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Functions, EXT1 F1 2011 HSC 1c

Solve  `(4-x)/x <1`.  (3 marks)

Show Answers Only

 `x<0\ \ text(or)\ \ x>2`

Show Worked Solution

`text(Solution 1)`

`(4-x)/x < 1`

`text(If)\ x<0,\ \ \ \ \ 4-x` `> x`
`2x` `< 4`
`x` `<2`

`=> x<0\ \ \ text{(satisfies both)}`
 

`text(If)\ x>0,\ \ \ \ \ 4-x` `<x`
`2x` `>4`
`x` `>2`

`=> x>2\ \ \ text{(satisfies both)}`

`:.\ x < 0\ \ text(or)\ \ x > 2`

 
`text(Solution 2)`

`text(Multiply both sides by)\ \ x^2`

`x(4-x)` `< x^2`
`4x-x^2` `< x^2`
`2x^2-4x` `>0`
`2x(x-2)` `>0`

 

 EXT1 2011 1c

`text(From graph)`

`x<0\ \ text(or)\ \ x >2`

Probability, 2ADV S1 2009 HSC 5b

On each working day James parks his car in a parking station which has three levels. He parks his car on a randomly chosen level. He always forgets where he has parked, so when he leaves work he chooses a level at random and searches for his car. If his car is not on that level, he chooses a different level and continues in this way until he finds his car.    

  1. What is the probability that his car is on the first level he searches?     (1 mark)

  2. What is the probability that he must search all three levels before he finds his car?   (1 mark)

  3. What is the probability that on every one of the five working days in a week, his car is not on the first level he searches?    (1 mark)

Show Answers Only
  1. `1/3`
  2. `1/3`
  3. `32/243`
Show Worked Solution

i.  `P text{(1st chosen)} = 1/3`
 

ii.  `P text{(search 3 levels)}`

♦♦ Mean marks of 31% and 39% for part (ii) and (iii) respectively.

`= P text{(not 1st)} xx P text{(not 2nd)}`

`= 2/3 xx 1/2`

`= 1/3`
 

iii.  `P text{(not 1st for 5 days)}`

`= 2/3 xx 2/3 xx 2/3 xx 2/3 xx 2/3`

`= 32/243`

Plane Geometry, 2UA 2009 HSC 4c

In the diagram,  `Delta ABC`  is a right-angled triangle, with the right angle at  `C`. The midpoint of  `AB`  is  `M`, and  `MP _|_ AC`.

Plane Geometry, 2UA 2009 HSC 4c_1

Copy or trace the diagram into your writing booklet. 

  1. Prove that  `Delta AMP`  is similar to  `Delta ABC`.   (2 marks)
  2. What is the ratio of  `AP`  to  `AC`?    (1 mark)
  3. Prove that  `Delta AMC`  is isosceles.   (2 marks)
  4. Show that  `Delta ABC`  can be divided into two isosceles triangles.   (1 mark)
  5. Plane Geometry, 2UA 2009 HSC 4c_2
  6. Copy or trace this triangle into your writing booklet and show how to divide it into four isosceles triangles.   (1 mark)
  7.  
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Ratio)\ AP:AC = 1:2`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. Plane Geometry, 2UA 2009 HSC 4c_2 Answer1
Show Worked Solution
(i)    `text(Need to prove)\ Delta AMP\  text(|||)\  Delta ABC`
  `/_ PAM\ text(is common)`
  `/_ MPA = /_BCA = 90°\ \ \ text{(given)}`
  `:.\ Delta AMP \ text(|||) \ Delta ABC\ \ \ text{(equiangular)}`

 

(ii)   `(AP)/(AC) = (AM)/(AB)\ \ \ ` `text{(corresponding sides of}`
    `text{similar triangles)}`

 

`text(S)text(ince)\ \ AB = 2 xx AM`
`(AP)/(AC) = 1/2`
`:.\ text(Ratio)\ AP:AC = 1:2`

 

(iii)

Plane Geometry, 2UA 2009 HSC 4c_1 Answer
`AP = PC \ \ \ text{(from part (ii))}`
`PM\ text(is common)`
`/_APM = /_CPM = 90°\ \ text{(∠}\ APC\ text{is a straight angle)}`
`:.\ Delta AMP ~= Delta CMP\ text{(SAS)}`
`=> AM = CM\ \ ` `text{(corresponding sides of}`
  `text{congruent triangles)}`

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

 

(iv)   `text(S)text(ince)\ \ AM` `= MC\ \ \ text{(part (iii))}`
  `AM`  `= MB\ text{(given)}`
  `:. MC`  `= MB`
`:. Delta MCB\ text(is isosceles)`
`:. Delta ABC\ text(can be divided into 2 isosceles)`
`text(triangles)\ (Delta AMC\ text(and)\ Delta MCB text{)}`

 

(v)

Plane Geometry, 2UA 2009 HSC 4c_2 Answer1
♦♦ Mean mark 25%.
MARKER’S COMMENT: Use a ruler, draw large diagrams and always pay careful attention to previous parts of the question!
`/_AFB = /_CFB = 90°`
`DF\ text(bisects)\ AB`
`EF\ text(bisects)\ BC`
`text(From part)\ text{(iv)}\ text(we get 4 isosceles)`
`text(triangles as shown.)`

Functions, 2ADV F1 2009 HSC 3b

2009 3b
 

The circle in the diagram has centre  `N`. The line  `LM`  is tangent to the circle at  `P`. 

  1. Find the equation of  `LM`  in the form  `ax + by + c = 0`.   (2 marks)
  2. Find the distance  `NP`.    (2 marks)
  3. Find the equation of the circle.    (1 mark)
Show Answers Only
  1. `4x – 3y – 5 = 0`
  2. `2\ text(units)`
  3. `text(Equation is)\ \ (x – 1)^2 + (y – 3)^2 = 4`
Show Worked Solution

(i)  `text(Need to find the gradient of)\ LM`

`m_(LM)` `= (y_2\ – y_1)/(x_2\ – x_1)`
  `= (5\ – 1)/(5\ – 2)`
  `= 4/3`

 
`text(Equation of)\ LM\ text(has)\ m = 4/3\ text(through)\ (2,1)`

`y\ – y_1` `= m (x\ – x_1)`
`y\ – 1` `= 4/3 (x\ – 2)`
`3y\ – 3` `= 4x\ – 8`
`4x\ – 3y\ – 5` `= 0`

 

(ii)  `NP\ text(is)  _|_ text(distance of)\ \ N(1,3)\ \ text(from)\ \ LM`

`_|_\ text(dist)` `= |(ax_1 + by_1 + c)/sqrt(a^2 + b^2)|`
  `= |(4(1)\ – 3(3)\ – 5)/sqrt(4^2 + (-3)^2)|`
  `= |-10/5|`
  `= 2\ text(units)`

 

(iii)  `text{The circle has centre (1,3) and radius 2 units:`

`:.\ text(Equation is)\ \ (x\ – 1)^2 + (y\ – 3)^2 = 4`

Calculus, 2ADV C4 2009 HSC 2b

  1. Find  `int 5\ dx`.   (1 mark)

  2. Find  `int 3/((x - 6)^2)\ dx`.    (2 marks)

  3. Evaluate  `int_1^4 x^2 + sqrtx\ dx`.   (3 marks)

Show Answers Only
  1. `5x + C`
  2. `(-3)/((x – 6)) + C`
  3. `77/3`
Show Worked Solution

i.  `int 5\ dx= 5x + C`

 

ii.  `int 3/((x – 6)^2)\ dx`

`= 3 int (x – 6)^(-2)\ dx`

`= 3 xx 1/(-1) xx (x – 6)^(-1) + c`

`= (-3)/((x – 6)) + c`

 

iii.  `int_1^4 x^2 + sqrtx\ \ dx`

`= int_1^4 (x^2 + x^(1/2))\ dx`

`= [1/3 x^3 + 1/(3/2) x^(3/2)]_1^4`

`= [(x^3)/3 + 2/3x^(3/2)]_1^4`

`= [((4^3)/3 + 2/3 xx 4^(3/2))\ – (1/3 + 2/3)]`

`= [(64/3 + 16/3) – 3/3]`

`= [80/3 – 3/3]`

`= 77/3`

Functions, 2ADV F1 2009 HSC 1a

Sketch the graph of  `y-2x = 3`, showing the intercepts on both axes.   (2 marks)

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Show Worked Solution

`y-2x=3\ \ =>\ \ y=2x+3`

`ytext{-intercept}\ = 3`

`text{Find}\ x\ text{when}\ y=0:`

`0-2x=3\ \ =>\ \ x=-3/2`
 

Calculus, 2ADV C1 2010 HSC 7b

The parabola shown in the diagram is the graph  `y = x^2`. The points  `A (–1,1)`  and  `B (2, 4)`  are on the parabola.
 

 
 

  1.  Find the equation of the tangent to the parabola at `A`.   (2 marks)

  2.  Let `M` be the midpoint of  `AB`.

     

    There is a point `C` on the parabola such that the tangent at `C` is parallel to  `AB`.

     

    Show that the line  `MC`  is vertical.   (2 marks)  

  3. The tangent at `A` meets the line `MC` at `T`.

     

    Show that the line `BT` is a tangent to the parabola.  (2 marks)

Show Answers Only
  1. `2x + y + 1 = 0`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `text(Proof)  text{(See Worked Solutions)}`
Show Worked Solution
i.
`y` `=x^2`
`dy/dx` `= 2x` 

 
`text(At)\ \ A text{(–1,1)}\ => dy/dx = -2`
 

`text(T)text(angent has)\ \ m=text(–2),\ text(through)\ text{(–1,1):}`

`y – y_1` `= m(x\ – x_1)`
`y – 1` `= -2 (x + 1)`
`y – 1` `= -2x -2`
`2x + y + 1` `= 0`

 

 `:.\ text(T)text(angent at)\ A\ text(is)\ \ 2x + y + 1 = 0`

 

♦ Mean mark 37%.
IMPORTANT: The critical understanding required for this question is that the gradient of `AB` needs to be equated to the gradient function (i.e. `dy/dx`).

ii.   `Atext{(–1,1)}\ \ \ B(2,4)`

`M` `= ((-1+2)/2 , (1+4)/2)`
  `= (1/2, 5/2)`

 

`m_(AB)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (4 – 1)/(2 + 1)=1`

 

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

 
`:.\ C \ (1/2, 1/4)`
 
`=>M\ text(and)\ C\ text(both have)\ x text(-value)=1/2`

`:. MC\ text(is vertical  … as required)`

 

iii.   `T\ text(is point on tangent when) \ x=1/2` 

♦♦ Mean mark 29%.

`text(T)text(angent)\ \ \ 2x + y + 1 = 0`

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

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

`=> y=–2`

`:.\ T (1/2, –2)`

 
`text (Given)\ \ B (2, 4)`

`m_(BT)` `= (4+2)/(2\ – 1/2)`
  `=4`

 
`text(At)\ \ B(2,4),\ text(find gradient of tangent:)`

`dy/dx = 2x=2 xx2=4`

`:.m_text(tangent) = 4=m_(BT)`

`:.BT\ text(is a tangent)`

Linear Functions, 2UA 2010 HSC 3a

In the diagram  `A`,  `B`  and  `C`  are the points  `( –2, –4 ),\ (12,6)`  and  `(6,8)`  respectively.

The point  `N (2,2)`  is the midpoint of   `AC`. The point  `M`  is the midpoint of  `AB`.
 

2010 3a
 

  1. Find the coordinates of  `M`.   (1 mark)
  2. Find the gradient of  `BC`.   (1 mark)
  3. Prove that  `Delta ABC`  is similar to  `Delta AMN`.   (2 marks)
  4. Find the equation of  `MN`.   (2 marks)
  5. Find the exact length of  `BC`.   (1 mark)
  6. Given that the area of  `Delta ABC`  is  `44`  square units, find the perpendicular distance  from  `A`  to  `BC`.    (1 mark)

 

Show Answers Only
  1. `M (5,1)`
  2. `-1/3`
  3. `text{Proof (See Worked Solutions)}`
  4. `x + 3y  – 8 = 0`
  5. `2 sqrt 10 \ text(units)`
  6. `(22 sqrt 10)/5 \ text(units)`
  7.  
Show Worked Solution

(i)   `M \ text(is the midpoint of) \ AB`

`A text{(–2,–4),}\ \ \ \ B(12,6)`

`:.M` `=((x_1+x_2)/2 ‘ (y_1+y_2)/2)`
  `= ((-2 + 12)/2  ,  (-4 + 6)/2)`
  `= (5,1)`

 

(ii)   `B(12,6),     C(6,8)`

`m_(BC)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (8– 6)/(6– 12)`
  `= -1/3`

 

(iii)  `text(Prove)\ Delta ABC \ text(|||) \ Delta AMN`

`text(Find gradient of) \ NM`

`N(2,2)   M(5,1)`

`m_(NM)` `= (1- 2)/(5– 2)`
  `= -1/3`

`:. NM\ text(||)\ BC \ \ \ text{(gradients are equal)}`

`angle NAM \ text(is common)`

`angle ANM = angle ACB\ \ \ text{(corresponding angles},\ NM\ text(||)\ BC text{)}`

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

 

(iv)   `text(Equation of) \ MN \ text(has) \ m=-1/3 \ \ text(through) \ (2,2)`

`text(Using) \ \ \ y- y_1` `= m(x  – x_1)`
`y  – 2` `= -1/3 (x  – 2)`
`3y  – 6` `= -x + 2`
`x +3y  – 8` `= 0`

 

`:. \ text(Equation of) \ MN \ text(is) \ \ x + 3y  – 8 = 0`

 

(v)   `B(12,6)   C(6,8)` 

`BC` `=sqrt{(x_2 – x_1)^2 + (y_2– y_1)^2}`
  `=sqrt{(6– 12)^2 + (8– 6)^2}`
  `=sqrt(36 +4)`
  `=sqrt(40)`
  `= 2 sqrt(10) \ text(units)`

 

(vi)   `text(Given the Area) \  Delta ABC=44\ text(u²)`

`1/2 xx b xx h` `= 44`
`1/2 xx BC xx h`  `= 44`
`1/2 xx 2 sqrt 10 xx h` `= 44`
`:. h` `= 44/ sqrt 10 xx sqrt 10 / sqrt 10`
  `= (44 sqrt 10 )/ 10`
  `= (22 sqrt 10) / 5`

 

`:. _|_ text(distance from) \ A \ text(to) \ BC \ text(is) \ (22 sqrt 10) / 5  \ text(units)`.

Functions, EXT1* F1 2010 HSC 2b

Solve the inequality  `x^2 − x − 12 < 0`.   (2 marks) 

Show Answers Only

 `-3 < x < 4`

Show Worked Solution
 MARKER’S COMMENT: Drawing the parabola proved the most efficient way of answering this question for students.

`x^2\ – x\ – 12<0`

`text(Solve)\ \ \ ` `x^2\ – x\ – 12` `=0`
  `(x\ – 4)(x + 3)` `=0`
`=>x = 4\ text(or)\ -3`

 

 

`x^2 – x – 12 < 0`

`text(when)\ \ -3 < x < 4`

Calculus, 2ADV C2 2010 HSC 2a

Differentiate  `cosx/x`  with respect to  `x`.   (2 marks) 

Show Answers Only

 `(-x sinx – cos x)/(x^2)`

Show Worked Solution
MARKER’S COMMENT: A significant number of students did not know the quotient rule and many found assistance by writing out `u,\ u prime,\ v,\ v prime\ ` before going into calculations.

`y = cos x/x`

`text(Let)` `\ \ u = cos x` `u prime = – sin x`
  `\ \ v = x` `v prime = 1`

 

`text(Using quotient rule:)`

`dy/dx` `= (u prime v – u v prime)/(v^2)`
  `= (-sinx *x  – cos x*1)/x^2`
  `= (-x sin x – cos x)/(x^2)`

Calculus, 2ADV C2 2010 HSC 1e

Differentiate  `x^2 tan x`  with respect to  `x`.   (2 marks)

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`2x tanx + x^2 sec^2 x`

Show Worked Solution
COMMENT: There is no necessity to take out the common factor `x` in this case.

`y = x^2 tan x`

`text(Using product rule:)`

`d/dx (uv)` ` = u prime v + u v prime`
`dy/dx` `=2x tanx + x^2 sec^2 x`

Financial Maths, 2ADV M1 2011 HSC 9d

 

  1. Rationalise the denominator in the expression  `1/(sqrtn + sqrt(n+1))`  where  `n`  is an integer and  `n >= 1`.   (1 mark)

  2. Using your result from part (i), or otherwise, find the value of the sum
     
         `1/(sqrt1+ sqrt2) + 1/(sqrt2 + sqrt3) + 1/(sqrt3 + sqrt4) + ... + 1/(sqrt99 + sqrt100)`   (2 marks)

Show Answers Only
  1. `sqrt(n+1)\ – sqrtn`
  2. `9`
Show Worked Solution

i.  `1/(sqrtn + sqrt(n+1)) xx (sqrtn\ – sqrt(n+1))/(sqrtn\ – sqrt(n+1))`

MARKER’S COMMENT: Students connecting the information between parts (i) and (ii) were easily the most successful. AGAIN, the information from earlier parts of multi-part questions is gold plated information for directing your answer.

`= (sqrtn\ – sqrt(n+1))/((sqrtn)^2\ – (sqrt(n+1))^2)`

`= (sqrtn\ – sqrt(n+1))/(n\ – (n + 1))`

`= (sqrtn\ – sqrt(n+1))/-1`

`= sqrt(n+1)\ – sqrtn`

 

ii.  `1/(sqrt1 + sqrt2) + 1/(sqrt2 + sqrt3) + 1/(sqrt3 + sqrt4) + … + 1/(sqrt99 + sqrt100)`

`= (sqrt2\ – sqrt1) + (sqrt3\ – sqrt2) + (sqrt4\ – sqrt3) + … + (sqrt100\ – sqrt99)`

`= – sqrt1 + sqrt 100`

♦♦♦ Mean mark 15%.

`= -1 + 10`

`= 9`

 

Calculus, 2ADV C2 2012 HSC 12aii

Differentiate with respect to  `x`. 

`(cos x)/(x^2)`.    (2 marks)

Show Answers Only

`(-x sin x\ – 2 cos x)/(x^3)`

 

Show Worked Solution

`y = cosx/(x^2)`

`u = cos x` `\ \ \ \ \ \ u prime = – sin x`
`v = x^2` `\ \ \ \ \ \ v prime = 2x`

 

`text(Using the quotient rule,)` 

`dy/dx` `= (u prime v\ – v prime u)/(v^2)`
  `= (- sin x * x^2\ – 2x * cos x)/(x^4)`
  `= (- x sin x\ – 2 cos x)/(x^3)`

Functions, 2ADV F1 2010 HSC 1b

Find integers  `a`  and  `b`  such that  `1/(sqrt5\ - 2) = a + b sqrt5`.  (2 marks)

Show Answers Only

 `a = 2, \ b = 1`

Show Worked Solution

`text(Given)\ 1/(sqrt5\ – 2) = a + b sqrt5` 

`1/(sqrt5\ – 2) xx (sqrt5 + 2)/(sqrt5 + 2)` `= (sqrt5 + 2)/((sqrt5)^2\ – 2^2)`
  `= (sqrt5 + 2)/(5\ – 4)`
  `= 2 + sqrt5`

`:.\ a = 2, \ b = 1`

Probability, 2ADV S1 2012 HSC 13c

Two buckets each contain red marbles and white marbles. Bucket `A` contains 3 red and 2 white marbles. Bucket `B`  contains 3 red and 4 white marbles.

Chris randomly chooses one marble from each bucket. 

  1. What is the probability that both marbles are red?   (1 mark)

  2. What is the probability that at least one of the marbles is white?   (1 mark)

  3. What is the probability that both marbles are the same colour?   (2 marks)

Show Answers Only
  1. `9/35`
  2. `26/35`
  3. ` 17/35`
Show Worked Solution
i.    `P{(both red)}` `= P(R_1) xx P(R_2)`
  `= 3/5 xx 3/7`
  `= 9/35`

 

STRATEGY: When the term “at least” appears in a probability question, it is likely that `1-P(text{complement})` will solve the question more efficiently and with less chance of error, as shown in part (ii).
ii.    `Ptext{(at least one white)}` `= 1 – Ptext{(none white)}`
  `= 1 – P(R_1) xx P(R_2)`
  `= 1 – 9/35`
  `= 26/35`

 

iii.    `Ptext{(same colour)}` `= P(R_1 R_2) + P(W_1 W_2)`
  `= 9/35 + (2/5 xx 4/7)`
  `= 9/35 + 8/35`
  `= 17/35`

Calculus, 2ADV C3 2011 HSC 7a

Let  `f(x) = x^3-3x + 2`. 

  1. Find the coordinates of the stationary points of  `y = f(x)`, and determine their nature.   (3 marks)

  2. Hence, sketch the graph  `y = f(x)`  showing all stationary points and the  `y`-intercept.   (2 marks)

Show Answers Only
  1. `text(MIN at)\ \ (1,0);\ text(MAX at)\ \ (–1,4)`
  2.  
    Geometry and Calculus, 2UA 2011 HSC 7a
Show Worked Solution
i.    `f(x)` `= x^3-3x + 2`
  `f^{′}(x)` `= 3x^2-3`
  `f^{″}(x)` `=6x`

 

`text(Stationary points when)\ f prime (x) = 0`

`3x^2-3` `=0`
`3 (x^2-1)` `= 0`
`:. x^2` `=1`
`x` `=+- 1`

 
`text(When)\ x = 1`

`f(1)` `= 1-3 + 2 = 0`
`f^{″}(1)` `= 6 > 0` 
`:.\ text(MIN S.P. at)\ \ (1,0)`

 

`text(When)\ \ x= -1`

`f(–1)` `= -1 + 3 + 2 = 4`
`f^{″}(–1)` `= –6 < 0`
`:.\ text(MAX S.P. at)\ \ (–1,4)`
MARKER’S COMMENT: Graphs should be large (around ½ page), axes drawn with a ruler, with intercepts and turning points clearly shown. The scale can be different on each axe for clarity, as shown in the Worked Solution.

 

ii.    `y = x^3-3x + 2`
  `y text(-intercept) = 2`

Geometry and Calculus, 2UA 2011 HSC 7a

Calculus, 2ADV C4 2011 HSC 6c

The diagram shows the graph  `y = 2 cos x` . 
  

2011 6c 
  

  1. State the coordinates of  `P`.   (1 mark)

  2. Evaluate the integral  `int_0^(pi/2) 2 cos x\ dx`.    (2 marks)

  3. Indicate which area in the diagram,  `A`,  `B`,  `C` or  `D`, is represented by the integral
     
           `int_((3pi)/2)^(2pi) 2 cos x\ dx`.   (1 mark)

  4. Using parts (ii) and (iii), or otherwise, find the area of the region bounded by the curve  `y = 2 cos x`  and the  `x`-axis, between  `x = 0`  and  `x = 2pi` .   (1 mark)

  5. Using the parts above, write down the value of
     
         `int_(pi/2)^(2pi) 2 cos x\ dx`.   (1 mark)

Show Answers Only
  1. `P(0,2)`
  2. `2`
  3. `C`
  4. `8\ text(u²)`
  5. `-2`
Show Worked Solution

i.  `y = 2 cos x`

`text(At)\ x = 0`

`y = 2 cos 0 = 2`

`:.\ P(0,2)`

 

ii.  `int_0^(pi/2) 2 cos x\ dx`

`= [2 sin x]_0^(pi/2)`

`= [2 sin (pi/2)\ – 2 sin 0]`

`= 2\ – 0`

`= 2`
 

iii.  `C`
 

iv.    `text(S)text(ince Area)\ A` `=\ text(Area)\ C,\ \ text(and)`
  `text(Area)\ B` `= 2 xx text(Area)\ A`

 

`:.\ text(Total Area)` `= 2 + (2xx2) + 2`
  `= 8\ text(u²)`

 

MARKER’S COMMENT: “Using the parts above” in part (v) was ignored by many students. Important to know when finding an area and evaluating an integral may differ.

v.  `int_(pi/2)^(2pi) 2 cos x\ dx`

`= text(Area)\ C\ – text(Area)\ B`

`= 2\ – (2 xx 2)`

`= -2`

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)}`

Calculus, 2ADV C2 2011 HSC 4a

Differentiate  `x/sinx`  with respect to  `x`.   (2 marks)

Show Answers Only

 `(sin x\ – x cos x)/(sin^2x)`

Show Worked Solution

`y = x/sinx`

`u = x` `\ \ \ \ \ u prime = 1`
`v = sin x` `\ \ \ \ \ v prime = cos x`

 

`text(Using)\ \ d/dx (uv) = (u prime v\ – uv prime)/(v^2),`

`dy/dx` `= (1 * sinx \ – x * cos x)/((sin x)^2)`
  `= (sin x\ – x cos x)/(sin^2x)`

Linear Functions, 2UA 2011 HSC 3c

The diagram shows a line  `l_1`, with equation  `3x + 4y - 12 = 0`, which intersects the  `y`-axis at  `B`.

A second line  `l_2`, with equation  `4x - 3y = 0`, passes through the origin  `O`  and intersects `l_1`  at  `E`.  
  

2011 3c
 

  1. Show that the coordinates of  `B`  are  `(0, 3)`.    (1 mark)
  2. Show that  `l_1`  is perpendicular to  `l_2`.   (2 marks)
  3. Show that the perpendicular distance from  `O`  to  `l_1`  is  `12/5`.   (1 mark) 
  4. Using Pythagoras’ theorem, or otherwise, find the length of the interval  `BE`.   (1 mark)
  5. Hence, or otherwise, find the area of  `Delta BOE`.  (1 mark)
Show Answers Only
  1. `B(0,3)`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `text(Proof)  text{(See Worked Solutions)}`
  4. `BE = 9/5\ \ text(or)\ \ 1.8\ text(units)`
  5. `54/25\ \ text(or)\ \ 2.16\ text(u²)`
Show Worked Solution
(i)    `B\ text(is)\ y text(-intercept of)\ l_1`
  `text(When)\ x = 0`
`(3 xx 0) + 4y\ – 12` `= 0`
`4y` `=12`
`y` `=3`

`:.\ B\ text(is)\ (0,3)`

 

MARKER’S COMMENT: Better responses involved turning the equations into the “gradient intercept” form as shown in the worked solutions. Many students who tried to use the general formula `m=- b/a` made errors in the process. 
(ii)    `l_1\ text(is)\ 3x + 4y -12`  `= 0`
  `4y` `= -3x + 12`
  `y` `= -3/4x + 3`
  `=> m\ text(of)\  l_1` `= -3/4`
`l_2\ text(is)\ 4x -3y` `= 0`
`3y` `= 4x`
`y` `= 4/3 x`
`=> m\  text(of)\ l_2` `= 4/3`
`m (l_1) xx m (l_2)` `= -3/4 xx 4/3`
  `= -1`

`:.\ l_1\ text(and)\ l_2\ text(are perpendicular)`

 

(iii)   `text(Show)\ _|_\ text(distance)\ O\ (0,0)\ text(to)\ l_1 = 12/5`
`_|_\ text(dist)` `= | (ax_1 + by_1 + c)/sqrt(a^2 + b^2) |`
  `= | (0 + 0\ – 12)/sqrt(3^2 + 4^2) |`
  ` = | -12/5 |`
  `= 12/5\ text(units    … as required)`

 

(iv)   `text(S)text(ince)\ Delta OBE\ text(is right angled)`
`OB^2` `= OE^2 + BE^2`
`3^2` `= (12/5)^2 + BE^2`
`BE^2` `= 9\ – 144/25`
  `= 81/25`
`:.\ BE` `= 9/5`
  `=1.8\ text(units)`

 

(v)    `text(Area)\ Delta BOE` `= 1/2 bh`
    `= 1/2 xx 9/5 xx 12/5`
    `= 108/50`
    `= 54/25`
    `=2.16\  text(u²)`

Calculus, 2ADV C1 2011 HSC 2c

Find the equation of the tangent to the curve  `y = (2x + 1)^4`   at the point where  `x = –1`.   (3 marks)

Show Answers Only

`8x + y + 7 = 0`

Show Worked Solution

`y = (2x + 1)^4`

`text{Using the Function of a Function Rule  (or Chain Rule)}`

`dy/dx` `= 4 xx (2x + 1)^3 xx d/dx (2x + 1)`
  `= 8 (2x + 1)^3`

 
`text(At)\ \ x = –1,\ y = 1`

MARKER’S COMMENT: The best setting out clearly showed the derivative function, the gradient, the point and then finally, calculations for the equation of the tangent, as per the Worked Solution.
`dy/dx` `= 8 (2(–1) + 1)^3`
  `= 8 (–1)^3`
  `= -8`

 
`text(T)text(angent has)\ m = –8\ text(through)\ (–1,1)`

`text(Using)\ \ \ y – y_1` `= m (x – x_1)`
`y -1` `= -8 (x + 1)`
`y – 1` `= -8x -8`
`8x + y + 7` `= 0`

 
`:.\ text(Equation of tangent is)\ 8x + y + 7 = 0`

Functions, 2ADV F1 2011 HSC 1f

Rationalise the denominator of  `4/(sqrt5\ - sqrt3)`.

Give your answer in the simplest form.   (2 marks)

Show Answers Only

`2(sqrt5 + sqrt3)`

Show Worked Solution

`4/(sqrt5\ – sqrt3) xx (sqrt5 + sqrt3)/(sqrt5 + sqrt3)`

`= (4(sqrt5 + sqrt3))/(5\ – 3)`

`= 2 (sqrt5 + sqrt3)`

Functions, 2ADV F1 2011 HSC 1a

Evaluate  `root(3)(651/(4pi))`  correct to four significant figures.  (2 marks)

Show Answers Only

 `3.728\ text{(to 4 sig. figures)}`

Show Worked Solution
 MARKER’S COMMENT: Show answer to 5 or 6 decimals before rounding. Correct rounding of a wrong answer still receives half marks.
`root(3)(651/(4pi))` `=3.72783…`
  `=3.728\ text{(to 4 sig. figures)}`

Trigonometry, 2ADV T1 2012 HSC 13a

The diagram shows a triangle  `ABC`. The line  `2x + y = 8`  meets the `x` and `y` axes at the points `A` and `B` respectively. The point `C` has coordinates  `(7, 4)`. 
 

2012 13a
 

  1. Calculate the distance  ` AB `.   (2 marks)

  2. It is known that  `AC = 5`  and  `BC = sqrt 65 \ \ \ `(Do NOT prove this)  

     

    Calculate the size of  `angle ABC` to the nearest degree.    (2 marks)

  3. The point `N` lies on  `AB`  such that  `CN`  is perpendicular to  `AB`. 

     

    Find the coordinates of `N`.    (3 marks)

Show Answers Only
  1. `4 sqrt 5  \ text(units)`
  2. ` 34°  \ text{(nearest degree)}`
  3. `N (3,2)`
Show Worked Solution

i.   `text(Find distance) \  AB:`

`text(Find A),\ \ y=0`

`2x + 0` `= 8`
`x` `= 4 \ => A (4,0)`

 
`text(Find B),\ \ x=0`

` 0 + y = 8  \ => B(0,8)`
  

`text(Using Pythagoras:)`

`AB^2` `= OB^2 + OA^2`
  `= 8^2 + 4^2`
  `= 80`
`:. \ AB` `= sqrt 80`
  ` = 4 sqrt 5  \ text(units)`

 

ii.   `text(Find)\  angle ABC:`

`text(Using cosine rule)`

`cos angle ABC` `=  (AB^2 + BC^2 – AC^2)/(2 xx AB xx BC)`
  `= ((4 sqrt 5)^2 + (sqrt 65)^2 – 5^2)/(2 xx 4 sqrt 5 xx sqrt 65)` 
  `= (80 + 65 – 25) / (8 xx sqrt 325)`
  `= 120/(40 sqrt 13)`
  `= 3/ sqrt 13`
  `= 0.83205…`
`:. angle ABC` `= 33.690…`
  `= 34°\ \  text{(nearest degree)}`

 

iii.  `text(Find) \ N:`

`AB   text(is)  \ 2x +y = 8`

`=>  \ text(Gradient)  \ AB = -2`

`:.\ text(Gradient of) \  CN = ½ \ \ \ (m_1 m_2 = -1\ \ text(for ⊥ lines))`

 

`text(Equation of) \ CN, \ m = ½\ text(through)\ (7,4)`

MARKER’S COMMENT: Many students could not find the correct equation on `CN` because they took its gradient to be the reciprocal of `AB` and not the negative reciprocal. 
`y  – 4` `= ½ (x  – 7)`
`2y  – 8` `= x  – 7`
`x  – 2y + 1` `= 0`

 

` N \ text(is intersection of) \ AB \  text(and) \ CN`

`2x + y  – 8` `= 0\ \ …\ (1)`
`x  – 2y + 1` `= 0\ \ …\ (2)`

 
`text(Multiply)  \ (1) xx 2`

`4x +2y  – 16 = 0\ \ …\ (3)`

 
`text(Add) \  (2) + (3)`

`5x  – 15` `= 0`
`x` `=3`

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

`2(3) + y  – 8 = 0   =>  y = 2`

`:. N (3,2)`

Integration, 2UA 2012 HSC 12d

At a certain location a river is `12` metres wide. At this location the depth of the river, in metres, has been measured at `3` metre intervals. The cross-section is shown below.
 

2012 12d
 

  1. Use Simpson's rule with the five depth measurements to calculate the approximate area of the cross-section   (3 marks)
  2. The river flows at 0.4 metres per second.
  3. Calculate the approximate volume of water flowing through the cross-section in 10 seconds.   (1 mark)
  4.  
Show Answers Only
  1. `32.8 \ text(m²)`
  2. `131.2 \ text(m³)`  
Show Worked Solution
(i)    `A` `~~ h/3 [y_0 + 4y_1 + 2y_2 +4y_3 +y_4 ]`
  `~~3/3 [ 0.5 + (4xx2.3) + (2xx2.9) + (4xx3.8) + 2.1 ]`
  `~~ [ 0.5 + 9.2 + 5.8 + 15.2 + 2.1 ]`
  `~~ 32.8\  text(m²)`

 

♦ Over half of the students examined answered part (ii) incorrectly. 
(ii)  `text(Distance water flows)`  `= 0.4 xx 10`
  `= 4 \ text(metres)`

 

`text(Volume flow in 10 seconds)` `~~ 4 xx 32.8`
  `~~ 131.2  text(m³)`

Functions, EXT1* F1 2012 HSC 11b

 Solve  `|\ 3x -1\ | < 2`   (2 marks)

Show Answers Only

 ` -1/3 < x < 1 `

Show Worked Solution
 MARKER’S COMMENT: Note that both conditions must be satisfied! Dealing with negative signs and division for inequalities produced many errors.

`|\ 3x -1\ | < 2`

`3x -1` `<2`  `\ \ \ \ \-(3x -1)` `< 2`
`3x`  `<3` `-3x + 1` `< 2`
`x` `< 1`  `3x` `> -1`
    `x` `> -1/3`

`:. -1/3 < x < 1`

Calculus, 2ADV C1 2012 HSC 11c

Find the equation of the tangent to the curve  `y = x^2`  at the point where  `x = 3`.   (2 marks)

Show Answers Only

`6x-y-9 = 0 `

Show Worked Solution
`y` `=x^2`
`dy/dx` `=2x`

 

`text{Need to find equation with m = 6, through (3,9) }`

`text(When) \  x = 3, y = 9, dy/dx = 6`

`text(Using)\ \ \ y-y_1`  `= m (x-x_1)`
`y -9`  `= 6(x -3)`
`y -9`  `= 6x -18`
`6x -y -9` `=0`

 

 `:.\ text( Equation of the tangent is 6x-y-9 = 0)`

 

Functions, 2ADV F1 2012 HSC 2 MC

Which of the following is equal to  `1/(2sqrt5\-sqrt3)`? 

  1. `(2sqrt5\-sqrt3)/7`  
  2. `(2sqrt5 + sqrt3)/7`
  3. `(2sqrt5\-sqrt3)/17` 
  4. `(2sqrt5 + sqrt3)/17` 
Show Answers Only

`D`

Show Worked Solution

`1/(2sqrt5\-sqrt3) xx (2sqrt5 + sqrt3)/(2sqrt5 + sqrt3)`

`= (2sqrt5 + sqrt3)/( (2sqrt5\-sqrt3)(2sqrt5 + sqrt3) )`

`= (2sqrt5 + sqrt3)/{(2sqrt5)^2-(sqrt3)^2)`

`= (2sqrt5 + sqrt3)/17`

`=>  D`

Linear Functions, 2UA 2013 HSC 12b

The points  `A(–2, –1)`,  `B(–2, 24)`,  `C(22, 42)` and  `D(22, 17)` form a parallelogram as shown. The point  `E(18, 39)` lies on  `BC`. The point  `F`  is the midpoint of  `AD`.
 

2013 12b
 

  1. Show that the equation of the line through  `A`  and  `D`  is  `3x- 4y + 2 = 0`.    (2 marks)
  2. Show that the perpendicular distance from  `B`  to the line through  `A`  and  `D`  is  `20`  units.   (1 mark)
  3. Find the length of  `EC`.    (1 mark)
  4. Find the area of the trapezium  `EFDC`.    (2 marks)
Show Answers Only
  1. `3x-4y+2=0`
  2. `text(Proof)\  text{(See Worked Solutions)}`
  3. `text(5 units)`
  4. `200\  text(u²)`
Show Worked Solution
(i)    `text(Need to find the equation of)\ AD`
`m_(AD)` `= (y_2\-y_1)/(x_2\-x_1)`
  `= (17+1)/(22+2)`
  `= 18/24`
  `= 3/4`

`text(Equation of)\ AD\ text(has)\ m=3/4 text(, through)\ A text{(–2,–1)}`

`text(Using)\ y\-y_1` `= m (x\-x_1)`
`y+1` `=3/4 (x +2)`
`4y+4` `=3x+6`
`3x-4y+2` `=0`
MARKER’S COMMENT: Students must know the perpendicular distance formula which was again the cause of many errors such that it was flagged by markers.
 

(ii)    `B(–2,24)`
  `AD\ text(is)\  \ 3x -4y+2 =0`
`_|_ text(dist)` `= | (ax_1 + by_1 + c)/sqrt(a^2 + b^2) |`
  `= | (3(–2)\-4(24) + 2)/sqrt(3^2 + (–4)^2 )|`
  `= | (-6\ -96 +2)/sqrt25 |`
  `= | -100/5 |`
  `= 20\ text(units)`

 

`:.\ _|_ text(dist of)\ B\ text(from)\ AD = 20\ text(units   … as required)`

 

(iii)   `E(18,39)\ \ \ C(22,42)`
`EC` `=sqrt ( (x_2\-x_1)^2 + (y_2\-y_1)^2 )`
  `= sqrt ( (22\-18)^2 + (42\-39)^2 )`
  `= sqrt (4^2 + 3^2)`
  `= sqrt25`
  `= 5\ text(units)`

 

`:.\ text(The length of)\ EC\ text(is 5 units.)`

 

(iv)    `text(Area of trapezium)` `=1/2 h (a+b)`
    `=1/2 h (EC + FD)`

`EC = 5\ text(units)`

 MARKER’S COMMENT: One of the most common cause of errors in this part occurred when students failed to use the result found in part (ii).

`text(Need to find)\ FD`

`text(S)text(ince)\ FD = 1/2 xx AD\ \ \ ( F\ text(is midpoint) )`

`A (–2,–1)\ \ D(22,17)`

`AD` `=sqrt( (22 + 2) + (17 + 1)^2 )`
  `= sqrt (24^2 + 18^2)`
  `= sqrt (900)`
  `= 30`

`=> FD = 1/2 xx 30 = 15`

`text(S)text(ince)\ ABCD\ text(is a parallelogram)`

`h` `= _|_ text(distance in part)\ text{(i)}`
  `= 20`
`:.\ text(Area of trapezium)` `=1/2 xx 20 (5 + 15)`
  `=200\  text(u²)`

Calculus, 2ADV C3 2013 HSC 12a

The cubic  `y = ax^3 + bx^2 + cx + d`  has a point of inflection at  `x = p`. 

Show that  `p= - b/(3a)`.   (2 marks)

Show Answers Only

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

Show Worked Solution

`text(Show)\ \ p= – b/(3a)`

`y` `=ax^3 + bx^2 + cx + d`
`y prime` `=3ax^2 + 2bx + c`
`y″` `=6ax + 2b`

 
`text(Given P.I. occurs when)\ \ x = p`

`=> y″=0\ \ text(when)\ \ x=p`

`:.\ 6ap + 2b` `=0`
`6ap` `=-2b`
`p` `= -(2b)/(6a)`
  `=-b/(3a)\ \ \ text(… as required)`

Functions, 2ADV F1 2013 HSC 1 MC

What are the solutions of   `2x^2-5x-1 = 0`? 

  1. `x = (-5 +-sqrt17)/4` 
  2. `x = (5 +-sqrt17)/4`
  3. `x = (-5 +-sqrt33)/4`
  4. `x = (5 +-sqrt33)/4`
Show Answers Only

`D`

Show Worked Solution

`2x^2-5x-1 = 0`

`text(Using)\ x = (-b +- sqrt( b^2-4ac) )/(2a)`

`x` `= (5 +- sqrt{\ \ (-5)^2-4 xx 2 xx(-1) })/ (2 xx 2)`
  `= (5 +- sqrt(25 + 8) )/4`
  `= (5 +- sqrt(33) )/4`

 
`=>  D`

Financial Maths, STD2 F1 2010 HSC 23d

Warrick has a net income of $590 per week. He has created a budget to help manage his money.

       2010 23d

  1. Find the value of `X`, the amount that Warrick allocates towards electricity each week.    (1 mark)

  2. Warrick has an unexpectedly high telephone and internet bill. For the last three weeks, he has put aside his savings as well as his telephone and internet money to pay the bill.

     

    How much money has he put aside altogether to pay the bill?    (1 mark)

  3. The bill for the telephone and internet is $620. It is due in two weeks time. Warrick realises he has not put aside enough money to pay the bill.

     

    How could Warrick reallocate non-essential funds in his budget so he has enough money to pay the bill? Justify your answer with suitable reasons and calculations.   (3 marks)

Show Answers Only
  1. `$25`
  2. `text(Warwick has put aside $240 to pay the bill.)`
  3. `text(Warwick could reallocate funds for Entertainment and`

  4.  

    `text(Clothes and Gifts to pay the bill.)`

Show Worked Solution
i.    `X` `= 590 ` `-(175 + 45 + 10 + 15 + 90`
      `+ 40 + 30 + 70 + 50 + 40)`
    `= 590-565`
    `= $25`

 

ii.    `text(Weekly Amount)` `= 40 + 40`
    `= 80`
`text{Total (3 weeks)}` `= 3 xx 80`
  `= 240`

 
`:.\ text(Warwick has put aside $240 to pay the bill.)`

 

(iii)   `text(Amount required less amount put aside)`
  `= 620\-240`
  `= $380`

 
`text(Extra 2 weeks of savings and telephone)`

`= 2 xx (40 + 40)`

`= $160`
 

`:.\ text(Funds to be reallocated)`

`= 380\-160`

`= 220\ text(over 2 weeks)`

`= $110\ text(per week)`
 

`text(Non essential items are Entertainment)`

`text(and Clothes and Gifts)`

`text(Amount)` `= 70 + 50`
  `= $120\ text(per week)`

 

`:.\ text(Warwick could reallocate funds for Entertainment)`

`text(and Clothes and Gifts to pay the bill.)`

Measurement, STD2 M7 2011 HSC 24a

Part of the floor plan of a house is shown. The plan is drawn to scale.

2011 24a

  1. What is the width of the stairwell, in millimetres?   (1 mark)

  2. What are the internal dimensions of the bathroom, in millimetres?    (1 mark)

  3. What is the length `AB`, the internal length of the rumpus room, in millimetres?   (1 mark)

Show Answers Only
  1. `900\  text(mm)`
  2. `2000\  text(mm) xx 2000\  text(mm)`
  3. `9695\  text(mm)`
Show Worked Solution
i.      `900\  text(mm)`

 

ii.     `2000\  text(mm) xx 2000\  text(mm)`

 

iii.    `text(Length of Rumpus Room) = AB`
 `AB` `= 3600 + 90 + 2000 + 90 + 3915`
  `=9695\ text(mm)`

Probability, 2UG 2011 HSC 26a

The two spinners shown are used in a game.

2UG 2011 26a1

Each arrow is spun once. The score is the total of the two numbers shown by the arrows.
A table is drawn up to show all scores that can be obtained in this game.

2UG 2011 26a2

  1. What is the value of `X` in the table?   (1 mark)
  2. What is the probability of obtaining a score less than 4?   (1 mark)
  3. On Spinner `B`, a 2 is obtained. What is the probability of obtaining a score of 3?   (1 mark)
  4. Elise plays a game using the spinners with the following financial outcomes.  

⇒ Win `$12` for a score of `4`

⇒ Win nothing for a score of less than `4`

⇒ Lose `$3` for a score of more than `4`

It costs `$5` to play this game. Will Elise expect a gain or a loss and how much will it be?

Justify your answer with suitable calculations.   (3 marks)

Show Answers Only
  1. `5`
  2. `1/2`
  3.  
  4. `2/3`
  5.  
  6. `text(Loss of)\ $1.50`
  7.  
Show Worked Solution

(i)   `X=3+2=5`

 

(ii)   `P(text{score}<4)=6/12=1/2`

 

(iii)   `P(3)=2/3`

 

(iv)   `P(4)=4/12=1/3`

♦ Mean mark 34%
MARKER’S COMMENT: Better responses remembered to deduct the $5 cost to play and recognised the negative result as a loss.
`P(text{score}<4)` `=6/12=1/2`
`P(text{score}>4)` `=2/12=1/6`

 

`text(Financial Expectation)`

`=(1/3xx12)+(1/2xx0)-(1/6xx3)-5`
`=4-0.5-5`
`=-1.50`

 

`:.\ text(Elise should expect a loss of $1.50) `

Probability, STD2 S2 2011 HSC 25c

At another school, students who use mobile phones were surveyed. The set of data is shown in the table.

2UG 2011 25c

  1. How many students were surveyed at this school?  (1 mark)

  2. Of the female students surveyed, one is chosen at random. What is the probability that she uses pre-paid?  (1 mark)

Ten new male students are surveyed and all ten are on a plan. The set of data is updated to include this information.

  1. What percentage of the male students surveyed are now on a plan? Give your answer to the nearest per cent.  (1 mark)

Show Answers Only
  1. `580`
  2. `text(54%)`
  3. `text(42%)`
Show Worked Solution

i.   `text(# Students surveyed)=319+261=580`

 

ii.   `Ptext{(Female uses prepaid)}=text(# Females on prepaid)/text(Total females)`

  `=172/319`
  `=0.53918…`
  `=\ text{54%  (nearest %)}`

 

iii.   `text(% Males on plan)` `=text(# Males on plan + 10)/text(Total males + 10)`
  `=(103+10)/(261+10)`
  `=113/271`
  `=0.4169…`
  `=\ text{42%  (nearest %)}`

Statistics, STD2 S1 2011 HSC 25b

The graph below displays data collected at a school on the number of students
in each Year group, who own a mobile phone.
 

2UG 2011 25b
 

  1. Which Year group has the highest percentage of students with mobile phones? (1 mark)

  2. Two students are chosen at random, one from Year 9 and one from Year 10.

     

    Which student is more likely to own a mobile phone?

     

    Justify your answer with suitable calculations. (2 marks)

  3. Identify a trend in the data shown in the graph. (1 mark)

Show Answers Only
  1. `text{Year 12 (100%)}`
  2. `text(Year 10)`
  3. `text(See Worked Solutions for detail)`
Show Worked Solution

i.   `text(Year 12 (100%))`

MARKER’S COMMENT: Many students ignore the instruction to use calculations and lose marks as a result.
 

ii.     `text(% Ownership in Year 9)` `=55/70`
    `=\ text{78.6%  (1d.p.)}`
      `text(% Ownership in Year 10)` `=50/60`
    `=\ text{83.3%  (1d.p.)}`

  
`:.\ text(The Year 10 student is more likely to own a mobile phone.)`

 

iii.   `text(% Ownership increases as students)`

 `text(progress from Year 7 to Year 12.)`

Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.

2010 27c

  1. Use the graph to find the tax payable on a taxable income of $21 000.  (1 mark)

  2. Use suitable points from the graph to show that the gradient of the section of the graph marked  `A`  is  `1/3`.    (1 mark)

  3. How much of each dollar earned between  $21 000  and  $39 000  is payable in tax?   (1 mark)

  4. Write an equation that could be used to calculate the tax payable, `T`, in terms of the taxable income, `I`, for taxable incomes between  $21 000  and  $39 000.   (2 marks)

Show Answers Only
  1. `$3000\ \ \ text{(from graph)}`
  2. `1/3`
  3. `33 1/3\ text(cents per dollar earned)`
  4. `text(Tax payable on)\ I = 1/3 I\-4000`
Show Worked Solution
i.

 `text(Income on)\ $21\ 000=$3000\ \ \ text{(from graph)}`

 

ii.  `text(Using the points)\ (21,3)\ text(and)\ (39,9)`

♦♦ Mean mark 25%
`text(Gradient at)\ A` `= (y_2\-y_1)/(x_2\ -x_1)`
  `= (9000-3000)/(39\ 000 -21\ 000)`
  `= 6000/(18\ 000)`
  `= 1/3\ \ \ \ \ text(… as required)`

 

iii.  `text(The gradient represents the tax applicable to each dollar)`

♦♦♦ Mean mark 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.
`text(Tax)` ` = 1/3\ text(of each dollar earned)`
  ` = 33 1/3\ text(cents per dollar earned)`

 

iv.  `text( Tax payable up to $21 000 = $3000)`

`text(Tax payable on income between $21 000 and $39 000)`

♦♦♦ Mean mark 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.

` = 1/3 (I\-21\ 000)`

`:.\ text(Tax payable on)\ \ I` `= 3000 + 1/3 (I\-21\ 000)`
  `= 3000 + 1/3 I\-7000`
  `= 1/3 I\-4000`

Statistics, STD2 S1 2010 HSC 26b

A new shopping centre has opened near a primary school. A survey is conducted to determine the number of motor vehicles that pass the school each afternoon between 2.30 pm and 4.00 pm.

The results for 60 days have been recorded in the table and are displayed in the cumulative frequency histogram.
 

2010 26b

  1. Find the value of  Χ  in the table.   (1 mark)

  2. On the cumulative frequency histogram above, draw a cumulative frequency polygon (ogive) for this data.   (1 mark)
  3. Use your graph to determine the median. Show, by drawing lines on your graph, how you arrived at your answer.   (1 mark)
  4. Prior to the opening of the new shopping centre, the median number of motor vehicles passing the school between  2.30 pm  and  4.00 pm  was 57 vehicles per day.

     

    What problem could arise from the change in the median number of motor vehicles passing the school before and after the opening of the new shopping centre?

     

    Briefly recommend a solution to this problem.   (2 marks)

Show Answers Only
  1. `15`
  2.  
  3.  
  4. `text(Problems)`
  5. `text(- increased traffic delays)`
  6. `text(- increased danger to students leaving school)`

  7.  

    `text(Solutions)`

  8. `text(- signpost alternative routes around school)`
  9. `text(- decrease the speed limit in the area)`
Show Worked Solution
i. `X` `= 25\ -10`
    `= 15`

 

♦♦♦ Mean mark 18%
MARKER’S COMMENT: The ogive was poorly drawn with many students incorrectly joining the middle of each column rather than from corner to corner.
ii.
♦♦ Mean mark 25%
MARKER’S COMMENT: Many students did not “show by drawing lines on the graph” as the question asked.

iii.  `text(Median)\ ~~155`

♦ Mean mark 47%
MARKER’S COMMENT: Short answers were often the best. Be concise when you can.
iv. `text(Problems)`
  `text(- increased traffic delays)`
  `text(- increased danger to students leaving school)`
   
  `text(Solutions)`
  `text(- signpost alternative routes around school)`
  `text(- decrease the speed limit in the area)`

Algebra, STD2 A4 2013 HSC 30a

Wind turbines, such as those shown, are used to generate power.

2013 30a

In theory, the power that could be generated by a wind turbine is modelled using the equation

`T = 20\ 000w^3`

where `T` is the theoretical power generated, in watts 
  `w` is the speed of the wind, in metres per second.

 

  1. Using this equation, what is the theoretical power generated by a wind turbine if the wind speed is 7.3 m/s ?  (1 mark)

In practice, the actual power generated by a wind turbine is only 40% of the theoretical power.

  1. If `A` is the actual power generated, in watts, write an equation for `A` in terms of `w`.    (1 mark)

The graph shows both the theoretical power generated and the actual power generated by a particular wind turbine.
 
        2013 30a2

  1. Using the graph, or otherwise, find the difference between the theoretical power and the actual power generated when the wind speed is 9 m/s.   (1 mark)

A particular farm requires at least 4.4 million watts of actual power in order to be self-sufficient.

  1. What is the minimum wind speed required for the farm to be self-sufficient?    (1 mark)

A more accurate formula to calculate the power (`P`) generated by a wind turbine is

`P = 0.61 xx pi xx r^2 × w^3` 

where     `r` is the length of each blade, in metres
  `w` is the speed of the wind, in metres per second. 

 
Each blade of a particular wind turbine has a length of 43 metres.The turbine operates at a wind speed of 8 m/s.

  1. Using the formula above, if the wind speed increased by 10%, what would be the percentage increase in the power generated by this wind turbine?   (3 marks)

Show Answers Only
  1. `7\ 780\ 340\ text(watts)`
  2. `8000w^3`
  3. `8.8\ text(million watts, or 8 748 000 watts)`
  4. `w = 8.2\ text(m/s)\ \ \ text{(1 d.p.)}`
  5. `text(33%)`
Show Worked Solution
i.    `T=20\ 000w^3`
  `text(If)\ \ w = 7.3`
`T` `=20\ 000 xx (7.3)^3`
  `= 7\ 780\ 340\ text(watts)`

  

♦ Mean mark 34%
ii.    `text(We know)\ A = 40% xx T`
`=> A` `=0.4 xx 20\ 000 xx w^3`
  `=8000w^3`

  

♦ Mean mark 38%
iii.    `text(Solution 1)`
  `text(At)\ w=9`
  `A = text(5.8 million watts)\ \ \ text{(from graph)}`
  `T = text(14.6 million watts)\ \ \ text{(from graph)}`
`text(Difference)` `= text(14.6 million)\-text(5.8 million)`
  `= text(8.8 million watts)`

 

`text(Alternative Solution)`
`text(At)\ w=9`
`T` `= 20\ 000 xx 9^3`
  `= 14\ 580\ 000\ text(watts)`
`A` `= 8000 xx 9^3`
  `= 5\ 832\ 000\ text(watts)`
`text(Difference)` `=14\ 580\ 000\-5\ 832\ 000`
  `=8\ 748\ 000\ \ text(watts)`

  

♦♦ Mean mark 25%
COMMENT: Students need to be comfortable in finding the cube roots of values – a calculation that can be required in a number of topic areas and is regularly examined.
iv.    `text(Find)\ w\ text(if)\ A=4.4\ text(million)`
`8000w^3` `= 4\ 400\ 000`
`w^3` `= (4\ 400\ 000)/8000`
  `= 550`
`:. w` `= root(3)(550)`
  `=8.1932…`
  `=8.2\ text(m/s)\ \ \ text{(1 d.p.)}`

 

`:.\ text(The minimum wind speed required is 8.2 m/s)`

 

♦ Mean mark 41%
MARKER’S COMMENT: Students are reminded that a % increase requires them to find the difference in power generated at different wind speeds and divide this result by the original power output, as shown in the Worked Solution.
v.    `text(Find)\ P\ text(when)\ w=8\ text(and)\ r=43`
`P` `= 0.61 xx pi xx r^2 xx w^3`
  `= 0.61 xx pi xx 43^2 xx 8^3`
  `= 1\ 814\ 205.92\ text(watts)`

 

`text(When speed of wind)\ uarr10%`
`w′ = 8 xx 110text(%) = 8.8\ text(m/s)`
 

`text(Find)\ P\ text(when)\ w′ = 8.8`

`P` `=0.61 xx pi xx 43^2 xx 8.8^3`
  `= 2\ 414\ 708.08\ text(watts)`

 

`text(Increase in Power)` `=2\ 414\ 708.08\-1\ 814\ 205.92`
  `= 600\ 502.16`

 

`:.\ text(% Power increase)` `= (600\ 502.16)/(1\ 814\ 205.92)`
  `= 0.331`
  `= text(33%)\ \ \ text{(nearest %)}`

 

Financial Maths, STD2 F1 2011 HSC 23a

Sri has a gross salary of  $56 350. She has tax deductions of $350 for union fees, $2000 in work-related expenses and $250 in donations to charities.

The Medicare levy is 1.5% of her taxable income.

Calculate Sri’s Medicare levy.   (3 marks)

Show Answers Only

 `$806.25`

Show Worked Solution
MARKER’S COMMENT: Clear separate steps results in less calculation errors and more marks for working when errors are made.
`text(Deductions)` `=350+2000+250`
  `=$2600`
`text(Taxable)` `=\ text(Gross Income – deduct)`
  `=56\ 350-2600`
  `=$53\ 750`
`text(Medicare Levy)` `=\ text(1.5%)\ xx 53\ 750`
  `=$806.25`

 

`:.\ text(Sri’s Medicare levy is)\ \ $806.25`

Probability, STD2 S2 2011 HSC 15 MC

An unbiased coin is tossed 10 times.

A tail is obtained on each of the first 9 tosses.

What is the probability that a tail is obtained on the 10th toss?

  1. `1/2^10`
  2. `1/2`
  3. `1/10`
  4. `9/10`
Show Answers Only

`B`

Show Worked Solution

`text(Each toss is an independent event and has an even chance)`

`text(of being a head or tail.)`

`=> B`

Probability, STD2 S2 2010 HSC 23c

On Saturday, Jonty recorded the colour of T-shirts worn by the people at his gym. The results are shown in the graph.

 

  1. How many people were at the gym on Saturday? (Assume everyone was wearing a T-shirt).   (1 mark)

  2. What is the probability that a person selected at random at the gym on Saturday, would be wearing either a blue or green T-shirt?   (1 mark)

Show Answers Only
  1. `34`
  2. `15/34`
Show Worked Solution
i.   `text(# People)` `=5+15+10+3+1`
  `=34`

 

ii.   `P (B\ text{or}\ G)` `=P(B)+P(G)`
  `=5/34+10/34`
  `=15/34`

Financial Maths, STD2 F1 2010 HSC 23a

This advertisement appeared in a newspaper

2010 23a 

 What is the maximum possible salary per annum for this civil engineer, correct to the nearest dollar?   (2 marks)

Show Answers Only

`$86\ 396`

Show Worked Solution

`text(Base wage)=1586.70\ text(pw)`

`text(Max weekly base)` `=$1586.70+(text{3.5%}\ xx 1586.70)`
  `=1586.70+55.53`
  `=1642.23\ text((nearest cent))`
`:.\ text(Max annual base)` `=1642.23xx52`
  `=85\ 395.96`
  `=$85\ 396\ text{(nearest dollar)}`

Financial Maths, STD2 F4 2009 HSC 24e

Jay bought a computer for $3600. His friend Julie said that all computers are worth nothing (i.e. the value is $0) after 3 years.

  1. Find the amount that the computer would depreciate each year to be worth nothing after 3 years, if the straight line method of depreciation is used.   (1 mark)

  2. Explain why the computer would never be worth nothing if the declining balance method of depreciation is used, with 30% per annum rate of depreciation. Use suitable calculations to support your answer.    (2 marks)

Show Answers Only
  1. `$1200`
  2. `text(See Worked Solutions.)`
Show Worked Solution
i.    `S` `= V_0-Dn`
  `0` `= 3600-D xx 3`
  `3D` `= 3600`
  `D` `= 3600/3`
    `= 1200`

 
`:.\ text(Annual depreciation = $1200`

 

♦ Mean mark 45%
ii   `text(Using)\ \ S = V_0 (1-r)^n`
  `text(where)\ r = text(30%)\ \ text(and)\ \ V_0 = 3600`

 

`S` `=3600 (1-30/100)^n`  
  `= 3600 (0.7)^n`  

 
`(0.7)^n > 0\ text(for all)\ n`

`:.\ text(Salvage value is always)\ >0`

Statistics, STD2 S1 2009 HSC 24b

Tayvan is an international company that reports its profits in the USA, Belgium and India at the end of each quarter. The profits for 2008 are shown in the area chart.

2009 24b

  1. What was the total profit for Tayvan on June 30?   (1 mark)
  2. What was Tayvan’s profit in Belgium on March 31?     (1 mark)
Show Answers Only
  1. `$8\ 000\ 000\ \ text{(from graph)}`
  2. `$4\ 000\ 000`
Show Worked Solution
MARKER’S COMMENT: Area charts provide cumulative totals. Many students did not know this and incorrectly answered 14 million.
(i)     `$8\ 000\ 000\ \ \ text{(from graph)}`

 

(ii)    `text{Belgium profit (30 March)}`
  `= $5\ 000\ 000-$1\ 000\ 000` 
  `= $4\ 000\ 000`

Statistics, STD2 S1 2009 HSC 24a

The diagram below shows a stem-and-leaf plot for 22 scores. 
 

2UG-2009-24a
 

  1.  What is the mode for this data?   (1 mark)

  2.  What is the median for this data?     (1 mark)

Show Answers Only
  1. `78`
  2. `46`
Show Worked Solution

i.   `text(Mode) = 78`

 

ii.    `22\ text(scores)`

`=>\ text(Median is the average of 11th and 12th scores)`
 

`:.\ text(Median)` `= (45 + 47)/2`
  `= 46`

Financial Maths, STD2 F1 2009 HSC 10 MC

Billy worked for 35 hours at the normal hourly rate of pay and for five hours at double time. He earned $561.60 in total for this work.

What was the normal hourly rate of pay?

  1. $7.02
  2. $12.48
  3. $14.04
  4. $16.05
Show Answers Only

`B`

Show Worked Solution

`text(Let hourly rate) = $X\ text(per hour)`

`35X + 5(2X)` `= 561.60`
`45X` `=561.60`
`X` `= 561.60/45` 
  `= $12.48`

 
`=>  B`

Financial Maths, STD2 F4 2009 HSC 6 MC

A house was purchased in 1984 for $35 000. Assume that the value of the house has increased by 3% per annum since then. 

Which expression gives the value of the house in 2009?  

  1. `35\ 000(1 + 0.03)^25`
  2. `35\ 000(1 + 3)^25` 
  3. `35\ 000 xx 25 xx 0.03`
  4. `35\ 000 xx 25 xx 3`
Show Answers Only

`A`

Show Worked Solution

`r =\ text(3%)\ = 0.03`

`n = 25\ text(years)`

`text(Using)\ \ FV = PV(1 + r)^n`

` :.\ text(Value in 2009) = 35\ 000(1+0.03)^25` 

`=>  A`

Statistics, STD2 S3 2009 HSC 5 MC

Jamie wants to know how many songs were downloaded legally from the internet in the last 12 months by people aged 18–25 years. He has decided to conduct a statistical inquiry.

After he collects the data, which of the following shows the best order for the steps he should take with the data to complete his inquiry?

  1.    Display, organise, conclude, analyse
  2.    Organise, display, conclude, analyse
  3.    Display, organise, analyse, conclude
  4.    Organise, display, analyse, conclude
Show Answers Only

`D`

Show Worked Solution

`text(Process of statistical enquiry requirements)`

`=>  D`

Statistics, STD2 S1 2009 HSC 3 MC

The eye colours of a sample of children were recorded.

When analysing this data, which of the following could be found?

  1. Mean
  2. Median
  3. Mode
  4. Range
Show Answers Only

`C`

Show Worked Solution

`text(Eye colour is categorical data)`

`:.\ text(Only the mode can be found)`

`=>  C`