SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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)

Show Answers Only

`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`

Statistics, STD2 S4 2014* HSC 30b

The scatterplot shows the relationship between expenditure per primary school student, as a percentage of a country’s Gross Domestic Product (GDP), and the life expectancy in years for 15 countries.
 

 
 

  1. For the given data, the correlation coefficient,  `r`, is 0.83. What does this indicate about the relationship between expenditure per primary school student and life expectancy for the 15 countries?   (1 mark)

  2. For the data representing expenditure per primary school student,  `Q_L`  is 8.4 and  `Q_U`  is 22.5.

     

    What is the interquartile range?   (1 mark)

  3. Another country has an expenditure per primary school student of 47.6% of its GDP.

     

    Would this country be an outlier for this set of data? Justify your answer with calculations.   (2 marks)

  4. On the scatterplot, draw the least-squares line of best fit  `y = 1.29x + 49.9`.    (2 marks)

  5. Using this line, or otherwise, estimate the life expectancy in a country which has an expenditure per primary school student of 18% of its GDP.   (1 mark)

  6. Why is this line NOT useful for predicting life expectancy in a country which has expenditure per primary school student of 60% of its GDP?   (1 mark)

Show Answers Only
  1. `text(It indicates there is a strong positive)`

     

    `text(correlation between the two variables.)`

  2. `14.1`
  3. `text(Yes, because it’s > 43.65%)`
  4.  
  5. `73.1\ text(years)`
  6. `text(At 60% GDP, the line predicts a life expectancy)`

  7.  

    `text(of 127.3. This line of best fit is only accurate)`

  8.  

    `text(in a lower range of GDP expediture.)`

Show Worked Solution
i. `text(It indicates there is a strong positive)`
  `text(correlation between the two variables)`

 

ii. `text(IQR)` `= Q_U\ – Q_L`
    `= 22.5\ – 8.4`
    `= 14.1`

 

♦ Mean mark 35% 

iii.  `text(An outlier on the upper side must be more than)` 

`Q_u\ +1.5xxIQR`

`=22.5+(1.5xx14.1)`

`=\ text(43.65%)`

`:.\ text(A country with an expenditure of 47.6% is an outlier).`

 

iv.  

v.  `text(Life expectancy) ~~ 73.1\ text{years (see dotted line)}`

♦♦ Mean mark 39%

 

`text(Alternative Solution)`

`text(When)\ x=18`

`y=1.29(18)+49.9=73.12\ \ text(years)`

  

♦♦♦ Mean mark 0%. The toughest question on the 2014 paper.
COMMENT: Examiners regularly ask students to identify and comment on outliers where linear relationships break down.
vi.   `text(At 60% GDP, the line predicts a life)`
  `text(expectancy of 127.3. This line of best)`
  `text(fit is only predictive in a lower range)`
  `text(of GDP expenditure.)`

Algebra, STD2 A1 2014 HSC 29b

Blood alcohol content of males can be calculated using the following formula

`BAC_text(Male) = (10N - 7.5H)/(6.8M)`

where    `N` is the number of standard drinks consumed

`H` is the number of hours drinking

`M` is the person's mass in kilograms 

What is the maximum number of standard drinks that a male weighing 84 kg can consume over 4 hours in order to maintain a blood alcohol content (BAC) of less than 0.05?   (3 marks)

Show Answers Only

`5`

Show Worked Solution

`text(BAC)_text(male) = (10N\ – 7.5H)/(6.8M)`

`text(Find)\ \ N\ \ text(for BAC)<0.05,\ \ text(given)\ \ H = 4\ \ text(and)\ \ M = 84`
 

` (10N – 7.5(4))/(6.8(84))` `< 0.05`
`10N – 30` `< 0.05 (571.2)`
`10N` `< 28.56 + 30`
  `< 58.56`
`N` `< 5.856`

 

`:.\ text(Max number of drinks is 5.)`

Algebra, STD2 A4 2014 HSC 29a

The cost of hiring an open space for a music festival is  $120 000. The cost will be shared equally by the people attending the festival, so that  `C`  (in dollars) is the cost per person when  `n`  people attend the festival.

  1. Complete the table below by filling in the THREE missing values.   (1 mark)
    \begin{array} {|l|c|c|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
    \hline
    \rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} &  &  &  & 60 & 48\ & 40 \ \\
    \hline
    \end{array}
  2. Using the values from the table, draw the graph showing the relationship between  `n`  and  `C`.   (2 marks)
     
  3. What equation represents the relationship between `n` and `C`?   (1 mark)

  4. Give ONE limitation of this equation in relation to this context.   (1 mark)

  5. Is it possible for the cost per person to be $94? Support your answer with appropriate calculations.   (1 mark)

Show Answers Only

i.   

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
 

ii. 

iii.   `C = (120\ 000)/n`

`n\ text(must be a whole number)`
 

iv.   `text(Limitations can include:)`

  `•\ n\ text(must be a whole number)`

  `•\ C > 0`
 

v.   `text(If)\ C = 94:`

`94` `= (120\ 000)/n`
`94n` `= 120\ 000`
`n` `= (120\ 000)/94`
  `= 1276.595…`

 
`:.\ text(C)text(ost cannot be $94 per person,)`

`text(because)\ n\ text(isn’t a whole number.)`

Show Worked Solution

i.   

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
 

ii. 

 

♦ Mean mark (iii) 48%

iii.   `C = (120\ 000)/n`

 

♦♦♦ Mean mark (iv) 7%
COMMENT: When asked for limitations of an equation, look carefully at potential restrictions with respect to both the domain and range.

iv.   `text(Limitations can include:)`

  `•\ n\ text(must be a whole number)`

  `•\ C > 0`

 

v.   `text(If)\ C = 94`

`=> 94` `= (120\ 000)/n`
`94n` `= 120\ 000`
`n` `= (120\ 000)/94`
  `= 1276.595…`
♦ Mean mark (v) 38%

 

`:.\ text(C)text(ost cannot be $94 per person,)`

`text(because)\ n\ text(isn’t a whole number.)`

FS Resources, 2UG 2014 HSC 28d

An aerial diagram of a swimming pool is shown. 

The swimming pool is a standard length of 50 metres but is not in the shape of a rectangle.

(i)   Given  `AB=8\ text(cm)`, determine the scale of the diagram such that

1 cm = `x` m   (1 mark)

(ii)  If the length of a carpark next to the pool measured 5 cm (not shown), how long would it be in real life?   (1 mark)

(iii)  In the diagram of the swimming pool, the five widths are measured to be: 

`CD = 21.88\ text(m)`

`EF = 25.63\ text(m)`

`GH = 31.88\ text(m)`

`IJ = 36.25\ text(m)`

`KL = 21.88\ text(m)` 

 

The average depth of the pool is 1.2 m

Calculate the approximate volume of the swimming pool, in cubic metres. In your calculations, use TWO applications of Simpson’s Rule.   (3 marks)

Show Answers Only

(i)   `x=6.25\ text(m)`

(ii)  `31.25\ text(m)`

(iii) `1775\ text(m³)`

Show Worked Solution
(i) `\ \ \ \ 8\ text(cm)` `=50\ text(m)`
  `1\ text(cm)` `=50/8`
    `=6.25\ text(m)`

`:.x=6.25\ text(m)`

 

(ii)  `text{Using scale from (i)}`

`5\ text(cm)` `=5 xx 6.25`
  `=31.25\ text(m)`

`:.\ text(The carpark would be 31.25 m long)`

 

(iii)

`h = 50/4 = 12.5\ text(m)`

♦ Mean mark 50%. Be careful not to give away easy marks! 
`A` `~~ h/3 [y_0 + 4(y_1) + y_2]\ \ text(… applied twice)`
  `~~ 12.5/3 [21.88 + 4(25.63) + 31.88]`
  `+ 12.5/3 [31.88 + 4(36.25) + 21.88]`
  `~~ 12.5/3 [156.28] + 12.5/3 [198.76]`
  `~~ 1,479.33\ text(m²)`

 

`V` `= Ah`
  `~~ 1479.33 xx 1.2`
  `~~ 1775.2`
  `~~ 1775\ text(m³)`

Probability, STD2 S2 2014 HSC 28c

A fair coin is tossed three times. Using a tree diagram, or otherwise, calculate the probability of obtaining two heads and a tail in any order.   (2 marks)

Show Answers Only

`3/8`

Show Worked Solution

 
`P text{(2 heads, 1 tail)}`

`= P(HHT) + P(HTH) + P(THH)`

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

`= 1/8 + 1/8 + 1/8`

`= 3/8`

Measurement, STD2 M6 2014 HSC 28b

A radial compass survey of a sports centre is shown in the diagram. 
 

 
 

  1. Show that the size of angle  `AOB`  is  114°.   (1 mark)

  2. Calculate the length of the boundary `AB`, to the nearest metre.     (2 marks)

  3. Find the area of triangle `AOB` in hectares, correct to two significant figures.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `420\ text(m)\ text{(nearest m)}`
  3. `2.8\ text(ha)\ text{(2 sig. figures)}`
Show Worked Solution
i.   

 

`text(Let)\ D\ text(be directly north of)\ O`

`/_AOD = 360\ – 320 = 40^@`

`:.\ /_AOB = 40 + 74 = 114^@\ \ \ text(… as required)`

 

ii.  `text(Using cosine rule)`

`AB^2` `= AO^2 + BO^2\ – 2 xx AO xx BO xx cos /_AOB`
  `= 287^2 + 211^2\ – 2 xx 287 xx 211 xx cos 114^@`
  `= 126\ 890\ – 121\ 114 (-0.4067…)`
  `= 176\ 151.50…`
`AB` `= 419.704…`
  `= 420\ text(m)\ text{(nearest m)}`
♦ Mean marks of 42% and 41% for parts (ii) and (iii) respectively.
 

iii.  `text(Using)\ A = 1/2 ab sin C,`

`text(Area)\ Delta AOB` `= 1/2 xx 287 xx 211 xx sin 114^@`
  `= 27\ 660.786…\ text(m²)`
  `= 2.7660…\ text(ha)\ \ \ \ text{(1 ha = 10 000 m²)}`
  `= 2.8\ text(ha)\ text{(2 sig figures)}`

FS Driving, 2UG 2014 HSC 27a

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

2014 27a1

  1. Stamp Duty for this car is calculated at $3 for every $100, or part thereof, of the sale price.  
  2. Calculate the Stamp Duty payable.   (1 mark)
  3.  
  4. Alex borrows the total amount to be paid for the car including Stamp Duty and transfer of registration. Interest on the loan is charged at a flat rate of  7.5%  per annum. The loan is to be repaid in equal monthly instalments over 3 years.  
  5.  
    Calculate Alex’s monthly repayments.   (4 marks)
  6.  
  7. Alex wishes to take out comprehensive insurance for the car for 12 months. The cost of comprehensive insurance is calculated using the following: 
    1.  
    2. 2014 27a2
  8. Find the total amount that Alex will need to pay for comprehensive insurance.   (3 marks)
  9.  
  10. Alex has decided he will take out the comprehensive car insurance rather than the less expensive non-compulsory third-party car insurance.
  11. What extra cover is provided by the comprehensive car insurance?   (1 mark)

 

Show Answers Only
  1. `$402`
  2. `$470\ text{(nearest dollar)}`
  3. `$985.74`
  4. `text(Comprehensive insurance covers Alex)`
  5. `text(for damage done to his own car as well.)`
  6.  
Show Worked Solution
♦♦♦ Mean mark 12%
IMPORTANT: “or part thereof ..” in the question requires students to round up to 134 to get the right multiple of $3 for their calculation.
(i)    `($13\ 380)/100 = 133.8`
`:.\ text(Stamp duty)` `= 134 xx $3`
  `= $402`

 

(ii)    `text(Total loan)` `= $13\ 380 + 30 + 402`
    `= $13\ 812`

 

`text(Total interest)\ (I)` `= Prn`
  `= 13\ 812 xx 7.5/100 xx 3`
  `= 3107.70`

 

`text(Total to repay)` `= 3107.70 + 13\ 812`
  `= 16\ 919.70`

 

`text(# Repayments) = 3 xx 12 = 36`

`:.\ text(Monthly repayment)` `= (16\ 919.70)/36`
  `= 469.9916…`
  `= $470\ text{(nearest dollar)}`

 

(iii)   `text(Base rate) = $845`

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

`text(Stamp)` `=\ text(5.5%) xx(845 + 8.45)`
  `= 46.9397…`
  `= $46.94\ text{(nearest cent)}`
`text(GST)` `= 10 text(%) xx(845 + 8.45)`
  `= 85.345`
  `= $85.35`

 

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

 

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

Algebra, STD2 A2 2014 HSC 26f

The weight of an object on the moon varies directly with its weight on Earth.  An astronaut who weighs 84 kg on Earth weighs only 14 kg on the moon.

A lunar landing craft weighs 2449 kg when on the moon. Calculate the weight of this landing craft when on Earth.   (2 marks)

Show Answers Only

 `14\ 694\ text(kg)`

Show Worked Solution

`W_text(moon) prop W_text(earth)`

`=> W_text(m) = k xx W_text(e)`

`text(Find)\ k\ text{given}\  W_text(e) = 84\ text{when}\ W_text(m) = 14`

`14` `= k xx 84`
`k` `= 14/84 = 1/6`

 

`text(If)\ W_text(m) = 2449\ text(kg),\ text(find)\ W_text(e):`

`2449` `= 1/6  xx W_text(e)`
`W_text(e)` `= 14\ 694\ text(kg)`

 

`:.\ text(Landing craft weighs)\ 14\ 694\ text(kg on earth)`

Algebra, STD2 A1 2014 HSC 26c

Solve the equation  `(5x + 1)/3-4 = 5-7x`.   (3 marks)

Show Answers Only

 `x = 1`

Show Worked Solution
`(5x + 1)/3-4` `= 5-7x`
`5x + 1-3(4)` `= 3(5-7x)`
`5x + 1-12` `= 15-21x`
`26x` `= 26`
`:. x` `= 1`

Algebra, STD2 A2 2014 HSC 22 MC

Heather’s car uses fuel at the rate of 6.6 L per 100 km for long-distance driving and  8.9 L per 100 km for short-distance driving.

She used the car to make a journey of 560 km, which included 65 km of short-distance driving.  

Approximately how much fuel did Heather’s car use on the journey?

  1.  37 L
  2. 38 L
  3. 48 L
  4. 50 L
Show Answers Only

`B`

Show Worked Solution

`text(Fuel used in short distance)`

`= 65/100 xx 8.9\ text(L) = 5.785\ text(L)`

`text(Fuel used in long distance)`

`= 495/100 xx 6.6\ text(L) = 32.67\ text(L)`
 

`:.\ text(Total Fuel)` `= 5.785 + 32.67`
  `= 38.455\ text(L)`

`=>  B`

Probability, STD2 S2 2014 HSC 19 MC

Jaz has 2 bags of apples.

Bag A contains 4 red apples and 3 green apples. 

Bag B contains 3 red apples and 1 green apple.

Jaz chooses an apple from one of the bags. 

Which tree diagram could be used to determine the probability that Jaz chooses a red apple?
 

2014 19 mc1

2014 19 mc2

Show Answers Only

`A`

Show Worked Solution

`text(The tree diagram needs to identify 2 separate events.)`

`text(1st event – which bag is chosen)`

`text(2nd event – choosing a red apple from a particular bag)`

`=>  A`

Measurement, STD2 M7 2014 HSC 17 MC

A child who weighs 14 kg needs to be given 15 mg of paracetamol for every 2 kg of body weight.

Every 10 mL of a particular medicine contains 120 mg of paracetamol.

What is the correct dosage of this medicine for the child?

  1. 5.6 mL
  2. 8.75 mL
  3. 11.43 mL
  4. 17.5 mL
Show Answers Only

`B`

Show Worked Solution

`text(Paracetamol needed)`

`= 14/2 xx 15\ text(mg)`

`= 105\ text(mg)`

 

`text(S)text(ince 120 mg is contained in 10 mL,)`

`=> 105\ text(mg is contained in)`

`105/120 xx 10\ text(mL)= 8.75\ text(mL)`

`=>  B`

Measurement, 2UG 2014 HSC 15 MC

Which expression will give the shortest distance, in kilometres, between Mount Isa  `(20^@ text(S)\ 140^@ text(E))`  and Tokyo  `(35^@ text(N)\ 140^@ text(E))`?

(A)   `15/360 xx 2 xx pi xx 6400`

(B)   `55/360 xx 2 xx pi xx 6400`

(C)   `140/360 xx 2 xx pi xx 6400`

(D)   `305/360 xx 2 xx pi xx 6400`

Show Answers Only

`B`

Show Worked Solution

`text(Tokyo is 35° North of the equator, Mt Isa 20° South)`

`text(Arc length) = 55/360 xx 2 xx pi xx 6400`

`=>  B`

Measurement, STD2 M1 2014 HSC 12 MC

A path 1.5  metres wide surrounds a circular lawn of radius 3 metres. 

What is the approximate area of the path?

  1. 7.1 m²
  2. 21.2 m²
  3. 35.3 m²
  4. 56.5 m²
Show Answers Only

`C`

Show Worked Solution

`text(Area of annulus)`

`= pi (R^2-r^2)`

`= pi (4.5^2-3^2)`

`= pi (11.25)`

`=35.3\ text{m²  (1 d.p.)}`
 

`=>  C`

Algebra, STD2 A4 2014 HSC 3 MC

The diagram shows the graph of an equation.
  

 Which of the following equations does the graph best represent?

  1. `y = 3/x + 1`
  2. `y = 3^x + 1`
  3. `y = 3x^2 + 1`
  4. `y = 3x^3 + 1`
Show Answers Only

`C`

Show Worked Solution

`text(Graph is a parabola that passes through)\ (0, 1).`

`=>  C`

Proof, EXT2 P2 EQ-Bank 10

Use mathematical induction to prove that

`sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2`

`text(for integers)\ n>=1`  (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Need to prove)\  sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2\ \ text(integral)\ n>=1 `

`text(i.e.)\ 1^3 + 2^3 + 3^3 + … + n^3 = 1/4 n^2 (n + 1)^2`

`text(When)\ n = 1`

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

`text(RHS) = 1/4 1^2 (1 + 1)^2 = 1`

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

 
`text(Assume true for)\ n = k`

`text(i.e.)\ 1^3 + 2^3 + … + k^3 = 1/4 k^2 (k + 1)^2`

`text(Need to prove true for)\ n = k + 1`

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

`text(LHS)` `= 1/4 k^2 (k + 1)^2 + (k + 1)^3`
  `= 1/4 (k + 1)^2 [k^2 + 4(k + 1)]`
  `= 1/4 (k + 1)^2 (k^2 + 4k + 4)`
  `= 1/4 (k + 1)^2 (k + 2)^2`
  `=\ text(RHS)`

 
`=>text(True for)\ n = k + 1`

`:.text(S)text(ince true for)\ n = 1,\ text(true for integral)\ n >= 1`

Quadratic, EXT1 2014 HSC 13c

The point  `P(2at, at^2)`  lies on the parabola  `x^2 = 4ay`  with focus  `S`.

The point  `Q`  divides the interval  `PS`  internally in the ratio  `t^2 :1`.

2014 13c 

  1. Show that the coordinates of  `Q`  are  
  2. `x = (2at)/(1 + t^2)`  and  `y = (2at^2)/(1 + t^2)`.  (2 marks)
  4. Express the slope of  `OQ`  in terms of  `t`.    (1 mark)
  5. Using the result from part (ii), or otherwise, show that  `Q`  lies on a fixed circle of radius  `a`.   (3 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `m_(OQ) = t`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)   `text(Show)\ \ Q = ((2at)/(1 + t^2), (2at^2)/(1 + t^2))`

`P (2at, at^2),\ S (0, a)`

`PS\ text(is divided internally in ratio)\ t^2: 1`

`Q` `= ((nx_1 + mx_2)/(m + n), (ny_1 + my_2)/(m + n))`
  `= ((1(2at) + t^2(0))/(t^2 + 1), (1(at^2) + t^2 (a))/(t^2 + 1))`
  `= ((2at)/(1 + t^2), (2at^2)/(1 + t^2))\ \ text(… as required.)`

 

(ii)    `m_(OQ)` `= (y_2\ – y_1)/(x_2\ – x_1)`
    `= ((2at^2)/(1 + t^2))/((2at)/(1 + t^2)) xx (1+t^2)/(1+t^2)`
    `= (2at^2)/(2at)`
    `= t`

 

(iii)   `text(Show)\ Q\ text(lies on a fixed circle radius)\ a`
  `text(S)text(ince)\ Q\ text(passes through)\ (0, 0)`

 

`=>\ text(If locus of)\ Q\ text(is a circle, it has)`

`text(diameter)\ QT\ text(where)\ T(0, 2a)`

 

`text(Show)\ \ QT _|_ OQ`

♦♦ Mean mark 22%

`text{(} text(angles on circum. subtended by)`

  `text(a diameter are)\ 90^@ text{)}`

 

`m_(OQ) = t\ \ \ \ text{(see part (ii))}`

`text(Find)\ m_(QT),\ \ text(where:)`

`Q((2at)/(1 + t^2), (2at^2)/(1 + t^2)),\ \ \ \ \ T (0,2a)`

`m_(QT)` `= (y_2\ – y_1)/(x_2\ – x_1)`
  `= ((2at^2)/(1 + t^2)\ – 2a)/((2at)/(1 + t^2)\ – 0)`
  `= (2at^2\ – 2a (1 + t^2))/(2at)`
  `= – (2a)/(2at)`
  `= – 1/t`

 

`m_(QT) xx m_(OT) = -1/t xx t = -1`

`=> QT _|_ OQ`

`=>O,\ T,\ Q\ text(lie on a circle.)`

`:.\ text(Locus of)\ Q\ text(is a fixed circle,)`

`text(centre)\ (0, a),\ text(radius)\ a`

Quadratic, EXT1 2009 HSC 2c

The diagram shows points  `P(2t, t^2)`  and  `Q(4t, 4t^2)`  which move along the parabola  `x^2 = 4y`. The tangents to the parabola at  `P`  and  `Q`  meet at  `R`. 

2009 2c

  1. Show that the equation of the tangent at  `P`  is  `y = tx\ - t^2.`   (2 marks)
  2. Write down the equation of the tangent at  `Q`, and find the coordinates of the point  `R`  in terms of  `t`.     (2 marks)
  3. Find the Cartesian equation of the locus of  `R`.   (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `y = 2tx\ – 4t^2,\ R (3t, 2t^2)`
  3. `y = 2/9 x^2`
Show Worked Solution
(i)    `x^2` `= 4y`
  `y` `= (x^2)/4`
  `dy/dx` `= x/2`
IMPORTANT: Deriving this equation was specifically asked in 2009 and 2011, as well as being required in 2014. KNOW IT! Completing this in 60 seconds gains 2 full minutes on allocated time.

 

`text(At)\ \ x = 2t,`

`dy/dx = (2t)/2 = t`

 

`:.\ text(T)text(angent has)\ \ m = t\ \ text(and passes)`

`text(through)\ \ (2t, t^2).`

`y – t^2` `= t (x – 2t)`
`y` `= tx\ – 2t^2 + t^2`
  `= tx\ – t^2\ \ text(… as required)`

 

(ii)    `text(T)text(angent at)\ \ P (2t, t^2)\ \ text(is)\ \ y = tx\ – t^2`

`=>Q(4t, 4t^2) -= Q(2(2t), (2t)^2)`

`:.\ text(Equation of the tangent at)\ \ Q,`

`y` `= 2tx\ – (2t)^2`
  `= 2tx\ – 4t^2`

 

`text(T)text(angents intersect at)\ \ R`

`y` `= tx\ – t^2\ \ \ \ \ …\ text{(1)}`
`y` `= 2tx\ – 4t^2\ \ \ \ \ …\ text{(2)}`

 

`text(Intersection when)\ text{(1)} = text{(2)}`

`tx\ – t^2` `= 2tx\ – 4t^2`
`tx` `= 3t^2`
`x` `= 3t`

 

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

`y` `= t (3t)\ – t^2`
  `= 2t^2`

 

`:.\ R (3t, 2t^2)`

 

(iii)   `text(Find locus of)\ \ R`
`x` `= 3t` `\ \ \ \ \   …\ text{(1)}`
`y` `= 2t^2\ \ \ \ \ …\ text{(2)}`

 

`text{From (1),}\ \ \ \ t = x/3`

`text(Substitute)\ \ t = x/3\ text(into)\ text{(2)}`

`y` `= 2 (x/3)^2`
  `= 2/9 x^2`

 

`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = 2/9 x^2`

Polynomials, EXT1 2013 HSC 14c

The equation  `e^t = 1/t`   has an approximate solution  `t_0 = 0.5`

  1. Use one application of Newton’s method to show that  `t_1 = 0.56`  is another approximate solution of  `e^t = 1/t`.   (2 marks)
  3. Hence, or otherwise, find an approximation to the value of  `r`  for which the graphs  `y = e^(rx)`  and  `y = log_e x`  have a common tangent at their point of intersection.     (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `0.097\ text{(to 3 d.p.)}`
Show Worked Solution
MARKER’S COMMENT: Many students had difficulty expressing a valid function while too many others carelessly used `t_0=0.56` rather than 0.5.
(i) `e^t` `= 1/t`
  `e^t\ – 1/t` `= 0`
`text(Let)\ \ f(t)` `= e^t\ – 1/t`
`f prime (t)` `= e^t + 1/(t^2)`
`f(0.5)` `=e^0.5-1/0.5`
  `=–0.3512…`
`f′(0.5)` `=e^0.5 +1/0.5^2`
  `=5.6487…`

 

`text(Applying Newton where)\ \ t_0 = 0.5`

`t_1` `= 0.5\ – (f(0.5))/(f prime (0.5)`
  `= 0.5\ – (–0.3512…)/(5.6487…)`
  `= 0.5\ – (- 0.062)`
  `= 0.56\ text{(2 d.p.)}\ \ text(… as required.)`

 

(ii) `y_1` `= e^(rx),` `\ \ \ \ \ (dy_1)/(dx) = re^(rx)`
  `y_2` `= log_e x,` `\ \ \ \ \ (dy_2)/(dx) = 1/x`
♦♦♦ 2nd toughest question on 2013 paper with mean mark 15%.
MARKER’S COMMENT: Few students used the common tangent information to equate the derivative functions.
 

`text(S)text(ince tangent is common),\ \ (dy_1)/(dx)=(dy_2)/(dx)`

`re^(rx)` `= 1/x`
`e^(rx)` `= 1/(rx)`

 

`text(Using part)\ text{(i)},\ rx ~~ 0.56`

`text(At intersection of curves)\ \ y_1 = y_2`

`e^(rx)` `= log_e x`
`e^0.56` `= log_e x`
`x` `= e^(e^0.56)`
  `= 5.758 …`
`text(S)text(ince)\ rx` `~~ 0.56`
`=> r` `~~ 0.56/(5.758 … )`
  `~~ 0.0972 …`
  `~~ 0.097\ text{(to 3 d.p.)}`

Binomial, EXT1 2013 HSC 14b

  1. Write down the coefficient of  `x^(2n)`  in the binomial expansion of  `(1 + x)^(4n)`.    (1 mark)
  2. Show that  
    1. `(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ - k)(x + 2)^(2n\ - k)`.   (2 marks)
       
  3. It is known that  
         `x^(2n\ - k) (x + 2)^(2n\ - k) = ((2n\ - k),(0)) 2^(2n\ - k) x^(2n\ - k) + ((2n\ - k),(1)) 2^(2n\ - k\ - 1) x^(2n\ - k + 1)`

    1.  
      1.     `+ ... + ((2n\ - k),(2n\ - k)) 2^0 x^(4n\ - 2k)`.   (Do NOT prove this.)
      2.  
  4. Show that
  5.  
    1. `((4n),(2n)) = sum_(k = 0)^(n) 2^(2n\ - 2k) ((2n),(k))((2n\ - k),(k))`.   (3 marks)

 

 

Show Answers Only
  1. `((4n),(2n))`
  2.  
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)  `text(Find co-efficient of)\ \ x^(2n)`

`text(Expanding)\ \ (1+x)^(4n)`

`((4n),(0)) + ((4n),(1))x + ((4n),(2))x^2 + … + ((4n),(2n))x^(2n) + …`

`:.\ text(Co-efficient of)\ \ x^(2n)\ text(is)\ ((4n),(2n))`

 

(ii)  `text(Show)\ (1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)`

♦♦ Mean mark 30%.

`text(Using)\ (1 + x^2 + 2x)^(2n) = [x(x + 2) + 1]^(2n)`

`[x (x + 2) + 1]^(2n)`
`= ((2n),(0)) (x(x + 2))^(2n) + ((2n),(1)) (x(x + 2))^(2n\ – 1) + … + ((2n),(2n))` 
`= ((2n),(0)) x^(2n)(x + 2)^(2n) + ((2n),(1)) x^(2n\ – 1) (x + 2)^(2n\ – 1) + … + ((2n),(2n))` 
`= sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)\ text(… as required.)`

 

(iii)  `((4n),(2n))\ text(is the co-eff of)\ x^(2n)\ text(in expansion)\ (1+x)^(4x)`

`text(S)text(ince)\ (1 + x^2 + 2x)^(2n) = ((x+1)^2)^(2n) = (1 + x)^(4n)`

`=> ((4n),(2n))\ text(is co-efficient of)\ x^(2n)\ text(in expansion)\ (1 + x^2 + 2x)^(2n)`

 
`text(Using part)\ text{(ii)}`

`(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k))\ x^(2n\ – k) (x + 2)^(2n\ – k)`

 

`text(Using the given identity,)\ x^(2n)\ text(co-efficients are)`

♦♦♦ Toughest question in the 2013 exam with Mean mark 12%!
MARKER’S COMMENT: Simply stating `(1+x^2+2x)^(2n)“=(1+x)^(4n)` received 1 full mark, showing that even initial working can gain marks.
`k = 0,` `\ ((2n),(0))((2n\ – 0),(0)) 2^(2n\ – 0)`
`k = 1,` `\ ((2n),(1))((2n\ – 1),(1)) 2^(2n\ – 1\ – 1)`
`vdots`  
`k = n,` `\ ((2n),(n))((2n\ – n),(n)) 2^(2n\ – n\ – n)`

 

`:.\ ((4n),(2n))`
 `= ((2n),(0))((2n),(0))2^(2n) + ((2n),(1))((2n\ – 1),(1))2^(2n\ – 2) + … + ((2n),(n))((n),(n)) 2^0`
` = sum_(k=0)^(n)\ ((2n),(k))((2n\ – k),(k)) 2^(2n\ – 2k)\ \ \ \ text(… as required)`

Calculus, EXT1 C1 2013 HSC 13a

A spherical raindrop of radius `r` metres loses water through evaporation at a rate that depends on its surface area.  The rate of change of the volume `V` of the raindrop is given by

`(dV)/(dt) = -10^(-4) A`, 

where `t` is time in seconds and `A` is the surface area of the raindrop. The surface area and the volume of the raindrop are given by  `A = 4pir^2`  and  `V = 4/3 pi r^3`  respectively.

  1. Show that  `(dr)/(dt)`  is constant.   (1 mark)

  2. How long does it take for a raindrop of volume  `10^(–6)` m3  to completely evaporate?     (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `62\ text(seconds)`
Show Worked Solution
i.    `text(Show)\ \ (dr)/(dt)\ \ text(is a constant)`

 `(dV)/(dt) = (dV)/(dr) * (dr)/(dt)\ \ \ …\ text{(1)}`

`V` `= 4/3 pi r^3`
`:. (dV)/(dr)` `= 4 pi r^2`
`(dV)/(dt)` `= -10^(-4) A\ \ text{(given)}`

 
`text(Substituting into)\ text{(1)}`

`-10^(-4) A` `= 4 pi r^2 xx (dr)/(dt)`
  `= A xx (dr)/(dt)`
`:.\ (dr)/(dt)` `= -10^(-4)\ \ text(… as required)`

 

ii.    `V` `= 10^(-6)\ text(m³)`
  `4/3 pi r^3` `= 10^(-6)`
  `r^3` `= (3 xx 10^(-6))/(4pi)`
  `r` `= root(3)((3 xx 10^(-6))/(4pi))`

 

`text(S)text(ince the radius decreases at a constant rate,)`

♦♦ Mean mark 31%

`t=(root(3)((3 xx 10^(-6))/(4pi)))/(10^(-4))`

`\ \ =62.035 …`

`\ \ =62\ text(seconds)\ text{(nearest whole)}`

 

`:.\ text(It takes 62 seconds for the raindrop)`

`text(to evaporate.)`

Mechanics, EXT2* M1 2013 HSC 12e

A particle moves along a straight line. The displacement of the particle from the origin is `x`, and its velocity is `v`. The particle is moving so that  `v^2 + 9x^2 = k`, where `k` is a constant.

Show that the particle moves in simple harmonic motion with period  `(2pi)/3`.   (2 marks)

Show Answers Only

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

Show Worked Solution
`v^2 + 9x^2` `= k`
`v^2` `= k\ – 9x^2`
`1/2 v^2` `= 1/2k\ – 9/2 x^2`

 

`text(For SHM,)\ \ ddot x = -n^2x`

`ddot x` `= d/(dx) (1/2v^2)`
  `= -9x`
  `= -3^2 x \ \ \ text(… as required)`

 

`text(Period)\ (T)\ text(of SHM) = (2pi)/n`

`text(Here,)\ \ n=3`

`:.T= (2pi)/3\ \ \ text(… as required)`

Geometry and Calculus, EXT1 2013 HSC 12d

The point  `P(t, t^2 + 3)`  lies on the curve  `y = x^2 + 3`. The line  `l`  has equation  `y = 2x\ – 1`. The perpendicular distance from  `P`  to the line  `l`  is  `D(t)`.

2013 12d

  1. Show that  
    1. `D(t) = (t^2\ - 2t + 4)/sqrt5`.   (2 marks)
  3. Find the value of  `t`  when  `P`  is closest to  `l`.     (1 mark)
  4. Show that, when  `P`  is closest to  `l`, the tangent to the curve at  `P`  is parallel to  `l`.   (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `t=1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)    `text(Show)\ \ D(t) = (t^2\ – 2t + 4)/sqrt5`
`D(t)` `= _|_\ text(dist of)\ (t, t^2 + 3)\ text(from)\ 2x\ – y\ – 1 = 0`
`D` `= |(ax_1 + by_1 + c)/sqrt(a^2 + b^2)|`
  `= |(2(t)\ – 1 (t^2 + 3)\ – 1)/sqrt(2^2 + (-1)^2)|`
  `= |(2t\ – t^2\ – 3\ – 1)/sqrt5|`
  `= |(-(t^2\ – 2t + 4))/sqrt5|`
Mean mark 40%
MARKER’S COMMENT: The biggest difficulty in part (i) was to explain the removal of the absolute value sign.

 

`text(S)text(ince)\ \ t^2\ – 2t + 4\ \ text(is positive definite)`

`text{(i.e.} \ Delta < 0\ text(and co-efficient of)\ t^2 > 0 text{)},`

`t^2\ – 2t + 4\ \ text(is positive for all)\ t.`

`:.\ D(t) = (t^2\ – 2t + 4)/sqrt5\ \ text(… as required)`

 

(ii)    `text(MAX or MIN when)\ \ (dD)/(dt) = 0`
`(dD)/(dt) = 1/sqrt5 (2t\ – 2)` `= 0`
`2t\ – 2` `= 0`
`t` `= 1`

`(d^2D)/(dt^2) = 2/sqrt5 > 0\  =>\  text(MIN)`

`:.\ P\ text(is closest to)\ l\ text(when)\ t = 1`

 

(iii)   `y` `= x^2 + 3`
  `dy/dx` `= 2x`
`text(When)\ x = t`

`dy/dx = 2t`

`text(At)\ t = 1`

`dy/dx = 2,`

`=> m_text(tangent) = 2\ \ and\ \ m_l = 2`

 

`:.\ text(T)text(angent is parallel to)\ \ l\ \ \ \ text(… as required)`

Calculus, EXT1 C3 2013 HSC 12b

The region bounded by the graph  `y = 3 sin\ x/2`  and the  `x`-axis between  `x = 0`  and  `x = (3pi)/2`  is rotated about the  `x`-axis to form a solid.  
 

2013 12b
 

Find the exact volume of the solid.   (3 marks)

Show Answers Only

`(9pi)/2 ((3pi)/2 + 1) text(u³)`

Show Worked Solution
`y` `= 3 sin\ x/2`
`y^2` `= 9 sin^2\ x/2`

 
`text(Using:)\ \ sin^2x= 1/2 (1 – cos 2x)`
 

`:. V` `= pi int_0^((3pi)/2) 9 sin^2\ x/2\ dx`
  `= (9pi)/2 int_0^((3pi)/2) (1\ – cosx)\ dx`
  `= (9pi)/2 [x\ – sinx]_0^((3pi)/2)`
  `= (9pi)/2 [((3pi)/2\ – sin\ (3pi)/2)\ – 0]`
  `= (9pi)/2 ((3pi)/2 + 1)\ text(u³)`

Trigonometry, EXT1 T3 2013 HSC 12a

 

  1. Write  `sqrt3cos x - sin x`  in the form  `2 cos (x + alpha)`, where  `0 < alpha < pi/2`.   (1 mark)

  2. Hence, or otherwise, solve  `sqrt3 cos x = 1 + sin x`,  where  `0 < x < 2pi`.     (2 marks)

Show Answers Only
  1. `2cos (x + pi/6)`
  2. `pi/6,\ (3pi)/2`
Show Worked Solution

i.  `text(Write)\ sqrt3 cosx\ – sinx\ text(in form)`

`2cos (x + alpha),\ \ \ 0 < alpha < pi/2`

`2 (cosx cos alpha\ – sinx sin alpha)` `= sqrt 3 cosx\ – sinx`
`cosx cos alpha\ – sinx sin alpha` `= sqrt3/2 cos x\ – 1/2 sinx`
`=> cos alpha` `= sqrt3/2`
`=> sin alpha` `= 1/2`
`alpha` `= pi/6`

 
`:.\ 2cos (x + pi/6) = sqrt3 cosx\ – sinx`
 

ii.  `text(Solve)\ sqrt3 cosx = 1 + sinx,\ \ \ 0 < x < 2pi`

`sqrt3 cosx\ – sinx` `= 1`
`2 cos (x + pi/6)` `= 1\ \ \ text{(from part (i))}`
`cos (x + pi/6)` `= 1/2`
`cos(pi/3)` `=1/2`

 
`text(S)text(ince cos is positive in)\  1^text(st) // 4^text(th)\ text(quadrants)`

`x + pi/6` `= pi/3,\ 2pi\ – pi/3\ \ \ \ \ (0 < x < 2pi)`
`:.\ x` `= pi/6,\ (3pi)/2`

Calculus, EXT1 C2 2013 HSC 11f

Use the substitution  `u = e^(3x)`  to evaluate  `int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`.   (3 marks)

Show Answers Only

`1/3 (tan^(-1)e\ – pi/4)`

Show Worked Solution
 MARKER’S COMMENT: Many students did not calculate in radians and incorrectly got an answer of 8.2. BE CAREFUL!
Note that converting your answer to 0.14 is also correct but not required.

`text(Let)\ \ u = e^(3x)`

`(du)/(dx)` `= 3e^(3x)`
`:.dx` `= (du)/(3e^(3x))`

 

`text(When)` `\ x = 1/3,` `\ u = e^(3 xx 1/3) = e`
  `\ x = 0,` `\ u = e^0 = 1`

 
`:.int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`

`=int_1^e (e^(3x))/(u^2 + 1) xx (du)/(3e^(3x))`

`= 1/3 int_1^e 1/(u^2 + 1)\ du`

`= 1/3 [tan^(-1)u]_1^e`

`= 1/3 [tan^(-1) e\ – tan^(-1) 1]`

`= 1/3 (tan^(-1)e\ – pi/4)`

Geometry and Calculus, EXT1 2013 HSC 11d

Consider the function  `f(x) = x/(4\ - x^2)`.

  1. Show that  `f prime (x) > 0`  for all  `x`  in the domain of  `f(x)`.     (2 marks)
  2. Sketch the graph  `y = f(x)`, showing all asymptotes.     (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`

 

 

 

 

 

 

 

Show Worked Solution

(i)   `f(x) = x/(4\ – x^2)`

`u` `= x` `\ \ \ \ \ v` `= 4\ – x^2`
`u prime` `= 1` `\ \ \ \ \ \ \ \ \ \ v prime` `= -2x` 
`f prime (x)` `= (u prime v\ – u v prime)/(v^2)`
  `= (1* (4\ – x^2)\ – x(-2x))/((4\ – x^2)^2)`
  `= (4 + x^2)/((4\ – x^2)^2)`

 

`text(Domain is all)\ x,\ x != +- 2`

`text(Consider)\ f prime (x)`

`4 + x^2` `> 0\ \ \ text{(in domain)}`
`(4\ – x^2)^2` `> 0\ \ \ text{(in domain)}`

 

`:.\ f prime (x) > 0\ text(for all)\ x,\ x != +- 2`

 

(ii)  EXT1 2013 11d

`text(Asymptotes at)\ x = +- 2`

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

COMMENT: Note that `x->2(–)` is a short way of writing as `x` approaches 2 from the negative (or left-hand) side. This notation can save time when required.
`text(As)` `\ x -> 2 (-),` `\ y -> oo`
  `\ x -> 2 (+),` `\ y -> – oo`
`text(As)` `\ x -> -2 (-),` `\ y -> oo`
  `\ x -> -2 (+),` `\ y -> -oo`
`text(As)` `\ x -> oo,` `\ y -> 0`
  `\ x -> – oo,` `\ y -> 0`