.)`

Plane Geometry, EXT1 2007 HSC 4c

EXT1 2007 4c

The diagram shows points `A`, `B`, `C` and `D` on a circle. The lines  `AC`  and  `BD`  are perpendicular and intersect at  `X`. The perpendicular to  `AD`  through  `X`  meets  `AD`  at  `P`  and  `BC`  at  `Q`.

Copy or trace this diagram into your writing booklet.

  1. Prove that  `∠QXB =∠QBX`.  (3 marks)
  2. Prove that  `Q`  bisects  `BC`.  (2 marks)
Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution
(i)   

 `text(Prove)\ \ ∠QXB =∠QBX`

MARKER’S COMMENT: Many students did not copy the diagram into their answer! Efficient solutions labelled angles with a symbol.

`∠ADX = ∠ACB = theta`

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

 

`text(In)\ ΔDPX`

`∠DPX` `= 90^@\ \ (PQ ⊥ AD)`
`∠PXD` `= (90 – theta)^@\ \ text{(angle sum of}\ Δ DPX)`
`∠QXB` `= (90 – theta)^@\ \ text{(vertically opposite angle)}`

 

`text(In)\ Δ BXC`

`∠BXC` `= 90^@\ \ (∠AXC\ text{is a straight angle)}`
`∠QBX` `= (90 – theta)^@\ \ text{(angle sum of}\ ΔBXC)`
`:.∠QXB` `= ∠QBX`

 

(ii)  `text(Prove)\ \ Q\ \ text(bisects)\ \ BC`

`BQ = QX\ \ ` `text{(sides opposite equal angles}`
  `text{of isosceles}\ Delta BXQ text{)}`
`∠QXC` `= 180 − 90 − (90 − theta)\ \ (∠AXC\ text{is a straight angle)}`
  `= theta`
`∠XCB` `=theta\ \ \ text{(from part (i))}`
`:. ΔXQC\ text(is isosceles)`

 

`QX = QC\ \ ` `text{(sides opposite base angles}`
  `text{of isosceles}\ ΔQXC)`

`:. BQ = QC`

`:. Q\ text(bisects)\ BC`

Quadratic, EXT1 2005 HSC 4c

The points  `P (2ap, ap^2)`  and  `Q (2aq, aq^2)`  lie on the parabola  `x^2 = 4ay`.

The equation of the normal to the parabola at  `P`  is  `x + py = 2ap + ap^3`  and the equation of the normal at  `Q`  is similarly given by  `x + qy = 2aq + aq^3.`

  1. Show that the normals at  `P`  and  `Q`  intersect at the point  `R`  whose coordinates are
    1. `(–apq[p + q], a[p^2 + pq + q^2 + 2]).`  (2 marks)

  2. The equation of the chord  `PQ`  is
    1. `y = 1/2 (p + q) x - apq.` (Do NOT show this.)
  4. If the chord  `PQ`  passes through  `(0, a)`, show that  `pq = –1.`  (1 mark)

  6. Find the equation of the locus of  `R`  if the chord  `PQ`  passes through  `(0, a).`  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `y = x^2/a + 3a`
Show Worked Solution

(i)   `text(Show)\ \ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`

 

ext1 2005 4c

`text(Equations of normals through)\ \  P and Q`

`x + py` `= 2ap + ap^3` `\ \ text{… (1)}`
`x + qy` `= 2aq + aq^3` `\ \ text{… (2)}`

 

`text{Multiply (1)} xx q\ ,\ \ text{(2)} xx p`

`qx + pqy` `= 2apq + ap^3q` `\ \ text{… (3)}`
`px + pqy` `= 2apq + apq^3` `\ \ text{… (4)}`

 

`text{Subtract  (4) – (3)}`

`x (p – q)` `= apq^3 – ap^3q`
  `= apq (q^2 – p^2)`
  `= apq (q – p) (q + p)`
  `= -apq (p -q) (p + q)`
`x` `= -apq [p + q]`

 

`text(Substitute)\ \ x = -apq [p + q]\ \ text{into  (1)}`

`py – apq[p + q]` `= 2ap + ap^3`
`py` `= 2ap + ap^3 + apq (p + q)`
`y` `= 2a + ap^2 + aq (p + q)`
  `= 2a + ap^2 + apq + aq^2`
  `= a[p^2 + pq + q^2 + 2]`

 

`:.\ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`

`text(…  as required.)`

 

(ii)  `PQ\ \ text(is)\ \ y = 1/2 (p + q)x – apq`

`text(Passes)\ \ (0, a)`

`a` `= 1/2 (p + q)0 – apq`
`a` `= -apq`
`:. pq` `= -1\ \ text(…  as required)`

 

(iii)   `R\ \ text(has coordinates)`

`x` `= -apq [p + q]`
`x` `= a(p + q)\ \ \ text{(using part (ii))}`
 `x/a` `=(p + q)`

 

`y` `= a [p^2 + pq + q^2 + 2]`
  `= a (p^2 – 1 + q^2 + 2)`
  `= a (p^2 + q^2 +1)`
  `= a [(p + q)^2 – 2pq + 1]`
  `= a [(p + q)^2 + 3]`
  `= a [(x/a)^2 + 3]`
  `= x^2/a + 3a`

 

`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = x^2/a + 3a`

Differentiation, EXT1 2005 HSC 3c

Use the definition of the derivative,

`f prime (x) = lim_(h -> 0) (f (x + h) - f(x))/h,`

to find `f prime (x)` when

`f(x) = x^2 + 5x.`  (2 marks)

Show Answers Only

`2x + 5`

Show Worked Solution

`f(x) = x^2 + 5x`

`f(x + h)` `= (x + h)^2 + 5(x + h)`
  `= x^2 + 2xh + h^2 + 5x + 5h`

 

`f prime (x)` `= lim_(h -> 0) (f (x + h) – f(x))/h`
  `= lim_(h -> 0) (x^2 + 2xh + 5x + h^2 + 5h – x^2 – 5x)/h`
  `= lim_(h -> 0) (2xh + h^2 + 5h)/h`
  `= lim_(h -> 0) 2x + h + 5`
  `= 2x + 5`

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

Calculus, EXT1* C3 2007 HSC 9a

2007 9a
  

The shaded region in the diagram is bounded by the curve  `y = x^2 + 1`, the `x`-axis, and the lines  `x = 0`  and  `x = 1.`

Find the volume of the solid of revolution formed when the shaded region is rotated about the `x`-axis.  (3 marks)

Show Answers Only

`(28 pi)/15\ \ text(u³)`

Show Worked Solution
`V` `= pi int_0^1 y^2\ dx`
  `= pi int_0^1 (x^2 + 1)^2\ dx`
  `= pi int_0^1 x^4 + 2x^2 + 1\ dx`
  `= pi [1/5 x^5 + 2/3 x^3 + x]_0^1`
  `= pi[(1/5 + 2/3 + 1) – 0]`
  `= pi [3/15 + 10/15 + 1]`
  `= (28 pi)/15\ \ text(u³)`

Plane Geometry, 2UA 2007 HSC 8b

In the diagram, `AE` is parallel to `BD`, `AE = 27`, `CD = 8`, `BD = p`, `BE = q` and `/_ABE`, `/_BCD` and `/_BDE` are equal.

Copy or trace this diagram into your writing booklet.

  1. Prove  that `Delta ABE\ text(|||)\ Delta BCD`.  (2 marks)
  2. Prove  that `Delta EDB\ text(|||)\ Delta BCD`.  (2 marks)
  3. Show that `8`, `p`, `q`, `27` are the first four terms of a geometric series.  (1 mark)
  4. Hence find the values of `p` and `q`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solution)}`
  2. `text(Proof)\ \ text{(See Worked Solution)}`
  3. `text(Proof)\ \ text{(See Worked Solution)}`
  4. `p = 12, q = 18`
Show Worked Solution

(i)

`text(Prove)\ Delta ABE\ text(|||)\  Delta BCD`

`/_ ABE = /_ BCD\ \ \ text{(given)}`

`/_ EAB = /_ DBC\ \ \ text{(corresponding angles,}\ AE\ text(||)\ BD text{)}`

`:. Delta ABE\ text(|||)\ Delta BCD\ \ \ text{(equiangular)}`

 

(ii)  `text(Prove)\ Delta EDB\ text(|||)\ Delta BCD`

`/_ EDB = /_ BCD\ \ \ text{(given)}`

`/_ CDB = 180° – (/_ BCD + /_ DBC)\ \ \ text{(Angle sum of}\ Delta BCD text{)}`

`/_ EBD` `= 180° – (/_ ABE + /_ DBC)\ \ \ text{(}/_ ABC\ text{is a straight angle)}`
  `= 180° – (/_ BCD + /_ DBC)\ \ \ text{(}/_ ABE = /_ BCD,\ text{given)}`
  `= /_ CDB`

`:. Delta EDB\ text(|||)\ Delta BCD\ \ \ text{(equiangular)}`

 

(iii)  `text(In a GP,)\ \ r = T_n/T_(n-1)`

`text(If)\ \ \ 8, p, q, 27\ \ \ text(are 1st 4 terms of a GP)`

`=> p/8 = q/p = 27/q .`

`text(S)text(ince corresponding sides of similar)`

`text(triangles are in the same ratio)`

`(BD)/(DC) = (EB)/(BD) = (AE)/(EB)`

`p/8 = q/p = 27/q .`

 

`:. 8, p, q, 27\ \ text(are 1st 4 terms of a GP.)`

 

(iv)  `8, p, q, 27`

`a = 8`

`text(Using)\ \ T_n = ar^(n-1)`

`T_4=ar^3`

`27` `= 8 xx r^3`
`r^3` `= 27/8`
`r` `= 3/2`

 

`T_2` `=ar`
`:.p` `= 8 * 3/2`
  `= 12`

 

`T_3` `=ar^2`
`:.q` `= 8 * (3/2)^2`
  `= 18`

Quadratic, 2UA 2007 HSC 7a

  1. Find the coordinates of the focus, `S`, of the parabola  `y = x^2 + 4`.  (2 marks)
  2. The graphs of  `y = x^2 + 4`  and the line  `y = x + k`  have only one point of intersection, `P`. Show that the `x`-coordinate of `P` satisfies.
    1. `x^2 - x + 4 - k = 0`.  (1 mark)
  3. Using the discriminant, or otherwise, find the value of `k`.  (1 mark)
  4. Find the coordinates of `P`.  (2 marks)
  5. Show that `SP` is parallel to the directrix of the parabola.  (1 mark)
Show Answers Only
  1. `(0, 4 1/4)`

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

  3. `15/4`

  4. `(1/2, 4 1/4)`

  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   `y` `= x^2 + 4`
`x^2` `= y – 4`

`text(Using)\ (x – x_0) = 4a (y – y_0)`

`x_0` `= 0`
`a` `= 1/4`
`y_0` `= 4`

`(x – 0) = 4 xx 1/4 xx (y – 4)`

`text(Vertex)` `= (0, 4)`
`:.\ text(Focus)` `= (0, 4 1/4)`

 

(ii)  `y` `= x^2 + 4` `\ \ …\ \ (1)`
`y` `= x + k` `\ \ …\ \ (2)`

 

`text(Solving simultaneously)`

`x^2 + 4` `= x + k`
`x^2 – x + 4 – k` `= 0`

 

(iii)  `text(If only 1 point of intersection,)`

`b^2 – 4ac` `= 0`
`(–1)^2 – 4 xx 1 xx (4 – k)` `= 0`
`1 – 16 + 4k` `= 0`
`4k` `= 15`
`k` `= 15/4`

 

(iv)  `text(Finding)\ \ P`

`x^2 – x + 4 – 15/4 = 0`

`x^2 – x + 1/4 = 0`

`(x – 1/2)^2 = 0`

`x = 1/2`

`text(When)\ x = 1/2`

`y` `= (1/2)^2 + 4`
  `= 4 1/4`

`:. P\ text(has coordinates)\ \  (1/2, 4 1/4)`

 

(v)  `text(Focus)\ (S)` `= (0, 4 1/4)`
`P` `= (1/2, 4 1/4)`

 

`:.\ text(S)text(ince)\ y text(-values are the same,)\ SP\ text(is parallel)`

`text(with the)\ x text(-axis.)`

`=>text(Directrix has the equation)\ \ y = 3 3/4`

`:. SP\ text(is parallel with the directrix.)`

Plane Geometry, 2UA 2007 HSC 5a

In the diagram, `ABCDE` is a regular pentagon. The diagonals `AC` and `BD` intersect at `F`.

Copy or trace this diagram into your writing booklet.

  1. Show that the size of `/_ABC` is `108°`.  (1 mark)
  2. Find the size of `/_BAC`. Give reasons for your answer.  (2 marks)
  3. By considering the sizes of angles, show that `Delta ABF` is isosceles.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `36°`
  3. `text(See Worked Solutions)`
Show Worked Solution
(i)

`text(Sum of all internal angles`

`= (n – 2) xx 180°`

`= (5 – 2) xx 180°`

`= 540°`

`:. /_ABC` `= 540/5= 108°`

 

(ii)  `BA = BC`

`text{(equal sides of a regular pentagon)}`

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

`/_BAC` `= 1/2 (180 – 108)\ \ \ text{(base angle of}\ Delta BAC text{)}`
  `= 36°`

 

(iii)  `text(Consider)\ Delta BCD and Delta ABC`

`BC = CD = BA`

`text{(equal sides of a regular pentagon)}`

`/_BCD = /_ABC = 108°`

`text{(internal angles of a regular pentagon)}`

`:. Delta BCD -= Delta ABC\ \ \ text{(SAS)}`

`:. CBF` `= 36°` `text{(corresponding angles in}`
     `text{congruent triangles)}`
`/_FBA` `= 108 – 36`
  `= 72°`
`/_BFA` `= 180 – (72 + 36)\ \ \ \ text{(angle sum of}\ Delta ABF text{)}`
  `= 72°`

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

Linear Functions, 2UA 2007 HSC 3a

2007 3a

In the diagram, `A`, `B` and `C` are the points `(10, 5)`, `(12, 16)` and `(2, 11)` respectively.

Copy or trace this diagram into your writing booklet.

  1. Find the distance `AC`.  (1 mark)
  2. Find the midpoint of `AC`.  (1 mark)
  3. Show that `OB_|_AC`.  (2 marks)
  4. Find the midpoint of `OB` and hence explain why `OABC` is a rhombus.  (2 marks)
  5. Hence, or otherwise, find the area of `OABC`.  (1 mark)
Show Answers Only
  1. `10\ text(units)`
  2. `(6, 8)`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `(6, 8)\ \ text(See Worked Solutions)`
  5. `100\ text(u²)`
Show Worked Solution

(i)   `A (10, 5)\ \ \ C (2, 11)`

`text(dist)\ AC` `= sqrt {(x_2 – x_1)^2 + (y_2 – y_1)^2}`
  `= sqrt {(2 – 10)^2 + (11 – 5)^2}`
  `= sqrt (64 + 36)`
  `= sqrt 100`
  `= 10\ text(units)`

 

(ii)  `text(Midpoint)\ AC` `= ((x_1 + x _2)/2 , (y_1 + y_2)/2)`
  `= ((10+2)/2 , (5 + 11)/2)`
  `= (6, 8)`

 

(iii)  `B (12, 16)`

`M_(OB)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (16 – 0)/(12 – 0)`
  `= 4/3`
`M_(AC)` `= (11 – 5)/(2 – 10)`
  `= 6/-8`
  `= -3/4`

`M_(OB) xx M_(AC) = 4/3 xx -3/4 = -1`

`:. OB_|_ AC`

 

(iv)  `text(Midpoint)\ OB` `= ((12 + 0)/2 , (16 + 0)/2)`
  `= (6, 8)`

`text(S)text(ince midpoint)\ OB = text(midpoint)\ AC`

`text(and)\ OB_|_AC`

`=>  text(Diagonals of)\ OABC\ text(are perpendicular)`

`text(bisectors)`

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

 

(v)  `text(dist)\ OB`

`= sqrt ((12 – 0)^2 + (16 – 0)^2)`

`= sqrt (144 + 256)`

`= sqrt 400`

`= 20\ text(units)`

`text(Area of)\ OABC`

`= 1/2 xx AC xx OB`

`= 1/2 xx 10 xx 20`

`= 100\ text(u²)`

Functions, 2ADV F1 2007 HSC 1c

Rationalise the denominator of  `1/(sqrt 3 - 1)`.  (2 marks)

Show Answers Only

`(sqrt 3 + 1)/2`

Show Worked Solution

`1/(sqrt 3 -1) xx (sqrt 3 + 1)/(sqrt 3 + 1)`

`= (sqrt 3 + 1)/{(sqrt 3)^2 – 1^2}`

`= (sqrt 3 + 1)/2`

Measurement, 2UG 2015 HSC 28c

Three equally spaced cross-sectional areas of a vase are shown.

2UG 2015 29c

Use Simpson’s rule to find the approximate capacity of the vase in litres.  (3 marks)

Show Answers Only

`4\ text(litres)`

Show Worked Solution

2UG 2015 29c Answer 

`V` `≈ h/3[y_0 + 4y_1 + y_2]`
  `≈ 15/3[45 + 4(180) + 35]`
  `≈ 5[800]`
  `≈ 4000\ text(cm³)`
  `~~4\ text(litres)\ \ \ text{(1 cm³ = 1 mL)}`

FS Comm, 2UG 2015 HSC 27e

A `42` megabyte (MB) file is to be downloaded at a rate of `500` kilobits per second (kbps), where `1` kilobit = `1000` bits.

How long would it take to download this file? Give your answer in minutes and seconds, correct to the nearest second.  (3 marks)

Show Answers Only

`text{11 mins 45 seconds  (nearest second)}`

Show Worked Solution
`42\ text(MB)` `= 42 xx 2^20`
  `= 44\ 040\ 192\ \ text(bytes)`
  `= 352\ 321\ 536\ \ text{bits  (1 byte = 8 bits)}`

 

♦ Mean mark 50%.

`text(Download speed = 500 000 bits per second)`

`:.\ text(Download time)` `= (352\ 321\ 536)/(500\ 000)`
  `= 704.64…`
  `=11\ text{mins 45 seconds  (nearest second)}`

FS Comm, 2UG 2015 HSC 26g

Pat’s mobile phone plan is shown.

2015 26g

Last month Pat:

• made calls to the value of `$561`
• sent `152` SMS messages
• sent `37` MMS messages
• used `1.7` GB of data.

What was the total of Pat’s phone bill for last month?  (3 marks)

Show Answers Only

`$96.36`

Show Worked Solution

`text(C)text(ost of excess calls)`

`= 561 − 550`

`= $11`

`text(Excess data)`

`1.7\ text(GB)` `= 1.7 xx 2^(10)\ text(MB)`
`text(Excess)` `= 1.7 xx 2^(10) − 500`
  `= 1240.8\ text(MB)`

 

`:.\ text(C)text(ost of excess data)`

`= 1240.8 xx $0.0293`

`= $36.355…`

`= $36.36`

 

`text(SMS and MMS messages within plan)`

`:.\ text(Total phone bill)`

`= $49 + $11 + $36.36`

`= $96.36`

Data, 2UG 2015 HSC 26a

A farmer used the ‘capture‑recapture’ technique to estimate the number of chickens he had on his farm. He captured, tagged and released 18 of the chickens. Later, he caught 26 chickens at random and found that 4 had been tagged.

What is the estimate for the total number of chickens on this farm?  (2 marks)

Show Answers Only

`117`

Show Worked Solution

`text(Let)\ C = text(total chickens on farm)`

`text(Capture)` `= 18/C`
`text(Recapture)` `= 4/26`
`⇒ 18/C` `= 4/26`
`4C` `= 18 xx 26`
`C` `= (18 xx 26)/4`
  `= 117`

 

`:.\ text(There are an estimated 117 chickens)`

`text(on the farm.)`

Probability, 2UG 2015 HSC 18 MC

A Student Representative Council (SRC) consists of five members. Three of the members are being selected to attend a conference.

In how many ways can the three members be selected?

(A)   `10`

(B)   `20`

(C)   `30`

(D)   `60`

Show Answers Only

`A`

Show Worked Solution
♦ Mean mark 37%.
`#\ text(Combinations)` `= (5 xx 4 xx 3)/(3 xx 2 xx 1)`
  `= 10`

`⇒ A`

Measurement, 2UG 2015 HSC 14 MC

Stockholm is located at `text(59°N 18°E)` and Darwin is located at `text(13°S 131°E)`. 

What is the time difference between Stockholm and Darwin? (Ignore time zones and daylight saving.)

(A)   `184` minutes

(B)   `288` minutes

(C)   `452` minutes

(D)   `596` minutes

Show Answers Only

`C`

Show Worked Solution

`text(Stockholm is 59°N 18°E,  Darwin is 13°S 131°E)`

`text{Angular difference (longitude)}`

`= 131^@− 18^@`

`= 113^@`
 

`:.\ text(Time difference)` `= 113 xx 4`
  `= 452\ text(minutes)`

`⇒ C`

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, EXT1* C1 2015 HSC 15a

The amount of caffeine, `C`, in the human body decreases according to the equation

`(dC)/(dt) = -0.14C,` 

where `C` is measured in mg and `t` is the time in hours.

  1. Show that  `C = Ae^(-0.14t)`  is a solution to  `(dC)/(dt) = -0.14C,` where ` A` is a constant.

     

    When `t = 0`, there are 130 mg of caffeine in Lee’s body.  (1 mark)

  2. Find the value of `A.`  (1 mark)

  3. What is the amount of caffeine in Lee’s body after 7 hours?   (1 mark)

  4. What is the time taken for the amount of caffeine in Lee’s body to halve?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `130`
  3. `48.8\ text(mg)`
  4. `4.95\ text(hours)`
Show Worked Solution
i. `C` `= Ae^(-0.14t)`
  `(dC)/(dt)` `= d/(dt) (Ae^(-0.14t))`
    `= -0.14 xx Ae^(-0.14t)`
    `= -0.14\ C`

 
`:.\ C = Ae^(-0.14t)\ \ text(is a solution)`

 

ii.  `text(When)\ \ t = 0,\ C = 130`

`130` `= Ae^(-0.14 xx 0)`
`:.\ A` `= 130`

 

iii.  `text(Find)\ \ C\ \ text(when)\ \ t = 7`

`C` `= 130\ e^(-0.14 xx 7)`
  `= 130\ e^(-0.98)`
  `= 48.79…`
  `= 48.8\ text{mg  (to 1 d.p.)}`

 
`:.\ text(After 7 hours, Lee will have 48.8 mg)`

`text(of caffeine left in her body.)`

 

iv.  `text(Find)\ \ t\ \ text(when caffeine has halved.)`

`text(When)\ \ t = 0,\ \ C = 130`

`:.\ text(Find)\ \ t\ \ text(when)\ \ C = 65`

`65` `= 130 e^(-0.14 xx t)`
`e^(-0.14t)` `= 65/130`
`ln e^(-0.14t)` `= ln\ 65/130`
`-0.14t xx ln e` `= ln\ 65/130`
`t` `= (ln\ 65/130)/-0.14`
  `= 4.951…`
  `= 4.95\ text{hours  (to 2 d.p.)}`

 

`:.\ text(It will take 4.95 hours for Lee’s)`

`text(caffeine to halve.)`

Quadratic, 2UA 2015 HSC 12e

The diagram shows the parabola `y = x^2/2` with focus `S (0, 1/2).` A tangent to the parabola is drawn at `P (1, 1/2).`

  1. Find the equation of the tangent at the point `P`.   (2 marks)
  2. What is the equation of the directrix of the parabola?   (1 mark)
  3. The tangent and directrix intersect at `Q`.
    Show that `Q` lies on the `y`-axis.   (1 mark)

  4. Show that `Delta PQS` is isosceles.   (1 mark)
Show Answers Only
  1. `y = x – 1/2`
  2. `y = -1/2`
  3.  
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   

`y = 1/2 x^2`

`(dy)/(dx) = x`
 

`text(When)\ \ x = 1,\ \ (dy)/(dx) = 1`

`text(Equation of tangent,)\ m = 1,\ text(through)\ (1, 1/2):`

`y – y_1` `= m (x – x_1)`
`y – 1/2` `= 1 (x – 1)`
`y – 1/2` `= x – 1`
`y` `= x – 1/2`

 

(ii)  `text(Directrix is)\ \ y = -1/2`
 

(iii)  `Q\ text(is at the intersection of)`

`y = x – 1/2\ \ …\ \ text{(1)}`

`y = -1/2\ \ \ \ …\ \ text{(2)}`

`text{(1) = (2)}`

`x – 1/2` `= -1/2`
`x` `= 0`

 
`:.\ Q\ text(lies on the)\ y text(-axis)\ \ …\ \ text(as required)`

 

(iv)  `text(Show)\ Delta PQS\ text(is isosceles.)`

`text(Distance)\ PS = 1 – 0 = 1`

`Q\ text(has coordinates)\ (0, -1/2)`

`text(Distance)\ SQ = 1/2 + 1/2 = 1`

`:. PS = SQ = 1`

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

Quadratic, 2UA 2015 HSC 12d

For what values of `k` does the quadratic equation `x^2 – 8x + k = 0` have real roots?  (2 marks)

Show Answers Only

`k <= 16`

Show Worked Solution

`x^2 – 8x + k = 0`

`text(Real roots when)\ Delta >= 0`

`b^2 – 4ac` `>= 0`
`(-8)^2 – 4 xx 1 xx k` `>= 0`
`64 – 4k` `>= 0`
`4k` `<= 64`
`k` `<= 16`

`:.\ text(Real roots exists when)\ k <= 16`

Functions, 2ADV F1 2015 HSC 12b

The diagram shows the rhombus  `OABC`.

The diagonal from the point  `A (7, 11)`  to the point `C` lies on the line `l_1`.

The other diagonal, from the origin `O` to the point `B`, lies on the line `l_2` which has equation  `y = -x/3`.

2015 2ua 12b

  1. Show that the equation of the line  `l_1`  is  `y = 3x - 10`.  (2 marks)
  2. The lines  `l_1`  and  `l_2`  intersect at the point `D`.
  3. Find the coordinates of  `D`.  (2 marks)

 

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

(i)  `text(Show)\ \ l_1\ \ text(is)\ \ y = 3x – 10`

`l_2\ \ text(is)\ \ y = -x/3`

`m_(l_2)` `= -1/3`  
`:.\ m_(l_1)` `= 3\ \ \ ` `text{(diagonals of rhombus}\ OABC\ text(are)`
    `text{perpendicular bisectors.)}`

 
`l_1\ text(has)\ m = 3,\ text(through)\ \ A(7, 11)`

`y – y_1` `= m (x – x_1)`
`y – 11` `= 3 (x – 7)`
`y-11` `= 3x – 21`
`y` `= 3x – 10\ \ text(… as required)`

 

(ii)  `D\ text(occurs at the intersection of)\ \ l_1 and l_2`

`y` `= -1/3\ x` `\ text{…  (1)}`
`y` `= 3x – 10` `\ text{…  (2)}`
`text{(1)}` `=\ text{(2)}`
`-1/3\ x` `= 3x – 10`
`10/3\ x` `= 10`
`:.\ x` `= 3`

 

`text(Substitute)\ \ x = 3\ \ text{into   (1)}`

`y` `= -1/3 xx 3`
  `= -1`
`:.\ D\ text{has coordinates  (3, –1)}`

Trig Calculus, 2UA 2015 HSC 6 MC

What is the value of the derivative of  `y = 2 sin 3x - 3 tan x`  at  `x = 0`?

(A)  `-1`

(B)     `0`

(C)     `3`

(D)  `-9`

Show Answers Only

`C`

Show Worked Solution
`y` `= 2 sin 3x – 3 tan x`
`(dy)/(dx)` `= 6 cos 3x – 3 sec^2 x`

 

`text(At)\ x = 0`

`(dy)/(dx)` `= 6 cos 0 – 3 sec^2 0`
  `= 6 (1) – 3/(cos^2 0)`
  `= 6 – 3`
  `= 3`

 `=> C`

Integration, 2UA 2006 HSC 10a

Use Simpson’s rule with three function values to find an approximation to the value of
 
          `int_0.5^1.5 (log_e x )^3\ dx`.
 

Give your answer correct to three decimal places.  (2 marks)

Show Answers Only

`text(–0.044)\ \ \ text{(to 3 d.p.)}`

Show Worked Solution

`int_0.5^1.5 (log_e x )^3 dx`

`A` `~~ h/3 [y_0 + y_2 + 4(y_1)]`
  `~~0.5/3 [-0.3330… + 0.0666… + 0]`
  `~~ -0.04439…`
  `~~-0.044\ \ \ text{(to 3 d.p.)}`

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)`

Quadratic, 2UA 2006 HSC 9a

Find the coordinates of the focus of the parabola  `12y = x^2 - 6x - 3`.  (2 marks)

Show Answers Only

`(3, 2)`

Show Worked Solution
`12y` `= x^2 – 6x – 3`
  `= x^2 – 6x + 9 – 12`
`12y` `= (x – 3)^2 – 12`
`(x – 3)^2` `= 12y + 12`
  `= 12(y + 1)`

 

`:.\ text{Vertex is (3, –1)}`

`4a` `= 12`
`a` `= 3`

 

`:.\ text(Coordinates of focus are)\ (3, 2).`

Quadratic, 2UA 2006 HSC 7c

  1. Write down the discriminant of  `2x^2 + (k - 2)x + 8`  where  `k`  is a constant.  (1 mark)
  2. Hence, or otherwise, find the values of  `k`  for which the parabola  `y = 2x^2 + kx + 9` does not intersect the line  `y = 2x + 1`.  (2 marks)

 

Show Answers Only
  1. `k^2 – 4k – 60`
  2. `-6 < k < 10`
Show Worked Solution

(i)  `2x^2 + (k – 2)x + 8`

`Delta` `= b^2 – 4ac`
  `= (k – 2)^2 – 4 xx 2 xx 8`
  `= k^2 – 4k + 4 – 64`
  `= k^2 – 4k – 60`

 

(ii)  `y` `= 2x^2 + kx + 9` `\ \ text{…  (1)}`
`y` `= 2x + 1` `\ \ text{…  (2)}`

`text(Substitute)\ y = 2x + 1\ text{into (1)}`

`2x + 1 = 2x^2 + kx + 9`

`2x^2 + kx – 2x + 8 = 0`

`2x^2 + (k – 2)x + 8 = 0\ …\ text{(∗)}`

 

`text{The graphs will not intercept if (∗) has}`

`text(no roots, i.e.)\ \ Delta <0`

`k^2 – 4k – 60` `< 0`
`(k – 10) (k + 6)` `< 0`

HSC quadratic

`text(From the graph, no intersection when)`

`-6 < k < 10`

Quadratic, 2UA 2006 HSC 7a

Let  `alpha`  and  `beta`  be the solutions of  `x^2 - 3x + 1 = 0`.

  1. Find  `alpha beta`.  (1 mark)
  2. Hence find  `alpha + 1/alpha`.  (1 mark)  

 

Show Answers Only
  1. `1`
  2. `3`
Show Worked Solution

(i)  `x^2 – 3x + 1 = 0`

`alpha beta` `= c/a`
  `= 1`

 

(ii)  `alpha + 1/alpha` `= alpha + beta\ \ \ text{(using}\ beta = 1/alpha\ text{from part (i))}`
  `= -b/a`
  `= (-(-3))/1`
  `= 3`

Plane Geometry, 2UA 2006 HSC 6a

In the diagram, `AD` is parallel to `BC`, `AC` bisects `/_BAD` and `BD` bisects `/_ABC`. The lines `AC` and `BD` intersect at `P`.

Copy or trace the diagram into your writing booklet.

  1. Prove that `/_BAC = /_BCA`.  (1 mark)
  2. Prove that `Delta ABP ≡ Delta CBP`.  (2 marks)
  3. Prove that `ABCD` is a rhombus.  (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)}`
Show Worked Solution
(i)  

`text(Prove)\ /_BAC = /_BCA`

`/_BCA` `= /_CAD\ \ \ text{(alternate angles,}\ BC\ text(||)\ AD text{)}`
`/_CAD` `= /_BAC\ \ \ text{(}AC\ text(bisects)\ /_BAD text{)}`
`:. /_BAC` `= /_BCA\ …\ text(as required)`

 

(ii)  `text(Prove)\ Delta ABP ≡ Delta CBP`

`/_BAC` `= /_BCA\ \ \ text{(from part (i))}`
`/_ABP` `= /_CBP\ text{(}BD\ text(bisects)\ /_ABC text{)}`
`BP\ text(is common)`

 

`:. Delta ABP ≡ Delta CBP\ \ text{(AAS)}`

 

(iii)  `text(Using)\ Delta ABP ≡ Delta CBP`

`AP = PC\ \ text{(corresponding sides of congruent triangles)}`

`/_BPA = /_BPC\ \ text{(corresponding angles of congruent triangles)}`

`text(Also,)\ /_BPA = /_BPC = 90^@`

`text{(}/_APC\ text{is a straight angle)}`

 

`text(Considering)\ Delta APD and Delta APB`

`/_DAP` `= /_BAP\ text{(}AC\ text(bisects)\ /_BAD text{)}`
`/_DPA` `= 90^@\ text{(vertically opposite angles)}`
`:. /_DPA` `= /_BPA = 90^@`

`PA\ text(is common)`

`:. Delta APD ≡ Delta APB\ \ text{(AAS)}`

`BP = PD\ \ text{(corresponding sides of congruent triangles)}`

 

 `:. ABCD\ text(is a rhombus as its diagonals are)`

`text(perpendicular bisectors.)`

Measurement, 2UG 2015 HSC 26c

Two cities lie on the same meridian of longitude. One is 40° north of the other.

What is the distance between the two cities, correct to the nearest kilometre? (2 marks)

Show Answers Only

`4468\ text{km   (nearest km)}`

Show Worked Solution
♦ Mean mark 46%.

2UG 2015 23c Answer

`text(Distance between two cities)`

`= text(Arc length)\  AB`

`= 40/360 xx 2 xx pi xx r`

`= 1/9 xx 2 xx pi xx 6400`

`= 4468.04…`

`= 4468\ text{km  (nearest km)}`

Linear Functions, 2UA 2006 HSC 3a

In the diagram, `A, B and C` are the points  `(1, 4), (5, –4) and (–3, –1)`  respectively. The line  `AB`  meets the y-axis at `D`.

  1. Show that the equation of the line  `AB`  is  `2x + y - 6 = 0`.  (2 marks)
  2. Find the coordinates of the point `D`.  (1 mark)
  3. Find the perpendicular distance of the point `C` from the line  `AB`.  (1 mark)
  4. Hence, or otherwise, find the area of the triangle  `ADC`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `D(0, 6)`
  3. `13/sqrt 5\ text(units)`
  4.  
  5. `text(6.5 u²)`
Show Worked Solution

(i)  `text(Show)\ \ AB\ \ text(is)\ \ 2x + y – 6 = 0`

`A (1, 4)\ \ \ B (5, text(–4))`

`m_(AB)` `= (y_2 – y_1) / (x_2 – x_1)`
  `= (-4 – 4) / (5 – 1)`
  `= (-8)/4`
  `= – 2`

 
`:.\ text(Equation of)\ AB, m = -2,\ text(through)\ \ (1,4)`

`y-y_1` `=m(x-x_1)`
`y – 4` `= -2 (x – 1)`
`y – 4` `= -2x + 2`
`2x + y – 6` `= 0\ …\ text(as required)`

 

(ii)  `AB\ text(intersects y-axis at)\ D`

`0 + y – 6` `= 0`
`y` `= 6`

`:. D\ text(has coordinates)\ (0, 6)`

 

(iii)  `C\ text{(–3, –1)}`

`AB\ text(is)\ 2x + y – 6 = 0`

`_|_ text(dist)` `= |\ (ax_1 + by_1 + c)/sqrt (a^2 + b^2)\ |`
  `= |\ (2(−3) + 1(-1) – 6)/sqrt(2^2 + 1^2)\ |`
  `= |\ (-13)/sqrt 5\ |`
  `= 13/sqrt 5\ text(units)`

 

(iv)

 
`text(dist)\ AD` `= sqrt((x_2 – x_1)^2 + (y_2 – y_1)^2)`
  `= sqrt((0 – 1)^2 + (6 – 4)^2`
  `= sqrt (1 + 4)`
  `= sqrt 5`
`text(Area of)\ Delta ADC` `= 1/2 xx b xx h`
  `= 1/2 xx sqrt 5 xx 13/sqrt 5`
  `= 6.5\ text(u²)`

Integration, 2UA 2005 HSC 6a

Five values of the function `f(x)` are shown in the table.

Integration, 2UA 2005 HSC 6a

Use Simpson’s rule with the five values given in the table to estimate

`int_0^20 f(x)\ dx`.  (3 marks)

Show Answers Only

`401 2/3`

Show Worked Solution

Integration, 2UA 2005 HSC 6a Answer

`int_0^20 f(x)\ dx` `= h/3[y_0 + y_4 + 4text{(odds)} + 2text{(evens)}]`
  `= 5/3[y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]`
  `= 5/3[15 + 10 + 4(25 + 18)+ 2(22)]`
  `= 5/3[25 + 172 + 44]`
  `= 401 2/3`

Data, 2UG 2006 HSC 23b

This radar chart was used to display the average daily temperatures each month for two different towns.

2UG-2006-23b

  1. What is the average daily temperature of Town `B` for April?  (1 mark)
  2. In which month do the average daily temperatures of the two towns have the greatest difference?  (1 mark)
  4. In which months is the average daily temperature in Town `B` higher than in Town `A`?  (1 mark)
Show Answers Only
  1. `18.5^@\ \ text{(approx)}`
  2. `text(February)`
  3. `text(May, June, July, August and September)`
Show Worked Solution

(i)   `18.5^@\ \ \ text{(approx)}`

 

(ii)   `text(February)`

 

(iii)  `text(May, June, July, August and September)`

Algebra, STD2 A4 2004 HSC 26a

  1. The number of bacteria in a culture grows from 100 to 114 in one hour.

     

    What is the percentage increase in the number of bacteria?   (1 mark)

  2. The bacteria continue to grow according to the formula  `n = 100(1.14)^t`, where `n` is the number of bacteria after `t` hours.

     

    What is the number of bacteria after 15 hours?   (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\;  &  \;\; 5 \;\;  & \;\; 10 \;\;  & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\;  &  \;\; 193 \;\;  & \;\; 371 \;\;  & \;\; ? \;\; \\
\hline
\end{array}

  1. Use the values of `n` from  `t = 0`  to  `t = 15`  to draw a graph of  `n = 100(1.14)^t`.

     

    Use about half a page for your graph and mark a scale on each axis.   (4 marks)

  2. Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300.   (1 mark)

Show Answers Only
  1. `text(14%)`
  2. `714`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(8.4 hours)`
Show Worked Solution

i.   `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`

 

ii.  `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n` `= 100(1.14)^15`
  `= 713.793\ …`
  `= 714\ \ \ text{(nearest whole)}`

 

iii. 

 

iv.  `text(Using the graph)`

`text(The number of bacteria reaches 300 after)`

`text(approximately 8.4 hours.)`

Measurement, 2UG MM6 SM-Bank 01 MC

 

Mapupu and Minoha are two towns on the equator.

The longitude of Mapupu is `text(16°E)` and the longitude of Minoha is `text(52°W)`.

How far apart are these two towns if the radius of Earth is approximately `6400\ text(km)`?

(A)  `4000\ text(km)`

(B)  `7600\ text(km)`

(C)  `1\ 447\ 600\ text(km)`

(D)  `2\ 734\ 400\ text(km)`

Show Answers Only

`B`

Show Worked Solution
`text(Difference in longitude)` `=52+16`
  `=68º`

 

`:.\ text(Distance between the two towns)`

`=\ text(Length of arc)`

`=68/360 xx 2 xx pi xx r`

`=68/360 xx 2 xx pi xx 6400`

`=7595.67…\ text(km)`

`=> B`

Functions, EXT1 F2 2008 HSC 2c

The polynomial  `p(x)`  is given by  `p(x) = ax^3 + 16x^2 + cx - 120`, where  `a`  and  `c`  are constants.

The three zeros of  `p(x)`  are  `– 2`,  `3`  and  `beta`.

Find the value of  `beta`.   (3 marks) 

Show Answers Only

`- 5`

Show Worked Solution

`p(x) = ax^3 + 16x^2 + cx – 120`

`text(Roots:)\ \ – 2, \ 3, \ beta`

`-2 + 3 + beta` `= -B/A`
`beta + 1` `= -16/a`
`beta` `= -16/a – 1\ \ \ \ \ …\ (1)`

 

`-2 xx 3 xx beta` `= -D/A`
`-6 beta` `= 120/a`
`beta` `= -20/a\ \ \ \ \ …\ (2)`

 

MARKER’S COMMENT: Many students displayed significant inefficiencies in solving simultaneous equations.
`- 16/a – 1` `= -20/beta`
`-16 – a` `= -20`
`a` `= 4`

 
`text(Substitute)\ \ a = 4\ \ text(into)\ (1)`

`:. beta` `= – 16/4 – 1`
  `= -5`

Measurement, 2UG 2008 HSC 28b

A tunnel is excavated with a cross-section as shown.
 

 
 

  1. Find an expression for the area of the cross-section using TWO applications of Simpson’s rule.  (2 marks)
  2. The area of the cross-section must be 600 m2. The tunnel is 80 m wide.   
  3. If the value of `a` increases by 2 metres, by how much will `b` change?   (2 marks)

 

Show Answers Only
  1. `(2h)/3 (4a + b)`
  2.  
  3. `b\ text(decreases by 8 m.)`
Show Worked Solution
(i)    `A` `~~ h/3 [y_0 + 4y_1 + y_2] + h/3 [y_0 + 4y_1 + y_2]`
    `~~ h/3 [0 + 4a + b] + h/3 [b + 4a + 0]`
    `~~ (2h)/3 (4a + b)`

 

(ii)    `text(Given)\ \ A = 600\ text(m²)`
  `text(If 80 m wide) \ => h = 20`

 

`A` `= (2h)/3 (4a + b)`
`600` `= ((2 xx 20))/3 (4a + b)`
`4a + b` `= (600 xx 3)/40`
`b` `= 45 – 4a`

 
`:.\ text(If)\ a\ text(increases by 2 m,)\ b\ text(will)`

`text(decrease by 8 m.)`

Linear Functions, 2UA 2008 HSC 3a

2008 3a

In the diagram,  `ABCD`  is a quadrilateral. The equation of the line  `AD`  is  `2x- y- 1 = 0`. 

  1. Show that  `ABCD`  is a trapezium by showing that  `BC`  is parallel to  `AD`.  (2 marks)
  2. The line  `CD`  is parallel to the  `x`-axis. Find the coordinates of  `D`.   (1 mark)
  3. Find the length of  `BC`.   (1 mark)
  4. Show that the perpendicular distance from  `B`  to  `AD`  is  `4/sqrt5`.   (2 marks)
  5. Hence, or otherwise, find the area of the trapezium  `ABCD`.   (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(3,5)`
  3. `sqrt 5\ text(units)`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `8\ text(u²)`
Show Worked Solution
(i)    `text(Show)\ BC \ text(||)\ AD`

`B(0,3),\ \ C(1,5)`

`m_(BC)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (5 – 3)/(1 – 0)`
  `= 2`

 

`text(Equation)\ \ AD\ \ text(is)\ \ 2x – y – 1 = 0`

`y` `= 2x – 1`
`m_(AD)` `= 2`

`:. BC\  text(||) \ AD`

`:. ABCD\ text(is a trapezium)`

 

(ii)    `text(Given)\ CD\  text(||) \ x text(-axis)`
  `text(Equation)\ CD\ text(is)\ y = 5`
  `D\ text(is intersection of)`
`y` `= 5,\ \ and`
`2x – y – 1` `= 0`
`:. 2x – 5 – 1` `=0`
`2x` `=6`
`x` `=3`
`:.\ D` `= (3,5)`

 

(iii)   `B(0,3),\ \ C(1,5)`
`text(dist)\ BC` `= sqrt ( (x_2 – x_1)^2 + (y_2 – y_1)^2 )`
  `= sqrt ( (1-0)^2 + (5-3)^2 )`
  `= sqrt (1 + 4)`
  `= sqrt 5\ text(units)`

 

(iv)   `text(Show)\ _|_\ text(dist of)\ B\ text(to)\ AD\ text(is)\ 4/sqrt5`

`B (0,3)\ \ \ \ \ 2x – y – 1 = 0` 

`_|_\ text(dist)` `= | (ax_1 + by_1 + c)/sqrt (a^2 + b^2) |`
  `= |( 2(0) – 1(3) -1 )/sqrt (2^2 + (-1)^2) |`
  `= | -4/sqrt5 |`
  `= 4/sqrt 5\ \ \ text(… as required.)`

 

(v)    `text(Area)` `= 1/2 h (a + b)`
    `= 1/2 xx 4/sqrt5 (BC + AD)`

 
`BC = sqrt5\ \ text{(part (iii))}`
 

`A(0,–1),\ \ D(3,5)`

`text(dist)\ AD` `= sqrt ( (3-0)^2 + (5+1)^2 )`
  `= sqrt (9 + 36)`
  `= sqrt 45`
  `= 3 sqrt 5`
`:.\ text(Area)\ ABCD` `= 1/2 xx 4/sqrt5 (sqrt5 + 3 sqrt 5)`
  `= 2 / sqrt5 (4 sqrt 5)`
  `= 8\ text(u²)`

Binomial, EXT1 2009 HSC 6b

  1. Sum the geometric series  
    1. `(1 + x)^r + (1 + x)^(r + 1) + ... + (1 + x)^n`
  3.  
  4. and hence show that
     
    1. `((r),(r)) + ((r + 1),(r)) + ... + ((n),(r)) = ((n + 1),(r + 1))`.   (3 marks)
    2.  
  6. Consider a square grid with  `n`  rows and  `n`  columns of equally spaced points.
  7.  
    1. 2009 6b
  8. The diagram illustrates such a grid. Several intervals of gradient  `1`, whose endpoints are a pair of points in the grid, are shown. 
  9.  
  10. (1)   Explain why the number of such intervals on the line  `y = x`  is equal to  `((n),(2))`.   (1 mark)
  11. (2)   Explain why the total number,  `S_n`, of such intervals in the grid is given by
  12.  
    1. `S_n = ((2),(2)) + ((3),(2)) + ... + ((n - 1),(2)) + ((n),(2)) + ((n - 1),(2)) +`
      1. `... + ((3),(2)) + ((2),(2))`.   (1 mark)
    2.  
  14. Using the result in part (i), show that 
    1. `S_n = (n(n - 1)(2n - 1))/6`.   (3 marks)

 

 

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

(i)   `(1 + x)^r + (1 + x)^(r + 1) + … + (1 + x)^n`

♦♦ Exact data unavailable although mean marks for all Q6 was < 25%.
MARKER’S COMMENT: A common mistake was to assume the number of terms in the series was  `n`.
`=> text(GP where)\ \ a` `= (1 + x)^r`
`r` `= (1 + x)`
`#\ text(Terms)` `= n – r + 1`

 

`text(Sum)` `= (a (r^n – 1))/(r – 1)`
  `= ((1 + x)^r [(1 + x)^(n\ – r + 1) – 1])/((1 + x) – 1)`
  `= ((1 + x)^(n\ – r + 1 + r) – (1 + x)^r)/x`
  `= ((1 + x)^(n + 1) – (1 + x)^r)/x`

 

`text(Show)`

`((r),(r)) + ((r + 1),(r)) + … + ((n),(r)) = ((n + 1),(r + 1))`

 

`text(Consider series)\ (1 + x)^r + (1 + x)^(r + 1) + … + (1 + x)^n`

`text(Co-efficient of)\ \ x^r = ((r),(r)) + ((r + 1),(r)) + … + ((n),(r))`

 

`text(Consider sum of series)\ \ \ ((1 + x)^(n + 1)\ – (1 + x)^r)/x`

`text(Co-efficient of)\ \ x^r` `=\ text(co-efficient of)\ \ x^(r + 1)\ \ text(in numerator)`
  `= ((n +1),(r + 1))`

 

`:.\ text(S)text(ince co-efficients are equal)`

`((r),(r)) + ((r + 1),(r)) + … + ((n),(r)) = ((n + 1),(r + 1))`

`text(… as required.)`

 

(ii)(1)   `text(All intervals start and finish at different points.)`
  `text(Any)\ n xx x\ text(grid has)\ n\ text(points on the)\ y = x\ text(diagonal.)`
  `:.\ text(Possible intervals) = ((n),(2))`
   
     (2)   `text(The longest diagonal of)\ n xx n\ text(grid is)`
  `text(the)\ y = x\ text(diagonal with)\ ((n),(2))\ text(intervals.)`
  `text(Each side of this, there is one less point)`
  `text(on the diagonals, with)\ ((n – 1),(2))\ text(intervals.)`
  `text(This continues until there are only 2 points)`
  `text(on the diagonal with)\ ((2),(2))\ text(intervals.)`

 

`:.\ S_n` `= ((2),(2)) + ((3),(2)) + … + ((n – 1),(2)) + ((n),(2))`
  `+ ((n – 1),(2)) + … + ((2),(2))`
  `text(… as required.)`

 

♦♦♦ Exact data unavailable although very few completed part (iii) correctly. 
MARKER’S COMMENT: Note that even if part (i) wasn’t solved, the proof as stated in the question can be used to solve part (iii).
(iii) `text(Show)\ \ S_n = (n(n – 1)(2n – 1))/6`
   `S_n` `= { ((2),(2)) + ((3),(2)) + … + ((n – 1),(2)) + ((n),(2))}`
  `+ {((n – 1),(2)) + … + ((2),(2))}`
  `= ((n + 1),(3)) + ((n),(3))\ \ \ text{(from part (i))}`
  `= ((n + 1)*n*(n – 1))/(3 * 2 * 1) + (n* (n – 1)*(n – 2))/(3 * 2 * 1)`
  `= 1/6 n (n – 1) (n + 1 + n – 2)`
  `= (n(n – 1)(2n – 1))/6\ \ \ text(… as required.)`

Trig Calculus, EXT1 2009 HSC 3b

  1. On the same set of axes, sketch the graphs of  
    1. `y = cos 2x`  and  `y = (x + 1)/2`, for  `–pi <= x <= pi`.    (2 marks)
  3. Use your graph to determine how many solutions there are to the equation  `2 cos 2x = x + 1`  for  `–pi <= x <= pi`.     (1 mark)
  5. One solution of the equation  `2 cos 2x = x + 1`  is close to  `x = 0.4`. Use one application of Newton’s method to find another approximation to this solution. Give your answer correct to three decimal places.   (3 marks)

 

Show Answers Only
  1. `text(See sketch in Worked Solutions)`
  2. `text(3 solutions)`
  3. `0.398\ text{(3 d.p.)}`
Show Worked Solution
(i)   

 

(ii)  `text(3 solutions)`

 

(iii)  `2 cos 2x = x + 1`

MARKER’S COMMENT: Better responses defined `f(x)` and `f′(x)` and evaluated each for `x=0.4` before calculating Newton’s formula, as done in the Worked Solution.
`f(x)` `= 2 cos 2x\ – x\ – 1`
`f prime (x)` `= -4 sin 2x\ – 1`

 

`=>f(0.4)` `= 2 cos 0.8\ – 0.4\ – 1`
  `=-0.0065865 …`
`=> f prime(0.4)` `= -4 sin 0.8\ – 1`
  `=-3.869424 …`

 

`text(Find)\ x_1\ text(where)`

 `x_1` `= 0.4\ – (f(0.4))/(f prime(0.4))`
  `= 0.4\ – ((-0.0065865 …)/(-3.869424 …))`
  `= 0.39829…`
  `= 0.398\ \ text{(3 d.p.)}`

Plane Geometry, 2UA 2010 HSC 10a

In the diagram  `ABC`  is an isosceles triangle with  `AC = BC = x`. The point  `D`  on the interval  `AB`  is chosen so that  `AD = CD`. Let  `AD = a`,  `DB = y`  and  `/_ADC = theta`.
 

 
 

  1. Show that  `Delta ABC`  is similar to  `Delta ACD`.    (2 marks)
  2. Show that  `x^2 = a^2 + ay`.     (1 mark)
  3. Show that  `y = a(1 − 2 cos theta )`.   (2 marks)
  4. Deduce that  `y <= 3a`.   (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)}`
Show Worked Solution
MARKER’S COMMENT: When dealing with multiple triangles, the best performing students label all angles clearly and unambiguously.
(i)

`/_CAD\ text(is common)`

`/_CAD = /_ACD = /_DBC`

`text{(Angles opposite equal sides in}`

`\ \ text{isosceles}\ Delta ACD\ text(and)\ Delta ABC text{)}`
 

`/_ADC = /_ACB\ \ (180^@\ text(in)\ Delta text{)}`

`:.\ Delta ABC\  text(|||) \ Delta ACD\ \ \ text{(AAA)}\ \ \ text(… as required)`

 

♦♦ Mean mark 25%.
MARKER’S COMMENT: The similarity proof in part (i) should have immediately alerted students to using similar ratios of sides to solve part (ii). Many did not follow this obvious hint.

(ii)  `text(Using similarity)`

`(AC)/(AD)` `= (AB)/(CB)` `\ \ text{(corresponding sides of}`
`\ \ \ \ text{similar triangles)}`
`x/a` `= (a + y)/x`  

 

`:.\ x^2 = a^2 + ay\ \ \ text(… as required)`

 

(iii)  `text(Show)\ \ y = a(1\ – 2 cos theta)`

`text(Using Cosine Rule:)`

`cos theta` `= (a^2 + a^2\ – x^2)/(2 xx a xx a)`
`2a^2costheta` `= 2a^2\ – x^2`
  `= 2a^2\ – (a^2 + ay)\ \ text{(from part (ii))}`
  `= a^2\ – ay`
`ay` `= a^2\ – 2a^2cos theta`
`y` `= a\ – 2a cos theta`
  `= a (1\ – 2 cos theta)\ \ \ \ text(… as required)`

 

(iv)  `text(S)text(ince)\  1 <= cos theta <= 1`

♦♦♦ Parts (iii) and (iv) mean marks – 18% and just 4% respectively.
IMPORTANT: Limits of trig functions are often extremely useful in proving inequalities (see Worked Solutions).

` -2 <= 2 cos theta <= 2`

` -1 <= 1\ – 2 cos theta <= 3`

`y` `=a(1\ – 2 cos theta)\ \ \ text{(from part (iii))}`
`:.\ y` `<= a(3)`
  `<= 3a\ \ …\ text(as required)`

Plane Geometry, 2UA 2011 HSC 6a

The diagram shows a regular pentagon `ABCDE`. Sides `ED` and `BC` are produced to meet at `P`.
  

  1. Find the size of `/_CDE`.    (1 mark)

  2. Hence, show that `Delta EPC` is isosceles.    (2 marks)

Show Answers Only
  1. `108°`
  2. `text(Proof)\ \ text{(see Worked Solutions)}`
Show Worked Solution
i.  

`text(Angle sum of pentagon)=(5-2) xx 180°=540°`

`:.\ /_CDE` `= 540/5\ \ \ text{(regular pentagon has equal angles)}`
  `= 108°`
MARKER’S COMMENT: Very few students solved part (i) efficiently. Remember the general formula for the sum of internal angles equals (# sides – 2) x 90°.

 
ii.  `text(Show)\ Delta EPC\ text(is isosceles)`

`text(S)text(ince)\ ED=CD\ \ text{(sides of a regular pentagon)}`

`Delta ECD\ text(is isosceles)`

`/_DEC=1/2 xx (180-108)= 36^{\circ}\ \ \ text{(Angle sum of}\ Delta DEC text{)}`

`/_CDP=72^@\ \ \ (\angle PDE\ \text{is a straight angle})`

`/_DCP=72^@\ \ \ (\angle PCB\ \text{is a straight angle})`

`=> /_CPD= 180-(72 + 72)=36^{\circ}\ \ \ text{(angle sum of}\ Delta CPD text{)}`

`:.\ Delta EPC\ \text(is isosceles)\ \ \ text{(2 equal angles)}`

Statistics, 2ADV 2011 HSC 5b

Kim has three red shirts and two yellow shirts. On each of the three days, Monday, Tuesday and Wednesday, she selects one shirt at random to wear. Kim wears each shirt that she selects only once.

  1. What is the probability that Kim wears a red shirt on Monday?    (1 mark)
  2. What is the probability that Kim wears a shirt of the same colour on all three days?     (1 mark)
  3. What is the probability that Kim does not wear a shirt of the same colour on consecutive days?   (2 marks)
Show Answers Only
  1. `3/5`
  2. `1/10`
  3. `3/10`
Show Worked Solution
i.    `P (R\ text(on Monday) text{)}` `= text(# Red)/text(# Shirts)`
    `= 3/5`

 

MARKER’S COMMENT: Students could also have drawn a probability tree setting out this problem over 3 different days (stages).
ii.    `text(S)text(ince not enough yellow shirts)`
  `P text{(same colour each day)}`

`= P (R,R,R)`

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

`= 1/10`

 

iii.   `P text{(not wearing same colour 2 days in a row)}`

`= P (Y,R,Y) + P (R,Y,R)`

♦ Mean mark 47%.
MARKER’S COMMENT: Students should clearly write what probability they are finding in words or symbols before showing calculations.

`=  (2/5 xx 3/4 xx 1/3)+(3/5 xx 2/4 xx 2/3)`

`= 6/60 + 12/60 `

`= 3/10`

Probability, 2UG 2010 HSC 26c

Tai plays a game of chance with the following outcomes. 

• `1/5`  chance of winning  `$10` 

• `1/2`  chance of winning  `$3` 

• `3/10`  chance of losing  `$8` 

The game has a  `$2`  entry fee. 

What is his financial expectation from this game?    (2 marks)

Show Answers Only

 `text(A loss of $0.90.)`

Show Worked Solution
♦♦ Mean mark 31%
MARKER’S COMMENT: A common error was not to deduct the chance of losing or the entry fee. BE CAREFUL!

`text(Financial Expectation)`

`= (1/5 xx 10) + (1/2 xx 3)\-(3/10 xx 8)\-2`

`= 2 + 1.5\-2.4\-2`

`=-0.9`
 

`:.\ text(The financial expectation is a loss of $0.90.)`

Measurement, 2UG 2011 HSC 24c

A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`.  The
size of angle `ABC` is 114°.

2UG 2011 24c

Copy the diagram into your writing booklet and show all the information on it.

  1. What is the bearing of `C` from `B`?   (1 mark)
  2. Find the distance `AC`. Give your answer correct to the nearest kilometre.   (2 marks)
  3. What is the bearing of `A` from `C`? Give your answer correct to the nearest degree.   (3 marks)

 

Show Answers Only
  1.  `055^@`
  2.  `13\ text(km)`
  3.  `261^@`
Show Worked Solution
♦♦ Mean mark 24% 
STRATEGY: This deserves repeating again: Draw North-South parallel lines through major points to make the angle calculations easier.
(i)     2011 HSC 24c

 `text(Let point)\ D\ text(be due North of point)\ B`

`/_ABD` `=180-121\ text{(cointerior with}\ \ /_A text{)}`
  `=59^@`
`/_DBC` `=114-59`
  `=55^@`

  

`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`

 

(ii)    `text(Using cosine rule:)`

♦ Mean mark 39%
`AC^2` `=AB^2+BC^2-2xxABxxBCxxcos/_ABC`
  `=6^2+9^2-2xx6xx9xxcos114^@`
  `=160.9275…`
`:.AC` `=12.685…\ \ \ text{(Noting}\ AC>0 text{)}`
  `=13\ text(km)\ text{(nearest km)}`

 

(iii)    `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`

♦♦♦ Mean mark 15%
MARKER’S COMMENT: The best responses clearly showed what steps were taken with working on the diagram. Note that all North/South lines are parallel.
`cos/_ACB` `=(AC^2+BC^2-AB^2)/(2xxACxxBC)`
  `=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)`
  `=0.9018…`
`/_ACB` `=25.6^@\ text{(to 1 d.p.)}`

 

`text(From diagram,)`

`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`

`:.\ text(Bearing of)\ A\ text(from)\ C`

  `=180+55+25.6`
  `=260.6`
  `=261^@\ text{(nearest degree)}`

Calculus in the Physical World, 2UA 2008 HSC 6b

The graph shows the velocity of a particle,  `v`  metres per second, as a function of time,  `t`  seconds.
 


 

  1. What is the initial velocity of the particle?   (1 mark)
  2. When is the velocity of the particle equal to zero?    (1 mark)
  3. When is the acceleration of the particle equal to zero?    (1 mark)
  4. By using Simpson's Rule with five function values, estimate the distance travelled by the particle between  `t=0`  and  `t=8`.   (3 marks)
  5.  
Show Answers Only
  1. `20\ text(m/s)`
  2. `t=10\ text(seconds)`
  3. `t=6\ text(seconds)`
  4. `493 1/3\ text(metres)`
Show Worked Solution

(i)    `text(Find)\   v  \ text(when)  t=0`

`v=20\ \ text(m/s)`

 

(ii)    `text(Particle comes to rest at)\  t=10\ text{seconds  (from graph)}`

 

(iii)  `text(Acceleration is zero when)\ t=6\ text{seconds  (from graph)}`

 

(iv)   

MARKER’S COMMENT: Less errors were made by students using a table and the given formula. Note however, that the formula `A~~(b-a)/6xx` `[f(a)+4f((a+b)/2)+f(b)]` produced the most errors.
`text(Area)` `~~h/3[y_0+y_n+4text{(odds)}+2text{(evens)}]`
  `~~h/3[y_0+y_4+4(y_1+y_3)+2(y_2)]`
  `~~2/3[20+60+4(50+80)+2(70)]`
  `~~2/3[740]`
  `~~493 1/3`

 

`:.\ text{Distance travelled is 493 1/3 m (approx.)}`