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Geometry and Calculus, EXT1 2009 HSC 4b

Consider the function  `f(x) = (x^4 + 3x^2)/(x^4 + 3)`. 

  1. Show that  `f(x)`  is an even function.    (1 mark)
  2. What is the equation of the horizontal asymptote to the graph  `y = f(x)`?    (1 mark)
  3. Find the  `x`-coordinates of all stationary points for the graph  `y = f(x)`.   (3 marks)
  4. Sketch the graph  `y = f(x)`. You are not required to find any points of inflection.    (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `y = 1`
  3. `x = 0, – sqrt 3, sqrt 3`
  5.  
Show Worked Solution
(i)    `f(x)` `= (x^4 + 3x^2)/(x^4 + 3)`
  `f(–x)` `= ((–x)^4 + 3(-x)^2)/((–x)^4 + 3)`
    `= (x^4 + 3x^2)/(x^4 + 3)`
    `= f(x)`

 

`:.\ text(Even function.)`

 

(ii)    `y` `= (x^4 + 3x^2)/(x^4 + 3)`
    `= (1 + 3/(x^2))/(1 + 3/(x^4))`
`text(As)\ \ ` `x` `-> oo`
  `y` `-> 1`

 

`:.\ text(Horizontal asymptote at)\ \ y = 1`

 

(iii)   `f(x) = (x^4 + 3x^2)/(x^4 + 3)`
`u` `= x^4 + 3x^2\ \ \ \ \ ` `v` `= x^4 + 3`
`u prime` `= 4x^3 + 6x\ \ \ \ \ ` `v prime` `= 4x^3`
`f prime (x)` `= (u prime v\ – u v prime)/(v^2)`
  `= ((4x^3 + 6x)(x^4 + 3)\ – (x^4 + 3x^2)4x^3)/((x^4 + 3)^2)`
  `= (4x^7 + 12x^3 + 6x^5 + 18x\ – 4x^7\ – 12x^5)/((x^4 + 3)^2)`
  `= (-6x^5 + 12x^3 + 18x)/((x^4 + 3)^2)`
  `= (-6x(x^4\ – 2x^2\ – 3))/((x^4 + 3)^2)`

 

`text(S.P. when)\ x = 0\ \ \ text(or) \ \ x^4\ – 2x^2\ – 3 = 0`

`text(Let)\ X = x^2`

MARKER’S COMMENT: Many students did not realise the denominator of  `f′(x)` could be ignored when equating  `f′(x)=0`.
`X^2\ – 2X\ – 3` `= 0`
`(X\ – 3)(X + 1)` `= 0`

`X = 3\ \ text(or)\ \ -1`

`:. x^2` `= 3` `text(or)\ \ \ \ \ ` `x^2 = -1`
`x` `= +- sqrt3\ \ \ \ `   `text{(no solution)}`

 

`:.\ text(SPs occur when)\ \ x = 0, – sqrt3, sqrt3`

 

(iv)    `text(When)\ x = 0,\ ` `y = 0`
  `text(When)\ x = sqrt3,\ \ \ ` `y = ((sqrt3)^4 + 3(sqrt3)^2)/((sqrt3)^4 + 3) = 3/2`

 

EXT1 2009 4b

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.)}`

Functions, EXT1 F2 2009 HSC 2a

The polynomial  `p(x) = x^3-ax + b`  has a remainder of  `2`  when divided by  `(x-1)`  and a remainder of  `5`  when divided by  `(x + 2)`.  

Find the values of  `a`  and  `b`.   (3 marks)

Show Answers Only
`a` `= 4`
`b` `= 5`
Show Worked Solution
`p(x)` `= x^3-ax + b`
`P(1)` `= 2`
`1-a + b` `= 2`
`b` `= a+1\ \ \ …\ text{(1)}`
`P (-2)` `= 5`
`-8 + 2a + b` `= 5`
`2a + b` `= 13\ \ \ …\ text{(2)}`

 

`text(Substitute)\ \ b = a+1\ \ text(into)\ \ text{(2)}`

`2a + a+1` `= 13`
`3a` `= 12`
`:. a` `= 4`
`:. b` `= 5`

Plane Geometry, 2UA 2014 HSC 15b

In  `Delta DEF`, a point  `S`  is chosen on the side  `DE`. The length of  `DS`  is  `x`, and the length of  `ES`  is  `y`. The line through  `S`  parallel to  `DF`  meets  `EF`  at  `Q`. The line through  `S`  parallel to  `EF`  meets  `DF`  at  `R`.

The area of  `Delta DEF`  is  `A`. The areas of  `Delta DSR`  and  `Delta SEQ`  are  `A_1`  and  `A_2`  respectively.

  1. Show that  `Delta DEF`  is similar to  `Delta DSR`.    (2 marks)
  2. Explain why  `(DR)/(DF) = x/(x + y)`.    (1 mark)
  3.  
  4. Show that  
    1. `sqrt ((A_1)/A) = x/(x + y)`.  (2 marks)
    2.  
  5. Using the result from part (iii) and a similar expression for  
    1. `sqrt ((A_2)/A)`, deduce that  `sqrt A = sqrt (A_1) + sqrt (A_2)`.   (2 marks)

 

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Corresponding sides of similar)\ Delta text(s)`
  3. `text(are in the same ratio.)`
  4.  
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
  6. `text(Proof)\ \ text{(See Worked Solutions)}`
  7.  
Show Worked Solution

(i)   `text(Need to show)\ Delta DEF\  text(|||) \ Delta DSR`

`/_FDE\ text(is common)`

`/_DSR = /_DEF = theta\ \ text{(corresponding angles,}\ RS\ text(||)\ FE text{)}`

`:.\ Delta DEF \ text(|||) \ Delta DSR\ \ text{(equiangular)}`

 

(ii)  `(DR)/(DF) = (DS)/(DE) = x/(x + y)`

`text{(Corresponding sides of similar triangles)}`

 

♦♦ Mean mark 27%
COMMENT: The critical step in solving part (iii) is realising that you need the areas of 2 non-right angled triangles and therefore the formula  `A=½\ ab sin C` is required.

(iii)  `text(Show)\ sqrt((A_1)/A) = x/(x + y)`

`text(Using Area)` `= 1/2 ab sin C`
`A_1` `= 1/2 xx DR xx x xx sin alpha`
`A` `= 1/2 xx DF xx (x + y) xx sin alpha`
`(A_1)/A` `= (1/2 * DR * x * sin alpha)/(1/2 * DF * (x + y) * sin alpha)`
  `= (DR * x)/(DF * (x + y)`
  `= (x * x)/((x + y)(x + y))\ \ \ \ text{(using part(ii))}`
  `= (x^2)/((x + y)^2)`
`:.\ sqrt ((A_1)/A)` `= x/((x + y))\ \ \ text(… as required.)`

 

(iv)  `text(Consider)\ Delta DFE\ text(and)\ Delta SQE`

`/_FED` `= theta\ text(is common)`
`/_FDE` `= /_QSE = alpha\ \ ` `text{(corresponding angles,}\ DF\ text(||)\ QS text{)}`

 

`:.\ Delta DFE\ text(|||)\ Delta SQE\ \ text{(equiangular)}`

`(QE)/(FE) = (SE)/(DE) = y/(x +y)`

`text{(corresponding sides of similar triangles)}` 

♦♦ Mean mark 26%
`A_2` `= 1/2 xx QE xx y xx sin theta`
`A` `= 1/2 xx FE xx (x + y) xx sin theta`
`(A_2)/A` `= (QE * y)/(FE * (x + y))`
  `= (y^2)/((x + y)^2)`
`sqrt ((A_2)/A)` `= y/((x + y))`

 

`text(Need to show)\ sqrt A = sqrt (A_1) + sqrt (A_2)`

`sqrt(A_2)/sqrtA` `= y/((x + y))`
`sqrt (A_2)` `= (sqrtA * y)/((x + y))`

 

`text(Similarly, from part)\ text{(iii)}`

`sqrt (A_1) = (sqrtA * x)/((x + y))`

`sqrt (A_1) + sqrt (A_2)` `= (sqrt A * x)/((x + y)) + (sqrt A * y)/((x + y))`
  `= (sqrt A (x + y))/((x + y))`
  `= sqrt A\ \ \ text(… as required.)`

Calculus, 2ADV C4 2014 HSC 12d

The parabola  `y = −2x^2 + 8x`  and the line  `y = 2x`  intersect at the origin and at the point  `A`.
  


  

  1. Find the  `x`-coordinate of the point  `A`.    (1 mark)

  2. Calculate the area enclosed by the parabola and the line.     (3 marks)

Show Answers Only
  1. `x=3`
  2. `9\ text(u²)`
Show Worked Solution
i.  

`text(Need to find)\ x\ text(-co-ord of)\ A`

`y` `= 2x\ \ \ \ \ …\ text{(i)}`
`y` `= -2x^2 + 8x\ \ \ \ \ …\ text{(ii)}`

 
`text(Subst)\ y = 2x\ text(from)\ text{(i)}\ text(into)\ text{(ii)}`

`-2x^2 + 8x` `= 2x`
`-2x^2 + 6x` `= 0`
`-2x (x\ – 3)` `= 0`

  
`:.\ x = 0\ text(or)\ 3`

`:.\ x\ text(-coordinate of)\ A\ text(is 3)`

 

ii. `text(Area)` `= int_0^3 (-2x^2 + 8x)\ dx\ – int_0^3 2x\ dx`
    `= int_0^3 (-2x^2 + 6x)\ dx`
    `= [-2/3x^3 + 3x^2]_0^3`
    `= [(-2/3(3^3) + 3(3^2))\ – (0 + 0)]`
    `= -18 + 27`
    `= 9\ text(u²)`

Probability, 2ADV S1 2014 HSC 12c

A packet of lollies contains 5 red lollies and 14 green lollies. Two lollies are selected at random without replacement.

  1. Draw a tree diagram to show the possible outcomes. Include the probability on each branch.   (2 marks)

  2. What is the probability that the two lollies are of different colours?     (1 mark)

Show Answers Only
  1.  
  2. `70/171`
Show Worked Solution
i.         2UA HSC 2014 12c

 

ii.  `P text{(different colours)}`

`= P(RG) + P(GR)`

`= 5/19 xx 14/18 + 14/19 xx 5/18`

`= 70/342 + 70/342`

`= 140/342`

`= 70/171`

Financial Maths, 2ADV M1 2014 HSC 14d

At the beginning of every 8-hour period, a patient is given 10 mL of a particular drug.

During each of these 8-hour periods, the patient’s body partially breaks down the drug. Only  `1/3`  of the total amount of the drug present in the patient’s body at the beginning of each 8-hour period remains at the end of that period.  

  1. How much of the drug is in the patient’s body immediately after the second dose is given?    (1 mark)

  2. Show that the total amount of the drug in the patient’s body never exceeds 15 mL.     (2 marks)

Show Answers Only
  1. `13.33\ text{mL  (2 d.p.)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `text(Let)\ \ A =\ text(Amount of drug in body)`

`text(Initially)\ A = 10`

`text(After 8 hours)\ \ \ A` `=1/3 xx 10`
`text(After 2nd dose)\ \ A` `= 10 + 1/3 xx 10\ text(mL)`
  `=13.33\ text{mL  (2 d.p.)}`

 

ii.   `text(After the 3rd dose)`

`A_3` `= 10 + 1/3 (10 + 1/3 xx 10)`
  `= 10 + 1/3 xx 10 + (1/3)^2 xx 10`

 
`  =>\ text(GP where)\ a = 10,\ r = 1/3`

`text(S)text(ince)\ \ |\ r\ | < 1:`

`S_oo` `= a/(1\ – r)`
  `= 10/(1\ – 1/3)`
  `= 10/(2/3)`
  `= 15`

 

 
`:.\ text(The amount of the drug will never exceed 15 mL.)`

Quadratic, 2UA 2014 HSC 14b

The roots of the quadratic equation  `2x^2 + 8x + k = 0`  are  `alpha`  and  `beta`.   

  1. Find the value of  `alpha + beta`.    (1 mark)
  2. Given that  `alpha^2 beta + alpha beta^2 = 6`, find the value of  `k`.     (2 marks)
Show Answers Only
  1. `-4`
  2. `-3`
Show Worked Solution
(i) `2x^2 + 8x + k = 0`
  `alpha + beta = (-b)/a = (-8)/2 = -4`

 

(ii) `alpha^2 beta + alpha beta^2` `= 6`
  `alpha beta (alpha + beta)` `= 6`

`text(S)text(ince)\ \ alpha beta = c/a = k/2`

`=> k/2 (–4)` `= 6`
`-2k` `= 6`
`k` `= -3`

Calculus, EXT1* C1 2014 HSC 13b

A quantity of radioactive material decays according to the equation 

`(dM)/(dt) = -kM`,

where  `M`  is the mass of the material in kg,  `t`  is the time in years and  `k`  is a constant.

  1. Show that  `M = Ae^(–kt)`  is a solution to the equation, where  `A`  is a constant.   (1 mark)

  2. The time for half of the material to decay is 300 years. If the initial amount of material is 20 kg, find the amount remaining after 1000 years.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.98\ text(kg)\ text{(2 d.p.)}`
Show Worked Solution
i. `M` `= Ae^(-kt)`
  `(dM)/(dt)` `= -k * Ae^(-kt)`
    `= -kM\ \ text(… as required)`

 

ii.    `text(At)\ \ t = 0,\ M = 20`
`=> 20` `= Ae^0`
`A` `= 20`
`:.\ M` `= 20 e^(-kt)`

`text(At)\ \ t = 300,\ M = 10`

TIP: Many students find it efficient to save the exact value of `k` in the memory function of their calculator for these questions.
`=> 10` `= 20 e^(-300 xx k)`
`e^(-300k)` `= 10/20`
`ln e^(-300k)` `= ln 0.5`
`-300 k` `= ln 0.5`
`k` `= – ln 0.5/300`
  `= 0.00231049…`

 
`text(Find)\ M\ text(when)\ t = 1000`

`M` `= 20 e^(-1000k)`
  `= 20 e^(-1000 xx 0.00231049…)`
  `= 20 xx 0.099212…`
  `= 1.9842…`
  `= 1.98\ text(kg)\ text{(2 d.p.)}`

Linear Functions, 2UA 2014 HSC 12b

The points  `A(0, 4)`,  `B(3, 0)`  and  `C(6, 1)`  form a triangle, as shown in the diagram.
 

2014 12b
 

  1. Show that the equation of  `AC`  is  `x + 2y − 8 = 0`.   (2 marks)
  2. Find the perpendicular distance from  `B`  to  `AC`.    (2 marks)
  3. Hence, or otherwise, find the area of  `Delta ABC`.   (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `sqrt5\ text(units)`
  3. `15/2\ text(u²)`
Show Worked Solution

(i)  `text(Need to show equine of)\ AC\ text(is)\ x + 2y\ – 8 = 0`

`A (0,4)\ \ \ \ \ C (6,1)`

`m_(AC)` `= (y_2\ – y_1)/(x_2\ – x_1)`
  `= (1\ – 4)/(6\ – 0)`
  `= – 1/2`

 `text(Equation of line)\ m = -1/2,\ text(through)\ (0,4)`

`text(Using)\ \ y\ – y_1` `= m (x\ – x_1)`
`y\ – 4` `= -1/2 (x\ – 0)`
`y\ – 4` `= -1/2 x`
`2y\ – 8` `= – x`

`x + 2y\ – 8 = 0\ \ \ text(… as required)`

 

(ii)  `_|_\ text(distance of)\ B (3,0)\ text(from)\ AC`

`_|_\ text(dist)` `= | (ax_1 + by_1 + c)/sqrt(a^2 + b^2) |`
  `= | (1(3) + 2(0)\ – 8)/sqrt(1^2 + 2^2) |`
  `= | (-5)/sqrt5 |`
  `= 5/sqrt5 xx sqrt5/sqrt5`
  `= sqrt 5\ text(units)`

 

(iii)  `text(Area)\ Delta ABC = 1/2 b h`

`b = AC`

`text(dist)\ AC` `= sqrt( (x_2\ – x_1)^2 + (y_2\ – y_1)^2 )`
  `= sqrt( (0\ – 6)^2 + (4\ – 1)^2 )`
  `= sqrt (36 + 9)`
  `= sqrt 45`
  `= 3 sqrt 5\ text(units)`

`h = sqrt 5\ \ \ text{(from part (ii))}`

`:.\ text(Area)\ Delta ABC` `= 1/2 xx 3 sqrt 5 xx sqrt 5`
  `= 15/2\ text(u²)`

Functions, 2ADV F1 2014 HSC 6 MC

Which expression is a factorisation of  `8x^3 + 27`? 

  1. `(2x - 3)(4x^2 + 12x - 9)`
  2. `(2x + 3)(4x^2 - 12x + 9)`
  3. `(2x - 3)(4x^2 + 6x - 9)`
  4. `(2x + 3)(4x^2 - 6x + 9)`
Show Answers Only

`D`

Show Worked Solution

`8x^3 + 27`

COMMENT: Factorising a cubic is only examinable with scaffolding, as provided here by expanding the answer options.

`= (2x)^3 + 3^3`

`= (2x + 3)(4x^2 – 6x + 9)`

 
`=>  D`

Trigonometry, 2ADV T1 2014 HSC 11g

The angle of a sector in a circle of radius 8 cm is  `pi/7`  radians, as shown in the diagram.  
  

2014 11g

 
Find the exact value of the perimeter of the sector.   (2 marks)

Show Answers Only

`(8pi)/7 + 16\ text(cm)`

Show Worked Solution
`text(Arc length)` `= theta/(2pi) xx 2 pi r`
  `= pi/7 xx 8`
  `= (8pi)/7\ text(cm)`

 

`text(S)text(ector perimeter)` `= text(arc) + 2 xx text(radius)`
  `= (8pi)/7 + 16\ text(cm)`

Calculus, 2ADV C1 2014 HSC 11c

Differentiate  `x^3/(x + 1)`.   (2 marks)

Show Answers Only

`(x^2(2x + 3))/((x + 1)^2)`

Show Worked Solution

`y = (x^3)/(x + 1)`

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

`u` `=x^3` `\ \ \ \ \ v` `=(x+1)`
`u ′`  `=3x^2` `v′` `=1`
`dy/dx` `= (3x^2 (x + 1)\ – x^3 (1))/((x + 1)^2)`
  `= (3x^3 + 3x^2\ – x^3)/((x + 1)^2)`
  `= (2x^3 + 3x^2)/((x + 1)^2)`
  `= (x^2 (2x + 3))/((x + 1)^2)`

Functions, 2ADV F1 2014 HSC 11a

Rationalise the denominator of   `1/(sqrt5\ - 2)`.   (2 marks)

Show Answers Only

`sqrt5 + 2`

Show Worked Solution

`1/(sqrt5\ – 2) xx (sqrt5 + 2)/(sqrt5 + 2)`

`= (sqrt5 + 2)/((sqrt5)^2\ – 2^2)`

`= sqrt5 + 2`

Statistics, STD2 S1 2014 HSC 29c

Terry and Kim each sat twenty class tests. Terry’s results on the tests are displayed in the box-and-whisker plot shown in part (i).
 

  1. Kim’s  5-number summary for the tests is  67,  69,  71,  73,  75.

     

    Draw a box-and-whisker plot to display Kim’s results below that of Terry’s results.   (1 mark)
     
         

  2. What percentage of Terry’s results were below 69?     (1 mark)

  3. Terry claims that his results were better than Kim’s. Is he correct?

     

    Justify your answer by referring to the summary statistics and the skewness of the distributions.    (4 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(50%)`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    

 

♦ Mean mark 39%

ii.  `text(50%)`

 

iii.  `text(Terry’s results are more positively skewed than)`

♦♦ Mean mark 29%
COMMENT: Examiners look favourably on using language of location in answers, particularly the areas they have specifically pointed students towards (skewness in this example).

`text(Kim’s and also have a higher limit high.)`

`text(However, Kim’s results are more consistent,)`

`text(showing a tighter IQR. They also have a)`

`text(significantly higher median than Terry’s and)`

`text(are evenly skewed.)`

`:.\ text(Kim’s results were better.)`

Algebra, STD2 A2 2014 HSC 27b

Xuso is comparing the costs of two different ways of travelling to university.

Xuso’s motorcycle uses one litre of fuel for every 17 km travelled. The cost of fuel is $1.67/L and the distance from her home to the university car park is 34 km. The cost of travelling by bus is  $36.40 for 10 single trips.

Which way of travelling is cheaper and by how much? Support your answer with calculations.  (2 marks)

Show Answers Only

`text(Motorcycle is $0.30 cheaper per 1-way trip)`

Show Worked Solution

`text(Compare cost of a 1-way trip)`

`text(Motorcycle)`

`text(Fuel used) = 34/17 = 2\ text(L)`

`text(C)text(ost) = 2 xx $1.67 = $3.34`

`text(Bus)`

`text(C)text(ost) = 36.40/10 = $3.64`

`text(Difference) = $3.64\ – 3.34\ = $0.30`

 

`:.\ text(Motorcycle is $0.30 cheaper per 1-way trip.)`

Financial Maths, STD2 F1 2014 HSC 13 MC

Jane sells jewellery. Her commission is based on a sliding scale of 6% on the first $2000 of her sales, 3.5% on the next $1000, and 2% thereafter.

What is Jane’s commission when her total sales are $5670? 

  1. $188.40
  2. $208.40
  3. $321.85
  4. $652.05
Show Answers Only

`B`

Show Worked Solution

`text(Commission)`

`= (2000 xx text(6%)) + (1000 xx text(3.5%)) + (5670-3000) xx text(2%)`

`= (2000 xx 0.06) + (1000 xx 0.035) + (2670 xx 0.02)`

`= 120 + 35 + 53.40`

`= 208.40`
 

`=>  B`

Financial Maths, STD2 F4 2014 HSC 9 MC

A car is bought for  $19 990. It will depreciate at 18% per annum. 

Using the declining balance method, what will be the salvage value of the car after 3 years, to the nearest dollar? 

  1. $8968
  2. $9195
  3. $11 022
  4. $16 392
Show Answers Only

`C`

Show Worked Solution
`S` `= V_0 (1-r)^n`
  `= 19\ 990 (1-18/100)^3`
  `= 19\ 990 (0.82)^3`
  `= $11\ 021.85`

 
`=>  C`

Probability, STD2 S2 2014 HSC 8 MC

A group of 150 people was surveyed and the results recorded.
  

A person is selected at random from the surveyed group. 

What is the probability that the person selected is a male who does not own a mobile?

  1. `28/150`
  2. `45/150` 
  3. `28/70` 
  4. `45/70` 
Show Answers Only

`A`

Show Worked Solution
`P` `= text(number of males without mobile)/text(number in group)`
  `= 28/150`

 
`=>  A`

Probability, STD2 S2 2014 HSC 6 MC

A cafe menu has 3 entrees, 5 main courses and 2 desserts. Ariana is choosing a three-course meal consisting of an entree, a main course and a dessert.

How many different three-course meals can Ariana choose?

  1.    3
  2.    10
  3.    15
  4.    30
Show Answers Only

`D`

Show Worked Solution
`text(# Combinations)` `= 3 xx 5 xx 2`
  `= 30`

`=>  D`

Measurement, STD2 M1 2014 HSC 2 MC

A measurement of  72 cm is increased by 20% and then the result is decreased by 20%. 

What is the new measurement, correct to the nearest centimetre?

  1. 46 cm
  2. 69 cm
  3. 72 cm
  4. 104 cm
Show Answers Only

`B`

Show Worked Solution
STRATEGY: Students confident in this area could save time in the calculations as follows: `72 xx 1.2 xx 0.8“ = 69.12`

`text(72 increased by 20%)`

`= 72 + (text(20%) xx 72) = 86.4\ text(cm)`

`text(86.4 decreased by 20%)`

`= 86.4\ – (text(20%) xx 86.4) = 69.12\ text(cm)`

`=>  B`

Quadratic, EXT1 2014 HSC 13c

The point  `P(2at, at^2)`  lies on the parabola  `x^2 = 4ay`  with focus  `S`.

The point  `Q`  divides the interval  `PS`  internally in the ratio  `t^2 :1`.

2014 13c 

  1. Show that the coordinates of  `Q`  are  
  2. `x = (2at)/(1 + t^2)`  and  `y = (2at^2)/(1 + t^2)`.  (2 marks)
  4. Express the slope of  `OQ`  in terms of  `t`.    (1 mark)
  5. Using the result from part (ii), or otherwise, show that  `Q`  lies on a fixed circle of radius  `a`.   (3 marks)

 

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

(i)   `text(Show)\ \ Q = ((2at)/(1 + t^2), (2at^2)/(1 + t^2))`

`P (2at, at^2),\ S (0, a)`

`PS\ text(is divided internally in ratio)\ t^2: 1`

`Q` `= ((nx_1 + mx_2)/(m + n), (ny_1 + my_2)/(m + n))`
  `= ((1(2at) + t^2(0))/(t^2 + 1), (1(at^2) + t^2 (a))/(t^2 + 1))`
  `= ((2at)/(1 + t^2), (2at^2)/(1 + t^2))\ \ text(… as required.)`

 

(ii)    `m_(OQ)` `= (y_2\ – y_1)/(x_2\ – x_1)`
    `= ((2at^2)/(1 + t^2))/((2at)/(1 + t^2)) xx (1+t^2)/(1+t^2)`
    `= (2at^2)/(2at)`
    `= t`

 

(iii)   `text(Show)\ Q\ text(lies on a fixed circle radius)\ a`
  `text(S)text(ince)\ Q\ text(passes through)\ (0, 0)`

 

`=>\ text(If locus of)\ Q\ text(is a circle, it has)`

`text(diameter)\ QT\ text(where)\ T(0, 2a)`

 

`text(Show)\ \ QT _|_ OQ`

♦♦ Mean mark 22%

`text{(} text(angles on circum. subtended by)`

  `text(a diameter are)\ 90^@ text{)}`

 

`m_(OQ) = t\ \ \ \ text{(see part (ii))}`

`text(Find)\ m_(QT),\ \ text(where:)`

`Q((2at)/(1 + t^2), (2at^2)/(1 + t^2)),\ \ \ \ \ T (0,2a)`

`m_(QT)` `= (y_2\ – y_1)/(x_2\ – x_1)`
  `= ((2at^2)/(1 + t^2)\ – 2a)/((2at)/(1 + t^2)\ – 0)`
  `= (2at^2\ – 2a (1 + t^2))/(2at)`
  `= – (2a)/(2at)`
  `= – 1/t`

 

`m_(QT) xx m_(OT) = -1/t xx t = -1`

`=> QT _|_ OQ`

`=>O,\ T,\ Q\ text(lie on a circle.)`

`:.\ text(Locus of)\ Q\ text(is a fixed circle,)`

`text(centre)\ (0, a),\ text(radius)\ a`

Quadratic, EXT1 2009 HSC 2c

The diagram shows points  `P(2t, t^2)`  and  `Q(4t, 4t^2)`  which move along the parabola  `x^2 = 4y`. The tangents to the parabola at  `P`  and  `Q`  meet at  `R`. 

2009 2c

  1. Show that the equation of the tangent at  `P`  is  `y = tx\ - t^2.`   (2 marks)
  2. Write down the equation of the tangent at  `Q`, and find the coordinates of the point  `R`  in terms of  `t`.     (2 marks)
  3. Find the Cartesian equation of the locus of  `R`.   (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `y = 2tx\ – 4t^2,\ R (3t, 2t^2)`
  3. `y = 2/9 x^2`
Show Worked Solution
(i)    `x^2` `= 4y`
  `y` `= (x^2)/4`
  `dy/dx` `= x/2`
IMPORTANT: Deriving this equation was specifically asked in 2009 and 2011, as well as being required in 2014. KNOW IT! Completing this in 60 seconds gains 2 full minutes on allocated time.

 

`text(At)\ \ x = 2t,`

`dy/dx = (2t)/2 = t`

 

`:.\ text(T)text(angent has)\ \ m = t\ \ text(and passes)`

`text(through)\ \ (2t, t^2).`

`y – t^2` `= t (x – 2t)`
`y` `= tx\ – 2t^2 + t^2`
  `= tx\ – t^2\ \ text(… as required)`

 

(ii)    `text(T)text(angent at)\ \ P (2t, t^2)\ \ text(is)\ \ y = tx\ – t^2`

`=>Q(4t, 4t^2) -= Q(2(2t), (2t)^2)`

`:.\ text(Equation of the tangent at)\ \ Q,`

`y` `= 2tx\ – (2t)^2`
  `= 2tx\ – 4t^2`

 

`text(T)text(angents intersect at)\ \ R`

`y` `= tx\ – t^2\ \ \ \ \ …\ text{(1)}`
`y` `= 2tx\ – 4t^2\ \ \ \ \ …\ text{(2)}`

 

`text(Intersection when)\ text{(1)} = text{(2)}`

`tx\ – t^2` `= 2tx\ – 4t^2`
`tx` `= 3t^2`
`x` `= 3t`

 

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

`y` `= t (3t)\ – t^2`
  `= 2t^2`

 

`:.\ R (3t, 2t^2)`

 

(iii)   `text(Find locus of)\ \ R`
`x` `= 3t` `\ \ \ \ \   …\ text{(1)}`
`y` `= 2t^2\ \ \ \ \ …\ text{(2)}`

 

`text{From (1),}\ \ \ \ t = x/3`

`text(Substitute)\ \ t = x/3\ text(into)\ text{(2)}`

`y` `= 2 (x/3)^2`
  `= 2/9 x^2`

 

`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = 2/9 x^2`

Proof, EXT1 P1 2014 HSC 13a

Use mathematical induction to prove that  `2^n + (− 1) ^(n + 1)`  is divisible by 3 for all integers  `n >= 1`.   (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Prove)\ \ 2^n + (-1)^(n + 1)`

`text(is divisible by 3 for integers)\ n >= 1`

`text(If)\ n = 1`

`2^1 + (-1)^2` `= 2 + 1`
  `= 3\ \ text{(which is divisible by 3)}`

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

 
`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ \ 2^k + (-1)^(k + 1)` `= 3P\ text{(} P\ text(integer) text{)}`
`2^k` `= 3P\ – (-1)^(k + 1)\ \ \ \ \ … \ text{(∗)}`

 
`text(Prove true for)\ \ n = k + 1`

`2^(k + 1) + (-1)^(k + 1 + 1)` `= 2*2^k + (-1)^(k + 2)`
  `= 2 (3P\ – (-1)^(k + 1)) + (-1)^(k + 2)\ \ text(… from)\ text{(∗)}`
  `= 6P\ – 2 (-1)^(k + 1)\ – 1 (-1)^(k + 1)`
  `= 6P\ – 3 (-1)^(k + 1)`
  `= 3 (2P\ – (-1)^(k + 1))`
`text(… which is divisible by 3)`

 
`=>\ text(True for)\ n = k + 1`

`:.text(S)text(ince true for)\ n = 1,\ text(by PMI, true for integral)\ n >= 1.`

Quadratic, EXT1 2013 HSC 13b

The point  `P(2ap, ap^2)`  lies on the parabola  `x^2 = 4ay`.  The tangent to the parabola at  `P`  meets the  `x`-axis at  `T (ap, 0)`.  The normal to the tangent at  `P`  meets the  `y`-axis at  `N(0, 2a + ap^2)`.

2013 13b

The point  `G`  divides  `NT`  externally in the ratio  `2 :1`. 

  1. Show that the coordinates of  `G`  are  `(2ap, –2a – ap^2)`.    (2 marks)
  2. Show that  `G`  lies on a parabola with the same directrix and focal length as the original parabola.    (2 marks)

 

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

(i)   `N (0, 2a + ap^2)`

`T (ap, 0)`

`G\ \ text(divides)\ \ NT\ \ text(externally in ratio)\ \ 2:1,`

`text(i.e.)\ \ (m:n = 2:–1)`

`:.\ G` `= ((nx_1 + mx_2)/(m + n), (ny_1 + my_2)/(m + n))`
  `= ((0 + 2ap)/(2\ – 1), (-2a\ – ap^2 + 0)/(2\ – 1))`
  `= (2ap, -2a\ – ap^2)\ text(… as required)`

 

♦♦ Mean mark 31%.
IMPORTANT: Remember that finding the locus involves eliminating the parameter. Expressing the locus in the form `x^2=4ay` is critical to finding its directrix and focal length.

(ii)  `x^2 = 4ay\ text(has focal length)\ \  a`

`text(and directrix)\ \ y = -a`

`text(Locus of)\ G`

`x = 2ap \ \ \ \  \ \ \ \ \ \ …\ (1)`

`y = -2a\ – ap^2\ \ \ \ \ \ …\ (2)`

`text{From (1)},\ \ \ \ p = x/(2a)`

`text(Substitute)\ \ p = x/(2a)\ \ text{into (2)}`

`y` `= -2a\ – a (x/(2a))^2`
`y` `= -2a\ – (ax^2)/(4a^2)`
`(x^2)/(4a)` `= -y\ – 2a`
`x^2` `= -4a (y + 2a)`

 

`text(Focal length) = a`

`text(Vertex at)\ (0, –2a)`

`text(Directrix)\ \ \ y = -2a + a = -a`

 

`:.\ text(Locus of)\ G\ text(has same focal length and)`

`text(directrix as)\ \ x^2 = 4ay.`

Trigonometry, EXT1 T3 2013 HSC 12a

 

  1. Write  `sqrt3cos x - sin x`  in the form  `2 cos (x + alpha)`, where  `0 < alpha < pi/2`.   (1 mark)

  2. Hence, or otherwise, solve  `sqrt3 cos x = 1 + sin x`,  where  `0 < x < 2pi`.     (2 marks)

Show Answers Only
  1. `2cos (x + pi/6)`
  2. `pi/6,\ (3pi)/2`
Show Worked Solution

i.  `text(Write)\ sqrt3 cosx\ – sinx\ text(in form)`

`2cos (x + alpha),\ \ \ 0 < alpha < pi/2`

`2 (cosx cos alpha\ – sinx sin alpha)` `= sqrt 3 cosx\ – sinx`
`cosx cos alpha\ – sinx sin alpha` `= sqrt3/2 cos x\ – 1/2 sinx`
`=> cos alpha` `= sqrt3/2`
`=> sin alpha` `= 1/2`
`alpha` `= pi/6`

 
`:.\ 2cos (x + pi/6) = sqrt3 cosx\ – sinx`
 

ii.  `text(Solve)\ sqrt3 cosx = 1 + sinx,\ \ \ 0 < x < 2pi`

`sqrt3 cosx\ – sinx` `= 1`
`2 cos (x + pi/6)` `= 1\ \ \ text{(from part (i))}`
`cos (x + pi/6)` `= 1/2`
`cos(pi/3)` `=1/2`

 
`text(S)text(ince cos is positive in)\  1^text(st) // 4^text(th)\ text(quadrants)`

`x + pi/6` `= pi/3,\ 2pi\ – pi/3\ \ \ \ \ (0 < x < 2pi)`
`:.\ x` `= pi/6,\ (3pi)/2`

Calculus, EXT1 C2 2013 HSC 11g

Differentiate  `x^2 sin^(–1) 5x`.   (2 marks)

Show Answers Only

`(5x^2)/sqrt(1\ – 25x^2) + 2x sin^(-1) 5x`

Show Worked Solution
`y` `= x^2 sin^(-1) 5x`
`text(Using the product rule)`
`(dy)/(dx)` `= x^2 xx 1/sqrt(1\ – (5x)^2) xx d/(dx)(5x) + 2x sin^(-1) 5x`
  `= (5x^2)/sqrt(1\ – 25x^2) + 2x sin^(-1) 5x`

Statistics, EXT1 S1 2013 HSC 11c

An examination has 10 multiple-choice questions, each with 4 options. In each question, only one option is correct. For each question a student chooses one option at random.

Write an expression for the probability that the student chooses the correct option for exactly 7 questions.   (2 marks)

Show Answers Only

`\ ^10C_7 (1/4)^7 (3/4)^3`

Show Worked Solution
`P text{(Correct)}` `= 1/4`
`P text{(Incorrect)}` `= 3/4`

 
`:. P text{(exactly 7 correct)} =\ ^10C_7 (1/4)^7 (3/4)^3`

Plane Geometry, EXT1 2010 HSC 5c

In the diagram,  `ST`  is tangent to both the circles at  `A`.

The points  `B`  and  `C`  are on the larger circle, and the line  `BC`  is tangent to the smaller circle at  `D`. The line  `AB`  intersects the smaller circle at  `X`.

 

Plane Geometry, EXT1 2010 HSC 5c

Copy or trace the diagram into your writing booklet 

  1. Explain why  `/_AXD = /_ABD + /_XDB.`   (1 mark)
  2. Explain why  `/_AXD = /_TAC + /_CAD.`   (1 mark)
  3. Hence show that  `AD`  bisects  `/_BAC`.    (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)    Plane Geometry, EXT1 2010 HSC 5c Answer
`/_ AXD = /_ABD + /_XDB`
`text{(exterior angle of}\ Delta BXD text{)}`

 

♦ Mean mark 47%
(ii)    `/_AXD` `= /_TAD \ \ text{(angle in alternate segment)}`
    `= /_TAC + /_CAD\ \ text(… as required)`

 

(iii)  `text(Show)\ \ /_XAD = /_CAD`

`/_ABD + /_XDB` `=/_TAC + /_CAD\ \ text{(} text(from part)\ text{(i)} text(,)\ text{(ii)} text{)}`

 

`text(S)text(ince)\ /_TAC= /_ABD\ text{(angle in alternate segment)}`

♦♦ Mean mark 22%
IMPORTANT: Recognising that angles in the alternate segment are equal is an examiner favourite. LOOK FOR IT!
`=>/_XDB` `=/_CAD`
`/_XDB` `= /_XAD\ text{(angle in alternate segment)}`
`:. /_XAD` `= /_CAD`

 

`:.\ AD\ \ text(bisects)\ /_BAC.`

Trig Ratios, EXT1 2010 HSC 5a

A boat is sailing due north from a point  `A`  towards a point  `P`  on the shore line.

The shore line runs from west to east.

In the diagram,  `T`  represents a tree on a cliff vertically above  `P`, and  `L`  represents a landmark on the shore. The distance  `PL`  is 1 km.

From  `A`  the point  `L`  is on a bearing of 020°, and the angle of elevation to  `T`  is 3°.

After sailing for some time the boat reaches a point  `B`, from which the angle of elevation to  `T`  is 30°.
 

5a
 

  1. Show that 
     
    `qquad BP = (sqrt3 tan 3°)/(tan20°)`.   (3 marks) 
     
  2. Find the distance  `AB`. Give your answer to 1 decimal place.   (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2.5\ text(km)\ \ text{(to 1 d.p.)}`
Show Worked Solution

(i)   `text(Show)\ \ BP = (sqrt3 tan 3°)/(tan 20°)` 

`text(In)\ Delta ATP`
`tan 3°` `= (TP)/(AP)`
`=> AP` `= (TP)/(tan 3)`

 
`text(In)\ Delta APL`

`tan 20°` `= 1/(AP)`
`=> AP` `= 1/tan 20`

 

`:. (TP)/(tan3)` `= 1/(tan20)`
`TP` `= (tan3°)/(tan20°)\ \ \ text(…)\ text{(} text(1)text{)}`

 

`text(In)\ \ Delta BTP`

`tan 30°` `= (TP)/(BP)`
`1/sqrt3` `= (TP)/(BP)`
`BP` `= sqrt3 xx TP\ \ \ \ \ text{(using (1) above)}`
  `= (sqrt3 tan3°)/(tan20°)\ \ \ text(… as required)`

 

(ii)    `AB` `= AP\ – BP`
  `AP` `= 1/(tan20°)\ \ \ text{(} text(from part)\ text{(i)} text{)}`
`:.\ AB` `= 1/(tan 20°)\ – (sqrt3 tan3°)/(tan20°)`
  `= (1\ – sqrt3 tan 3)/(tan20°)`
  `= 2.4980…`
  `= 2.5\ text(km)\ text{(to 1 d.p.)`

Mechanics, EXT2* M1 2010 HSC 4a

A particle is moving in simple harmonic motion along the `x`-axis. 

Its velocity `v`, at `x`, is given by  `v^2 = 24 − 8x − 2x^2`. 

  1. Find all values of `x` for which the particle is at rest.   (1 mark)

  2. Find an expression for the acceleration of the particle, in terms of `x`.   (1 mark)

  3. Find the maximum speed of the particle.    (2 marks)

Show Answers Only
  1. `x = –6\ \ text(or)\ \ 2`
  2. `-4\ – 2x`
  3. `4 sqrt 2`
Show Worked Solution
(i)    `v^2 = 24\ – 8x\ – 2x^2`

`text(Find)\ \ x\ \ text(when)\ \ v=0`

`24 – 8x – 2x^2` `= 0`
`x^2 + 4x – 12` `= 0`
`(x + 6)(x – 2)` `= 0`
`x= -6\ \ text(or)\ \ 2`

 

`:.\ text(Particle at rest when)\ \ x = –6\ \ text(or)\ \ 2`

 

(ii)    `ddot x` `= d/(dx) (1/2 v^2)`
    `= d/(dx) (12 – 4x – x^2)`
    `= -4 – 2x`

 

(iii)   `text(Solution 1)`

`text(Max speed when)\ \ ddot x = 0`

`-4 – 2x` `= 0`
`2x` `= -4`
`x` `= -2`

`text(At)\ \ x = -2`

MARKER’S COMMENT: While most students found `x=-2` correctly, too many made errors substituting this back in or failed to finish the answer by taking the square root. BE CAREFUL! .
`v^2` `= 24 – 8(–2) – 2(–2)^2`
  `= 24 + 16 – 8`
  `= 32`
`v` `= +- sqrt32`
  `= +- 4 sqrt2`

`:.\ text(Maximum speed is)\ \ 4 sqrt2`

 

`text(Alternate Answer)`

`text(Maximum speed occurs when)`

`x = (-6+2)/2 = –2`

`text(At)\ \ x = –2`

`v^2` `= 32\ \ \ text{(see working above)}`
`v` `= +- 4 sqrt 2`

 

`:.\ text(Maximum speed is)\ 4sqrt2`

Inverse Functions, EXT1 2010 HSC 3b

Let  `f(x) = e^(-x^2)`.  The diagram shows the graph  `y = f(x)`.
 

 Inverse Functions, EXT1 2010 HSC 3b
 

  1. The graph has two points of inflection.  
  2. Find the  `x`  coordinates of these points.   (3 marks)
  3.  
  4. Explain why the domain of  `f(x)`  must be restricted if  `f(x)`  is to have an inverse function.    (1 mark)
  5. Find a formula for  `f^(-1) (x)`  if the domain of  `f(x)`  is restricted to  `x ≥ 0`.   (2 marks)
  6. State the domain of  `f^(-1) (x)`.    (1 mark)
  7. Sketch the curve  `y = f^(-1) (x)`.    (1 mark)
  8. (1)   Show that there is a solution to the equation  `x = e^(-x^2)`  between  `x = 0.6`  and  `x = 0.7`.   (1 mark)
  9. (2)   By halving the interval, find the solution correct to one decimal place.   (1 mark)

 

Show Answers Only
  1. `x = +- 1/sqrt2` 
  2.  
  3. `text(There can only be 1 value of)\ y`
  4. `text(for each value of)\ x.`
  5.  
  6. `f^(-1)x = sqrt(ln(1/x))`
     
  7. `0 <= x <= 1`
  8.  
  9. Inverse Functions, EXT1 2010 HSC 3b Answer
  10. (1)   `text(Proof)\ \ text{(See Worked Solutions)}`
  11. (2)   `0.7\ text{(1 d.p.)}`
Show Worked Solution
(i)    `y` `= e^(-x^2)`
  `dy/dx` `= -2x * e^(-x^2)`
  `(d^2y)/(dx^2)` `= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)`
    `= 4x^2 e^(-x^2)\ – 2e^(-x^2)`
    `= 2e^(-x^2) (2x^2\ – 1)`

 

`text(P.I. when)\ \ (d^2y)/(dx^2) = 0`

`2e^(-x^2) (2x^2\ – 1)` `= 0` 
 `2x^2\ – 1` `= 0` 
 `x^2` `= 1/2`
 `x` `= +- 1/sqrt2` 
COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.
`text(When)\ \ ` `x < 1/sqrt2,` `\ (d^2y)/(dx^2) < 0`
  `x > 1/sqrt2,` `\ (d^2y)/(dx^2) > 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = 1/sqrt2`

`text(When)\ \ ` `x < – 1/sqrt2,` `\ (d^2y)/(dx^2) > 0`
  `x > – 1/sqrt2,` `\ (d^2y)/(dx^2) < 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = – 1/sqrt2`

 

(ii)   `text(In)\ f(x), text(there are 2 values of)\ y\ text(for)`
  `text(each value of)\ x.`
  `:.\ text(The domain of)\ f(x)\ text(must be restricted)`
  `text(for)\ \ f^(-1) (x)\ text(to exist).`

 

(iii)   `y = e^(-x^2)`

`text(Inverse function can be written)` 

`x` `= e^(-y^2),\ \ \ x >= 0`
`lnx` `= ln e^(-y^2)`
`-y^2` `= lnx`
`y^2` `= -lnx`
  `=ln(1/x)`
`y` `= +- sqrt(ln (1/x))`

 

`text(Restricting)\ \ x>=0,\ \ =>y>=0`

`:.  f^(-1) (x)=sqrt(ln (1/x))`
 

♦ Parts (iv) and (v) were poorly answered with mean marks of 39% and 49% respectively.
(iv)   `f(0) = e^0 = 1`

`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`

`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`

 

(v) 

Inverse Functions, EXT1 2010 HSC 3b Answer

 

(vi)(1)  `x = e^(-x^2)`

`text(Let)\ g(x) = x\ – e^(-x^2)`

`g(0.6)` `=0.6\ – e^(-0.6^2)`
  `=0.6\ – 0.6977 < 0`
`g(0.7)` `=0.7\ – e^(-0.7^2)`
  `=0.7\ – 0.6126 > 0`
`=>g(x)\ text(changes sign)`

 

`:.\ g(x)\ \ text(has a root between  0.6  and  0.7)`

`:.\ x = e^(-x^2)\ \ text(has a solution between  0.6  and  0.7)`

♦ Mean mark 37%.
MARKER’S COMMENT: Better responses showed the change in sign between `g(0.65)` and `g(0.7)` as shown in the solution.

 

(vi)(2)   `g(0.65)` `=0.65\ – e^(-0.65^2)`
    `=0.65\ – 0.655 < 0`

 

`:.\ text(A solution lies between 0.65 and 0.7)`

`:.\ x = 0.7\ \ text{(1 d.p.)}`

Functions, EXT1 F2 2010 HSC 2c

Let  `P(x) = (x + 1)(x-3) Q(x) + ax + b`, 

where  `Q(x)`  is a polynomial and  `a`  and  `b`  are real numbers.

The polynomial  `P(x)`  has a factor of  `x-3`.

When  `P(x)`  is divided by  `x + 1`  the remainder is  `8`. 

  1. Find the values of  `a`  and  `b`.  (2 marks)

  2. Find the remainder when  `P(x)`  is divided by  `(x + 1)(x-3)`.     (1 mark)

Show Answers Only
  1. `a = -2,\ b = 6`
  2. ` -2x + 6`
Show Worked Solution

i.  `P(x) = (x+1)(x-3)Q(x) + ax + b`

`(x-3)\ \ text{is a factor   (given)}`

`:. P (3)` `= 0`
`3a + b` `= 0\ \ \ …\ text{(1)}`

 
`P(x) ÷ (x+1)=8\ \ \ text{(given)}`

`:.P(-1)` `= 8`
`-a + b` `= 8\ \ \ …\ text{(2)}`

 
`text{Subtract  (1) – (2)}`

`4a` `= -8`
`a` `= -2`

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

`-6 + b` `= 0`
`b` `= 6`

 
`:. a= – 2, \ b=6` 
 

ii.  `P(x) -: (x + 1)(x-3)`

`= ((x+1)(x-3)Q(x)-2x + 6)/((x+1)(x-3))`

`= Q(x) + (-2x + 6)/((x+1)(x-3))`

 
`:.\ text(Remainder is)\ \ -2x + 6`

COMMENT: This question requires a fundamental understanding of the remainder theorem.

Calculus, EXT1 C2 2010 HSC 1e

Use the substitution  `u = 1 - x`  to evaluate  `int_0^1 x sqrt(1 - x)\ dx`.   (3 marks)

Show Answers Only

`4/15`

Show Worked Solution
`u = 1 – x` `\ \ \ \ \ => x = 1 – u`
`(du)/(dx) = -1` `\ \ \ \ \ => du = – dx`

 

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

 
`:. int_0^1 x sqrt(1 – x)\ dx`

`= – int_1^0 (1 – u) u^(1/2)\ du`

`= int_1^0 (u – 1) u^(1/2)\ du`

`= int_1^0 (u^(3/2) – u^(1/2))\ du`

`= [2/5 u^(5/2) – 2/3 u^(3/2)]_1^0`

`= [0 – (2/5 – 2/3)]`

`= – (6/15 – 10/15)`

`= 4/15`

Functions, EXT1 F1 2010 HSC 1d

Solve  `3/(x+2) < 4`.   (3 marks)

Show Answers Only

 `x < -2\ \ text(or)\ \ x > -5/4`

Show Worked Solution

`text(Solution 1)`

`3/(x + 2) < 4`

`text(Multiply b.s. by)\ \ (x + 2)^2`

`3(x + 2)` `< 4(x + 2)^2`
`3x + 6` `< 4 (x^2 + 4x + 4)`
`3x + 6` `< 4x^2 + 16x + 16`
`4x^2 + 13x + 10` `> 0`
`(4x + 5)(x + 2)` `> 0`

 
`text(LHS)\ = 0\ \ text(when)\ \ x = -5/4\ \ text(or)\ \ -2`

Algebra, EXT1 2010 HSC 1d Answer

`text(From graph)`

`x < -2\ \ text(or)\ \  x > -5/4`

 
`text(Alternate Solution)`

`text(If)\ \ x + 2 > 0\ \ \ \ text{(i.e.}\ \ x > –2 text{)}`

`3` `< 4(x + 2)`
`3` `< 4x + 8`
`4x` `> -5`
`x` `> -5/4`

 
`text(If)\ \ x + 2 < 0\ \ \ \ text{(i.e.}\ x < –2 text{)}`

`3` `> 4 (x + 2)`
`3` `> 4x + 8`
`4x` `< -5`
`x` `< -5/4`
`:. x` `< –2\ \ \ \ text{(satisfies both)}`

 
`:.\ x < –2\ \ text(or)\ \ x > –5/4`

Plane Geometry, EXT1 2011 HSC 4b

In the diagram, the vertices of  `Delta ABC`  lie on the circle with centre  `O`. The point  `D`  lies on  `BC`  such that  `Delta ABD`  is isosceles and  `/_ABC = x`.

Copy or trace the diagram into your writing booklet. 

  1. Explain why  `/_AOC = 2x`.    (1 mark)
  2. Prove that  `ACDO`  is a cyclic quadrilateral.    (2 marks)
  3. Let  `M`  be the midpoint of  `AC`  and  `P`  the centre of the circle through `A, C, D`  and  `O`. 
  4. Show that  `P, M`  and  `O`  are collinear.   (1 mark)
Show Answers Only
  1. `text(Angle at the centre of a circle is twice)`
  2. `text(angle of circumference on same arc)`
  3.  
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)

`/_AOC = 2x`

`text{(angles at circumference and}`

`text{centre on arc}\ AC text{)}`

 

♦ Mean mark 38%
STRATEGY: Proving part (ii) by showing opposite angles are supplementary also worked but was more time consuming.

(ii)  `text(Prove)\ ACDO\ text(is cyclic)`

`text(S)text(ince)\ Delta ADB\ text(is isosceles)`

`/_DAB = x\ \ \ text{(opposite equal sides in}\ Delta DBA text{)}`

`=> /_ADB` `= 180\ – 2x\ \ \ text{(angle sum of}\ Delta DAB text{)}`
`=> /_CDA` `= 2x\ \ \ text{(}CDB\ text{is a straight angle)}`

 

`text(S)text(ince chord)\ AC\ text(subtends)\ /_CDA = 2x`

`text(and)\ /_COA = 2x,`

`:.\ text(Quadrilateral)\ ACDO\ text(must be cyclic.)`

 

(iii)

 `text(Need to show)\ OM\ text(passes through)\ P,\ text(centre)`

♦♦♦ Mean mark 7%. 2nd hardest question in the 2011 paper!

`text(of circle through)\ ACDO.`

`AM` `= CM\ text{(} M\ text(is midpoint) text{)}`
`OC` `= OA\ text{(radii)}`

`OM\ text(is common)`

`:.\ Delta OAM -= Delta OCM\ \ \ text{(SSS)}`

`:. /_CMO = /_AMO\ \ \ ` `text{(corresponding angles of}`
  `\ \ text{congruent triangles)}`

 

`text(S)text(ince)\ ∠AMC\ text(is straight angle)`

`/_CMO = /_AMO = 90°`

`:.OM\ text(is perpendicular bisector)`

`text(of chord)\ AC.`

`:. OM\ text(passes through)\ P.`

`:.\ P, M,\ text(and)\ O\ text(are collinear.)`

Geometry and Calculus, EXT1 2011 HSC 4a

Consider the function  `f(x) = e^(-x)\ - 2e^(-2x)`. 

  1. Find  `f prime (x)`.   (1 mark)
  2. The graph  `y = f(x)`  has one maximum turning point. 
  3. Find the coordinates of the maximum turning point.   (2 marks)
  4.  
  5. Evaluate  `f(ln2)`.   (1 mark)
  6. Describe the behaviour of  `f(x)`  as  `x -> oo`.   (1 mark)
  7. Find the  `y`-intercept of the graph  `y = f(x)`.   (1 mark)
  8. Sketch the graph  `y = f(x)`  showing the features from parts  (ii) - (v).
  9. You are not required to find any points of inflection.    (2 marks)
  10.  
  11.  
  12.  
  13.  
  14.  
  15.  
  16.  
Show Answers Only
  1. `-e^(-x) + 4e^(-2x)`
  2. `text(Max T.P. at)\ (ln4, 1/8)`
  3. `0`
  4. `text(As)\ x -> oo, f(x) -> 0`
  5. `-1`
  7.  
  8.  
  9.  
  10.  
  11.  
Show Worked Solution
(i)    `f(x)` `= e^(-x)\ – 2e^(-2x)`
  `f prime (x)` `= -e^(-x) + 4e^(-2x)`

 

(ii)   `text(T.P. when)\ f prime (x) = 0`

`-e^(-x) + 4e^(-2x) = 0`

`text(Let)\ e^(-x) = X`

`- X + 4X^2` `= 0`
`X (4X\ – 1)` `= 0`

`X = 0\ \ text(or)\ \ 1/4`

`text(If)\ \ e^(-x) = 0,\ \ \ text(No solution)`

`text(If)\ \ e^(-x) = 1/4`

`ln e^(-x)` `= ln (1/4)`
`-x` `= ln 4^(-1)`
  `= -ln 4`
`x` `= ln4`

 

ALGEBRA TIP: Note that `e^ln x=x` can often help calculations. This is easily proven by simply taking the `ln` of both sides of the equation.
`f″(x)` `= e^(-x)\ – 8e^(-2x)`
`f″(ln4)` `< 0 => text(MAX)`
`f (ln4)` `= e^(-ln4)\ – 2e^(-2ln4)`
  `= e^(ln(1/4))\ – 2 e^(ln(1/16))`
  `= 1/4\ – 2(1/16)`
  `= 1/8`

`:.\ text(Maximum T.P. at)\ \ (ln4, 1/8)`

 

(iii)   `f(ln2)` `= e^(-ln2)\ – 2e^(-2ln2)`
    `= e^(ln(1/2))\ – 2e^(ln(1/4))`
    `= 1/2\ – 2 xx 1/4`
    `= 0`

 

(iv)   `f(x)` `= e^(-x)\ – 2e^(-2x)`
    `= e^(-x) (1\ – 2e^(-x))`
  `text(As)\ x -> oo, \ e^(-x) ->0,\ f(x) -> 0`

 

(v)    `y text(-intercept at)\ \ f(0)`
`f(0)` `= e^0\ – 2e^0`
  `= -1`

 

(vi)

EXT1 2011 4a

Quadratic, EXT1 2011 HSC 3b

The diagram shows two distinct points  `P(t, t^2)`  and  `Q(1\ - t, (1\ - t)^2)`  on the parabola  `y = x^2`.  The point  `R`  is the intersection of the tangents to the parabola at  `P`  and  `Q`. 

 

  1. Show that the equation of the tangent to the parabola at  `P`  is  `y = 2tx\ – t^2`.   (2 marks)
  2. Using part  `text{(i)}`, write down the equation of the tangent to the parabola at  `Q`.     (1 mark)
  3. Show that the tangents at  `P`  and  `Q`  intersect at
    `R (1/2, t\ - t^2)`.   (2 marks)
  5. Describe the locus of  `R`  as  `t`  varies, stating any restriction on the  `y`-coordinate.   (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `y = 2 (1 -t)x\ – (1\ – t)^2`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Locus of)\ R\ text(is vertical line)`
    `x = 1/2,\ y<1/4`
Show Worked Solution
(i)

`text(Show tangent at)\ P\ text(is)\ y = 2tx\ – t^2`

`y` `= x^2`
`dy/dx` `= 2x`

`x=t\ \ \ \ text(at)\ P`

`dy/dx = 2t`

`text(Find equation with)\ \ m = 2t\ \ text(through)\ \ P(t, t^2)`

`y\ – y_1` `= m(x\ – x_1)`
`y\ – t^2` `= 2t ( x\ – t)`
`y` `= 2tx\ – 2t^2 + t^2`
  `= 2tx\ – t^2\ text(… as required)`

 

(ii)  `text(T)text(angent at)\ Q\ text(has equation)`

MARKER’S COMMENT: Many students derived this equation rather than substituting the new parameter, costing them valuable time. This is a benefit of using the parametric approach.

`y = 2(1\ – t)x\ – (1\ – t)^2`

 

(iii)  `text(Need to show)\ R(1/2,\ t\ – t^2)`

`R\ text(is at intersection of tangents)`

`2tx\ – t^2` `= 2(1\ – t)x\ – (1\ – t)^2`
`2tx\ – t^2` `= 2x\ – 2tx\ – 1 + 2t\ – t^2`
`4tx\ – 2x` `= -1 + 2t\ – t^2 + t^2`
`2x(2t\ – 1)` `= 2t\ – 1`
`2x` `= 1`
`x` `= 1/2`

`text(Using)\ \ y = 2tx – t^2\ \ text(when)\ \ x = 1/2`

`y` `= 2t(1/2)\ – t^2`
  `= t\ – t^2`

`:.\ R(1/2, t\ – t^2)\ text(… as required)`

 

(iv)  `text(Locus of)\ R`

♦♦ Mean mark of 22%. 
MARKER’S COMMENT: Many students stated the locus as `y=t-t^2` rather than realising it had to be a straight line since `x=½`, and that `y=t-t^2` simply restricted the values of `y`.

`text(S)text(ince)\ x = 1/2\ text(is a constant)`

`R\ text(is a vertical line)`

`text(Now,)\ y = t\ – t^2 = t(1\ – t)`

`text(Graphically,)\ \ y\ \ text(has a maximum at)\ \ t = 1/2`

`text(Max)\ \ y = 1/2\ – (1/2)^2 = 1/4`

`=> y < 1/4\ \ text{(tangents can’t meet on parabola)}`

`:.\ text(Locus of)\ R\ text(is vertical line)\ x = 1/2,\ \ y<1/4`

Functions, EXT1 F2 2011 HSC 2a

Let  `P(x) = x^3-ax^2 + x`  be a polynomial, where  `a`  is a real number.

When  `P(x)`  is divided by  `x-3`  the remainder is  `12`.

Find the remainder when  `P(x)`  is divided by  `x + 1`.    (3 marks)

Show Answers Only

`-4`

Show Worked Solution

`P(x) = x^3 – ax^2 + x`

`text(S)text(ince)\ \ P(x) -: (x – 3)\ \ text(has remainder 12,)`

`P(3) = 3^3-a xx 3^2 + 3` `=12`
`27-9a + 3` `= 12`
`9a` `= 18`
`a` `=2`

 
`:.\ P(x) = x^3-2x^2 + x`

 

`text(Remainder)\ \ P(x) -: (x + 1)\ \ text(is)\ \ P(–1)`

`P(-1)` `= (-1)^3-2(-1)^2-1`
  `= – 4`

Calculus, EXT1 C2 2011 HSC 1d

Using the substitution  `u = sqrtx`, evaluate  `int_1^4 (e^(sqrtx))/(sqrtx)\ dx`.   (3 marks) 

Show Answers Only

`2e (e – 1)`

Show Worked Solution
`u` `= sqrtx = x^(1/2)`
`(du)/(dx)` `= 1/2 x^(-1/2) = 1/(2 sqrtx)`
 `du` `= (dx)/(2sqrtx)`
`:.2du` `= (dx)/(sqrtx)`

 

`text(When)\ \ \ ` `x=4,\ \ ` `u = 2`
  `x = 1,` `\ x = 1`

 

`:. int_1^4 (e^(sqrtx))/(sqrtx)\ dx`
`= int_1^2 e^u xx 2\ du`
`= 2 [e^u]_1^2`
`= 2 [e^2 – e^1]`
`= 2e (e – 1)`

Trig Calculus, EXT1 2011 HSC 1b

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

Show Answers Only

 `(sinx (2x cosx\ – sinx))/(x^2)`

Show Worked Solution

`y = (sin^2 x)/x`

`u`  `= sin^2 x,\ \ \ \ \ ` `u prime = 2 sinx cosx`
`v`  `= x,`  `v prime = 1`

 

`dy/dx` `= (u prime v\ – uv prime)/(v^2)`
  `= (2sinxcosx * x – sin^2x * 1)/(x^2)`
  `= (sinx (2x cosx\ – sinx))/(x^2)`

Linear Functions, EXT1 2011 HSC 1a

The point  `P`  divides the interval joining  `A(–1, –2)`  to  `B(9, 3)`  internally in the ratio  `4 : 1`. Find the coordinates of  `P`.   (2 marks)

Show Answers Only

 `(7, 2)`

Show Worked Solution

`A(–1,–2),\ \ \ B(9,3)`

`text(Internal division)\ \ 4:1\ \ text(at)\ \ P`

`P` `≡ ( (nx_1 + mx_2)/(m + n),\ (ny_1 + my_2)/(n + m) )`
  `≡ ( ((1 xx –1) + (4 xx 9))/(4 + 1),\ ((1 xx –2) + (4 xx 3))/(4 + 1))`
  `≡ (35/5, 10/5)`
  `≡ (7,2)`

Mechanics, EXT2* M1 2012 HSC 13c

A particle is moving in a straight line according to the equation

`x = 5 + 6 cos 2t + 8 sin 2t`, 

where `x` is the displacement in metres and `t` is the time in seconds.

  1. Prove that the particle is moving in simple harmonic motion by showing that `x`  satisfies an equation of the form  `ddot x = -n^2 (x\ - c)`.  (2 marks)

  2. When is the displacement of the particle zero for the first time?    (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.5\ text(seconds)\ text{(1 d.p.)}`
Show Worked Solution

i.  `text(Prove)\ ddot x = -n^2(x\ – c)`

`x` `= 5 + 6 cos 2t + 8 sin 2t`
`dot x` `= -12 sin 2t + 16 cos 2t`
`ddot x` `= – 24 cos 2t\ – 32 sin 2t`
  `= -4 (6 cos 2t + 8 sin 2t)`
  `= -2^2 (5 + 6 cos 2t + 8 sin 2t\ – 5)`
  `= -2^2 (x\ – 5)\ \ \ text(… as required)`

 

ii.  `text(Find)\ \ t\ \ text(when)\ \ x=0\ \ text(for 1st time:)`

♦ Mean mark 42%.
IMPORTANT: The critical insight required to solve `x=0` is to realise that the cosine of the difference between 2 angles, i.e. `cos (2t- theta)`, applies.
`5 + 6 cos 2t + 8 sin 2t` `= 0`
`6 cos 2t + 8 sin 2t` `= -5`
`6/10 cos 2t+ 8/10 sin 2t` `=-1/2`

 

 Calculus in the Physical World, EXT1 2012 HSC 13c Answer

`=>cos theta=6/10\ \ text(and)\ \ sin theta=8/10`
`cos 2t cos theta+sin 2t sin theta` `=- 1/2`
`cos(2t\ – theta)` `= – 1/2`

`text(S)text(ince)\ \ cos\ pi/3 = 1/2\ \ text(and)\ cos\ text(is negative)`

`text(in the 2nd and 3rd quadrants,)`

`=>2t\ – theta = pi\ – pi/3,\ pi + pi/3`

 

`text(We need the 1st time)\ \ x = 0`

`text(S)text(ince)\ \ cos theta` `= 6/10`
`theta` `= 0.9273…`

 

`:.\ 2t\ – 0.9273…` `= (2pi)/3`
`2t` `= (2pi)/3 + 0.9273…`
`t` `= 1/2 ((2pi)/3 + 0.9273…)`
  `= 1.5108…`
  `= 1.5\ text(seconds)\ text{(1 d.p.)}`

Geometry and Calculus, EXT1 2012 HSC 13b

  1. Find the horizontal asymptote of the graph
     
    `qquad qquad y=(2x^2)/(x^2 + 9)`.   (1 mark)
  2. Without the use of calculus, sketch the graph 
     
    `qquad qquad y=(2x^2)/(x^2 + 9)`,
     
    showing the asymptote found in part (i).    (2 marks)
Show Answers Only
  1. `text(Horizontal asymptote at)\ y = 2`
    1. Geometry and Calculus, EXT1 2012 HSC 13b Answer
Show Worked Solution
(i)    `y` `= (2x^2)/(x^2 +9)`
    `= 2/(1 + 9/(x^2))`

 

`text(As)\ \ x -> oo,\ y ->2`

`text(As)\ \ x -> – oo,\ y -> 2`

`:.\ text(Horizontal asymptote at)\ y = 2`

 

(ii)    `text(At)\ \ x = 0,\ y = 0`

`f(x) = (2x^2)/(x^2 + 9) >= 0\ text(for all)\ x`

`f(–x) = (2(–x)^2)/((–x)^2 + 9) = (2x^2)/(x^2 + 9) = f(x)`

`text(S)text(ince)\ \ f(x) = f(–x) \ \ =>\ text(EVEN function)`

Geometry and Calculus, EXT1 2012 HSC 13b Answer

Functions, EXT1 F1 2012 HSC 12b

Let  `f(x) = sqrt(4x-3)` 

  1.  Find the domain of  `f(x)`.   (1 mark)

  2.  Find an expression for the inverse function  `f^(-1) (x)`.    (2 marks)

  3.  Find the points where the graphs  `y = f(x)`  and  `y=x`  intersect.   (1 mark)

  4.  On the same set of axes, sketch the graphs  `y = f(x)`  and  `y = f^(-1) (x)`  showing the information found in part (iii).   (2 marks)

Show Answers Only
  1. `x >= 3/4`
  2. `f^(-1) (x) = 1/4 x^2 + 3/4,\ x >= 0`
  3. `(1,1),\ (3,3)`
  4.   
    1. Inverse Functions, EXT1 2012 HSC 12b Answer
Show Worked Solution

i.  `f(x) = sqrt(4x-3)`

`text(Domain exists when)\ \ 4x-3` `>= 0`
`4x` `>= 3`
`x` `>= 3/4`

 

ii.  `text(Inverse function when)`

`x` `= sqrt(4y-3)`
`x^2` `= 4y-3`
`4y` `= x^2 + 3`
`y` `= 1/4 x^2 + 3/4`

 

`text(S)text(ince range of)\ \ f(x)\ \ text(is)\ \ y >= 0`

`=> text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ x >= 0`

`:.\ f^(-1) (x) = 1/4 x^2 + 3/4,\ \ \ \ x >= 0`

 

iii.  `text(Find intersection)`

`y` ` = sqrt(4x-3)\ \ …\ text{(1)}`
`y ` `=x\ \ …\ text{(2)}`

 

`text{Intersection occurs when}`

`x` `= sqrt(4x-3)`
`x^2` `= 4x-3`
`x^2-4x + 3` `= 0`
`(x-3)(x-1)` `= 0`
`x` `=1\ \ text(or 3)`

 

`:.\ text(Intersection at)\ \ (1,1)\ \ text(and)\ \ (3,3)`

 

iv.

`qquad`Inverse Functions, EXT1 2012 HSC 12b Answer

Calculus, EXT1 C2 2012 HSC 11d

Use the substitution  `u = 2 - x`  to evaluate  `int_1^2 x (2 - x)^5\ dx`.   (3 marks)  

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 `4/21`

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`u` `=2-x`
`:. x` `= 2 – u`
`(du)/dx` `= -1`
`:.dx` `=-du`

 

`text(When)\ \ x = 2,` `\ \ u = 0`
`text(When)\ \ x = 1,` `\ \ u = 1`

 

`:. int_1^2 x (2 – x)^5\ dx` `= int_1^0 – (2 – u) u^5\ du`
  `= int_1^0 u^6 – 2u^5\ du`
  `= [1/7 u^7 – 2/6 u^6]_1^0`
  `= [0 – (1/7 – 1/3)]`
  `= – (3/21 – 7/21)`
  `= 4/21`

Functions, EXT1 F1 2012 HSC 11c

Solve  `x/(x - 3) < 2`.  (3 marks) 

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 `x<3\ \ text(or)\ \ x>6`

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`text(Solution 1)`

`x/(x – 3) < 2`

`text(When)\ \ x – 3 > 0,\ \ \ (text(i.e.)\ \ x > 3)`

`x` `< 2 ( x – 3)`
`x` `< 2x\ – 6`
`=>x` `>6\ \ \ text{(satisfies both)}`

 
`text(When)\ \ x\ – 3 < 0,\ \ \ (text(i.e.)\ \ x<3)`

`x` `>2 (x – 3)`
`x` `> 2x – 6`
`x` `<6`
`=> x` `<3\ \ \ text{(satisfies both)}`

 
`:.\ text(Equation is correct for)\ x<3\ text(or)\ x>6`

 
`text(Solution 2)`

`x/((x\ – 3)) < 2`
 

`text(Multiply b.s. by)\ \  (x-3)^2`

`x(x – 3)` `< 2(x^2 – 6x + 9)`
`x^2 – 3x` `< 2x^2 – 12x + 18`
`x^2 – 9x + 18` `>0`
`(x – 3)(x – 6)` `>0`

 

 Algebra, EXT1 2012 HSC 11c Answer

`:.\ text(Equation correct for)\ x<3\ \ text(or)\ \ x>6`

Calculus, EXT1 C2 2012 HSC 9 MC

What is the derivative of  `cos^(–1) (3x)`?

  1. `1/(3 sqrt(1 - 9x^2))`  
  2. `(-1)/(3 sqrt(1 - 9x^2))`  
  3. `3/sqrt(1 - 9x^2)`  
  4. `(-3)/sqrt(1 - 9x^2)`  
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`D`

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`y = cos^(-1) (3x)`

`dy/dx` `= (-1)/sqrt(1\ – (3x)^2) xx d/dx (3x)`
  `= (-3)/sqrt(1\ – 9x^2)`

`=>  D`

Functions, EXT1 F2 2012 HSC 3 MC

A polynomial equation has roots  `alpha`,  `beta`,  `gamma`  where

`alpha + beta + gamma = -2`,    `alphabeta + alphagamma + betagamma = 3`,    `alphabetagamma = 1`.

Which polynomial equation has the roots  `alpha`,  `beta`, and  `gamma`?

  1. `x^3 + 2x^2 + 3x + 1 = 0`
  2. `x^3 + 2x^2 + 3x- 1 = 0`
  3. `x^3- 2x^2 + 3x + 1 = 0`
  4. `x^3- 2x^2 + 3x- 1 = 0`
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`B`

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`text(Using)\ \ ax^3 + bx^2 + cx + d = 0`

`text(By elimination)`

`alpha beta gamma = -d/a  = 1`

`:.\ text(Cannot be)\ A\ text(or)\ C\ \ (text(where)\ alphabetagamma = -1 text{)}`

`alpha + beta + gamma = -b/a = -2`

`:.\ text(Cannot be)\ D\ \ (text(where)\ alpha + beta + gamma = 2 text{)}`

`=> B`

Algebra, EXT1 2012 HSC 1 MC

Which expression is a correct factorisation of  `x^3 - 27`? 

  1. ` (x - 3)( x^2 - 3x + 9)`
  2. `(x - 3)( x^2 - 6x + 9)`
  3. `(x - 3)( x^2 + 3x + 9)`
  4. `(x - 3)( x^2 + 6x + 9)`
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`C`

Show Worked Solution
`x^3 – 27` `= x^3 – 3^3`
  `= (x – 3)(x^2 + 3x + 9)`
`=> C`