SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Vectors, SPEC1 2017 VCAA 9

A particle of mass 2 kg with initial velocity  `3underset~i + 2underset~j`  ms−1 experiences a constant force for 10 seconds.

The particle's velocity at the end of the 10-second period  `43underset~i-18underset~j`  ms−1 .

  1. Find the magnitude of the constant force in newtons.   (2 marks)

  2. Find the displacement of the particle from its initial position after 10 seconds.   (3 marks)

Show Answers Only
  1. `4sqrt5 N`
  2. `230underset~i-80underset~j`
Show Worked Solution
a.    `u_i` `= 3` `v_i` `= 43`
  `u_j` `= 2` `v_j` `= −18`
`43` `= 3 + 10a`
`40` `= 10a`
`a_i` `= 4`

♦ Mean mark (a) 38%.

`−18` `= 2 + 10a_j`
`−20` `= 10a_j`
`a_j` `= −2`

 

`underset~a` `= 4underset~i-2underset~j`
`|underset~a|` `= sqrt(4^2 + (−2)^2)`
  `= sqrt(20)`
  `= 2sqrt5`

 

`:.|F|` `= 2 xx 2sqrt5`
  `= 4sqrt5 N`

 

b.  `underset~a = 4underset~i-2underset~j`

`underset~v` `= int4underset~i-2underset~j\ dt`
  `= (4t + C_i)underset~i + (−2t + C_j)underset~j\ \ \ (C_i, C_j ∈ R)`

 
`underset~v(0) = C_i underset~i + C_j underset~j = 3underset~i + 2underset~j`

`=> C_i = 3, C_j = 2`
 

`underset~v(t) = (4t + 3)underset~i + (2-2t)underset~j`

`underset~x(t)` `= int(4t + 3)underset~i + (2-2t)underset~j\ dt`
  `= (((4t^2)/2 +3t) + b_i)underset~i + ((2t-(2t^2)/2) + b_j)underset~j \ \ \ (b_i, b_j ∈ R)`
  `= (2t^2 + 3t + b_i)underset~i + (2t-t^2 + b_j)underset~j`

 
`underset~x(0) = b_iunderset~i + b_junderset~j`

`underset~0 => b_i = b_j = 0`

`underset~x(t) = (2t^2 + 3t)underset~i + (2t-t^2)underset~j`

`underset~x(10)` `= (2(10)^2 + 3(10))underset~i + (2(10)-10^2)underset~j`
  `= 230underset~i-80underset~j`

Calculus, SPEC1 2017 VCAA 8

A slope field representing the differential equation  `dy/dx = −x/(1 + y^2)`  is shown below.

  1. Sketch the solution curve of the differential equation corresponding to the condition  `y(−1) = 1`  on the slope field above and, hence, estimate the positive value of `x` when  `y = 0`. Give your answer correct to one decimal place.   (2 marks)

  2. Solve the differential equation  `(dy)/(dx) = (−x)/(1 + y^2)`  with the condition  `y(−1) = 1`. Express your answer in the form  `ay^3 + by + cx^2 + d = 0`, where `a`, `b`, `c` and `d` are integers.   (2 marks)

Show Answers Only
  1.  

  2. `2y^3 + 6y + 3x^2 – 11 = 0`
Show Worked Solution
a.   

♦♦ Mean mark part (a) 32%.
MARKER’S COMMENT: Solution curve should follow slope ticks and not cross them.

 

b.    `(1 + y^2)(dy)/(dx)` `= −x`
  `int 1 + y^2 dy` `= −int x\ dx`
  `y + (y^3)/3` `= −(x^2)/2 + C, C ∈ R`

 
`text(Substituting)\ (-1,1):`

`1 + (1^3)/3` `= −((−1)^2)/2 + C`
`1 + 1/3` `= −1/2 + C`
`:. C` `= 1/2 + 4/3`
  `= 11/6`

 

`y + 1/3y^3` `= −1/2x^2 + 11/6`
`6y + 2y^3` `= −3x^2 + 11`

 

`:. 2y^3 + 6y + 3x^2 – 11 = 0`

Complex Numbers, SPEC1 VCAA 2017 3

Let  `z^3 + az^2 + 6z + a = 0, \ z ∈ C`, where `a` is a real constant.

Given that  `z = 1 - i`  is a solution to the equation, find all other solutions.  (3 marks)

Show Answers Only
`z_2` `= 1 + i`
`z_3` `= 2`
Show Worked Solution

`z_1 = 1 – i`

`=>\ z_2 = 1 + i\ \ text{(conjugate pair)}`

`=>\ z_3 ∈ R`
 

`z^3 + az^a + 6z + a`

  `= (z – z_1)(z – z_2)(z – z_3)`
  `= (z – (1 – i))(z – (1 + i))(z – z_3)`
  `= ((z – 1) + i)((z – 1) – i)(z – z_3)`
  `= ((z – 1)^2 – i^2)(z – z_3)`
  `= (z^2 – 2z + 1 + 1)(z – z_3)`
  `= (z^2 – 2z + 2)(z – z_3)`
  `= z^3 – (z + z_3)z^2 + (2z_3 + 2)z – 2z_3`

 
`text(Equating the co-efficients of)\ \ z:`

`6` `= 2z_3 + 2`
`4` `= 2z_3`
`z_3` `= 2`

 
`:.\ text(Other solutions are:)`

`z_2` `= 1 + i`
`z_3` `= 2`

Calculus, SPEC1 2017 VCAA 1

Find the equation of the tangent to the curve given by  `3xy^2 - 2y = x`  at the point (1, –1).  (3 marks)

Show Answers Only

`y = 1/2x – 3/2`

Show Worked Solution
`d/dx(3xy^2) + d/dx(2y)` `= d/dx(x)`
`d/(dx)(3x) · y^2 + d/(dx)(y^2)(3x) + 2(dy)/(dx)` `= 1`
`3y^2 + 6xy*(dy)/(dx) + 2(dy)/(dx)` `= 1`
`text(At)\ \ (1, -1):`  
`3(−1)^2 + 6(1)(−1)m + 2m` `= 1`
`3 – 6m + 2m` `= 1`
`−4m` `= −2`
`m` `= 1/2`
`y+1` `=1/2(x-1)`  
`:. y` `=1/2x -3/2`  

Statistics, SPEC2-NHT 2018 VCAA 19 MC

A local supermarket sells apples in bags that have negligible mass. The stated mass of a bag of apples is 1 kg.

The mass of this particular type of apple is known to be normally distributed with a mean of 115 grams and a standard deviation of 7 grams. A particular bag contains nine randomly selected apples.

The probability that the nine apples in this bag have a total mass of less than 1 kg is

  1. 0.0478
  2. 0.1132
  3. 0.4265
  4. 0.5373
  5. 0.9522
Show Answers Only

`A`

Show Worked Solution

`M~N (115, 7^2)`

`T = M + M + M + M + M + M + M + M + M`

`T~N (9 xx 115, 9 xx 7^2)~N(1035, 21^2)`

`\text{Pr} (T < 1000) ~~ 0.0478`

 
`=>  A`

Statistics, SPEC2-NHT 2018 VCAA 18 MC

The heights of all six-year-old children in a given population are normally distributed. The mean height of a random sample of 144 six-year-old children from this population is found to be 115 cm.

If a 95% confidence interval for the mean height of all six-year-old children is calculated to be (113.8, 116.2) cm, the standard deviation used in this calculation is closest to

A.     1.20

B.     7.35

C.   15.09

D.   54.02

E.   88.13

Show Answers Only

`B`

Show Worked Solution
`M` `=(116.2 – 113.8)/2`
`(1.96 xx sigma)/sqrt 144` `= 1.2`
`sigma` `= (1.2 xx 12)/1.96`
  `= 7.35`

 
`=>  B`

Calculus, SPEC2-NHT 2018 VCAA 17 MC

An object travels in a straight line relative to an origin `O`.

At time `t` seconds its velocity, `v` metres per second, is given by
 

`v(t) = {(sqrt(4 - (t - 2)^2), text(,) quad 0 <= t <= 4), (-sqrt(9 - (t - 7)^2), text(,) quad 4 < t <= 10):}`
 

The graph of  `v(t)`  is shown below.

The object will be back at its initial position when `t` is closest to

A.   4.0

B.   6.5

C.   6.7

D.   6.9

E.   7.0

Show Answers Only

`C`

Show Worked Solution

`text(Solve for)\ k:`

`int_0^4 sqrt(4 – (t – 2)^2)\ dt= int_4^k sqrt(9 -(t – 7)^2)\ dt`

 
`k~~ 6.7`

`=>  C`

Mechanics, SPEC2-NHT 2018 VCAA 16 MC

A body of mass 2 kg is moving in a straight line with constant velocity when an external force of `8 N` is applied in the direction of motion for `t` seconds.

If the body experiences a change in momentum of 40 kg ms¯¹, then `t` is

A.   3

B.   4

C.   5

D.   6

E.   7

Show Answers Only

`C`

Show Worked Solution

`sum F = 8 = 2a\ \ =>\ \ a = 4`

`m(v – u)` `= 40`
`2(v – u)` `= 40`
`v – u` `= 20`

 

`v` `= u + at`
`v – u` `= at`
`20` `= 4t`
`:. t` `= 5`

 
`=>  C`

Mechanics, SPEC2-NHT 2018 VCAA 15 MC

An 80 kg person stands in an elevator that is accelerating downwards at `1.2\ text(ms)^(-2)`.

The reaction force of the elevator floor on the person, in newtons, is

A.   688

B.   704

C.   784

D.   880

E.   896

Show Answers Only

`A`

Show Worked Solution

`sum F` `=80 text(g) – R`
`80 text(g) – R` `= 80 xx 1.2`
`:. R` `= 80xx9.8 – 80 xx 1.2`
  `= 688`

 
`=>  A`

Mechanics, SPEC2-NHT 2018 VCAA 14 MC

The diagram above shows a particle at `O` in equilibrium in a plane under the action of three forces of magnitudes `P, Q` and `R`.

Which one of the following statements is false?

A.   `R = Q sin (60^@)`

B.   `Q = R sin (60^@)`

C.   `P = R sin(30^@)`

D.   `Q cos (60^@) = P cos (30^@)`

E.   `P cos (60^@) + Q cos (30^@) = R`

Show Answers Only

`A`

Show Worked Solution

 

`text(By elimination:)`

`sin 60^@` `= Q/R`
`Q` `= R\ sin 60^@ \ \ text{(B correct)}`
`sin30^@` `= P/R`
`P` `= R\ sin30^@\ \ text{(C correct)}`

 

`text(Equating horizontal forces:)`

`Q\ cos 60^@ = P\ cos 30^@\ \ text{(D correct)}`

 

`text(Equating vertical forces:)`

`P\ cos 60^@ + Q\ cos 30^@ = R\ \ text{(E correct)}`

 
`=>  A`

Vectors, SPEC2-NHT 2018 VCAA 12 MC


 

In the diagram above, `LOM` is a diameter of the circle with centre `O`.

`N` is a point on the circumference of the circle.

If  `underset~r = vec(ON)`  and  `underset ~s = vec(MN)`, then  `vec(LN)`  is equal to

  1. `2 underset ~r - 2 underset ~s`
  2. `underset ~r - 2 underset ~s`
  3. `underset ~r + 2 underset ~s`
  4. `2 underset ~r + underset ~s`
  5. `2 underset ~r - underset ~s`
Show Answers Only

`E`

Show Worked Solution

`vec(LN)` `= vec(LM) + vec(MN)`
`vec(LN)` `= vec(LO) + vec(ON)`
  `= 1/2\ vec(LM) + vec(ON)`

 

`:. vec(LM) + underset~s` `= 1/2\ vec(LM) + underset~r`
`1/2\ vec(LM)` `= underset ~r – underset~s`
`vec(LM)` `= 2 underset~r – 2 underset~s`

 

`:. vec(LN)` `= 2 underset~r – 2 underset~s + underset~s`
  `= 2 underset~r –  underset~s`

 
`=>  E`

Calculus, SPEC2-NHT 2018 VCAA 8 MC

Using a suitable substitution,  `int_1^2(3/(2 + (4x + 1)^2))\ dx`  can be expressed as

A.  `3/4 int_1^2 (1/(2 + u^2))\ du`

B.  `3/4 int_5^9 (1/(2 + u^2))\ du`

C.   `3 int_5^9 (1/(2 + u^2))\ du`

D.   `3 int_1^2 (1/(2 + u^2))\ du`

E.   `-12 int_9^5 (1/(2 + u^2))\ du`

Show Answers Only

`B`

Show Worked Solution

`u = 4x + 1 qquad qquad qquad u(1) = 5`

`(du)/(dx) = 4 qquad qquad qquad qquad \ \ u(2) = 9`

`3/4 \ du =3\ dx`
 

`int_1^2(3/(2 + (4x + 1)^2))\ dx`

`=int_5^9 ((3/4)/(2 + u^2))\ du`

`= 3/4 int_5^9 (1/(2 + u^2))\ du`

 
`=>  B`

Complex Numbers, SPEC2-NHT 2018 VCAA 6 MC

Given that  `(z - 3i)`  is a factor of  `P(z) = z^3 + 2z^2 + 9z + 18`, which one of the following statements is false?

  1. `P(3i) = 0`
  2. `P(-3i) = 0`
  3. `P(z)`  has three linear factors over `C`
  4. `P(z)`  has no real roots
  5. `P(z)`  has two complex conjugate roots
Show Answers Only

`D`

Show Worked Solution

`z – 3i\ \ text(is a factor) \ => \ z + 3i\ text(is a factor)`

`text(deg) (P) = 3`

`text(Complex roots occur in conjugate pairs.)`

`text(S)text(ince 3 roots → 3rd root must be real.)`

`=>   D`

Graphs, SPEC2-NHT 2018 VCAA 3 MC

The implied domain of the function with rule  `f(x) = (3x)/(pi/2 - arccos (2 - x))`  is

A.   `[1, 3]`

B.   `[-1, 1]`

C.   `[0, 1) uu (1, 2]`

D.   `[-1, 0) uu (0, 1]`

E.   `[1, 2) uu (2, 3]`

Show Answers Only

`E`

Show Worked Solution

`-1 <= 2 -x <= 1`

`-3 <= – x <= – 1`

   `3 >= x >= 1\ \ …\ (1)`
 

`(2):`

`pi/2 != cos^(-1) (2 – x)`

`0 != 2 – x`

`x != 2\ \ …\ (2)`
 

`(1) nn (2)`

`1 <= x <= 3 nn x != 2`

`:.  x in [1, 2) uu (2, 3]`
 

`=>   E`

Algebra, SPEC2-NHT 2018 VCAA 2 MC

Let  `f(x) = (sqrt(x + 1))/x`  and  `g(x) = tan^2 (x)`, where  `0 < x < pi/2`.

`f(g(x))` is equal to

  1. `sin (x) sec^2 (x)`
  2. `sec (x) tan^2 (x)`
  3. `cos (x) cot^2 (x)`
  4. `cos (x) text(cosec)^2 (x)`
  5. `text(cosec) (x) cos^2 (x)`
Show Answers Only

`D`

Show Worked Solution
`f(tan^2 (x))` `= (sqrt (tan^2(x) + 1))/(tan^2(x))`
  `= (sqrt(sec^2 (x)))/(tan^2(x))`

 
`0 < x < pi/2 => sqrt (sec^2(x)) = sec(x)`

`f(g(x))` `= (sec (x))/(tan^2(x))`
  `= (cos^2(x))/(sin^2(x) cos(x))`
  `= (cos(x))/(sin^2(x))`
  `= cos(x) text(cosec)^2 (x)`

 
`=>   D`

Graphs, SPEC2-NHT 2018 VCAA 1 MC

Let  `f(x) = text(cosec) (x)`. The graph of  `f` is transformed by:

  • a dilation by a factor of 3 from the `x`-axis, followed by
  • translation of 1 unit horizontally to the right, followed by
  • a dilation by a factor of `1/2` from the `y`-axis

The rule of the transformed graph is

  1. `g(x) = 2\ text(cosec) (3x + 1)`
  2. `g(x) = 3\ text(cosec) (2x - 1)`
  3. `g(x) = 3\ text(cosec) (2 (x - 1))`
  4. `g(x) = 2\ text(cosec) (x/3 - 1)`
  5. `g(x) = 3\ text(cosec) ((x - 1)/2)`
Show Answers Only

`B`

Show Worked Solution

`f_1 (x) = 3\ text(cosec) (x)`

`f_2 (x) = 3\ text(cosec) (x – 1)`

`f_3 (x) = g(x) = 3\ text(cosec) (2x – 1)`
 

`=>   B`

Calculus, SPEC1-NHT 2018 VCAA 9

  1.    i. Given that  `cot(2 theta) = a`, show that  `tan^2(theta) + 2a tan(theta)-1 = 0`.   (2 marks)

  2.  ii. Show that  `tan(theta) = -a +- sqrt(a^2 + 1)`.  (1 mark)

  3. iii. Hence, show that  `tan(pi/12) = 2-sqrt 3`, given that  `cot(2 theta) = sqrt 3`, where  `theta in (0, pi)`.   (1 mark)

  4.  Find the gradient of the tangent to the curve  `y = tan (theta)`  at  `theta = pi/12`.   (2 marks)

  5.  A solid of revolution is formed by rotating the region between the graph of  `y = tan(theta)`, the horizontal axis, and the lines  `theta = pi/12`  and  `theta = pi/3`  about the horizontal axis.
  6. Find the volume of the solid of revolution.   (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)}`
  2. `8-4 sqrt 3`
  3. `pi (2 sqrt 3-2-pi/4)`

Show Worked Solution

a.i.   `1/(tan 2 theta)` `= a`
  `1` `= a tan (2 theta)`
  `1` `= a ((2 tan (theta))/(1-tan^2(theta)))`
  `1/a (1-tan^2 (theta))` `= 2 tan (theta)`
  `1-tan^2 (theta)` `= 2a tan (theta)`
  `:. tan^2 (theta) + 2 a tan (theta)-1` `=0\ \ text(… as required)`

 

a.ii.   `[tan^2 (theta) + 2a tan (theta) + a^2]-a^2-1` `= 0`
  `(tan (theta) + a)^2` `= a^2 + 1`
  `tan (theta) + a` `= +- sqrt(a^2 + 1)`
  `:. tan (theta)` `= -a +- sqrt(a^2 + 1)`

 

a.iii.   `theta in (0, pi) \ => \ 2 theta in (0, 2 pi)`

`text(S)text(ince)\ \ cot(2theta)=sqrt3\ \ \ =>\ \ \ tan(2theta)=1/sqrt3`

  `:. 2 theta` `= pi/6, pi + pi/6`
  `theta` `= pi/12, (7 pi)/12`
  `tan (theta)` `=-sqrt 3 +- sqrt((sqrt 3)^2 + 1)`
    `=-sqrt 3 +- sqrt(3 + 1)`
    `=-sqrt 3 +- 2`

 
`:. tan (pi/12) = 2-sqrt 3,\ \ \ \ (tan (pi/12) > 0)`

 

b.   `y` `=tan(theta)`
  `y prime` `= sec^2 (theta)`
    `= 1 + tan^2 (theta)`

 

`y prime (pi/12)` `= 1 + tan^2 (pi/12)`
  `= 1 + (2-sqrt 3)^2`
  `= 1 + 4-4 sqrt 3 + 3`
  `= 8-4 sqrt 3`

 

c.   `V` `= pi int_(pi/12)^(pi/3) y^2\ d theta`
    `= pi int_(pi/12)^(pi/3) tan^2 (theta)\ d theta`
    `= pi int_(pi/12)^(pi/3) (1 + tan^2 (theta)-1)\ d theta`
    `= pi int_(pi/12)^(pi/3) (sec^2 (theta)-1)\ d theta`
    `= pi [tan (theta)-theta]_(pi/12)^(pi/3)`
    `= pi (tan (pi/3)-pi/3-(tan (pi/12)-pi/12))`
    `= pi (sqrt 3-(4 pi)/12-(2-sqrt 3) + pi/12)`
    `= pi (2 sqrt 3-2-pi/4)\ \ text(u³)`

Complex Numbers, SPEC1-NHT 2018 VCAA 8

A circle in the complex plane is given by the relation  `|z-1-i| = 2, \ z in C`.

  1. Sketch the circle on the Argand diagram below.   (1 mark)

 

  1. i.  Write the equation of the circle in the form  `(x-a)^2 + (y-b)^2 = c`  and show that the gradient of a tangent to the circle can be expressed as  `(dy)/(dx) = (1-x)/(y-1)`.   (2 marks)

  2. ii. Find the gradient of the tangent to the circle where  `x = 2`  in the first quadrant of the complex plane.   (1 mark)

  3. Find the equations of all rays that are perpendicular to the circle in the form  `text(Arg) (z) = alpha`.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
    1.  `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `(-1)/sqrt 3`
  3. `text(Arg) (z) = pi/4; quad text(Arg) (z) = (-3 pi)/4`
Show Worked Solution
a.  

 

b.i.   `(x-1)^2 + (y-1)^2` `= 4`
  `2(x-1) + d/(dx) ((y-1)^2)` `= 0`
  `2(x-1) + 2(y-1)*(dy)/(dx)` `= 0`
  `2 (y-1)*(dy)/(dx)` `= -2(x-1)`
  `(dy)/(dx)` `= (-(x-1))/(y-1)`
    `= (1-x)/(y-1)`

 

b.ii.   `(2-1)^2 + (y-1)^2` `= 4`
  `1 + (y-1)^2` `= 4`
  `(y-1)^2` `= 3`
  `y` `= 1 + sqrt 3\ \ \ (y > 0)`
     
  `(dy)/(dx)|_{(2, 1 + sqrt 3)}` `= (1-2)/(1 + sqrt 3-1`
    `= (-1)/sqrt 3`

 

c.   `P (0, 0) quad C(1, 1)`
  `-> y = x`
  `:. alpha = pi/4, quad (-3 pi)/4` 

 
`text(Arg) (z) = pi/4`

`text(Arg) (z) = (-3 pi)/4`

Calculus, SPEC1-NHT 2018 VCAA 7

  1. Find  `d/(dx) ((1-x^2)^(1/2))`.  (2 marks)

  2. Hence, find the length of the curve specified by  `y = sqrt (1-x^2)`  from  `x = 1/2`  to  `x = sqrt 3/2`.
  3. Give your answer in the form  `k pi, k in R`.  (2 marks)

Show Answers Only

  1. `(-x)/sqrt(1-x^2)`
  2. `pi/6`

Show Worked Solution

a.  `d/(dx) ((1-x^2)^(1/2))`

`= (1/2 (-2x) (1-x^2)^(-1/2))`

`= -x(1-x^2)^(-1/2)`

`= (-x)/sqrt (1-x^2)`

 

b.   `l` `= int_(1/2)^(sqrt 3/2) sqrt(1 + ((-x)/sqrt(1-x^2))^2)\ dx`
    `= int_(1/2)^(sqrt 3/2) sqrt((1-x^2 + x^2)/(1-x^2))\ dx`
    `= int_(1/2)^(sqrt 3/2) sqrt(1/(1-x^2))\ dx`
    `= int_(1/2)^(sqrt 3/2) 1/sqrt(1-x^2)\ dx`
    `= [sin^(-1) (x)]_(1/2)^(sqrt 3/2)`
    `= sin^(-1) (sqrt 3/2)-sin^(-1) (1/2)`
    `= pi/3-pi/6`
    `= pi/6`

Calculus, SPEC1-NHT 2018 VCAA 6

Given that  `y = (x-1)e^(2x)`  is a solution to the differential equation  `a(d^2y)/(dx^2) + b (dy)/(dx) = y`, find the values of `a` and `b`, where `a` and `b` are real constants.   (4 marks)

Show Answers Only

`a = -1/4, b = 1`

Show Worked Solution
`y` `= (x-1)e^(2x)`
`y^{′}` `= e^(2x) + 2e^(2x) (x-1)`
  `= e^(2x) + 2y`
`y^{″}` `= 2e^(2x) + 2y^{′}`
  `= 2e^(2x) + 2e^(2x) + 4e^(2x) (x-1)`
  `= 4e^(2x) + 4e^(2x) (x-1)`
  `= 4e^(2x) (1 + x-1)`
  `= 4xe^(2x)`

 
`ay^{″} + by^{′} = y`

`4axe^(2x) + be^(2x) (1 + 2(x-1))` `= (x-1)e^(2x)`  
`e^(2x)(4ax + b (2x-1)-(x-1))` `= 0`  
`4ax + b + 2bx-2b-x + 1` `= 0,\ \ \ (e^(2x) != 0)`  
`(4a + 2b-1) x + 1-b` `= 0x + 0`  

 
`:.b = 1`
 

`4a + 2-1 = 0`

`:. a = -1/4`

Calculus, SPEC1-NHT 2018 VCAA 5

Evaluate  `int_1^(2 sqrt(3) - 1) (1/(x^2 + 2x + 5))\ dx`.  (4 marks)

Show Answers Only

`pi/24`

Show Worked Solution

` int_1^(2 sqrt(3) – 1) (1/(x^2 + 2x + 5))\ dx`

`= int_1^(2 sqrt(3) – 1) 1/((x + 1)^2 + 4)\ dx`
 

`text(Let)\ \ u = x + 1, quad (du)/(dx) = 1\ \ =>\ \ du=dx`

`text(Limits:)\ \ u(1) = 2, quad u(2 sqrt 3 – 1) = 2 sqrt 3`
 

`:. int_1^(2 sqrt(3) – 1) 1/(x^2 + 2x + 5)\ dx`

`= int_2^(2 sqrt 3) (1/(u^2 + 4))\ du`

`= 1/2 int_2^(2 sqrt 3) 2/(u^2 + 4)\ du`

`= 1/2 [tan^(-1) (u/2)]_2^(2 sqrt 3)`

`= 1/2(tan^(-1) (sqrt 3) – tan^(-1)(1))`

`= 1/2 (pi/3 – pi/4)`

`= 1/2 ((4 pi – 3 pi)/12)`

`= pi/24`

Statistics, SPEC1-NHT 2018 VCAA 4

Throughout this question, use an integer multiple of standard deviations in calculations.

The standard deviation of all scores on a particular test is 21.0

  1. From the results of a random sample of `n` students, a 95% confidence interval for the mean score for all students was calculated to be  `(44.7, 51.7)`.

     

    Calculate the mean score and the size of this random sample.  (2 marks)

  2. Determine the size of another random sample for which the endpoints of the 95% confidence interval for the population mean of the particular test would be 1.0 either side of the sample mean.  (2 marks)

Show Answers Only
  1.  `bar x = 48.2; quad n = 144`
  2.  `1764`
Show Worked Solution

a.  `bar x = (51.7 + 44.7)/2 = 96.4/2 = 48.2`
 

`text{Difference: z-score –2 to 0 = 3.5}`

`2 xx 21/sqrt n` `= 7/2`
`7 sqrt n` `= 2 xx 2 xx 21`
`sqrt n` `= (2 xx 2 xx 7 xx 3)/7`
  `=12`
`:.n` `=144`

 

b.   `2 xx 21/sqrtn` `= 1`
  `sqrt n` `= 42`
  `n` `= 42^2`
    `= (40 + 2)^2`
    `= 1600 + 160 + 4`
    `= 1764`

Algebra, SPEC1-NHT 2018 VCAA 3

Find  `sin(t)`  given that  `t = arccos (12/13) + arctan (3/4)`.  (3 marks)

Show Answers Only

`56/65`

Show Worked Solution
`sin (t)` `= sin (cos^(-1) (12/13) + tan^(-1)(3/4))`
  `= sin (cos^(-1) (12/13)) cos (tan^(-1) (3/4)) + sin (tan^(-1)(3/4)) cos (cos^(-1)(12/13))`

 

 

`cos^(-1) (12/13) in (0, pi/2), quad tan^(-1)(3/4) in (0, pi/2)`

`:. sin (t)` `= 5/13 xx 4/5 + 3/5 xx 12/13`
  `= 20/65 + 36/65`
  `= 56/65`

Vectors, SPEC1-NHT 2018 VCAA 2

Let  `underset ~a = 3 underset ~i - 2 underset ~j + m underset ~k`  and  `underset ~b = 2 underset ~i - underset ~j + 3 underset ~k`, where  `m in R`.

Find the value(s) of `m` such that the magnitude of the vector resolute of  `underset ~a` parallel to `underset ~b`  is equal to  `sqrt 14`.  (3 marks)

Show Answers Only

`m = (-22)/3, 2`

Show Worked Solution

`underset ~a ⋅ underset ~ hat b = (3 xx 2 +(-2 xx – 1) + m xx 3)/sqrt(2^2 + (-1)^2 + 3^2) = +- sqrt 14`

`(8 + 3m)/sqrt 14` `= +- sqrt 14`
`8 + 3m` `= +- 14`
`3m` `= -8 +- 14`
`m` `= (-8 +- 14)/3`
`:. m` `= – 22/3, 2`

Mechanics, SPEC1-NHT 2018 VCAA 1

A light inextensible string hangs over a frictionless pulley connecting masses of 3 kg and 7 kg, as shown below.
 


 

  1.  Draw all of the forces acting on the two masses on the diagram above. (1 mark)
  2.  Calculate the tension in the string.  (2 marks)
Show Answers Only
  1.  `text(See Worked Solutions)`
  2. `(21g)/5 N`
Show Worked Solution
a.   

 

b.    `sum F = 7g – 3g = 4g` ` = (7 + 3)a`
  `a = (4g)/10= (2g)/5`  
  `sum F_3 = T – 3g` `= 3a`
  `T – 3g` `= (6g)/5`
  `T` `= 3g + (6g)/5`
    `= (21g)/5\ \ N`

Statistics, SPEC2 2018 VCAA 20 MC

The scores on the Mathematics and Statistics tests, expressed as percentages, in a particular year were both normally distributed. The mean and the standard deviation of the Mathematics test scores were 71 and 10 respectively, while the mean and the standard deviation of the statistics test scores were 75 and 7 respectively.

Assuming the sets of tests scores were independent of each other, the probability, correct to four decimal places, that a randomly chosen Mathematics score is higher than a randomly chosen Statistics score is

  1. 0.2877
  2. 0.3716
  3. 0.4070
  4. 0.7123
  5. 0.9088
Show Answers Only

`B`

Show Worked Solution

`M~N (71 , 10^2)`

`S~N (75, 7^2)`

`text(Pr)(M > S) = text(Pr)(M-S > 0)`

`M-S = X~N(71-75, 1^2 xx 10^2 + (-1)^2 xx 7^2)`

`X~N(-4, 149)`
 

`text(Pr)(M > S) = text(Pr)(X > 0) ~~ 0.3716`

`=>  B`

Statistics, SPEC2 2018 VCAA 19 MC

The gestation period of cats is normally distributed with mean  `mu = 66`  days and variance  `sigma^2 = 16/9`.

The probability that a sample of five cats chosen at random has an average gestation period greater than 65 days is closest to

  1. 0.5000
  2. 0.7131
  3. 0.7734
  4. 0.8958
  5. 0.9532
Show Answers Only

`E`

Show Worked Solution

`G~N (66 , 16/9)`

`bar G~N (66, 16/(9 xx 5))`

`text(Pr)(bar G > 65) ~~ 0.9532`

`=>  E`

Mechanics, SPEC2 2018 VCAA 16 MC

The diagram below shows a mass being acted on by a number of forces whose magnitudes are labelled. All forces are measured in newtons and the system is in equilibrium.
 

The value of `F_2` is

A.   `sqrt 2/2 (8 + 3 sqrt 3)`

B.   `(11 sqrt 2)/2`

C.   `(3 sqrt 2)/2`

D.   `7.78`

E.   `7.0`

Show Answers Only

`B`

Show Worked Solution

`text(vertical:)\ \ sum F_y = 4 + 3 sin 30^@ +\ ^(−)F_2 sin 45^@ = 0`

`4 + 3/2 – F_2/sqrt 2` `= 0`
`F_2/sqrt 2` `= 4 + 3/2`
`:. F_2` `= sqrt 2 (4 + 3/2)`
  `= (11 sqrt 2)/2`

 
`=>  B`

Statistics, SPEC2 2018 VCAA 18 MC

A 95% confidence interval for the mean height `mu`, in centimetres, of a random sample of 36 Irish setter dogs is  `58.42 < mu < 67.31`

The standard deviation of the height of the population of Irish setter dogs, in centimetres, correct to two decimal places, is

A.     2.26

B.     2.27

C.   13.60

D.   13.61

E.   62.87

Show Answers Only

`D`

Show Worked Solution
`M` `= (67.31 – 58.42)/2`
  `= 8.89/2`
  `= 4.445`

 

`M` `= (1.96 sigma)/sqrt 36`
`4.445` `= (1.96 sigma)/6`
`:. sigma` `=(6 xx 4.445)/1.96`
  `= 13.61`

 
`=>  D`

Mechanics, SPEC2 2018 VCAA 15 MC

A constant force of magnitude `P` newtons accelerates a particle of mass 8 kg in a straight line from a speed of 4 ms¯¹ to a speed of 20 ms¯¹ over a distance of 15 m.

The magnitude of `P` is

A.       9.8

B.     12.5

C.     12.8

D.   100

E.   102.4 

Show Answers Only

`E`

Show Worked Solution

`u = 4, quad v = 20, quad s = 15`

`v^2` `= u^2 + 2as`
`20^2` `= 4^2 + 30a`
`400` `= 16 + 30a`
`384` `= 30a`
`a` `= 64/5`

 

`:. P` `= 8a`
  `= (8 xx 64)/5`
  `= 102.4`

 
`=>  E`

Vectors, SPEC2 2018 VCAA 14 MC

The scalar resolute of  `underset ~a = 3 underset ~i - 2 underset ~k`  in the direction of  `underset ~b = - underset ~i + 2 underset ~j + 3 underset ~k`  is

A.   `-(9 sqrt 13)/13`

B.   `-9/14(-underset ~i + 2 underset ~j + 3 underset ~k)`

C.   `-(9 sqrt 14)/14`

D.   `-9/13 (3 underset ~i - 2 underset ~k)`

E.   `- sqrt 14/2` 

Show Answers Only

`C`

Show Worked Solution
`underset ~a ⋅ underset ~hat b`

 `= (3 xx (-1) + 0 xx 2 + (-2) xx 3)/sqrt((-1)^2 + 2^2 + 3^2)`
  `= (-3-6)/sqrt 14`
  `= -9/sqrt 14 xx sqrt 14/sqrt 14`
  `= -(9 sqrt 14)/14`

 
`=>  C`

Vectors, SPEC2 2018 VCAA 13 MC

The position vector of a particle that is moving along a curve at time `t` is given by  `underset ~r(t) = 3 cos (t) underset ~i + 4 sin (t) underset ~j, \ t >= 0`.

The first time when the speed of the particle is a minimum is

  1. `3`
  2. `pi/2`
  3. `tan^(-1) (4/3)`
  4. `(3 pi)/2`
  5. `9` 
Show Answers Only

`B`

Show Worked Solution
`underset ~dot r(t)` `= -3 sin(t) underset ~i + 4 cos (t) underset ~j`
`|underset ~dot r(t)|` `= sqrt (9 sin^2(t) + 16 cos^2(t))`
  `= sqrt (9 sin^2(t) + 9 cos^2(t) + 7 cos^2(t))`
  `= sqrt (9 + 7 cos^2(t))`

 
`text(Minimised for)\ cos(t) = 0`

`:.  t_1 = pi/2`

`=>  B`

Calculus, SPEC2 2018 VCAA 10 MC

The differential equation that best represents the direction field above is

A.  `(dy)/(dx) = (2x + y)/(y - 2x)`

B.  `(dy)/(dx) = (x + 2y)/(2x - y)`

C.  `(dy)/(dx) = (2x - y)/(x + 2y)`

D.  `(dy)/(dx) = (x - 2y)/(y - 2x)`

E.  `(dy)/(dx) = (2x + y)/(2y - x)` 

Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=0, \ m=1`

`text(When)\ \ y=0, \ m=-1`

`=>  A`

Calculus, SPEC2 2018 VCAA 9 MC

A solution to the differential equation  `(dy)/(dx) = 2/{sin(x + y)-sin(x-y)}`  can be obtained from

  1. `int 1\ dx = int 2 sin(y)\ dy`
  2. `int cos(y)\ dy = int text{cosec}(x)\ dx`
  3. `int cos(x)\ dx = int text{cosec}(y)\ dy`
  4. `int sec(x)\ dx = int sin(y)\ dy`
  5. `int sec(x)\ dx = int text{cosec}(y)\ dy` 
Show Answers Only

`D`

Show Worked Solution
`(dy)/(dx)` `= 2/{sin(x) cos(y) + sin(y) cos(x)-(sin(x) cos(y)-sin(y) cos(x))}`
  `= 2/{2 sin(y) cos(x)}`
  `= 1/{sin(y) cos(x)}`

 
`sin(y) *(dy)/(dx)= sec(x)`

`int sin (y)\ dy= int sec(x)\ dx`

`=>  D`

Calculus, SPEC2 2018 VCAA 8 MC

Using a suitable substitution, `int_0^(pi/6) tan^2 (x) sec^2(x)\ dx`  can be expressed as

A.  `int_0^(1/sqrt 3) (u^4 + u^2) du`

B.  `int_1^(2/sqrt 3) (u^4 + u^2) du`

C. `int_0^(1/(sqrt 3)) u\ du`

D. `int_0^(pi/6) u^2\ du`

E. `int_0^(1/sqrt 3) u^2\ du` 

Show Answers Only

`E`

Show Worked Solution
`text(Let)\ \ u` `= tan(x),`
`(du)/(dx)` `= sec^2 (x)\ \ =>\ du = sec^2(x)\ dx`
`u(0)` `= tan(0) = 0`
`u (pi/6)` `= tan (pi/6) = 1/sqrt 3`

 
`int_0^(pi/6) tan^2 (x)\ sec^2 (x)\ dx = int_0^(1/sqrt 3) u^2 du`

`=>  E`

Complex Numbers, SPEC2 2018 VCAA 6 MC

The complex numbers  `z, iz `  and  `z + iz`, where  `z in C\ text(\{0})`, are plotted in the Argand plane, forming the vertices of a triangle.

The area of this triangle is given by

  1. `|z|`
  2. `|z| + |z|^2`
  3. `(|z|^2)/2`
  4. `|z|^2`
  5. `(sqrt 3 |z|^2)/2` 
Show Answers Only

`C`

Show Worked Solution

`text(Area)` `= 1/2 b h`
  `= 1/2 |z + iz – z| |z + iz – iz|`
  `= 1/2 |iz| |z|`
  `= 1/2 |i| |z|^2`
  `= (|z|^2)/2`

 
`=>  C`

Calculus, SPEC1 2018 VCAA 9

A curve is specified parametrically by  `underset ~r(t) = sec(t) underset ~i + sqrt 2/2 tan(t) underset ~j, \ t in R`.

  1.  Show that the cartesian equation of the curve is  `x^2-2y^2 = 1`.   (2 marks)

  2.  Find the `x`-coordinates of the points of intersection of the curve  `x^2-2y^2 = 1`  and the line  `y = x-1`.   (1 mark)

  3.  Find the volume of the solid of revolution formed when the region bounded by the curve and the line is rotated about the `x`-axis.   (2 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x = 1 or x = 3`
  3. `(2 pi)/3`

Show Worked Solution

a.     `x = sec(t), qquad y = sqrt 2/2 tan(t)`

`x^2 = sec^2(t), qquad y^2 = 1/2 tan^2(t)`

`x^2 = sec^2(t), qquad 2y^2 = tan^2(t)`

`1 + tan^2(t)` `= sec^2(t)`
`1 + 2y^2` `= x^2`
`:.x^2-2y^2` `=1\ \ text(.. as required)`

 

b.    `x^2-2(x-1)^2` `= 1`
  `x^2-2(x^2-2x + 1)` `=1`
  `x^2-2x^2 + 4x-2` `=1`
  `-x^2 + 4x-2-1` `=0`
  `x^2-4x + 3` `=0`
  `(x-3) (x-1)` `=0`

 
`:. x = 1 or x = 3`

♦♦ Mean mark 30%.

 

c.   `x^2-{:2y_1:}^2` `=1`
  `{:2y_1:}^2` `=x^2-1`
  `{:y_1:}^2` `= (x^2-1)/2`
  `{:y_2:}^2` `= (x-1)^2`

 

`V` `=pi int_1^3 {:y_1:}^2-{:y_2:}^2 \ dx`
  `= pi int_1^3 (x^2-1)/2-(x-1)^2\ dx`
  `= pi [x^3/6-x/2-(x-1)^3/3]_1^3`
  `= pi [(3^3/6-3/2-2^3/3)-(1^3/6-1/2-0)]`
  `= pi (9/2-3/2-8/3-1/6 + 1/2)`
  `= pi (7/2-8/3-1/6)`
  `= pi ((21-16-1)/6)`
  `= (2 pi)/3`

Calculus, SPEC1 2018 VCAA 8

A tank initially holds 16 L of water in which 0.5 kg of salt has been dissolved. Pure water then flows into the tank at a rate of 5 L per minute. The mixture is stirred continuously and flows out of the tank at a rate of 3 L per minute.

  1.  Show that the differential equation for `Q`, the number of kilograms of salt in the tank after `t` minutes, is given by
  2. `qquad (dQ)/(dt) = -(3Q)/(16 + 2t)`  (1 mark)

  3. Solve the differential equation given in part a. to find `Q` as a function of `t`.
  4. Express your answer in the form  `Q = a/(16 + 2t)^(b/c)`, where `a, b` and `c` are positive integers.  (3 marks)

Show Answers Only
  1.  `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  `Q = 32/(16 + 2t)^(3/2)`
Show Worked Solution

a. `Q_0 = 0.5, \ V_0 = 16`

♦ Net mean mark of both parts 44%.

`V(t)= 16 + (5-3) t= 16 + 2t`

`text(Concentration)\ (C)= Q/V= Q/(16 + 2t)\ text(kg/L)`

`(dQ)/(dt)= 0 xx 5-3C= -(3Q)/(16 + 2t)`
 

MARKER’S COMMENT: Many students took the common factor of 2 from  `16+2t`. This wasn’t necessary and complicated the arithmetic in part b.

b.   `-1/(3Q) * (dQ)/(dt) = 1/(16 + 2t)`

`int -1/(3Q)\ dQ` `= int 1/(16 + 2t) dt`
`-1/3 int 1/Q\ dQ` `= 1/2 int 2/(16 + 2t)\ dt`
`-1/3 ln Q` ` = [1/2 ln(16 + 2t)] + c`

 
`text(When)\ \ t=0,\ \ Q=0.5,`

`-1/3 ln (1/2)` `= 1/2 ln (16) +c`
`c` `= -1/2 ln (16) -1/3 ln (1/2)`

 

`-1/3 ln Q` `= 1/2 ln(16 + 2t) -1/2 ln(16)-1/3 ln (1/2)`
`-1/3 ln Q` `= 1/2 ln ((16 + 2t)/16)-1/3 ln (1/2)`
`-1/3 ln Q` `= ln (((16 + 2t)^(1/2))/4)-ln (2^(-1/3))`
`ln (Q^(-1/3))` `= ln (((16 + 2t)^(1/2))/(2^2 ⋅ 2^(-1/3)))`
`Q^(-1/3)` `= ((16 + 2t)^(1/2))/(2^(5/3))`
`Q` `= (((16 + 2t)^(1/2))/(2^(5/3)))^-3`
  `= (16 + 2t)^(- 3/2)/(2^(-5))`
`:. Q` `= 32/((16 + 2t)^(3/2))`

Vectors, SPEC1 2018 VCAA 6

A particle of mass 2 kg moves under a force  `underset ~ F`  so that its position vector  `underset ~ r`  at anytime `t` is given by  `underset ~r = sin(t) underset ~i + cos(t) underset ~j + t^2 underset ~k`.  Distances are measured in metres and time is measured in seconds.

Find the change in momentum, in kg ms¯², from  `t = pi/2`  to  `t = pi`.   (3 marks)

Show Answers Only

`2 (- underset ~i + underset ~j +  pi underset ~k)`

Show Worked Solution
`underset ~ dot r (t)` `= cos(t) underset ~i-sin(t) underset ~j + 2t underset~k`
`underset ~ dot r (pi)` `= cos(pi) underset ~i-sin(pi) underset ~j + 2 pi underset ~k`
  `=-underset ~i + 2 pi underset ~k`
   
`underset ~ dot r (pi/2)` `= cos(pi/2) underset ~i-sin (pi/2) underset ~j + pi underset ~k`
  `=-underset ~j + pi underset ~k`
   
`:. m Delta r` `= 2(underset ~ dot r (pi)-underset ~ dot r (pi/2))`
  `= 2(- underset ~i + 2 pi underset ~k-(- underset ~j + pi underset ~k))`
  `= 2 (- underset ~i + underset ~j +  pi underset ~k)`

Graphs, SPEC1 2018 VCAA 5

Sketch the graph of  `f(x) = (x + 1)/(x^2 - 4)`  on the axes provided below, labelling any asymptotes with their equations and any intercepts with their coordinates.  (4 marks)

Show Answers Only

Show Worked Solution

`text(Using partial fractions:)`

`(x + 1)/(x^2 – 4)= A/(x – 2) + B/(x + 2)`

`A(x + 2) + B (x – 2) = x + 1`

 
`text(When)\ \ x = 2,`

`4A` `= 3 => \ A=3/4`

 
`text(When)\ \ x = -2`

`-4B` `= -1 =>\ B=1/4`

 
`f(x)= 1/(4(x + 2)) + 3/(4(x – 2))`
 

`text(As)\ \ x->oo, \ f(x)->0^+`

`text(As)\ \ x->- oo, \ f(x)->0^-`

`f(0)= – 1/4`
 

`text(Find)\ x\ text(when)\ \ f(x)=0:`

`x + 1` `= 0`
`x` `= -1`

Complex Numbers, SPEC1 2018 VCAA 2

  1. Show that  `1 + i = sqrt 2\ text(cis)(pi/4)`.   (1 mark)

  2. Evaluate  `(sqrt 3-i)^10/(1 + i)^12`, giving your answer in the form  `a + bi`, where  `a, b ∈ R`.   (3 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `-8-8 sqrt 3 i`

Show Worked Solution

a.   `r` `= sqrt(1^2 + 1^2)`
    `= sqrt 2`
  `theta` `= tan^(-1) (1/1) = pi/4`
     
  `:. 1 + i` `= sqrt 2\ text(cis)(pi/4)`

 

b.   
 `r_2` `= sqrt((sqrt 3)^2 + (-1)^2)`
    `= sqrt (3 + 1)`
    `= 2`

 
`theta_2 = -tan^(-1) (1/sqrt 3)=-pi/6`

`sqrt 3-i` `= 2\ text(cis)(-pi/6)`
`(sqrt 3-i)^10` `= 2^10\ text(cis) ((-10 pi)/6)`
   
`=>(1 + i)^12` `= (sqrt 2)^12text(cis)((12pi)/4)`
  `=2^6 text(cis)(3pi)`

 

`:. (sqrt 3-i)^10/(1 + i)^12` `= (2^10 text(cis)((-5pi)/3))/(2^6\text(cis)(3pi))`
  `= 2^4 text(cis)((-5pi)/3-(9pi)/3)`
  `= 16 text(cis)((-14pi)/3)`
  `= 16 text(cis) ((-2pi)/3)`
  `= 16(-1/2 + (-sqrt3)/2 i)`
  `= -8-8 sqrt 3 i`

Mechanics, SPEC1 2018 VCAA 1

Two objects of masses 5 kg and 8 kg are attached by a light inextensible string that passes over a smooth pulley. The 8 kg mass is on a smooth plane inclined at 30° to the horizontal. The 5 kg mass is hanging vertically, as shown in the diagram below.
 


 

  1. On the diagram above, show all forces acting on both masses.  (1 mark)
  2. Find the magnitude, in `text(ms)^(-2)`, and state the direction of the acceleration of the 8 kg mass.  (3 marks)
Show Answers Only
a.  

b.   `a = 9/13\ text(ms)^(-2)\ text(up the incline)`

Show Worked Solution
a.  

 

b.   
 `sum F` `= 5 text(g) – 8 text(g)\ sin (30^@) = (5 + 8)a\ \ text{(up slope → +)}`
  `5 text(g) – 4 text(g)` `= 13a`
  `:. a` `= 9/13\ text(ms)^(-2)\ text(up the incline)`

Calculus, MET1 2018 VCAA 8

Let  `f: R -> R, \ f(x) = x^2e^(kx)`, where `k` is a positive real constant.

  1.  Show that  `f^{′}(x) = xe^(kx)(kx + 2)`.   (1 mark)

  2. Find the value of `k` for which the graphs of  `y = f(x)`  and  `y = f^{′}(x)`  have exactly one point of intersection.   (3 marks)

Let  `g(x) = −(2xe^(kx))/k`. The diagram below shows sections of the graphs of `f` and `g` for  `x >= 0`.
 


 

Let `A` be the area of the region bounded by the curves  `y = f(x), \ y = g(x)` and the line  `x = 2`.

  1. Write down a definite integral that gives the value of `A`.   (1 mark)

  2. Using your result from part a., or otherwise, find the value of `k` such that  `A = 16/k`.   (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(One point of intersection)\ x = 0\ text(when)\ k = 1.`
  3. `int_0^2 (x^2e^(kx) + (2xe^(kx))/k)dx`
  4. `1/2 ln 4\ \ text(or)\ \ ln2`
Show Worked Solution

a.   `f(x) = x^2e^(kx)`

`f^{′}(x)` `= 2x · e^(kx) + x^2 · k · e^(kx)`
  `= xe^(kx)(kx + 2)\ \ …text(as required)`

 

b.  `text(Intersection occurs when:)`

♦♦♦ Mean mark 7%.
MARKER’S COMMENT: Most students found the correct quadratic equation but solved for `k`.

`x^2e^(kx)` `= xe^(kx)(kx + 2)`
`x^2-x(kx + 2)` `= 0,\ \ e^(kx) != 0`
`x^2-kx^2-2x` `= 0`
`x^2(1-k)-2x` `= 0`
`x[x(1-k)-2]` `= 0`

 

`:.x = 0\ \ text(or)\ \ x(1-k)-2` `= 0`
`x` `= 2/(1-k)`

 
`text(S)text(ince)\ \ x = 2/(1-k)\ \ text(is undefined when)\ \ k = 1,`

`=>\ text(One point of intersection only at)\ \ x = 0\ \ text(when)\ \ k = 1.`

 

c.    `A` `= int_0^2 f(x)\ dx-int_0^2 g(x)\ dx`
    `= int_0^2 x^2e^(kx)\ dx-int_0^2 −(2xe^(kx))/k \ dx`
    `= int_0^2 (x^2e^(kx) + (2xe^(kx))/k)dx`

♦♦ Mean mark 29%.

 

d.    `int_0^2(x^2e^(kx) + (2xe^(kx))/k)\ dx` `= 16/k`
  `1/k int_0^2(kx^2e^(kx) + 2xe^(kx))\ dx` `= 16/k`
  `1/k int_0^2(xe^(kx)(kx + 2))\ dx` `= 16/k`
  `1/k[x^2e^(kx)]_0^2` `= 16/k\ \ \ text{(using part a)}`
  `[2^2 · e^(2k)-0]` `= 16`
  `e^(2k)` `= 4`
  `2k` `= ln 4`
  `k` `= 1/2 ln 4\ \ text(or)\ \ ln2`

Probability, MET1 2018 VCAA 6

Two boxes each contain four stones that differ only in colour.

Box 1 contains four black stones.

Box 2 contains two black stones and two white stones.

A box is chosen randomly and one stone is drawn randomly from it.

Each box is equally likely to be chosen, as is each stone.

  1.  What is the probability that the randomly drawn stone is black?  (2 marks)

  2.  It is not known from which box the stone has been drawn.
  3. Given that the stone that is drawn is black, what is the probability that it was drawn from Box 1?  (2 marks)

Show Answers Only

  1. `3/4`
  2. `2/3`

Show Worked Solution

a.    `text(Pr)(text(Black))` `= text{Pr(Box 1)} · text{Pr(B)} + text{Pr(Box 2)} · text{Pr(B)}`
    `= 0.5 xx 4/4 + 0.5 xx 2/4`
    `= 3/4`

 

b.    `text(Pr)(text(Box 1 | black))` `= (text{Pr(Box 1)}\ ∩\ text{Pr(Black)})/text{Pr(Black)}`
    `= (1/2)/(3/4)`
    `= 2/3`

Algebra, STD2 A2 2007 HSC 27b*

A clubhouse uses four long-life light globes for five hours every night of the year. The purchase price of each light globe is $6.00 and they each cost  `$d`  per hour to run.

  1. Write an equation for the total cost (`$c`) of purchasing and running these four light globes for one year in terms of  `d`.    (2 marks)

  2. Find the value of  `d`  (correct to three decimal places) if the total cost of running these four light globes for one year is $250.   (1 mark)

  3. If the use of the light globes increases to ten hours per night every night of the year, does the total cost double? Justify your answer with appropriate calculations.   (1 mark)

Show Answers Only
  1. `$c = 24 + 7300d`
  2. `0.031\ $ text(/hr)\ text{(3 d.p.)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.   `text(Purchase price) = 4 xx 6 = $24`

`text(Running cost)` `= text(# Hours) xx text(Cost per hour)`
  `= 4 xx 5 xx 365 xx d`
  `= 7300d`
 
`:.\ $c = 24 + 7300d`

 

b.   `text(Given)\ $c = $250`

`=>250` `= 24 + 7300d`
`7300d` `= 226`
`d` `= 226/7300`
  `= 0.03095…`
  `= 0.031\ $ text(/hr)\ text{(3 d.p.)}`

 

c.   `text(If)\ d\ text(doubles to 0.062)\ \ $text(/hr)`

`$c` `= 24 + 7300 xx 0.062`
  `= $476.60`
   
`text(S) text(ince $476.60 is less than)\ 2 xx $250\ ($500),`
`text(the total cost increases to less than double)`
`text(the original cost.)`

 

Measurement, STD2 M6 SM-Bank 4

The diagram shows three checkpoints A, B and C. Checkpoint C is due east of Checkpoint A. The bearing of Checkpoint B from Checkpoint A is N22°E and the bearing of Checkpoint C from Checkpoint B is S68°E. The distance between Checkpoint A and Checkpoint B is 42 kilometres.
 


 

  1. Mark the given information on the diagram and explain why `angleABC` is 90°.  (2 marks)

  2. Find the distance, to the nearest kilometre, between Checkpoint A and Checkpoint C.  (2 marks)

  3. If a runner is travelling 12.6 km/h, how long does it take her to travel between Checkpoint A and Checkpoint B, in hours and minutes?  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `112\ text{km}`
  3. `3text(h 20 mins)`
Show Worked Solution
i.   
`angle ABC` `= 22 + 68`
  `= 90°`

 

ii.  `text(In)\ \ DeltaABC,`

`cosangleBAC` `= (AB)/(AC)`
`cos68°` `= 42/(AC)`
`AC` `= 42/(cos68°)`
  `= 112.11…`
  `= 112\ text{km  (nearest km)}`

 

iii.    `text(Travel time)` `= text(dist)/text(speed)`
    `= 42/12.6`
    `= 3.333…`
    `= 3text(h 20 mins)`

Statistics, STD2 S1 SM-Bank 1

Write down the five-number summary for the dataset 

`3, \ 7, \ 8, \ 11, \ 13, \ 18.`  (2 marks)

Show Answers Only
`text(Minimum value:)`   `3`
`text(First quartile:)`   `7`
`text(Median:)`   `(11 + 8)/2 = 9.5`
`text(Third quartile:)`   `13`
`text(Maximum value:)`   `18`
Show Worked Solution
`text(Minimum value:)`   `3`
`text(First quartile:)`   `7`
`text(Median:)`   `(11 + 8)/2 = 9.5`
`text(Third quartile:)`   `13`
`text(Maximum value:)`   `18`

Statistics, STD2 S5 SM-Bank 5

The diastolic measurement for blood pressure in 30-year-old people is normally distributed, with a mean of 82 and standard deviation of 16.

  1.   A person is considered to have low blood pressure if the diastolic measurement is 66 or less.

     

      What percentage of 30-year-old people have low blood pressure?  (1 mark)

  2.   Calculate the `z`-score for a diastolic measurement of 70.  (1 mark)

Show Answers Only
  1. `16text(%)`
  2. `−0.75`
Show Worked Solution
a.    `ztext(-score)\ (66)` `= (66 – 82)/16`
    `= −1`

 

`:.\ text(Percentage)` `= 100 – (50+34)`
  `= 16text(%)`

 

b.    `ztext(-score)\ (70)` `= (70 – 82)/16`
    `= −0.75`

Measurement, STD2 M6 SM-Bank 7 MC

Jeet walks 5 km from his home on a bearing of 153°. He then walks due north until he arrives a point which is due east of his home.

How far east, to the nearest 0.1 km, is Jeet from home?

  1.  2.3 km
  2.  2.5 km
  3.  4.9 km
  4.  9.8 km
Show Answers Only

`A`

Show Worked Solution

`text(Jeet finishes at)\ P`

`text(Find)\ \ OP:`

`cos 63°` `= (OP)/5`
`:. OP` `= 5 xx cos 63°`
  `= 2.26…`

 
`=> A`

Measurement, STD2 M6 SM-Bank 3 MC

Which of the following expresses S65°W as a true bearing?

  1. 065°
  2. 155°
  3. 245°
  4. 295°
Show Answers Only

`C`

Show Worked Solution

`text(True bearing)` `= 180 + 65`
  `= 245°`

`=> C`

Algebra, STD2 A4 SM-Bank 7 MC

A computer application was used to draw the graphs of the equations

`x + y = 4`  and  `x - y = 4`

Part of the screen is shown.
 

Which row of the table correctly matches the equations with the lines drawn and identifies the solution when the equations are solved simultaneously?

Show Answers Only

`A`

Show Worked Solution

`text(Line 1:) \ \ x + y = 4`

`text(Line 2:) \ \ x – y = 4`

`text(Intersection at) \ (4, 0).`

`=> A`

Algebra, STD2 A2 SM-Bank 5

The diagram shows the graph of a line.
 

 
What is the equation of this line?    (2 marks)

Show Answers Only

`y = 3/5x + 3`

Show Worked Solution

`ytext(-intercept = 3)`

`text(Gradient)` `= text(rise)/text(run)`
  `= 3/5`

 
`:. text(Equation:)\ \ y = 3/5x + 3`

Algebra, STD2 A4 2014 HSC 26d

Draw each graph on the grid below and hence solve the simultaneous equations.  (3 marks)

`y = 2x + 1`

`x - 2y - 4 = 0`   (3 marks)
 

Show Answers Only

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

Show Worked Solution

`text(Solution is at the intersection:)\ \ x=-2,\ y=-3`

Statistics, STD2 S1 2018 HSC 17 MC

The area chart shows the number of students involved in tennis or cricket at a school over a number of years.
 


 

In which year was the number of students involved in tennis equal to the number of students involved in cricket?

  1. 2013
  2. 2014
  3. 2015
  4. 2016
Show Answers Only

`C`

Show Worked Solution

`text(Area Charts are cumulative,)`

`text(Consider 2015,)`

`text(Cricket players = 30)`

`text(Tennis players) = 60 – 30 = 30`

`=> C`

Algebra, STD2 A4 2011 HSC 26b

Jack needs to find the number of years, `t`, it will take for a population of bats to first exceed 18 000.

He uses a ‘guess-and-check’ method to estimate `t` in the following equation

`5 xx 3^t = 18\ 000.`

Here is his working:

  1. Jack’s next guess is  `t = 6`. Show Jack’s correct working for this guess, including the calculation and conclusion.  (1 mark)

  2. Continue using the ‘guess-and-check’ method to find the number of years, `t`, it will take for the population to first exceed 18 000, if `t` is a whole number. Include the calculations and conclusions.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `8`
Show Worked Solution

i.   `text(When)\ t = 6,`

`5 xx 3^6 = 3645`

`=>\ text(Too small)`

 

ii.   `text(When)\ \ t = 7,`

`5 xx 3^7 = 10\ 935`

`=>\ text(Too small)`
 

`text(When)\ \ t = 8,`

`5 xx 3^8 = 32\ 805`

`=>\ text(exceeds 18 000)`

`:. t = 8`

Probability, STD2 S2 2018 HSC 30d

A game consists of two tokens being drawn at random from a barrel containing 20 tokens. There are 17 tokens labelled 10 cents and 3 tokens labelled $2. The player wins the total value of the two tokens drawn.

Complete the probability tree by writing the missing probabilities in the boxes.  (2 marks)

 


Show Answers Only

Show Worked Solution

Financial Maths, STD2 2014 HSC 27a

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

2014 27a1

  1. Stamp Duty for this car is calculated at $3 for every $100, or part thereof, of the sale price.  
  2. Calculate the Stamp Duty payable.   (1 mark)
  3.  
  4. Alex wishes to take out comprehensive insurance for the car for 12 months. The cost of comprehensive insurance is calculated using the following: 
    1.  
    2. 2014 27a2
  5. Find the total amount that Alex will need to pay for comprehensive insurance.   (3 marks)
  6.  
  7. Alex has decided he will take out the comprehensive car insurance rather than the less expensive non-compulsory third-party car insurance.
  8. What extra cover is provided by the comprehensive car insurance?   (1 mark)

 

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

 

(ii)   `text(Base rate) = $845`

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

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

 

`text(GST)` `= 10 text(%) xx(845 + 8.45)`
  `= 85.345`
  `= $85.35`

 

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

 

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

Measurement, STD2 M2 SM-Bank 4

Lee wants to call home to Sydney (34°S, 151°E) from Pateete, Tahiti (17°S, 149°W) when it is 7 pm on Friday in Sydney.

Sydney is eleven hours ahead of UTC. Papeete is ten hours behind UTC.

Give the time and day in Papeete when she should make the call. (Ignore time zones.)  (2 marks)

Show Answers Only

`10text{pm (Thurs)}`

Show Worked Solution

`text(Sydney is 21 hours ahead.)`

`:.\ text(Time in Papeete)`

`= 7text{pm (Fri) less 21 hours}`

`= 10text{pm (Thurs)}`