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Number and Algebra, NAP-I1-10

Andrew is travelling from Brisbane to Mackay.

He knows that it is more 968 kilometres but less than 986 kilometres.

Which of these could be the number of kilometres that Andrew has to travel?

`946` `964` `984` `988`
 
 
 
 
Show Answers Only

`984`

Show Worked Solution

`text(S)text(ince)\ 968 < 984 < 986,`

`:. 984\ text(km could be the number.)`

Measurement, NAP-B2-01

The minute hand is not shown.
 


 

What time could this clock be displaying?

`text(10 o'clock)` `text(half past 10)` `text(11 o'clock)` `text(half past 11)`
 
 
 
 
Show Answers Only

`text(half past 10)`

Show Worked Solution

`text(The hour hand is halfway between 10 and 11.)`

`:.\ text(The time could be half past 10.)`

Measurement, NAP-C2-01

Peter is packing soup cans into a box for charity.

The box is the same height as a soup can.
 

 

 
How many cans of soup will fit into the box?

`8`    `9`  `15` `20`
 
 
 
 
Show Answers Only

`20\ text(cans)`

Show Worked Solution

`text(4 rows of cans will fit inside the box)`

`text(and each row will have 5 cans.)`

`:.\ text(Number of cans)` `=4 xx 5`
  `=20`

Number and Algebra, NAP-I2-3

Andrew is travelling from Brisbane to Mackay.

He knows that it is more 968 kilometres but less than 986 kilometres.

Which of these could be the number of kilometres that Andrew has to travel?

`946` `964` `984` `988`
 
 
 
 
Show Answers Only

`984`

Show Worked Solution

`text(S)text(ince)\ \ \ 968 < 984 < 986,`

`:. 984\ text(km could be the number.)`

Binomial, EXT1 2016 HSC 14b

Consider the expansion of  `(1 + x)^n`, where `n` is a positive integer.

  1. Show that  `2^n = ((n),(0)) + ((n),(1)) + ((n),(2)) + ((n),(3)) + … + ((n),(n))`.  (1 mark)
  2. Show that  `n2^(n - 1) = ((n),(1)) + 2((n),(2)) + 3((n),(3)) + … + n((n),(n))`.  (1 mark)
  3. Hence, or otherwise, show that  `sum_(r = 1)^n ((n),(r))(2r - n) = n`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text[(See Worked Solutions)}`
  2. `text(Proof)\ \ text[(See Worked Solutions)}`
  3. `text(Proof)\ \ text[(See Worked Solutions)}`
Show Worked Solution

i.     `text(Show)`

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

`text(Using binomial expansion)`

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

`text(Let)\ \ x = 1,`

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

`text(… as required.)`

 

ii.   `text(Differentiate both sides of expansion,)`

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

`text(Let)\ \ x = 1,`

`n2^(n – 1) = ((n), (1)) + 2 ((n), (2)) + 3 ((n), (3)) + … + n ((n), (n))`

`text(… as required.)`

 

iii.  `text{Multiply part (i)} xx n`

♦♦♦ Mean mark 14%.
`n2^n` `= n[((n), (0)) + ((n), (1)) + … + ((n), (n))]`
  `= sum_(r = 0)^n ((n), (r)) n\ \ text{…  (1)}`

 

`text{Multiply part (ii)} xx 2`

`2 xx n2^(n – 1)` `= 2[((n), (1)) + 2 ((n), (2)) + … + n ((n), (n))]`
`n2^n` `= sum_(r = 1)^n ((n), (r)) 2r\ \ text{…  (2)}`

 

 `text(Subtract) qquad (2) – (1)`

`sum_(r = 1)^n ((n), (r)) 2r – sum_(r = 0)^n ((n), (r))n` `= n2^n – n2^n`
`sum_(r = 1)^n ((n), (r)) 2r – sum_(r = 1)^n ((n), (r)) n – n` `= 0`
`sum_(r = 1)^n ((n), (r)) (2r – n)` `= n\ \ text(…  as required)`

Proof, EXT1 P1 2016 HSC 14a

  1. Show that  `4n^3 + 18n^2 + 23n + 9`  can be written as
     

     

    `qquad (n + 1)(4n^2 + 14n + 9)`.  (1 marks)

  2. Using the result in part (i), or otherwise, prove by mathematical induction that, for  `n >= 1`,

     

    `qquad 1 × 3 + 3 × 5 + 5 × 7 + … + (2n - 1)(2n + 1) = 1/3 n(4n^2 + 6n - 1)`.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.   `text(RHS)` `= (n + 1) (4n^2 + 14n + 9)`
    `= 4n^3 + 14n^2 + 9n + 4n^2 + 14n + 9`
    `= 4n^3 + 18n^2 + 23n + 9`

 

ii.   `text(Prove)\ \ 1 xx 3 + 3 xx 5 + 5 xx 7 + … + (2n – 1) (2n + 1)`

`= 1/3n (4n^2 + 6n – 1)`

`text(If)\ \ n = 1,`

`text(LHS) = (1) (3) = 3`

`text(RHS) = 1/3 (1) (4 + 6 – 1) = 3`

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

 

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

`text(i.e.)\ \ 1 xx 3 + 3 xx 5 + … + (2k – 1) (2k + 1)`

`= 1/3 k (4k^2 + 6k – 1)`

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

`text(i.e.)\ \ 1 xx 3 + … + (2k + 1) (2k + 3)`

`= 1/3 (k + 1) [4 (k + 1)^2 + 6(k + 1) – 1]`

`= 1/3 (k + 1) (4k^2 + 14k + 9)`

`= 1/3 (4k^3 + 18k^2 + 23k + 9)`

 

`text(LHS)` `= 1 xx 3 + … + (2k – 1) (2k + 1) + (2k + 1) (2k + 3)`
  `= 1/3k (4k^2 + 6k – 1) + (2k + 1) (2k + 3)`
  `= 1/3 (4k^3 + 6k^2 – k) + (4k^2 + 8k + 3)`
  `= 1/3 (4k^3 + 6k^2 – k + 12k^2 + 24k + 9)`
  `= 1/3 (4k^3 + 18k^2 + 23k + 9)`
  `=\ text(RHS …)`

 

`=> 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 2016 HSC 13b

The trajectory of a projectile fired with speed  `u\ text(ms)^-1`  at an angle  `theta`  to the horizontal is represented by the parametric equations

`x = utcostheta`   and   `y = utsintheta - 5t^2`,

where `t` is the time in seconds.

  1. Prove that the greatest height reached by the projectile is  `(u^2 sin^2 theta)/20`.  (2 marks)

A ball is thrown from a point `20\ text(m)` above the horizontal ground. It is thrown with speed `30\ text(ms)^-1` at an angle of `30^@` to the horizontal. At its highest point the ball hits a wall, as shown in the diagram.
 

     ext1-2016-hsc-q13
 

  1. Show that the ball hits the wall at a height of `125/4\ text(m)` above the ground.  (2 marks)

The ball then rebounds horizontally from the wall with speed `10\ text(ms)^-1`. You may assume that the acceleration due to gravity is `10\ text(ms)^-2`.

  1. How long does it take the ball to reach the ground after it rebounds from the wall?  (2 marks)

  2. How far from the wall is the ball when it hits the ground?  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `2.5\ text(seconds)`
  4. `25\ text(m)`
Show Worked Solution
i.    `y` `= u t sin theta – 5t^2`
  `y prime` `= u sin theta – 10t`

 

`text(Maximum height when)\ \ y prime = 0`

`10 t` `= u sin theta`
`t` `= (u sin theta)/10`

 

`:.\ text(Maximum height)`

`= u ((u sin theta)/10) · sin theta – 5 ((u sin theta)/10)^2`

`= (u^2 sin^2 theta)/10 – (u^2 sin^2 theta)/20`

`= (u^2 sin^2 theta)/20\ text(… as required)`

 

ii.   `text{Using part (i)},`

`text(Height that ball hits wall)`

`= (30^2 · (sin 30)^2)/20 + 20`

`= (30^2 · (1/2)^2)/20 + 20`

`= 11 1/4 + 20`

`= 125/4\ text(m … as required)`

 

♦♦ Mean mark part (iii) 35%.
iii.   ext1-hsc-2016-13bi
`y ″` `= -10`
`y prime` `= -10 t`
`y` `= 125/4 – 5t^2`

 

`text(Ball hits ground when)\ \ y = 0,`

MARKER’S COMMENT: Many students struggled to solve: `5t^2=125/4`.
`5t^2` `= 125/4`
`t^2` `= 25/4`
`:. t` `= 5/2,\ \ t > 0`

 

`:.\ text(It takes the ball 2.5 seconds to hit the ground.)`

 

iv.   `text(Distance from wall)`

♦ Mean mark 43%.

`= 2.5 xx 10`

`= 25\ text(m)`

Calculus, EXT1 C1 2016 HSC 12a

The diagram shows a conical soap dispenser of radius 5 cm and height 20 cm.
 

     ext1-2016-hsc-q12
 

At any time `t` seconds, the top surface of the soap in the container is a circle of radius `r` cm and its height is `h` cm.

The volume of the soap is given by  `v = 1/3 pir^2h`.

  1.  Explain why  `r = h/4`.  (1 mark)

  2.  Show that  `(dv)/(dh) = pi/16 h^2`.  (1 mark)

The dispenser has a leak which causes soap to drip from the container. The area of the circle formed by the top surface of the soap is decreasing at a constant rate of  `0.04\ text(cm² s)^-1`.
 

  1.  Show that  `(dh)/(dt) = (−0.32)/(pih)`.  (2 marks)

  2.  What is the rate of change of the volume of the soap, with respect to time, when `h = 10`?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `-0.2\ text(cm³ s)^-1`
Show Worked Solution
i.   ext1-hsc-2016-12a

`text(Using similar triangles,)`

`r/h` `= 5/20`
 `:. r` `= h/4\ text(… as required)`

 

ii.   `v` `= 1/3 pi r^2 h`
    `= 1/3 pi · (h/4)^2 h`
    `= (pi h^3)/48`
  `:. (dv)/(dh)` `= 3 xx (pi h^2)/48`
    `= (pi h^2)/16\ text(… as required.)`

 

iii.   `(dA)/(dt)` `= -0.04\ text(cm² s)^-1`
  `A` `= pi r^2`
    `= (pi h^2)/16`
  `:. (dA)/(dh)` `= (pi h)/8`

 

`(dA)/(dt)` `= (dA)/(dh) xx (dh)/(dt)`
`-0.04` `= (pi h)/8 xx (dh)/(dt)`
`:. (dh)/(dt)` `= (-0.32)/(pi h)\ text(… as required.)`

 

iv.   `(dv)/(dt)` `= (dv)/(dh) · (dh)/(dt)`
    `= (pi h^2)/16 · (-0.32)/(pi h)`
    `= (-0.32 h)/16`

 

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

`(dv)/(dt)` `= (-0.32 xx 10)/16`
  `= -0.2\ text(cm³ s)^-1`

Statistics, EXT1 S1 2016 HSC 11f

A darts player calculates that when she aims for the bullseye the probability of her hitting the bullseye is  `3/5`  with each throw.

  1. Find the probability that she hits the bullseye with exactly one of her first three throws.  (1 mark)

  2. Find the probability that she hits the bullseye with at least two of her first six throws.  (2 marks)

Show Answers Only
  1. `36/125`
  2. `2997/3125`
Show Worked Solution

i.   `P text{(exactly 1 bullseye)}`

`=\ ^3C_1 · (3/5)^1 (2/5)^2`

`= 3 · (3/5) · (4/25)`

`= 36/125`

 

ii.   `P text{(at least 2 from 6 throws)}`

`= 1 – [P(0) + P(1)]`

`= 1 – [(2/5)^6 + \ ^6C_1 · (3/5)^1· (2/5)^5]`

`= 1 – [128/3125]`

`= 2997/3125`

Calculus, EXT1 C2 2016 HSC 11b

Use the substitution  `u = x - 4`  to find  `int xsqrt(x - 4)\ dx`.  (3 marks)

Show Answers Only

`2/5 (x – 4)^(5/2) + 8/3 (x – 4)^(3/2) + c`

Show Worked Solution

`u = x – 4\ \ => \ x = u + 4`

`(du)/(dx) = 1\ \ => \ dx = du`
 

`:. int x sqrt (x – 4)\ dx`

`= int (u + 4) · u^(1/2)\ du`

`= int u^(3/2) + 4u^(1/2)\ du`

`= 2/5 u^(5/2) + 4 · 2/3 u^(3/2) + c`

`= 2/5 (x – 4)^(5/2) + 8/3 (x – 4)^(3/2) + c`

Trig Ratios, EXT1 2016 HSC 6 MC

What is the general solution of the equation  `2sin^2x - 7sinx + 3 = 0`?

  1. `npi - (−1)^n pi/3`
  2. `npi + (−1)^n pi/3`
  3. `npi - (−1)^n pi/6`
  4. `npi + (−1)^n pi/6`
Show Answers Only

`D`

Show Worked Solution

`2 sin^2 x – 7 sin x + 3 = 0`

`(2 sin x – 1) (sin x – 3) = 0`

`sin x = 1/2\  or\  sin x = 3\ \ text{(no solution)}`

`:. x` `= pi/6, \ pi – pi/6, \ 2pi + pi/6, …`
  `= pi/6, \ (5 pi)/6, \ (13 pi)/6, …`
  `= n pi + (-1)^n pi/6`

 

`=>   D`

Financial Maths, 2ADV M1 2016 HSC 14b

A gardener develops an eco-friendly spray that will kill harmful insects on fruit trees without contaminating the fruit. A trial is to be conducted with 100 000 insects. The gardener expects the spray to kill 35% of the insects each day and that exactly 5000 new insects will be produced each day.

The number of insects expected at the end of the `n`th day of the trial is `A_n.`

  1. Show that  `A_2 = 0.65 (0.65 xx 100\ 000 + 5000) + 5000`.  (2 marks)

  2. Show that  `A_n = 0.65^n xx 100\ 000 + 5000 ((1 - 0.65^n))/0.35`.  (1 mark)

  3. Find the expected insect population at the end of the fourteenth day, correct to the nearest 100.  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `14\ 500\ text{(nearest 100)}`
Show Worked Solution
i.   `A_1` `= 0.65 xx 100\ 000 + 5000`
  `A_2` `= 0.65 xx A_1 + 5000`
    `= 0.65 (0.65 xx 100\ 000 + 5000) + 5000`
    `qquad qquad text(… as required)`

 

ii.  `A_2` `= 0.65^2 xx 100\ 000 + 0.65 xx 5000 + 5000`
  `A_3` `= 0.65^3 xx 100\ 000 + 0.65^2 xx 5000 + 0.65 xx 5000 + 5000`
  `vdots`  
  `A_n` `= 0.65^n xx 100\ 000 + 0.65^(n – 1) xx 5000 + 0.65^(n – 2) xx 5000 + … + 5000`
    `= 0.65^n xx 100\ 000 + 5000 (1 + 0.65 + … + 0.65^(n – 1))`
    `qquad qquad => text(GP where)\ \ a = 1,\ \ r = 0.65`
    `= 0.65^n xx 100\ 000 + 5000 ({(1 – r^n)}/(1 – r))`
    `= 0.65^n xx 100\ 000 + 5000 ((1 – 0.65^n))/0.35`

 

iii.  `A_14` `= 0.65^14 xx 100\ 000 + 5000 ((1 – 0.65^14)/0.35)`
    `= 14\ 491.70…`
    `= 14\ 500\ text{(nearest 100)}`

Integration, 2UA 2016 HSC 14a

The diagram shows the cross-section of a tunnel and a proposed enlargement.
 

hsc-2016-14a

The heights, in metres, of the existing section at 1 metre intervals are shown in Table `A.`
 

hsc-2016-14ai

The heights, in metres, of the proposed enlargement are shown in Table `B.`
 

hsc-2016-14aii

Use Simpson’s rule with the measurements given to calculate the approximate increase in area.  (3 marks)

Show Answers Only

`1.4\ text(m²)`

Show Worked Solution

`text(Shaded Area distances)`

2ua-hsc-2016-14ai

`A` `~~ h/3 [y_0 + 4 (y_1 + y_3) + 2y_2 + y_4]`
  `~~ 1/3 [0 + 4 (0.4 + 0.4) + 2 (0.5) + 0]`
  `~~ 1/3 (4.2)`
  `~~ 1.4\ text(m²)`

Polynomials, EXT2 2016 HSC 15c

  1. Use partial fractions to show that

    `qquad (3!)/(x(x + 1) (x + 2) (x + 3)) = 1/x - 3/(x + 1) + 3/(x + 2) - 1/(x + 3).`  (2 marks)

  2. Suppose that for `n` a positive integer

     

    `qquad qquad (n!)/(x(x + 1) … (x + n)) = a_0/x + a_1/(x + 1) + … + a_k/(x + k) + … + a_n/(x + n).`

    Show that  `a_k = (-1)^k ((n), (k)).`  (3 marks)

  3. Hence, or otherwise, find the limiting sum of

     

    `qquad qquad 1 - 1/2 ((n), (1)) + 1/3 ((n), (2)) - 1/4 ((n), (3)) + … + (-1)^n/(n + 1).`  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `1/(n + 1)`
Show Worked Solution

i.   `text(Show)`

`(3!)/(x(x + 1)(x + 2)(x + 3)) = 1/x – 3/(x + 1) + 3/(x + 2) – 1/(x + 3)`

`text(LHS)` `= (a_0)/x + (a_1)/(x + 1) + (a_2)/(x + 2) + (a_3)/(x + 3)`
  `= [(a_0(x + 1)(x + 2)(x + 3) + a_1x(x + 2)(x + 3) + a_2x(x + 1)(x + 3) +`
  ` a_3x(x + 1)(x + 2)]//(x(x + 1)(x + 2)(x + 3))`

 

`a_0 = lim_(x -> 0) (3!)/((x + 1)(x + 2)(x + 3)) = 6/6 = 1`

`a_1 = lim_(x -> −1) (3!)/(x(x + 2)(x + 3)) = 6/((−1)(1)(2)) = −3`

`a_2 = lim_(x -> −2) (3!)/(x(x + 1)(x + 3)) = 6/((−2)(−1)(1)) = 3`

`a_3 = lim_(x -> − 3)(3!)/(x(x + 1)(x + 2)) = 6/((−3)(−2)(−1)) = −1`

 

ii.   `text(Given that)`

`(n!)/(x(x + 1) … (x + n)) = (a_0)/x + (a_1)/(x + 1) + … + (a_k)/(x + k) + … +  (a_n)/(x + n)`

 

`text(Show)\ a_k = (−1)^k((n),(k))`

`a_k` `= lim_(x -> −k) (n!(x + k))/(x(x + 1) … (x + k – 1)(x + k)(x + k + 1) …(x + n))`
  `= (n!)/((−k)(−k + 1) … (−k + k – 1)(−k + k + 1) … (−k + n))`
   
  `=>\ text(S)text(ince negative for)\ k\ text(odd,)`
 `a_k` `= ((−1)^kn!)/(k(k – 1) … (2)(1)(1)(2) … (n – k))`
  `= ((−1)^kn!)/(k!(n – k)!)`
  `= (−1)^k ((n),(k))`

 

iii.   `(n!)/(x(x + 1) … (x + n))`

`= (a_0)/x + (a_1)/(x + 1) + … + (a_n)/(x + n)`

`= ((n),(0)) – 1/2((n),(1)) + 1/3((n),(2)) – … + ((−1)^n)/(x + n)((n),(n))`

 

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

`(n!)/((1)(2)(3) … (n + 1))`

`= ((n),(0)) – 1/2((n),(1)) + 1/3((n),(2)) – … + ((−1)^n)/(n + 1)`

`text(LHS)` `= (n!)/((n + 1)!)`
  `= 1/(n + 1)`

 

`:.\ text(Limiting sum) = 1/(n + 1)`

Functions, EXT1′ F2 2016 HSC 15a

The equation  `x^3 - 3x + 1 = 0`  has roots  `alpha, beta`  and  `gamma.`

Find a cubic equation with integer coefficients that has roots  `alpha^2, beta^2`  and  `gamma^2.`  (2 marks)

Show Answers Only

`x^3 – 6x^2 + 9x – 1 = 0`

Show Worked Solution

`x^3 – 3x + 1 = 0\ …\ (1)`

`text(Find cubic with roots)\ alpha^2, beta^2, gamma^2`

`x = alpha^2`

`:. alpha = ±sqrtx\ \ \ text{is a root of  (1)}`

`(±sqrtx)^3 – 3(±sqrtx) + 1` `= 0`
`(sqrtx)^3 – 3sqrtx` `= ±1`
`sqrtx(x – 3)` `= ±1`

 

`text(Square both sides:)`

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

 

`:. x^3 – 6x^2 + 9x – 1 = 0\ \ \ text(has roots)\ alpha^2, beta^2, gamma^2`

Calculus, EXT1* C1 2016 HSC 13c

A radioactive isotope of Curium has a half-life of 163 days. Initially there are 10 mg of Curium in a container.

The mass `M(t)` in milligrams of Curium, after `t` days, is given by

`M(t) = Ae^(-kt),`

where `A` and `k` are constants.

  1. State the value of `A`.  (1 mark)

  2. Given that after 163 days only 5 mg of Curium remain, find the value of `k`.  (2 marks)

Show Answers Only
  1. `10`
  2. `(ln 2)/163`
Show Worked Solution

i.  `text(When)\ \ t = 0,\ \ M = 10`

`10` `= Ae°`
`:. A` `= 10`

 

ii.  `text(When)\ \ t = 163,\ \ M = 5`

`5` `= 10 e^(-163k)`
`e^(-163k)` `= 1/2`
`e^(163k)` `= 2`
`163 k` `= ln 2`
`:. k` `= (ln 2)/163`

Calculus, 2ADV C3 2016 HSC 13a

Consider the function  `y = 4x^3 - x^4.`

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

  2. Sketch the graph of the function, clearly showing the stationary points and the `x` and `y` intercepts.  (2 marks)

Show Answers Only

i.   `text{P.I. at (0, 0) and Max at (3, 27)}`

ii.
 ext2-hsc-2016-13bi

Show Worked Solution
i.    `y` `= 4x^3 – x^4`
  `y prime` `= 12x^2 – 4x^3`
  `y″` `= 24x – 12x^2`

 

`text(S.P.’s when)\ \ y prime = 0,`

`12x^2 – 4x^3` `= 0`
 `4x^2 (3 – x)` `= 0`
`:. x = 0 or 3`

 

`text(When)\ \ x = 0,\ \ y″ (0) = 0`

`:.\ text(P.I. at)\ \ (0, 0)`

 

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

`y″ (3) = 24(3) – 12 (9) = -36 < 0`

`:.\ text(MAX)\ \ text(at)\ \ (3, 27)`

 

ii.  ext2-hsc-2016-13bi

Calculus, 2ADV C4 2016 HSC 12d

  1. Differentiate  `y = xe^(3x)` . (1 mark)

  2. Hence find the exact value of  `int_0^2 e^(3x) (3 + 9x)\ dx`.  (2 marks)

Show Answers Only
  1. `e^(3x) (1 + 3x)`
  2. `6e^6`
Show Worked Solution

i.  `y = xe^(3x)`

`text(Using product rule:)`

`(dy)/(dx)` `= x · 3e^(3x) + 1 · e^(3x)`
  `= e^(3x) (1 + 3x)`

 

ii.  `int_0^2 e^(3x) (3 + 9x)\ dx`

`= 3 int_0^2 e^(3x) (1 + 3x)\ dx`

`= 3 [x e^(3x)]_0^2`

`= 3 (2e^6 – 0)`

`= 6e^6`

Trigonometry, 2ADV T1 2016 HSC 12c

Square tiles of side length 20 cm are being used to tile a bathroom.

The tiler needs to drill a hole in one of the tiles at a point `P` which is 8 cm from one corner and 15 cm from an adjacent corner.

To locate the point `P` the tiler needs to know the size of the angle `theta` shown in the diagram.
 

 hsc-2016-12c

 
Find the size of the angle `theta` to the nearest degree.  (3 marks)

Show Answers Only

`69°\ text{(nearest degree)}`

Show Worked Solution

`α + theta = 90`

`text(Using the cosine rule,)`

`cos alpha` `= (20^2 + 15^2 – 8^2)/(2 xx 20 xx 15)`
  `= 0.935`
`alpha` `= 20.7…°`

 

`:. theta` `= 90 – 20.7…`
  `= 69.22…`
  `= 69°\ text{(nearest degree)}`

Plane Geometry, 2UA 2016 HSC 12b

The diagram shows a semicircle with centre `O.` It is given that  `AB = OB,\ \ /_ COD = 87^@ and /_ BAO = x^@.`

hsc-2016-12b

  1. Show that  `/_ CBO = 2x^@,\ text(giving reasons)`.  (1 mark)
  2. Find the value of `x`, giving reasons.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `29`
Show Worked Solution
i.   `/_ AOB` `= x\ \ (Delta AOB\ text{is isosceles})`
  `/_ ABO` `= 180 – 2x\ \ (text{Angle sum of}\ Delta AOB)`
  `/_ CBO` `= 180 – (180 – 2x)` `qquad text{(}/_ CBA\ \ text{is a straight}`
    `= 2x^@` `qquad \ \ text{angle)}`

 

ii   `/_ CBO = /_ BCO = 2x\ \ \ (Delta CBO\ \ text{is isosceles})`

`/_ BOC = 180 – 4x\ \ \ (text{Angle sum of}\ \ Delta CBO)`

 

`x + (180 – 4x) + 87` `= 180\ \ text{(}/_ AOD\ \ text{is a straight}`
  `qquad qquad qquad qquad text{angle)}`
`180 – 3x + 87` `= 180`
`3x` `= 87`
`:. x` `= 29`

Algebra, STD2 A2 2016 HSC 29e

The graph shows the life expectancy of people born between 1900 and 2000.
 


  1. According to the graph, what is the life expectancy of a person born in 1932?  (1 mark)

  2. With reference to the value of the gradient, explain the meaning of the gradient in this context.  (2 marks)

Show Answers Only
  1. `text(68 years)`
  2. `text(After 1900, life expectancy increases 0.25 years for each later year someone is born.)`
Show Worked Solution

i.    \(\text{68 years}\)

ii.    \(\text{Using (1900,60), (1980,80):}\)

\(\text{Gradient}\) \(= \dfrac{y_2-y_1}{x_2-x_1}\)
  \(= \dfrac{80-60}{1980-1900}\)
  \(= 0.25\)

 
\(\text{After 1900, life expectancy increases by 0.25 years for}\)

\(\text{each year later that someone is born.}\)

♦♦ Mean mark (ii) 33%.

Complex Numbers, EXT2 N2 2016 HSC 12c

Let  `z = cos theta + i sin theta.`

  1. By considering the real part of  `z^4`, show that  `cos 4 theta`  is
     
    `qquad cos^4 theta - 6 cos^2 theta sin^2 theta + sin^4 theta.`  (2 marks)

     

  2. Hence, or otherwise, find an expression for  `cos 4 theta`  involving only powers of `cos theta.`  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `8cos^4theta – 8cos^2theta + 1`
Show Worked Solution

i.   `z = costheta + isintheta`

`z^4` `= (costheta + isintheta)^4`
 

`= cos^4theta + 4cos^3theta*(isintheta) + 6cos^2theta*(isintheta)^2 +`

`4costheta*(isintheta)^3 + (isintheta)^4`
 

`= cos^4theta + 4icos^3thetasintheta – 6cos^2thetasin^2theta -`

`4icosthetasin^3theta + sin^4theta`

 

`z^4 = cos4theta + isin4theta\ \ text{(by De Moivre)}`
 

`text(Equating real parts:)`

`cos4theta = cos^4theta – 6cos^2thetasin^2theta + sin^4theta`

`…\ text(as required)`

 

ii.    `cos4theta` `= cos^4theta – 6cos^2theta(1 – cos^2theta) + (1 – cos^2theta)^2`
    `= cos^4theta – 6cos^2theta + 6cos^4theta + 1 – 2cos^2theta + cos^4theta`
    `= 8cos^4theta – 8cos^2theta + 1`

Calculus, EXT2 C1 2016 HSC 12b

  1. Differentiate  `x\ f(x)-int x\ f^(′)(x)\ dx.`  (1 mark)

  2. Hence, or otherwise, find  `int tan^-1 x\ dx.`  (2 marks)

Show Answers Only
  1. `f(x)`
  2. `x tan^(−1)x-1/2 ln(1 + x^2)`
Show Worked Solution

i.   `d/dx (x\ f(x)-int x\ f^(′)(x)\ dx)`

`= x\ f^(′)(x) + f(x)-x\ fprime(x)`

`= f(x)`

 

ii.    `int tan^(−1)x\ dx` `= x\ tan^(−1)x-int x/(1 + x^2)\ dx`
    `= x\ tan^(−1)x-1/2 ln(1 + x^2)+c`

Conics, EXT2 2016 HSC 12a

The diagram shows an ellipse.

ext2-hsc-2016-12a

  1. Write an equation for the ellipse.  (1 mark)
  2. Find the eccentricity of the ellipse.  (1 mark)
  3. Write the coordinates of the foci of the ellipse.  (1 mark)
  4. Write the equations of the directrices of the ellipse.  (1 mark)
Show Answers Only
  1. `(x^2)/9 + (y^2)/4 = 1`
  2. `sqrt5/3`
  3. `(−sqrt5,0)\ text(and)\ (sqrt5,0)`
  4. `x = −(9sqrt5)/5\ text(and)\ x = (9sqrt5)/5`
Show Worked Solution

i.   `(x^2)/9 + (y^2)/4 = 1`

 

ii.    `4` `= 9(1 – e^2)`
  `9e^2` `= 5`
  `e^2` `= 5/9`
  `:. e` `= sqrt5/3`

 

iii.   `text(Foci are)`

`(−sqrt5,0)\ text(and)\ (sqrt5,0)`

 

iv.   `text(Ellipse directrices:)`

`x = −(9sqrt5)/5\ \ text(and)\ \ x = (9sqrt5)/5`

Functions, EXT1′ F1 2016 HSC 11d

The diagram shows the graph of  `y = f(x).`
 

ext2-hsc-2016-11d

 
Draw a separate half-page diagram for each of the following functions, showing all asymptotes and intercepts.

  1.  `y = sqrt (f(x))`  (2 marks)

  2.  `y = 1/(f(x))`  (2 marks)

Show Answers Only
i.   

ext2-hsc-2016-11d-answer2
ii.   

ext2-hsc-2016-11d-answer4
Show Worked Solution
i.    ext2-hsc-2016-11d-answer2

 

ii.    ext2-hsc-2016-11d-answer4

Calculus, EXT2 C1 2016 HSC 11b

Find  `int x e^(-2x)\ dx.`  (3 marks)

Show Answers Only

`-1/2 xe^(-2x)-1/4 e^(-2x) + c`

Show Worked Solution

`text(Integrating by parts:)`

`text(Let)` `u` `= x` `v^{′}` `= e^(-2x)`
  `u^{′}` `= 1` `v` `= -1/2e^(-2x)`

 

`int xe^(-2x)\ dx` `= x · -1/2 e^(-2x) + 1/2int e^(-2x)\ dx`
  `= -1/2 xe^(-2x)-1/4 e^(-2x) + c`

Complex Numbers, EXT2 N1 2016 HSC 11a

Let  `z = sqrt 3 - i.`

  1.  Express  `z`  in modulus-argument form.  (2 marks)

  2.  Show that  `z^6`  is real.  (1 mark)

  3.  Find a positive integer `n` such that  `z^n`  is purely imaginary.  (1 mark)

Show Answers Only
  1. `z = sqrt3 – i = 2 text(cis)((−pi)/6)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `n = 3`
Show Worked Solution

i.   `z = sqrt3 – i`

`|\ z\ | = sqrt((sqrt3)^2 + 1^2) = 2`

`:. z = sqrt3 – i` `= 2(sqrt3/2 – 1/2 i)`
  `= 2(cos(− pi/6) + isin(− pi/6))`
  `= 2 text(cis)(− pi/6)`

 

ii.    `z^6` `= 2^6(cos(− pi/6) + isin(− pi/6))^6`
    `= 64\ text(cis)(−pi)quadquadtext{(by De Moivre)}`
    `= −64`

 
`:. z^6\ text(is real.)`

 

iii.   `z^n = 2^n (cos(−(npi)/6) + isin(−(npi)/6))`

`z^n\ text(is purely imaginary when:)`

`cos(−(npi)/6)=0`

`text(Or more generally,)`

`(npi)/6` `= pi/2 + kpi`
`n` `= 6k + 3,quad(k ∈ ZZ)`
`:.n` `=3,\ \ (n>0)`

Mechanics, EXT2 2016 HSC 8 MC

A small object of mass `m` kg sits on a rotating conical surface at `C, r` metres from the axis `OA` and with `/_ OCB = theta,` as shown in the diagram.

The surface is rotating about its axis with angular velocity `omega.` The forces acting on the object are gravity, a normal reaction force `N` and a frictional force `F`, which prevents the object from sliding down the surface.

 ext2-hsc-2016-8a

Which pair of statements is correct?

A. `F cos theta + N sin theta = mr omega^2`
  `F sin theta + N cos theta = mg`
B. `F cos theta + N sin theta = mr omega^2`
  `F sin theta - N cos theta = mg`
C. `F cos theta - N sin theta = mr omega^2`
  `F sin theta + N cos theta = mg`
D.  `F cos theta - N sin theta = mr omega^2`
  `F sin theta - N cos theta = mg`
Show Answers Only

`=> C`

Show Worked Solution

`text(Resolving the forces,)`

`text(Horizontally:)`

`Fcostheta – Nsintheta = mromega^2`

`text(Vertically:)`

`Fsintheta + Ncostheta = mg`

`=> C`

Measurement, STD2 M1 2016 HSC 28b

The cost of buying a new heater is $990. It has an energy consumption of 505 kWh per year.

Energy is charged at the rate of $0.35 kWh.

How much will it cost in total to purchase and then run this heater for five years?  (2 marks)

Show Answers Only

`$1873.75`

Show Worked Solution

`text(C)text(ost to run heater for 5 years)`

`= 5 xx 505 xx 0.35`

`= $883.75`
 

`:.\ text(Total purchase and running cost)`

`= 883.75 + 990`

`= $1873.75`

Financial Maths, STD2 F4 2016 HSC 27d

Marge borrowed $19 000 to buy a used car. Interest on the loan was charged at 4.8% pa at the end of each month. She made a repayment of $436 at the end of every month. The table below sets out her monthly repayment schedule for the first four months of the loan. 
 

2ug-2016-hsc-q27_1

  1. Some values in the table are missing. Write down the values for `A` and `B`.  (2 marks)

  2. Calculate the value of `X`.  (2 marks)

  3. Marge repaid this loan over four years.

     

    What is the total amount that Marge repaid?  (1 mark)

Show Answers Only
  1. `A = $19\ 000,quadB = $17\ 551.33`
  2. `$74.56`
  3. `$20\ 928`
Show Worked Solution
i.    `A + 76-436` `= 18\ 640`
  `:. A` `= $19\ 000`

 

`17\ 915.67 + 71.66-436 = B`

`:. B = $17\ 551.33`

 

ii.   `18\ 640 + X-436 = 18\ 278.56`

`:. X` `= 18\ 278.56 + 436-18\ 640`
  `= $74.56`
♦♦ Mean mark part (iii) 28%.
COMMENT: Read carefully whether total paid or total interest paid is required.

 

iii.   `text(Total amount repaid)`

`= 48 xx 436`

`= $20\ 928`

Algebra, STD2 A2 2016 HSC 26c

Peta’s car uses fuel at the rate of 5.9 L /100 km for country driving and 7.3 L /100 km for city driving. On a trip, she drives 170 km in the country and 25 km in the city.

Calculate the amount of fuel she used on this trip.  (2 marks)

Show Answers Only

`11.855\ text(L)`

Show Worked Solution

`text(Fuel used in country)`

`= 170 xx 5.9/100`

`= 10.03\ text(L)`

`text(Fuel used in city)`

`= 25 xx 7.3/100`

`= 1.825\ text(L)`

 

`:.\ text(Total fuel used)`

`= 10.03 + 1.825`

`= 11.855\ text(L)`