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Binomial, EXT1 2009 HSC 6b

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

 

 

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

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

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

 

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

 

`text(Show)`

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

 

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

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

 

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

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

 

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

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

`text(… as required.)`

 

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

 

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

 

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

Calculus, EXT1 C1 2009 HSC 5b

The cross-section of a 10 metre long tank is an isosceles triangle, as shown in the diagram. The top of the tank is horizontal.
 

 
 

When the tank is full, the depth of water is 3 m. The depth of water at time `t` days is `h` metres.   

  1. Find the volume, `V`, of water in the tank when the depth of water is `h` metres.     (1 mark)

  2. Show that the area, `A`, of the top surface of the water is given by  `A = 20 sqrt3 h`.   (1 mark)

  3. The rate of evaporation of the water is given by  `(dV)/(dt) = - kA`, where `k` is a positive constant. 

     

    Find the rate at which the depth of water is changing at time `t`.   (2 marks)

  4. It takes 100 days for the depth to fall from 3 m to 2 m. Find the time taken for the depth to fall from 2 m to 1 m.   (1 mark)

Show Answers Only
  1. `10 sqrt 3 h^2\ \ text(m³)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `-k\ \ \ text(metres per day)`
  4. `100\ text(days)`
Show Worked Solution
MARKER’S COMMENT: Students who drew a diagram and included their working calculations on it were the most successful.
(i) 
  `text(Let)\ A = text(area of front)`
`tan 30^@` `= h/x`
`x` `= h/(tan 30^@)`
  `= sqrt3 h`
`:.\ A` `= 2 xx 1/2 xx sqrt 3 h xx h`
  `= sqrt 3 h^2\ \ text(m²)`

 

`V` `= Ah`
  `= sqrt 3 h^2 xx 10`
  `= 10 sqrt3 h^2\ \ text(m³)`

 

(ii)    `text(Area of surface)`
  `= 10 xx 2 sqrt 3 h`
  `= 20 sqrt 3 h\ \ text(m²)`

 

(iii)    `(dV)/(dt)` `= -kA`
    `= -k\ 20 sqrt3 h`
`V` `= 10 sqrt3 h^2`
`(dV)/(dh)` `= 20 sqrt 3 h`

 

`text(Find)\ (dh)/(dt)`

MARKER’S COMMENT: Half marks awarded for stating an appropriate chain rule, even if the following calculations were incorrect. Show your working!
`(dV)/(dt)` `= (dV)/(dh) * (dh)/(dt)`
`(dh)/(dt)` `= ((dV)/(dt))/((dV)/(dh))`
  `= (-k * 20 sqrt 3 h)/(20 sqrt 3 h)`
  `= -k`

 

`:.\ text(The water depth is changing at a rate)`

`text(of)\ -k\ text(metres per day.)`

 

♦♦♦ Exact data for part (iv) not available.
COMMENT: Interpreting a constant rate of change was very poorly understood!
(iv)    `text(S)text(ince)\ \ (dh)/(dt)\ \ text(is a constant, each metre)`
  `text(takes the same time.)`

 
`:.\ text(It takes 100 days to fall from 2 m to 1 m.)`

Mechanics, EXT2* M1 2009 HSC 5a

The equation of motion for a particle moving in simple harmonic motion is given by

`(d^2x)/(dt^2) = -n^2x`

where  `n`  is a positive constant,  `x`  is the displacement of the particle and  `t`  is time.  

  1. Show that the square of the velocity of the particle is given by
     
         `v^2 = n^2 (a^2\ - x^2)`

     

    where  `v = (dx)/(dt)`  and  `a`  is the amplitude of the motion.   (3 marks)

  2. Find the maximum speed of the particle.     (1 mark)

  3. Find the maximum acceleration of the particle.    (1 mark)

  4. The particle is initially at the origin. Write down a formula for  `x`  as a function of  `t`, and hence find the first time that the particle’s speed is half its maximum speed.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `na`
  3. `n^2 a`
  4. `pi/(3n)`
Show Worked Solution
i.    `text(Show)\ \ v^2 = n^2 (a^2\ – x^2)`

`(d^2x)/(dt^2) = -n^2x`

`d/(dx) (1/2 v^2)` `= -n^2 x`
`1/2 v^2` `= int -n^2x\ dx`
  `= (-n^2 x^2)/2 + c`
`v^2` `= -n^2 x^2 + c`

 
`text(When)\ \ v = 0,\ \ x = a`

`0` `= -n^2a + c`
`c` `= n^2 a^2`
`:.\ v^2` `= -n^2 x^2 + n^2 a^2`
  `= n^2 (a^2\ – x^2)\ \ text(… as required)`

 

ii.    `text(Max speed when)\ \ x = 0`
`v^2` `= n^2 (a^2\ – 0)`
  `= n^2 a^2`
`:.v_text(max)` `= na`

 

iii.   `(d^2x)/(dt^2)\ \ text(is maximum at limits)\ \ (x = +-a)`
`(d^2x)/(dt^2)` `= |n^2 (a)|`
  `= n^2 a`

 

iv.   `x` `= a sin nt`
  `dot x` `= an cos nt`

 

`text(Find)\ \ t\ \ text(when)\ \ dot x = (na)/2`

`(na)/2` `= an cos nt`
`cos nt` `= 1/2`
`nt` `= pi/3`
`:.t` `= pi/(3n)`

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

Statistics, EXT1 S1 2009 HSC 4a

A test consists of five multiple-choice questions. Each question has four alternative answers. For each question only one of the alternative answers is correct.

Huong randomly selects an answer to each of the five questions. 

  1. What is the probability that Huong selects three correct and two incorrect answers?   (2 marks)

  2. What is the probability that Huong selects three or more correct answers?    (2 marks)

  3. What is the probability that Huong selects at least one incorrect answer?  (1 mark)

Show Answers Only
  1. `45/512`
  2. `53/512`
  3. `1023/1024`
Show Worked Solution

i.   `P text{(correct)} = 1/4`

`P text{(wrong)} = 3/4`

`P text{(3 correct, 2 wrong)}`

`=\ ^5C_3 * (1/4)^3 (3/4)^2`

`= (5!)/(3!2!) * (1/64) * (9/16)`

`= 90/1024`

`= 45/512`
 

ii.  `P text{(3 or more correct)}`

`= P text{(3 correct)} + P text{(4 correct)} + P text{(5 correct)}`

`=\ ^5C_3 * (1/4)^3 (3/4)^2 +\ ^5C_4 * (1/4)^4 (3/4)^1 +\ ^5C_5 (1/4)^5 (3/4)^0`

`= 90/1024 + 15/1024 + 1/1024`

`= 53/512`
 

iii.  `P text{(at least 1 incorrect)}`

TIP: The use of “at least” should flag a good chance of applying `1-P text{(complement)}` to solve.

`= 1\ – P text{(0 incorrect)}`

`= 1 -\ ^5C_5 (1/4)^5 (3/4)^0`

`= 1\ – 1/1024`

`= 1023/1024`

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 F1 2009 HSC 3a

Let  `f(x) = (3 + e^(2x))/4`. 

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

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

Show Answers Only
  1. `y > 3/4`
  2. `f^(-1) (x) = 1/2 ln(4x\ – 3)`
Show Worked Solution
i.   `f(x) = (3 + e^(2x))/4`

`text(As)\ \ x -> oo, \ e^(2x) ->oo, \ f(x)->oo`

`text(As)\ \ x -> -oo, \ e^(2x) ->0, \ f(x)->3/4`

`:.\ text(Range is)\ \ y > 3/4`

 

ii.   `text(Inverse function: swap)\ \ x↔y` 

`x` `= (3 + e^(2y))/4`
`4x` `= 3 + e^(2y)`
`e^(2y)` `= 4x\ – 3`
`ln e^(2y)` `= ln (4x\ – 3)`
`2y` `= ln (4x\ – 3)`
`y` `= 1/2 ln (4x\ – 3)`

 

`:.\ f^(-1) (x) = 1/2 ln (4x\ – 3)`

Calculus, 2ADV C3 2014 HSC 16c

The diagram shows a window consisting of two sections. The top section is a semicircle of diameter  `x`  m. The bottom section is a rectangle of width  `x`  m and height  `y`  m.

The entire frame of the window, including the piece that separates the two sections, is made using 10 m of thin metal.

The semicircular section is made of coloured glass and the rectangular section is made of clear glass.

Under test conditions the amount of light coming through one square metre of the coloured glass is 1 unit and the amount of light coming through one square metre of the clear glass is 3 units.

The total amount of light coming through the window under test conditions is  `L`  units.

  1. Show that  `y = 5 - x(1 + pi/4)`.   (2 marks)

  2. Show that  `L = 15x - x^2 (3 + (5pi)/8)`.    (2 marks)

  3. Find the values of  `x`  and  `y`  that maximise the amount of light coming through the window under test conditions.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `x = 1.511\ text(m  and)\ \ y = 2.302\ text(m)`
Show Worked Solution
i.   

`text(Frame is 10m)`

`10` `= 2x + 2y + 1/2 pi x`
`2y` `= 10\ – 2x\ – 1/2 pi x`
`:.y` `= 5\ – x\ – pi/4 x`
  `= 5\ – x (1 + pi/4)\ \ \ text(… as required.)`

 

♦ Mean mark 35%
ii.    `text(Area)\ text{(clear)}` `= x xx y`
  `text(Area)\ text{(colour)}` `= 1/2 xx pi r^2`
    `= 1/2 xx pi (x/2)^2`
    `= (pi x^2)/8`

 

`:.\ L` `= 3xy + ((pix^2)/8 xx 1)`
  `= 3x [5\ – x (1 + pi/4)] + (pix^2)/8\ \ \ text{(see part (i))}`
  `= 15x\ – 3x^2\ – (3x^2pi)/4 + (x^2 pi)/8`
  `= 15x\ – 3x^2\ – (5x^2 pi)/8`
  `= 15x\ – x^2 (3 + (5pi)/8)\ \ \ text(… as required)`

 

♦ Mean mark 38%
COMMENT: A sanity check for your answer could be to compare your answers to the perimeter restriction of 10m.
iii.   `L` `= 15x\ – x^2 (3 + (5pi)/8)`
  `(dL)/(dx)` `= 15\ – 2x (3 + (5pi)/8)`
  `(d^2L)/(dx^2)` `= -2 (3 + (5pi)/8)`

 

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

`15\ – 2x (3x + (5pi)/8)=` `0`
`2x (3 + (5pi)/8)=` `15`
`x=` `15/(2 (3 + (5pi)/8)`
 `=` `1.51103…`
`=` `1.511\ \ \ text{(3 d.p.)}`

 

`text(S)text(ince)\ (d^2L)/(dx^2) < 0\ \ \ => text(MAX)`

`text(When)\ \ x` `= 1.511`
`y` `= 5\ – 1.511 (1 + pi/4)`
  `= 2.3022…`
  `= 2.302\ text{(3 d.p.)}`

 

`:.\ text(MAX light when)\ x = 1.511\ text(m)`

`text(and)\ y = 2.302\ text(m.)`

Financial Maths, 2ADV M1 2014 HSC 16b

At the start of a month, Jo opens a bank account and makes a deposit of $500.  At the start of each subsequent month, Jo makes a deposit which is 1% more than the previous deposit.

At the end of every month, the bank pays interest of 0.3% (per month) on the balance of the account.  

  1. Explain why the balance of the account at the end of the second month is 

     

      `$500 (1.003)^2 + $500 (1.01) (1.003)`.    (2 marks)

  2. Find the balance of the account at the end of the 60th month, correct to the nearest dollar.     (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `$44\ 404`
Show Worked Solution

i.  `text(Let)\ A_n=text(balance at end of month)\ n`

`A_1` `= 500 (1.003)`
`A_2` `= A_1 (1.003) + 500 (1.01) (1.003)`
  `= 500 (1.003) (1.003) + 500 (1.01)(1.003)`
  `= 500 (1.003)^2 + 500 (1.01) (1.003)`

 

♦ Mean mark 38%
 

ii.  `text(Need to find)\ A_60`

`A_3` `= A_2 (1.003) + 500 (1.01)^2 (1.003)`
  `= [500 (1.003)^2 + 500 (1.01) (1.003)] (1.003)`
  `+ 500 (1.01)^2 (1.003)`
  `= 500 (1.003)^3 + 500 (1.003)^2 (1.01) + 500 (1.003) (1.01)^2`
  `= 500 [1.003 (1.01)^2 + (1.003)^2 (1.01) + (1.003)^3]`

`vdots`

`A_60` `= 500 [1.003 (1.01)^59 + (1.003)^2 (1.01)^58 + … `
  `… + (1.003)^60]`
  `=> text(GP where)\ a = 1.003 (1.01)^59,\ \ r = T_2/T_1=1.003/1.01`
 `:. A_60` `= 500 [ (a(r^n\ – 1))/(r\ – 1)]`
  `= 500 [ (1.003 (1.01)^59 ((1.003/1.01)^60\ – 1))/(1.003/1.01\ – 1) ]`
  `= 500 [ (1.8041057(-0.3411697))/-0.00693069 ]`
  `= 44\ 404.396…`
  `= $44\ 404\ \ text{(nearest dollar)}`

 

`:.\ text(At the end of 60 months, the balance would)`

`text(be)\ $44\ 404.`

Integration, 2UA 2014 HSC 16a

Use Simpson’s Rule with five function values to show that 

`int_(- pi/3)^(pi/3) sec x\ dx ~~ pi/9 (3 + 8/sqrt3)`.   (3 marks)

Show Answers Only

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

Show Worked Solution
♦ Mean mark below 50%. BE CAREFUL! 

`A` `~~ h/3 [y_0 + 4(y_1 + y_3) + 2y_2 + y_4]`
  `~~ pi/18 [2 + 4(2/sqrt3 + 2/sqrt3) + 2(1) + 2]`
  `~~ pi/18 [6 + 16/sqrt3]`
  `~~ pi/9 (3 + 8/sqrt3)\ text(u² … as required)`

Calculus, 2ADV C3 2014 HSC 15c

The line  `y = mx`  is a tangent to the curve  `y = e^(2x)`  at a point  `P`. 

  1. Sketch the line and the curve on one diagram.   (1 mark)

  2. Find the coordinates of  `P`.     (3 marks)

  3. Find the value of  `m`.   (1 mark)

Show Answers Only
  1.   
  2. `P(1/2 ln (m/2), m/2)`
  3. `2e`
Show Worked Solution
i. 

 

ii. `y` `= e^(2x)`
  `dy/dx` `= 2e^(2x)`

 
`text(Gradient of)\ \ y = mx\ \ text(is)\ \ m`

♦ Mean mark 40%
COMMENT: Given `y= e^(ln(m/2))`, it follows `y=m/2`. Make sure you understand the arithmetic behind this (NB. Simply take the `ln` of both sides).

`text(Gradients equal when)`

`2e^(2x)` `= m`
`e^(2x)` `= m/2`
`ln e^(2x)` `= ln (m/2)`
`2x` `= ln (m/2)`
`x` `= 1/2 ln (m/2)`

 
`text(When)\ \ x = 1/2 ln (m/2)`

`y` `= e^(2 xx 1/2 ln (m/2))`
  `= e^(ln(m/2))`
  `= m/2`

 
`:.\ P (1/2 ln (m/2), m/2)`

 

iii.   `y=mx\ \ text(passes through)\ \ (0,0)\ text(and)\ (1/2 ln (m/2), m/2)`

♦♦ Mean mark 30%.

`text(Equating gradients:)`

`(m/2 – 0)/(1/2 ln (m/2) – 0)`  `=m`
`m/2` `=m xx 1/2 ln(m/2)`
`ln (m/2)` `= 1`
`m/2` `= e^1`
`m` `= 2e`

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.)`

Trigonometry, 2ADV T2 2014 HSC 15a

Find all solutions of  `2 sin^2 x + cos x − 2 = 0`, where  `0 <= x <= 2pi`.   (3 marks)

Show Answers Only

`x = pi/3,\ pi/2,\ (3pi)/2,\ (5pi)/3`

Show Worked Solution
♦ Mean mark 42%
`2 sin^2 x + cos x\ – 2` `= 0`
`2(1\ – cos^2x) + cos x\ – 2` `= 0`
`2\ – 2cos^2x + cosx\ – 2` `= 0`
`-2cos^2x + cosx` `= 0`
`cosx (-2 cosx + 1)` `= 0`

 

`:. -2 cosx + 1` `= 0` `\ text(or)\ \ \ \ \ \ \ ` `cos x` `= 0`
`2 cos x` `= 1`   `x` `= pi/2,\ (3pi)/2`
`cos x` `= 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/3,\ 2 pi \  – pi/3`
  `= pi/3,\ (5pi)/3`

 

`:. x = pi/3,\ pi/2,\ (3pi)/2,\ (5pi)/3\ \ text(for)\ \ 0 <= x <= 2pi`

Probability, 2ADV S1 2014 HSC 10 MC

Three runners compete in a race. The probabilities that the three runners finish the race in under  10  seconds are  `1/4`, `1/6`  and  `2/5`  respectively.

What is the probability that at least one of the three runners will finish the race in under 10 seconds? 

  1. `1/60`
  2. `37/60`
  3. `3/8`
  4. `5/8`
Show Answers Only

`D`

Show Worked Solution
♦♦ Mean mark 26%

`text{P} (R_1 < 10\ text(secs) ) = 1/4\ \ =>text{P} (bar R_1) = 3/4`

`text{P} (R_2 < 10\ text(secs) ) = 1/6\ \ =>text{P} (bar R_2) = 5/6`

`text{P} (R_3 < 10\ text(secs) ) = 2/5\ \ =>text{P} (bar R_3) = 3/5`
 

`:.\text{P} ( text(at least)\ 1 < 10\ text(secs) )`

`= 1\ -text{P} ( text(all) >= 10\ text(secs) )`

`= 1\ – 3/4 xx 5/6 xx 3/5`

`= 1\ – 45/120`

`= 5/8`

`=>  D`

Statistics, STD2 S4 2014* HSC 30b

The scatterplot shows the relationship between expenditure per primary school student, as a percentage of a country’s Gross Domestic Product (GDP), and the life expectancy in years for 15 countries.
 

 
 

  1. For the given data, the correlation coefficient,  `r`, is 0.83. What does this indicate about the relationship between expenditure per primary school student and life expectancy for the 15 countries?   (1 mark)

  2. For the data representing expenditure per primary school student,  `Q_L`  is 8.4 and  `Q_U`  is 22.5.

     

    What is the interquartile range?   (1 mark)

  3. Another country has an expenditure per primary school student of 47.6% of its GDP.

     

    Would this country be an outlier for this set of data? Justify your answer with calculations.   (2 marks)

  4. On the scatterplot, draw the least-squares line of best fit  `y = 1.29x + 49.9`.    (2 marks)

  5. Using this line, or otherwise, estimate the life expectancy in a country which has an expenditure per primary school student of 18% of its GDP.   (1 mark)

  6. Why is this line NOT useful for predicting life expectancy in a country which has expenditure per primary school student of 60% of its GDP?   (1 mark)

Show Answers Only
  1. `text(It indicates there is a strong positive)`

     

    `text(correlation between the two variables.)`

  2. `14.1`
  3. `text(Yes, because it’s > 43.65%)`
  4.  
  5. `73.1\ text(years)`
  6. `text(At 60% GDP, the line predicts a life expectancy)`

  7.  

    `text(of 127.3. This line of best fit is only accurate)`

  8.  

    `text(in a lower range of GDP expediture.)`

Show Worked Solution
i. `text(It indicates there is a strong positive)`
  `text(correlation between the two variables)`

 

ii. `text(IQR)` `= Q_U\ – Q_L`
    `= 22.5\ – 8.4`
    `= 14.1`

 

♦ Mean mark 35% 

iii.  `text(An outlier on the upper side must be more than)` 

`Q_u\ +1.5xxIQR`

`=22.5+(1.5xx14.1)`

`=\ text(43.65%)`

`:.\ text(A country with an expenditure of 47.6% is an outlier).`

 

iv.  

v.  `text(Life expectancy) ~~ 73.1\ text{years (see dotted line)}`

♦♦ Mean mark 39%

 

`text(Alternative Solution)`

`text(When)\ x=18`

`y=1.29(18)+49.9=73.12\ \ text(years)`

  

♦♦♦ Mean mark 0%. The toughest question on the 2014 paper.
COMMENT: Examiners regularly ask students to identify and comment on outliers where linear relationships break down.
vi.   `text(At 60% GDP, the line predicts a life)`
  `text(expectancy of 127.3. This line of best)`
  `text(fit is only predictive in a lower range)`
  `text(of GDP expenditure.)`

Financial Maths, STD2 F4 2014 HSC 30a

Chandra and Sascha plan to have $20 000 in an investment account in 15 years time for their grandchild’s university fees.

The interest rate for the investment account will be fixed at 3% per annum compounded monthly.

Calculate the amount that they will need to deposit into the account now in order to achieve their plan.   (3 marks)

Show Answers Only

`$12\ 760\ \ text{(nearest $)}`

Show Worked Solution
♦ Mean mark 49%

`FV = $20\ 000,\ \ n = 15xx 12=180,`

`r = 0.03 /12=0.0025`
 

`FV` `= PV (1 + r)^n`
`20\ 000` `=PV (1 + 0.0025)^180`
`PV` `=(20\ 000)/(1.0025)^180`
  `=12\ 759.73…`

 

`:.\ text(They need to deposit) \ \ $12\ 760\ \ text{(nearest $)}`

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 A4 2014 HSC 29a

The cost of hiring an open space for a music festival is  $120 000. The cost will be shared equally by the people attending the festival, so that  `C`  (in dollars) is the cost per person when  `n`  people attend the festival.

  1. Complete the table below by filling in the THREE missing values.   (1 mark)
    \begin{array} {|l|c|c|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
    \hline
    \rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} &  &  &  & 60 & 48\ & 40 \ \\
    \hline
    \end{array}
  2. Using the values from the table, draw the graph showing the relationship between  `n`  and  `C`.   (2 marks)
     
  3. What equation represents the relationship between `n` and `C`?   (1 mark)

  4. Give ONE limitation of this equation in relation to this context.   (1 mark)

  5. Is it possible for the cost per person to be $94? Support your answer with appropriate calculations.   (1 mark)

Show Answers Only

i.   

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
 

ii. 

iii.   `C = (120\ 000)/n`

`n\ text(must be a whole number)`
 

iv.   `text(Limitations can include:)`

  `•\ n\ text(must be a whole number)`

  `•\ C > 0`
 

v.   `text(If)\ C = 94:`

`94` `= (120\ 000)/n`
`94n` `= 120\ 000`
`n` `= (120\ 000)/94`
  `= 1276.595…`

 
`:.\ text(C)text(ost cannot be $94 per person,)`

`text(because)\ n\ text(isn’t a whole number.)`

Show Worked Solution

i.   

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
 

ii. 

 

♦ Mean mark (iii) 48%

iii.   `C = (120\ 000)/n`

 

♦♦♦ Mean mark (iv) 7%
COMMENT: When asked for limitations of an equation, look carefully at potential restrictions with respect to both the domain and range.

iv.   `text(Limitations can include:)`

  `•\ n\ text(must be a whole number)`

  `•\ C > 0`

 

v.   `text(If)\ C = 94`

`=> 94` `= (120\ 000)/n`
`94n` `= 120\ 000`
`n` `= (120\ 000)/94`
  `= 1276.595…`
♦ Mean mark (v) 38%

 

`:.\ text(C)text(ost cannot be $94 per person,)`

`text(because)\ n\ text(isn’t a whole number.)`

FS Resources, 2UG 2014 HSC 28d

An aerial diagram of a swimming pool is shown. 

The swimming pool is a standard length of 50 metres but is not in the shape of a rectangle.

(i)   Given  `AB=8\ text(cm)`, determine the scale of the diagram such that

1 cm = `x` m   (1 mark)

(ii)  If the length of a carpark next to the pool measured 5 cm (not shown), how long would it be in real life?   (1 mark)

(iii)  In the diagram of the swimming pool, the five widths are measured to be: 

`CD = 21.88\ text(m)`

`EF = 25.63\ text(m)`

`GH = 31.88\ text(m)`

`IJ = 36.25\ text(m)`

`KL = 21.88\ text(m)` 

 

The average depth of the pool is 1.2 m

Calculate the approximate volume of the swimming pool, in cubic metres. In your calculations, use TWO applications of Simpson’s Rule.   (3 marks)

Show Answers Only

(i)   `x=6.25\ text(m)`

(ii)  `31.25\ text(m)`

(iii) `1775\ text(m³)`

Show Worked Solution
(i) `\ \ \ \ 8\ text(cm)` `=50\ text(m)`
  `1\ text(cm)` `=50/8`
    `=6.25\ text(m)`

`:.x=6.25\ text(m)`

 

(ii)  `text{Using scale from (i)}`

`5\ text(cm)` `=5 xx 6.25`
  `=31.25\ text(m)`

`:.\ text(The carpark would be 31.25 m long)`

 

(iii)

`h = 50/4 = 12.5\ text(m)`

♦ Mean mark 50%. Be careful not to give away easy marks! 
`A` `~~ h/3 [y_0 + 4(y_1) + y_2]\ \ text(… applied twice)`
  `~~ 12.5/3 [21.88 + 4(25.63) + 31.88]`
  `+ 12.5/3 [31.88 + 4(36.25) + 21.88]`
  `~~ 12.5/3 [156.28] + 12.5/3 [198.76]`
  `~~ 1,479.33\ text(m²)`

 

`V` `= Ah`
  `~~ 1479.33 xx 1.2`
  `~~ 1775.2`
  `~~ 1775\ text(m³)`

Measurement, STD2 M6 2014 HSC 28b

A radial compass survey of a sports centre is shown in the diagram. 
 

 
 

  1. Show that the size of angle  `AOB`  is  114°.   (1 mark)

  2. Calculate the length of the boundary `AB`, to the nearest metre.     (2 marks)

  3. Find the area of triangle `AOB` in hectares, correct to two significant figures.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `420\ text(m)\ text{(nearest m)}`
  3. `2.8\ text(ha)\ text{(2 sig. figures)}`
Show Worked Solution
i.   

 

`text(Let)\ D\ text(be directly north of)\ O`

`/_AOD = 360\ – 320 = 40^@`

`:.\ /_AOB = 40 + 74 = 114^@\ \ \ text(… as required)`

 

ii.  `text(Using cosine rule)`

`AB^2` `= AO^2 + BO^2\ – 2 xx AO xx BO xx cos /_AOB`
  `= 287^2 + 211^2\ – 2 xx 287 xx 211 xx cos 114^@`
  `= 126\ 890\ – 121\ 114 (-0.4067…)`
  `= 176\ 151.50…`
`AB` `= 419.704…`
  `= 420\ text(m)\ text{(nearest m)}`
♦ Mean marks of 42% and 41% for parts (ii) and (iii) respectively.
 

iii.  `text(Using)\ A = 1/2 ab sin C,`

`text(Area)\ Delta AOB` `= 1/2 xx 287 xx 211 xx sin 114^@`
  `= 27\ 660.786…\ text(m²)`
  `= 2.7660…\ text(ha)\ \ \ \ text{(1 ha = 10 000 m²)}`
  `= 2.8\ text(ha)\ text{(2 sig figures)}`

Probability, 2UG 2014 HSC 28a

James plays a game involving a spinner with sectors of equal size labelled `A`, `B` and `C`, as shown.

2014 28a

He pays  `$2`  to play the game. He wins  `$5`  if the spinner stops in `A` and  `50`  cents if it stops in `B` or `C`.

Calculate James’s financial expectation for the game.  (2 marks)

Show Answers Only

`text(James should expect to breakeven.)`

Show Worked Solution
♦ Mean mark 49%

`P(A) = 1/3`

`P(B\ text(or)\ C) = 2/3`

`text(Financial expectation)`

`=(1/3 xx 5) + (2/3 xx 0.50)\ – 2`

`= 5/3 + 1/3\ – 2`

`= 0`

`:.\ text(James should expect to breakeven.)`

Measurement, STD2 M1 2014 HSC 27c

The base of a water tank is in the shape of a rectangle with a semicircle at each end, as shown.

The tank is 1400 mm long, 560 mm wide, and has a height of 810 mm.  
  

What is the capacity of the tank, to the nearest litre?   (4 marks) 

Show Answers Only

`581\ text(L)`

Show Worked Solution

`V = Ah` 

♦ Mean mark 41%
STRATEGY: Adjusting measurements to metres makes the final conversion to litres simple.

`text(Finding Area of base)`

`text(Semi-circles have radius 280 mm) = 0.28\ text(m)`

`:.\ text(Area of 2 semicircles)`

`=2 xx 1/2 xx pi r^2`

`= pi xx (0.28)^2`

`= 0.2463…\ text(m)^2`
 

`text(Area of rectangle)`

`= l xx b`

`= (1.4-2 xx 0.28) xx 0.56`

`= 0.4704\ text(m)^2`

 

`:.\ text(Volume)` `= Ah`
  `= (0.2463… + 0.4704) xx 0.810`
  `= 0.580527…\ text(m)^3`
  `= 580.527…\ text(L)\ \ text{(using 1m³} = 1000\ text{L)}`
  `= 581\ text(L)\ text{(nearest L)}`

FS Driving, 2UG 2014 HSC 27a

Alex is buying a used car which has a sale price of  $13 380. In addition to the sale price there are the following costs:

2014 27a1

  1. Stamp Duty for this car is calculated at $3 for every $100, or part thereof, of the sale price.  
  2. Calculate the Stamp Duty payable.   (1 mark)
  3.  
  4. Alex borrows the total amount to be paid for the car including Stamp Duty and transfer of registration. Interest on the loan is charged at a flat rate of  7.5%  per annum. The loan is to be repaid in equal monthly instalments over 3 years.  
  5.  
    Calculate Alex’s monthly repayments.   (4 marks)
  6.  
  7. Alex wishes to take out comprehensive insurance for the car for 12 months. The cost of comprehensive insurance is calculated using the following: 
    1.  
    2. 2014 27a2
  8. Find the total amount that Alex will need to pay for comprehensive insurance.   (3 marks)
  9.  
  10. Alex has decided he will take out the comprehensive car insurance rather than the less expensive non-compulsory third-party car insurance.
  11. What extra cover is provided by the comprehensive car insurance?   (1 mark)

 

Show Answers Only
  1. `$402`
  2. `$470\ text{(nearest dollar)}`
  3. `$985.74`
  4. `text(Comprehensive insurance covers Alex)`
  5. `text(for damage done to his own car as well.)`
  6.  
Show Worked Solution
♦♦♦ Mean mark 12%
IMPORTANT: “or part thereof ..” in the question requires students to round up to 134 to get the right multiple of $3 for their calculation.
(i)    `($13\ 380)/100 = 133.8`
`:.\ text(Stamp duty)` `= 134 xx $3`
  `= $402`

 

(ii)    `text(Total loan)` `= $13\ 380 + 30 + 402`
    `= $13\ 812`

 

`text(Total interest)\ (I)` `= Prn`
  `= 13\ 812 xx 7.5/100 xx 3`
  `= 3107.70`

 

`text(Total to repay)` `= 3107.70 + 13\ 812`
  `= 16\ 919.70`

 

`text(# Repayments) = 3 xx 12 = 36`

`:.\ text(Monthly repayment)` `= (16\ 919.70)/36`
  `= 469.9916…`
  `= $470\ text{(nearest dollar)}`

 

(iii)   `text(Base rate) = $845`

`text(FSL) =\ text(1%) xx 845 = $8.45`

`text(Stamp)` `=\ text(5.5%) xx(845 + 8.45)`
  `= 46.9397…`
  `= $46.94\ text{(nearest cent)}`
`text(GST)` `= 10 text(%) xx(845 + 8.45)`
  `= 85.345`
  `= $85.35`

 

`:.\ text(Total cost)` `= 845 + 8.45 + 46.94 + 85.35`
  `= $985.74`

 

♦ Mean mark 34%.
(iv)   `text(Comprehensive insurance covers Alex)`
  `text(for damage done to his own car as well.)`

Measurement, 2UG 2014 HSC 26g

Singapore is located at  1°N 104°E  and has a UTC of +8. Sydney is located at 34°S 151°E and has a UTC of +11. 

What is the time difference between Singapore and Sydney? (Ignore daylight saving.)   (2 marks)

Show Answers Only

`text(Sydney is 3hrs 8mins ahead)`

Show Worked Solution

`text(Longitude angular difference)`

♦ Mean mark 50%

`= 151\ – 104`

`= 47^@`
 

`text(Time difference)` `= 47 xx 4`
  `= 188\ text(mins)`
  `=\ text(3hrs 8mins)`

 
`=>text(Sydney is East of)\ text(S)text(ingapore)`

`:.\ text(Sydney is 3hrs 8mins ahead.)`

Algebra, 2UG 2014 HSC 26d

Solve these simultaneous equations to find the values of  `x`  and  `y`.    (3 marks)

`y = 2x + 1`

`x − 2y − 4 = 0`

Show Answers Only

`x = -2,\ y = -3`

Show Worked Solution
♦ Mean mark 36%

`text(Solution 1 – Substitution)`

`y = 2x + 1\ \ \ \ \ …\ text{(i)}`

`x\ – 2y\ – 4 = 0\ \ \ \ \ …\ text{(ii)}`

 

`text(Subst.)\ \ y = 2x + 1\ \ text(into)\ text{(ii)}`

`x\ – 2(2x + 1)\ – 4` `= 0`
`x\ – 4x\ – 2\ – 4` `= 0`
`-3x\ – 6` `= 0`
`3x` `= -6`
`x` `= -2`

`text(Subst.)\ \ x = –2\ \ text(into)\ text{(i)}`

`y = 2(–2) + 1 = -3`

 

`:.\ text(Solution is)\ x = -2,\ y = -3`

 

`text(Alternative Solution – Elimination)`

`y = 2x + 1\ \ \ \ \ …\ text{(i)}`

`x\ – 2y\ – 4 = 0\ \ \ \ \ …\ text{(ii)}`

`text(Mult)\ text{(i)} xx 2`

`2y` `= 4x + 2`
`-4x + 2y\ – 2` `= 0\ \ \ \ \ …\ text{(iii)}`

 

`text{Add  (ii) + (iii)}`

`-3x\ – 6` `= 0`
`x` `= -2`
`y` `= -3\ \ text{(see Solution 1)}`

 

Measurement, STD2 M1 2014 HSC 25 MC

A grain silo is made up of a cylinder with a hemisphere (half a sphere) on top. The outside of the silo is to be painted.
  

 What is the area to be painted?

  1. `8143\ text(m²)`
  2. `11\ 762\ text(m²)`
  3. `12\ 667\ text(m²)`
  4. `23\ 524\ text(m²)`
Show Answers Only

`A`

Show Worked Solution

`text(Total Area) = text(Area of cylinder) + text(½ sphere)`

♦ Mean mark 40%
`text(Area of cylinder)` `= 2 pi rh`
  `= 2pi xx 24 xx 30`
  `= 4523.9`
`text(Area of ½ sphere)` `= 1/2 xx 4 pi r^2`
  `= 1/2 xx 4 pi xx 24^2`
  `= 3619.1`
`:.\ text(Total area)` `= 4523.9 + 3619.1`
  `= 8143\ text(m²)`

`=>  A`

Statistics, STD2 S5 2014 HSC 24 MC

The weights of  10 000 newborn babies in NSW are normally distributed. These weights have a mean of 3.1 kg and a standard deviation of 0.35 kg.

How many of these newborn babies have a weight between 2.75 kg and 4.15 kg?

  1. `4985`
  2. `6570`
  3. `8370`
  4. `8385`
Show Answers Only

`D`

Show Worked Solution
♦ Mean mark 46%

`text(Find)\ z text(-scores of 2.75 and 4.15 kg)`

`z\ (2.75)` `= (x – mu)/5 = (2.75 – 3.1)/0.35 = -1`
`z\ (4.15)` `= (4.15 – 3.1)/0.35 = 3`

 
`text(68% between)\ z=–1\ text(and 1)`

`=> text(34% between)\ z=–1\ text(and 0)`

`text(99.7% between)\ z=–3\ text(and 3)`

`=> text(49.85% between)\ z=0\ text(and)\ 3`

 

`:.%\ text(with)\ z text(-scores between)\ –1\ text(and 3)`

`= 34 + 49.85`

`=\ text(83.85%)`

 

`:.\ text(# Babies between 2.75 kg and 4.15 kg)`

`= text(83.85%) xx 10\ 000`

`= 8385`

`=>  D`

Measurement, STD2 M6 2014 HSC 23 MC

The following information is given about the locations of three towns `X`, `Y` and `Z`: 

• `X` is due east of  `Z`

• `X` is on a bearing of  145°  from  `Y` 

• `Y` is on a bearing of  060°  from  `Z`. 

Which diagram best represents this information?
 

HSC 2014 23mci

Show Answers Only

`C`

Show Worked Solution
♦ Mean mark 38%
COMMENT: Drawing a parallel North/South line through `Y` makes this question much simpler to solve.

`text(S)text(ince)\ X\ text(is due east of)\ Z`

`=> text(Cannot be)\ B\ text(or)\ D`
 

 
`text(The diagram shows we can find)`

`/_ZYX = 60 + 35^@ = 95^@`

`text(Using alternate angles)\ (60^@)\ text(and)`

`text(the)\ 145^@\ text(bearing of)\ X\ text(from)\ Y`

`=>  C`

Financial Maths, STD2 F5 2014 HSC 21 MC

A table of future value interest factors is shown.

2014 21 mc

A certain annuity involves making equal contributions of $25 000 into an account every 6 months for 2 years at an interest rate of 4% per annum.

Based on the information provided, what is the future value of this annuity? 

  1.    `$50\ 500`
  2.    `$51\ 000`
  3.    `$103\ 040`
  4.    `$106\ 162`
Show Answers Only

`C`

Show Worked Solution

`text(4 contributions of $25 000 made.)`

♦ Mean mark 43%

`text(Annuity period = 6 months)`

`text{Rate (per annuity period)}=(text(4%))/2=text(2%)`

`text{# Periods = 4     (4 x 6 months = 2 years)}`

`text(Table value = 4.1216)`
 

`:.\ text(Annuity Value)=4.1216 xx 25\ 000=$103\ 040`

`=>  C`

Measurement, STD2 M1 2014 HSC 20 MC

In a household of 4, each member uses an average of 13 minutes of hot water per day.

The household uses a 9 kW hot water unit.

Electricity is charged at 11.97 c/kWh when the hot water unit is being used.

What is the electricity cost for the hot water used by this household in one week?

  1. $1.63
  2. $6.54
  3. $392.14
  4. $653.56
Show Answers Only

`B`

Show Worked Solution

`text(Usage per day) = 4 xx 13 = 52\ text(mins)`

♦ Mean mark 39%.

`text(Usage per week) = 7 xx 52 = 364\ text(mins)`

`text(Converting to kWh)`

`= text{(hours of usage)} xx 9\ text(kW)`

`= 364/60 xx 9`

`= 54.6\ text(kWh)`
 

`:.\ text(C)text(ost)` `= 54.6 xx 11.97 text(c)`
  `~~ 654 text(c)`
  `~~$6.54`

`=>  B`

Financial Maths, STD2 F1 2014 HSC 18 MC

The average NSW annual water consumption from the residential sector is equal to  90 340  litres per person per year. The Building Sustainability Index (BASIX) uses this as the benchmark to set a target for reducing water consumption by up to 40%.

A new building, planned to house 50 people, has been designed to meet a 25% reduction on this water consumption benchmark.

How much water per year, to the nearest kilolitre, is this building designed to save when fully occupied?

  1.    1129
  2.    1807
  3.    2710
  4.    3388
Show Answers Only

`A`

Show Worked Solution

`text(Benchmark) = 90\ 340\ text(L)`

`text(Water saved if usage)\ darr text(25%)`

`= 90\ 340 xx text(25%)`

`= 22\ 585\ text(L per person)`

 
`:.\ text(Water saved for 50 people)`

`= 50 xx 22\ 585`

`= 1\ 129\ 250\ text(L)`

`~~ 1129\ text(kL)`

`=>  A`

Statistics, STD2 S1 2014 HSC 14 MC

Twenty Year 12 students were surveyed. These students were asked how many hours of sport they play per week, to the nearest hour.

The results are shown in the frequency table. 

2014 14 mc

 What is the mean number of hours of sport played by the students per week?

  1.    3.3
  2.    4.3
  3.    5.0
  4.    5.3
Show Answers Only

`B`

Show Worked Solution

`text(Using the class centres)`

`text(Total hours)` `= (1 xx 5) + (4 xx 10) + (7 xx 3) + (10 xx 2)`
  `= 5 + 40 + 21 + 20`
  `= 86`
♦ Mean mark 45%
COMMENT: The mean is calculated using “class centres” in grouped data.
`text(Mean hours)` `= 86/20 = 4.3`

`=>  B`

Measurement, STD2 M1 2014 HSC 10 MC

The top of the Sydney Harbour Bridge is measured to be 138.4 m above sea level. 

What is the percentage error in this measurement?

  1. 0.036%
  2. 0.050%
  3. 0.072%
  4. 0.289%
Show Answers Only

`A`

Show Worked Solution
 
♦ Mean mark 48%

`text{Absolute error}\ =1/2 xx text{precision}\ = 1/2 xx 0.1 = 0.05\ text{m}`

`text{% error}` `=\ frac{text{absolute error}}{text{measurement}} xx 100%`  
  `=0.05/138.4 xx 100%`  
  `=0.036%`  

 
`=>  A`

Algebra, STD2 A2 2014 HSC 7 MC

Which of the following is the graph of   `y = 2x-2`? 
  


  

Show Answers Only

`D`

Show Worked Solution
♦ Mean mark 46%

`y = 2x-2`

`text(By elimination)`

`text(It has a)\ y\ text(intercept of)\ -2`

`=> text(Cannot be)\ B\ text(or)\ C`

 

`(-1, 0)text{ from}\ A\ text(doesn’t satisfy equation)`

`text(but)\ (1,0)\ text(from)\ D\ text(does)`

`=>  D`

Algebra, STD2 A1 2014 HSC 4 MC

Young’s formula below is used to calculate the required dosages of medicine for children aged 1–12 years.

 `text(Dosage) = (text(age of child)\ text{(in years)} xx text(adult dosage))/(text(age of child)\ text{(in years)} + 12)`

How much of the medicine should be given to an 18-month-old child in a 24-hour period if each adult dosage is 45 mL? The medicine is to be taken every 6 hours by both adults and children.

  1.       5 mL
  2.    20 mL
  3.    27 mL
  4.    30 mL
Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 42%
`text(Dosage)` `= (1.5 xx 45)/(1.5 + 12)`
  `= 5\ text(mL)`

 
`text(S)text(ince 1 dosage every 6 hrs)`

`text(In 24 hours,)`

`text(Medicine given) = 4 xx 5 = 20\ text(mL)`

`=>  B`

Proof, EXT2 P2 EQ-Bank 8

Prove by mathematical induction that  `2^n > n^2`  for integral  `n > 4`.  (3 marks)

Show Answers Only

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

Show Worked Solution
IMPORTANT: Inequality induction is challenging. Students need to state clearly what they need to prove and then use logic to advance their answer toward the desired result.

`text(Prove)\ \ 2^n > n^2\ \ text(for integral)\ \ n > 4`

`text(If)\ \ n = 5,`

`text(LHS)` `= 2^5 = 32`
`text(RHS)` `= 5^2 = 25 < text(LHS)`

 
`:.\ text(True for)\ n = 5`

 
`text(Assume true for)\ n = k:`

`text(i.e.)\ \ \ 2^k > k^2`

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

`text(i.e.)\ \ \ ` `2^(k + 1)` `> (k + 1)^2`
  `2^(k + 1)` `> k^2 + 2k + 1`

 

`text(LHS)` `= 2^(k + 1)`
  `= 2*2^k`
  `> 2k^2`
  `> k^2 + k^2`
  `> k^2 + 4k\ \ text{(} text(noting)\ k^2>4k,\ k>4 text{)}`
  `> k^2 + 2k + 8\ \ text{(} text(noting)\ 4k > 2k + 8,\ k > 4 text{)}`
  `> k^2 + 2k + 1`
  `> (k + 1)^2`

 

`=>\ text(True for)\ n = k + 1`

`:.\ text(S)text(ince true for)\ \ n = 5, text(by PMI, true for integral)\ \ n > 4.`

Proof, EXT1 P1 SM-Bank 3

Prove by mathematical induction that  `(3n + 1)7^n  - 1`  is divisible by 9 for integral  `n >= 1`.  (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Prove)\ \ (3n + 1)7^n\ – 1\ \ text(is divisible)`

`text(by 9 for integers)\ n >= 1.`

`text(If)\ \ n = 1`

`(3n + 1)7^n\ – 1` `= (3 xx 1 + 1)7^1\ – 1`
  `= 27`

`=>\ text(Divisible by 9)\ \ (27//9=3)`
 

`text(Assume true for)\ n = k`

IMPORTANT: In cases where substitution doesn’t work, remember that `f(k+1)-f(k)` can be used to illustrate divisibility as well (Tony Lantry special).

`text(i.e.)\ (3k + 1)7^k\ – 1 = 9P\ \ \ text{(} P\ text(integer) text{)}`

`f(k)` `= (3k + 1)7^k\ – 1`
`f(k + 1)` `= (3 (k + 1) + 1)7^(k+1)\ – 1`
  `= (3k + 4)7^(k+1)\ – 1`

 
`f(k + 1)\ – f(k)`

`= (3k + 4) 7^(k + 1)\ -1- [(3k + 1)7^k\ – 1]`
`= 3k * 7^(k + 1) + 4 * 7^(k + 1)\ – 1\ – [3k*7^k + 7^k\ – 1]`
`= 21k*7^k + 28*7^k\ – 1\ – 3k*7^k\ – 7^k + 1`
`= 18k*7^k + 27*7^k`
`= 9 (2k*7^k + 3*7^k)\ \ text{(divisible)}`
 
`text(S)text(ince)\ \ ` `f(k + 1)\ – f(k)\ text(is divisible by 9 and)`
  `\ \ \ \ \ \ f(k)\ text(is divisible by 9,)`

 
`=> f(k + 1)\ text(is divisible by 9)`

`text(S)text(ince true for)\ n = 1,\ text(true for integral)\ n >= 1`

Proof, EXT2 P2 EQ-Bank 5

Use mathematical induction to prove that  `7^n + 15^n`  is divisible by 11 where  `n`  is an odd integer.  (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Prove)\ \ 7^n + 15^n\ \ text(is divisible by 11, where)\ n\ text(is odd)`

`text(If)\ \ n = 1`

`7^n + 15^n` `= 7 + 15 = 22`
`22/11` `= 2\ \ \ :.\ text(Divisible by 11)`

 
`text(Assume true for)\ \ n = k`

`text(i.e.)\ 7^k + 15^k` `= 11P\ \ \ \ text{(}P\ text(integer) text{)}`
`7^k` `= 11P\ – 15^k\ \ \ \ \ …\ text{(∗)}`

 
`text(Prove true for)\ \ n = k + 2`

`7^(k + 2) + 15^(k + 2)` `= 7^2 * 7^k + 15^2 * 15^k`
  `= 49 (11P\ – 15^k) + 225 (15^k)\ \ \ text(… from)\ text{(∗)}`
  `= 49 * 11P\ – 49 * 15^k + 225 (15^k)`
  `= 49 * 11P + 176 * 15^k`
  `= 11 (49P + 16 * 15^k)`

  
`=>text(True for)\ n = k + 2`

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

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`

Binomial, EXT1 2011 HSC 7b

The binomial theorem states that 
 

`(1 + x)^n = sum_(r = 0)^(n) ((n),(r)) x^r`  for all integers  `n >= 1`. 
 

  1. Show that    
    1. `sum_(r=1)^n ((n),(r)) rx^r = nx (1 + x)^(n\ - 1)`.   (2 marks)
    2.  
  2. By differentiating the result from part (i), or otherwise, show that
    1. `sum_(r = 1)^n ((n),(r)) r^2 = n (n + 1) 2^(n\ - 2)`.    (2 marks)
    2.  
  3. Assume now that  `n`  is even. Show that, for  `n >= 4`, 
    1. `((n),(2)) 2^2 + ((n),(4)) 4^2 + ((n),(6)) 6^2 + ... + ((n),(n)) n^2 = n(n+1)2^(n\ - 3)`.  
    2. (3 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
♦ Mean mark 36%. 
MARKER’S COMMENT: Better responses provided a narrative through the solution such as “Differentiate both sides” and “Mult. by `x`”, as seen in the solutions.

(i) `text(Show)\ sum_(r=1)^n ((n),(r)) rx^r = nx (1 + x)^(n\ – 1)`

`text(Using)\ (1 + x)^n = sum_(r=0)^n ((n),(r)) x^r :`

`(1 + x)^n = ((n),(0)) + ((n),(1)) x + ((n),(2)) x^2 + … + ((n),(n)) x^n`
 

`text(Differentiate both sides)`

`n(1 + x)^(n\ – 1) = ((n),(1)) + ((n),(2)) 2x + … + ((n),(n)) nx^(n\ – 1)`
 

`text(Multiply both sides by)\ x`

`nx (1 + x)^(n\ – 1)` `= ((n),(1)) x + ((n),(2)) 2x^2 + … + ((n),(n)) nx^n`
  `= sum_(r=1)^n ((n),(r)) rx^r\ \ text(… as required.)`

 

(ii)  `text(Show)\ \ sum_(r=1)^n ((n),(r)) r^2 = n (n+1) 2^(n\ – 2)`

♦♦ Mean mark 36%.

 
`text(Using part)\ text{(i)}`

`nx (1 + x)^(n\ – 1) = ((n),(1)) x + ((n),(2)) 2x^2 + … + ((n),(n)) nx^n`
 

`text(Differentiate LHS)`

`nx (n\ – 1)(1 + x)^(n\ – 2) + n(1 + x)^(n\ – 1)`

`= n (1 + x)^(n\ – 2) [x (n\ – 1) + (1 + x)]`

`= n (1 + x)^(n\ – 2) (xn\ – x + 1 + x)`

`= n (xn + 1)(1 + x)^(n\ – 2)`
 

`text(Differentiate RHS)`

`((n),(1)) + ((n),(2)) 2^2 x + ((n),(3)) 3^2 x^2 + … + ((n),(n)) n^2 x^(n\ – 1)\ text{… (*)}`

 
`text(Substitute)\ \x = 1\ text(into both sides)`

`((n),(1)) + ((n),(2)) 2^2 + … + ((n),(n)) n^2` `= n (n + 1)(1 + 1)^(n\ – 2)`
`sum_(r=1)^n ((n),(r)) r^2` `= n (n + 1) 2^(n\ – 2)`

 

(iii)  `text(Show for)\ \ n >= 4`

`((n),(2)) 2^2 + ((n),(4)) 4^2 + ((n),(6)) 6^2 + … + ((n),(n)) n^2 = n (n + 1) 2^(n\ – 3)`

♦♦♦ Toughest question in the 2011 exam. Mean mark of 2%. The lowest mean of any Ext1 question since that data became available post-2009!

 
`text{From part(ii) … (*) we know}`

`((n),(1)) + ((n),(2)) 2^2 x + … + ((n),(n)) n^2 x^(n\ – 1) = n (xn + 1)(1 + x)^(n\ – 2)`

 
`text(When)\ \ x = 1`

`((n),(1)) + ((n),(2)) 2^2 + … + ((n),(n)) n^2 = n (n + 1) 2^(n\ – 2)\ \ \ \ \ …\ (1)`

 
`text(When)\ x = –1`

`((n),(1))\ – ((n),(2)) 2^2 + ((n),(3)) 3^2\ – …\ – ((n),(n)) n^2`
 `= n (-n + 1)* 0^(n\ – 2)=0\ \ \ \ \ …\ (2)`
`text{(Note that}\ n≠2\ \ text{because}\ 0^0\ text{is undefined)}`

 
`(1)\ – (2)`

`2[((n),(2)) 2^2 + ((n),(4)) 4^2 + … + ((n),(n)) n^2]` `= n (n + 1) 2^(n\ – 2) – 0`
`((n),(2)) 2^2 + ((n),(4)) 4^2 + … + ((n),(n)) n^2` `= 1/2 n (n + 1) 2^(n\ – 2)`
  `= n (n + 1) 2^(n\ – 3)`

 
`:.\ text(Above statement true for integers)\ n >= 4.`

Binomial, EXT1 2013 HSC 14b

  1. Write down the coefficient of  `x^(2n)`  in the binomial expansion of  `(1 + x)^(4n)`.    (1 mark)
  2. Show that  
    1. `(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ - k)(x + 2)^(2n\ - k)`.   (2 marks)
       
  3. It is known that  
         `x^(2n\ - k) (x + 2)^(2n\ - k) = ((2n\ - k),(0)) 2^(2n\ - k) x^(2n\ - k) + ((2n\ - k),(1)) 2^(2n\ - k\ - 1) x^(2n\ - k + 1)`

    1.  
      1.     `+ ... + ((2n\ - k),(2n\ - k)) 2^0 x^(4n\ - 2k)`.   (Do NOT prove this.)
      2.  
  4. Show that
  5.  
    1. `((4n),(2n)) = sum_(k = 0)^(n) 2^(2n\ - 2k) ((2n),(k))((2n\ - k),(k))`.   (3 marks)

 

 

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

(i)  `text(Find co-efficient of)\ \ x^(2n)`

`text(Expanding)\ \ (1+x)^(4n)`

`((4n),(0)) + ((4n),(1))x + ((4n),(2))x^2 + … + ((4n),(2n))x^(2n) + …`

`:.\ text(Co-efficient of)\ \ x^(2n)\ text(is)\ ((4n),(2n))`

 

(ii)  `text(Show)\ (1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)`

♦♦ Mean mark 30%.

`text(Using)\ (1 + x^2 + 2x)^(2n) = [x(x + 2) + 1]^(2n)`

`[x (x + 2) + 1]^(2n)`
`= ((2n),(0)) (x(x + 2))^(2n) + ((2n),(1)) (x(x + 2))^(2n\ – 1) + … + ((2n),(2n))` 
`= ((2n),(0)) x^(2n)(x + 2)^(2n) + ((2n),(1)) x^(2n\ – 1) (x + 2)^(2n\ – 1) + … + ((2n),(2n))` 
`= sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)\ text(… as required.)`

 

(iii)  `((4n),(2n))\ text(is the co-eff of)\ x^(2n)\ text(in expansion)\ (1+x)^(4x)`

`text(S)text(ince)\ (1 + x^2 + 2x)^(2n) = ((x+1)^2)^(2n) = (1 + x)^(4n)`

`=> ((4n),(2n))\ text(is co-efficient of)\ x^(2n)\ text(in expansion)\ (1 + x^2 + 2x)^(2n)`

 
`text(Using part)\ text{(ii)}`

`(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k))\ x^(2n\ – k) (x + 2)^(2n\ – k)`

 

`text(Using the given identity,)\ x^(2n)\ text(co-efficients are)`

♦♦♦ Toughest question in the 2013 exam with Mean mark 12%!
MARKER’S COMMENT: Simply stating `(1+x^2+2x)^(2n)“=(1+x)^(4n)` received 1 full mark, showing that even initial working can gain marks.
`k = 0,` `\ ((2n),(0))((2n\ – 0),(0)) 2^(2n\ – 0)`
`k = 1,` `\ ((2n),(1))((2n\ – 1),(1)) 2^(2n\ – 1\ – 1)`
`vdots`  
`k = n,` `\ ((2n),(n))((2n\ – n),(n)) 2^(2n\ – n\ – n)`

 

`:.\ ((4n),(2n))`
 `= ((2n),(0))((2n),(0))2^(2n) + ((2n),(1))((2n\ – 1),(1))2^(2n\ – 2) + … + ((2n),(n))((n),(n)) 2^0`
` = sum_(k=0)^(n)\ ((2n),(k))((2n\ – k),(k)) 2^(2n\ – 2k)\ \ \ \ text(… as required)`

Plane Geometry, EXT1 2013 HSC 13d

The circles  `C_1`  and  `C_2`  touch at the point  `T`. The points  `A`  and  `P`  are on  `C_1`. The line  `AT`  intersects  `C_2`  at  `B`. The point  `Q`  on  `C_2`  is chosen so that  `BQ`  is parallel to  `PA`.

Copy or trace the diagram into your writing booklet.

Prove that the points  `Q`,  `T`  and  `P`  are collinear.   (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Need to prove)\ Q,\ T,\ P\ text(are collinear)`

♦♦ Mean mark 31%
COMMENT: Proving or using collinearity is often examined in this topic area.

`CD\ text(is a common tangent)`

`text(Let)\ /_CTQ` `= alpha`
`/_ QBT` `= alpha\ \ text{(angle in alternate segment)}`
`text(Let)\ \ ` `/_DTP` `= beta`
  `/_PAT` `= beta\ \ text{(angle in alternate segment)}`

 

`text(S)text(ince)\ /_PAT = /_QBT\ text{(} text(alternate,)\ BQ\  text(||) \ AP text{)}`

`=> alpha = beta`

`:.\ /_QTC = /_DTP`

`QP\ text(must be a straight line since)`

`/_QTC\ text(and)\ /_DTP\ text(are vertically opposite and equal)`

`:.\ P,\ T,\ Q\ text(are collinear   … as required)`

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.`

Calculus, EXT1 C1 2013 HSC 13a

A spherical raindrop of radius `r` metres loses water through evaporation at a rate that depends on its surface area.  The rate of change of the volume `V` of the raindrop is given by

`(dV)/(dt) = -10^(-4) A`, 

where `t` is time in seconds and `A` is the surface area of the raindrop. The surface area and the volume of the raindrop are given by  `A = 4pir^2`  and  `V = 4/3 pi r^3`  respectively.

  1. Show that  `(dr)/(dt)`  is constant.   (1 mark)

  2. How long does it take for a raindrop of volume  `10^(–6)` m3  to completely evaporate?     (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `62\ text(seconds)`
Show Worked Solution
i.    `text(Show)\ \ (dr)/(dt)\ \ text(is a constant)`

 `(dV)/(dt) = (dV)/(dr) * (dr)/(dt)\ \ \ …\ text{(1)}`

`V` `= 4/3 pi r^3`
`:. (dV)/(dr)` `= 4 pi r^2`
`(dV)/(dt)` `= -10^(-4) A\ \ text{(given)}`

 
`text(Substituting into)\ text{(1)}`

`-10^(-4) A` `= 4 pi r^2 xx (dr)/(dt)`
  `= A xx (dr)/(dt)`
`:.\ (dr)/(dt)` `= -10^(-4)\ \ text(… as required)`

 

ii.    `V` `= 10^(-6)\ text(m³)`
  `4/3 pi r^3` `= 10^(-6)`
  `r^3` `= (3 xx 10^(-6))/(4pi)`
  `r` `= root(3)((3 xx 10^(-6))/(4pi))`

 

`text(S)text(ince the radius decreases at a constant rate,)`

♦♦ Mean mark 31%

`t=(root(3)((3 xx 10^(-6))/(4pi)))/(10^(-4))`

`\ \ =62.035 …`

`\ \ =62\ text(seconds)\ text{(nearest whole)}`

 

`:.\ text(It takes 62 seconds for the raindrop)`

`text(to evaporate.)`

Geometry and Calculus, EXT1 2013 HSC 12d

The point  `P(t, t^2 + 3)`  lies on the curve  `y = x^2 + 3`. The line  `l`  has equation  `y = 2x\ – 1`. The perpendicular distance from  `P`  to the line  `l`  is  `D(t)`.

2013 12d

  1. Show that  
    1. `D(t) = (t^2\ - 2t + 4)/sqrt5`.   (2 marks)
  3. Find the value of  `t`  when  `P`  is closest to  `l`.     (1 mark)
  4. Show that, when  `P`  is closest to  `l`, the tangent to the curve at  `P`  is parallel to  `l`.   (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `t=1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)    `text(Show)\ \ D(t) = (t^2\ – 2t + 4)/sqrt5`
`D(t)` `= _|_\ text(dist of)\ (t, t^2 + 3)\ text(from)\ 2x\ – y\ – 1 = 0`
`D` `= |(ax_1 + by_1 + c)/sqrt(a^2 + b^2)|`
  `= |(2(t)\ – 1 (t^2 + 3)\ – 1)/sqrt(2^2 + (-1)^2)|`
  `= |(2t\ – t^2\ – 3\ – 1)/sqrt5|`
  `= |(-(t^2\ – 2t + 4))/sqrt5|`
Mean mark 40%
MARKER’S COMMENT: The biggest difficulty in part (i) was to explain the removal of the absolute value sign.

 

`text(S)text(ince)\ \ t^2\ – 2t + 4\ \ text(is positive definite)`

`text{(i.e.} \ Delta < 0\ text(and co-efficient of)\ t^2 > 0 text{)},`

`t^2\ – 2t + 4\ \ text(is positive for all)\ t.`

`:.\ D(t) = (t^2\ – 2t + 4)/sqrt5\ \ text(… as required)`

 

(ii)    `text(MAX or MIN when)\ \ (dD)/(dt) = 0`
`(dD)/(dt) = 1/sqrt5 (2t\ – 2)` `= 0`
`2t\ – 2` `= 0`
`t` `= 1`

`(d^2D)/(dt^2) = 2/sqrt5 > 0\  =>\  text(MIN)`

`:.\ P\ text(is closest to)\ l\ text(when)\ t = 1`

 

(iii)   `y` `= x^2 + 3`
  `dy/dx` `= 2x`
`text(When)\ x = t`

`dy/dx = 2t`

`text(At)\ t = 1`

`dy/dx = 2,`

`=> m_text(tangent) = 2\ \ and\ \ m_l = 2`

 

`:.\ text(T)text(angent is parallel to)\ \ l\ \ \ \ text(… as required)`

Functions, EXT1 F1 2013 HSC 10 MC

Which inequality has the same solution as  `|\ x + 2\ | + |\ x- 3\ | = 5`?

  1. `5/(3 - x) >= 1`
  2. `1/(x - 3)\ - 1/(x + 2) <= 0`
  3. `x^2 - x - 6 <= 0`
  4. `|\ 2x - 1\ | >= 5`
Show Answers Only

`C`

Show Worked Solution
♦♦ Mean mark 39%
COMMENT: Note that the quick elimination of `A, B` and `D` is sufficient to get to the correct answer without proving `C` (although we have done this in the solution).

`text(In)\ A\ text(and)\ B, \ x ≠ 3\ text(but when)\ x=3,`

`|\ 3 + 2\ | + |\ 3 – 3\ | = 5\ \ text(is correct.)`

`:.\ text(Not)\ \ A\ \ text(or)\ \ B.`

 

`text(Consider)\ D`

`x -> oo\ text(satisfies)\ |\ 2x – 1\ | >= 5,\ \ text(but)`

`text(obviously not)\ |\ x + 2\ | + |\ x – 3\ | = 5.`

 
`text(Consider)\ C`

`x^2 – x – 6` `<= 0`
`(x – 3)(x + 2)` `<= 0`

 

`text(True when)\ \ \ -2 <= x <= 3.`

 
`text(In this range,)`

`(x + 2) >= 0\ \ text(and)\ \ (x – 3)<= 0`

`:.\ text(We can write)`

`|\ x + 2\ | + |\ x – 3\ |` `= (x + 2)\ – (x – 3)`
  `= x + 2 – x + 3`
  `= 5`

 
`:. C\ text(has the same solution)`

`=>  C\ text(is correct.)`

Trigonometry, EXT1 T2 2013 HSC 8 MC

The angle  `theta`  satisfies  `sin theta = 5/13`  and  `pi/2 < theta < pi`.

What is the value of  `sin 2 theta`?

  1. `10/13`
  2. `- 10/13` 
  3. `120/169`
  4. `- 120/169`
Show Answers Only

`D`

Show Worked Solution
 
♦ Mean mark 44% 

`sin theta = 5/13\ \ \ (pi/2 < theta < pi)`

`text(S)text(ince)\ \ theta\ \ text(is in the 2nd quadrant,)`

`cos theta = -12/13`

`:.sin 2theta` `= 2 sin theta cos theta`
  `= 2 xx 5/13 xx -12/13`
  `= -120/169`

`=>  D`

Trig Ratios, EXT1 2013 HSC 6 MC

Let  `|\ a\ | <= 1`. What is the general solution of  `sin 2x = a`?

  1. `x = n pi + (-1)^n (sin^(-1) a)/2,\ n\ text(is an integer)`
  2. `x = (n pi + (-1)^n sin^(-1) a)/2,\ n\ text(is an integer)`
  3. `x = 2n pi +- (sin^(-1) a)/2,\ n\ text(is an integer)`
  4. `x = (2n pi +- sin^(-1) a)/2,\ n\ text(is an integer)`
Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 42%

`text(Find general solution:)`

`sin 2x = a,\ \ \ |\ a\ | <= 1`

`2x` `= sin^(-1)a,\ pi\ – sin^(-1)a,\ 2pi + sin^(-1)a,\ 3pi\ – sin^(-1) a,\ …`
  `= npi + (-1)^n sin^(-1)a`
`x` `= (npi + (-1)^n sin^(-1)a)/2`

`=>  B`

Plane Geometry, EXT1 2013 HSC 3 MC

The points  `A`,  `B`  and  `C`  lie on a circle with centre  `O`, as shown in the diagram.

The size of  `/_AOC`  is  `(3pi)/5`  radians. 

What is the size of  `/_ABC`  in radians? 

  1. `(3pi)/10`
  2. `(2pi)/5`
  3. `(7pi)/10`
  4. `(4pi)/5`
Show Answers Only

`C`

Show Worked Solution
♦ Mean mark 49%

`text(Reflex)\ /_AOC` `= 2pi\ – (3pi)/5`
  `= (7pi)/5`

`:. /_ABC = 1/2 xx (7pi)/5 = (7pi)/10`

`text{(angle at circumference and}`

`text{centre on arc}\ AC text{)}`

`=>  C`

Binomial, EXT1 2010 HSC 7b

The binomial theorem states that 
 

`(1 + x)^n = ((n),(0)) + ((n),(1))x + ((n),(2))x^2 + ((n),(3))x^3 + ... + ((n),(n))x^n.` 
  

  1. Show that  
    1. `2^n = sum_(k = 0)^n ((n),(k))`.   (1 mark)
  2.  
  3. Hence, or otherwise, find the value of
     

    1. `((100),(0)) + ((100),(1)) + ((100),(2)) + ... + ((100),(100))`.   (1 mark)
    2.  
  4. Show that
    1.  `n2^(n\ - 1) = sum_(k = 1)^n k ((n),(k))`.   (2 marks)
    2.  
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2^100`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)   `(1 + x)^n = ((n),(0)) + ((n),(1))x + … + ((n),(n))x^n`

`text(Let)\ \ x = 1`

`(1 + 1)^n` `= ((n),(0)) + ((n),(1))1 + ((n),(2))1^2 + … + ((n),(n))1^n`
`2^n` `= ((n),(0)) + ((n),(1)) + … + ((n),(n))`
  `= sum_(k=0)^n ((n),(k))\ text(… as required)`

 

(ii)   `((100),(0)) + ((100),(1)) + ((100),(2)) + … + ((100),(100)) = 2^100`

`text{(} text(from part)\ text{(i)} text{)}`

♦ Mean mark 37%
 

(iii)  `(1 + x)^n = ((n),(0)) + ((n),(1))x + ((n),(2))x^2 + … + ((n),(n))x^n`
 

`text(Differentiate both sides)`

`n(1 + x)^(n\ – 1) = 1((n),(1)) + 2x ((n),(2)) + 3x^2 ((n),(3)) + … + nx^(n\ – 1) ((n),(n))`
 

`text(Let)\ \ x = 1`

`n (1 + 1)^(n\ – 1)` `= 1 ((n),(1)) + 2 ((n),(2)) + 3 ((n),(3)) + … + n ((n),(n))`
`n 2^(n\ – 1)` `= sum_(k = 1)^n k ((n),(k))\ text(… as required)`

Proof, EXT1 P1 2010 HSC 7a

Prove by induction that 

`47^n + 53 xx 147^(n-1)`

is divisible by  `100`  for all integers  `n >= 1`.    (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Prove)\ \ 47^n + 53 xx 147^(n – 1)\ \ text(is divisible)`

`text(by 100 for)\ n >= 1.`

`text(If)\ n = 1`

`47^1 + 53 xx 147^(1 – 1)` `= 47 + 53`
  `= 100\ \ text{(divisible by 100)}`

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

♦ Mean mark 51%.
IMPORTANT: Making `47^k` the subject helps the substitution in the proof for `n+1`. Students must state that `P` is an integer. NOTE: using `147^k=147xx147^(k-1)` develops proof for `n=k+1`.

 
`text(Assume true for)\ \ n = k`

`text(i.e.)\ 47^k + 53 xx 147^(k – 1) = 100P\ \ \ text{(} P\ text(integer) text{)}`

`47^k = 100P\ – (53 xx 147^(k – 1))\ \ \ text(…)\ text{(1)}`

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

`47^(k + 1) + 53 xx 147^(k)`
`= 47 xx 47^k + 53 xx 147^k`
`text{(substitute from (1) above)}`
`= 47 [100P\ – (53 xx 147^(k – 1))] + 53 xx 147 xx 147^(k – 1)`
`= 4700P\ – 2491 xx 147^(k – 1) + 7791 xx 147^(k – 1)`
`= 4700P + 5300 xx 147^(k – 1)`
`= 100 (47P + 53 xx 147^(k – 1))`

 
`=>\ text(True for)\ n = k + 1`

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

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.`

Calculus, EXT1 C2 2010 HSC 5b

Let  `f(x) = tan^(-1)(x) + tan^(-1)(1/x)`  for  `x != 0`. 

  1. By differentiating  `f(x)`, or otherwise, show that  `f(x) = pi/2`  for  `x > 0`.   (3 marks)

  2. Given that  `f(x)`  is an odd function, sketch the graph  `y = f(x)`.     (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. Inverse Functions, EXT1 2010 HSC 5b Answer

 

 

 

 

 

 

Show Worked Solution
♦ Mean mark 48%
MARKER’S COMMENT: The most common errors were not being able to differentiate `tan^(-1) (1/x)` and not recognising the significance of `f′(x)=0`.
i.    `f(x)` `= tan^(-1) (x) + tan^(-1) (1/x)\ text(for)\ x != 0`
  `f prime (x)` `= 1/(1 + x^2) + 1/(1 + (1/x)^2) xx d/(dx) (1/x)`
    `= 1/(1 + x^2) + 1/(1 + 1/(x^2)) xx -1/(x^2)`
    `= 1/(1 + x^2)\ – 1/(x^2 + 1)`
    `= 0`

 

`text(S)text(ince)\ \ f prime (x) = 0`

`=> f(x)\ text(is a constant)`

 

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

`f(1)` `= tan^(-1) 1 + tan^(-1) (1/1)`
  `= pi/4 + pi/4`
  `= pi/2`

 

`:.\ f(x) = pi/2\ \ text(for)\ \ x > 0\ \ \ …\ text(as required)`

 

♦ Mean mark 35%
ii.    `text(Given)\ \ f(x)\ \ text(is odd)`
  `f(–x) = -f(x)`

Inverse Functions, EXT1 2010 HSC 5b Answer

Trigonometry, EXT1 T3 2010 HSC 4b

  1. Express  `2 cos theta + 2 cos (theta + pi/3)`  in the form  `R cos (theta + alpha)`,
     
    where  `R > 0`  and  `0 < alpha < pi/2`.    (3 marks)

  2. Hence, or otherwise, solve  `2 cos theta + 2 cos (theta + pi/3) = 3`,   
  3. for  `0 < theta < 2pi`.    (2 marks)

Show Answers Only
  1. `2 sqrt 3 cos (theta + pi/6)`
  2. `(5pi)/3`
Show Worked Solution
i.    `2 cos theta + 2 cos (theta + pi/3)`
  `= 2 cos theta + 2 (cos theta cos (pi/3)\ – sin theta sin (pi/3))`
  `= 2 cos theta + 2 cos theta xx 1/2\ – 2 sin theta xx sqrt3/2`
  `= 2 cos theta + cos theta\ – sqrt3 sin theta`
  `= 3 cos theta\ – sqrt3 sin theta`

 
`R cos (theta + alpha) = R cos theta cos alpha – R sin theta sin alpha`

`R cos alpha` `= 3\ \ \ \ \ ` `R sin alpha` `= sqrt3`
`cos alpha` `= 3/R\ \ \ \ \ ` `sin alpha` `= sqrt3/R`

 

`tan alpha` `= sin alpha/cos alpha = sqrt3/3 = 1/sqrt3`
`tan\ pi/6` `=1/sqrt3`
`:. alpha` `=pi/6\ \ \ \ \ (0 < alpha < pi/2)`
`R^2` `= 3^2 + (sqrt3)^2`
  `= 9 + 3`
`R` `= sqrt 12 = 2 sqrt3`

 
`:.\ 2 cos theta + 2 cos (theta + pi/3) = 2 sqrt 3 cos (theta + pi/6)`

 

♦ Mean mark part (ii) 49%
MARKER’S COMMENT: Many students did not check their answers against the stated domain for `theta`.
ii.    `2 cos theta + 2 cos(theta + pi/3)` `= 3`
  `2 sqrt 3 cos (theta + pi/6)` `= 3`
  `cos (theta + pi/6)` `= 3/(2sqrt3) = sqrt3/2`
  `cos^(-1) (sqrt3/2)` `= pi/6`

 
`text(S)text(ince cos is positive in 1st and 4th quadrants,)`

`theta + pi/6` `= pi/6,\ 2pi\ – pi/6`
`:. theta` `= (5pi)/3\ \ \ \ \ (0 < theta < 2pi)`

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.)}`

Combinatorics, EXT1 A1 2010 HSC 3a

At the front of a building there are five garage doors. Two of the doors are to be painted red, one is to be painted green, one blue and one orange. 

  1. How many possible arrangements are there for the colours on the doors?   (1 mark)

  2. How many possible arrangements are there for the colours on the doors if the two red doors are next to each other?    (1 mark)

Show Answers Only
  1. `60`
  2. `24`
Show Worked Solution
i.    `text(# Arrangements)` `= (5!)/(2!)`
    `= 60`

 

♦ Mean mark 50%
MARKER’S COMMENT: Drawing a diagram was a successful strategy for many students in this part.
ii.    `text(When 2 red doors are side-by-side,)`
`text(# Arrangements)` `= 4!`
  `= 24`

Calculus, EXT1 C1 2011 HSC 7a

The diagram shows two identical circular cones with a common vertical axis.  Each cone has height `h` cm and semi-vertical angle 45°.
 

2011 7a

The lower cone is completely filled with water. The upper cone is lowered vertically into the water as shown in the diagram. The rate at which it is lowered  is given by  

`(dl)/(dt) = 10`,

where `l` cm is the distance the upper cone has descended into the water after `t` seconds.

As the upper cone is lowered, water spills from the lower cone. The volume of water remaining in the lower cone at time `t` is  `V` cm³.

  1. Show that  `V = pi/3(h^3\ - l^3)`.   (1 mark)

  2. Find the rate at which `V` is changing with respect to time when  `l = 2`.     (2 marks)

  3. Find the rate at which `V` is changing with respect to time when the lower cone has lost  `1/8`  of its water. Give your answer in terms of `h`.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `-40 pi\ \ text(cm³)// text(sec)`
  3. `(-5pih^2)/2\ text(cm³)// text(sec)`
Show Worked Solution

i.   `text(Show that)\ V = pi/3 (h^3\ – l^3)`

 ♦ Mean mark 42% 

`text(S)text(ince)\ \ tan45° = r/h=1`

`=>r=h`

`=>\ text(Radius of lower cone) = h`

`:.\ V text{(lower cone)}` `= 1/3 pi r^2 h`
  `= 1/3 pi h^3`

 
 `text(Similarly,)`

`V text{(submerged upper cone)} = 1/3 pi l^3`

`V text{(water left)}` `= 1/3 pi h^3\ – 1/3 pi l^3`
  `= pi/3 (h^3\ – l^3)\ \ \ text(… as required)`

 

ii.  `text(Find)\ (dV)/(dt)\ text(at)\ l = 2`

`(dV)/(dt)= (dV)/(dl) xx (dl)/(dt)\ …\ text{(1)}`

`=>(dl)/(dt)` `= 10\ text{(given)}`
`text(Using)\ \ V` `= pi/3 (h^3\ – l^3)\ \ \ \ text(from part)\ text{(i)}`
`=>(dV)/(dl)` `= -3 xx pi/3 l^2`
  `= -pi l^2`

  
`text(At)\ \ l = 2,`

`text{Substitute into (1) above}`

`(dV)/(dt)` `= -pi xx 2^2 xx 10`
  `= -40 pi\ \ \ text(cm³)//text (sec)`

 

iii.  `text(Find)\ \ (dV)/(dt)\ \ text(when lower cone has lost)\ 1/8 :`

♦♦♦ Mean mark 12%
MARKER’S COMMENT: Many unsuccessful answers attempted to find an alternate version of `(dV)/(dt)`. Part (ii) directed students directly toward the correct strategy.

`text(Find)\ \ l\ \ text(when)\ \ V = 7/8 xx 1/3 pi h^3`

`pi/3 (h^3\ – l^3)` `= 7/8 xx 1/3 pi h^3`
`h^3 -l^3` `= 7/8 h^3`
`l^3` `= 1/8 h^3`
`l` `= h/2`

   
`text(When)\ \ l = h/2 ,`

`(dV)/(dt)` `= -pi (h/2)^2 xx 10\ \ …\ text{(*)}`
  `= (-5pih^2)/2\ text(cm³)// text(sec)`

 
`:. V\ text(is decreasing at the rate of)\ \  (5 pi h^2)/2\ text(cm³)//text(sec).`

Statistics, EXT1 S1 2011 HSC 6c

A game is played by throwing darts at a target. A player can choose to throw two or three darts.

Darcy plays two games. In Game 1, he chooses to throw two darts, and wins if he hits the target at least once. In Game 2, he chooses to throw three darts, and wins if he hits the target at least twice.

The probability that Darcy hits the target on any throw is  `p`, where  `0 < p < 1`.

  1. Show that the probability that Darcy wins Game 1 is  `2p- p^2`.    (1 mark)

  2. Show that the probability that Darcy wins Game 2 is  `3p^2- 2p^3`.     (1 mark)

  3. Prove that Darcy is more likely to win Game 1 than Game 2.    (2 marks)

  4. Find the value of  `p`  for which Darcy is twice as likely to win Game 1 as he is to win Game 2.    (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `p = (7 – sqrt17)/8`
Show Worked Solution

i.  `text(Show)\ P text{(wins Game 1)} = 2p\ – p^2`

♦ Mean mark 47%.

`P text{(Hits)} = P text{(H)} = p`

`P text{(Miss)} = P text{(M)} = 1 – p`
 

`P text{(Wins G1)}` `= 1 – P text{(MM)}`
  `= 1 – (1 – p)^2`
  `= 1 – 1 + 2p – p^2`
  `= 2p – p^2\ \ \ text(… as required)`

 

ii.  `text(Show)\ \ P text{(wins Game 2)} = 3p – 2p^3 :`

♦ Mean mark 40%.

`text(Darcy wins G2 if he hits 3 times, or twice.)`

`Ptext{(Hits 3 times)}=\ ^3C_3 (p)^3=p^3`

`Ptext{(Hits twice)}=\ ^3C_2 (p)^2 (1-p)=3p^2-3p^3`

 `P text{(Wins G2)}` `= p^3 + 3p^2-3p^3`
  `= 3p^2 – 2p^3\ \ \ text(… as required)`

 

♦♦♦ Mean mark 17%.
COMMENT: It is critical for students to utilise the limits of  `0<p<1`  to answer part (iii).

iii.  `text(Prove more likely to win G1 vs G2)`

`text(i.e.  Show)\ \ \ ` `2p – p^2 > 3p^2 – 2p^3`
  `2p^3 – 4p^2 + 2p` `> 0`
  `2p (p^2 – 2p + 1)` `> 0` 
  `2p (p – 1)^2` `> 0` 

 
`=> text(TRUE since)\ \ (p-1)^2>0\ \ text(and)\ \ 0 < p < 1`

`:.\ text(More likely for Darcy to win Game 1.)`
 

iv.  `text(If twice as likely to win G1 vs G2)`

♦♦ Mean mark 21%.
MARKER’S COMMENT: BE CAREFUL in formulating your equation here. Many students multiplied the wrong side by 2.
`2p\ – p^2` `= 2(3p^2 – 2p^3)`
  `= 6p^2 – 4p^3`
`4p^3 – 7p^2 + 2p` `= 0`
`p(4p^2 – 7p + 2)` `= 0`

  

`p` `= (–(–7) +- sqrt((–7)^2\ – 4 xx 4 xx 2))/(2 xx 4)`
  `= (7 +- sqrt(49\ – 32))/8`
  `= (7 +- sqrt17)/8`

 
`text(S)text(ince)\ \ 0<p<1,`

`p = (7\ – sqrt17)/8`

Trig Ratios, EXT1 2011 HSC 5a

In the diagram,  `Q(x_0, y_0)`  is a point on the unit circle  `x^2 + y^2 = 1`  at an angle  `theta`  from the positive  `x`-axis, where  `− pi/2 < theta < pi/2`. The line through  `N(0, 1)`  and  `Q`  intersects the line  `y = –1`  at  `P`. The points  `T(0, y_0)`  and  `S(0, –1)` are on the  `y`-axis.
 

 
 

  1. Use the fact that  `Delta TQN`  and  `Delta SPN`  are similar to show that
     
  2.      `SP = (2costheta)/(1\ - sin theta)`.  (2 marks)
  3.  
  4. Show that  `(costheta)/(1\ - sin theta) = sec theta + tan theta`.    (1 mark)
  5.  
  6. Show that  `/_ SNP = theta/2 + pi/4`.   (1 mark)
  7.  
  8. Hence, or otherwise, show that  `sectheta + tantheta = tan(theta/2 + pi/4)`.   (1 mark)
  9.  
  10. Hence, or otherwise, solve  `sec theta + tan theta = sqrt3`, where  `-pi/2 < theta < pi/2`.   (2 marks)

 

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

 

♦ Mean mark 41%
MARKER’S COMMENT: When questions direct you to use a certain fact for a proof, use that fact!

`text(Show)\ SP = (2 cos theta)/(1\ – sin theta)`

`Delta SPN \ text(|||) \ Delta TQN\ \ \ text{(given)}`

`(SP)/(SN) = (TQ)/(TN)\ \ \ text{(corresponding sides of similar triangles)}`

`/_TQO = theta\ \ \ text{(alternate,}\ TQ\ text(||)\  x text{-axis)}`

  `sin theta` `= (OT)/1 => OT = sin theta`
`=>` `TN` `= 1\ – sin theta`
  `cos theta` `= (TQ)/1`
`=>` `TQ` `= cos theta`
`SN` `= 2\ \ \ text{(diameter of unit circle)}`
`:. (SP)/2` `= cos theta/(1\ – sin theta)`
`SP` `= (2 cos theta)/(1\ – sin theta)\ \ \ text(… as required)`

 

(ii)   `text(Show)\ \ costheta/(1\ – sin theta) = sec theta + tan theta`

♦ Mean mark 35%
`text(RHS)` `= 1/(costheta) + (sintheta)/(costheta)`
  `=(1 + sin theta)/cos theta xx cos theta/cos theta`
  `= (costheta(1 + sintheta))/(cos^2theta)`
  `= (costheta(1 + sin theta))/(1\ – sin^2 theta)`
  `= (costheta(1 + sin theta))/((1 + sin theta)(1\ – sin theta)`
  `= costheta/(1\ – sin theta)\ \ \ text(… as required)`

 

(iii)  `text(Show)\ \ /_SNP = theta/2 + pi/4`

♦♦♦ Mean mark 11%

`/_TOQ = 90\ – theta`

`text(S)text(ince)\ ON = OQ = 1\ text{(unit circle)}`

`=> Delta ONQ\ text(is isosceles)`

`:.\ /_SNP` `= 1/2 (180\ – (90\ – theta))\ \ \ \ text{(angle sum of}\ Delta ONQ text{)}`
  `= 90\ – 45 + theta/2`
  `= 45 + theta/2`
  `= pi/4 + theta/2\ \ \ text(… as required)`

 

(iv)  `text(Show)\ sec theta + tan theta = tan (theta/2 + pi/4)`

♦♦ Mean mark 28%

`text(S)text(ince)\ /_SNP = pi/4 + theta/2\ \ \ \ text{(} text(part)\ text{(iii)} text{)}`

`=> tan\  /_SNP = tan (pi/4 + theta/2)`

 

`text(Also,)\ tan\   /_SNP` `= (SP)/(SN)`
  `= ((2costheta)/(1\ – sin theta))/2`
  `= (cos theta)/(1\ – sin theta)`
  `= sec theta + tan theta\ \ \ \ text{(part (ii))}`

 

`:.\ sec theta + tan theta = tan (pi/4 + theta/2)\ \ \ text(… as required)`

 

♦ Mean mark 45%
(v)    `sec theta + tan theta` `= sqrt3,\ \ \ (-pi/2 < theta < pi/2)`
  `tan (pi/4 + theta/2)` `= sqrt3\ \ \ \ text{(part (iv))}`
  `tan (pi/3)` `= sqrt 3`

`text(S)text(ince tan is positive in)\  1^text(st) // 3^text(rd)\ text(quads)`

`theta/2 + pi/4` `= pi/3,\ (4pi)/3`
`theta/2` `= pi/12\ \ \ \ (-pi/2 < theta < pi/2)`
`:.\ theta` `= pi/6\ text(radians)`