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Quadratic, EXT1 2017 HSC 14b

Let  `P(2p, p^2)`  be a point on the parabola  `x^2 = 4y`.

The tangent to the parabola at `P` meets the parabola  `x^2 = −4ay`, `a > 0`, at `Q` and `R`. Let `M` be the midpoint of `QR`.

  1. Show that the `x` coordinates of `R` and `Q` satisfy
  2. `qquadx^2 + 4apx - 4ap^2 = 0`.  (2 marks)

  3. Show that the coordination of `M` are  `(−2ap, −p^2(2a + 1))`.  (2 marks)
  4. Find the value of `a` so that the point `M` always lies on the parabola  `x^2 = −4y`.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `1 + sqrt2, a > 0`
Show Worked Solution

(i)   `x^2 = 4y,\ \ =>y = (x^2)/4`

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

`text(At)\ P(2p, p^2),`

`(dy)/(dx) = p`

`text(Equation of tangent:)`

`y = px – p^2`

 

`R and Q\ text(at intersection)`

`y` `= px – p^2\ …\ (1)`
`y` `= −(x^2)/(4a)\ …\ (2)`

 

`text(Subtract)\ (1) – (2)`

`px – p^2 + (x^2)/(4a)` `= 0`
`x^2 + 4apx – 4ap^2` `= 0\ …\ text(as required)`

 

(ii)   `x^2 + 4apx – 4ap^2 = 0`

♦♦ Mean mark 30%.
`x=` `\ (−4ap ± sqrt((4ap)^2 – 4 · 1 · (−4ap^2)))/2`
`=`  `\ (−4ap ± sqrt(16ap^2(a + 1)))/2`
`=`  `\ −2ap ± 2psqrt(a(a + 1))`

 

`=> xtext(-coordinate of)\ M\ text(is)\ −2ap.`

 

`M\ text(lies on tangent)\ \ y = px – p^2`

`:.y` `= p(−2ap) – p^2`
  `= −2ap^2 – p^2`
  `= −p^2(2a + 1)`

`:. M\ text(has coordinates)\ \ (−2ap, −p^2(2a + 1))`

 

(iii)   `text(If)\ M\ text(always lies on)\ \ x^2 = −4y`

♦♦ Mean mark 23%.
`(−2ap)^2` `= −4(−p^2(2a + 1))`
`4a^2p^2` `= 4p^2(2a + 1)`
`a^2` `= 2a + 1`
`a^2 – 2a – 1` `= 0`
`:. a` `= (+2 ± sqrt(4 + 4 · 1· 1))/2`
  `= 1 ± sqrt2`
  `= 1 + sqrt2, \ a > 0`

Proof, EXT1 P1 2017 HSC 14a

Prove by mathematical induction that  `8^(2n + 1) + 6^(2n − 1)`  is divisible by 7, for any integer  `n ≥ 1`.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Prove)\ \ 8^(2n + 1) + 6^(2n − 1)\ \ text(is divisible)`

`text(by 7 for integers)\ \ n >= 1`

`text(If)\ n = 1,`

`8^3 + 6^1 = 518 = 74 xx 7`

`:. text(True for)\ n = 1`
 

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

`text(i.e.)\ \ 8^(2k + 1) + 6^(2k – 1)` `= 7P\ \ (text(where)\ P\ text(is an integer))`
`8^(2k + 1)` `= 7P – 6^(2k – 1)`

 
`text(Prove true for)\ n = k + 1`

`8^(2k + 3) + 6^(2k − 1)` `= 64 · 8^(2k + 1) + 36*6^(2k – 1)`
  `= 64(7P – 6^(2k – 1)) + 36 · 6^(2k – 1)`
  `= 64 · 7P – 64 · 6^(2k – 1) + 36 · 6^(2k – 1)`
  `= 64 · 7P – 28 · 6^(2k – 1)`
  `= 7(64P – 4 · 6^(2k – 1))`

 
`…\ text(which is divisible by 7.)`
 

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

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

Mechanics, EXT2* M1 2017 HSC 13c

A golfer hits a golf ball with initial speed `V\ text(ms)^(−1)` at an angle `theta` to the horizontal. The golf ball is hit from one side of a lake and must have a horizontal range of 100 m or more to avoid landing in the lake.
 

     

Neglecting the effects of air resistance, the equations describing the motion of the ball are

`x = Vt costheta`

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

where `t` is the time in seconds after the ball is hit and `g` is the acceleration due to gravity in `text(ms)^(−2)`. Do NOT prove these equations.

  1. Show that the horizontal range of the golf ball is
     
         `(V^2sin 2theta)/g` metres.  (2 marks)

  2. Show that if  `V^2 < 100 g`  then the horizontal range of the ball is less than 100 m.  (1 mark)

It is now given that  `V^2 = 200 g`  and that the horizontal range of the ball is 100 m or more.

  1. Show that  `pi/12 <= theta <= (5pi)/12`.  (2 marks)

  2. Find the greatest height the ball can achieve.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `text(See Worked Solution)`
  4. `25(2 – sqrt 3)\ text(metres)`
Show Worked Solution

i.   `text(Find)\ \ t\ \ text(when)\ \ y = 0:`

`1/2 g t^2` `= Vtsintheta`
`1/2 g t` `= Vsintheta`
`t` `= (2Vsintheta)/g`

 

`text(Horizontal range)\ (x)\ text(when)\ \ t = (2Vsintheta)/g :`

`x` `= V · (2Vsintheta)/g costheta`
  `= (V^2 2sintheta costheta)/g`
  `= (V^2sin2theta)/g\ … text(as required)`

 

ii.   `text(If)\ \ V^2 < 100 g`

♦ Mean mark 44%.
`x` `< (100 g sin2theta)/g`
`x` `< 100 sin2theta`

 

`text(S)text(ince)\ −1 <= 2theta <= 1,`

`x < 100\ text(metres)`

 

iii.   `V^2 = 200g,\ \ x >= 100`

`(200 g · sin2theta)/g` `>= 100`
`sin2theta` `>= 1/2`

`:. pi/6 <= 2theta <= (5pi)/6`

`:. pi/12 <= theta <= (5pi)/12\ …\ text(as required)`

 

iv.   `text(Max height occurs when)`

♦♦ Mean mark 35%.
`t` `= 1/2 xx text(time of flight)`
  `= (Vsintheta)/g`

 
`text(Find)\ \ y\ \ text(when)\ \ t = (Vsintheta)/g`

`y` `= V · (Vsintheta)/g · sintheta – 1/2 g ((Vsintheta)/g)^2`
  `= (V^2 sin^2theta)/g – 1/2 · (V^2 sin^2 theta)/g`
  `= (V^2 sin^2theta)/(2g)`

 

`text(Max height when)\ theta = (5pi)/12\ (text(steepest angle)), V^2 = 200 g\ (text(given))`

`y_text(max)` `= (200 g · sin^2 ((5pi)/12))/(2g)`
  `= 100 sin^2 ((5pi)/12)`
  `= 50(1 – cos((5pi)/6))`
  `= 50(1 + sqrt3/2)`
  `= 25(2 + sqrt 3)\ text(metres)`

Binomial, EXT1 2017 HSC 13b

Let `n` be a positive EVEN integer.

  1. Show that
  2. `(1 + x)^n + (1 - x)^n = 2[((n),(0)) + ((n),(2))x^2 + … + ((n),(n))x^n]`.  (2 marks)

  3. Hence show that
  4. `n[(1 + n)^(n - 1) - (1 - x)^(n - 1)] = 2[2((n),(2))x + 4((n),(4))x^3 + … + n((n),(n))x^(n - 1)]`.  (1 mark)

  5. Hence show that
  6. `((n),(2)) + 2((n),(4)) + 3((n),(6)) + … + n/2((n),(n)) = n2^(n - 3)`.  (2 marks)

 

 

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

(i)   `text(Show)\ (1 + x)^n + (1 – x)^n = 2[\ ^nC_0 + \ ^nC_2x^2 + … + \ ^nC_nx^n]`

`text(Using binomial expansion)`

`(1 + x)^n + (1 – x)^n`

`= \ ^nC_0 + \ ^nC_1x + \ ^nC_2x^2 + … + \ ^nC_nx^n`

`+ \ ^nC_0 – \ ^nC_1x + \ ^nC_2x^2 + … + \ ^nC_nx^n`

`= 2[\ ^nC_0 + \ ^nC_2x^2 + \ ^nC_4x^4 + … + \ ^nC_nx^n]`

`…\ text(as required)`

 

(ii)   `text{Differentiate both sides of part (i):}`

`n(1 + x)^(n-1) – n(1 – x)^(n-1)`

`= 2[2\ ^nC_2x + 4\ ^nC_4x^3 + … + n \ ^nC_nx^(n – 1)]`

`n[(1 + x)^(n – 1) + (1 – x)^(n – 1)]`

`= 2[2\ ^nC_2x + 4\ ^nC_4x^3 + … + n \ ^nC_nx^(n – 1)]`

`…\ text(as required.)`

♦ Mean mark part (iii) 48%.

 

(iii)   `text(Let)\ \ x = 1\ \ text{in part (ii)}`

`n(2^(n – 1) + 0) = 2[2\ ^nC_2 + 4\ ^nC_4 + 6\ ^nC_6 + … + n\ ^nC_n]`

`text(Divide both sides by)\ 2^2:`

`(n2^(n – 1))/(2^2)` `= 2/(2^2)[2\ ^nC_2 + 4\ ^nC_4 + 6\ ^nC_6 + … + n\ ^nC_n]`
`n2^(n – 3)`

`= \ ^nC_2 + 2\ ^nC_4 + 3\ ^nC_6 + … + n/2\ ^nC_ n`

`…\ text(as required)`

Mechanics, EXT2* M1 2017 HSC 12d

At time `t` the displacement, `x`, of a particle satisfies  `t=4-e^(-2x)`.

Find the acceleration of the particle as a function of `x`.  (3 marks)

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`(e^(4x))/2`

Show Worked Solution
`t` `= 4 – e^(−2x)`
`(dt)/(dx)` `= 2e^(−2x)`
`(dx)/(dt)` `= (e^(2x))/2`

 

`a` `= {:d/(dx):}^(1/2v^2)`
  `= d/(dx)(1/2 · ((e^(2x))/2)^2)`
  `= d/(dx)((e^(4x))/8)`
  `= 4 · (e^(4x))/8`
  `= (e^(4x))/2`

Integration, EXT1 2017 HSC 12c

The region enclosed by the semicircle  `y = sqrt(1 - x^2)`  and the `x`-axis is to be divided into two pieces by the line  `x = h`, when  `0 <= h <1`.
 


 

The two pieces are rotated about the `x`-axis to form solids of revolution. The value of `h` is chosen so that the volumes of the solids are in the ratio `2 : 1`.

  1. Show that `h` satisfies the equation  `3h^3 - 9h + 2 = 0`.  (3 marks)
  2. Given  `h_1 = 0`  as the first approximation for `h`, use one application of Newton’s method to find a second approximation for `h`.  (1 mark)

 

Show Answers Only
  1. `text(Show Worked Solution)`
  2. `2/9`
Show Worked Solution

(i)   `text(Volume of smaller solid)`

`= pi int_h^1 (sqrt(1 – x^2))^2\ dx`

`= pi int_h^1 1 – x^2\ dx`

`= pi[x – (x^3)/3]_h^1`

`= pi[(1 – 1/3) – (h – (h^3)/3)]`

`= pi(2/3 – h + (h^3)/3)`

 
`text(S)text(ince smaller solid is)\ 1/3\ text(volume of sphere,)`

`pi(2/3 – h + (h^3)/3)` `= 1/3 xx 4/3 · pi · 1^3`
`(h^3)/3 – h + 2/3` `= 4/9`
`3h^3 – 9h + 6` `= 4`
`:. 3h^3 – 9h + 2` `= 0\ \ text(… as required)`

 

(ii)    `f(h)` `= 3h^3 – 9h + 2`
  `f′(h)` `= 6h^2 – 9`
  `f(0)` `= 2`
  `f′(0)` `= −9`

 

`h_2` `= h_1 – (f(0))/(f′(0))`
  `= 0 + 2/9`
  `= 2/9`

Functions, EXT1 F1 2017 HSC 12b

  1. Carefully sketch the graphs of  `y = |\ x + 1\ |`  and  `y = 3 - |\ x - 2\ |`  on the same axes, showing all intercepts.  (3 marks)

  2. Using the graphs from part (i), or otherwise, find the range of values of `x` for which
     
    `qquad qquad |\ x + 1\ | + |\ x - 2\ | = 3`.  (1 mark)

Show Answers Only
  1.  
  2. `−1 <= x <= 2`
Show Worked Solution
i.   

 

ii.    `|\ x + 1\ | + |\ x – 2\ |` `= 3`
  `|\ x + 1\ |` `= 3 – |\ x – 2\ |`

 

`:.\ text(Solution is intersection of the graphs)`

`:. −1 <= x <= 2`

Calculus, 2ADV C3 2017 HSC 16a

John’s home is at point `A` and his school is at point `B`. A straight river runs nearby.

The point on the river closest to `A` is point `C`, which is 5 km from `A`.

The point on the river closest to `B` is point `D`, which is 7 km from `B`.

The distance from `C` to `D` is 9 km.

To get some exercise, John cycles from home directly to point `E` on the river, `x` km from `C`, before cycling directly to school at `B`, as shown in the diagram.
 


  

The total distance John cycles from home to school is `L` km.

  1. Show that  `L = sqrt (x^2 + 25) + sqrt (49 + (9 - x)^2)`.   (1 mark)

  2. Show that if  `(dL)/(dx) = 0`, then  `sin alpha = sin beta`.   (3 marks)

  3. Find the value of `x` that makes  `sin alpha = sin beta`.   (2 marks)

  4. Explain why this value of `x` gives a minimum for `L`.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `45/12`
  4. `text(See Worked Solutions)`
Show Worked Solution
i.  

`text(Using Pythagoras:)`

`L` `= AE + EB`
  `= sqrt (5^2 + x^2) + sqrt (7^2 + (9 – x)^2)`
  `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)\ text(… as required)`

 

ii.  `text(From diagram):`

♦ Mean mark 41%.

`sin alpha = x/sqrt(25 + x^2) and sin beta = (9 – x)/sqrt(49 + (9 – x)^2)`

`L` `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)`
`(dL)/(dx)` `= (2x)/sqrt(25 + x^2) – (2(9 – x))/sqrt(49 + (9 – x)^2)`

 

`text(If)\ \ (dL)/(dx) = 0,`

`=> (2x)/sqrt(25 + x^2)` `= (2(9 – x))/sqrt(49 + (9 – x)^2)`
`x/sqrt(25 + x^2)` `= (9 – x)/sqrt(49 + (9 – x)^2)`
`sin alpha` `= sin beta\ text(… as required)`

 

iii.  `text(If)\ sin alpha = sin beta,\ text(then)\ alpha = beta and`

♦♦ Mean mark 29%.

`Delta ACE\ text(|||)\ Delta BDE`

`text(Using corresponding sides of similar triangles:)`

`x/5` `= (9 – x)/7`
`7x` `= 45 – 5x`
`12x` `= 45`
`:. x` `= 45/12\ text(km)`
♦♦♦ Mean mark 9%.

 

iv.  

`text(If point)\ B\ text(is reflected across the)`

`text(river),\ AEB\ text(will be a straight line.)`

`text(If any other point is chosen,)\ AEB`

`text(would not be straight and the distance)`

`text(would be longer.)`

Calculus, EXT1* C1 2017 HSC 15c

Two particles move along the `x`-axis.

When  `t = 0`, particle `P_1` is at the origin and moving with velocity 3.

For  `t >= 0`, particle `P_1` has acceleration given by  `a_1 = 6t + e^(-t)`.

  1. Show that the velocity of particle `P_1` is given by  `v_1 = 3t^2 + 4-e^(-t)`  (2 marks)

When  `t = 0`, particle `P_2` is also at the origin.

For  `t >= 0`, particle `P_2` has velocity given by  `v_2 = 6t + 1-e^(-t)`.

  1. When do the two particles have the same velocity?  (2 marks)

  2. Show that the two particles do not meet for  `t > 0`.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `t = 1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.   `a_1` `= 6t + e^(-t)`
  `v_1` `= int a_1\ dt`
    `= int 6t + e^(-t)\ dt`
    `= 3t^2-e^(-t) + c`

 

`text(When)\ t = 0,\ v_1 = 3`

`3` `= 0-1 + c`
`c` `= 4`
`:. v_1` `= 3t^2 + 4-e^(-t) …\ text(as required)`

 

ii.  `v_2 = 6t + 1-e^(-t)`

`text(Find)\ \ t\ \ text(when)\ \ v_1 = v_2`

`3t^2 + 4-e^(-t)` `= 6t + 1-e^(-t)`
`3t^2-6t + 3` `= 0`
`t^2-2t + 1` `= 0`
`(t-1)^2` `= 0`
`:. t` `=1`

 

iii.   `x_1` `= int v_1\ dt`
    `= int 3t^2 + 4-e^(-t)\ dt`
    `= t^3 + 4t + e^(-t) + c`

 

`text(When)\ \ t = 0,\ \ x_1 = 0`

♦ Mean mark (iii) 39%.
`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_1` `= t^3 + 4t + e^(-t)-1`

 

`x_2` `= int 6t + 1-e^(-t)\ dt`
  `= 3t^2 + t + e^(-t) + c`

 
`text(When)\ \ t = 0,\ \ x_2 = 0`

`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_2` `= 3t^2 + t + e^(-t)-1`

 

`text(Find)\ \ t\ \ text(when)\ \ x_1 = x_2`

`t^3 + 4t + e^(-t)-1` `= 3t^2 + t + e^(-t)-1`
`t^3-3t^2 + 3t` `= 0`
`t(t^2-3t + 3)` `= 0`

 

`text(S)text(ince)\ \ Delta < 0\ \ text(for)\ \ t^2-3t + 3`

`=>\ text(No real solution)`

 

`:.\ text(The particles do not meet)`

`(x_1 != x_2)\ \ text(for)\ \ t > 0.`

Financial Maths, 2ADV M1 2017 HSC 15b

Anita opens a savings account. At the start of each month she deposits `$X` into the savings account. At the end of each month, after interest is added into the savings account, the bank withdraws $2500 from the savings account as a loan repayment. Let `M_n` be the amount in the savings account after the `n`th  withdrawal.

The savings account pays interest of 4.2% per annum compounded monthly.

  1. Show that after the second withdrawal the amount in the savings account is given by
    `qquad qquad M_2 = X(1.0035^2 + 1.0035) - 2500 (1.0035 + 1)`.  (2 marks)

  2. Find the value of  `X`  so that the amount in the savings account is $80 000 after the last withdrawal of the fourth year.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `$ 4019.42\ text{(nearest cent)}`
Show Worked Solution
i.   `M_1` `= X (1 + 4.2/(12 xx 100)) – 2500`
    `= X (1.0035) – 2500`
  `M_2` `= X (1.0035) + M_1 (1.0035) – 2500`
    `=X(1.0035)+(1.0035)[X(1.0035)-2500]-2500`
    `= X (1.0035) + X (1.0035^2) – 2500 (1.0035) – 2500`
    `= X (1.0035^2 + 1.0035) – 2500 (1.0035 + 1)`
♦ Mean mark part (ii) 48%.

 

ii.  

`M_3 = X (1.0035^3 + … + 1.0035) – 2500 (1.0035^2 + 1.0035 + 1)`

`vdots`

`M_n = X (1.0035^n + … + 1.0035) – 2500 (1.0035^(n – 1) + … + 1)`

`text(Find)\ X\ text(such that)\ \ M_n = 80\ 000\ \ text(when)\ n = 48`

`80\ 000 = X (1.0035^48 + …\ 1.0035) – 2500 (1.0035^47 + … + 1)`

`text(Using)\ S_n = (a(r^n – 1))/(r – 1)`

`80\ 000` `= X [(1.0035(1.0035^48 – 1))/(1.0035 – 1)] – 2500 [(1(1.0035^48 – 1))/(1.0035 – 1)]`
`80\ 000` `= X (52.351…) – 130\ 421.20…`
   
`:. X` `= (80\ 000 + 130\ 421.20…)/(52.351…)`
  `= $4019.42\ text{(nearest cent)}`

Plane Geometry, 2UA 2017 HSC 15a

The triangle `ABC` is isosceles with  `AB = AC`  and the size of `/_BAC` is `x^@`.

Points `D` and `E` are chosen so that `Delta ABC, Delta ACD` and `Delta ADE` are congruent, as shown in the diagram.
 

Find the value of `x` for which `AB` is parallel to `ED`, giving reasons.  (3 marks)

Show Answers Only

`x = 36`

Show Worked Solution

`text(All base angles) = 1/2(180-x)\ \ text{(Angle sum of}\ Delta text{)}`

`text(If)\ \ AB\ text(||)\ ED,`

`/_ BAD` `= /_ ADE\ \ \ text{(alternate angles)}`
`2x` `= 1/2 (180-x)`
`4x` `= 180-x`
`5x` `= 180`
`x` `= 36^{\circ}`

Calculus, 2ADV C4 2017 HSC 14d

The shaded region shown is enclosed by two parabolas, each with `x`-intercepts at  `x = –1`  and  `x = 1`.

The parabolas have equations  `y = 2k (x^2 - 1)`  and  `y = k(1 - x^2)`, where  `k > 0`.
  


 

Given that the area of the shaded region is 8, find the value of `k`.  (3 marks)

Show Answers Only

`k = 2`

Show Worked Solution
`text(Area)` `= int_(-1)^1 k (1 – x^2) dx – int_(-1)^1 2k (x^2 – 1) dx`
`8` `= 2 int_0^1 k – kx^2 – 2kx^2 + 2k\ dx`
`8` `= 2 int_0^1 3k – 3kx^2\ dx`
`8` `= 2 [3kx – kx^3]_0^1`
`8` `= 2 [(3k – k) – 0]`
`8` `= 4k`
`:. k` `= 2`

Calculus, EXT1* C1 2017 HSC 14c

Carbon-14 is a radioactive substance that decays over time. The amount of carbon-14 present in a kangaroo bone is given by

`C(t) = Ae^(kt),`

where `A` and `k` are constants, and `t` is the number of years since the kangaroo died.

  1. Show that `C(t)` satisfies  `(dC)/(dt) = kC`.  (1 mark)

  2. After 5730 years, half of the original amount of carbon-14 is present.

     

    Show that the value of `k`, correct to 2 significant figures, is – 0.00012.  (2 marks)

  3. The amount of carbon-14 now present in a kangaroo bone is 90% of the original amount.

     

    Find the number of years since the kangaroo died. Give your answer correct to 2 significant  figures.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `870\ text{years (2 sig. fig.)}`
Show Worked Solution
i.   `C` `= Ae^(kt)`
   `(dC)/(dt)` `= k * Ae^(kt)`
    `= kC …\ text(as required)`

 

ii.  `text(When)\ \ t = 5730, qquad A = 0.5 A_0`

`0.5 A_0` `= A_0 * e^(5730 k)`
`e^(5730 k)` `= 0.5`
`text(ln)\ e^(5730 k)` `= text(ln)\ 0.5`
`5730 k` `= text(ln)\ 0.5`
`k` `= {text(ln)\ 0.5}/5730`
  `= -0.0001209…`
  `= -0.00012\ text{(2 sig fig) … as required}`

 

iii.  `text(Find)\ t\ text(when)\ A = 0.9 A_0`

`0.9 A_0` `= A_0 e^(kt)`
`e^(kt)` `= 0.9`
`kt` `= text(ln)\ 0.9`
`t` `= (text(ln)\ 0.9)/k`
  `= (5730 xx text(ln)\ 0.9)/(text(ln)\ 0.5)`
  `= 870.97…`
  `= 870\ text{years (2 sig.fig.)}`

Combinatorics, EXT1 A1 2017 HSC 10 MC

Three squares are chosen at random from the 3 × 3 grid below, and a cross is placed in each chosen square.
 


 

What is the probability that all three crosses lie in the same row, column or diagonal?

A.     `1/28`

B.     `2/21`

C.     `1/3`

D.     `8/9`

Show Answers Only

`B`

Show Worked Solution
`P` `= text(favourable events)/text(total possible events)`
  `= (3 + 3 + 2)/(\ ^9C_3)`
  `= 2/21`

 
`=>B`

Combinatorics, EXT1 A1 2017 HSC 9 MC

When expanded, which expression has a non-zero constant term?

A.     `(x + 1/(x^2))^7`

B.     `(x^2 + 1/(x^3))^7`

C.     `(x^3 + 1/(x^4))^7`

D.     `(x^4 + 1/(x^5))^7`

Show Answers Only

`C`

Show Worked Solution

`text(Consider the general term for option)\ A:`

`T_k` `= \ ^7C_k · x^(7 – k) · x^(−2k)`
  `= \ ^7C_k · x^(7 – 3k)`

 
`text(Non zero constant term occurs when)`

`7 – 3k` `= 0`
`k` `= 7/3 => text(no terms exists)\ (k\ text{not integer)}`

 
`text(Consider option)\ C:`

`T_k` `= \ ^7C_k · x^(3(7 – k)) · x^(−4k)`
  `= \ ^7C_k · x^(21 – 7k)`
`21 – 7k` `= 0`
`k` `= 3`

 
`:.\ text(Non-zero constant term exists)`

`text(since)\ k\ text(is an integer)`

`⇒C`

Calculus, 2ADV C4 2017 HSC 14b

  1. Find the exact value of
     
  2. `qquad int_0^(pi/3) cos x\ dx`.  (1 mark)

  3. Using Simpson’s rule with one application, find an approximation to the integral
  4.  
    `qquad int_0^(pi/3) cos x\ dx,`
     
  5. leaving your answer in terms of `pi` and `sqrt 3`.  (2 marks)
     

  6. Using parts (i) and (ii), show that
     
  7. `qquad pi ~~ (18 sqrt 3)/(3 + 4 sqrt 3)`.  (1 mark)

 

 

Show Answers Only
(i)   `sqrt 3/2`
(ii)   `((4 sqrt 3 + 3)pi)/36`
(iii)   `text{Proof (See Worked Solutions)}`
Show Worked Solution
(i)   `int_0^(pi/3) cos x\ dx` `= [sin x]_0^(pi/3)`
    `= sin\ pi/3 – 0`
    `= sqrt 3/2`

 

(ii)  
    `x`     `0`    `overset(pi) underset(6) _`     `overset(pi) underset(3) _`  
    `y`     `1`     `overset(sqrt 3) underset(2) _`     `overset(1) underset(2) _`  
      `y_0`     `y_1`     `y_2`  
`int_0^(pi/3) cos x\ dx` `~~ h/3 [y_0 + 4y_1 + y_2]`
  `~~ pi/6 ⋅ 1/3 [1 + 4 ⋅ sqrt 3/2 + 1/2]`
  `~~ pi/18 ((4 sqrt 3 + 3)/2)`
  `~~ ((4 sqrt 3 + 3) pi)/36`

 

(iii)  `text{Using parts (i) and (ii)}`

♦ Mean mark 49%.
`((4 sqrt 3 + 3) pi)/36` `~~ sqrt 3/2`
`:. pi` `~~ (36 sqrt 3)/(2(3 + 4 sqrt 3))`
  `~~ (18 sqrt 3)/(3 + 4 sqrt 3) … text( as required)`

Quadratic, EXT1 2017 HSC 6 MC

The point  `P(2/p, 1/(p^2))`, where  `p != 0`  lies on the parabola  `x^2 = 4y`.

What is the equation of the normal at `P`?

A.     `py - x = −p`

B.     `p^2y + px = −1`

C.     `p^2y - p^3x = 1 − 2p^2`

D.     `p^2y + p^3x = 1 + 2p^2`

Show Answers Only

`D`

Show Worked Solution

`x^2 = 4y,\ \ => a = 1`

`text(Normal equation at)\ (2t,t^2):`

`=> x + ty = t^3 + 2t`

`text(When)\ (2/p, 1/(p^2)) ≡ (2t,t^2),`

`t = 1/p`

`:.\ text(Equation of normal:)`

`x + y/p` `= 1/(p^3) + 2/p`
`p^3x + p^2y` `= 1 + 2p^2`

 

`⇒D`

Trigonometry, EXT1 T3 2017 HSC 4 MC

What is the value of  `tan alpha`  when the expression  `2sinx - cosx`  is written in the form  `sqrt5 sin(x - alpha)`?

A.     `−2`

B.     `−1/2`

C.     `1/2`

D.     `2`

Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ sqrt5 sin(x – alpha) = sqrt5 sinx cosalpha – sqrt5cosxsinalpha,`

`=>sqrt5 sinx cosalpha – sqrt5 cosx sinalpha = 2sinx – cosx`

`=> sqrt5 cosalpha = 2, sqrt5 sinalpha = 1`

`(sinalpha)/(cosalpha)` `= 1/2`
`:. tan alpha` `= 1/2`

`⇒C`

Quadratic, 2UA 2017 HSC 13c

By letting  `m = t^(1/3)`, or otherwise, solve  `t^(2/3) + t^(1/3) - 6 = 0`.  (2 marks)

Show Answers Only

`t = -27 or 8`

Show Worked Solution

`t^(2/3) + t^(1/3) – 6 = 0`

`text(Let)\ \ m = t^(1/3),`

`m^2 + m – 6 = 0`

`(m + 3) (m – 2) = 0`

`m = -3 or 2`

`t^(1/3)` `= -3` `or` `t^(1/3)` `= 2`
`t` `= (-3)^3`   `t` `= 2^3`
  `= -27`     `= 8`

 

`:. t = -27 or 8`

Calculus, 2ADV C3 2017 HSC 13b

Consider the curve  `y = 2x^3 + 3x^2 - 12x + 7`.

  1. Find the stationary points of the curve and determine their nature.  (4 marks)

  2. Sketch the curve, labelling the stationary points.  (2 marks)

  3. Hence, or otherwise, find the values of `x` for which `(dy)/(dx)` is positive.  (1 mark)

Show Answers Only
  1. `text(maximum at)\ (-2, 27)`

     

    `text(minimum at)\ (1, 0)`

  2.    
  3. `x < -2 and x > 1`
Show Worked Solution
i.   `y` `= 2x^3 + 3x^2 – 12x + 7`
  `(dy)/(dx)` `= 6x^2 + 6x – 12`
  `(d^2y)/(dx^2)` `= 12x + 6`

 

`text(S.P. when)\ (dy)/(dx)` `= 0`
`6x^2 + 6x – 12` `= 0`
`x^2 + x – 2` `= 0`
`(x + 2) (x – 1)` `= 0`

 
`x = -2 or 1`
 

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

`:.\ text(MAX at)\ (–2, 27)`
 

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

`:.\ text(MIN at)\ (1, 0)`

 

ii.  

 

iii.  `text(Solution 1)`

`text(From graph, gradient is positive for)`

`x < –2 and x > 1`

`:. (dy)/(dx) > 0\ \ text(for)\ \ x < –2 and x > 1`

 

`text(Solution 2)`

`(dy)/(dx) > 0`

`6x^2 + 6x – 12` `> 0`
`(x + 2) (x – 1)` `> 0`

 
 
`:. x < –2 and x > 1`

Trigonometry, 2ADV T1 2017 HSC 13a

Using the cosine rule, find the value of `x` in the following diagram.  (3 marks)
 

Show Answers Only

`x = 11`

Show Worked Solution

`text(C)text(osine Rule:)`

`c^2` `= a^2 + b^2 – 2ab cos C`
`13^2` `= (x – 4)^2 + (x + 4)^2 – 2 (x – 4) (x + 4) cos 60^@`
`169` `= x^2 – 8x + 16 + x^2 + 8x + 16 – (x^2 – 16)`
`169` `= x^2 + 48`
`x^2` `= 121`
`:. x` `= 11, qquad (x != –11)`

Statistics, 2ADV 2017 HSC 12e

A spinner is marked with the numbers 1, 2, 3, 4 and 5. When it is spun, each of the five numbers is equally likely to occur.
 

 
The spinner is spun three times.

  1. What is the probability that an even number occurs on the first spin?  (1 mark)
  2. What is the probability that an even number occurs on at least one of the three spins?  (1 mark)
  3. What is the probability that an even number occurs on the first spin and odd numbers occur on the second and third spins?  (1 mark)
  4. What is the probability that an even number occurs on exactly one of the three spins?  (1 mark)
Show Answers Only
  1. `2/5`
  2. `98/125`
  3. `18/125`
  4. `54/125`
Show Worked Solution

i.   `Ptext{(even)} = 2/5`
 

ii.  `Ptext{(at least 1 even)}`

`= 1 – Ptext{(no evens)}`

`= 1 – 3/5 ⋅ 3/5 ⋅ 3/5`

`= 1 – 27/125`

`= 98/125`
 

iii.  `Ptext{(even, odd, odd)}`

`= 2/5 ⋅ 3/5 ⋅ 3/5`

`= 18/125`
 

iv.  `Ptext{(even occurs exactly once)}`

`= Ptext{(e, o, o)} + P text{(o, e, o)} + P text{(o, o, e)}`

`= 2/5 ⋅ 3/5 ⋅ 3/5 + 3/5 ⋅ 2/5 ⋅ 3/5 + 3/5 ⋅ 3/5 ⋅ 2/5`

`= 54/125`

Linear Functions, 2UA 2017 HSC 12d

The points  `A(–4, 0)`  and  `B(1, 5)`  lie on the line  `y = x + 4`.

The length of  `AB`  is  `5 sqrt 2`.

The points  `C(0, –2)`  and  `D(3, 1)`  lie on the line  `x - y - 2 = 0`.

The points `A, B, D, C` form a trapezium as shown.
 


 

  1. Find the perpendicular distance from point `A(–4, 0)` to the line  `x - y - 2 = 0`.  (1 mark)
  2. Calculate the area of the trapezium.  (2 marks)
Show Answers Only

(i)   `3sqrt2\ text(units)`

(ii)  `24\ text(u²)`

Show Worked Solution

(i)   `A(-4, 0), qquad qquad x – y – 2 = 0`

`_|_\ text(dist)` `= |ax_1 + by_1 + c|/sqrt(a^2 + b^2)`
  `= |-4 + 0 – 2|/sqrt (1 + 1)`
  `= 6/sqrt 2 xx sqrt 2/sqrt 2`
  `= 3 sqrt 2\ text(units)`

 

(ii)  `text(Area) = 1/2 ⋅ h ⋅ (AB + CD)`

`AB` `= 5 sqrt 2\ text{(given)}`
`CD` `= sqrt((3 – 0)^2 + (1 + 2)^2)`
  `= sqrt 18`
  `= 3 sqrt 2`

 

`:.\ text(Area)` `= 1/2 ⋅ 3 sqrt 2\ \ (5 sqrt 2 + 3 sqrt 2)`
  `= 1/2 ⋅ 3 sqrt 2 ⋅ 8 sqrt 2`
  `= 24\ text(u)²`

Financial Maths, 2ADV M1 2017 HSC 12c

In an arithmetic series, the fifth term is 200 and the sum of the first four terms is 1200.

Find the value of the tenth term.  (3 marks)

Show Answers Only

`0`

Show Worked Solution

`T_n = a + (n – 1) d`

`=>a + 4d = 200 …\ (1)`

 

`S_n = n/2 [2a + (n – 1) d]`

`=>4a + 6d = 1200 …\ (2)`

 

`text(Multiply)\ (1) xx 4`

`=>4a + 16d = 800 …\ (1 prime)`

 

`text(Subtract)\ \ (1 prime) – (2)`

`10d` `= -400`
`d` `= -40`

 

`text(Substitute)\ \ d = -40\ \ text(into)\ (1)`

`a – 160` `= 200`
`a` `= 360`
`:. T_10` `= 360 – 9(-40)`
  `= 0`

Functions, 2ADV F1 2017 HSC 11h

Find the domain of the function  `f(x) = sqrt (3-x)`.  (2 marks)

Show Answers Only

`x <= 3 or (-oo,3].`

Show Worked Solution

`text(Domain of)\ \ f(x) = sqrt (3-x)`

`3-x` `>= 0`
`x` `<= 3`

 

`text(Note domain can also be expressed as:)\ \ (-oo,3]`

Quadratic, 2UA 2017 HSC 11f

Determine the equation of the parabola shown. Write your answer in the form  `(x - h)^2 = 4a (y - k)`.  (2 marks)

Show Answers Only

`(x – 2)^2 = 4 (1/3) (y – 1)`

Show Worked Solution

`text(Vertex)\ (2, 1)`

COMMENT: Many students struggled with a basic concept here – mean mark of just 56%.

`=> (x – 2)^2 = 4a (y – 1)`

 

`text(Passes through)\ (0, 4)`

`(0 – 2)^2` `= 4a (4 – 1)`
`12a` `=4`
`a` `= 1/3`

`:. (x – 2)^2 = 4(1/3) (y – 1)`

Trigonometry, 2ADV T1 2017 HSC 11e

In the diagram, `OAB` is a sector of the circle with centre `O` and radius 6 cm, where  `/_ AOB = 30^@`.


 

  1.  Find the exact value of the area of the triangle `OAB`.  (1 mark)

  2. Find the exact value of the area of the shaded segment.  (1 mark)

Show Answers Only
  1.  `9\ text(cm²)`
  2. `3 pi – 9\ text(cm²)`
Show Worked Solution
i.   `text(Area)\ Delta OAB` `= 1/2 ab sin C`
    `= 1/2 xx 6^2 xx sin 30^@`
    `= 9\ text(cm²)`

 

ii.  `text(Area segment)` `= text(Area sector) – text(Area)\ Delta OAB`
    `= 30/360 xx pi xx 6^2 – 9`
    `= 3 pi – 9\ \ text(cm²)`

Functions, EXT1* F1 2017 HSC 8 MC

The region enclosed by  `y = 4 - x,\ \ y = x`  and  `y = 2x + 1`  is shaded in the diagram below.
 

Which of the following defines the shaded region?

A.   `y <= 2x + 1, qquad` `y <= 4-x, qquad` `y >= x`
B.   `y >= 2x + 1, qquad` `y <= 4-x, qquad` `y >= x`
C.   `y <= 2x + 1, qquad` `y >= 4-x, qquad` `y >= x`
D.   `y >= 2x + 1, qquad` `y >= 4-x, qquad` `y >= x`
Show Answers Only

`A`

Show Worked Solution

`text(Consider)\ \ y = 2x + 1,`

`text(Shading is below graph)`

`=> y <= 2x + 1`

`text(Consider)\ \ y = 4-x,`

`text(Shading is below graph)`

`=> y <= 4-x`

`=>  A`

Trigonometry, 2ADV T2 2017 HSC 7 MC

Which expression is equivalent to  `tan theta + cot theta`?

  1. `text(cosec)\ theta + sec theta`
  2. `sec theta\ text(cosec)\ theta`
  3. `2`
  4. `1`
Show Answers Only

`B`

Show Worked Solution
`tan theta + cot theta` `= (sin theta)/(cos theta) + (cos theta)/(sin theta)`
  `= (sin^2 theta + cos^2 theta)/(cos theta sin theta)`
  `= 1/(cos theta sin theta)`
  `= sec theta\ text(cosec)\ theta`

`=>  B`

Functions, 2ADV F1 2017 HSC 6 MC

The point `P` moves so that it is always equidistant from two fixed points, `A` and `B.`

What best describes the locus of `P`?

(A)  A point

(B)  A circle

(C)  A parabola

(D)  A straight line

Show Answers Only

`D`

Show Worked Solution

`text(An example of)\ \ P\ \ text(is drawn below:)`
 

`=>  D`

L&E, 2ADV E1 2017 HSC 5 MC

It is given that  `ln a = ln b-ln c`, where  `a, b, c > 0.`

Which statement is true?

  1. `a = b-c`
  2. `a = b/c`
  3. `text(ln)\ a = b/c`
  4. `text(ln)\ a = (text(ln)\ b)/(text(ln)\ c)`
Show Answers Only

`B`

Show Worked Solution
Mean mark 51%.
COMMENT: Use of log laws here proved difficult for many students.
`ln a` `= ln b-ln c`
`ln a` `= ln (b/c)`
`:. a` `= b/c`

 
`=>  B`

Calculus, 2ADV C3 2017 HSC 4 MC

The function  `f(x)` is defined for  `a <= x <= b.`

On this interval,  `f ^{′}(x) > 0 and f^{″}(x) < 0.`

Which graph best represents  `y = f(x)`?
 

(A)   (B)  
(C)   (D)  
Show Answers Only

`A`

Show Worked Solution

`text(Interpreting)\ \ f^{′}(x) > 0,`

`=>\ text(gradient is always positive)`

`text(Interpreting)\ \ f^{″}(x) < 0,`

`=>\ text(Curve is concave down)`

`=>  A`

Algebra, 2UG 2017 HSC 30d

In an investigation, students used different numbers of identical small solar panels to power model cars. The cars were then tested and their speed measured in km/h. The results are summarised in the table.
 


 

The equation of the least-squares line of best fit, relating the speed and the number of solar panels, has been calculated to be
 

`y = 2.125x + 2.0375`
 

  1. What would be the speed of a car powered by 5 solar panels, based on this equation?  (1 mark)
  2. Calculate the correlation coefficient, `r`, between the number of solar panels and the speed of a car.  (2 marks)

 

Show Answers Only
  1. `12.6625\ text(km/h)`
  2. `0.85`
Show Worked Solution
(i)    `text(Speed)(y)` `= 2.125(5) + 2.0375`
    `= 12.6625\ text(km/h)`

 

(ii)   `text(Using the formula sheet:)`

♦♦ Mean mark 23%.
`text(gradient)` `= r xx (text(std dev)(y))/(text(std dev)(x))`
`2.125` `= r xx 2/0.8`
`:. r` `= (2.125 xx 0.8)/2`
  `= 0.85`

Probability, STD2 S2 2017 HSC 29c

A group of Year 12 students was surveyed. The students were asked whether they live in the city or the country. They were also asked if they have ever waterskied.

The results are recorded in the table.
  


  

  1. A person is selected at random from the group surveyed. Calculate the probability that the person lives in the city and has never waterskied.  (2 marks)

  2. A newspaper article claimed that Year 12 students who live in the country are more likely to have waterskied than those who live in the city.

     

    Is this true, based on the survey results? Justify your answer with relevant calculations.  (2 marks)

Show Answers Only
  1. `125/176`
  2. `text(See Worked Solutions)`
Show Worked Solution
i.     `P` `= text(live in city, not skied)/text(total surveyed)`
    `= 2500/3520`
    `= 125/176`
♦ Mean mark part (ii) 47%.

 

ii.   `P(text(live in country, skied))` `= 70/((70 + 800))`
    `= 0.0804…`
    `= 8text(%)`

 

`P(text(live in city, skied))` `= 150/((150 + 2500))`
  `= 0.0566`
  `= 6text(%)`

 
`text(S)text(ince 8% > 6%, the statement is true.)`

Financial Maths, STD2 F1 2017 HSC 29b

Sabrina’s taxable income is $86 725 in a particular year.

The table below is used to calculate her tax payable. In addition, she pays the Medicare levy, which is 2% of her taxable income.
 

 
Calculate Sabrina’s net income in that year.  (3 marks)

Show Answers Only

`$65\ 257.87`

Show Worked Solution
`text(Tax payable)` `= 3572 + 0.325(86\ 725 – 37\ 000)`
  `= 19\ 732.625`
  `= $19\ 732.63`
`text(Medicare levy)` `= 2text(%) xx 86\ 725`
  `= $1734.50`

 

`:.\ text(Net Income)` `= 86\ 725 – (19\ 732.63 + 1734.50)`
  `= 65\ 257.87`
  `= $65\ 257.87`

Measurement, 2UG 2017 HSC 29a

A new 200-metre long dam is to be built.
The plan for the new dam shows evenly spaced cross-sectional areas.
 


 

  1. Using TWO applications of Simpson’s rule, show that the volume of the dam is approximately  44 333 m³.  (2 marks)
  2. It is known that the catchment area for this dam is 2 km².
    Calculate how much rainfall is needed, to the nearest mm, to fill the dam.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solution)`
  2. `22\ text{mm  (nearest mm)}`
Show Worked Solution
(i)    `text(Volume)` `~~ h/3 (y_0 + 4y_1 + y_2)\ \ …\ text(applied twice)`
    `~~ 50/3(0 + 4 xx 140 + 270) + 50/3(270 + 4 xx 300 + 360)`
    `~~ 50/3(830) + 50/3(1830)`
    `~~ 44\ 333\ text(m³)\ …\ \ text(as required)`

 

(ii)   `text(Convert 2km² to m²:)`

♦♦♦ Mean mark 11%.
`text(2 km²)` `= 2 xx 1000 xx 1000`
  `= 2\ 000\ 000\ text(m²)`
`:.\ text(Rainfall needed)` `= (44\ 333)/(2\ 000\ 000)`
  `= 0.0221…\ text(m)`
  `= 22.1…\ text(mm)`
  `= 22\ text{mm  (nearest mm)}`

Algebra, STD2 A4 2017 HSC 28e

A movie theatre has 200 seats. Each ticket currently costs $8.

The theatre owners are currently selling all 200 tickets for each session. They decide to increase the price of tickets to see if they can increase the income earned from each movie session.

It is assumed that for each one dollar increase in ticket price, there will be 10 fewer tickets sold.

A graph showing the relationship between an increase in ticket price and the income is shown below.
 


 

  1. What ticket price should be charged to maximise the income from a movie session?  (1 mark)

  2. What is the number of tickets sold when the income is maximised?  (1 mark)

  3. The cost to the theatre owners of running each session is $500 plus $2 per ticket sold.

     

    Calculate the profit earned by the theatre owners when the income earned from a session is maximised.  (2 marks)

Show Answers Only
  1. `$14`
  2. `140`
  3. `$1180`
Show Worked Solution

i.   `text(Graph is highest when increase = $6)`

♦ Mean mark 50%.

`:.\ text(Ticket price)\ = 8 + 6= $14`
 

ii.   `text(Solution 1)`

♦ Mean mark 45%.

`text(Tickets sold)\ =200-(4 xx 10)=140`
 

`text(Solution 2)`

`text(Tickets)\ = text(max income)/text(ticket price) = 1960/14= 140`
 

iii.  `text{Cost}\ = 140 xx $2 + $500= $780`

`:.\ text(Profit when income is maximised)`

`= 1960-780`

`= $1180`

Probability, 2UG 2017 HSC 28b

Five people are in a team. Two of them are selected at random to attend a competition.

  1. How many different groups of two can be selected?  (1 mark)
  2. If Mary is one of the five people in the team, what is the probability that she is selected to attend the competition?  (1 mark)

 

Show Answers Only
  1. `10`
  2. `2/5`
Show Worked Solution
(i)   `text(# Groups of 2)` `= (5 xx 4)/(2 xx 1)`
    `=10`
♦♦ Mean mark part (ii) 33%.

 

(ii)   `text(Number of pairs with Mary = 4)`

`:. Ptext{(Mary attends)}` `= 4/10`
  `= 2/5`

Algebra, STD2 A1 2017 HSC 27e

Rhys is drinking low alcohol beer at a party over a five-hour period. He reads on the label of the low alcohol beer bottle that it is equivalent to 0.8 of a standard drink.

Rhys weighs 90 kg.

The formula below  can be used to calculate a Rhys's blood alcohol content.
 

`BAC_text(Male) = (10N - 7.5H)/(6.8M)`

where    `N` is the number of standard drinks consumed

`H` is the number of hours drinking

`M` is the person's mass in kilograms
 

What is the maximum number of complete bottles of the low alcohol beer he can drink to remain under a Blood Alcohol Content (BAC) of 0.05?  (4 marks)

Show Answers Only

`8`

Show Worked Solution
`text(BAC)_text(male)` `= (10N – 7.5H)/(6.8M)`
`0.05` `= (10N – 7.5 xx 5)/(6.8 xx 90)`
`10N` `= (0.05 xx 6.8 xx 90) + 7.5 xx 5`
  `= 68.1`
`N` `= 6.81\ \ text(standard drinks)`

 

`:.\ text(Number of low alcohol bottles)`

`= 6.81/0.8`

`= 8.51`
 

`:.\ text(Max complete bottles to stay under 0.05)`

`= 8`

Financial Maths, STD2 F5 2017 HSC 27c

A table of future value interest factors for an annuity of $1 is shown.
 


 

An annuity involves contributions of $12 000 per annum for 5 years. The interest rate is 4% per annum, compounded annually.

  1. Calculate the future value of this annuity.  (1 mark)

  2. Calculate the interest earned on this annuity.  (1 mark)

Show Answers Only
  1. `$64\ 995.60`
  2. `$4995.60`
Show Worked Solution

i.   `text(FV factor = 5.4163)`

`:.\ text(FV of Annuity)` `= 12\ 000 xx 5.4163`
  `= $64\ 995.60`
♦♦ Mean mark part (ii) 22%.
COMMENT: A very poorly answered question dealing with a core concept in this area.

 

ii.   `text(Interest earned)` `=\ text(FV − total repayments)`
    `= 64\ 995.60 – (5 xx 12\ 000)`
    `= $4995.60`

Financial Maths, 2UG 2017 HSC 26g

Rachel bought a motorcycle advertised for $7990. She paid a $500 deposit and took out a flat-rate loan to repay the balance. Simple interest was charged at a rate of 7% per annum on the amount borrowed. She repaid the loan over 2 years, making equal weekly repayments.

Calculate the weekly repayment.  (3 marks)

Show Answers Only

`$82.10\ \ text{(nearest cent)}`

Show Worked Solution
`text(Loan amount)` `= 7990 – 500`
  `= $7490`
`text(Interest)` `= Prn`
  `= 7490 xx 0.07 xx 2`
  `= $1048.60`

 

`:.\ text(Weekly repayment)` `= text(total repayments)/text(total weeks)`
  `= (7490 + 1048.60)/((52 xx 2))`
  `= 82.1019…`
  `= $82.10\ \ text{(nearest cent)}`

Statistics, STD2 S1 2017 HSC 26f

The area chart shows the number of goals scored by three hockey teams, `A`, `B` and `C`, in the first 4 rounds.
 


 

  1. How many goals were scored by team `C` in round 1?  (1 mark)

  2. In which round did all three teams score the same number of goals?  (1 mark)

Show Answers Only
  1. `4`
  2. `text(Round 3)`
Show Worked Solution
i.   `text(Goals by team C)` `= 9 – 5`
    `= 4`

 

ii.   `text{In Round 3 (all teams scored 2 goals).}`

Financial Maths, STD2 F4 2017 HSC 26e

Sam purchased 500 company shares at $3.20 per share. Brokerage fees were 1.5% of the purchase price.

Sam is paid a dividend of 26 cents per share, then immediately sells the shares for $4.80 each.

If he pays no further brokerage fees, what is Sam’s total profit?  (3 marks)

Show Answers Only

`$906`

Show Worked Solution
`text(Purchase price)` `= 500 xx 3.20`
  `= $1600`
`text(Brokerage fees)` `= 1.5text(%) xx 1600`
  `= $24`
`text(Dividends)` `= 500 xx 0.26`
  `= $130`
`text(Sale price)` `= 500 xx 4.80`
  `= $2400`

 

`:.\ text(Total profit)` `= 2400 + 130 – (1600 + 24)`
  `= $906`

Measurement, STD2 M6 2017 HSC 26d

A sewer pipe needs to be placed into the ground so that it has a 2° angle of depression. The length of the pipe is 15 000 mm.
 


 

How much deeper should one end of the pipe be compared to the other end? Answer to the nearest mm.  (2 marks)

Show Answers Only

`523\ text{mm  (nearest mm)}`

Show Worked Solution

`text(Let)\ \ x = text(depth needed)`

`sin 2^@` `= x/(15\ 000)`
`x` `= 15\ 000 xx sin 2^@`
  `= 523.49…`
  `= 523\ text{mm  (nearest mm)}`

Data, 2UG 2017 HSC 26c

A farmer needed to estimate the number of goats on his property. He tagged 80 of his goats. Later, he collected a random sample of 45 goats and found that 16 of these had tags.

Estimate the number of goats the farmer has on his property.  (2 marks)

Show Answers Only

`225`

Show Worked Solution

`text(Let)\ \ P =\ text(total number of goats)`

`text(Capture) = 80/P`

`text(Recapture) = 16/45`

`:. 80/P` `= 16/45`
`16P` `= 80 xx 45`
`P` `= (80 xx 45)/16`
  `= 225`

Measurement, STD2 M7 2017 HSC 26b

Toby’s mobile phone plan costs $20 per month, plus the cost of all calls. Calls are charged at the rate of 70 cents per 30 seconds, or part thereof. There is also a call connection fee of 50c per call.

Here is a record of all his calls in July.
 

How much is Toby’s mobile phone bill for July?  (2 marks)

Show Answers Only

`$27.10`

Show Worked Solution

`text(Call cost) = 0.70 + (2 xx 0.70) + (5 xx 0.70) = $5.60`

`text(Connection fees) = 3 xx 0.50 = $1.50`

`:.\ text(Total bill)` `= 5.60 + 1.50 + 20`
  `= $27.10`

Algebra, STD2 A4 2017 HSC 17 MC

The graph of the line with equation  `y = 6 - 2x`  is shown.
 

When the graph of the line with equation  `y = x + 3`  is also drawn on this number plane, what will be the point of intersection of the two lines?

A.     `(0, 6)`

B.     `(1, 4)`

C.     `(2, 2)`

D.     `(3, 0)`

Show Answers Only

`text(B)`

Show Worked Solution

`text(Solution 1)`

`text(From graph, intersection at (1,4))`
 

`=>B`
 

`text(Solution 2)`

`y` `= 6 – 2x` `…\ (1)`
`y` `= x + 3` `…\ (2)`

 
`text{Substitute (2) into (1)}`

`x + 3` `= 6 – 2x`
`3x` `= 3`
`x` `= 1`

 
`text(When)\ \ x = 1, y = 4`

`=>B`

Measurement, STD2 M1 2017 HSC 22 MC

A concrete water pipe is manufactured in the shape of an annular cylinder. The dimensions are shown in the diagrams.
 


 

What is the approximate volume of concrete needed to make the water pipe?

  1. `text(0.06 m)³`
  2. `text(0.09 m)³`
  3. `text(0.70 m)³`
  4. `text(0.99 m)³`
Show Answers Only

`C`

Show Worked Solution
`text(Volume)` `= text(Area of annulus) xx h`
  `= (piR^2 – pir^2) xx 2.8`
  `= (pi xx 0.45^2 – pi xx 0.35^2) xx 2.8`
  `= 0.7037…`
  `= 0.70\ text(m)³`

  
`=>C`

Algebra, STD2 A1 2017 HSC 19 MC

Young’s formula, shown below, is used to calculate the dosage of medication for children aged 1−12 years based on the adult dosage.

`D = (yA)/(y + 12)`

where     `D`   = dosage for children aged 1−12 years
`y`   = age of child (in years)
`A`   = Adult dosage

 
A child’s dosage is calculated to be 20 mg, based on an adult dosage of 40 mg.

How old is the child in years?

A.     `6`

B.     `8`

C.     `10`

D.     `12`

Show Answers Only

`text(D)`

Show Worked Solution
`D` `= (yA)/(y + 12)`
`20` `= (40y)/(y + 12)`
`20(y + 12)` `= 40y`
`20y + 240` `= 40y`
`20y` `= 240`
`y` `= 12`

`=>\ text(D)`

Measurement, STD2 M1 2017 HSC 18 MC

A skip bin is in the shape of a trapezoidal prism, with dimensions as shown.
 

What is the volume of the skip bin?

  1. `5.4\ text(m)^3`
  2. `7.776\ text(m)^3`
  3. `10.8\ text(m)^3`
  4. `15.552\ text(m)^3`
Show Answers Only

`A`

Show Worked Solution
`text(Area of trapezoid)` `= 1/2h (a + b)`
  `= 1/2 xx 1.2 xx (3.6 + 2.4)`
  `= 3.6\ text(m)^2`

 

`:.\ text(Volume)` `= Ah`
  `= 3.6 xx 1.5`
  `= 5.4\ text(m)^3`

 
`=>A`

Measurement, STD2 M1 2017 HSC 16 MC

The benchmark for annual greenhouse gas emissions from the residential sector is 3292 kg of carbon dioxide per person per year.

A new building, planned to house 6 people, has been designed to achieve a 25% reduction on this benchmark.

What is the maximum amount of carbon dioxide per year, to the nearest kilogram, that this building is designed to emit when fully occupied?

A.     823 kg

B.     2469 kg

C.     4938 kg

D.     14 814 kg

Show Answers Only

`D`

Show Worked Solution
`text(Benchmark emissions)` `= 6 xx 3292`
  `= 19\ 752\ text(kg)`

 
`:.\ text(Max emissions of new building)`

`= 75text(%) xx 19\ 752`

`= 14\ 814\ text(kg)`

`=> D`

Financial Maths, STD2 F1 2017 HSC 11 MC

A new car was bought for $19 900 and one year later its value had depreciated to $16 300.

What is the approximate depreciation, expressed as a percentage of the purchase price?

  1. 18%
  2. 22%
  3. 78%
  4. 82%
Show Answers Only

`A`

Show Worked Solution
`text(Net Depreciation)` `= 19\ 900-16\ 300`
  `= $3600`

 

`:. %\ text(Depreciation)` `= 3600/(19\ 900) xx 100`
  `= 18.09…text(%)`

`=>A`

Financial Maths, STD2 F4 2017 HSC 10 MC

A single amount of $10 000 is invested for 4 years, earning interest at the rate of 3% per annum, compounded monthly.

Which expression will give the future value of the investment?

  1. `10\ 000 xx (1 + 0.03)^4`
  2. `10\ 000 xx (1 + 0.03)^48`
  3. `10\ 000 xx (1 + 0.03/12)^4`
  4. `10\ 000 xx (1 + 0.03/12)^48`
Show Answers Only

`D`

Show Worked Solution

`text(Compounding rate)\ = 3/100 ÷ 12= 0.03/12`

`text(Compounding periods)` `= 4 xx 12=48`

 
`:.\ text(FV) = 10\ 000 xx (1 + 0.03/12)^48`

\(\Rightarrow D\)

Measurement, STD2 M6 2017 HSC 8 MC

The diagram shows a right-angled triangle.
 


 

What is the value of  `theta`, to the nearest minute?

  1. `70°16^{′}`
  2. ` 70°17^{′}`
  3. `70°27^{′}`
  4. `70°28^{′}`
Show Answers Only

`B`

Show Worked Solution
`tan theta` `= text(opp)/text(adj)`
  `= 5.3/1.9`
  `= 2.789…`
COMMENT: An angle that has over 30″ (seconds) is rounded up to the next minute (i.e. rounded up to 70°17′).

 

`:. theta` `= 70.277…^@`
  `=70°16^{′}39.8^{″}`
  `= 70^@17^{′}`

 
`=>B`

Statistics, STD2 S1 2017 HSC 4 MC

A factory’s quality control department has tested every 50th item produced for possible defects.

What type of sampling has been used?

A.     Random

B.     Stratified

C.     Systematic

D.     Numerical

Show Answers Only

`C`

Show Worked Solution

`text(A systematic sample divides a population)`

`text(into equal sample sizes and then selects)`

`text(equally among them.)`

`=> C`