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Mechanics, SPEC1 2011 VCAA 7

A flowerpot of mass `m` kg is held in equilibrium by two light ropes, both of which are connected to a ceiling. The first rope makes an angle of 30° to the vertical and has tension `T_1` newtons. The second makes an angle of 60° to the vertical and has tension `T_2` newtons.
 

VCAA 2011 spec 7b
 

  1. Show that `T_2 = T_1/sqrt 3.`  (1 mark)
  2. The first rope is strong, but the second rope will break if the tension in it exceeds 98 newtons.

     

    Find the maximum value of `m` for which the flowerpot will remain in equilibrium.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `m = 20\ text(kg)`
Show Worked Solution

a.   `text(Forces in equilibrium:)`

`tan 30^@` `=T_2/T_1`
`T_2` `= T_1 xx tan30^@`
`T_2` `= T_1/sqrt3\ \ text(.. as required)`

 

b.    `sin30^@` `= (T_2)/(mg)`
  `T_2` `= (mg)/2`
  `98` `=(m_text(max) xx 9.8)/2`
  `m_text(max)` `=(2 xx 98)/9.8`
    `=20\ text(kg)`

Calculus, SPEC1 2011 VCAA 5

For the curve with parametric equations

`x = 4 sin (t) - 1`

`y = 2 cos (t) + 3`

Find  `(dy)/(dx)`  at the point  `(1, sqrt 3 + 3).`  (3 marks)

Show Answers Only

`– sqrt 3/6`

Show Worked Solution

`(dx)/(dt) = 4cos(t)`

`(dy)/(dt) = −2sin(t)`

`(dy)/(dx)` `= ((dy)/(dt))/((dx)/(dt))`
  `= (-sin t)/(2cos t)`

 
`x = 4 sin (t) – 1\ \ text{(given)}\ \ =>\ \ sin(t)=(x+1)/4`

`y = 2 cos (t) + 3\ \ text{(given)}\ \ =>\ \ 2cos(t)=(y-3)`

`=> dy/dx= (-(x + 1)/4)/(y – 3)`
 

`text(At)\ \ (1, sqrt 3 + 3):`

`:. (dy)/(dx)` `= (-1/2)/(sqrt3+3 -3)`
  `= -1/(2sqrt3)`
  `= – sqrt3/6`

Complex Numbers, SPEC2 2012 VCAA 8 MC

If  `z = a + bi`, where both `a` and `b` are non-zero real numbers and  `z in C`, which of the following does not represent a real number?

  1. `z + barz`
  2. `|\ z\ |`
  3. `zbarz`
  4. `z^2-2abi`
  5. `(z-bar z) (z + bar z)`
Show Answers Only

`E`

Show Worked Solution

`text(Consider each option:)`

`text(A.) quad a + bi + a-bi = 2a`  ✔

`text(B.) quad sqrt(a^2 + b^2)`  ✔

`text(C.) quad (a + bi) (a-bi) = a^2 + b^2`  ✔

`text(D.) quad (a + bi)^2-2abi`

`= a^2 +2abi-b^2-2abi`

`= a^2-b^2`  ✔

`text(E.) quad (a + bi-(a-bi))((a + bi + a-bi))`

`= (2bi) (2a)`

`= 4abi`  ✖

 
`=> E`

Complex Numbers, SPEC2 2012 VCAA 7 MC

The set of points in the complex plane defined by  `|\ z + 2i\ | = |\ z\ |`  corresponds to

A.   the point given by  `z = −i`

B.   the line  `text(Im)(z) = −1`

C.   the line  `text(Im)(z) = −i`

D.   the line  `text(Re)(z) = −1`

E.   the circle with centre  `−2i`  and radius `1`

Show Answers Only

`B`

Show Worked Solution

`|\ z + 2i\ | = |\ z\ |\ \ =>\ \  |\ z – (−2i)\ | = |\ z\ |`

`text{Find all points equidistant from (0, −2) and (0, 0):}`

`text(Midpoint):\ ((0 + 0)/2, (0 – 2)/2) = (0, -1)`

`m= (-2 – 0)/(0 – 0) = oo`

`:. m_(_|_) = 0\ \ text{(i.e. horizontal line)}`

`:. y = -1 \ \ or\ \ text{Im} (z) = -1`

 
`=> B`

Complex Numbers, SPEC2 2012 VCAA 5 MC

If  `z = sqrt 2 text(cis)(-(4 pi)/5)`  and  `w = z^9`, then

A.   `w = 16 sqrt 2 text(cis)((36 pi)/5)`

B.   `w = 16 sqrt 2 text(cis)(−pi/5)`

C.   `w = 16 sqrt 2 text(cis)((4pi)/5)`

D.   `w = 9 sqrt 2 text(cis)(-pi/5)`

E.   `w = 9 sqrt 2 text(cis)((4pi)/5)`

Show Answers Only

`C`

Show Worked Solution

`z = sqrt2 text(cis) (−(4pi)/5)`

`text(By De Moivre:)`

`w` `= z^9`
  `= (sqrt 2)^9 text(cis)(-(4pi)/5 xx 9)`
  `= 16 sqrt2 text(cis)(-(36pi)/5)`
  `= 16 sqrt 2 text(cis)((4pi)/5 – 8pi)`
  `= 16 sqrt 2 text(cis)((4pi)/5)`

 
`=> C`

Calculus, SPEC1 2011 VCAA 3

  1. Show that  `f(x) = (2x^2 + 3)/(x^2 + 1)`  can be written in the form  `f(x) = 2 + 1/(x^2 + 1).`  (1 mark)

  2. Sketch the graph of the relation  `y = (2x^2 + 3)/(x^2 + 1)`  on the axes below.
  3. Label any asymptotes with their equations and label any intercepts with the axes, writing them as coordinates.  (3 marks)

     
    VCAA 2011 spec 3b
     

  4. Find the area enclosed by the graph of the relation  `y = (2x^2 + 3)/(x^2 + 1)`, the `x`-axis, and the lines  `x = -1`  and  `x = 1.`  (3 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. “
    VCAA 2011 spec 3bi
  3. `4 + pi/2`

Show Worked Solution

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

 

b.   `underset (x→oo) (lim y) = 2`

`text(S)text(ince)\ \ x^2>0\ \ text(for all)\ x,`

`=> f(x)_text(max)\ \ text(occurs when)\ \ x=0\ \ text(at)\ \ (0,3)`
 

VCAA 2011 spec 3bi

 

c.   `f(x)\ \ text(is an even function.)`

`:.\ text(Area)` `= 2 int_0^1 2 + 1/(x^2 + 1)\ dx`  
  `= 2[2x + tan^(−1)(x)]_0^1`  
  `=2[(2+pi/4)-0]`  
  `= 4 + pi/2`  

Calculus, SPEC1 2011 VCAA 2

Find the value of the real constant `k` given that  `kxe^(2x)`  is a solution of the differential equation

`(d^2y)/(dx^2)-2 (dy)/(dx) + 5y = e^(2x) (15x + 6).`  (3 marks)

Show Answers Only

`k = 3`

Show Worked Solution
`y` `=kxe^(2x)`
`(dy)/(dx)` `= ke^(2x) + kx(2e^(2x))`
  `= ke^(2x)(1 + 2x)`

 

`(d^2 y)/(dx)` `= 2ke^(2x)(1 + 2x) + 2ke^(2x)`
  `= 2ke^(2x)(1 + 2x + 1)`
  `= 4ke^(2x)(1 + x)`

 

`e^(2x)(15x + 6)` `= (d^2 y)/(dx^2)-2(dy)/(dx) + 5y`
`e^(2x)(15x + 6)` `= 4ke^(2x) + 4kxe^(2x)-2ke^(2x)-4kxe^(2k) + 5kxe^(2x)`
`e^(2x)(15x + 6)` `= 2ke^(2x) + 5kxe^(2x)`
` e^(2x)(15x + 6)` `=e^(2x)(5kx + 2k)`

 
`=> 5kx = 15x, \ \ 2k = 6`

`:. k = 3`

Calculus, SPEC1 2011 VCAA 1

Find an antiderivative of  `(1 + x)/(9-x^2),\ \ x in R \ text(\{– 3, 3}).`  (3 marks)

Show Answers Only

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

Show Worked Solution
`int (1 + x)/(9-x^2)\ dx` `= int (1-x)/((3-x)(3 + x))\ dx`
  `= int A/(3-x) + B/(3 + x)\ dx`

 
`text(Using partial fractions:)`

`A(3 + x) + B(3-x) = 1 + x`

`text(When)\ \ x=3, \ 6A=4\ \ =>\ \ A=2/3`

`text(When)\ \ x=-3, \ 6B=-2\ \ =>\ \ B=-1/3`
 

`:. int (1 + x)/(9-x^2)\ dx`

`int 2/3 (1/(3-x))-1/3 (1/(3 + x))\ dx`

`= -2/3ln|3-x|-1/3ln|3 + x|,\ \ \ (c=0)`

Graphs, SPEC2 2012 VCAA 3 MC

The graph of  `y = f(x)`  is shown below.
 

All the axes below have the same scale as the axes in the diagram above.

The graph of  `y = 1/(f(x))`  is best represented by

A.    B.   
C.    D.   
E.       

Show Answers Only

`E`

Show Worked Solution

`y(-1) = 0`

`-> 1/(y(-1)):\ text(asymptote)\ x = -1`

`text(Eliminate A and B.)`

`text(Local max at)\ \ x = 1`

`-> 1/(y(1)):\ text(local min)`
 

`=> E`

Graphs, SPEC1 2012 VCAA 10

Consider the functions with rules  `f(x) = arcsin (x/2) + 3/sqrt (25 x^2-1)`  and  `g(x) = arcsin (3x)-3/sqrt (25x^2-1).`

    1. Find the maximal domain of  `f_1(x) = arcsin (x/2).`   (1 mark)

    2. Find the maximal domain of  `f_2(x) = 3/sqrt (25x^2-1).`   (1 mark)

    3. Find the largest set of values of  `x in R`  for which  `f(x)`  is defined.   (1 mark)

  2. Given that  `h(x) = f(x) + g(x)`  and that  `theta = h(1/4)`, evaluate  `sin (theta).`

     

    Give your answer in the form  `(a sqrt b)/c, \ a, b, c in Z.`   (3 marks)

Show Answers Only
    1. `-2 <= x <= 2`
    2. `x < -1/5 uu x > 1/5`
    3. `-2 <= x < -1/5 uu 1/5 < x <= 2`
  2. `(5 sqrt 7)/16`
Show Worked Solution

a.i.   `text(Maximal Domain occurs when:)`

  `-1 <= x/2 <= 1`
  `{x: -2 <= x <= 2}`

 
a.ii.   `text(Maximal Domain occurs when:)`

♦♦ Mean mark part (a)(ii) 33%.

  `25x^2-1 > 0`
  `x^2 > 1/25`
  `{x: x < -1/5 uu x > 1/5}`

 

a.iii.  `text(Max domain for which)\ \ f(x)\ \ text(is defined:)`

♦♦ Mean mark part (a)(iii) 26%.

  `(-2 <= x <= 2) nn (x < -1/5 uu x > 1/5)`
  `{x: -2 <= x < -1/5 \ uu \ 1/5 < x <= 2}`

 

b.    `h(x)` `= sin^(-1) (x/2) + 3/sqrt(25x^2-1) + sin^(-1)(3x)-3/sqrt(25x^2-1)`
    `= sin^(-1)(x/2) + sin^(-1)(3x)`

 
`text(When)\ \ x=1/4,\ \ h(x)=theta`

♦♦♦ Mean mark part (b) 20%.

`theta=sin^(-1) (1/8) + sin^(-1) (3/4)`

`text(Let)\ \ theta_1 = sin^(-1) (1/8),\ \ theta_2 = sin^(-1) (3/4)`

`text(Using)\ \ sin(theta_1 + theta_2)=sin theta_1 cos theta_2 + cos theta_1 sin theta_2:`

`sin(theta)` `= 1/8 * sqrt7/4 + sqrt63/8 * 3/4`
  `=sqrt7/32 + (9sqrt7)/32`
  `=(5 sqrt7)/16`

Vectors, SPEC1 2012 VCAA 9

The position of a particle at time  `t`  is given by

`underset ~r (t) = (2 sqrt (t^2 + 2) - t^2) underset ~i + (2 sqrt (t^2 + 2) + 2t) underset ~j,\ \ t >= 0.`

  1. Find the velocity of the particle at time  `t.`   (1 mark)

  2. Find the speed of the particle at time  `t = 1`  in the form  `(a sqrt b)/c`, where `a, b` and `c` are positive integers.   (2 marks)

  3. Show that at time  `t = 1,\ \ (dy)/(dx) = (1 + sqrt 3)/(1 - sqrt 3).`   (2 marks)

  4. Find the angle in terms of `pi`, between the vector  `-sqrt 3 underset ~i + underset ~j`  and the vector  `underset ~r (t)`  at time  `t = 0.`   (2 marks)

Show Answers Only
  1. `((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`
  2. `(4 sqrt 6)/3`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `(7 pi)/12`
Show Worked Solution

a.  `underset ~r (t) = (2 sqrt (t^2 + 2) – t^2) underset ~i + (2 sqrt (t^2 + 2) +2t) underset ~j`

  `underset ~v(t)` `=dot underset ~r(t)`
    `= (2(2t)(1/2)(t^2 + 2)^(-1/2) – 2t) underset ~i + (2(2t)(1/2)(t^2 + 2)^(-1/2) + 2)underset ~j`
    `= ((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`

 

b.    `underset ~v(1)` `= (2/sqrt(1 + 2) – 2)underset ~i + (2/sqrt(1 + 2) + 2) underset ~j`
    `= (2/sqrt 3 – 2) underset ~i + (2/sqrt 3 + 2)underset ~j`
  `|\ underset ~v(1)\ |` `= sqrt((2/sqrt 3 – 2)^2 + (2/sqrt 3 + 2)^2)`
    `= sqrt(4/3 – 8/sqrt 3 + 4 + 4/3 + 8/sqrt 3 + 4)`
    `= sqrt(8/3 + 8)`
    `= sqrt(32/3)`
    `= (4sqrt2)/sqrt3`
    `= (4 sqrt 6)/3`

♦ Mean mark part (c) 41%.

c.    `(dy)/(dt)` `= (dy)/(dx) xx (dx)/(dt)`
  `(dy)/(dx)` `=((dy)/(dt))/((dx)/(dt))`
  `:. (dy)/(dx)|_(t=1)` `= (2/sqrt 3 + 2)/(2/sqrt 3 – 2)`
    `= ((2 + 2 sqrt 3)/sqrt 3) xx (sqrt 3/(2 – 2 sqrt 3))`
    `= (2 + 2 sqrt 3)/(2 – 2 sqrt 3)`
    `= (1 + sqrt 3)/(1 – sqrt 3)`

 

d.   `text(At)\ \ t=0:` 

♦♦ Mean mark part (d) 32%.

`underset ~r(0) = 2 sqrt 2 underset ~i + 2 sqrt 2 underset ~j`

`theta_1` `= tan^(-1)((2 sqrt 2)/(2 sqrt 2)) = pi/4`
`theta_2` `= tan^(-1)(1/sqrt 3)=pi/6`

 
`text(Let)\ \ theta=\ text(angle between the vectors:)`

`theta` `= pi – theta_1 – theta_2`
  `= pi – pi/4 – pi/6`
  `= (12 pi – 3 pi – 2 pi)/12`
  `= (7 pi)/12`

Calculus, SPEC1 2012 VCAA 8

The velocity, `v` m/s, of a body when it is `x` metres from a fixed point `O` is given by

`v = (2x)/sqrt(1 + x^2).`

Find an expression for the acceleration of the body in terms of `x` in simplest form.  (3 marks)

Show Answers Only

`(4x)/(1 + x^2)^2`

Show Worked Solution
`v` `= (2x)/sqrt (1 + x^2)`  
`v^2` `=(4x^2)/(1+x^2)`  
`1/2 v^2` `= (2x^2)/(1 + x^2)`  

 

`a` `= d/(dx) (1/2 v^2)`
  `= d/(dx) ((2x^2)/(1 + x^2))`
  `= ((1 + x^2) (4x) – (2x^2) (2x))/(1 + x^2)^2`
  `= (4x)/(1 + x^2)^2`

Calculus, SPEC1 2012 VCAA 7

Consider the curve with equation  `y = (x - 1) sqrt (2 - x),\ \ 1 <= x <= 2.`

Calculate the area of the region enclosed by the curve and the `x`-axis.  (3 marks)

Show Answers Only

`4/15`

Show Worked Solution

`x – 1 >= 0\ \ text(for)\ \ \ 1 <= x <= 2`

`sqrt(2 – x) >= 0\ \ text(for)\ \ \ 1 <= x <= 2`

`:. (x – 1) sqrt(2 – x) >=0\ \ text(for)\ \ \ 1 <= x <= 2`
 

`text(Let)\ \ u=2-x\ \ =>\ \ x=2-u`

`(du)/dx = -1\ \ =>\ \ dx = -du`

`text(When)\ \ x=2,\ \ u=0`

`text(When)\ \ x=1,\ \ u=1`
 

`A` `= int_1^2 (x – 1) sqrt(2 – x)\ dx`
  `= int_1^0 -(2 – u -1) u^(1/2)\ du`
  `= int_1^0 (u-1) u^(1/2)\ du`
  `= int_1^0 u^(3/2) – u^(1/2)\ du`
  `= [2/5u^(5/2)-2/3 u^(3/2)]_1^0`
  `= [0-(2/5-2/3)]`
  `= -((6-10))/15`
  `= 4/15`

Calculus, SPEC1 2012 VCAA 6

Find the gradient of the tangent to the curve  `xy^2 + y + (log_e (x - 2))^2 = 14`  at the point  `(3, 2).`  (3 marks)

Show Answers Only

`(dy)/(dx) = -4/13`

Show Worked Solution

`xy^2 + y + (log_e (x – 2))^2 = 14`

`text(Using implicit differentiation:)`

`d/(dx) (xy^2) + d/(dx) (y) + d/(dx) ((ln (x – 2))^2) = d/(dx)(14)`

`d/(dx) (x) ⋅ y^2 + d/(dx) (y^2) ⋅ x + (dy)/(dx) + 2 xx 1/(x – 2) xx ln(x – 2) = 0`

`y^2 + 2xy* (dy)/(dx) + (dy)/(dx) + (2 ln (x – 2))/(x – 2) = 0`

`text{At (3,2):}`

`2^2 + 2(2)(3) m + m + (2 ln(1))/1` `= 0`
`4 + 12m + m` `= 0`
`13m` `= -4`
`:. m` `= -4/13`

Calculus, SPEC1 2012 VCAA 5

Let  `y = arctan (2x).`

Find the value of `a` given that  `(d^2y)/(dx^2) = ax ((dy)/(dx))^2`, where `a` is a real constant.  (3 marks)

Show Answers Only

`a = -4`

Show Worked Solution
`(dy)/(dx)` `= 2x xx (1/(1 + (2x)^2))`
  `= 2/(1 + 4x^2)`
  `= 2(1 + 4x^2)^{-1}`
   
`(d^2y)/(dx^2)` `= 2(8x)(-1)(1 + 4x^2)^(-2)`
  `= (-16x)/(1 + 4x^2)^2`

 

`(d^2y)/(dx^2)` `= ax ((dy)/(dx))^2`
`(-16x)/(1 + 4x^2)^2` `= ax (2/(1 + 4x^2))^2`
  `= (4ax)/(1 + 4x^2)^2`

 

`4a` `= -16`
`:. a` `= -4`

Calculus, SPEC1 VCE SM-Bank 5

Find  `int(3x^2 + 8)/(x(x^2 +4))\ dx`.  (3 marks)

Show Answers Only

` 2log_e x + 1/2log_e(x^2 + 4) + c` 

Show Worked Solution
`(3x^2 + 8)/(x(x^2 +4))` `=  a/x + (bx + c)/(x^2 + 4)`
`3x^2 +8` `=ax^2 + 4a+ bx^2 + cx`
  `=(a+b)x^2+cx+4a`
   

 `a + b = 3, \ c = 0, \ 4a = 8`

`:.a = 2, b = 1, c = 0`

`int(3x^2 + 8)/(x(x^2 +4))\ dx` `= int2/x\ dx + int x/(x^2 + 4)\ dx`
  `= 2log_e x + 1/2log_e(x^2 + 4) + c` 

Complex Numbers, SPEC1 2012 VCAA 3

Consider the equation  `z^3-z^2-2z-12 = 0, \ z in C.`

  1. Given that  `z = 2 text(cis) ((2pi)/3)`  is a root of the equation, find the other two roots in the form  `a + ib`, where  `a, b in R.`   (3 marks)

  2. Plot all of the roots clearly on the Argand diagram below.   (1 mark)

     

     


Show Answers Only
  1. `z_2 = -1-i sqrt 3`
    `z_3 = 3 + 0i`
     

  2.  

Show Worked Solution

a.   `text(Given)\ \  z = 2 text(cis)((2 pi)/3)\ \ text(is a root, and)`

`z^3-z^2-2z-12 = 0\ \ text(has real coefficients,)`

`=> z= 2 text(cis)(-(2 pi)/3)\ \ text{is a root (conjugate).}`

`=> text(Roots:)\ \ -1+sqrt3i,\ \ -1-sqrt3i`
 

`(z-alpha) (z-beta) (z-gamma) = z^3-z^2-2z-12`
 

`text(Using)\ \ alpha beta gamma = 12:`

`(-1 + i sqrt 3) (-1-i sqrt 3) gamma` `=12`  
`((-1)^2-(sqrt3i)^2)gamma` `=12`  
`4gamma` `=12`  
`gamma` `=3`  

 
`:.\ text(Other two roots:)`

♦♦ Mean mark part (b) 33%.

`z_2` `= -1-i sqrt 3`
`z_3` `= 3 + 0i`

 

b.   

Mechanics, SPEC2 2013 VCAA 20 MC

A 5 kg parcel is on the floor of a lift that is accelerating downwards at 3 m/s².

The reaction, in newtons, of the floor of the lift on the parcel is

A.  `−15 + 5g`

B.   `15 + 5g`

C.  `−15 + 3g`

D.  `−15 − 5g`

E.   `15 + 3g`

Show Answers Only

`A`

Show Worked Solution

`text(Let)\ R\ text(be the reaction force of the floor on the 5kg parcel.)`

`sum F` `= 5g – R`
`5xx3` `= 5g-R`
`:. R` `=-15 +5g`

 
`=> A`

Calculus, SPEC2 2013 VCAA 19 MC

A tourist in a hot air balloon, which is rising at 2 m/s, accidentally drops a camera over the side and it falls 100 m to the ground.

Neglecting the effect of air resistance on the camera, the time taken for the camera to hit the ground, correct to the nearest tenth of a second, is

A.   4.3 s

B.   4.5 s

C.   4.7 s

D.   4.9 s

E.   5.0 s

Show Answers Only

`C`

Show Worked Solution

`u = 2, s = −100, a = -9.8`

`s` `= ut + 1/2at^2`
`-100` `= 2t – 4.9t^2`
`0` `= 4.9t^2 – 2t – 100`

 
`text(Solve:)\ \ 4.9t^2 – 2t – 100=0\ \ text(for)\ \ t`

`t~~4.7\ \ text(sec)`

`=> C`

Calculus, SPEC2 2013 VCAA 18 MC

A particle moves in a straight line such that its acceleration is given by  `a = sqrt(v^2 - 1)` , where `v` is its velocity and `x` is its displacement from a fixed point.

Given that  `v = sqrt2`  when  `x = 0`, the velocity `v` in terms of `x` is

A.   `v = sqrt(2 + x)`

B.   `v = 1 + |\ x + 1\ |`

C.   `v = sqrt(2 + x^2)`

D.   `v = sqrt(1 + (1 + x)^2)`

E.   `v = sqrt(1 + (x - 1)^2)`

Show Answers Only

`D`

Show Worked Solution
`v(dv)/(dx)` `= sqrt(v^2 – 1)`
`(dv)/(dx)` `= sqrt(v^2 – 1)/v`
`(dx)/(dv)` `= v/sqrt(v^2 – 1)`
`x` `=int v/sqrt(v^2 – 1)\ dv`
  `= sqrt(v^2 – 1) + c`

 
`text(When)\ \ x=0, \ v = sqrt2\ \ =>\ \ c = −1`

`x` `= sqrt(v^2 – 1) -1`
`v^2` `= 1 + (1 + x)^2`

 
`:. v = sqrt(1 + (1 + x)^2)`

`=> D`

Vectors, SPEC2 2013 VCAA 17 MC

Consider the four vectors  `underset~a = underset~j + 3underset~k`,  `underset~b = underset~i - 4underset~k`,  `underset~c = 3underset~j - k`  and  `underset~d = 2underset~j + underset~k`.

Which one of the following is a linearly dependent set of vectors?

  1. `{underset~a, underset~b, underset~c}`
  2. `{underset~a, underset~c, underset~d}`
  3. `{underset~a, underset~b, underset~d}`
  4. `{underset~b, underset~c, underset~d}`
  5. `{underset~a, underset~b}`
Show Answers Only

`B`

Show Worked Solution

`underset ~b\ \ text(is the only vector which has an)\ \ i\ \ text(component.)`
 

`text(The other three vectors are on the same plane are therefore`

`text(linearly dependent.)`

`=> B`

Mechanics, SPEC2 2013 VCAA 16 MC

Forces of magnitude 5 N, 7 N and `Q` N act on a particle that is in equilibrium, as shown in the diagram below.
 

SPEC2 2013 VCAA 16 MC
 

The magnitude of `Q`, in newtons, can be found by evaluating

A.   `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(70^@))`

B.   `5^2 + 7^2 - 2 xx 5 xx 7 cos(110^@)`

C.   `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(110^@))`

D.   `5^2 + 7^2 - 2 xx 5 xx 7 cos(70^@)`

E.   `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(20^@))`

Show Answers Only

`A`

Show Worked Solution

`text(Using cosine rule:)`
 


 

`Q^2` `= 5^2 + 7^2 – 2(5)(7)cos(70^@)`
`Q` `= sqrt(5^2 + 7^2 – (5)(7)cos(70^@))`

 
`=> A`

Vectors, SPEC2 2013 VCAA 15 MC

Let  `underset~u = 4underset~i - underset~j + underset~k`,  `underset~v = 3underset~j + 3underset~k`  and  `underset~w = −4underset~i + underset~j + underset~k`.

Which one of the following statements is not true?

  1. `|\ underset~u\ | = |\ underset~v\ |`
  2. `|\ underset~u\ | = |\ −underset~w\ |`
  3. `underset~u`, `underset~v` and `underset~w` are linearly independent
  4. `underset~u.underset~v = 0`
  5. `(underset~u + underset~w).underset~v = 12`
Show Answers Only

`E`

Show Worked Solution
`|underset~u|` `= sqrt(16 + 1 + 1)=sqrt18`
`|underset~v|` `= sqrt(9 + 9)=sqrt18`
`|underset~w|` `= sqrt(16+1+1)=sqrt18`

 
`underset~u · underset~v= 4 xx 0 + (−1) xx 3 + 1 xx 3=0`
 

`(underset~u + underset~w) · underset~v` `= ((4 – 4)underset~i + (−1 + 1)underset~j + (1 + 1)underset~k) · underset~v`
  `= 2underset~k · underset~v`
  `= 6`

 
`=> E`

Calculus, SPEC2 2013 VCAA 13 MC

Water containing 2 grams of salt per litre flows at the rate of 10 litres per minute into a tank that initially contained 50 litres of pure water. The concentration of salt in the tank is kept uniform by stirring and the mixture flows out of the tank at the rate of 6 litres per minute.

If `Q` grams is the amount of salt in the tank `t` minutes after the water begins to flow, the differential equation relating `Q` to `t` is

A.   `(dQ)/(dt) = 20 - (3Q)/(25 + 2t)`

B.   `(dQ)/(dt) = 10 - (3Q)/(25 + 2t)`

C.   `(dQ)/(dt) = 20 - (3Q)/(25 - 2t)`

D.   `(dQ)/(dt) = 10 - (3Q)/(25 - 2t)`

E.   `(dQ)/(dt) = 20 - (3Q)/25`

Show Answers Only

`A`

Show Worked Solution
`text(Volume)` `= 50 + (10 – 6)t`
  `= 50 + 4t`

 
`text(Salt in tank at time)\ \ t=Q\ text(grams)`

`:.\ text(Concentration)\ = Q/(50 + 4t)\ text(grams per litre)`
 

`(dQ)/(dt)text(in) = 2 xx 10 = 20\ \ text(g/min)`

`(dQ)/(dt)text(out)` `= 6 xx Q/(50 + 4t)`
  `= (3Q)/(25 + 2t)`

 
`:. (dQ)/(dt) = 20 – (3Q)/(25 + 2t)`

`=> A`

Calculus, SPEC2 2013 VCAA 12 MC

SPEC2 2013 VCAA 12 MC

The differential equation that best represents the above direction field is

A.   `(dy)/(dx) = x^2 - y^2`

B.   `(dy)/(dx) = y^2 - x^2`

C.   `(dy)/(dx) = y/x`

D.   `(dy)/(dx) = −x/y`

E.   `(dy)/(dx) = x/y`

Show Answers Only

`E`

Show Worked Solution

`text(By inspection:)`

`text(When)\ \ x=0\ \ =>\ \ (dy)/(dx) = 0`

`text(When)\ \ y=0\ \ => (dy)/(dx) -> oo`

`:.\ text(Eliminate A, B and C)`
 

`text(Along)\ \ y = x\ \ =>\ \ (dy)/(dx) > 0`

`:.\ text(Eliminate D)`

`=> E`

Calculus, SPEC2 2013 VCAA 11 MC

Consider the differential equation  `(dy)/(dx) = 1/(3 + 3x + x^2)`, with  `y_0 = 1`  when  `x_0 = 0`.

Using Euler's method with a step size of 0.1, the value of  `y_2` correct to three decimal places, is

A.   1.033

B.   1.063

C.   1.064

D.   1.065

E.   1.066

Show Answers Only

`C`

Show Worked Solution
`y_1` `= y_0 + h xx (dy)/(dx)|_{(x_0,y_0)}`
  `= 1 + 0.1 xx 1/(3 + 3 xx 0 + 0)`
  `= 1.03`

 

`y_2` `= y_1 + h xx (dy)/(dx)|_{(x_1,y_1)}`
  `= 1.03 + 0.1(1/(3 + 0.3 + 0.01))`
  `~~ 1.064`

 
`=> C`

Calculus, SPEC2 2013 VCAA 10 MC

The region bounded by the lines  `x = 0`, `y = 3`  and the graph of  `y = x^(4/3)`  where  `x ≥ 0`  is rotated about the `y`-axis to form a solid of revolution.

The volume of this solid is

A.   `(81pi3^(2/3))/11`

B.   `(12pi3^(3/4))/7`

C.   `(27pi3^(1/3))/7`

D.   `(18pi3^(1/2))/5`

E.   `(6pi3^(1/2))/5`

Show Answers Only

`D`

Show Worked Solution
`y` `= (x^(2/3))^2`
`y` `= (x^2)^(2/3)`
`y^(3/2)` `= x^2`

 

`:. V` `= pi int_0^3 x^2\ dy`
  `= pi int_0^3 y^(3/2)\ dy\ \ \ text{(by CAS)}`
  `= (18pisqrt3)/5`

 
`=> D`

Complex Numbers, SPEC2 2013 VCAA 8 MC

The principal arguments of the solutions to the equation  `z^2 = 1 + i`  are

  1. `pi/8`  and  `(9pi)/8`
  2. `−pi/8`  and  `(7pi)/8`
  3. `−(7pi)/8`  and  `pi/8`
  4. `(7pi)/8`  and  `(15pi)/8`
  5. `−(3pi)/4`  and  `pi/4`
Show Answers Only

`C`

Show Worked Solution
`z^2` `= 1 + i`
  `= sqrt2 text(cis)(pi/4 + 2kpi)`
`z` `= 2^(1/4) text(cis) (pi/8 + kpi)`

 
`:.\ text(Principal arguments are)\ −(7pi)/8\ text(and)\ pi/8.`

`=> C`

Complex Numbers, SPEC2 2013 VCAA 5 MC

The region in the complex plane that is outside the circle of radius `b` centred at the origin is given by the set of points `z`, where  `z ∈ C`, such that

A.   `|\ z\ | < b`

B.   `|\ z\ | > b`

C.   `|\ z\ | > b^2`

D.   `|\ z\ | = b`

E.   `|\ z\ | < b^2`

Show Answers Only

`B`

Show Worked Solution

`text(Circle:)\ |z| = b`

`text(Outside:)\ |z| > b`

`=> B`

Calculus, SPEC1 2013 VCAA 9

The shaded region below is enclosed by the graph of  `y = sin(x)`  and the lines  `y = 3x`  and  `x = pi/3.`

This region is rotated about the `x`-axis.

VCAA 2013 spec 9

Find the volume of the resulting solid of revolution.  (4 marks)

Show Answers Only

`pi^4/9 – pi^2/6 + (sqrt 3 pi)/8\ \ text(u³)`

Show Worked Solution

`text{Volume of large (outer) cone}`

`= 1/3 pi r^2 xx h`

`=1/3 pi xx pi^2 xx pi/3`

`=pi^4/9`
 

`text{Volume of smaller (inner) cone}`

`= pi int_0^(pi/3) sin^2 x\ dx`

`= pi/2 int_0^(pi/3) (1 – cos 2x)\ dx`

`= pi/2[x – 1/2 (sin 2x)]_0^(pi/3)`

`=pi/2 [pi/3 – 1/2 sin ((2pi)/3)]`

`=pi/2(pi/3 + 1/2 xx sqrt3/2)`

`= pi^2/6 – (sqrt3 pi)/8`
 

`:.\ text(Volume of solid)`

`=pi^4/9 – pi^2/6 + (sqrt3 pi)/8\ \ text(u³)`

Calculus, SPEC1 2013 VCAA 5

A container of water is heated to boiling point (100°C) and then placed in a room that has a constant temperature of 20°C. After five minutes the temperature of the water is 80°C.

  1. Use Newton’s law of cooling  `(dT)/(dt) = -k (T-20)`, where `T text(°C)` is the temperature of the water at the time `t` minutes after the water is placed in the room, to show that  `e^(-5k) = 3/4.`   (2 marks)

  2. Find the temperature of the water 10 minutes after it is placed in the room.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `65`
Show Worked Solution
a.    `(dT)/(dt)` `= -k (T-20)`
  `(dt)/(dT)` `= -1/k * 1/(T-20)`
  `t` `= -1/k int 1/(T-20)\ dT`
  `-kt` `= log_e (T-20) + c`
     

`text(When)\ \ t = 0,\ \ T = 100,`

`=>c = -log_e 80,`

`:. -kt` `= log_e (T-20)-log_e 80`
  `= log_e ((T-20)/80)`

 
`text(When)\ \ t = 5,\ \ T = 80,`

`-5k = log_e (3/4)`

`:. e^(-5k) = 3/4`

 

(b)   `text(Find)\ \ T\ \ text(when)\ \ t = 10:`

`-10k` `= log_e ((T-20)/80)`
`(T-20)/80` `= e^(-10k)`
`(T-20)/80` `= (e^(-5k))^2`
`(T-20)/80` `= (3/4)^2`
`:. T` `= (3/4)^2 xx 80 + 20`
  `=(9 xx 80)/16 +20`
  `= 65^@ text(C)`

Graphs, SPEC1 2013 VCAA 4

  1. State the maximal domain and the range of  `y = arccos(1-2x).`   (2 marks)

  2. Sketch the graph of  `y = arccos(1-2x)`  over its maximal domain. Label the endpoints with their coordinates.   (2 marks)

     

     
              VCAA 2013 spec 4b
     

  3. Find the gradient of the tangent to the graph of  `y = arccos (1 – 2x)`  at  `x = 1/4.`   (2 marks)

Show Answers Only
  1. `text(Maximal domain)\ \ [0, 1];\ \ \ text(Range)\ \ [0, pi]`
  2.  

  3. `4/sqrt3`
Show Worked Solution

a.  `text(Maximal domain:)`

`1-2x` `∈ [−1, 1]`
`-2x` `∈ [−2, 0]`
`x` `∈ [0, 1]`

 
`text(When)\ \ x = 0,\ \ y = cos^-1 1= 0;`

`text(When)\ \ x = 1,\ \ y = cos^-1 (-1)= pi`

`:.\ text(Range is)\ \ [0, pi]`

 

b.   

 

c.   `y = cos^-1 (1-2x),`

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

 
`text(At)\ \ x = 1/4,`

`m_T` `= 2/sqrt(1-(1-1/2)^2)`
  `=2/sqrt(3/4)`
  `= 4/sqrt 3`
  `= (4 sqrt 3)/3`

Vectors, SPEC1 2013 VCAA 3

The coordinates of three points are  `A (– 1, 2, 4), \ B(1, 0, 5) and C(3, 5, 2).`

  1. Find  `vec (AB)`  (1 mark)

  2. The points `A, B` and `C` are the vertices of a triangle.
  3. Prove that the triangle has a right angle at `A.`  (2 marks)

  4. Find the length of the hypotenuse of the triangle.  (1 mark)

Show Answers Only

  1. `2underset~i – 2underset~j + underset~k`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `sqrt 38`

Show Worked Solution

a.   `vec(AB)` `=(1 – −1)underset~i + (0 – 2)underset~j + (5 – 4)underset~k`
    `= 2underset~i – 2underset~j + underset~k`

 

b.    `overset(->)(AC)` `= (3 – −1)underset~i + (5 – 2)underset~j + (2 – 4)underset~k`
    `= 4underset~i + 3underset~j – 2underset~k`

 

`overset(->)(AB) · overset(->)(AC)` `= 2 xx 4 + (−2) xx 3 + 1 xx (−2)`
  `= 8 – 6 – 2`
  `= 0`

 
`=>  overset(->)(AB) ⊥ overset(->)(AC)`

`:. DeltaABC\ text(has a right angle at)\ A.`
 

c.    `overset(->)(BC)` `= (3 – 1)underset~i + (5 – 0)underset~j + (2 – 5)underset~k`
    `= 2underset~i + 5underset~j – 3underset~k`

 

`|overset(->)(BC)|` `= sqrt(2^2 + 5^2 + (−3)^2)`
  `= sqrt(4 + 25 + 9)`
  `= sqrt38`

Calculus, SPEC1 2013 VCAA 2

Evaluate  `int_0^1 (x-5)/(x^2-5x + 6)\ dx.`  (4 marks)

Show Answers Only

`ln(9/32)`

Show Worked Solution
`int_0^1 (x-5)/(x^2-5x + 6)\ dx` `= int_0^1 (x-5)/((x-3)(x-2))\ dx`
  `= int_0^1 A/(x-3) + B/(x-2)\ dx`

 
`text(Using partial fractions:)` 

`A(x-2) + B(x-3) = x-5`

`text(If)\ \ x = 2,` `-B` `= −3`
  `B` `= 3`
`text(If)\ \ x = 3,` `A` `=-2`

 
`:. int_0^1 (x-5)/(x^2-5x + 6)\ dx`

  `= int_0^1 −2/(x-3) + 3/(x-2)\ dx`
  `= [−2ln|x-3| + 3ln|x-2|]_0^1`
  `= −2ln(2) + 3ln(1) + 2ln(3)-3ln(2)`
  `= 2ln(3)-5ln(2)`
  `= ln(3^2/2^5)`
  `= ln(9/32)`

Calculus, SPEC1 2013 VCAA 6

Find the value of `c`, where  `c in R`, such that the curve defined by

`y^2 + (3e^{(x - 1)})/(x - 2) = c`

has a gradient of 2 where  `x = 1.`   (4 marks)

Show Answers Only

`-3/4`

Show Worked Solution

`text(Using implicit differentiation:)`

`d/(dx)(y^2) + d/(dx)((3e^(x – 1))/(x – 2))` `= d/(dx)(c)`
`2y(dy)/(dx) + ((x – 2)3e^(x – 1) – 3e^(x – 1))/(x – 2)^2` `= 0`
`2y(dy)/(dx) +(3e^(x – 1) (x – 3))/(2y(x – 2)^2)` `=0`

 
`:. (dy)/(dx) = -(3e^(x – 1) (x – 3))/(2y(x – 2)^2)`

`text(When)\ \ x=1,\ \ dy/dx=2`

`2=(-6)/(-2y)\ \ =>\ \ y=3/2`

`text(Substituting into curve equation:)`

`(3/2)^2 + (3e^0)/(1 – 2)` `= c`
`9/4 – 3` `= c`
`c` `= -3/4`

Calculus, SPEC2 2014 VCAA 21 MC

The acceleration, in `text(ms)^(-2)`, of a particle moving in a straight line is given by  `–4x`, where `x` metres is its displacement from a fixed origin `O`.

If the particle is at rest where  `x = 5`, the speed of the particle, in `text(ms)^(−1)`, where  `x = 3`  is

A.           `8`

B.     `8 sqrt 2`

C.         `12`

D.     `4 sqrt 2`

E.   `2 sqrt 34`

Show Answers Only

`A`

Show Worked Solution
`a` `= -4x`
`(d(1/2 v^2))/(dx)` `=-4x`
`1/2 v^2` `=-2x^2 + c_0`
`v^2` `=-4x^2 + c_1`

 
`text(When)\ \ x=5,\ \ v=0:`

`=> c_1 = 100`
 

`text(Find)\ \ |\ v\ |\ \ text(when)\ \ x=3:`

`v^2 =  -4 xx 3^2 + 100`

`:. |\ v\ | = 8\ \ text(m/s)`

`=> A`

Mechanics, SPEC2 2014 VCAA 20 MC

Particles of mass 3 kg and 5 kg are attached to the ends of a light inextensible string that passes over a fixed smooth pulley, as shown above. The system is released from rest.

Assuming the system remains connected, the speed of the 5 kg mass after two seconds is

A.     4.0 m/s

B.     4.9 m/s

C.     9.8 m/s

D.   10.0 m/s

E.   19.6 m/s

Show Answers Only

`B`

Show Worked Solution
`sum F` `= 5g – 3g = 2g`
  `=2 xx 9.8`
  `=19.6\ \ text(N)`
`(5+3)a` `= 19.6`
`a` `=2.45`

  
`u = 0, quad t = 2, quad a = 2.45`

`v` `= u + at`
  `= 0 + 2.45 xx 2`
  `=4.9\ \ text(m/s)`

 
`=> B`

Vectors, SPEC2 2014 VCAA 17 MC

The acceleration vector of a particle that starts from rest is given by

`underset ~a(t) = −4 sin(2t) underset ~i + 20 cos (2t) underset ~j - 20 e^(−2t) underset ~k`, where `t >= 0`.

The velocity vector of the particle, `underset ~v(t)`, is given by

  1. `−8 cos(2t) underset ~i - 40 sin(2t) underset ~j + 40e^(−2t) underset ~k`
  2. `2 cos(2t) underset ~i + 10 sin(2t) underset ~j + 10e^(−2t) underset ~k`
  3. `(8 - 8 cos(2t)) underset ~i - 40 sin(2t) underset ~j + (40e^(−2t) - 40) underset ~k`
  4. `(2 cos(2t) - 2) underset ~i + 10 sin(2t) underset ~j + (10e^(−2t) - 10) underset ~k`
  5. `(4 cos(2t) - 4) underset ~i + 20 sin(2t) underset ~j + (20 - 20e^(−2t)) underset ~k`
Show Answers Only

`D`

Show Worked Solution
`underset ~ dot v(t)` `= underset ~a(t)`
`underset ~v(t)` `= int underset ~a (t)\ dt`
  `= (2 cos (2t) + c_0) underset ~i + (10 sin (2t) + c_1) underset ~j + (10e^(-2t) + c_2) underset ~k`

 
`text(S)text(ince)\ \ v=0\ \ text(when)\ \ t=0:`

`0` `= (2 cos (0) + c_0) underset ~i + (10 sin (0) + c_1) underset ~j + (10e^0 + c_2) underset ~k`
`0` `= (2 + c_0) underset ~i + c_1 underset ~j + (10 + c_2) underset ~k `

 
`=> c_0 = -2, \ \  c_1 = 0, \ \  c_2 = -10`
 

`:. underset ~v(t) = (2 cos (2t) – 2) underset ~i + 10 sin(2t) underset ~j + (10e^(-2t) – 10) underset ~k`

`=> D`

Vectors, SPEC2 2014 VCAA 15 MC

If  `theta`  is the angle between  `underset ~a = sqrt 3 underset ~i + 4 underset ~j - underset ~k`  and  `underset ~b = underset ~i - 4 underset ~j + sqrt 3 underset ~k`, then  `cos(2 theta)`  is

  1. `−4/5`
  2.    `7/25`
  3. `−7/25`
  4.    `14/25`
  5. `−24/25`
Show Answers Only

`B`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= sqrt 3 xx 1 + 4 xx (-4) + (-1) xx sqrt 3`
  `= -16`

 
`|\ underset~a\ | = sqrt20,\ \ |\ underset~b\ | = sqrt20,`

`cos(theta)` `=(underset ~a ⋅ underset ~b)/(|\ underset~a\ |\ |\ underset~b\ | `
  `= (-16)/(sqrt 20 xx sqrt 20)`
  `= -4/5`

 

`cos (2 theta)` `= 2cos^2theta – 1`
  `= 2(-4/5)^2 – 1`
  `= 7/25`

 
`=> B`

Calculus, SPEC2 2014 VCAA 14 MC

The differential equation that is best represented by the above direction field is

A.   `(dy)/(dx) = 1/(x - y)`

B.   `(dy)/(dx) = y - x`

C.   `(dy)/(dx) = 1/(y - x)`

D.   `(dy)/(dx) = x - y`

E.   `(dy)/(dx) = 1/(y + x)`

Show Answers Only

`C`

Show Worked Solution

`text(Method 1:)`

`text(Draw direction field of each option by CAS.)`
 

`text(Method 2:)`

`text(Consider quadrant 2,)`

`x < 0, \ \ y > 0, \ \  m > 0\ \ => text(Eliminate A, D and E)`
 

`text(Consider vertical gradients where)\ \ m=oo\ \ => text(Eliminate B)`

 
`=> C`

Calculus, SPEC2 2015 VCAA 20 MC

An object is moving in a straight line, initially at `5\ text(ms)^(–1)`. Sixteen seconds later, it is moving at `11\ text(ms)^(–1)` in the opposite direction to its initial velocity.

Assuming that the acceleration of the object is constant, after 16 seconds the distance, in metres, of the object from its starting point is

A.   24

B.   48

C.   73

D.   96

E.   128

Show Answers Only

`B`

Show Worked Solution

`u = 5, v = −11, t = 16`

`v=u+at`

`-11=5+a xx 16\ \ =>\ \ a=-1`

`s` `= ut+1/2 at^2`
  `= 5xx16 – 1 xx 16^2`
  `=-48`

 
`:.\ text(Object is 48 metres from strating point.)`

`=> B`

Mechanics, SPEC2 2015 VCAA 19 MC

A light inextensible string passes over a smooth pulley, as shown below, with particles of mass 1 kg and `m` kg attached to the ends of the string.
 

SPEC2 2015 VCAA 19 MC

If the acceleration of the 1 kg particle is 4.9 `text(ms)^(-2)` upwards, then `m` is equal to

A.   1

B.   2

C.   3

D.   4

E.   5

Show Answers Only

`C`

Show Worked Solution

`T – (9.8 xx 1) = 4.9 xx 1`

`T=14.7`
 

`m xx 9.8 – T` `=m xx 4.9`
`4.9m` `= 14.7“
`:. m` `= 14.7/4.9`
  `= 3`

`=> C`

Vectors, SPEC2 2015 VCAA 18 MC

The position vectors of two moving particles are given by  `underset~r_1(t) = (2 + 4t^2)underset~i + (3t + 2)underset~j`  and  `underset~r(t) = (6t)underset~i + (4 + t)underset~j`, where  `t >=0`.

The particles will collide at

  1. `3underset~i + 3.5underset~j`
  2. `6underset~i + 5underset~j`
  3. `3underset~i + 4.5underset~j`
  4. `0.5underset~i + underset~j`
  5. `5underset~i + 6underset~j`
Show Answers Only

`B`

Show Worked Solution

`text(Equating)\ \ underset~i\ \ text(components:)`

`2 + 4t^2 = 6t`

`t=1\ \ text(or)\ \ t = 1/2`
 

`text(Equating)\ \ underset~j\ \ text(components:)`

`3t + 2=4+t`

`t = 1`
 

`underset~r_1(1) = 6underset~i + 5underset~j`

`=> B`

Calculus, SPEC2 2015 VCAA 11 MC

The velocity–time graph for a body moving along a straight line is shown below.
 

SPEC2 2015 VCAA 11 MC
 

The body first returns to its initial position within the time interval

A.   (0, 0.5)

B.   (0.5, 1.5)

C.   (1.5, 2.5)

D.   (2.5, 3.5)

E.   (3.5, 5)

Show Answers Only

`D`

Show Worked Solution

`text(Body moves forward between)\ \ t = 0 and t=2.`

`text(Body moves in reverse direction between)\ \ t = 2 and t=4.`

`text(By approximation, between)\ \ t=0 and t=4,\ text(the area below)`

`text(the)\ xtext(-axis equals the area above the)\ x text(-axis occurs when)\ \ t~~3.25`

`text(seconds.)`

 
`=> D`

Statistics, SPEC2 2016 VCAA 20 MC

The lifetime of a certain brand of batteries is normally distributed with a mean lifetime of 20 hours and a standard deviation of two hours. A random sample of 25 batteries is selected.

The probability that the mean lifetime of this sample of 25 batteries exceeds 19.3 hours is

A.   0.0401

B.   0.1368

C.   0.6103

D.   0.8632

E.   0.9599

Show Answers Only

`E`

Show Worked Solution

`X\ ~\ N (20, 2^2)`

`bar X\ ~\ N (20, (2^2)/25)`

`text(Pr) (bar X > 19.3) ~~ 0.9599`

`=>  E`

Statistics, SPEC2 2016 VCAA 18 MC

Oranges grown on a citrus farm have a mean mass of 204 grams with a standard deviation of 9 grams

Lemons grown on the same farm have a mean mass of 76 grams with a standard deviation of 3 grams.

The masses of the lemons are independent of the masses of the oranges.

The mean mass and standard deviation, in grams, respectively of a set of three of these oranges and two of these lemons are

  1. `764, 3 sqrt 29`
  2. `636, 12`
  3. `764, sqrt 33`
  4. `636, 3 sqrt 10`
  5. `636, 33`
Show Answers Only

`A`

Show Worked Solution

`O~N (204, 9^2)`

`L~N (76, 3^2)`

`T = O+O+O + L + L`

`T~N (3 xx 204 + 2 xx 76, 9^2 + 9^2 + 9^2 + 3^2 + 3^2)`

`T~N (764, 261)`

`:. mu_T= 764, quad sigma_T = sqrt 261 = 3 sqrt 29`

`=>  A`

Mechanics, SPEC2 2016 VCAA 17 MC

A body of mass 3 kg is moving to the left in a straight line at 2 `text(ms)^(-1)`. It experiences a force for a period of time, after which it is then moving to the right at 2 `text(ms)^(-1)`.

The change in momentum of the particle, in kg `text(ms)^(-1)`, in the direction of the final motion is

A.   − 6

B.     0

C.     4

D.     6

E.   12

Show Answers Only

`E`

Show Worked Solution
`m_1 v_1` `= 3 xx -2 = -6`
`m_2 v_2` `= 3 xx 2 = 6`

 

`Deltap` `=m_2 v_2 – m_1 v_1`
  `= 6 – (-6)`
  `= 12`

 
`=>  E`

Mechanics, SPEC2 2016 VCAA 15 MC

A variable force of  `F`  newtons acts on a 3 kg mass so that it moves in a straight line. At time  `t`  seconds,  `t >= 0`, its velocity  `v`  metres per second and position  `x`  metres from the origin are given by  `v = 3 - x^2`.

It follows that

A.   `F = -2x`

B.   `F = -6x`

C.   `F = 2x^3 - 6x`

D.   `F = 6x^3 - 18x`

E.   `F = 9x - 3x^3`

Show Answers Only

`D`

Show Worked Solution

`(dv)/(dx) = -2x`

`a = v (dv)/(dx) = -2x(3 – x^2)`
 

`:. F` `=ma`
  `= -6x(3 – x^2)`
  `= 6x^3-18x`

 
`=>  D`

Vectors, SPEC2 2016 VCAA 13 MC

A particle of mass 5 kg is subject to forces  `12 underset ~i`  newtons and  `9 underset ~j`  newtons.

If no other forces act on the particle, the magnitude of the particle’s acceleration, in ms¯², is

  1. `3`
  2. `2.4 underset ~i + 1.8 underset ~j`
  3. `4.2`
  4. `9`
  5. `60 underset ~i + 45 underset ~j`
Show Answers Only

`A`

Show Worked Solution
`underset ~F` `= 12 underset ~i + 9 underset ~j`
`5 underset ~a` `= 12 underset ~i + 9 underset ~j`
`underset ~a` `= 12/5 underset ~i + 9/5 underset ~j`
   
`|underset ~a|` `= sqrt(144/25 + 81/25)`
  `= 3`

 
`=>  A`

Vectors, SPEC2 2016 VCAA 12 MC

If  `underset ~a = -2 underset ~i - underset ~j + 3 underset ~k`  and  `underset ~b = -m underset ~i + underset ~j + 2 underset ~k`, where  `m`  is a real constant, the vector  `underset ~a - underset ~b`  will be perpendicular to vector  `underset ~b`  where  `m`  equals

A.   0 only

B.   2 only

C.   0 or 2

D.   4.5

E.   0 or −2

Show Answers Only

`C`

Show Worked Solution
`underset ~a – underset ~b` `= (m – 2) underset ~i – 2 underset ~j + underset ~k `
`(underset ~a – underset ~b) ⋅ underset ~b` `= -m(m – 2) – 2(1) + 1(2) = 0`
`0` `= -m^2 + 2m – 2 + 2`
`0` `= 2m – m^2`
`0` `= m(2 – m) \ => \ m = 0 \ or \ 2`

`=>  C`

Vectors, SPEC2 2016 VCAA 11 MC

Let  `underset ~a = 3 underset ~i + 2 underset ~j + alpha underset ~k`  and  `underset ~b = 4 underset ~i - underset ~j + alpha^2 underset ~k`, where  `alpha`  is a real constant.

If the scalar resolute of  `underset ~a`  in the direction of  `underset ~b`  is  `74/sqrt 273`,  then  `alpha`  equals

A.   1

B.   2

C.   3

D.   4

E.   5 

Show Answers Only

`D`

Show Worked Solution
`underset ~a ⋅ underset ~ hat b` `= (3 xx 4 + 2 xx -1 + alpha xx alpha^2)/sqrt(4^2 + 1^2 + alpha^4)`
`74/sqrt 273` `= (12 – 2 + alpha^3)/ sqrt(17 + alpha^4)`
`74 sqrt(17 + alpha^4)` `=sqrt 273 (10 + alpha^3)`

 

`alpha = 1:\ \ text(LHS:)\ sqrt 273 xx 11 = 74 sqrt 18\ \ text(RHS)`  ✖

`alpha = 2:\ \ sqrt 273(18) = 74 sqrt 33`  ✖

`alpha = 3:\ \ sqrt 273 xx 37 = 74 sqrt 98`  ✖

`alpha = 4:\ \ sqrt 273 xx 74 = 74 sqrt 273`  ✔

`=>  D`

Calculus, SPEC2 2014 VCAA 10 MC

A large tank initially holds 1500 L of water in which 100 kg of salt is dissolved. A solution containing 2 kg of salt per litre flows into the tank at a rate of 8 L per minute. The mixture is stirred continuously and flows out of the tank through a hole at a rate of 10 L per minute.

The differential equation for `Q`, the number of kilograms of salt in the tank after `t` minutes, is given by

A.   `(dQ)/(dt) = 16 - (5Q)/(750 - t)`

B.   `(dQ)/(dt) = 16 - (5Q)/(750 + t)`

C.   `(dQ)/(dt) = 16 + (5Q)/(750 - t)`

D.   `(dQ)/(dt) = (100Q)/(750 - t)`

E.   `(dQ)/(dt) = 8 - Q/(1500 - 2t)`

Show Answers Only

`A`

Show Worked Solution

`(dQ_text(in))/(dV)= 2\ text(kg/L),\ (dV_text(in))/(dt) = 8\ text(L/min)`

`V_0 = 1500,\ Q_0 = 100`

`(dV_text(out))/(dt) = 10\ text(L/min)`
  

`V(t)` `= 1500 + (8 – 10)t`
  `= 1500 – 2t`
  `= 2(750 – t)`

 

`(dQ_text(in))/(dt)` `= 2 xx 8 = 16 text(kg/min)`
`(dQ_text(out))/(dt)` `= Q/(v(t)) xx 10`
  `= (10Q)/(2(750 – t))`
  `= (5Q)/(750 – t)`

 
`:.(dQ)/(dt)= 16 – (5Q)/(150 – t)`

`=> A`

Calculus, SPEC2 2016 VCAA 9 MC

If  `f(x) = (dy)/(dx) = 2x^2 - x`, where  `y_0 = 0 = y(2)`, then  `y_3`  using Euler’s formula with step size 0.1 is

A.  `0.1\ f(2)`

B.  `0.6 + 0.1\ f(2.1)`

C.  `1.272 + 0.1\ f(2.2)`

D.  `2.02 + 0.1\ f(2.3)`

E.  `2.02 + 0.1\ f(2.2)` 

Show Answers Only

`C`

Show Worked Solution
`y_1` `= y_0 + 0.1(2(2)^2 – 2)`
  `= 0.6`
`y_2` `= 0.6 + 0.1 (2(2.1)^2 – 2.1)`
  `= 1.272`
`:.  y_3` `= 1.272 + 0.1\ f(2.2)`

 
`=>  C`

Calculus, SPEC2 2016 VCAA 8 MC

Using a suitable substitution,  `int_a^b (x^3e^(2x^4))\ dx`, where  `a`  and  `b`  are real constants, can be written as

A.  `int_a^b (e^(2u))\ du`

B.  `int_(a^4)^(b^4) (e^(2u))\ du`

C.  `1/8 int_a^b (e^u)\ du`

D.  `1/4 int_(a^4)^(b^4) (e^(2u))\ du`

E.  `1/8 int_(8a^3)^(8b^3) (e^u)\ du` 

Show Answers Only

`D`

Show Worked Solution
`u` `= x^4`
`(du)/(dx)` `= 4x^3\ \ =>\ \ 1/4\ du = x^3\ dx`

  
`text(When)\ \ x=a, \ u=a^4`

`text(When)\ \ x=b, \ u=b^4`
 

`:. int_a^b x^3 e^(2x^4)\ dx`

`= 1/4 int_(a^4)^(b^4) e^(2u)\ du`
 

`=>  D`

Complex Numbers, SPEC2 2016 VCAA 6 MC

The points corresponding to the four complex numbers given by
 

`z_1 = 2 text(cis) (pi/3),\ z_2 = text(cis) ((3 pi)/4),\ z_3 = 2 text(cis)(-(2 pi)/3),\ z_4 = text(cis) (-pi/4)`
 

are the vertices of a parallelogram in the complex plane.

Which one of the following statements is not true?

A.  The acute angle between the diagonals of the parallelogram is  `(5 pi)/12`

B.  The diagonals of the parallelogram have lengths 2 and 4

C.  `z_1 z_2 z_3 z_4 = 0 `

D.  `z_1 + z_2 + z_3 + z_4 = 0`

E.  `1 <= |z| <= 2`  for all four of  `z_1, z_2, z_3, z_4` 

Show Answers Only

`C`

Show Worked Solution

`text(Consider each option:)`

`A: qquad theta_1` `= pi/3 + pi/4=(7 pi)/12`
`theta_2` `= pi – (7 pi)/12= (5 pi)/12\ \ text{(correct)}`

 

`B: qquad l_1` `= 2 + 2 = 4`
`l_2` `= 1+ 1 = 2\ \ text{(correct)}`

 

`C: qquad 2 text(cis) (pi/3) ⋅ text(cis) ((3 pi)/4) ⋅ 2 text(cis) (-(2 pi)/3) ⋅ text(cis) (-pi/4)`

`!= 0\ \ text{(by CAS → incorrect)}`

 
`=>  C`

Complex Numbers, SPEC2 2016 VCAA 4 MC

One of the roots of   `z^3 + bz^2 + cz = 0`  is  `3 - 2i`, where `b` and `c` are real numbers.

The values of `b` and `c` respectively are

A.  `6, 13`

B.  `3, -2`

C.  `-3, 2`

D.  `2, 3`

E.  `-6, 13` 

Show Answers Only

`E`

Show Worked Solution

`z(z^2 + bz + c) = 0,\ b, c in RR`

`text(Given)\ \ z= 3 – 2i \ \ text(is a root,)`

`=> z = 3 + 2i\ \ text{is a root (conjugate).}`
 

`z(z^2 + bz + c)` `= 0`
`z((z – 3) + 2i)((z – 3) – 2i)` `= 0`
`z((z – 3)^2 – 4i^2)` `= 0`
`z(z^2 – 6z + 9 + 4)` `= 0`
`z^3 – 6z^2 + 13z` `= 0`

 
`b = -6,\ \ c = 13` 

`=>  E`