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Calculus, EXT1* C1 2015 HSC 15a

The amount of caffeine, `C`, in the human body decreases according to the equation

`(dC)/(dt) = -0.14C,` 

where `C` is measured in mg and `t` is the time in hours.

  1. Show that  `C = Ae^(-0.14t)`  is a solution to  `(dC)/(dt) = -0.14C,` where ` A` is a constant.

     

    When `t = 0`, there are 130 mg of caffeine in Lee’s body.  (1 mark)

  2. Find the value of `A.`  (1 mark)

  3. What is the amount of caffeine in Lee’s body after 7 hours?   (1 mark)

  4. What is the time taken for the amount of caffeine in Lee’s body to halve?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `130`
  3. `48.8\ text(mg)`
  4. `4.95\ text(hours)`
Show Worked Solution
i. `C` `= Ae^(-0.14t)`
  `(dC)/(dt)` `= d/(dt) (Ae^(-0.14t))`
    `= -0.14 xx Ae^(-0.14t)`
    `= -0.14\ C`

 
`:.\ C = Ae^(-0.14t)\ \ text(is a solution)`

 

ii.  `text(When)\ \ t = 0,\ C = 130`

`130` `= Ae^(-0.14 xx 0)`
`:.\ A` `= 130`

 

iii.  `text(Find)\ \ C\ \ text(when)\ \ t = 7`

`C` `= 130\ e^(-0.14 xx 7)`
  `= 130\ e^(-0.98)`
  `= 48.79…`
  `= 48.8\ text{mg  (to 1 d.p.)}`

 
`:.\ text(After 7 hours, Lee will have 48.8 mg)`

`text(of caffeine left in her body.)`

 

iv.  `text(Find)\ \ t\ \ text(when caffeine has halved.)`

`text(When)\ \ t = 0,\ \ C = 130`

`:.\ text(Find)\ \ t\ \ text(when)\ \ C = 65`

`65` `= 130 e^(-0.14 xx t)`
`e^(-0.14t)` `= 65/130`
`ln e^(-0.14t)` `= ln\ 65/130`
`-0.14t xx ln e` `= ln\ 65/130`
`t` `= (ln\ 65/130)/-0.14`
  `= 4.951…`
  `= 4.95\ text{hours  (to 2 d.p.)}`

 

`:.\ text(It will take 4.95 hours for Lee’s)`

`text(caffeine to halve.)`

Financial Maths, 2ADV M1 2015 HSC 14c

Sam borrows $100 000 to be repaid at a reducible interest rate of 0.6% per month. Let  `$A_n`  be the amount owing at the end of  `n`  months and  `$M`  be the monthly repayment.

  1. Show that  `A_2 = 100\ 000 (1.006)^2 - M (1 + 1.006).`  (1 mark)

  2. Show that  `A_n = 100\ 000 (1.006)^n - M (((1.006)^n - 1)/0.006).`  (2 marks)

  3. Sam makes monthly repayments of $780. Show that after making 120 monthly repayments the amount owing is $68 500 to the nearest $100.  (1 mark)

Immediately after making the 120th repayment, Sam makes a one-off payment, reducing the amount owing to $48 500. The interest rate and monthly repayment remain unchanged.

  1. After how many more months will the amount owing be completely repaid?  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(Show Worked Solutions)}`
  2. `text(Proof)\ \ text{(Show Worked Solutions)}`
  3. `text(Proof)\ \ text{(Show Worked Solutions)}`
  4. `79\ text(months)`
Show Worked Solution
i.  `A_1` `= 100\ 000 (1.006) – M`
`A_2` `= A_1 (1.006) – M`
  `= [100\ 000 (1.006) – M] (1.006) – M`
  `= 100\ 000 (1.006)^2 – M (1.006) – M`
  `= 100\ 000 (1.006)^2 – M (1 + 1.006)\ \ text(…  as required)`

 

ii.  `A_3 = 100\ 000 (1.006)^3 – M (1 + 1.006 + 1.006^2)`

`vdots`

`A_n = 100\ 000 (1.006)^n – M (1 + 1.006 + … + 1.006^(n-1))`

`=> text(S)text(ince)\ \ (1 + 1.006 + … + 1.006^(n-1))\ text(is a)`

`text(GP with)\ \ a = 1,\ r = 1.006`

`:.\ A_n` `= 100\ 000 (1.006)^n – M ((a (r^n – 1))/(r – 1))`
  `= 100\ 000 (1.006)^n – M ((1 (1.006^n – 1))/(1.006 – 1))`
  `= 100\ 000 (1.006)^n – M (((1.006)^n – 1)/0.006)`

`text(…  as required.)`

 

iii.  `text(If)\ \ M = 780 and n = 120`

`A_120` `= 100\ 000 (1.006)^120 – 780 ((1.006^120 – 1)/0.006)`
  `= 205\ 001.80… – 780 (175.0030…)`
  `= 205\ 001.80… – 136\ 502.34…`
  `= 68\ 499.45…`
  `= $68\ 500\ \ text{(to nearest $100)  …  as required}`

 

iv.  `text(After the one-off payment, amount owing)=$48\ 500`

`:. A_n = 48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006)`

`text(where)\ \ n\ \ text(is the number of months after)`

 `text(the one-off payment.)`
 

`text(Find)\ \ n\ \ text(when)\ \ A_n = 0`

`48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006) = 0`

`48\ 500 (1.006)^n` `= 780 ((1.006^n – 1)/0.006)`
`48\ 500 (1.006)^n` `= 130\ 000 (1.006^n – 1)`
  `= 130\ 000 (1.006)^n – 130\ 000`
`81\ 500 (1.006)^n` `= 130\ 000`
`1.006^n` `= (130\ 000)/(81\ 500)`
`n xx ln 1.006` `= ln\ 1300/815`
`n` `= (ln\ 1300/815)/(ln\ 1.006)`
  `= 78.055…`

 
`:.\ text(The amount owing will be completely repaid after)`

`text(another 79 months.)`

Probability, 2ADV S1 2015 HSC 14b

Weather records for a town suggest that:

  • if a particular day is wet `(W)`, the probability of the next day being dry is  `5/6`
  • if a particular day is dry `(D)`, the probability of the next day being dry is  `1/2`.

In a specific week Thursday is dry. The tree diagram shows the possible outcomes for the next three days: Friday, Saturday and Sunday.
 

2015 2ua 14b
 

  1. Show that the probability of Saturday being dry is `2/3`.  (1 mark)

  2. What is the probability of both Saturday and Sunday being wet?  (2 marks)

  3. What is the probability of at least one of Saturday and Sunday being dry?  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1/(18)`
  3. `(17)/(18)`
Show Worked Solution

i.  `text{Show}\ \ P text{(Sat dry)} = 2/3`

`P text{(Sat dry)}`

`= P (W,D) + P (D, D)`

`=(1/2 xx 5/6) + (1/2 xx 1/2)`

`= 5/(12) + 1/4`

`= 2/3\ \ text(…  as required)`
 

ii.  `Ptext{(Sat and Sun wet)}`

`= P (WWW) + P (DWW)`

`= (1/2 xx 1/6 xx 1/6) + (1/2 xx 1/2 xx 1/6)`

`= 1/(72) + 1/(24)`

`= 1/(18)`
 

iii.  `Ptext{(At least Sat or Sun dry)}`

`= 1 – Ptext{(Sat and Sun both wet)}`

`= 1 – 1/(18)`

`= (17)/(18)`

Calculus, 2ADV C3 2015 HSC 13c

Consider the curve  `y = x^3 − x^2 − x + 3`.

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

  2. Given that the point  `P (1/3, 70/27)`  lies on the curve, prove that there is a point of inflection at  `P`.  (2 marks)

  3. Sketch the curve, labelling the stationary points, point of inflection and `y`-intercept.  (2 marks)

Show Answers Only
  1. `text(MAX at)\ (-1/3, 86/27); \ text(MIN at)\ (1, 2)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3.  
Show Worked Solution
i. `y` `= x^3 – x^2 – x + 3`
  `(dy)/(dx)` `= 3x^2 – 2x – 1`
  `(d^2y)/(dx^2)` `= 6x – 2`

`text(S.P.’s when)\ (dy)/(dx) = 0`

`3x^2 – 2x – 1` `= 0`
`(3x + 1) (x – 1)` `= 0`

`x = -1/3 or 1`

 

`text(When)\ \ x = -1/3`

`f(-1/3)` `= (-1/3)^3 – (-1/3)^2 – (-1/3) + 3`
  `= -1/27 – 1/9 + 1/3 + 3`
  `= 86/27`
`f″(-1/3)` `= (6 xx -1/3) – 2 = -4 < 0`

`:.\ text(MAX at)\ \ (-1/3, 86/27)`

 

`text(When)\ \ x = 1`

`f(1)` `= 1^3 – 1^2 – 1 + 3 =2`
`f″(1)` `= (6 xx 1) – 2 = 4 > 0`

`:.\ text(MIN at)\ \ (1, 2)`

 

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

`6x-2` `=0`
`x` `=1/3`

 

`text(Checking change of concavity)`

`text(Concavity changes either side of)\ x = 1/3`

`:.\ (1/3, 70/27)\ \ text(is a P.I.)`

 

iii.  `text(When)\ \ x` `= 0`
`y` `= 3`

Trigonometry, 2ADV T1 2015 HSC 13a

The diagram shows `Delta ABC` with sides  `AB = 6` cm, `BC = 4` cm  and  `AC = 8` cm.
 

  1. Show that  `cos A = 7/8`.  (1 mark)

  2. By finding the exact value of `sin A`, determine the exact value of the area of  `Delta ABC`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `3 sqrt 15\ \ text(cm²)`
Show Worked Solution

i.  `text(Show)\ cos A = 7/8`

`text(Using the cosine rule)`

`cos A` `= (b^2 + c^2-a^2)/(2bc)`
  `= (8^2 + 6^2-4^2)/(2 xx 8 xx 6)`
  `= (64 + 36-16)/96`
  `= 84/96`
  `= 7/8\ \ text(…  as required)`

 

♦ Mean mark 40%.
ii.    2UA HSC 2015 13ai
`a^2 + 7^2` `= 8^2`
`a^2 + 49` `= 64`
`a^2` `= 15`
`a` `= sqrt 15`
`:.\ sin A` `= (sqrt 15)/8`

 

`:.\ text(Area)\ Delta ABC` `= 1/2 bc\ sin A`
  `= 1/2 xx 8 xx 6 xx (sqrt 15)/8`
  `= 3 sqrt 15\ \ text(cm²)`

Quadratic, 2UA 2015 HSC 12e

The diagram shows the parabola `y = x^2/2` with focus `S (0, 1/2).` A tangent to the parabola is drawn at `P (1, 1/2).`

  1. Find the equation of the tangent at the point `P`.   (2 marks)
  2. What is the equation of the directrix of the parabola?   (1 mark)
  3. The tangent and directrix intersect at `Q`.
    Show that `Q` lies on the `y`-axis.   (1 mark)

  4. Show that `Delta PQS` is isosceles.   (1 mark)
Show Answers Only
  1. `y = x – 1/2`
  2. `y = -1/2`
  3.  
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   

`y = 1/2 x^2`

`(dy)/(dx) = x`
 

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

`text(Equation of tangent,)\ m = 1,\ text(through)\ (1, 1/2):`

`y – y_1` `= m (x – x_1)`
`y – 1/2` `= 1 (x – 1)`
`y – 1/2` `= x – 1`
`y` `= x – 1/2`

 

(ii)  `text(Directrix is)\ \ y = -1/2`
 

(iii)  `Q\ text(is at the intersection of)`

`y = x – 1/2\ \ …\ \ text{(1)}`

`y = -1/2\ \ \ \ …\ \ text{(2)}`

`text{(1) = (2)}`

`x – 1/2` `= -1/2`
`x` `= 0`

 
`:.\ Q\ text(lies on the)\ y text(-axis)\ \ …\ \ text(as required)`

 

(iv)  `text(Show)\ Delta PQS\ text(is isosceles.)`

`text(Distance)\ PS = 1 – 0 = 1`

`Q\ text(has coordinates)\ (0, -1/2)`

`text(Distance)\ SQ = 1/2 + 1/2 = 1`

`:. PS = SQ = 1`

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

Calculus, 2ADV C1 2015 HSC 12c

Find  `f^{′}(x)`, where  `f(x) = (x^2 + 3)/(x-1).`   (2 marks)

Show Answers Only

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

Show Worked Solution

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

`text(Using the quotient rule:)`

`u` `= x^2 + 3` `\ \ \ \ \ \ v` `= x-1`
`u^{′}` `= 2x` `\ \ \ \ \ \ v^{′}` `= 1`
`f^{′}(x)` `= (u^{′} v-uv^{′})/v^2`
  `= (2x (x-1)-(x^2 + 3) xx 1)/(x-1)^2`
  `= (2x^2-2x-x^2-3)/(x-1)^2`
  `= (x^2-2x-3)/(x-1)^2`
  `= ((x-3) (x + 1))/(x-1)^2`

Functions, 2ADV F1 2015 HSC 12b

The diagram shows the rhombus  `OABC`.

The diagonal from the point  `A (7, 11)`  to the point `C` lies on the line `l_1`.

The other diagonal, from the origin `O` to the point `B`, lies on the line `l_2` which has equation  `y = -x/3`.

2015 2ua 12b

  1. Show that the equation of the line  `l_1`  is  `y = 3x - 10`.  (2 marks)
  2. The lines  `l_1`  and  `l_2`  intersect at the point `D`.
  3. Find the coordinates of  `D`.  (2 marks)

 

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

(i)  `text(Show)\ \ l_1\ \ text(is)\ \ y = 3x – 10`

`l_2\ \ text(is)\ \ y = -x/3`

`m_(l_2)` `= -1/3`  
`:.\ m_(l_1)` `= 3\ \ \ ` `text{(diagonals of rhombus}\ OABC\ text(are)`
    `text{perpendicular bisectors.)}`

 
`l_1\ text(has)\ m = 3,\ text(through)\ \ A(7, 11)`

`y – y_1` `= m (x – x_1)`
`y – 11` `= 3 (x – 7)`
`y-11` `= 3x – 21`
`y` `= 3x – 10\ \ text(… as required)`

 

(ii)  `D\ text(occurs at the intersection of)\ \ l_1 and l_2`

`y` `= -1/3\ x` `\ text{…  (1)}`
`y` `= 3x – 10` `\ text{…  (2)}`
`text{(1)}` `=\ text{(2)}`
`-1/3\ x` `= 3x – 10`
`10/3\ x` `= 10`
`:.\ x` `= 3`

 

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

`y` `= -1/3 xx 3`
  `= -1`
`:.\ D\ text{has coordinates  (3, –1)}`

Trigonometry, 2ADV T2 2015 HSC 12a

Find the solutions of  `2 sin theta = 1`  for  `0 <= theta <= 2 pi`.   (2 marks)

Show Answers Only

`pi/6, (5 pi)/6`

Show Worked Solution
`2 sin theta` `= 1,\ \ \ \ 0 <= theta <= 2 pi`
`sin theta` `= 1/2`

`=> sin (pi/6) = 1/2 \ \ text(and sin is positive)`

`text(in the 1st/2nd quadrants)`

`:. theta` `= pi/6, pi-pi/6`
  `= pi/6, (5 pi)/6`

Calculus, 2ADV C2 2015 HSC 11e

Differentiate  `(e^x + x)^5`.  (2 marks)

Show Answers Only

`5 (e^x + 1) (e^x + x)^4`

Show Worked Solution
`y` `= (e^x + x)^5`
 `(dy)/(dx)` `= 5 (e^x + x)^4 xx d/(dx) (e^x + x)`
  `= 5 (e^x + x)^4 xx (e^x + 1)`
  `= 5 (e^x + 1) (e^x + x)^4`

Financial Maths, 2ADV M1 2015 HSC 11d

Find the limiting sum of the geometric series  `1 - 1/4 + 1/16 - 1/64 + …`.  (2 marks)

Show Answers Only

`4/5`

Show Worked Solution

`1 – 1/4 + 1/16 – 1/64 + …`

`r = -1/4,\ \ a=1`

`text(S)text(ince)\ |\ r\ | = 1/4 < 1`

`S_oo` `= a/(1 – r)`
  `= 1/(1 – (-1/4))`
  `= 1/(5/4)`
  `= 4/5`

Probability, 2ADV S1 2015 HSC 4 MC

The probability that Mel’s soccer team wins this weekend is  `5/7`.

The probability that Mel’s rugby league team wins this weekend is  `2/3`.

What is the probability that neither team wins this weekend?

  1. `2/21`
  2. `10/21`
  3. `13/21`
  4. `19/21`
Show Answers Only

`A`

Show Worked Solution

`Ptext{(win at soccer)} = 5/7`

`:. Ptext{(not win at soccer)} = 1 – 5/7 = 2/7`

`Ptext{(win at league)} = 2/3`

`:. Ptext{(not win at league)} = 1/3`

`:. Ptext{(not win at both)}` `= 2/7 xx 1/3`
  `= 2/21`

`=> A`

Functions, 2ADV F1 2015 HSC 2 MC

What is the slope of the line with equation  `2x - 4y + 3 = 0`?

  1. `-2`
  2. `-1/2`
  3. `1/2`
  4. `2`
Show Answers Only

`C`

Show Worked Solution
`2x – 4y + 3` `= 0`
`4y` `= 2x + 3`
`y` `= 1/2 x + 3/4`

`:.\ text(Slope)\ = 1/2`

`=> C`

Quadratic, 2UA 2006 HSC 7c

  1. Write down the discriminant of  `2x^2 + (k - 2)x + 8`  where  `k`  is a constant.  (1 mark)
  2. Hence, or otherwise, find the values of  `k`  for which the parabola  `y = 2x^2 + kx + 9` does not intersect the line  `y = 2x + 1`.  (2 marks)

 

Show Answers Only
  1. `k^2 – 4k – 60`
  2. `-6 < k < 10`
Show Worked Solution

(i)  `2x^2 + (k – 2)x + 8`

`Delta` `= b^2 – 4ac`
  `= (k – 2)^2 – 4 xx 2 xx 8`
  `= k^2 – 4k + 4 – 64`
  `= k^2 – 4k – 60`

 

(ii)  `y` `= 2x^2 + kx + 9` `\ \ text{…  (1)}`
`y` `= 2x + 1` `\ \ text{…  (2)}`

`text(Substitute)\ y = 2x + 1\ text{into (1)}`

`2x + 1 = 2x^2 + kx + 9`

`2x^2 + kx – 2x + 8 = 0`

`2x^2 + (k – 2)x + 8 = 0\ …\ text{(∗)}`

 

`text{The graphs will not intercept if (∗) has}`

`text(no roots, i.e.)\ \ Delta <0`

`k^2 – 4k – 60` `< 0`
`(k – 10) (k + 6)` `< 0`

HSC quadratic

`text(From the graph, no intersection when)`

`-6 < k < 10`

Calculus, EXT1* C1 2006 HSC 6b

A rare species of bird lives only on a remote island. A mathematical model predicts that the bird population, `P`, is given by

`P = 150 + 300 e^(-0.05t)`

where `t` is the number of years after observations began.

  1. According to the model, how many birds were there when observations began?  (1 mark)

  2. According to the model, what will be the rate of change in the bird population ten years after observations began?  (2 marks)

  3. What does the model predict will be the limiting value of the bird population?  (1 mark)

  4. The species will become eligible for inclusion in the endangered species list when the population falls below `200`. When does the model predict that this will occur?  (2 marks)

Show Answers Only
  1. `450\ text(birds)`
  2. `-9.1\ text{(to 1 d.p.)}`
  3. `150\ text(birds)`
  4. `text(After 35.83… years)`
Show Worked Solution

i.  `P = 150 + 300 e^(-0.05t)`

`text(When)\ \ t = 0`

`P` `= 150 + 300 e^0`
  `= 450`

 
`:.\ text(There were 450 birds when observations began.)`
 

ii.  `(dP)/dt` `= 300 xx (-0.05) xx e^(-0.05t)`
  `= -15 e^(-0.05t)`

 
`text(When)\ \ t = 10`

`(dP)/dt` `= -15 e^(-0.05 xx 10)`
  `= -15 e^(-0.5)`
  `= -9.097…`
  `= -9.1\ text{(to 1 d.p.)}`

 

`:.\ text(After 10 years, the bird population will be)`

`text(decreasing at a rate of 9.1 birds per year.)`

 

iii.  `text(As)\ t rarr oo`

`300 e^(-0.05t) rarr 0`

`:. P = 150 + 300 e^(-0.05t) rarr 150`

`:.\ text(The model predicts a limiting population)`

`text(of 150 birds.)`

 

iv.  `text(Find)\ t\ text(when)\ P < 200`

`150 + 300 e^(-0.05t)` `< 200`
`300 e^(-0.05t)` `< 50`
`e^(-0.05t)` `< 50/300`
`ln e^(-0.05t)` `< ln­ 1/6`
`-0.05t` `< ln­ 1/6`
`t` `> (ln­ 1/6)/(-0.05t)`
`t` `> 35.83…`

 

`:.\ text(The model predicts the population)`

`text(will fall below 200 after 35.83… years.)`

Plane Geometry, 2UA 2006 HSC 6a

In the diagram, `AD` is parallel to `BC`, `AC` bisects `/_BAD` and `BD` bisects `/_ABC`. The lines `AC` and `BD` intersect at `P`.

Copy or trace the diagram into your writing booklet.

  1. Prove that `/_BAC = /_BCA`.  (1 mark)
  2. Prove that `Delta ABP ≡ Delta CBP`.  (2 marks)
  3. Prove that `ABCD` is a rhombus.  (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
(i)  

`text(Prove)\ /_BAC = /_BCA`

`/_BCA` `= /_CAD\ \ \ text{(alternate angles,}\ BC\ text(||)\ AD text{)}`
`/_CAD` `= /_BAC\ \ \ text{(}AC\ text(bisects)\ /_BAD text{)}`
`:. /_BAC` `= /_BCA\ …\ text(as required)`

 

(ii)  `text(Prove)\ Delta ABP ≡ Delta CBP`

`/_BAC` `= /_BCA\ \ \ text{(from part (i))}`
`/_ABP` `= /_CBP\ text{(}BD\ text(bisects)\ /_ABC text{)}`
`BP\ text(is common)`

 

`:. Delta ABP ≡ Delta CBP\ \ text{(AAS)}`

 

(iii)  `text(Using)\ Delta ABP ≡ Delta CBP`

`AP = PC\ \ text{(corresponding sides of congruent triangles)}`

`/_BPA = /_BPC\ \ text{(corresponding angles of congruent triangles)}`

`text(Also,)\ /_BPA = /_BPC = 90^@`

`text{(}/_APC\ text{is a straight angle)}`

 

`text(Considering)\ Delta APD and Delta APB`

`/_DAP` `= /_BAP\ text{(}AC\ text(bisects)\ /_BAD text{)}`
`/_DPA` `= 90^@\ text{(vertically opposite angles)}`
`:. /_DPA` `= /_BPA = 90^@`

`PA\ text(is common)`

`:. Delta APD ≡ Delta APB\ \ text{(AAS)}`

`BP = PD\ \ text{(corresponding sides of congruent triangles)}`

 

 `:. ABCD\ text(is a rhombus as its diagonals are)`

`text(perpendicular bisectors.)`

Calculus, 2ADV C4 2006 HSC 5b

  1. Show that `d/dx log_e (cos x) = -tan x.`  (1 mark)


  2.   
     
    2006 5b
     
    The shaded region in the diagram is bounded by the curve  `y =tan x`  and the lines  `y =x`  and  `x = pi/4.`

     

    Using the result of part (i), or otherwise, find the area of the shaded region.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(1/2 log_e 2 – pi^2/32)\ text(u²)`
Show Worked Solution
i.   `d/dx log_e (cos x)` `= (-sin x)/cos x`
  `= – tan x\ …\ text(as required)`

 

ii.  `text(Shaded Area)`

`= int_0^(pi/4) tan x\ dx – int_0^(pi/4) x\ dx`

`= int_0^(pi/4) tan x – x\ dx`

`= [-log_e (cos x) – 1/2 x^2]_0^(pi/4)`

`=[(-log_e (cos­ pi/4) – 1/2 xx (pi^2)/16) – (-log_e(cos 0) – 0)]`

`= -log_e­ 1/sqrt 2 – pi^2/32 + log_e1`

`= -log_e 2^(-1/2) – pi^2/32`

`= (1/2 log_e 2 – pi^2/32)\ text(u²)`

Calculus, 2ADV C3 2004 HSC 4b

Consider the function  `f(x) = x^3 − 3x^2`.

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

  2. Sketch the curve showing where it meets the axes.   (2 marks)

  3. Find the values of  `x`  for which the curve  `y = f(x)`  is concave up.   (2 marks)

Show Answers Only
  1. `text(MAX at)\ (0,0),\ \ text(MIN at)\ (2,-4)`
  2.  
    1. Geometry and Calculus, 2UA 2004 HSC 4b Answer
  3. `f(x)\ text(is concave up when)\ x>1`
Show Worked Solution
(i)    `f(x)` `= x^3 – 3x^2`
  `f'(x)` `= 3x^2 – 6x`
  `f″(x)` `= 6x – 6`

 

`text(S.P.’s  when)\ \ f'(x) = 0`

`3x^2 – 6x` `= 0`
`3x (x – 2)` `= 0`
`x` `= 0\ \ text(or)\ \ 2`

 

`text(When)\ x = 0`

`f(0)` `= 0`
`f″(0)` `= 0 – 6 = -6 < 0`
`:.\ text(MAX at)\ (0,0)`

 

`text(When)\ x = 2`

`f(2)` `= 2^3 – (3 xx 4) = -4`
`f″(2)` `= (6 xx 2) – 6 = 6 > 0`
`:.\ text(MIN at)\ (2, -4)`

 

(ii)   `f(x) = x^3 – 3x^2\ text(meets the)\ x text(-axis when)\ f(x) = 0`
`x^3 – 3x^2` `= 0`
`x^2 (x-3)` `= 0`
`x` `= 0\ \ text(or)\ \ 3`

 Geometry and Calculus, 2UA 2004 HSC 4b Answer

(iii)   `f(x)\ text(is concave up when)`
`f″(x)` `>0`
`6x – 6` `>0`
`6x` `>6`
`x` `>1`

 

`:. f(x)\ text(is concave up when)\ \ x>1`

Trigonometry, 2ADV T1 2004 HSC 3c

Trig Ratios, 2UA 2004 HSC 3c
 

The diagram shows a point  `P`  which is  30 km due west of the point  `Q`.

The point  `R`  is 12 km from  `P`  and has a bearing from  `P`  of  070°. 

  1. Find the distance of  `R`  from  `Q`.   (2 marks)

  2. Find the bearing of  `R`  from  `Q`.   (2 marks)

Show Answers Only
  1. `19.2\ text(km)\ \ \ text{(1 d.p.)}` 
  2. `282^@`
Show Worked Solution

i.   `text(Join)\ \ RQ\ \ text(to form)\ \ Delta RPQ`

Trig Ratios, 2UA 2004 HSC 3c Answer

`/_RPQ = 90 – 70 = 20^@`

`text(Using the cosine rule:)`

`RQ^2` `= PR^2 + PQ^2 – 2 xx PR xx PQ xx cos /_RPQ`
  `= 12^2 + 30^2 – 2 xx 12 xx 30 xx cos 20^@`
  `= 367.421…`
`:.\ RQ` `= 19.168…`
  `= 19.2\ text(km)\ \ text{(1 d.p.)}`

 

ii.   `text(Using sine rule:)`

`(sin /_RQP)/12` `= (sin 20^@)/(19.168…)`
`sin/_RQP` `= (12 xx sin 20^@)/(19.168…)`
  `= 0.214…`
`/_RQP` `= 12.36…^@`
  `= 12^@\ \ \ text{(nearest degree)}`

 

`:.\ text(Bearing of)\ R\ text(from)\ Q`

`=270+12`

`=282^@`

Quadratic, 2UA 2004 HSC 2c

For what values of  `k`  does  `x^2 − kx + 4 = 0`  have no real roots?   (2 marks)

Show Answers Only

`-4 < k < 4`

Show Worked Solution

`x^2 – kx + 4 = 0`

`text(No real roots when)\ Delta < 0`

`b^2 – 4ac` `< 0`
`(text(–) k^2) – 4 xx 1 xx 4` `< 0`
`k^2 – 16` `< 0`
`k^2` `< 16`
`:.\ -4 < k` `< 4`

 

`:.\ text(There are no real roots when)\ \ \ -4 < k < 4.`

Plane Geometry, 2UA 2004 HSC 2b

In the diagram, `ABC`  is an isosceles triangle with  `AB = AC`  and  `/_BAC = 38^@`. The line `BC` is produced to `D`. 

Find the size of `/_ACD`. Give reasons for your answer.   (2 marks)

Show Answers Only

`109^@`

Show Worked Solution

Plane Geometry, 2UA 2004 HSC 2b Answer

`/_ABC` `= 1/2 (180-38)\ \ \ text{(base angle of isosceles}\ Delta ABC text{)}`
  `= 71^@`

 

`:.\ /_ACD` `= 71 + 38\ \ \ text{(exterior angle of}\ Delta ABC text{)}`
  `= 109^@`

Probability, 2ADV S1 2006 HSC 4c

A chessboard has 32 black squares and 32 white squares. Tanya chooses three different squares at random.

  1. What is the probability that Tanya chooses three white squares?  (2 marks)

  2. What is the probability that the three squares Tanya chooses are the same colour?.  (1 mark)

  3. What is the probability that the three squares Tanya chooses are not the same colour?  (1 mark)

Show Answers Only
  1. `5/42`
  2. `5/21`
  3. `16/21`
Show Worked Solution
i.  `text(P)(WWW)` `= 32/64 xx 31/63 xx 30/62`
  `= 5/42`

 

ii.  `text{P(same colour)}`

`= P(WWW) + P(BBB)`

`= 5/42 + 32/64 xx 31/63 xx 30/62`

`= 5/42 + 5/42`

`= 5/21`

 

iii.  `text{P(not all the same colour)}`

`= 1 – text{P(same colour)}`

`= 1 – 5/21`

`= 16/21`

Calculus, EXT1* C3 2006 HSC 4b

2006 4b

In the diagram, the shaded region is bounded by the parabola  `y = x^2 + 1`, the `y`-axis and the line  `y = 5`.

Find the volume of the solid formed when the shaded region is rotated about the `y`-axis.  (3 marks)

Show Answers Only

`8 pi\ \ text(u³)`

Show Worked Solution

`y = x^2 + 1`

`x^2 = y – 1`

`V` `= pi int_1^5 x^2 \ dy`
  `= pi int_1^5 y-1 \ dy`
  `= pi [y^2/2 – y]_1^5`
  `= pi[(25/2 – 5) – (1/2 – 1)]`
  `= pi[15/2 – (-1/2)]`
  `= 8 pi\ \ text(u³)`

Trigonometry, 2ADV T1 2006 HSC 4a

In the diagram, `ABCD` represents a garden. The sector  `BCD`  has centre `B` and  `/_DBC = (5 pi)/6`

The points `A, B` and `C` lie on a straight line and  `AB = AD = 3` metres.

Copy or trace the diagram into your writing booklet.

  1. Show that  `/_DAB = (2 pi)/3.`  (1 mark)

  2. Find the length of  `BD`.  (2 marks)

  3. Find the area of the garden  `ABCD`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `3 sqrt 3\ \ text(m)`
  3. `(9 sqrt 3 + 45 pi) / 4\ \ text(m²)`
Show Worked Solution
i.   

`text(Show)\ /_DAB = (2 pi)/3`

`/_DBA` `= pi – (5 pi)/6\ \ \ text{(π radians in straight angle}\ ABC text{)}`
  `= pi/6\ text(radians)`

 
`:. /_BDA = pi/6\ text(radians)\ \ \ text{(base angles of isosceles}\ Delta ADB text{)}`

`:. /_DAB` `= pi – (pi/6 + pi/6)\ \ \ text{(angle sum of}\ Delta ADB text{)}`
  `= (2 pi)/3\  text(radians … as required)`

 

ii.  `text(Using the cosine rule:)`

`BD^2` `= AD^2 + AB^2 – 2 xx AD xx AB xx cos {:(2 pi)/3`
  `= 9 + 9 – (2 xx 3 xx 3 xx -0.5)`
  `= 27`
`:. BD` `= sqrt 27`
  `= 3 sqrt 3\ \ text(m)`

 

iii.  `text(Area of)\ Delta ADB` `= 1/2 ab sin C`
  `= 1/2 xx 3 xx 3 xx sin{:(2 pi)/3`
  `= 9/2 xx sqrt3/2`
  `= (9 sqrt 3)/4\ \ text(m²)`

 
`text(Area of sector)\ BCD`

`= {(5 pi)/6}/(2 pi) xx pi r^2`

`= (5 pi)/12 xx (3 sqrt 3)^2`

`= (45 pi)/4\ \ text(m²)`

 

`:.\ text(Area of garden)\ ABCD`

`= (9 sqrt 3)/4 + (45 pi)/4`

`= (9 sqrt 3 + 45 pi)/4\ \ text(m²)`

Financial Maths, 2ADV M1 2006 HSC 3c

On the first day of the harvest, an orchard produces 560 kg of fruit. On the next day, the orchard produces 543 kg, and the amount produced continues to decrease by the same amount each day.

  1. How much fruit is produced on the fourteenth day of the harvest?  (2 marks)

  2. What is the total amount of fruit that is produced in the first 14 days of the harvest?  (1 mark)

  3. On what day does the daily production first fall below 60 kg?  (2 marks)

Show Answers Only
  1. `text(339 kg of fruit is produced on 14th day.)`
  2. `text(6293 kg of fruit is produced in the 1st 14 days.)`
  3. `text(On the 31st day, production will first drop below 60 kg.)`
Show Worked Solution

i.   `T_1 = a = 560`

`T_2 = a + d = 543`

`=>\ text(AP where)\ a = 560, d = -17`

`vdots`

`T_14` `= a + 13d`
  `= 560 – (13 xx 17)`
  `= 339`

 

`:.\ text(339 kg of fruit is produced on 14th day.)`

 

ii.  `S_14 = text(total fruit produced in 1st 14 days)`

`S_14` `= n/2 [2a + (n – 1)d]`
  `= 14/2 [2 xx 560 – (14 – 1) xx 17]`
  `= 7 [1120 – 221]`
  `= 6293`

 

`:.\ text(6293 kg of fruit is produced in the 1st 14th days.)`

 

iii.  `text(Find)\ n\ text(such that)\ T_n < 60`

`T_n = a + (n – 1)d` `< 60`
`560 – 17(n – 1)` `< 60`
`560 – 17n + 17` `< 60`
`17n` `> 517`
`n` `> 30.41…`

 
`:.\ text(On the 31st day, production will first drop)`

`text(below 60 kg.)`

Calculus, 2ADV C3 2006 HSC 2c

Find the equation of the tangent to the curve  `y = cos 2x`  at the point whose `x`-coordinate is  `pi/6`.  (3 marks)

Show Answers Only

`y = – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

Show Worked Solution

`y = cos 2x`

`dy/dx = -2 sin 2x`

`text(When)\ \ x = pi/6`

`y` `= cos (2 xx pi/6)`
  `= cos (pi/3)`
  `= 1/2`

 

`dy / dx` `= -2 sin (pi/3)`
  `= -2 xx sqrt 3 / 2`
  `= – sqrt 3`

 
`text(Equation of tangent,)\ \ m = – sqrt 3, text(through)\ \ (pi/6, 1/2):`

`y – y_1` `= m(x – x_1)`
`y – 1/2` `= – sqrt 3 ( x – pi/6)`
`y – 1/2` `= – sqrt 3 x + (sqrt 3 pi)/6`
`y` `= – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

Linear Functions, 2UA 2006 HSC 3a

In the diagram, `A, B and C` are the points  `(1, 4), (5, –4) and (–3, –1)`  respectively. The line  `AB`  meets the y-axis at `D`.

  1. Show that the equation of the line  `AB`  is  `2x + y - 6 = 0`.  (2 marks)
  2. Find the coordinates of the point `D`.  (1 mark)
  3. Find the perpendicular distance of the point `C` from the line  `AB`.  (1 mark)
  4. Hence, or otherwise, find the area of the triangle  `ADC`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `D(0, 6)`
  3. `13/sqrt 5\ text(units)`
  4.  
  5. `text(6.5 u²)`
Show Worked Solution

(i)  `text(Show)\ \ AB\ \ text(is)\ \ 2x + y – 6 = 0`

`A (1, 4)\ \ \ B (5, text(–4))`

`m_(AB)` `= (y_2 – y_1) / (x_2 – x_1)`
  `= (-4 – 4) / (5 – 1)`
  `= (-8)/4`
  `= – 2`

 
`:.\ text(Equation of)\ AB, m = -2,\ text(through)\ \ (1,4)`

`y-y_1` `=m(x-x_1)`
`y – 4` `= -2 (x – 1)`
`y – 4` `= -2x + 2`
`2x + y – 6` `= 0\ …\ text(as required)`

 

(ii)  `AB\ text(intersects y-axis at)\ D`

`0 + y – 6` `= 0`
`y` `= 6`

`:. D\ text(has coordinates)\ (0, 6)`

 

(iii)  `C\ text{(–3, –1)}`

`AB\ text(is)\ 2x + y – 6 = 0`

`_|_ text(dist)` `= |\ (ax_1 + by_1 + c)/sqrt (a^2 + b^2)\ |`
  `= |\ (2(−3) + 1(-1) – 6)/sqrt(2^2 + 1^2)\ |`
  `= |\ (-13)/sqrt 5\ |`
  `= 13/sqrt 5\ text(units)`

 

(iv)

 
`text(dist)\ AD` `= sqrt((x_2 – x_1)^2 + (y_2 – y_1)^2)`
  `= sqrt((0 – 1)^2 + (6 – 4)^2`
  `= sqrt (1 + 4)`
  `= sqrt 5`
`text(Area of)\ Delta ADC` `= 1/2 xx b xx h`
  `= 1/2 xx sqrt 5 xx 13/sqrt 5`
  `= 6.5\ text(u²)`

Calculus, EXT1* C1 2005 HSC 9a

A particle is initially at rest at the origin. Its acceleration as a function of time, `t`, is given by

`ddot x = 4sin2t`

  1. Show that the velocity of the particle is given by  `dot x = 2 − 2\ cos\ 2t`.  (2 marks)

  2. Sketch the graph of the velocity for  `0 ≤ t ≤ 2π`  AND determine the time at which the particle first comes to rest after  `t = 0`.  (3 marks)

  3. Find the distance travelled by the particle between  `t = 0`  and the time at which the particle first comes to rest after  `t = 0`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `2pi\ \ text(units)`
Show Worked Solution

i.   `text(Show)\ \ dot x = 2 − 2\ cos\ 2t`

`ddot x` `= 4\ sin\ 2t`
`dot x` `= int ddot x\ dt`
  `= int4\ sin\ 2t\ dt`
  `= −2\ cos\ 2t + c`

 
`text(When)\ t = 0, \ x = 0`

`0 = −2\ cos\ 0 + c`

`c = 2`

`:.dot x = 2 − 2\ cos\ 2t\ \ …text(as required)`

 

ii.  `text(Considering the range)`

`−1` `≤ \ \ \ cos\ 2t` `≤ 1`
`−2` `≤ \ \ \ 2\ cos\ 2t`  `≤ 2` 
`0` `≤ 2 − 2\ cos\ 2t` `≤ 4` 

 
`text(Period) = (2pi)/n = (2pi)/2 = pi`

Calculus in the Physical World, 2UA 2005 HSC 9a Answer

`text(After)\ t = 0,\ text(the particle next comes)`

`text(to rest at)\ t = pi.`

 

iii.  `text(Distance travelled)`

`= int_0^pi dot x\ dt`

`= int_0^pi 2 − 2\ cos\ 2t\ dt`

`= [2t − sin\ 2t]_0^pi`

`= [(2pi − sin\ 2pi) − (0 − sin\ 0)]`

`= 2pi\ \ text(units)`

Financial Maths, 2ADV M1 2005 HSC 7a

Anne and Kay are employed by an accounting firm.

Anne accepts employment with an initial annual salary of  $50 000. In each of the following years her annual salary is increased by $2500.

Kay accepts employment with an initial annual salary of  $50 000. In each of the following years her annual salary is increased by 4%.

  1. What is Anne’s annual salary in her thirteenth year?  (2 marks)

  2. What is Kay’s annual salary in her thirteenth year?  (2 marks)

  3. By what amount does the total amount paid to Kay in her first twenty years exceed that paid to Anne in her first twenty years?  (3 marks)

Show Answers Only
  1. `$80\ 000`
  2. `text{$80 052 (nearest dollar)}`
  3. `text{$13 904 (nearest $)}`
Show Worked Solution

i.   `text(Let)\ T_n = text(Anne’s salary in year)\ n`

`T_1 = a = $50\ 000`

`T_2 = a + d = $52\ 500`

`⇒\ text(AP where)\ a = $50\ 000,\ \ d = $2500`

`T_n = a + (n − 1)d`

`T_13` `= 50\ 000 + (13 − 1) xx 2500`
  `=80\ 000`

 

`:.\ text(Anne’s salary in her 13th year is $80 000.)`

 

ii.  `text(Let)\ K_1 =text(Kay’s salary in year)\ n`

`K_1` `= a` `= 50\ 000`
`K_2` `= ar` `= 50\ 000 xx 1.04 = 52\ 000`
`⇒\ text(GP where)\ \ a = 50\ 000, \ \ r = 1.04`
`K_n` `= ar^(n − 1)`
 `K_13` `= 50\ 000 xx (1.04)^12`
  `= $80\ 051.61…`
  `= $80\ 052\ \ \ text{(nearest dollar)}`

 

iii.  `text(Anne)`

`S_n` `= n/2[2a + (n − 1)d]`
 `S_20` `= 20/2[2 xx 50\ 000 + (20 − 1)2500]`
  `= 10[100\ 000 + 47\ 500]`
  `= $1\ 475\ 000`

 

`text(Kay)`

`S_n` `= (a(r^n − 1))/(r − 1)`
`S_20`  `= (50\ 000(1.04^20 -1))/(1.04 − 1)`
  `= $1\ 488\ 903.929…`

 

`text(Difference)`

`= 1\ 488\ 903.929… − 1\ 475\ 000`

`= $13\ 903.928…`

`= $13\ 904\ \ \  text{(nearest $)}`

 

`:.\ text(Kay’s total salary exceeds Anne’s by)\ $13\ 904`

Trigonometry, 2ADV T1 2006 HSC 1d

2006 1d

 
Find the value of `theta` in the diagram. Give your answer to the nearest degree.  (2 marks)

Show Answers Only

`18°`

Show Worked Solution

`text(Using the sine rule)`

`sin theta / 5` `= (sin 33°)/9`
`sin theta` `= (5 xx sin 33°)/9`
  `= 0.30257…`
`:. theta` `= 17.612…`
  `= text{18°  (nearest degree)}`

Calculus, 2ADV C2 2007 HSC 2ai

Differentiate with respect to `x`:

`(2x)/(e^x + 1)`  (2 marks)

Show Answers Only

`{2(e^x + 1 – xe^x)}/(e^x + 1)^2`

Show Worked Solution

`y = (2x)/(e^x + 1)`

`u` `= 2x` `\ \ \ \ \v` `= e^x + 1`
`u prime` `= 2` `\ \ \ \ \ v prime` `= e^x`
`(dy)/(dx)` `= (u prime v – uv prime)/v^2`
  `= {2(e^x + 1) – 2x(e^x)}/(e^x + 1)^2`
  `= (2e^x + 2 – 2x * e^x)/(e^x + 1)^2`
  `= {2(e^x + 1 – xe^x)}/(e^x + 1)^2`

Financial Maths, STD2 F4 2006 HSC 27c

Kai purchased a new car for $30 000. It depreciated in value by $2000 per year for the first three years.

After the end of the third year, Kai changed the method of depreciation to the declining balance method at the rate of 25% per annum.

  1. Calculate the value of the car at the end of the third year.  (1 mark)

  2. Calculate the value of the car seven years after it was purchased.  (2 marks)

  3. Without further calculations, sketch a graph to show the value of the car over the seven years.

     

    Use the horizontal axis to represent time and the vertical axis to represent the value of the car.  (3 marks)

Show Answers Only
  1. `$24\ 000`
  2. `$7593.75`
  3. `text{See Worked Solutions}`
Show Worked Solution

i.  `text(Using)\ \ S = V_0 – Dn`

`S` `= 30\ 000 – (2000 xx 3)`
  `= $24\ 000`

 

ii.  `text(Using)\ \ S = V_0(1 – r)^n`

`text(where)\ V_0` `= 24\ 000`
`r` `= 0.25`
`n` `= 4`

 

`S` `= 24\ 000(1 – 0.25)^4`
  `= $7593.75`

 
`:.\ text(The value of the car after 7 years is $7593.75)`

 

iii.

Probability, STD2 S2 2006 HSC 26c

A new test has been developed for determining whether or not people are carriers of the Gaussian virus.

Two hundred people are tested. A two-way table is being used to record the results.
 

  1.  What is the value of  `A`?  (1 mark)

  2.  A person selected from the tested group is a carrier of the virus.

     

     What is the probability that the test results would show this?  (2 marks) 

  3.  For how many of the people tested were their test results inaccurate?  (1 mark)

Show Answers Only
  1. `98`
  2. `37/43`
  3. `28`
Show Worked Solution
i.  `A` `= 200-(74 + 12 + 16)`
  `= 98`

 

ii.  `P` `= text(# Positive carriers)/text(Total carriers)`
  `= 74/86`
  `= 37/43`

 

iii.  `text(# People with inaccurate results)`

`= 12 + 16`

`= 28`

Calculus, EXT1* C3 2005 HSC 6c

2005 6c
 

The graphs of the curves  `y = x^2`  and  `y = 12 - 2x^2`  are shown in the diagram.

  1. Find the points of intersection of the two curves.  (1 mark)

  2. The shaded region between the curves and the `y`-axis is rotated about the `y`-axis. By splitting the shaded region into two parts, or otherwise, find the volume of the solid formed.  (3 marks)

Show Answers Only
  1. `text{(2, 4), (–2, 4)`
  2. `24pi\ \ text(u³)`
Show Worked Solution
i.    `y` `= x^2` `\ …\ (1)`
  `y` `= 12 − 2x^2` `\ …\ (2)`

 

`text(Substitute)\ \ y = x^2\ \ text(into)\ (2)`

`x^2` `= 12 − 2x^2`
`3x^2 − 12` `= 0`
`3(x^2 − 4)` `= 0`
`x` `= ±2`
`text(When)` `\ x = 2,` `\ y = 4`
`text(When)` `\ x = text(−2),` `\ y = 4`

 
`:.\ text{Intersection at (2, 4), (−2, 4)}`
 

ii.  `text{In (1),}\ \ x^2=y`

`text{In (2),}\ \ \ y` `= 12 − 2x^2`
`2x^2` `= 12 − y`
`x^2` `= (12 − y)/2`
  `= 6 − 1/2y`

 
`:.\ text(Volume)`

`= pi int_0^4 y\ dy + pi int_4^12 6 − 1/2y\ dy`
`= pi[y^2/2]_0^4 + pi[6y − y^2/4]_4^12`
`= pi[16/2 − 0] + pi[(6 xx 12 − 12^2/4) − (6 xx 4 − 4^2/4)]`
`= 8pi + pi[36 − 20]`
`= 8pi + 16pi`
`= 24pi\ \ text(u³)`

Calculus, 2ADV C3 2005 HSC 6b

A tank initially holds 3600 litres of water. The water drains from the bottom of the tank. The tank takes 60 minutes to empty.

A mathematical model predicts that the volume, `V`  litres, of water that will remain in the tank after  `t`  minutes is given by
  

`V = 3600(1 − t/60)^2,\ \ text(where)\ \ 0 ≤ t ≤ 60`.
 

  1. What volume does the model predict will remain after ten minutes?  (1 mark)

  2. At what rate does the model predict that the water will drain from the tank after twenty minutes?  (2 marks)

  3. At what time does the model predict that the water will drain from the tank at its fastest rate?  (2 marks)

Show Answers Only
  1. `text(2500 L)`
  2. `80\ text(liters per minute)`
  3. `0`
Show Worked Solution

i.    `V = 3600(1 − t/60)^2`

`text(When)\ t = 10,`

`V` `= 3600(1 − 10/60)^2`
  `= 3600 xx (5/6)^2`
  `= 2500\ text(L)`

 

ii.   `V = 3600(1 -t/60)^2`

`text(Using chain rule:)`

`(dV)/dt` `= 3600 xx 2 xx (1 – t/60) xx d/dt(1 – t/60)`
  `= 7200(1 – t/60) xx -1/60`
  `= −120(1 – t/60)`

 

`text(When)\ \ t =20`

`(dV)/dt` `= −120(1 – 20/60)`
  `= −80`

 

`:.\ text(After 20 minutes, the water will drain)`

`text(at 80 litres per minute.)`

 

iii. `(dV)/dt` `= −120(1 − t/60)`
    `= −120 + 2t`
  `(d^2V)/dt^2` `= 2`

 
`text(S)text(ince)\ (d^2V)/dt^2\ text(is a constant, no S.P.’s)`

 

`text(Checking limits of)\ \ 0 ≤ t ≤ 60`

`text(At)\ t = 0,`

`(dV)/dt = −120(1-0) = −120\ text(L/min)`

`text(At)\ t = 60,`

`(dV)/dt = −120(1 − 60/60) = 0\ text(L/min)`

 

`:.\ text(The model predicts water will drain)`

`text(out the fastest when)\ \ t = 0.`

L&E, 2ADV E1 2005 HSC 5a

Use the change of base formula to evaluate  `log_3 7`, correct to two decimal places.  (1 mark)

Show Answers Only

`1.77\ \ text{(to 2 d.p.)}`

Show Worked Solution
`log_3 7` `= (log_10 7)/(log_10 3)`
  `= 1.771…`
  `= 1.77\ \ text{(to 2 d.p.)}`

Calculus, 2ADV C3 2005 HSC 4b

A function  `f(x)`  is defined by  `f(x) = (x + 3)(x^2- 9)`.

  1. Find all solutions of  `f(x) = 0`  (2 marks)

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

  3. Hence sketch the graph of  `y = f(x)`, showing the turning points and the points where the curve meets the `x`-axis.  (2 marks)

  4. For what values of `x` is the graph of  `y = f(x)`  concave down?  (1 mark)

Show Answers Only
  1. `−3 or 3`
  2. `text{53.2 cm  (to 1 d.p.)}`
  3. `text(See worked solutions)`
  4. `x < −1`
Show Worked Solutions
i.    `f(x)` `= (x + 3)(x^2 − 9)`
    `= (x + 3)(x +3)(x − 3)`
  `:. f(x)` `= 0\ text(when)\ \ x=–3\ text(or)\ 3`

 

ii.   `f (x)` `= (x +3)(x^2 − 9)`
    `= x^3 − 9x + 3x^2 − 27`
    `= x^3 + 3x^2 − 9x − 27`
  `f′(x)` `= 3x^2 + 6x − 9`
  `f″(x)` `= 6x + 6`

 

`text(S.P.’s  when)\ \ f′(x) = 0`

`3x^2 + 6x − 9` `= 0`
`3(x^2 + 2x − 3)` `= 0`
`3(x − 1)(x + 3)` `= 0`

 

`text(At)\ x =1`

`f(1)` `= (4)(−8)=−32`
 `f″(1)` `= 6 + 6=12>0`
`:.\ text(MIN at)\ (1, −32)` 

 

`text(At)\ x = −3`

`f(-3)` `= 0`
`f″(−3)` `= (6 xx −3) + 6 = −12 <0`
`:.\ text(MAX at)\ (−3, 0)`

 

iii.   Geometry and Calculus, 2UA 2005 HSC 4b Answer

 

iv.  `f(x)\ \ text(is concave down when)`

`f″(x)` `< 0`
`6x + 6` `< 0`
`6x` `< −6`
`x` `< −1`

Trigonometry, 2ADV T1 2005 HSC 4a

 Trig Calculus, 2UA 2005 HSC 4a
 

A pendulum is 90 cm long and swings through an angle of 0.6 radians. The extreme positions of the pendulum are indicated by the points `A` and `B` in the diagram.

  1. Find the length of the arc  `AB`.  (1 mark)

  2. Find the straight-line distance between the extreme positions of the pendulum.  (2 marks)

  3. Find the area of the sector swept out by the pendulum.  (1 mark)

Show Answers Only
  1. `text(54 cm)`
  2. `text{53.2 cm  (to 1 d.p.)}`
  3. `text(2430 cm²)`
Show Worked Solution
i.    `text(Arc)\ AB` `= theta/(2 pi) xx 2 pi r`
    `= rtheta`
    `= 90 xx 0.6`
    `= 54\ text(cm)`

 

ii. 

 Trig Calculus, 2UA 2005 HSC 4a Answer

`text(Using the cosine rule)`

`text(Distance)\ AB\ text(in straight line)`

`AB^2` `= 90^2 xx 90^2 − 2 xx 90 xx 90 xx cos\ 0.6`
  `= 2829.563…`
`:.AB` `= 53.193…`
  `= 53.2\ text{cm  (to 1 d.p.)}`

 

iii. `text(Area of Sector)`

`= 0.6/(2pi) xx pir^2`

`= 0.3 xx 90^2`

`= 2430\ text(cm²)`

Plane Geometry, 2UA 2005 HSC 3c

2005 3c

In the diagram, `A`, `B`  and  `C`  are the points  `(6, 0), (9, 0)`  and  `(12, 6)` respectively. The equation of the line  `OC`  is  `x - 2y = 0`. The point  `D`  on  `OC`  is chosen so that  `AD`  is parallel to  `BC`. The point  `E`  on  `BC`  is chosen so that  `DE`  is parallel to the `x`-axis.

  1.  Show that the equation of the line `AD` is `y = 2x - 12`.  (2 marks)
  2. Find the coordinates of the point `D`.  (2 marks)
  3. Find the coordinates of the point `E`.  (1 marks)
  4. Prove that  `ΔOAD\ text(|||)\ ΔDEC`.  (2 marks)
  5. Hence, or otherwise, find the ratio of the lengths `AD` and `EC`.  (1 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(8, 4)`
  3. `(11, 4)`
  4. `text(See Worked Solutions)`
  5. `2:1`
Show Worked Solution

(i)   `text(S)text(ince)\ DA\ text(||)\ CB`

`m_(DA)` `= (y_2 − y_1)/(x_2 − x_1)`
  `= (6 − 0)/(12 − 9)`
  `= 2`

 

`:.\ text(Equation of)\ AD, m = 2,\ text(through)\ A(6, 0)`

`y − y_1` `= m(x − x_1)`
 `y − 0` `= 2(x − 6)`
 `y` `= 2x − 12\ \ \ \ …\ text(as required)`

 

(ii)  `D\ text(is at the intersection of)`

`x − 2y` `= 0` `\ \ …\ (1)`
 `y` `= 2x − 12\ ` `\ \ …\ (2)` 

`text(Substitute)\ y = 2x − 12\ text{into (1)}`

`x − 2(2x − 12)` `= 0`
`x − 4x + 24` `= 0`
`text(−3)x + 24` `= 0`
`3x` `= 24`
`x` `= 8`

`text(Substitute)\ x = 8\ text{into (2)}`

`y` `= 2 xx 8 − 12 = 4`
`:.D` `= (8, 4)`

 

(iii)  `text(Distance)\ \ AB=3`

`:. E\ \ text(has coordinates)\ \ (8+3,4)-=(11,4)`

 

(iv)  `text(Prove)\ ΔOAD\ text(|||)\ ΔDEC`

`∠ODA = ∠DCE`

`text{(corresponding angles,}\ AD\ text(||)\ BC)`

`∠OAD = ∠ABE`

`text{(corresponding angles,}\ AD\ text(||)\ BC)`

`∠ABE = ∠DEC`

`text{(corresponding angles,}\ AB\ text(||)\ DE)`

`:.∠OAD = ∠DEC`

`:.ΔOAD\ text(|||)\ ΔDEC\ text{(equiangular)}`

 

(v)     `(AD)/(EC)` `= (OA)/(DE)` 
    `= 6/3` 
    `=2` 

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

 

`:.\ text(Ratio of lengths)\ AD:EC = 2:1` 

Calculus, 2ADV C3 2006 HSC 5a

A function  `f(x)`  is defined by  `f(x) =2x^2(3-x)`.

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

  2. Find the coordinates of the point of inflection.  (1 mark)

  3. Hence sketch the graph of  `y =f(x)`, showing the turning points, the point of inflection and the points where the curve meets the `x`-axis.  (3 marks)

  4. What is the minimum value of  `f(x)`  for  `–1 ≤ x ≤4`?  (1 mark)

Show Answers Only
  1. `text(Min)\ (0, 0),\ text(Max)\ (2, 8)`
  2. `text(P.I. at)\ (1, 4)`
  3.  
  4. `-32`
Show Worked Solution
i.       `f(x)` `= 2x^2 (3-x)`
  `= 6x^2-2x^3`
`f^{prime} (x)` `= 12x-6x^2`
`f^{″}(x)` `= 12-12x`

 

`text(S.P.’s when)\ f^{′}(x) = 0`

`12x-6x^2` `= 0`
`6x(2-x)` `= 0`

`x = 0 or 2`

`text(When)\ x = 0`

`f(0)` `= 0`
`f^{″}(0)` `= 12-0 = 12 > 0`
`:.\ text(MIN at)\ (0, 0)`

 

`text(When)\ x = 2`

`f(2)` `= 2 xx 2^2 (3-2)` `= 8`
`f^{″}(2)` `= 12-(12 xx 2)` `= -12 < 0`
`:.\ text(MAX at)\ (2, 8)`

 

ii.  `text(P.I. when)\ f^{″}(x) = 0`

`12-12x` `= 0`
`12x` `= 12`
`x` `= 1`
`f^{″}(0.5)` `=6>0`
`f^{″}(1.5)` `=-6<0`

`text(S)text(ince concavity changes)\ \ =>\  text(P.I. exists)` 

`f(1)` `= 2 xx 1^2(3-1)`
  `= 4`

`:.\ text(P.I. at)\ (1, 4)`

 

iii.  `f(x)\ text(meets)\ x text(-axis when)\ f(x) = 0`

`2x^2 xx (3-x) = 0`

`x = 0 or 3`

2UA HSC 2006 5a

 

(iv)  `text(The graph clearly shows that in the given range)`

`-1<= x<=4,\ text(the minimum will occur when)\ x = 4`

`:.\ text(Minimum` `= 2 xx 4^2 (3-4)`
  `= -32`

Calculus, 2ADV C3 2005 HSC 2d

Find the equation of the tangent to  `y = log_ex`  at the point  `(e, 1)`.  (2 marks)

Show Answers Only

`y = x/e`

Show Worked Solution

`y = log_ex`

`dy/dx = 1/x`

`text(At)\ \ (e, 1),`

`m = 1/e`
 

`text(Equation of tangent,)\ \ m = 1/e,\ text(through)\ (e, 1)`

`y – y_1` `= m(x – x_1)`
`y – 1`  `= 1/e(x -e)`
`y – 1`  `= x/e – 1`
`y`  `= x/e`

Trig Calculus, 2UA 2005 HSC 2cii

Evaluate `int_0^(pi/6) cos\ 3x\ dx`.  (2 marks)

Show Answers Only

`1/3`

Show Worked Solution
`int_0^(pi/6)\ cos\ 3x\ dx` `= 1/3\ [sin\ 3x]_0^(pi/6)`
  `= 1/3\ [sin\ (3 xx pi/6) − sin\ 0]`
  `= 1/3\ [sin\ pi/2 − sin\ 0]`
  `= 1/3\ [1 − 0]`
  `= 1/3`

Calculus, 2ADV C1 2005 HSC 2bii

Differentiate with respect to `x`:

`x^2/(x − 1).`  (2 marks)

Show Answers Only

`(x(x − 2))/(x − 1)^2`

Show Worked Solution

`y = x^2/(x − 1)`

`text(Using)\ \ dy/dx = (u′v − uv′)/v^2`

`u` `= x^2` `v` `= x − 1`
`u′` `= 2x` `v′` `= 1`

 

`dy/dx` `= (2x(x − 1) − x^2(1))/(x − 1)^2`
  `= (2x^2 − 2x − x^2)/(x − 1)^2`
  `= (x^2 − 2x)/(x − 1)^2`
  `= (x(x − 2))/(x − 1)^2`

Trigonometry, 2ADV T2 2005 HSC 2a

Solve  `cos\ theta =1/sqrt2`  for  `0 ≤ theta ≤ 2pi`.   (2 marks)

Show Answers Only

`pi/4, (7pi)/4`

Show Worked Solution

`cos\ theta = 1/sqrt2,\ \ \ 0 ≤ theta ≤ 2pi`

`text(S)text(ince)\ cos\ pi/4 = 1/sqrt2,\ \ text(and cos)`

`text(is positive in 1st/4th quadrants)`

`theta` `= pi/4, 2pi-pi/4`
  `= pi/4, (7pi)/4`

 

Functions, EXT1* F1 2005 HSC 1e

Find the values of  `x`  for which  `|\ x − 3\ | ≤ 1`.  (2 marks)

Show Answers Only

`2 ≤ x ≤ 4`

Show Worked Solution

`|\ x − 3\ | ≤ 1`

`text(Solution 1)`

`(x − 3)^2 ≤ 1`

`x^2 − 6x + 9 ≤ 1`

`x^2 − 6x +8 ≤ 0`

`(x − 4)(x − 2) ≤ 0`
 

Algebra, 2UA 2005 HSC 1e Answer  

 
`:. 2 ≤ x ≤ 4`

 

`text(Alternative Solution)`

`(x − 3)` `≤1` `–(x − 3)` ` ≤ 1`
`x` `≤4` `–x +3` `≤ 1`
    `–x` `≤−2`
    `x` `≥ 2`

`:. 2 ≤ x ≤ 4`

Functions, 2ADV F1 2005 HSC 1d

Express  `((2x-3))/2-((x-1))/5`  as a single fraction in its simplest form.  (2 marks)

Show Answers Only

`(8x-13)/10`

Show Worked Solution

`((2x-3))/2-((x-1))/5`

`= (5(2x-3)-2(x-1))/10`

`= (10x-15-2x + 2)/10`

`= (8x-13)/10`