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Probability, 2ADV S1 2008 HSC 9a

It is estimated that 85% of students in Australia own a mobile phone.

  1. Two students are selected at random. What is the probability that neither of them owns a mobile phone?  (2 marks)

  2. Based on a recent survey, 20% of the students who own a mobile phone have used their mobile phone during class time. A student is selected at random. What is the probability that the student owns a mobile phone and has used it during class time?   (1 mark)

Show Answers Only
  1. `9/400`
  2. `17/100`
Show Worked Solution

i.     `P(M) = 0.85`

COMMENT: `M^c` is syllabus notation for the complement of event `M`.

`P(M^c) = 1-0.85 = 0.15`

`:.\ P(M^c, M^c)` `= 15/100 * 15/100`
  `= 225/(10\ 000)`
  `= 9/400`

 

ii.  `text{P(owns mobile and used it)}`

`= P(M) xx  P\text{(used it)}`

`= 17/20 xx 20/100`

`= 17/100`

Plane Geometry, 2UA 2008 HSC 8b

In the diagram,  `ABCD`  is a parallelogram and  `ABEF`  and  `BCGH`  are both squares.

Copy or trace the diagram into your writing booklet.

  1. Prove that  `CD = BE`.  (1 mark)
  2. Prove that  `BD = EH`.   (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)  `text(Prove)\ CD = BE`

`CD` `= AB\ \ text{(opposite sides of parallelogram)}`
`AB` `= BE\ \ text{(sides of square)}`
`:.\ CD` `= BE\ \ \ text(… as required)`

 

(ii)

`text(Prove)\ BD = EH`

`CD` `= BE\ \ text{(from part (i))}`
`BC` `= BH\ \ text{(sides of square)}`

`/_BDC = /_ABD = alpha\ \ text{(} text(alternate,)\ AB\ text(||)\ CD text{)}`

`text(Let)\ /_DBC = beta`

`/_BCD` `= 180 – alpha – beta\ \ \ \ text{(angle sum of}\ Delta BDCtext{)}`
`/_EBH` `= 360 – 90 – 90 – alpha – beta\ \ \ `
   `text{(angles about a point)}`
  `= 180 – alpha – beta`
  `= /_BCD`

`=> Delta DCB ~= Delta EBH\ \ text{(SAS)}`

`:.\ BD = EH\ \ \ ` `text{(corresponding sides of)`
  `\ \ text{congruent triangles)}`

Calculus, 2ADV C3 2008 HSC 8a

Let  `f(x) = x^4 − 8x^2`.

  1. Find the coordinates of the points where the graph of  `y = f(x)`  crosses the axes.  (2 marks)

  2. Show that  `f(x)`  is an even function.   (1 mark)

  3. Find the coordinates of the stationary points of  `f(x)`  and determine their nature.   (4 marks)

  4. Sketch the graph of  `y = f(x)`.   (1 mark)

Show Answers Only
  1. `(-2 sqrt 2,0), (2 sqrt 2, 0)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(S.P.s at)\ (0,0), (2,-16), (-2,-16)`
  4.  
  5. 2UA HSC 2008 8ai
Show Worked Solution
i.    `f(x) = x^4 – 8x^2`

`y text(-intercept when)\ x = 0`

`:.\ text(Cuts)\ y\ text(axis at)\ (0,0)`

`x\ text(-intercept when)\ f(x) = 0`

`x^4 – 8x^2` `= 0`
`x^2 (x^2 – 8)` `= 0`

`x^2 – 8 = 0\ \ \ \ \ text(or)\ \ \ \ \ x = 0`

`x^2` `= 8`
`x` `= +- sqrt 8 = +- 2 sqrt 2`

 

`:.\ text(Cuts)\ x text(-axis at)\ (-2 sqrt 2,0),\ \ text(and)\ \ (2 sqrt 2, 0)`

`text{(Note that it only touches}\ xtext{-axis at (0,0))}`

 

ii.    `f(x)` `= x^4 – 8x^2`
  `f(–x)` `= (–x)^4 – 8(–x)^2`
    `= x^4 – 8x^2`
    `= f(x)`

 

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

 

iii.   `f(x)` `= x^4 – 8x^2`
  `f'(x)` `= 4x^3 – 16x`
  `f″(x)` `=12x^2-16`

 

`text(S.P.  when)\ \ f'(x) = 0`

`4x^3 – 16x` `= 0`
`4x (x^2 – 4)` `= 0`
`x^2 – 4` `=0\ \ \ \ \ x = 0`
`x^2` `= 4`
`x` `= +-2`
`text(At)\ x=2\ \ `  `f(x)`  `=(2)^4 – 8(2)^2 = -16`
  `f″(x)` `=12(2^2)-16>0`

`:.\ text{MIN at (2, –16)}`

`:.\ text{MIN at (–2, –16)},\ \ \ (f(x)\ text(is even))`

 

`text(At)\ x=0\ \ `  `f(x)`  `=(0)^4 – 8(0)^2 = 0`
  `f″(x)` `=12(0^2)-16<0`

`:.\ text{MAX at (0,0)}`

 

iv. 

Probability, 2ADV S1 2008 HSC 7c

Xena and Gabrielle compete in a series of games. The series finishes when one player has won two games. In any game, the probability that Xena wins is  `2/3`  and the probability that Gabrielle wins is  `1/3`.

Part of the tree diagram for this series of games is shown.
 

 
 

  1. Complete the tree diagram showing the possible outcomes.  (1 mark)
  2. What is the probability that Gabrielle wins the series?   (2 marks)

  3. What is the probability that three games are played in the series?   (2 marks)

Show Answers Only
  1.    
  2. `7/27`
  3. `4/9`
Show Worked Solution
i. 2UA HSC 2008 7ci

 

MARKER’S COMMENT: A tree diagram with 8 outcomes is incorrect (i.e. no third game is played if 1 player wins the first 2 games). If outcomes cannot occur, do not draw them on a tree diagram.

 

ii.  `P text{(} G\ text(wins) text{)}`

`= P(XGG) + P (GXG) + P (GG)`

`= 2/3 * 1/3 * 1/3 + 1/3 * 2/3 * 1/3 + 1/3 * 1/3`

`= 2/27 + 2/27 + 1/9`

`= 7/27`

 

iii.  `text(Method 1:)`

`P text{(3 games played)}`

`= P (XG) + P(GX)`

`= 2/3 * 1/3 + 1/3 * 2/3`

`= 4/9`

 

`text(Method 2:)`

`P text{(3 games)}`

`= 1 – [P(XX) + P(GG)]`

`= 1 – [2/3 * 2/3 + 1/3 * 1/3]`

`= 1 – 5/9`

`= 4/9`

Trigonometry, 2ADV T1 2008 HSC 7b

2008 7b

The diagram shows a sector with radius  `r`  and angle  `theta`  where  `0 < theta <= 2pi`.

The arc length is  `(10pi)/3`. 

  1.  Show that  `r >= 5/3`.   (2 marks)

  2.  Calculate the area of the sector when  `r = 4`.    (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(20pi)/3\ text(u²)`
Show Worked Solution
i.    `text(Show)\ r >= 5/3`
`text(Arc length)\ ` `= r theta\ \ text(where)\ \ 0 < theta <= 2pi`
`r theta` `= (10pi)/3`
`:.theta` `= (10pi)/(3r)`

 

`text(Using)\ \ \ 0 <= theta <= 2 pi`

`0 <= (10pi)/(3r)` `<= 2pi`
`(10pi)/3` `<= 2 pi r`
`5/3` `<= r`

 

`:.\ r >= 5/3\ \ \ text(… as required.)`

 

ii.   `text(Area)` `= 1/2 r^2 theta`
    `= 1/2 xx 4^2 xx (10pi)/(3 xx 4)`
    `= (20pi)/3\ text(u²)`

Calculus, EXT1* C3 2008 HSC 6c

The graph of  `y = 5/(x - 2)`  is shown below.
 

2008 6c
 

The shaded region in the diagram is bounded by the curve  `y = 5/(x - 2)`, the  `x`-axis and the lines  `x = 3`  and  `x = 6`.

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

Show Answers Only

`(75pi)/4\ text(u³)`

Show Worked Solution
`y` `= 5/(x – 2)`
`V` `= pi int_3^6 y^2\ dx`
  `= pi int_3^6 (5/(x – 2))^2\ dx`
  `= 25 pi int_3^6 1/((x – 2)^2)\ dx`
  `= 25 pi [(-1)/(x – 2)]_3^6`
  `= 25 pi [-1/4 – (-1)]`
  `=25 pi [3/4]`
  `= (75pi)/4\ text(u³)`

Calculus, 2ADV C4 2008 HSC 5a

The gradient of a curve is given by  `dy/dx = 1-6 sin 3x`. The curve passes through the point  `(0, 7)`.

What is the equation of the curve?   (3 marks)

Show Answers Only

`y = x + 2 cos 3x + 5`

Show Worked Solution
`dy/dx` `= 1-6 sin 3x`
`y` `= int 1-6 sin 3x\ dx`
  `= x + 2 cos 3x + c`

 

`text(Passes through)\ (0,7):`

`=> 0 + 2 cos 0 + c` `= 7`
`2 + c` `= 7`
`c` `= 5`

 

`:.\ text(Equation is)\ \ \ y = x + 2 cos 3x + 5`

Quadratic, 2UA 2008 HSC 4c

Consider the parabola  `x^2 = 8(y\ – 3)`.

  1. Write down the coordinates of the vertex.  (1 mark)
  2. Find the coordinates of the focus.   (1 mark)
  3. Sketch the parabola.   (1 mark)
  4. Calculate the area bounded by the parabola and the line  `y = 5`.   (3 marks)
Show Answers Only

(i)   `(0,3)`

(ii)   `(0,5)`

(iii)  

(iv)  `10 2/3\ text(u²)`

Show Worked Solution
(i)    `text(Vertex)\ = (0,3)`

 

(ii)   `text(Using)\ \ \ x^2` `= 4ay`
  `4a` `= 8`
  `a` `= 2`

`:.\ text(Focus) = (0,5)`

 

(iii) 2UA HSC 2008 4c 

 

(iv)   `text(Intersection when)\ y = 5`
`=> x^2` `= 8 (5-3)`
`x^2` `= 16`
`x` `= +- 4`

`text(Find shaded area)`

`x^2` `= 8 (y-3)`
`y – 3` `= x^2/8`
`y` `= x^2/8 +3`

 

`text(Area)` `= int_-4^4 5\ dx – int_-4^4 x^2/8 + 3\ dx`
  `= int_-4^4 5 – (x^2/8 + 3)\ dx`
  `= int_-4^4 2 – x^2/8\ dx`
  `= [2x – x^3/24]_-4^4`
  `= [(8 – 64/24) – (-8 + 64/24)]`
  `= 5 1/3 – (- 5 1/3)`
  `= 10 2/3\ text(u²)`

Plane Geometry, 2UA 2008 HSC 4a

In the diagram, `XR` bisects `/_PRQ` and `XY\ text(||)\ QR`.

Prove that `Delta XYR` is an isosceles triangle.   (2 marks)

Show Answers Only

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

Show Worked Solution

`text{Since}\ XR\ text{bisects}\ /_PRQ`

`/_XRQ` `= /_YRX = theta`
`/_RXY` `= theta\ \ \ text{(} text(alternate angles,)\ XY\ text(||)\ QR text{)}`

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

Calculus, 2ADV C4 2008 HSC 3b

  1. Differentiate  `log_e (cos x)`  with respect to  `x`.  (2 marks)

  2. Hence, or otherwise, evaluate  `int_0^(pi/4) tan x\ dx`.    (2 marks)

Show Answers Only
  1. `- tan x`
  2. `- log_e (1/sqrt2)\ \ text(or)\ \ 0.35\ \ text{(2 d.p.)}`
Show Worked Solution
i.    `y` `= log_e (cos x)`
  `dy/dx` `= (- sin x)/(cos x)`
    `= – tan x`

 

ii.    `int_0^(pi/4) tan x\ dx`
  `= – [log_e (cos x)]_0^(pi/4)`
  `= – [log_e(cos (pi/4)) – log_e (cos 0)]`
  `= – [log_e (1/sqrt2) – log_e 1]`
  `= – [log_e (1/sqrt2) – 0]`
  `= – log_e (1/sqrt2)`
  `= 0.346…`
  `= 0.35\ \ text{(2 d.p.)}`

Linear Functions, 2UA 2008 HSC 3a

2008 3a

In the diagram,  `ABCD`  is a quadrilateral. The equation of the line  `AD`  is  `2x- y- 1 = 0`. 

  1. Show that  `ABCD`  is a trapezium by showing that  `BC`  is parallel to  `AD`.  (2 marks)
  2. The line  `CD`  is parallel to the  `x`-axis. Find the coordinates of  `D`.   (1 mark)
  3. Find the length of  `BC`.   (1 mark)
  4. Show that the perpendicular distance from  `B`  to  `AD`  is  `4/sqrt5`.   (2 marks)
  5. Hence, or otherwise, find the area of the trapezium  `ABCD`.   (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(3,5)`
  3. `sqrt 5\ text(units)`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `8\ text(u²)`
Show Worked Solution
(i)    `text(Show)\ BC \ text(||)\ AD`

`B(0,3),\ \ C(1,5)`

`m_(BC)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (5 – 3)/(1 – 0)`
  `= 2`

 

`text(Equation)\ \ AD\ \ text(is)\ \ 2x – y – 1 = 0`

`y` `= 2x – 1`
`m_(AD)` `= 2`

`:. BC\  text(||) \ AD`

`:. ABCD\ text(is a trapezium)`

 

(ii)    `text(Given)\ CD\  text(||) \ x text(-axis)`
  `text(Equation)\ CD\ text(is)\ y = 5`
  `D\ text(is intersection of)`
`y` `= 5,\ \ and`
`2x – y – 1` `= 0`
`:. 2x – 5 – 1` `=0`
`2x` `=6`
`x` `=3`
`:.\ D` `= (3,5)`

 

(iii)   `B(0,3),\ \ C(1,5)`
`text(dist)\ BC` `= sqrt ( (x_2 – x_1)^2 + (y_2 – y_1)^2 )`
  `= sqrt ( (1-0)^2 + (5-3)^2 )`
  `= sqrt (1 + 4)`
  `= sqrt 5\ text(units)`

 

(iv)   `text(Show)\ _|_\ text(dist of)\ B\ text(to)\ AD\ text(is)\ 4/sqrt5`

`B (0,3)\ \ \ \ \ 2x – y – 1 = 0` 

`_|_\ text(dist)` `= | (ax_1 + by_1 + c)/sqrt (a^2 + b^2) |`
  `= |( 2(0) – 1(3) -1 )/sqrt (2^2 + (-1)^2) |`
  `= | -4/sqrt5 |`
  `= 4/sqrt 5\ \ \ text(… as required.)`

 

(v)    `text(Area)` `= 1/2 h (a + b)`
    `= 1/2 xx 4/sqrt5 (BC + AD)`

 
`BC = sqrt5\ \ text{(part (iii))}`
 

`A(0,–1),\ \ D(3,5)`

`text(dist)\ AD` `= sqrt ( (3-0)^2 + (5+1)^2 )`
  `= sqrt (9 + 36)`
  `= sqrt 45`
  `= 3 sqrt 5`
`:.\ text(Area)\ ABCD` `= 1/2 xx 4/sqrt5 (sqrt5 + 3 sqrt 5)`
  `= 2 / sqrt5 (4 sqrt 5)`
  `= 8\ text(u²)`

Functions, 2ADV F1 2008 HSC 1c

Simplify  `2/n-1/(n+1)`.   (2 marks)

Show Answers Only

`(n + 2)/(n(n+1))`

Show Worked Solution

`2/n-1/(n+1)`

`= (2(n+1)-1(n))/(n(n+1))`

`= (2n + 2-n)/(n(n+1))`

`= (n+2)/(n(n+1))`

Financial Maths, STD2 F5 SM-Bank 2

The table below shows the present value of an annuity with a contribution of  $1.
 

  1. Fiona pays $3000 into an annuity at the end of each year for 4 years at 2% p.a., compounded annually.   What is the present value of her annuity?  (1 mark)

  2. If John pays $6000 into an annuity at the end of each year for 2 years at 4% p.a., compounded annually, is he better off than Fiona?  Use calculations to justify your answer.    (2 marks)

Show Answers Only
  1. `$11\ 423.10`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text(Table factor when)\ \ n = 4,\ r = text(2%) \ => \ 3.8077`

`:.\ PVA\ text{(Fiona)}` `= 3000 xx 3.8077`
  `= $11\ 423.10`

 

ii.  `text(Table factor when)\ n = 2, r = text(4%)`

`=> 1.8861`

`:.\ PVA\ text{(John)}` `= 6000 xx 1.8861`
  `= $11\ 316.60`

 
`:.\ text(Fiona will be better off because her)\ PVA`

`text(is higher.)`

Mechanics, EXT2* M1 2014 HSC 14a

The take-off point `O` on a ski jump is located at the top of a downslope. The angle between the downslope and the horizontal is  `pi/4`.  A skier takes off from `O` with velocity `V` m s−1 at an angle `theta` to the horizontal, where  `0 <= theta < pi/2`.  The skier lands on the downslope at some point `P`, a distance `D` metres from `O`.
 

2014 14a
 

The flight path of the skier is given by

`x = Vtcos theta,\ y = -1/2 g t^2 + Vt sin theta`,      (Do NOT prove this.)

where  `t`  is the time in seconds after take-off.

  1. Show that the cartesian equation of the flight path of the skier is given by
     
         `y =  x tan theta - (gx^2)/(2V^2) sec^2 theta`.   (2 marks)

  2. Show that  
     
         `D = 2 sqrt 2 (V^2)/(g) cos theta (cos theta + sin theta)`.   (3 marks)

  3. Show that  
     
         `(dD)/(d theta) = 2 sqrt 2 (V^2)/(g) (cos 2 theta - sin 2 theta)`.   (2 marks)

  4. Show that `D` has a maximum value and find the value of `theta` for which this occurs.   (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)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text(Show)\ \ y = x tan theta – (gx^2)/(2V^2) sec^2 theta`

`x` `= Vt cos theta`
`t` `= x/(V cos theta)\ \ \ …\ text{(1)}`

`text(Subst)\ text{(1)}\ text(into)\ y = -1/2 g t^2 + Vt sin theta`

`y` `= -1/2 g (x/(Vcostheta))^2 + V sin theta (x/(Vcostheta))`
  `= (-gx^2)/(2V^2 cos^2 theta) + x * (sin theta)/(cos theta)`
  `= x tan theta – (gx^2)/(2V^2) sec^2 theta\ \ \ text(… as required.)`

 

ii.  `text(Show)\ D = 2 sqrt 2 (V^2)/g\ cos theta (cos theta + sin theta)`

♦♦ Mean mark 28%
MARKER’S COMMENT: The key to finding `P` and solving for `D` is to realise you need the intersection of the Cartesian equation in (i) and the line `y=–x`.

`text(S)text(ince)\ P\ text(lies on line)\ y = -x`

`-x` `=x tan theta – (gx^2)/(2V^2) sec^2 theta`
`-1` `=tan theta – (gx)/(2V^2) sec^2 theta`
`(gx)/(2V^2) sec^2 theta` `= tan theta + 1`
`x (g/(2V^2))` `=(sin theta)/(cos theta) * cos^2 theta + 1 * cos^2 theta`
`x` `=(2V^2)/g\ (sin theta cos theta + cos^2 theta)`
  `=(2V^2)/g\ cos theta (cos theta + sin theta)`

 

`text(Given that)\ \ cos(pi/4)` `= x/D = 1/sqrt2`
`text(i.e.)\ \ D` `= sqrt 2 x`

 
`:.\ D = 2 sqrt 2 (V^2)/g\ cos theta (cos theta + sin theta)`

`text(… as required.)`

 

iii.  `text(Show)\ (dD)/(d theta) = 2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)`

`D` `= 2 sqrt 2 (V^2)/g\ (cos^2 theta + cos theta sin theta)`
`(dD)/(d theta)` `= 2 sqrt 2 (V^2)/g\ [2cos theta (–sin theta) + cos theta cos theta + (– sin theta) sin theta]`
  `= 2 sqrt 2 (V^2)/(g) [(cos^2 theta – sin^2 theta) – 2 sin theta cos theta]`
  `= 2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)\ \ \ text(… as required)`

 

iv.  `text(Max/min when)\ (dD)/(d theta) = 0`

♦ Mean mark 47%
IMPORTANT: Note that students can attempt every part of this question, even if they couldn’t successfully prove earlier parts.
`2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)` `= 0`
`cos 2 theta – sin 2 theta` `= 0`
`sin 2 theta` `= cos 2 theta`
`tan 2 theta` `= 1`
`2 theta` `= pi/4`
`theta` `= pi/8`
`(d^2D)/(d theta^2)` `= 2 sqrt 2 (V^2)/g\ [-2 sin 2theta – 2 cos 2 theta]`
  `= 4 sqrt 2 (V^2)/g\ (-sin 2 theta – cos 2 theta)`

 

`text(When)\ \ theta = pi/8:`

`(d^2 D)/(d theta^2)` `= 4 sqrt 2 (V^2)/g\ (- sin (pi/4) – cos (pi/4))`
  `= 4 sqrt 2 (V^2)/g\ (- 1/sqrt2 – 1/sqrt2) < 0`
  ` =>\ text(MAX)`

 

`:.\ D\ text(has a maximum value when)\ theta = pi/8`

Plane Geometry, EXT1 2014 HSC 13d

In the diagram,  `AB`  is a diameter of a circle with centre  `O`. The point  `C`  is chosen such that  `Delta ABC`  is acute-angled. The circle intersects  `AC`  and  `BC`  at  `P`  and  `Q`  respectively.

Copy or trace the diagram into your writing booklet.

  1. Why is   `/_BAC = /_CQP`?   (1 mark)
  2. Show that the line  `OP`  is a tangent to the circle through  `P`,  `Q`  and  `C`.    (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   

`APQB\ text(is a cyclic quad)`

`=> /_BAC + /_BQP = 180°`

`text{(opposite angles of cyclic quadrilateral}\ BAPQ text{)}`

`/_CQP + /_BQP = 180°\ text{(}/_ CQB\ text{is a straight angle)}`

`:.\ /_BAC = /_CQP`

 

(ii)  `text(Let)\ /_BAC = alpha`

♦ Mean mark 37%
COMMENT: Labelling angles `alpha` and `beta` etc.. can often make a proof clearer and less messy.

`/_OPA = alpha\ \ \ \ text{(angles opposite equal sides}`

`text(in)\ Delta OAP,\ OA=OP\ text{radii)}`

`/_TPC = alpha\ \ \ text{(vertically opposite)}`

`text(S)text(ince)\ /_TPC = /_CQP = alpha`

 

`:. OP\ text(is a tangent to circle)\ CPQ`

`text{(angles in alternate segments equal)}`

Calculus, EXT1 C1 2014 HSC 13b

One end of a rope is attached to a truck and the other end to a weight. The rope passes over a small wheel located at a vertical distance of  40 m above the point where the rope is attached to the truck.

The distance from the truck to the small wheel is `L\ text(m)`, and the horizontal distance between them is  `x\ text(m)`. The rope makes an angle `theta` with the horizontal at the point where it is attached to the truck.

The truck moves to the right at a constant speed of   `text(3 m s)^(-1)`, as shown in the diagram.
 


 

  1. Using Pythagoras’ Theorem, or otherwise, show that  `(dL)/(dx) = cos theta`.   (2 marks)

  2.  Show that  `(dL)/(dt) = 3 cos theta`.    (1 mark)

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

`text(Show)\ \ (dL)/(dx) = cos theta`

`text(Using Pythagoras,)`

`L^2` `=40^2 + x^2`
`L` `=(40^2 + x^2)^(1/2)`
`(dL)/(dx)` `=1/2 * 2x * (40^2 + x^2)^(-1/2)`
  `=x/ sqrt((40^2 + x^2))`
  `=x/L`
  `=cos theta\ \ \ text(… as required.)`

 

ii.   `text(Show)\ \ (dL)/(dt) = 3 cos theta`

`(dL)/(dt)` `= (dL)/(dx) * (dx)/(dt)`
  `= cos theta * 3`
  `= 3 cos theta\ \ \ text(… as required)`

Binomial, EXT1 2014 HSC 12d

Use the binomial theorem to show that
 

 `0 = ((n),(0)) - ((n),(1)) + ((n),(2)) - ... + (-1)^n ((n),(n))`.   (2 marks) 

 

Show Answers Only

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

Show Worked Solution

`text(Prove)`

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

`:. 0 = ((n),(0)) – ((n),(1)) + ((n),(2)) – … + (-1)^n ((n),(n))`

Mechanics, EXT2* M1 2014 HSC 12c

A particle moves along a straight line with displacement  `x\ text(m)`  and velocity  `v\ \ text(ms)^(-1)`. The acceleration of the particle is given by

 `ddot x = 2 - e^(-x/2)`.

Given that  `v = 4`  when  `x = 0`, express  `v^2`  in terms of  `x`.   (3 marks)

Show Answers Only

`v^2 = 4x + 4e^(-x/2) + 12`

Show Worked Solution

`ddot x = d/(dx) (1/2 v^2) = 2 – e^(-x/2)`

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

 
`text(When)\ \ v = 4, x = 0:`

`:. 4^2` `= 0 + 4e^0 + c`
`16` `= 4 + c`
`c` `= 12`

 
`:.\ v^2 = 4x + 4e^(-x/2) + 12`

Calculus, EXT1 C3 2014 HSC 12b

The region bounded by  `y = cos 4x`  and the  `x`-axis, between  `x = 0`  and  `x = pi/8`, is rotated about the  `x`-axis to form a solid.   
 

2014 12b
 

Find the volume of the solid.   (3 marks)

Show Answers Only

`(pi^2)/16\ \ text(u³)`

Show Worked Solution
COMMENT: The identities `cos 2theta=` `cos^2 theta-sin^2 theta =` ` 2cos^2 theta-1=1-2sin^2 theta`  are tested every year – know them. 
`V` `= pi int_0^(pi/8) y^2\ dx`
  `= pi int_0^(pi/8) cos^2 4x\ dx`
  `= pi int_0^(pi/8) 1/2 (cos 8x + 1)\ dx`
  `= pi/2 [1/8 sin 8x + x]_0^(pi/8)`
  `= pi/2 [(1/8 sin pi + pi/8)] – 0]`
  `= (pi^2)/16\ \ text(u³)`

Mechanics, EXT2* M1 2014 HSC 12a

 A particle is moving in simple harmonic motion about the origin, with displacement `x` metres. The displacement is given by  `x = 2 sin 3t`, where `t` is time in seconds. The motion starts when  `t = 0`.

  1. What is the total distance travelled by the particle when it first returns to the origin?   (1 mark)

  2. What is the acceleration of the particle when it is first at rest?    (2 marks)

Show Answers Only
  1. `4\ text(m)`
  2. `-18\ text(m/s²)`
Show Worked Solution
i.    `x = 2 sin 3t`

 
`text(At)\ \ t = 0,\ x = 0`

`text(Amplitude) = 2`

`:.\ text(Distance travelled)`

`= 2 xx text(amplitude)`

`= 4\ text(m)`

 

ii.    `x` `= 2 sin 3t`
  `v` `= 6 cos 3t` 
  `ddot x` `= -18 sin 3t`

 
`text(Particle first comes to rest when)\ v = 0`

`6 cos 3t` `= 0`
`cos 3t` `= 0`
`3t` `= pi/2`
`t` `= pi/6`

 
`text(When)\ \  t = pi/6`

`ddot x` `= -18 * sin (3 xx pi/6)`
  `= -18 sin (pi/2)`
  `= -18\ text(m/s²)`

L&E, EXT1 2014 HSC 11f

Differentiate  `(e^x ln x)/x`.   (2 marks)

Show Answers Only

`(e^x (x * lnx + 1 – lnx))/(x^2)`

Show Worked Solution

`y = (e^x * lnx)/x`

`u` `= e^x lnx` `\ \ \ \ \ v` `= x`
`u prime` `= (e^x)/x + e^x lnx` `\ \ \ \ \ v prime` `= 1`
  `= e^x (1/x + lnx)`    

 

`dy/dx` `= (u prime v – u v prime)/(v^2)`
  `= (e^x (1/x + lnx) * x – e^x lnx * 1)/(x^2)`
  `= (e^x (x* lnx + 1 – lnx))/(x^2)`

Calculus, EXT1 C2 2014 HSC 11d

Evaluate  `int_2^5 x/(sqrt(x - 1))\ dx`  using the substitution  `x = u^2 + 1`.   (3 marks) 

Show Answers Only

`20/3`

Show Worked Solution
`x` `= u^2 + 1`
`u^2` `= x – 1`
`u` `= sqrt (x – 1)`
`(du)/(dx)` `= 1/2 (x – 1)^(-1/2)`
  `= 1/(2 sqrt(x – 1))`
`\ \ =>2du` `= dx/sqrt(x – 1)`

 

`text(When)\ \ \ x = 5,` `\ \ u = 2`
`x = 2,` `\ \ u = 1`

`:.\ int_2^5 x/(sqrt(x – 1))\ dx`

`= 2 int_1^2 u^2 + 1\ du`

`= 2 [ (u^3)/3 + u]_1^2`

`= 2 [(8/3 + 2) – (1/3 + 1)]`

`= 20/3`

Statistics, EXT1 S1 2014 HSC 11b

The probability that it rains on any particular day during the 30 days of November is 0.1.

Write an expression for the probability that it rains on fewer than 3 days in November.   (2 marks)

Show Answers Only

`\ ^30 C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1) (0.9)^29 + (0.9)^30`

Show Worked Solution
`P (R)` `= 0.1`
`P (bar R)` `= 1 – 0.1 = 0.9`

 
`text(Over 30 days:)`

`P (R<3)`

`= P (R=2) + P(R=1) + P (R=0)`

  
  `=\ ^30 C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1)^1 (0.9)^29`  
  `qquad  qquad +\ ^30C_0 (0.1)^0 (0.9)^30`  
  `=\ ^30C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1)(0.9)^29 + (0.9)^30`  

Functions, EXT1 F2 2014 HSC 9 MC

The remainder when the polynomial  `P(x) = x^4-8x^3-7x^2 + 3`  is divided by  `x^2 + x`  is  `ax + 3`.

What is the value of  `a`?

  1. `-14`
  2. `-11`
  3. `-2`
  4. `5`
Show Answers Only

`C`

Show Worked Solution

`P(x) = x^4-8x^3-7x^2 + 3`

`text(Given)\ \ P(x)` `= (x^2 + x) *Q(x) + ax + 3`
  `= x (x + 1) Q(x) + ax + 3`

 
`P(-1) = 1 + 8-7 + 3 = 5`

`:. -a + 3` `= 5`
`a` `= -2`

 
`=>  C`

Functions, EXT1 F2 2014 HSC 5 MC

Which group of three numbers could be the roots of the polynomial equation  `x^3 + ax^2 − 41x + 42 = 0`?

  1. `2, 3, 7`
  2. `1, −6, 7`
  3. `−1, −2, 21`
  4. `−1, −3, −14`
Show Answers Only

`B`

Show Worked Solution

`text(Let roots be)\ alpha, beta, gamma`

`alpha beta gamma = -d/a = -42`

`:.\ text(Cannot be)\ A\ text(or)\ C`

`alpha beta + beta gamma + gamma alpha = c/a = -41`

`:.\ text(Cannot be D)`

`=>  B`

Linear Functions, EXT1 2014 HSC 4 MC

The acute angle between the lines  `2x + 2y = 5`  and  `y = 3x + 1`  is  `theta`.

What is the value of  `tan theta`?

  1. `1/7` 
  2. `1/2`
  3. `1`
  4. `2`
Show Answers Only

`D`

Show Worked Solution

`y = 3x + 1`

`:. m_1 = 3`

`2x + 2y` `= 5`
`2y` `= -2x + 5`
`y` `= -x + 5/2` 

`:. m_2 = -1`

 

`tan theta` `= | (m_1 – m_2)/(1 + m_1 m_2)|`
  `= | (3 – (–1))/(1 + (3 xx –1))|`
  `= |4/(–2)|`
  `= 2`

`=>  D`

Calculus, EXT1 C1 2009 HSC 5b

The cross-section of a 10 metre long tank is an isosceles triangle, as shown in the diagram. The top of the tank is horizontal.
 

 
 

When the tank is full, the depth of water is 3 m. The depth of water at time `t` days is `h` metres.   

  1. Find the volume, `V`, of water in the tank when the depth of water is `h` metres.     (1 mark)

  2. Show that the area, `A`, of the top surface of the water is given by  `A = 20 sqrt3 h`.   (1 mark)

  3. The rate of evaporation of the water is given by  `(dV)/(dt) = - kA`, where `k` is a positive constant. 

     

    Find the rate at which the depth of water is changing at time `t`.   (2 marks)

  4. It takes 100 days for the depth to fall from 3 m to 2 m. Find the time taken for the depth to fall from 2 m to 1 m.   (1 mark)

Show Answers Only
  1. `10 sqrt 3 h^2\ \ text(m³)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `-k\ \ \ text(metres per day)`
  4. `100\ text(days)`
Show Worked Solution
MARKER’S COMMENT: Students who drew a diagram and included their working calculations on it were the most successful.
(i) 
  `text(Let)\ A = text(area of front)`
`tan 30^@` `= h/x`
`x` `= h/(tan 30^@)`
  `= sqrt3 h`
`:.\ A` `= 2 xx 1/2 xx sqrt 3 h xx h`
  `= sqrt 3 h^2\ \ text(m²)`

 

`V` `= Ah`
  `= sqrt 3 h^2 xx 10`
  `= 10 sqrt3 h^2\ \ text(m³)`

 

(ii)    `text(Area of surface)`
  `= 10 xx 2 sqrt 3 h`
  `= 20 sqrt 3 h\ \ text(m²)`

 

(iii)    `(dV)/(dt)` `= -kA`
    `= -k\ 20 sqrt3 h`
`V` `= 10 sqrt3 h^2`
`(dV)/(dh)` `= 20 sqrt 3 h`

 

`text(Find)\ (dh)/(dt)`

MARKER’S COMMENT: Half marks awarded for stating an appropriate chain rule, even if the following calculations were incorrect. Show your working!
`(dV)/(dt)` `= (dV)/(dh) * (dh)/(dt)`
`(dh)/(dt)` `= ((dV)/(dt))/((dV)/(dh))`
  `= (-k * 20 sqrt 3 h)/(20 sqrt 3 h)`
  `= -k`

 

`:.\ text(The water depth is changing at a rate)`

`text(of)\ -k\ text(metres per day.)`

 

♦♦♦ Exact data for part (iv) not available.
COMMENT: Interpreting a constant rate of change was very poorly understood!
(iv)    `text(S)text(ince)\ \ (dh)/(dt)\ \ text(is a constant, each metre)`
  `text(takes the same time.)`

 
`:.\ text(It takes 100 days to fall from 2 m to 1 m.)`

Mechanics, EXT2* M1 2009 HSC 5a

The equation of motion for a particle moving in simple harmonic motion is given by

`(d^2x)/(dt^2) = -n^2x`

where  `n`  is a positive constant,  `x`  is the displacement of the particle and  `t`  is time.  

  1. Show that the square of the velocity of the particle is given by
     
         `v^2 = n^2 (a^2\ - x^2)`

     

    where  `v = (dx)/(dt)`  and  `a`  is the amplitude of the motion.   (3 marks)

  2. Find the maximum speed of the particle.     (1 mark)

  3. Find the maximum acceleration of the particle.    (1 mark)

  4. The particle is initially at the origin. Write down a formula for  `x`  as a function of  `t`, and hence find the first time that the particle’s speed is half its maximum speed.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `na`
  3. `n^2 a`
  4. `pi/(3n)`
Show Worked Solution
i.    `text(Show)\ \ v^2 = n^2 (a^2\ – x^2)`

`(d^2x)/(dt^2) = -n^2x`

`d/(dx) (1/2 v^2)` `= -n^2 x`
`1/2 v^2` `= int -n^2x\ dx`
  `= (-n^2 x^2)/2 + c`
`v^2` `= -n^2 x^2 + c`

 
`text(When)\ \ v = 0,\ \ x = a`

`0` `= -n^2a + c`
`c` `= n^2 a^2`
`:.\ v^2` `= -n^2 x^2 + n^2 a^2`
  `= n^2 (a^2\ – x^2)\ \ text(… as required)`

 

ii.    `text(Max speed when)\ \ x = 0`
`v^2` `= n^2 (a^2\ – 0)`
  `= n^2 a^2`
`:.v_text(max)` `= na`

 

iii.   `(d^2x)/(dt^2)\ \ text(is maximum at limits)\ \ (x = +-a)`
`(d^2x)/(dt^2)` `= |n^2 (a)|`
  `= n^2 a`

 

iv.   `x` `= a sin nt`
  `dot x` `= an cos nt`

 

`text(Find)\ \ t\ \ text(when)\ \ dot x = (na)/2`

`(na)/2` `= an cos nt`
`cos nt` `= 1/2`
`nt` `= pi/3`
`:.t` `= pi/(3n)`

Geometry and Calculus, EXT1 2009 HSC 4b

Consider the function  `f(x) = (x^4 + 3x^2)/(x^4 + 3)`. 

  1. Show that  `f(x)`  is an even function.    (1 mark)
  2. What is the equation of the horizontal asymptote to the graph  `y = f(x)`?    (1 mark)
  3. Find the  `x`-coordinates of all stationary points for the graph  `y = f(x)`.   (3 marks)
  4. Sketch the graph  `y = f(x)`. You are not required to find any points of inflection.    (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `y = 1`
  3. `x = 0, – sqrt 3, sqrt 3`
  5.  
Show Worked Solution
(i)    `f(x)` `= (x^4 + 3x^2)/(x^4 + 3)`
  `f(–x)` `= ((–x)^4 + 3(-x)^2)/((–x)^4 + 3)`
    `= (x^4 + 3x^2)/(x^4 + 3)`
    `= f(x)`

 

`:.\ text(Even function.)`

 

(ii)    `y` `= (x^4 + 3x^2)/(x^4 + 3)`
    `= (1 + 3/(x^2))/(1 + 3/(x^4))`
`text(As)\ \ ` `x` `-> oo`
  `y` `-> 1`

 

`:.\ text(Horizontal asymptote at)\ \ y = 1`

 

(iii)   `f(x) = (x^4 + 3x^2)/(x^4 + 3)`
`u` `= x^4 + 3x^2\ \ \ \ \ ` `v` `= x^4 + 3`
`u prime` `= 4x^3 + 6x\ \ \ \ \ ` `v prime` `= 4x^3`
`f prime (x)` `= (u prime v\ – u v prime)/(v^2)`
  `= ((4x^3 + 6x)(x^4 + 3)\ – (x^4 + 3x^2)4x^3)/((x^4 + 3)^2)`
  `= (4x^7 + 12x^3 + 6x^5 + 18x\ – 4x^7\ – 12x^5)/((x^4 + 3)^2)`
  `= (-6x^5 + 12x^3 + 18x)/((x^4 + 3)^2)`
  `= (-6x(x^4\ – 2x^2\ – 3))/((x^4 + 3)^2)`

 

`text(S.P. when)\ x = 0\ \ \ text(or) \ \ x^4\ – 2x^2\ – 3 = 0`

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

MARKER’S COMMENT: Many students did not realise the denominator of  `f′(x)` could be ignored when equating  `f′(x)=0`.
`X^2\ – 2X\ – 3` `= 0`
`(X\ – 3)(X + 1)` `= 0`

`X = 3\ \ text(or)\ \ -1`

`:. x^2` `= 3` `text(or)\ \ \ \ \ ` `x^2 = -1`
`x` `= +- sqrt3\ \ \ \ `   `text{(no solution)}`

 

`:.\ text(SPs occur when)\ \ x = 0, – sqrt3, sqrt3`

 

(iv)    `text(When)\ x = 0,\ ` `y = 0`
  `text(When)\ x = sqrt3,\ \ \ ` `y = ((sqrt3)^4 + 3(sqrt3)^2)/((sqrt3)^4 + 3) = 3/2`

 

EXT1 2009 4b

Statistics, EXT1 S1 2009 HSC 4a

A test consists of five multiple-choice questions. Each question has four alternative answers. For each question only one of the alternative answers is correct.

Huong randomly selects an answer to each of the five questions. 

  1. What is the probability that Huong selects three correct and two incorrect answers?   (2 marks)

  2. What is the probability that Huong selects three or more correct answers?    (2 marks)

  3. What is the probability that Huong selects at least one incorrect answer?  (1 mark)

Show Answers Only
  1. `45/512`
  2. `53/512`
  3. `1023/1024`
Show Worked Solution

i.   `P text{(correct)} = 1/4`

`P text{(wrong)} = 3/4`

`P text{(3 correct, 2 wrong)}`

`=\ ^5C_3 * (1/4)^3 (3/4)^2`

`= (5!)/(3!2!) * (1/64) * (9/16)`

`= 90/1024`

`= 45/512`
 

ii.  `P text{(3 or more correct)}`

`= P text{(3 correct)} + P text{(4 correct)} + P text{(5 correct)}`

`=\ ^5C_3 * (1/4)^3 (3/4)^2 +\ ^5C_4 * (1/4)^4 (3/4)^1 +\ ^5C_5 (1/4)^5 (3/4)^0`

`= 90/1024 + 15/1024 + 1/1024`

`= 53/512`
 

iii.  `P text{(at least 1 incorrect)}`

TIP: The use of “at least” should flag a good chance of applying `1-P text{(complement)}` to solve.

`= 1\ – P text{(0 incorrect)}`

`= 1 -\ ^5C_5 (1/4)^5 (3/4)^0`

`= 1\ – 1/1024`

`= 1023/1024`

Trigonometry, EXT1 T3 2009 HSC 3c

  1. Prove that  `tan^2 theta = (1 - cos 2 theta)/(1 + cos 2 theta)`  provided that  `cos 2theta != -1`.   (2 marks)

  2. Hence find the exact value of   `tan\ pi/8`.    (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `sqrt 2\ – 1`
Show Worked Solution
i.    `text(Prove)\ \ tan^2 theta = (1 – cos 2 theta)/(1 + cos 2 theta),\ cos 2 theta != -1`
`text(Using)\ \ cos 2 theta` `= 2 cos^2 theta\ – 1`
  `= 1 – 2 sin^2 theta`

 

`text(RHS)` `= (1 – (1 – 2 sin^2 theta))/(1 + (2 cos^2 theta\ – 1))`
  `= (2 sin^2 theta)/(2 cos^2 theta)`
  `= tan^2 theta\ text(… as required.)`

 

ii.    `tan^2\ pi/8` `= (1 – cos (2 xx pi/8))/(1 + cos (2 xx pi/8))`
    `= (1 – 1/sqrt2)/(1 + 1/sqrt2) xx sqrt2/sqrt2`
    `= (sqrt2 – 1)/(sqrt2 + 1) xx (sqrt 2 – 1)/(sqrt2 – 1)`
    `= (sqrt 2 – 1)^2/(2\ – 1)`
    `= (sqrt 2 -1)^2`

 
`:.\ tan\ pi/8 = sqrt 2\ – 1\ \ \ \ \ (tan(pi/8)>0)`

`text{(}tan\ pi/8 = sqrt (3\ – 2 sqrt 2)\ \ text{is also a correct answer)}`

Trig Calculus, EXT1 2009 HSC 3b

  1. On the same set of axes, sketch the graphs of  
    1. `y = cos 2x`  and  `y = (x + 1)/2`, for  `–pi <= x <= pi`.    (2 marks)
  3. Use your graph to determine how many solutions there are to the equation  `2 cos 2x = x + 1`  for  `–pi <= x <= pi`.     (1 mark)
  5. One solution of the equation  `2 cos 2x = x + 1`  is close to  `x = 0.4`. Use one application of Newton’s method to find another approximation to this solution. Give your answer correct to three decimal places.   (3 marks)

 

Show Answers Only
  1. `text(See sketch in Worked Solutions)`
  2. `text(3 solutions)`
  3. `0.398\ text{(3 d.p.)}`
Show Worked Solution
(i)   

 

(ii)  `text(3 solutions)`

 

(iii)  `2 cos 2x = x + 1`

MARKER’S COMMENT: Better responses defined `f(x)` and `f′(x)` and evaluated each for `x=0.4` before calculating Newton’s formula, as done in the Worked Solution.
`f(x)` `= 2 cos 2x\ – x\ – 1`
`f prime (x)` `= -4 sin 2x\ – 1`

 

`=>f(0.4)` `= 2 cos 0.8\ – 0.4\ – 1`
  `=-0.0065865 …`
`=> f prime(0.4)` `= -4 sin 0.8\ – 1`
  `=-3.869424 …`

 

`text(Find)\ x_1\ text(where)`

 `x_1` `= 0.4\ – (f(0.4))/(f prime(0.4))`
  `= 0.4\ – ((-0.0065865 …)/(-3.869424 …))`
  `= 0.39829…`
  `= 0.398\ \ text{(3 d.p.)}`

Functions, EXT1 F1 2009 HSC 3a

Let  `f(x) = (3 + e^(2x))/4`. 

  1.  Find the range of  `f(x)`.   (1 mark)

  2.  Find the inverse function  `f^(-1) (x)`.    (2 marks)

Show Answers Only
  1. `y > 3/4`
  2. `f^(-1) (x) = 1/2 ln(4x\ – 3)`
Show Worked Solution
i.   `f(x) = (3 + e^(2x))/4`

`text(As)\ \ x -> oo, \ e^(2x) ->oo, \ f(x)->oo`

`text(As)\ \ x -> -oo, \ e^(2x) ->0, \ f(x)->3/4`

`:.\ text(Range is)\ \ y > 3/4`

 

ii.   `text(Inverse function: swap)\ \ x↔y` 

`x` `= (3 + e^(2y))/4`
`4x` `= 3 + e^(2y)`
`e^(2y)` `= 4x\ – 3`
`ln e^(2y)` `= ln (4x\ – 3)`
`2y` `= ln (4x\ – 3)`
`y` `= 1/2 ln (4x\ – 3)`

 

`:.\ f^(-1) (x) = 1/2 ln (4x\ – 3)`

Trigonometry, EXT1 T3 2009 HSC 2b

  1. Express  `3 sin x + 4 cos x`  in the form  `A sin(x + alpha)`  where  `0 <= alpha <= pi/2`.  (2 marks)

  2. Hence, or otherwise, solve  `3 sin x + 4 cos x = 5`  for  `0 <= x <= 2pi`.

     

    Give your answer, or answers, correct to two decimal places.    (2 marks)

Show Answers Only
  1. `5 sin (x + cos^-1 (3/5))\ \ text(or)`

     

    `5 sin (x+0.93)`

  2. `0.64\ text(radians)\ text{(2 d.p.)}`
Show Worked Solution
i.    `A sin (x + alpha) = 3 sin x + 4 cos x`
  `A sinx cos alpha + A cos x sin alpha = 3 sin x + 4 cos x`

 

`=> A cos alpha = 3\ \ \ \ \ A sin alpha = 4`

`A^2` `= 3^2 + 4^2 = 25`
`:. A` `= 5`
`=> 5 cos alpha` `= 3`
`cos alpha` `= 3/5`
`alpha` `= cos^(-1) (3/5)= 0.9272… \ \ text(radians)`

 

`:.\ 3 sin x + 4 cos x = 5 sin (x + cos^-1 (3/5)) `

 

ii.   `3 sin x + 4 cos x` `= 5`
  `5 sin (x + alpha)` `= 5`
  `sin (x + alpha)` `= 1`
  `x + alpha` `= pi/2, (5pi)/2, …`
  `x` `= pi/2\ – 0.9272…\ \ \ \ \ (0 <= x <= 2 pi)`
    `= 0.6435…`
    `= 0.64\ text(radians)\ text{(2 d.p.)}`

Calculus, EXT1 C2 2009 HSC 1f

Using the substitution  `u = x^3 + 1`, or otherwise, evaluate  `int_0^2 x^2 e^(x^3 + 1)\ dx`.   (3 marks)

Show Answers Only

`1/3 (e^9\ – e)`

Show Worked Solution
`u` `= x^3 + 1`
`(du)/(dx)` `= 3x^2`
`du` `= 3x^2\  dx`
`text(If)\ \ \ ` `x` `= 2,\ ` `u` `= 9`
  `x` `= 0,\ ` `u` `= 1`

 

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

`=1/3 int_0^2 e^(x^3 + 1) * 3x^2\ dx`

`= 1/3 int_1^9 e^u\ du`

`= 1/3 [e^u]_1^9`

`= 1/3 (e^9\ – e)`

Functions, EXT1 F1 2009 HSC 1d

Solve the inequality  `(x + 3)/(2x) > 1`.   (3 marks)

Show Answers Only

`0 < x < 3`

Show Worked Solution

`text(Solution 1)`

`(2x)^2 xx (x + 3)/(2x)` `> (2x)^2`
`2x (x + 3)` `> 4x^2`
`2x^2 + 6x` `> 4x^2`
`2x^2\ – 6x` `< 0`
`2x (x\ – 3)` `< 0`

 

`:.\ {x: 0 < x < 3}`

 

`text(Solution 2)`

`text(If)\ \ x > 0,`

`x + 3` `> 2x`
`x` `< 3`

`:.\ 0 < x < 3`

 

`text(If)\ \ x < 0,`

`x + 3` `< 2x`
`x` `> 3\ \ \ =>\ text(No solution)`

 

`:.\ {x: 0 < x < 3}`

Calculus, 2ADV C3 2014 HSC 16c

The diagram shows a window consisting of two sections. The top section is a semicircle of diameter  `x`  m. The bottom section is a rectangle of width  `x`  m and height  `y`  m.

The entire frame of the window, including the piece that separates the two sections, is made using 10 m of thin metal.

The semicircular section is made of coloured glass and the rectangular section is made of clear glass.

Under test conditions the amount of light coming through one square metre of the coloured glass is 1 unit and the amount of light coming through one square metre of the clear glass is 3 units.

The total amount of light coming through the window under test conditions is  `L`  units.

  1. Show that  `y = 5 - x(1 + pi/4)`.   (2 marks)

  2. Show that  `L = 15x - x^2 (3 + (5pi)/8)`.    (2 marks)

  3. Find the values of  `x`  and  `y`  that maximise the amount of light coming through the window under test conditions.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `x = 1.511\ text(m  and)\ \ y = 2.302\ text(m)`
Show Worked Solution
i.   

`text(Frame is 10m)`

`10` `= 2x + 2y + 1/2 pi x`
`2y` `= 10\ – 2x\ – 1/2 pi x`
`:.y` `= 5\ – x\ – pi/4 x`
  `= 5\ – x (1 + pi/4)\ \ \ text(… as required.)`

 

♦ Mean mark 35%
ii.    `text(Area)\ text{(clear)}` `= x xx y`
  `text(Area)\ text{(colour)}` `= 1/2 xx pi r^2`
    `= 1/2 xx pi (x/2)^2`
    `= (pi x^2)/8`

 

`:.\ L` `= 3xy + ((pix^2)/8 xx 1)`
  `= 3x [5\ – x (1 + pi/4)] + (pix^2)/8\ \ \ text{(see part (i))}`
  `= 15x\ – 3x^2\ – (3x^2pi)/4 + (x^2 pi)/8`
  `= 15x\ – 3x^2\ – (5x^2 pi)/8`
  `= 15x\ – x^2 (3 + (5pi)/8)\ \ \ text(… as required)`

 

♦ Mean mark 38%
COMMENT: A sanity check for your answer could be to compare your answers to the perimeter restriction of 10m.
iii.   `L` `= 15x\ – x^2 (3 + (5pi)/8)`
  `(dL)/(dx)` `= 15\ – 2x (3 + (5pi)/8)`
  `(d^2L)/(dx^2)` `= -2 (3 + (5pi)/8)`

 

`text(Max or min when)\ (dL)/(dx) = 0`

`15\ – 2x (3x + (5pi)/8)=` `0`
`2x (3 + (5pi)/8)=` `15`
`x=` `15/(2 (3 + (5pi)/8)`
 `=` `1.51103…`
`=` `1.511\ \ \ text{(3 d.p.)}`

 

`text(S)text(ince)\ (d^2L)/(dx^2) < 0\ \ \ => text(MAX)`

`text(When)\ \ x` `= 1.511`
`y` `= 5\ – 1.511 (1 + pi/4)`
  `= 2.3022…`
  `= 2.302\ text{(3 d.p.)}`

 

`:.\ text(MAX light when)\ x = 1.511\ text(m)`

`text(and)\ y = 2.302\ text(m.)`

Calculus, 2ADV C3 2014 HSC 15c

The line  `y = mx`  is a tangent to the curve  `y = e^(2x)`  at a point  `P`. 

  1. Sketch the line and the curve on one diagram.   (1 mark)

  2. Find the coordinates of  `P`.     (3 marks)

  3. Find the value of  `m`.   (1 mark)

Show Answers Only
  1.   
  2. `P(1/2 ln (m/2), m/2)`
  3. `2e`
Show Worked Solution
i. 

 

ii. `y` `= e^(2x)`
  `dy/dx` `= 2e^(2x)`

 
`text(Gradient of)\ \ y = mx\ \ text(is)\ \ m`

♦ Mean mark 40%
COMMENT: Given `y= e^(ln(m/2))`, it follows `y=m/2`. Make sure you understand the arithmetic behind this (NB. Simply take the `ln` of both sides).

`text(Gradients equal when)`

`2e^(2x)` `= m`
`e^(2x)` `= m/2`
`ln e^(2x)` `= ln (m/2)`
`2x` `= ln (m/2)`
`x` `= 1/2 ln (m/2)`

 
`text(When)\ \ x = 1/2 ln (m/2)`

`y` `= e^(2 xx 1/2 ln (m/2))`
  `= e^(ln(m/2))`
  `= m/2`

 
`:.\ P (1/2 ln (m/2), m/2)`

 

iii.   `y=mx\ \ text(passes through)\ \ (0,0)\ text(and)\ (1/2 ln (m/2), m/2)`

♦♦ Mean mark 30%.

`text(Equating gradients:)`

`(m/2 – 0)/(1/2 ln (m/2) – 0)`  `=m`
`m/2` `=m xx 1/2 ln(m/2)`
`ln (m/2)` `= 1`
`m/2` `= e^1`
`m` `= 2e`

Plane Geometry, 2UA 2014 HSC 15b

In  `Delta DEF`, a point  `S`  is chosen on the side  `DE`. The length of  `DS`  is  `x`, and the length of  `ES`  is  `y`. The line through  `S`  parallel to  `DF`  meets  `EF`  at  `Q`. The line through  `S`  parallel to  `EF`  meets  `DF`  at  `R`.

The area of  `Delta DEF`  is  `A`. The areas of  `Delta DSR`  and  `Delta SEQ`  are  `A_1`  and  `A_2`  respectively.

  1. Show that  `Delta DEF`  is similar to  `Delta DSR`.    (2 marks)
  2. Explain why  `(DR)/(DF) = x/(x + y)`.    (1 mark)
  3.  
  4. Show that  
    1. `sqrt ((A_1)/A) = x/(x + y)`.  (2 marks)
    2.  
  5. Using the result from part (iii) and a similar expression for  
    1. `sqrt ((A_2)/A)`, deduce that  `sqrt A = sqrt (A_1) + sqrt (A_2)`.   (2 marks)

 

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Corresponding sides of similar)\ Delta text(s)`
  3. `text(are in the same ratio.)`
  4.  
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
  6. `text(Proof)\ \ text{(See Worked Solutions)}`
  7.  
Show Worked Solution

(i)   `text(Need to show)\ Delta DEF\  text(|||) \ Delta DSR`

`/_FDE\ text(is common)`

`/_DSR = /_DEF = theta\ \ text{(corresponding angles,}\ RS\ text(||)\ FE text{)}`

`:.\ Delta DEF \ text(|||) \ Delta DSR\ \ text{(equiangular)}`

 

(ii)  `(DR)/(DF) = (DS)/(DE) = x/(x + y)`

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

 

♦♦ Mean mark 27%
COMMENT: The critical step in solving part (iii) is realising that you need the areas of 2 non-right angled triangles and therefore the formula  `A=½\ ab sin C` is required.

(iii)  `text(Show)\ sqrt((A_1)/A) = x/(x + y)`

`text(Using Area)` `= 1/2 ab sin C`
`A_1` `= 1/2 xx DR xx x xx sin alpha`
`A` `= 1/2 xx DF xx (x + y) xx sin alpha`
`(A_1)/A` `= (1/2 * DR * x * sin alpha)/(1/2 * DF * (x + y) * sin alpha)`
  `= (DR * x)/(DF * (x + y)`
  `= (x * x)/((x + y)(x + y))\ \ \ \ text{(using part(ii))}`
  `= (x^2)/((x + y)^2)`
`:.\ sqrt ((A_1)/A)` `= x/((x + y))\ \ \ text(… as required.)`

 

(iv)  `text(Consider)\ Delta DFE\ text(and)\ Delta SQE`

`/_FED` `= theta\ text(is common)`
`/_FDE` `= /_QSE = alpha\ \ ` `text{(corresponding angles,}\ DF\ text(||)\ QS text{)}`

 

`:.\ Delta DFE\ text(|||)\ Delta SQE\ \ text{(equiangular)}`

`(QE)/(FE) = (SE)/(DE) = y/(x +y)`

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

♦♦ Mean mark 26%
`A_2` `= 1/2 xx QE xx y xx sin theta`
`A` `= 1/2 xx FE xx (x + y) xx sin theta`
`(A_2)/A` `= (QE * y)/(FE * (x + y))`
  `= (y^2)/((x + y)^2)`
`sqrt ((A_2)/A)` `= y/((x + y))`

 

`text(Need to show)\ sqrt A = sqrt (A_1) + sqrt (A_2)`

`sqrt(A_2)/sqrtA` `= y/((x + y))`
`sqrt (A_2)` `= (sqrtA * y)/((x + y))`

 

`text(Similarly, from part)\ text{(iii)}`

`sqrt (A_1) = (sqrtA * x)/((x + y))`

`sqrt (A_1) + sqrt (A_2)` `= (sqrt A * x)/((x + y)) + (sqrt A * y)/((x + y))`
  `= (sqrt A (x + y))/((x + y))`
  `= sqrt A\ \ \ text(… as required.)`

Financial Maths, 2ADV M1 2014 HSC 14d

At the beginning of every 8-hour period, a patient is given 10 mL of a particular drug.

During each of these 8-hour periods, the patient’s body partially breaks down the drug. Only  `1/3`  of the total amount of the drug present in the patient’s body at the beginning of each 8-hour period remains at the end of that period.  

  1. How much of the drug is in the patient’s body immediately after the second dose is given?    (1 mark)

  2. Show that the total amount of the drug in the patient’s body never exceeds 15 mL.     (2 marks)

Show Answers Only
  1. `13.33\ text{mL  (2 d.p.)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `text(Let)\ \ A =\ text(Amount of drug in body)`

`text(Initially)\ A = 10`

`text(After 8 hours)\ \ \ A` `=1/3 xx 10`
`text(After 2nd dose)\ \ A` `= 10 + 1/3 xx 10\ text(mL)`
  `=13.33\ text{mL  (2 d.p.)}`

 

ii.   `text(After the 3rd dose)`

`A_3` `= 10 + 1/3 (10 + 1/3 xx 10)`
  `= 10 + 1/3 xx 10 + (1/3)^2 xx 10`

 
`  =>\ text(GP where)\ a = 10,\ r = 1/3`

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

`S_oo` `= a/(1\ – r)`
  `= 10/(1\ – 1/3)`
  `= 10/(2/3)`
  `= 15`

 

 
`:.\ text(The amount of the drug will never exceed 15 mL.)`

Calculus, EXT1* C3 2014 HSC 14c

The region bounded by the curve  `y = 1 + sqrtx`  and the  `x`-axis between  `x = 0`  and  `x = 4`  is rotated about the  `x`-axis to form a solid.
 

2014 14c
 

Find the volume of the solid.   (3 marks) 

Show Answers Only

`(68 pi)/3\ text(u³)`

Show Worked Solution

`y = 1 + sqrtx`

`V` `= pi int_0^4 y^2\ dx`
  `= pi int_0^4 (1 + sqrtx)^2\ dx`
  `= pi int_0^4 (1 + 2 sqrtx + x)\ dx`
  `= pi [x + 4/3 x^(3/2) + 1/2 x^2]_0^4`
  `= pi [(4 + 4/3 xx 4^(3/2) + 1/2 xx 4^2)\ – 0]`
  `= pi (4 + 32/3 + 8)`
  `= (68 pi)/3\ text(u³)`

Quadratic, 2UA 2014 HSC 14b

The roots of the quadratic equation  `2x^2 + 8x + k = 0`  are  `alpha`  and  `beta`.   

  1. Find the value of  `alpha + beta`.    (1 mark)
  2. Given that  `alpha^2 beta + alpha beta^2 = 6`, find the value of  `k`.     (2 marks)
Show Answers Only
  1. `-4`
  2. `-3`
Show Worked Solution
(i) `2x^2 + 8x + k = 0`
  `alpha + beta = (-b)/a = (-8)/2 = -4`

 

(ii) `alpha^2 beta + alpha beta^2` `= 6`
  `alpha beta (alpha + beta)` `= 6`

`text(S)text(ince)\ \ alpha beta = c/a = k/2`

`=> k/2 (–4)` `= 6`
`-2k` `= 6`
`k` `= -3`

Calculus, 2ADV C3 2014 HSC 14a

Find the coordinates of the stationary point on the graph  `y = e^x − ex`, and determine its nature.   (3 marks)

Show Answers Only

`(1,0)\ =>text(MINIMUM)`

Show Worked Solution
`y` `= e^x – ex`
`dy/dx` `= e^x – e`
`(d^2 y)/(dx^2)` `= e^x`

 
`text(S.P. when)\ \ dy/dx = 0`

`e^x – e` `= 0`
`e^x` `= e^1`
`x` `= 1`

 
`text(At)\ \ x = 1`

`y` `= e^1 – e = 0`
`(d^2 y)/(dx^2)` `= e > 0\ \  => text(MIN)`

 
`:.\ text(MINIMUM S.P. at)\ (1,0)`

Trigonometry, 2ADV T1 2014 HSC 13d

Chris leaves island  `A`  in a boat and sails 142 km on a bearing of 078° to island  `B`.  Chris then sails on a bearing of 191° for 220 km to island  `C`, as shown in the diagram.
 

 

  1. Show that the distance from island  `C`  to island  `A`  is approximately 210 km.    (2 marks)

  2. Chris wants to sail from island  `C`  directly to island  `A`. On what bearing should Chris sail? Give your answer correct to the nearest degree.    (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `333°`
Show Worked Solution
i.   

`text(Find)\ \ /_ABC`

 `text(Let)\ D\ text(be south of)\ B`

`:.\ /_CBD = 191\ – 180 = 11°`
 

`/_ DBA` `= 78°\ text{(alternate)}`
`/_ ABC` `= 78\ – 11`
  `= 67°`

 
`text(Using Cosine rule:)`

`AC^2` `= AB^2 + BC^2\ – 2 * AB * BC * cos /_ABC`
  `= 142^2 + 220^2\ – 2 xx 142 xx 220 xx cos 67°`
  `= 44\ 151.119…`
`:.\ AC` `= 210.121…`
  `~~ 210\ text(km)\ \ \ text(… as required)`

 

ii.  `text(Find)\ \ /_ ACB`

`text(Using Sine rule:)`

`(sin /_ ACB)/142` `= (sin /_ABC)/210`
`sin /_ ACB` `= (142 xx sin 67°)/210`
  `= 0.6224…`
`/_ ACB` `= 38.494…`
  `= 38°\ text{(nearest degree)}`

 

`text(Let)\ E\ text(be due North of)\ C`

`/_ECB = 11°\ text{(} text(alternate to)\ /_CBD text{)}`

`:.\ /_ECA` `= 38\ – 11`
  `= 27°`

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

`= 360\ – 27`

`= 333°`

Calculus, 2ADV C1 2014 HSC 13c

The displacement of a particle moving along the  `x`-axis is given by

 `x = t - 1/(1 + t)`,

where  `x`  is the displacement from the origin in metres,  `t`  is the time in seconds, and  `t >= 0`.

  1. Show that the acceleration of the particle is always negative.    (2 marks)

  2. What value does the velocity approach as  `t`  increases indefinitely?    (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1`
Show Worked Solution
i.    `x` `= t\ – 1/(1 + t)`
    `= t\ – (1 + t)^(-1)`

 

`dot x` `= 1\ – (-1) (1 + t)^(-2)`
  `= 1 + 1/((1 + t)^2)`

 

`ddot x` `= -2(1 + t)^(-3)`
  `= – 2/((1 + t)^3)`

 
`text(S)text(ince)\ \ t >= 0,`

`=> -2/((1 + t)^3) < 0`
 

`:.\ text(Acceleration is always negative.)`

 

ii.    `text(Velocity)\ (dot x) = 1 + 1/((1 + t)^2)`

 
`text(As)\ t -> oo,\ 1/((1 + t)^2) -> 0`

`:.\ text(As)\ t -> oo,\ dot x -> 1`

Calculus, EXT1* C1 2014 HSC 13b

A quantity of radioactive material decays according to the equation 

`(dM)/(dt) = -kM`,

where  `M`  is the mass of the material in kg,  `t`  is the time in years and  `k`  is a constant.

  1. Show that  `M = Ae^(–kt)`  is a solution to the equation, where  `A`  is a constant.   (1 mark)

  2. The time for half of the material to decay is 300 years. If the initial amount of material is 20 kg, find the amount remaining after 1000 years.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.98\ text(kg)\ text{(2 d.p.)}`
Show Worked Solution
i. `M` `= Ae^(-kt)`
  `(dM)/(dt)` `= -k * Ae^(-kt)`
    `= -kM\ \ text(… as required)`

 

ii.    `text(At)\ \ t = 0,\ M = 20`
`=> 20` `= Ae^0`
`A` `= 20`
`:.\ M` `= 20 e^(-kt)`

`text(At)\ \ t = 300,\ M = 10`

TIP: Many students find it efficient to save the exact value of `k` in the memory function of their calculator for these questions.
`=> 10` `= 20 e^(-300 xx k)`
`e^(-300k)` `= 10/20`
`ln e^(-300k)` `= ln 0.5`
`-300 k` `= ln 0.5`
`k` `= – ln 0.5/300`
  `= 0.00231049…`

 
`text(Find)\ M\ text(when)\ t = 1000`

`M` `= 20 e^(-1000k)`
  `= 20 e^(-1000 xx 0.00231049…)`
  `= 20 xx 0.099212…`
  `= 1.9842…`
  `= 1.98\ text(kg)\ text{(2 d.p.)}`

Calculus, 2ADV C4 2014 HSC 13a

  1.  Differentiate  `3 + sin 2x`.    (1 mark)

  2.  Hence, or otherwise, find  `int (cos2x)/(3 + sin 2x)\ dx`.    (2 marks)

Show Answers Only
  1. `2 cos 2x`
  2. `ln (3 + sin 2x)^(1/2) + C`
Show Worked Solution
i. `y` `= 3 + sin 2x`
  `dy/dx` `= 2 cos 2x`

 

ii. `int (cos 2x)/(3 + sin 2x)\ dx`
  `= 1/2 int (2 cos 2x)/(3 + sin 2x)\ dx`
  `= 1/2  ln (3 + sin 2x) + C\ \ \ \ \ text{(from part (i))}`

Financial Maths, 2ADV M1 2014 HSC 12a

Evaluate the arithmetic series  2 + 5 + 8 + 11 + ... + 1094.    (2 marks) 

Show Answers Only

`200\ 020`

Show Worked Solution

`2 + 5 + 8 + … + 1094`

`AP\ \ text(where)\ \ a = 2,\ \ \ d = 5-2 = 3`
 

`text(Find)\ n:`

`T_n` `= a + (n\ – 1) d`
`1094` `= 2 + (n\ – 1)3`
`3n\ – 3` `= 1092`
`3n` `= 1095`
`n` `= 365`

 

`:. S_365` `= n/2 (a + l)`
  `= 365/2 (2 + 1094)`
  `= 200\ 020`

Calculus, 2ADV C1 2014 HSC 9 MC

The graph shows the displacement  `x`  of a particle moving along a straight line as a function of time  `t`.

2014 9 mc

 Which statement describes the motion of the particle at the point  `P`? 

  1. The velocity is negative and the acceleration is positive.
  2. The velocity is negative and the acceleration is negative.
  3. The velocity is positive and the acceleration is positive.
  4. The velocity is positive and the acceleration is negative.
Show Answers Only

`A`

Show Worked Solution

`text(At)\ P,\ text(the particle is moving back towards)\ O`

`text{after hitting a max (positive) displacement}`

`:.\ text(Velocity is negative.)`

`text(Its displacement hits a minimum just after)\ P`

`text(and increases again.)`

`:.\ text{Acceleration is working against (negative) velocity}`

`text(and must be positive.)`

`=>  A` 

Financial Maths, 2ADV M1 2014 HSC 8 MC

Which expression is a term of the geometric series  `3x − 6x^2 + 12x^3 − ...` ?

  1. `3072 x^10`
  2. `–3072 x^10 `
  3. `3072 x^11`
  4. `–3072 x^11 `
Show Answers Only

`C`

Show Worked Solution

`3x\ – 6x^2 + 12x^3\ – …`

`a` `= 3x`
`r` `= (T_2)/(T_1) = (-6x^2)/(3x) = -2x`

 

`:.\ T_n = ar^n` `= 3x (-2x)^n` 
  `= 3(-2)^n x^(n + 1)`

`text(If)\ n = 9`

`T_9 = 3(–2)^9 x^(9 + 1) = -1536 x^10`

`text(If)\ n = 10`

`T_10 = 3(–2)^10 x^(10 + 1) = 3072 x^11`

`=>  C`

Functions, 2ADV F1 2014 HSC 5 MC

Which equation represents the line perpendicular to  `2x-3y = 8`, passing through the point  `(2, 0)`?

  1. `3x + 2y = 4`
  2. `3x + 2y = 6`
  3. `3x-2y = -4`
  4. `3x-2y = 6`
Show Answers Only

`B`

Show Worked Solution
`2x-3y` `= 8`
`3y` `= 2x-8`
`y` `= 2/3x-8/3`
`m` `= 2/3`
`:.\ m_text(perp)` `= -3/2\ \ \ (m_1 m_2=-1\text( for)_|_text{lines)}`

 

`text(Equation of line)\ \ m = -3/2\ \ text(through)\ \ (2,0):`

`y-y_1` `= m (x-x_1)`
`y-0` `= -3/2 (x-2)`
`y` `= -3/2x + 3`
`2y` `= -3x + 6`
`3x + 2y` `= 6`

 
`=>  B`

Calculus, 2ADV C4 2014 HSC 11f

The gradient function of a curve  `y = f(x)`  is given by  `f^{′}(x) = 4x-5`.  The curve passes through the point  `(2, 3)`.

Find the equation of the curve.  (2 marks)

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`f(x) = 2x^2-5x + 5`

Show Worked Solution
`f^{′}(x)` `= 4x-5`
`f(x)` `= int 4x-5\ dx`
  `= 2x^2-5x + C`

 

`text(Given)\ \ f(x)\ text(passes through)\ (2,3):`

`3` `= 2(2^2)-5(2) + C`
`3` `= 8-10 + C`
`C` `= 5`

 
`:.\ f(x) = 2x^2-5x + 5`