SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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 13a

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

When  `x = 2`  the velocity of the particle is 4.

When  `x = 5`  the velocity of the particle is 3.

Find the period of the motion.  (3 marks)

Show Answers Only

`2sqrt3pi`

Show Worked Solution

`text(S)text(ince motion centred at origin,)`

♦ Mean mark 42%.
`{:d/(dx):}^(1/2 v^2)` `= −n^2x`
`1/2 v^2`  `= int −n^2x\ dx`
  `= −1/2n^2x^2 + c`
 `v^2` `= c – n^2x^2`

 

`text(When)\ v = 4, x = 2`

`16 = c – 4n^2\ …\ (1)`

`text(When)\ v = 3, x = 5`

`9 = c – 25n^2\ …\ (2)`

`text(Subtract)\ (1) – (2)`

`7` `= 21n^2`
`n^2` `= 1/3`
`n` `= 1/sqrt3`
`:.\ text(Period)` `= (2pi)/n`
  `= 2sqrt3pi`

Plane Geometry, 2UA 2017 HSC 16c

In the triangle `ABC`, the point `M` is the mid-point of `BC`. The point `D` lies on `AB` and

`BD = DA + AC`.

The line that passes through the point `C` and is parallel to `MD` meets `BA` produced at `E`.

Copy or trace this diagram into your writing booklet.

  1. Prove that `Delta ACE` is isosceles.  (3 marks)
  2. The point `F` is chosen on `BC` so that `AF` is parallel to `DM`.
    `qquad` 
    Show that `AF` bisects `/_ BAC`.  (2 marks)

 

 

 

 

 

 

 

 

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

`text(Prove)\ Delta ACE\ text(is isosceles)`

♦♦♦ Mean mark 21%.

`text(S) text(ince)\ DM text(||) EC,`

`∠BDM = ∠BEC\ \ text{(corresponding angles)}`

`∠DBM\ \ text(is common)`

`:. Delta BDM\ text(|||)\ Delta BEC\ \ text{(AAA)}`

`text(Using ratios of similar)\ Delta text(s):`

`(BM)/(BD)` `= (BC)/(BE)`
`(BM)/(BC)` `= (BD)/(BE) = 1/2\ \ text{(}M\ text(is midpoint of)\ BC text{)}`

 

`=> D\ text(is the midpoint of)\ BE.`

`=> BD = DE`

`AE` `= DE – DA`
`AC` `= BD – DA\ text{(given)}`
  `= DE – DA`
`:.  AE` `= AC`

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

 

(ii)  

`text(Show)\ AF\ text(bisects)\ /_ BAC`

♦♦ Mean mark 31%.

`/_ BAF = /_ AEC\ \ \ text{(corresponding angles)}`

`text{From part (i)}`

`/_ AEC` `= /_ ECA\ \ \ text{(}Delta AEC\ text{is isosceles)}`
`/_ FAC` `= /_ ECA\ \ \ text{(alternate angles)}`
`:. /_ BAF` `= /_ FAC`

`:. AF\ text(bisects)\  /_ BAC\ \ \ text(… as required)`

Financial Maths, 2ADV M1 2017 HSC 16b

A geometric series has first term  `a`  and limiting sum 2.

Find all possible values for  `a`.  (3 marks)

Show Answers Only

`0 < a < 4`

Show Worked Solution

`S_oo = a/(1 – r)`

♦ Mean mark 39%.

`text(If)\ \ S_oo = 2:`

`2` `= a/(1 – r)`
`2(1 – r)` `= a`
`1 – r` `= a/2`
`r` `= 1 – a/2`

 

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

`|1 – a/2| < 1`

`1 – a/2` `< 1` `or qquad -(1 – a/2)` `< 1`
`a/2` `> 0` `a/2` `< 2`
`a` `> 0` `a` `< 4`

 

`:. 0 < a < 4`

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)}`

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)`

Calculus, 2ADV C3 2017 HSC 9 MC

The graph of  `y = f^{′}(x)`  is shown.
 

The curve  `y = f (x)`  has a maximum value of 12.

What is the equation of the curve  `y = f (x)`?

  1. `y = x^2-4x + 12`
  2. `y = 4 + 4x-x^2`
  3. `y = 8 + 4x-x^2`
  4. `y = x^2-4x + 16`
Show Answers Only

`C`

Show Worked Solution

`text(Find the equation of)\ \ f^{′}(x):`

`m = -2,\ \ y text(-int) = 4`

`y = -2x + 4`

`f(x)` `= int -2x + 4\ dx`
  `= -x^2 + 4x + c`

 

`text(Maximum)\ \ f(x) = 12\ \ text(when)\ \ f^{′}(x) = 0:`

`-2x + 4` `= 0`
`x` `= 2`

 
`text(Substitute)\ \ x=2\ \ text(into)\ \ f(x):`

`:. 12` `= -2^2 + 4 ⋅ 2 + c`
`c` `= 8`

 

`:. f(x) = 8 + 4x-x^2`

`=>  C`

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`

Measurement, STD2 M6 2017 HSC 30c

The diagram shows the location of three schools. School `A` is 5 km due north of school `B`, school `C` is 13 km from school `B` and `angleABC` is 135°.
 


 

  1. Calculate the shortest distance from school `A` to school `C`, to the nearest kilometre.  (2 marks)

  2. Determine the bearing of school `C` from school `A`, to the nearest degree.  (3 marks)

Show Answers Only
  1. `17\ text{km  (nearest km)}`
  2. `213^@`
Show Worked Solution

i.   `text(Using cosine rule:)`

`AC^2` `= AB^2 + BC^2 – 2 xx AB xx BC xx cos135^@`
  `= 5^2 + 13^2 – 2 xx 5 xx 13 xx cos135^@`
  `= 285.923…`
`:. AC` `= 16.909…`
  `= 17\ text{km  (nearest km)}`

 

ii.   

`text(Using sine rule, find)\ angleBAC:`

♦♦ Mean mark 31%.
`(sin angleBAC)/13` `= (sin 135^@)/17`
`sin angleBAC` `= (13 xx sin 135^@)/17`
  `= 0.5407…`
`angleBAC` `= 32.7^@`

 

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

`= 180 + 32.7`

`= 212.7^@`

`= 213^@`

Statistics, STD2 S1 2017 HSC 30a

A set of data has a lower quartile (`Q_L`) of 10 and an upper quartile (`Q_U`) of 16.

What is the maximum possible range for this set of data if there are no outliers?  (2 marks)

Show Answers Only

`24`

Show Worked Solution

`IQR = 16 – 10 = 6`

♦♦ Mean mark 34%.

`text(If no outliers,)`

`text(Upper limit)` `= Q_U + 1.5 xx IQR`
  `= 16 + 1.5 xx 6`
  `= 25`
`text(Lower limit)` `= Q_L – 1.5 xx IQR`
  `= 10 – 1.5 xx 6`
  `= 1`

 

`:.\ text(Maximum range)` `= 25 – 1`
  `= 24`

Statistics, STD2 S5 2017 HSC 29d

All the students in a class of 30 did a test.

The marks, out of 10, are shown in the dot plot.
 


 

  1. Find the median test mark.  (1 mark)

  2. The mean test mark is 5.4. The standard deviation of the test marks is 4.22.

     

    Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean.  (2 marks)

  3. A student states that for any data set, 68% of the scores should lie within one standard deviation of the mean. With reference to the dot plot, explain why the student’s statement is NOT relevant in this context.  (1 mark)

Show Answers Only
  1. `6`
  2. `text(43%)`
  3. `text(The statement assumes the data is normally)`
    `text(distributed which is not the case here.)`
Show Worked Solution
♦ Mean mark 50%.
i.    `text(Median)` `= text(15th + 16th score)/2`
    `= (4 + 8)/2`
    `= 6`

 

ii.   `text(Lower limit) = 5.4 – 4.22 = 1.18`

♦♦ Mean mark 34%.

`text(Upper limit) = 5.4 + 4.22 = 9.62`

`:.\ text(Percentage in between)`

`= 13/30 xx 100`

`= 43.33…`

`= 43text{%  (nearest %)}`

 

iii.   `text(The statement assumes the data is normally)`

♦♦♦ Mean mark 13%.

`text(distributed which is not the case here.)`

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.)`

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(C)text(ost)` `= 140 xx $2 + $500`
    `= $780`

 

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

`= 1960-780`

`= $1180`

Financial Maths, STD2 F4 2017 HSC 28c

Michelle borrows $100 000. The interest rate charged is 12% per annum compounded monthly. The monthly payment is $1029 and the first repayment is made after one month.

What is the amount outstanding immediately after the SECOND monthly repayment is made?  (3 marks)

Show Answers Only

`$99\ 941.71`

Show Worked Solution
`text(Interest per month)` `= text(12%)/12= 1text(%)`

`text(Let)\ \ A=\ text(amount owing)`

♦ Mean mark 41%.

 
`text(After 1st repayment:)`

`A_1` `= (100\ 000 + text(1%) xx 100\ 000) – 1029`
  `= $99\ 971`

 
`text(After 2nd repayment:)`

`A_2` `= (99\ 971 + text(1%) xx 99\ 971) – 1029`
  `= $99\ 941.71`

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, 2UG 2017 HSC 28a

Temperature can be measured in degrees Celsius (`C`) or degrees Fahrenheit (`F`).

The two temperature scales are related by the equation  `F = (9C)/5 + 32`.

  1. Calculate the temperature in degrees Fahrenheit when it is  −20 degrees Celsius.  (1 mark)
  2. Solve the following equations simultaneously, using either the substitution method or the elimination method.  (2 marks)
    `qquadF = (9C)/5 + 32`

    `qquadF = C`

     

  3. The graphs of  `F = (9C)/5 + 32`  and  `F = C`  are shown below.
  4.  


  5. What does the result from part (ii) mean in the context of the graph?  (1 mark)     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Show Answers Only
  1. `−4^@F`
  2. `C = −40, F = −40`
  3. `text(It means the two graphs intersect)`
    `text(at)\ (−40,−40).`

 

 

Show Worked Solution
(i)   `F` `= (9(−20))/5 + 32`
    `= −4^@F`

 

(ii)   `F` `= (9C)/5 + 32` `…\ (1)`
  `F` `= C` `…\ (2)`

 

♦♦ Mean mark 31%.
MARKER’S COMMENT: An area that requires attention.

`text(Substitute)\ \ F = C\ \ text{from (2) into (1)}`

`C` `= (9C)/5 + 32`
`(9C)/5 – C` `= −32`
`(4C)/5 – C` `= −32`
`C` `= −32 xx 5/4 = −40`

 

`text{From (2),}`

`F = −40`

`text{(i.e. when}\ C = −40, F = −40)`

♦♦♦ Mean mark 20%.

 

(iii)   `text(It means the two graphs intersect)`

`text{at (−40,−40).}`

Algebra, 2UG 2017 HSC 30b

The cost of a jewellery box varies directly with the cube of its height.

A jewellery box with a height of 10 cm costs $50.

Calculate the cost of a jewellery box with a height of 12 cm.  (2 marks)

Show Answers Only

`$86.40`

Show Worked Solution

`C\ prop\ h^3`

♦ Mean mark 49%.

`C = kh^3`

`text(When)\ h = 10, C = 50,`

`:. k = 50/(10^3) = 0.05`

 

`text(Find)\ \ C\ \ text(when)\ \ h = 12,`

`C` `= 0.05 xx 12^3`
  `= $86.40`

Measurement, 2UG 2017 HSC 27d

Island `A` and island `B` are both on the equator. Island `B` is west of island `A`. The longitude of island `A` is 5°E and the angle at the centre of Earth (`O`), between `A` and `B`, is 30°.
 


 

  1. What is the longitude of island `B`?  (1 mark)
  2. What time is it on island `B` when it is 10 am on island `A`?  (1 mark)
  3. A ship leaves island `A` and travels west along the equator to island `B`. It travels at a constant speed of 40 km/h.
    How long will the ship take to arrive at island `B`? Give your answer in days and hours to the nearest hour.  (3 marks)

 

 

Show Answers Only
  1. `25^@\ text(W)`
  2. `8\ text(am)`
  3. `3\ text(days 12 hours)`
Show Worked Solution
(i)    `text{Longitude (island}\ B)` `= 5 – 30`
    `= −25`
    `= 25^@\ text(W)`

 

(ii)   `text(Time difference) = 30^@ ÷ 15^@ = 2\ text(hours)`

♦ Mean marks of 40% and 45% for parts (i) and (ii) respectively.

`text(S)text(ince)\ B\ text(is west of)\ A,`

`text(Time on island)\ B` `= 10\ text(am less 2 hours)`
  `= 8\ text(am)`

 

(iii)    `text(Arc Distance)` `= theta/360 xx 2pir`
    `= 30/360 xx 2pi xx 6400`
    `= 3351.032…`

♦ Mean mark part (iii) 38%.
`:.\ text(Time to arrive)` `= (3351.032…)/40`
  `= 83.77…\ text(hours)`
  `= 3\ text(days 12 hours)`

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`

FS Comm, 2UG 2017 HSC 27b

How many 20 megabyte files can fit on a 3 terabyte external hard disc?  (2 marks)

Show Answers Only

`157\ 286\ text(files)`

Show Worked Solution
`text(3 TB)` `= 3 xx 2^10 xx 2^10\ text(MB)`
  `= 3\ 145\ 728\ text(MB)`

 

`:.\ text(Number of complete files that can fit)`

♦ Mean mark 49%.

`= (3\ 145\ 728)/20`

`= 157\ 286.4`

`= 157\ 286\ text(files)`

Statistics, STD2 S1 2017 HSC 27a

Jamal surveyed eight households in his street. He asked them how many kilolitres (kL) of water they used in the last year. Here are the results.

`220, 105, 101, 450, 37, 338, 151, 205`

  1. Calculate the mean of this set of data.  (1 mark)

  2. What is the standard deviation of this set of data, correct to one decimal place?  (1 mark)

Show Answers Only
  1. `200.875`
  2. `127.4\ \ text{(1 d.p.)}`
Show Worked Solution
i.   `text(Mean)` `= (220 + 105 + 101 + 450 + 37 + 338 + 151 + 205) ÷ 8`
    `= 200.875`
♦ Mean mark part (ii) 47%.
IMPORTANT: The population standard deviation is required here.

 

ii.   `text(Std Dev)` `= 127.357…\ \ text{(by calc)}`
    `= 127.4\ \ text{(1 d.p.)}`

Measurement, STD2 M1 2017 HSC 25 MC

In the circle, centre `O`, the area of the quadrant is 100 cm².
 


 

What is the arc length `l`, correct to one decimal place?

  1. 8.9 cm
  2. 11.3 cm
  3. 17.7 cm
  4. 25.1 cm
Show Answers Only

`C`

Show Worked Solution

`text(Find)\ r:`

♦ Mean mark 44%.
`text(Area)` `= 1/4 pir^2`
`100` `= 1/4 pir^2`
`r^2` `= 400/pi`
`:. r` `= 11.283…\ text(cm)`

 

`text(Arc length)` `= theta/360 xx 2pir`
  `= 90/360 xx 2pi xx 11.283…`
  `= 17.724…`
  `= 17.7\ text(cm)`

 
`=> C`

Measurement, STD2 M1 2017 HSC 21 MC

The length of a netball court is measured to be 30.50 metres, correct to the nearest centimetre.

What is the lower limit for the length of the netball court?

  1. 30.45 m
  2. 30.49 m
  3. 30.495 m
  4. 30.499 m
Show Answers Only

`C`

Show Worked Solution
`text(A)text(bsolute error)` `= 0.5\ text(cm)`
  `= 0.005\ text(m)`

 

`:.\ text(Lower limit)` `= 30.50-0.005`
  `= 30.495\ text(m)`

 
`=>C`

Algebra, STD2 A2 2017 HSC 20 MC

A pentagon is created using matches.

By adding more matches, a row of two pentagons is formed.

Continuing to add matches, a row of three pentagons can be formed.

Continuing this pattern, what is the maximum number of complete pentagons that can be formed if 100 matches in total are available?

A.     `25`

B.     `24`

C.     `21`

D.     `20`

Show Answers Only

`=>\ text(B)`

Show Worked Solution

`text(1 pentagon: 5 matches)`

`text(2 pentagons: 5 + 4 = 9)`

`text(3 pentagons: 5 + 4 × 2 = 13)`

`vdots`

`n\ text(pentagons:)\ 5 + 4(n – 1)`

`5 + 4(n – 1)` `= 100`
`4n – 4` `= 95`
`4n` `= 99`
`n` `= 24.75`

 

`:.\ text(Complete pentagons possible = 24)`

`=>\ text(B)`

Probability, STD2 S2 2017 HSC 15 MC

The faces on a twenty-sided die are labelled  $0.05, $0.10, $0.15, … , $1.00.

The die is rolled once.

What is the probability that the amount showing on the upper face is more than 50 cents but less than 80 cents?

A.     `1/4`

B.     `3/10`

C.     `7/20`

D.     `1/2`

Show Answers Only

`A`

Show Worked Solution

`text(Possible faces that satisfy are:)`

♦ Mean mark 50%.

`55text(c),60text(c),65text(c),70text(c),75text(c)`

`:.\ text(Probability)` `= 5/20`
  `= 1/4`

`=>A`

Statistics, NAP-J2-13

Ms Richards asked her class a question and recorded the results in the table below.
 


 

Which question could Ms Richards have asked?

 
 What type of farm animals do you own?
 
 How many farm animals do you have?
 
 What month was your farm animal born?
 
 How old are your farm animals?
Show Answers Only

`text(What type of farm animals do you own?)`

Show Worked Solution

`text(What type of farm animals do you own?)`

Number and Algebra, NAP-J2-11 SA

A school teacher allocates pieces of cardboard to class groups depending on the number of students in each group.

The table below is used.
 

 
Using the pattern in the table, how many pieces of cardboard should a group of 3 students receive?
Show Answers Only

`9`

Show Worked Solution

`text(The pattern shows that each student receives)`

`text(3 pieces of cardboard.)`

`:.\ text(A group of 3 will be given 9 pieces.)`

Quadratic, 2UA SM-Bank 10

Solve  `3e^t = 5 + 8e^(−t)`  for `t`.  (3 marks)

Show Answers Only

`log_e(8/3)`

Show Worked Solution

`3e^t – 5 – 8e^(−t) = 0`

`text(Multiply both sides by)\ e^t,`

`3e^(2t) – 5e^t – 8 = 0`

`text(Let)\ y = e^t`

`3y^2 – 5y – 8` `= 0`
`(3y – 8)(y + 1)` `= 0`

 

`y` `=8/3\ quadquadquadquadquad text(or)\ \ \ \ ` `y` `=-1\ \ text{(No soln)}`
`e^t` `= 8/3`  
`:. t` `= log_e(8/3)\ \ \ `    

Quadratic, 2UA SM-Bank 05

Solve the equation  `4^x - 15 × 2^x = 16`  for `x.`  (3 marks)

Show Answers Only

`x = 4`

Show Worked Solution
`4^x – 15 xx 2^x – 16` `= 0`
`2^(2x) – 15 xx 2^x – 16` `= 0`

 

COMMENT: Understand why this working is incorrect: if  `y=2^x`, then `2y=4^x`.

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

`y^2 – 15y – 16` `= 0`
`(y – 16) (y + 1)` `= 0`
`y` `= 16` `\ \ \ or\ \ \ ` `y` `= – 1`
`2^x` `= 16`   `2^x` `= – 1`
`:. x` `= 4`   `text(No solution)`

L&E, 2ADV E1 SM-Bank 6 MC

The expression

`log_c(a) + log_a(b) + log_b(c)`

is equal to

  1. `1/(log_c(a)) + 1/(log_a(b)) + 1/(log_b(c))`
  2. `1/(log_a(c)) + 1/(log_b(a)) + 1/(log_c(b))`
  3. `-1/(log_a(b))-1/(log_b(c))-1/(log_c(a))`
  4. `1/(log_a(a)) + 1/(log_b(b)) + 1/(log_c(c))`
Show Answers Only

`B`

Show Worked Solution

`text(Solution 1)`

`text(Using Change of Base:)`

`log_c(a) + log_a(b) + log_b(c)`

`=(log_a(a))/(log_a(c)) + (log_b(b))/(log_b(a)) + (log_c(c))/(log_c(b))`

`=1/(log_a(c)) + 1/(log_b(a)) + 1/(log_c(b))`

 
`=> B`

 

`text(Solution 2)`

`text(Let)\ \ x` `=log_c(a)`
`c^x` `=a`
`x log_a c` `=log_a a`
`x` `=1/log_a c`

 

`text(Apply similarly for the other terms.)`

`=> B`

GEOMETRY, FUR1 SM-Bank 35 MC

Kim lives in Perth (32°S, 115°E). He wants to watch an ice hockey game being played in Toronto  (44°N, 80°W)  starting at 10.00 pm on Wednesday.

What is the time in Perth when the game starts?

A.   9.00 am on Wednesday

B.   7.40 pm on Wednesday 

C.   9.00 pm on Wednesday

D.   12.20 am on Thursday

E.   11.00 am on Thursday

Show Answers Only

`E`

Show Worked Solution

`text(Perth is East of Toronto) =>\ text(Ahead)`

`text(Longitudinal difference)`

`= 115 + 80`

`= 195^@`
 

`text(Time difference)` `= 195/15`
  `= 13\ text(hours)`

 

`:.\ text(Time in Perth)`

`=\ text{10 pm (Wed) + 13 hours}`

`=\ text(11 am on Thursday)`
 

`=>  E`

GEOMETRY, FUR2 SM-Bank 26

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

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

Show Answers Only

`4468\ text{km   (nearest km)}`

Show Worked Solution

2UG 2015 23c Answer

`text(Distance between two cities)`

`=\ text(Arc length)\  AB`

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

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

`= 4468.04…`

`= 4468\ text{km  (nearest km)}`

GEOMETRY, FUR2 SM-Bank 15

Osaka is at  34°N, 135°E, and Denver is at  40°N, 105°W. 

  1. Show that there is a 16-hour time difference between the two cities.
    (Ignore time zones.)    (1 mark)
  2. John lives in Denver and wants to ring a friend in Osaka. In Denver it is 9 pm Monday.     

     

    What time and day is it in Osaka then?   (1 mark)

  3. John’s friend in Osaka sent him a text message which happened to take 14 hours to reach him. It was sent at 10 am Thursday, Osaka time.

     

    What was the time and day in Denver when John received the text?    (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `1\ text(pm Tuesday)`
  3. `8\ text(am Thursday)`
Show Worked Solution
a.   `text(Longitude difference)` `= 135 + 105`
    `= 240^@`

 

`text(Time difference)` `= 240/15`
  `= 16\ text(hours   … as required)`

 

b.  `text(Denver is behind Osaka time because it is further west.)`

`:.\ text(Time in Osaka)` `= 9\ text(pm Monday plus 16 hours)`
  `= 1\ text(pm Tuesday)`

 

c.  `text(Denver is 16 hours behind Osaka)`

`:.\ text(John will receive the text at 10 am Thursday less 16)`

`text{plus 14 hours (i.e. 8 am Thursday.)}`

GEOMETRY, FUR2 SM-Bank 14

Pontianak has a longitude of 109°E, and Jarvis Island has a longitude of 160°W.

Both places lie on the Equator

  1. Find the shortest great circle distance between these two places, to the nearest kilometre. You may assume that the radius of the Earth is 6400 km.    (2 marks)
  2. The position of Rabaul is 4° to the south and 48° to the west of Jarvis Island. What is the latitude and longitude of Rabaul?     (1 mark)
Show Answers Only
  1. `10\ 165\ text(km)\ \ \ text{(nearest km)}`
  2. `152^@ text(E)`
Show Worked Solution
a.   `text(Longitude difference)` `= 109 + 160`
    `= 269^@`

 

`=> text(Shortest distance)\ text{(by degree)}` `= 360 – 269`
  `= 91^@`

 

`:.\ text(Shortest distance)` `= 91/360 xx 2 pi r`
  `= 91/360 xx 2 xx pi xx 6400`
  `= 10\ 164.79…`
  `=10\ 165\ text(km)\ text{(nearest km)}`

 

b.   `text(Latitude)`
  `4^@\ text(South of Jarvis Island)`
  `text(S)text(ince Jarvis Island is on equator)`
  `=>\ text(Latitude is)\ 4^@ text(S)`
   
  `text(Longitude)`
  `text(Jarvis Island is)\ 160^@ text(W)`
  `text(Rubail is)\ 48^@\ text(West of Jarvis Island, or 208° West)`
  `text(which is)\ 28^@\ text{past meridian (180°)}`

 

`=>\ text(Longitude)` `= (180\ -28)^@ text(E)`
  `= 152^@ text(E)`

 

`:.\ text(Position is)\ (4^@text{S}, 152^@text{E})`

Algebra, MET2 2007 VCAA 17 MC

The function  `f`  satisfies the functional equation  `f (f (x)) = x`  for the maximal domain of  `f.`

The rule for the function is

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

`E`

Show Worked Solution

`text(Solution 1)`

`text(Define each specific function)`

♦ Mean mark 47%.

`(text{i.e. define}\ \ f(x) = (x + 1)/(x – 1))`

`f(f(x)) = x\ \ text(when)\ \ f(x) = (x + 1)/(x – 1)`

`=>   E`

 

`text(Solution 2)`

`text(Consider)\ \ f(x) = (x + 1)/(x – 1)`

`f(f(x))` `=((x + 1)/(x – 1) +1)/((x + 1)/(x – 1) -1)`
  `=(x+1+x-1)/(x+1-x+1)`
  `=x`

`=>E`

Algebra, MET2 2007 VCAA 5 MC

The simultaneous linear equations

`mx + 12y = 24`

`3x + my = m`

have a unique solution only for

  1. `m = 6 or m = – 6`
  2. `m = 12 or m = 3`
  3. `m in R\ text(\){– 6, 6}`
  4. `m = 2 or m = 1`
  5. `m in R\ text(\){– 12, – 3}`
Show Answers Only

`C`

Show Worked Solution
`mx + 12y` `=24`
`y` `=-m/12 x +2\ \ …\ (1)`
`3x + my` `=m`
`y` `= -3/m x+1\ \ …\ (2)`

 

`text(Unique solution occurs when)\ \ m_1!= m_2,`

♦♦ Mean mark 36%.
`-m/12` `!=-3/m`
`m^2` `!= 36`
`:. m` `!= +- 6`

`=>   C`

Graphs, MET2 2008 VCAA 20 MC

The function  `f: B -> R`  with rule  `f(x) = 4x^3 + 3x^2 + 1`  will have an inverse function for

  1. `B = R`
  2. `B = (1/2, oo)`
  3. `B = (text{−∞}, 1/2]`
  4. `B = (text{−∞}, 1/2)`
  5. `B = [−1/2, oo)`
Show Answers Only

`B`

Show Worked Solution

`text(Inverse exists if)\ \ f(x)\ \ text(is)\ \ 1 – 1:`

 

vcaa-2008-20i

`text(Option B’s domain ensures)\ \ f(x)\ \ text(is)\ \ 1- 1`

`=>   B`

Calculus, MET2 2008 VCAA 19 MC

The graph of a function  `f` is shown below.

VCAA 2008 19mc

The graph of an antiderivative of  `f` could be

VCAA 2008 19mci

VCAA 2008 19mcii 

VCAA 2008 19mciii

Show Answers Only

`B`

Show Worked Solution

`text(S)text(ince)\ \ f(x)>0\ \ text(for all)\ x,\ text(the)`

♦ Mean mark 50%.

`text(antiderivative function has no stationary)`

`text(points.)`

`=>\ text(Only B or E are possibilities.)`

 

`text(Also, the antiderivative function must have)`

`text(an increasing gradient for all)\ x.`

`=>   B`

Algebra, MET2 2008 VCAA 12 MC

Let  `f: R -> R,\ f(x) = e^x + e^(–x).`

For all  `u in R,\ f(2u)`  is equal to

  1. `f(u) + f(-u)`
  2. `2 f(u)`
  3. `(f(u))^2 - 2`
  4. `(f(u))^2`
  5. `(f(u))^2 + 2`
Show Answers Only

`C`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 44%.

`text(Define)\ \ f(x) = e^x + e^-x`

`text(Enter each functional equation)`

`[text(i.e.)\ \ f(2u) = (f(u))^2 – 2]`

`text(until CAS output is “true”)`

`=>   C`

 

`text(Solution 2)`

`f(2u)` `=e^(2u) + e^(-2u)`
`(f(u))^2` `=(e^u + e^(-u))^2`
  `=e^(2u) + 2 + e^(-2u)`
   

`:. f(2u) = (f(u))^2-2`

`=>C`

Graphs, MET2 2008 VCAA 9 MC

The transformation  `T: R^2 -> R^2` with rule

`T([(x), (y)]) = [(4, 0), (0, -2)] [(x), (y)] + [(1), (3)]`

maps the curve with equation  `y = x^3`  to the curve with equation

  1. `y = (-(x - 1)^3)/32 + 3`
  2. `y = (-(4x + 1)^3 + 3)/2`
  3. `y = (-(x + 1)^3)/32 - 3`
  4. `y = ((1 - x)^3)/64 - 3`
  5. `y = ((4x - 1)^3 + 3)/2`
Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ (x′,y′)\ \ text(be the transformation of)\ \ (x,y).`

♦ Mean mark 38%.
`x′` `= 4x + 1`
`x` `=(x′ – 1)/4`
`y′` `= – 2y + 3`
`y` `=(y′ – 3)/(- 2)`

 

`text(Substitute)\ \ x, y\ \ text(into)\ \ y = x^3:`

`(y′ – 3)/(- 2)` `= ((x′ – 1)/4)^3`
`y′ – 3` `= (- 2)/4^3 (x′ – 1)^3`
`y` `= (- (x – 1)^3)/32 + 3`

`=>   A`

Algebra, MET2 2008 VCAA 6 MC

The simultaneous linear equations

`ax + 3y = 0`

`2x + (a + 1)y = 0`

where `a` is a real constant, have infinitely many solutions for

  1. `a in R`
  2. `a in\ text({−3, 2})`
  3. `a in R\ text(\ {−3, 2})`
  4. `a in\ text({−2, 3})`
  5. `a in R\ text(\ {−2, 3})`
Show Answers Only

`B`

Show Worked Solution

`text(Infinite solution:)`

♦ Mean mark 45%.
`m_1` `= m_2` `and` `c_1` `= c_2`
`a/2` `= 3/(a + 1)`   `text(always true as both)`
`a` `= – 3, 2`   `text(lines have)\ y text(-int) = 0`

 
`=>   B`

Calculus, MET2 2008 VCAA 4 MC

If  `int_1^3 f(x)\ dx = 5`, then  `int_1^3 (2f(x) - 3)\ dx`  is equal to

A.   `4`

B.   `5`

C.   `7`

D.   `10`

E.   `16`

Show Answers Only

`A`

Show Worked Solution

`int_1^3 (2 f(x) – 3)\ dx`

♦ Mean mark 49%.
MARKER’S COMMENT: Over one third of students did not integrate –3.

`= 2 int_1^3 f(x)\ dx + int_1^3 (– 3)\ dx`

`=2(5) +[-3x]_1^3`

`= 10 + (– 6)`

`= 4`

 
`=>   A`

Calculus, MET1 SM-Bank 28

The function  `f`  has the rule  `f(x) = 1 + 2 cos x`.

  1. Show that the graph of  `y = f(x)`  cuts the `x`-axis at  `x = (2 pi)/3`.   (1 mark)
  2. Sketch the graph  `y = f(x)`  for  `x  in [-pi,pi]`  showing where the graph cuts each of the axes.   (2 marks)
  3. Find the area under the curve  `y = f(x)`  between  `x = -pi/2`  and  `x = (2 pi)/3`.   (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}` 
  2.  
       
  3. `((7 pi)/6 + sqrt 3 + 1)\ text(u²)`
Show Worked Solution

a.   `f(x) = 1 + 2 cos x`

`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`

`1 + 2 cos x` `= 0`
`2 cos x` `=-1`
 `cos x` `= -1/2`

 
`:.  x = (2 pi)/3\ …\ text(as required)`

 

b.   2UA HSC 2006 7b

 

c.  `text(Area)` `= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx`
  `= [x + 2 sin x]_(-pi/2)^((2 pi)/3)`
  `= [((2 pi)/3 + 2 sin­ (2 pi)/3) – ((-pi)/2 + 2 sin­ (-pi)/2)]`
  `= ((2 pi)/3 + 2 xx sqrt 3/2) – ((-pi)/2 +2(- 1))`
  `= (2 pi)/3 + sqrt(3) + pi/2 + 2`
  `= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)`

Algebra, MET1 SM-Bank 24

The rule for  `f`  is  `f(x) = e^x − e^(-x)`.

Show that the inverse function is given by

    `f^(-1)(x) = log_e((x + sqrt(x^2 + 4))/2)` (3 marks)

Show Answers Only

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

Show Worked Solution

`y = e^x − e^(-x)`

`text(Inverse: swap)\  x harr y`

`x` `= e^y − 1/(e^y)`
`xe^y` `= e^(2y) − 1`
`e^(2y) − xe^y − 1` `= 0`

 

`text(Let)\ \ A = e^y`

`:.A^2 − xA − 1 = 0`
 

`text(Using the quadratic formula)`

`A` `=(x ± sqrt((-x)^2 − 4 · 1 · (-1)))/(2 · 1)`
  `=(x ± sqrt(x^2 + 4))/2`

 

`text(S)text(ince)\ \ (x – sqrt(x^2 + 4))/2<0\ \ text(and)\ \ e^y>0,`

`:.e^y` `= (x + sqrt(x^2 + 4))/2`
`log_e e^y` ` = log_e((x + sqrt(x^2 + 4))/2)`
`y` `= log_e((x + sqrt(x^2 + 4))/2)`
`:.f^(-1)(x)` `= log_e((x + sqrt(x^2 + 4))/2)\ \  …\ text(as required)`

Algebra, MET1 SM-Bank 23

The function  `f: [0,oo) → R`  with rule  `f(x) = 1/(1 + x^2)`  is drawn below.

Inverse Functions, EXT1 2004 HSC 5b

  1. Copy or trace this diagram into your writing booklet.
  2. On the same set of axes, sketch  `y=f^(-1)(x)`  where  `f^(-1)` is the inverse function of  `f(x)`.  (1 mark)
  3. Find the domain of the inverse  `f^(-1)`.  (1 mark)
  4. Find an expression for  `y = f^(-1)(x)`  in terms of  `x`.  (2 marks)
  5. The graphs of  `y = f(x)`  and  `y = f^(-1)(x)`  meet at exactly one point  `P`.Let  `α`  be the `x`-coordinate of  `P`. Explain why  `α`  is a root of the equation
  6.      `x^3 + x-1 = 0`.  (1 mark)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0 < x ≤ 1`
  3. `y = sqrt((1-x)/x), y > 0`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.  

Inverse Functions, EXT1 2004 HSC 5b Answer

b.   `text(Range of)\ \ f(x):\ (0,1]`

`:.\ text(Domain of)\ \ f^(−1)(x):  (0,1]`

 

c.  `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\  x harr y`

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

 

`:.y = sqrt((1-x)/x), \ \ y >= 0`

 

d.   `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`

`text(i.e. where)`

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

 

`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`

`text(it is a root of)\ \ \ x^3 + x-1 = 0`

Calculus, MET1 SM-Bank 21

The rule for function  `f` is  `f(x) = e^(-x^2)`.  The diagram shows the graph  `y = f(x)`.

 Inverse Functions, EXT1 2010 HSC 3b

The graph has two points of inflection. 

  1. Find the  `x`  coordinates of these points.   (2 marks)
  2. Explain why the domain of  `f(x)`  must be restricted if  `f(x)`  is to have an inverse function.    (1 mark)
  3. Find the rule for the inverse function  `f^(-1)`  if the domain of  `f(x)`  is restricted to  `x ≥ 0`.   (2 marks)
  4. Find the domain for  `f^(-1)`.    (1 mark)
  5. Sketch the curve  `y = f^(-1) (x)`.    (1 mark)
Show Answers Only
  1. `x = +- 1/sqrt2`
  2. `text(There can only be 1 value of)\ y\ text(for each value of)\ x.`
  3. `f^(-1)x = sqrt(ln(1/x))`
  4. `0 <= x <= 1`
  5. Inverse Functions, EXT1 2010 HSC 3b Answer

Show Worked Solution
a.    `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`

 

b.   `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).`

 

c.  `y = e^(-x^2)`

`text(Inverse: swap)\  x harr y` 

`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))`

 

d.   `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`

 

e. 

Inverse Functions, EXT1 2010 HSC 3b Answer

Calculus, MET1 SM-Bank 1

Part of the graph of a function  `f: R→R, \ f(x) = 2/x`  is shown below.

The diagram shows the area under  `f(x)`  from  `x = 1`  to  `x = d`.

What value of `d` makes the shaded area equal to 2?  (2 marks)

 

Show Answers Only

`e`

Show Worked Solution

`int_1^d 2/x\ dx = 2`

`[2log_e x]_1^d = 2`

`2log_e d – 2log_e 1` `= 2`
`2log_e d` `= 2`
`log_e d` `= 1`
`:. d` `= e`

Algebra, MET1 SM-Bank 10

Solve the equation  `log_e x-3/log_ex=2`  for `x`.   (3 marks)

Show Answers Only

`x=e^3\ \ text(or)\ \ e^-1`

Show Worked Solution
 
IMPORTANT: Students should recognise this equation as a quadratic, and substitute `log_ex` with a variable such as `X`.
`log_e x-3/(log_ex)` `=2`
`(log_ex)^2-3` `=2log_e x`
`(log_ex)^2-2log_ex-3` `=0`
   
`text(Let)\  X=log_ex`  
`:.\ X^2-2X-3` `=0`
`(X-3)(X+1)` `=0`
COMMENT: Must know how to find `x` from the equations  `log_ex=-1`  and  `log_ex=3`.
`X` `=3` `\ \ \ \ \ \ \ \ \ \ ` `X` `=-1`
`log_ex` `=3` `\ \ \ \ \ \ \ \ \ \ ` `log_ex` `=-1`
`x` `=e^3` `\ \ \ \ \ \ \ \ \ \ ` `x` `=e^-1`

 

`:.x=e^3\ \ text(or)\ \ e^-1`