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Functions, EXT1 F1 2006 HSC 5b

Let  `f(x) = log_e (1 + e^x)`  for all  `x`.

Show that  `f(x)`  has an inverse.  (2 marks)

Show Answers Only

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

Show Worked Solution
`f(x)` `= log_e (1 + e^x)`
`f prime (x)` `= e^x/(1 + e^x)`

 

`=> e^x > 0\ \ text(for all)\ \ x`

`:.  e^x/(1 + e^x) > 0\ \ text(for all)\ \ x`

 

`:.\ f(x)\ \ text(is a monotonic increasing function)`

`text(for all)\ \ x.`

`:.\ f(x)\ \ text(has an inverse for all)\ \ x.`

Mechanics, EXT2* M1 2006 HSC 4c

A particle is moving so that  `ddot x = 18x^3 + 27x^2 + 9x.`

Initially  `x = – 2`  and the velocity, `v`, is  `– 6.`

  1. Show that  `v^2 = 9x^2 (1 + x)^2.`  (2 marks)

  2. Hence, or otherwise, show that 
     
         `int 1/(x(1 + x)) \ dx = -3t.`  (2 marks)

  3. It can be shown that for some constant  `c`,
     
         `log_e (1 + 1/x) = 3t + c.`       (Do NOT prove this.)
     
    Using this equation and the initial conditions, find  `x`  as a function of  `t.`  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `x = 2/(e^(3t) – 2)`
Show Worked Solution

i.   `text(Show)\ \ v^2 = 9x^2 (1 + x)^2`

`ddot x` `= d/(dx) (1/2 v^2)`
  `= 18x^3 + 27x^2 + 9x`
`1/2 v^2` `= int 18x^3 + 27x^2 + 9x\ dx`
  `= 18/4 x^4 + 27/3 x^3 + 9/2 x^2 + c`
`v^2` `= 9x^4 + 18x^3 + 9x^2 + c`

 

`text(When)\ \ t = 0,\ \ x = -2,\ \ v = -6`

`:.\ (-6)^2` `= 9 (-2)^4 + 18 (-2)^3 + 9 (-2)^2 + c`
`36` `= 144 – 144 + 36 + c`
`c` `= 0`

 

`:.\ v^2` `= 9x^4 + 18x^3 + 9x^2`
  `= 9x^2 (x^2 + 2x + 1)`
  `= 9x^2 (1 + x)^2\ \ text(…  as required)`

 

ii.   `text(Show)\ \ int 1/(x (1 + x)) \ dx = -3t`

`v^2 = 9x^2 (1 + x)^2`

`v = +- sqrt (9x^2 (1 + x)^2)`
 

`text(When)\ \ x = -2,\ \ v = -6:`

`:.\ v` `= -sqrt (9x^2 (1 + x)^2)`
  `= -3x (1 + x)`

 

`(dx)/(dt)` `= -3x (1 + x)`
`(dt)/(dx)` `= -1/(3x (1 + x))`
`t` `= -1/3 int 1/(x (1 + x)) \ dx`
`-3t` `= int 1/(x (1 + x)) \ dx\ \ text(…  as required)`

 

iii.  `text(Given)\ \ log_e (1 + 1/x) = 3t + c`

`text(When)\ \ t = 0,\ \ x = -2:`

`log_e (1 – 1/2)` `= 3(0) + c`
`log_e (1/2)` `= c`
`c` `= log_e 2^-1`
  `= -log_e 2`

 

`log_e (1 + 1/x)` `= 3t – log_e 2`
`1 + 1/x` `= e^(3t – log_e 2)`
`1/x` `= e^(3t – log_e 2) – 1`
  `= (e^(3t))/(e^(log_e 2)) – 1`
  `= e^(3t)/2 – 1`
  `= (e^(3t) – 2)/2`
`:.\ x` `= 2/(e^(3t) – 2)`

Functions, EXT1 F2 2006 HSC 4a

The cubic polynomial  `P(x) = x^3 + rx^2 + sx + t`. where  `r, \ s`  and  `t`  are real numbers, has three real zeros,  `1, alpha`  and  `-alpha.`

  1. Find the value of  `r.`  (1 mark)

  2. Find the value of  `s + t.`  (2 marks)

Show Answers Only
  1. `-1`
  2. `0`
Show Worked Solution

i.  `P(x) = x^3 + rx^2 + sx + t`

`text(Roots are)\ \ 1, alpha, -alpha`

`1 + alpha + -alpha` `= – b/a`
`1` `= – r/1`
`:.\ r` `= -1`

 

ii.  `P(x) = x^3 – x^2 + sx + t`

`P(1) = 0`

`0` `= 1 – 1 + (s xx 1) + t`
`0` `= s + t`

 
`:.\ text(Value of)\ \ s + t = 0`

Plane Geometry, EXT1 2006 HSC 3d

The points  `P, Q`  and  `T`  lie on a circle. The line  `MN`  is tangent to the circle at  `T`  with  `M`  chosen so that  `QM`  is perpendicular to  `MN`. The point  `K`  on  `PQ`  is chosen so that  `TK`  is perpendicular to  `PQ`  as shown in the diagram.

  1. Show that  `QKTM`  is a cyclic quadrilateral.  (1 mark)
  2. Show that  `/_KMT = /_KQT.`  (1 mark)
  3. Hence, or otherwise, show that  `MK`  is parallel to  `TP.`  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   

`/_ QMT = 90^@\ \ \ (QM _|_ MN)`

`/_ QKT = 90^@\ \ \ (PQ _|_ TK)`

`:.\ /_ QMT + /_ QKT = 180^@`

 

`:.\ QKTM\ \ text(is cyclic)\ \ text{(opposite angles are supplementary)}`

`text(…  as required.)`

 

(ii)  `text(Show)\ \ /_ KMT = /_ KQT`

`/_ KMQ = /_ KTQ = theta`

`text{(angles in the same segment on arc}\ \ KQ text{)}`

 

`/_ KQT` `= 90 – theta\ \ text{(angle sum of}\ \ Delta KQT text{)}`
`/_ KMT` `= /_ QMT – /_ KMQ`
  `= 90 – theta`

 

`:.\ /_ KMT = /_ KQT\ \ text(…  as required.)`

 

(iii)   `text(Show)\ \ MK\ text(||)\ TP`

`/_ NTP` `= /_ KQT\ \ text{(angle in alternate segment)`
  `= 90 – theta`

 

`:.\ /_ NTP = /_ KMT\ \ text{(from part (ii))}`

`:.\ MK\ text(||)\ TP\ \ text{(corresponding angles are equal)}`

Combinatorics, EXT1 A1 2006 HSC 3c

Sophie has five coloured blocks: one red, one blue, one green, one yellow and one white. She stacks two, three, four or five blocks on top of one another to form a vertical tower.

  1. How many different towers are there that she could form that are three blocks high?  (1 mark)

  2. How many different towers can she form in total?  (2 marks)

Show Answers Only
  1. `60`
  2. `320`
Show Worked Solution
i.   `text(Towers)` `= \ ^5P_3`
  `= 60`

 
ii.   `text(Number of different towers)`

`= \ ^5P_2 + \ ^5P_3 + \ ^5P_4 + \ ^5P_5`

`= 20 + 60 + 120 + 120`

`= 320`

Polynomials, EXT1 2006 HSC 3b

  1. By considering  `f(x) = 3log_e x - x`, show that the curve  `y = 3 log_e x`  and the line  `y = x`  meet at a point  `P`  whose `x`-coordinate is between  `1.5`  and  `2`.  (1 mark)

  2. Use one application of Newton’s method, starting at  `x = 1.5`, to find an approximation to the `x`-coordinate of  `P`. Give your answer correct to two decimal places.  (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.78\ \ text{(to 2 d.p.)}`
Show Worked Solution
(i)     `f(x)` `= 3 log_e x – x`
  `f(1.5)` `= 3 log_e 1.5 – 1.5`
    `= -0.283… <0`
  `f(2)` `= 3 log_e 2 – 2`
    `= 0.079… >0`

 

`:.\ text(S)text(ince the sign changes, a zero must)`

`text(exist between 1.5 and 2.)`

 

(ii) `f(x)` `= 3 log_e x – x`
  `f prime(x)` `= 3/x – 1`
  `f(1.5)` `= -0.283…`
  `f prime (1.5)` `= 3/1.5 – 1 = 1`

 

`x_1` `= x_0 – (f(x_0))/(f prime (x_0))`
  `= 1.5 – {(-0.283…)}/1`
  `= 1.783…`
  `= 1.78\ \ text{(to 2 d.p.)}`

Quadratic, EXT1 2006 HSC 2c

2006 2c

The points  `P(2ap, ap^2), Q(2aq, aq^2)`  and  `R(2ar, ar^2)`  lie on the parabola  `x^2 = 4ay`. The chord  `QR`  is perpendicular to the axis of the parabola. The chord  `PR`  meets the axis of the parabola at  `U`.

The equation of the chord  `PR`  is  `y = 1/2(p + r)x - apr.`     (Do NOT prove this.)

The equation of the tangent at  `P`  is  `y = px - ap^2.`            (Do NOT prove this.) 

  1. Find the coordinates of  `U.`  (1 mark)
  2. The tangents at  `P`  and  `Q`  meet at the point  `T`. Show that the coordinates of  `T`  are  `(a(p + q), apq).`  (2 marks)
  3. Show that  `TU`  is perpendicular to the axis of the parabola.  (1 mark)
Show Answers Only
  1. `(0, –apr)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)   `U\ \ text(is the)\ \ y\ \ text(intercept of)\ \ PR`

`y = 1/2 (p + r)x – apr`

`text(When)\ \ x = 0`

`y = -apr`

`:.\ U\ \ text(has coordinates)\ \ (0, –apr)`

 

(ii)   `text(Show)\ \ T\ \ text(is)\ \ (a(p + q), apq)`

`text(T)text(angents at)\ \ P and Q\ \ text(are)`

`y` `= px – ap^2\ \ \ \ text{…  (1)}`
`y` `= qx – aq^2\ \ \ \ text{…  (2)}`

 

`T\ \ text(occurs when)\ \ \ (1) = (2)`

`px – ap^2` `= qx – aq^2`
`px – qx` `= ap^2 – aq^2`
`x(p – q)` `= a (p^2 – q^2)`
  `= a (p – q) (p + q)`
`:.\ x` `= a (p + q)`

 

`text(Substitute)\ \ x = a (p + q)\ \ text{into  (1)}`

`y` `= p * a (p + q) – ap^2`
  `= ap^2 + apq – ap^2`
  `= apq`

 

`:.\ T (a (p + q), apq)\ \ text(…  as required.)`

 

(iii)  `text(Axis of parabola)\ \ x^2 = 4ay\ \ text(is)\ \ y text(-axis)`

`=>TU\ \ text(will be perpendicular to the)\ \ y text(-axis if it has a)`

`text(gradient of)\ \ 0\ \ text{(i.e.}\ \ T and U\ text{have same} y text{-values)}`

`:.\ text(Need to show)\ \ \ -apr = apq.`

 

`text(Consider)\ \ QR`

`text(S)text(ince it is perpendicular to)\ \ y text(-axis)`

`text(and the parabola is symmetrical)`

`=>\ \ 2aq` `= -2ar`
`q`  `= -r`
`:.\ apq` `= ap (-r)`
`apq` `= -apr`

 

`:.\ TU\ \ text(is perpendicular.)`

Binomial, EXT1 2006 HSC 2b

  1. By applying the binomial theorem to `(1 + x)^n` and differentiating, show that
  2. `n(1 + x)^(n - 1) = ((n), (1)) + 2((n), (2)) x + … + r((n), (r)) x^(r - 1) + … + n((n), (n)) x^(n - 1).`   (1 mark)
  4. Hence deduce that
  5. `n3^(n - 1) = ((n), (1)) + … + r((n), (r)) 2^(r - 1) + … + n((n), (n)) 2^(n - 1).`  (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)`

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

`+ … + n((n), (n)) x^(n – 1)`

`text(Using binomial expansion)`

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

`+ … + ((n), (n)) x^n`

`text(Differentiate both sides)`

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

`+ … + n((n), (n)) x^(n – 1)\ \ \ …\ text(as required)`

 

(ii)  `text(Substitute)\ \ x = 2\ \ text{into above equation}`

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

`+ … + n((n), (n)) 2^(n – 1)`

`n3^(n – 1) = ((n), (1)) + … + r((n), (r)) 2^(r – 1) + … + n((n), (n)) 2^(n – 1)`

`text(…  as required.)`

Calculus, EXT1 C2 2006 HSC 2a

Let  `f(x) = sin^-1 (x + 5).`

  1. State the domain and range of the function  `f(x).`  (2 marks)

  2. Find the gradient of the graph of  `y = f(x)`  at the point where  `x = -5.`  (2 marks)

  3. Sketch the graph of  `y = f(x).`  (2 marks)

Show Answers Only
  1. `text(Domain):\ -6 <= x <= -4, \ \ \ text(Range):\ -pi/2 <= y <= pi/2`
  2. `1`
  3.  

Show Worked Solution

i.  `f(x) = sin^-1 (x + 5)`

`text(Domain)`

`-1 <= x + 5 <= 1`

`-6 <= x <= -4`

`text(Range)`

`-pi/2 <= y <= pi/2`

 

ii.  `y = sin^-1 (x + 5)`

`(dy)/(dx) = 1/sqrt(1-(x + 5)^2)`
 

`text(When)\ \ x = -5`

`(dy)/(dx)` `= 1/sqrt(1-(-5 + 5)^2)`
  `= 1/sqrt(1-0)`
  `= 1`

 
`:.\ text(Gradient of)\ \ y = f(x)\ \ text(at)\ \ x = -5\ \ text(is)\ \ 1.`

 

iii.

 EXT1 2006 2a

Differentiation, EXT1 2006 HSC 1e

For what values of  `b`  is the line  `y =12x + b`  tangent to  `y =x^3`?  (3 marks)

Show Answers Only

`b = +- 16`

Show Worked Solution
`y_1` `= x^3`
`(dy_1)/(dx)` `= 3x^2`
`y_2` `= 12x + b`
`(dy_2)/(dx)` `= 12`

 

`text(T)text(angent occurs when)`

`(dy_1)/(dx)` `= (dy_2)/(dx)`
`3x^2` `= 12`
`x^2` `= 4`
`x` `= +- 2`

 

`text(When)\ \ x = +- 2,\ \ y_1 = +- 8` 

`:.\ y_2\ \ text(must satisfy)\ \ (2, 8) or (–2, –8)`

 

`text(Consider)\ \ (2, 8)`

`8 = (12 xx 2) + b`

`b = -16`

`text(Consider)\ \ (–2, –8)`

`-8 = (12 xx -2) + b`

`b = 16`

`:.\ b = +- 16`

Mechanics, EXT2* M1 2005 6b

An experimental rocket is at a height of  5000 m, ascending with a velocity of  ` 200 sqrt 2\ text(m s)^-1`  at an angle of  45°  to the horizontal, when its engine stops.
 

 
After this time, the equations of motion of the rocket are:

`x = 200t`

`y = -4.9t^2 + 200t + 5000,`

where `t` is measured in seconds after the engine stops. (Do NOT show this.)

  1. What is the maximum height the rocket will reach, and when will it reach this height?  (2 marks)

  2. The pilot can only operate the ejection seat when the rocket is descending at an angle between 45° and 60° to the horizontal. What are the earliest and latest times that the pilot can operate the ejection seat?  (3 marks)

  3. For the parachute to open safely, the pilot must eject when the speed of the rocket is no more than  `350\ text(m s)^-1`. What is the latest time at which the pilot can eject safely?  (2 marks)

Show Answers Only
  1. `7041\ text(metres)`

     

    `20.4\ text(seconds.)`

  2. `40.8\ text(seconds)\ ;\ 55.8\ text(seconds)`
  3. `49.7\ text(seconds)`
Show Worked Solution

i.

    

`y = -4.9t^2 + 200t + 5000`

`dot y = -9.8t + 200`

`text(Max height occurs when)\ \ dot y = 0`

`9.8t` `= 200`
`t` `= 20.408…`
  `= 20.4\ text{seconds  (to 1 d.p.)}`

 

`text(When)\ t = 20.408…`

`y` `= -4.9 (20.408…)^2 + 200 (20.408…) + 5000`
  `= 7040.816…`
  `= 7041\ text{m  (to nearest metre)}`

 

`:.\ text(The rocket will reach a maximum height of)`

`text(7041 metres when)\ \ t = 20.4\ text(seconds.)`

 

ii.  `text(The rocket is descending at 45° at point)\ A\ text(on the graph.)`

`text(The symmetry of the parabolic motion means that)\ \ A`

`text(occurs when)\ \ t = 2 xx text(time to reach max height, or)`

`40.8\ text(seconds.)`

 

`text(Point)\ B\ text(occurs when the rocket is descending at 60°)`

`text(to the horizontal.)`

 

`text(At)\ \ B,`

 

`-tan 60° = (dot y)/(dot x)`

`text{(The gradient of the projectile becomes}`

`text{negative after reaching its max height.)}`

`dot y = -9.8t + 200`

`dot x = d/(dt) (200t) = 200`

`:.\ – sqrt 3` `= (-9.8t + 200)/200`
`-200 sqrt 3` `= -9.8t + 200`
`9.8t` `= 200 + 200 sqrt 3`
`t` `= (200 + 200 sqrt 3)/9.8`
  `= (546.410…)/9.8`
  `= 55.756…`
  `= 55.8\ text{seconds  (nearest second)}`

 

`:.\ text(The pilot can operate the ejection seat)`

`text(between)\ \ t = 40.8 and t = 55.8\ text(seconds.)`

 

iii.

`v^2 = (dot x)^2 + (dot y)^2`

`text(When)\ \ v = 350`

`350^2` `= 200^2 + (-9.8t + 200)^2`
`122\ 500` `= 40\ 000 + (-9.8t + 200)^2`
`(-9.8t + 200)^2` `= 82\ 500`
`-9.8t + 200` `= +- sqrt (82\ 500)`
`9.8t` `= 200 +- sqrt (82\ 500)`
`t` `= (200 +- sqrt (82\ 500))/9.8`
  `= 49.717…\ \ \ (t > 0)`
  `= 49.7\ text{seconds  (to 1 d.p.)}`

 

`:.\ text(The latest time the pilot can eject safely)`

`text(is when)\ \ t = 49.7\ text(seconds.)`

Statistics, EXT1 S1 2005 HSC 6a

There are five matches on each weekend of a football season. Megan takes part in a competition in which she earns one point if she picks more than half of the winning teams for a weekend, and zero points otherwise. The probability that Megan correctly picks the team that wins any given match is `2/3`.

  1. Show that the probability that Megan earns one point for a given weekend is  0.7901, correct to four decimal places.  (2 marks)

  2. Hence find the probability that Megan earns one point every week of the eighteen-week season. Give your answer correct to two decimal places.  (1 mark)

  3. Find the probability that Megan earns at most 16 points during the eighteen-week season. Give your answer correct to two decimal places.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `0.01\ \ text{(to 2 d.p.)}`
  3. `0.92\ \ text{(to 2 d.p.)}`
Show Worked Solution

i.  `P\ text{(earns a point)}`

`= P\ text{(picks 3, 4 or 5 winners)}`

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

`+\ ^5 C_5 (2/3)^5*(1/3)^0`

`= 10 * 8/27 * 1/9 + 5 * 16/81 * 1/3 + 1 * 32/243*1`

`= 80/243 + 80/243 + 32/243`

`= 192/243`

`= 0.790123…`

`= 0.7901\ \ text{(to 4 d.p.)  …  as required.}`

 

ii.  `P\ text{(earns a point 18 weeks in a row)}`

`= (0.7901…)^18`

`= 0.01440…`

`= 0.01\ \ text{(to 2 d.p.)}`

 

iii.  `P\ text{(earns at most 16 points)}`

`= 1 – P\ text{(earns 17 or 18 points)}`
 

`P\ text{(earns 17)}`

`=\ ^18 C_17 * (0.7901…)^17 xx (1 – 0.7901…)`

`= 0.0688…`

`P\ text{(earns 18)} = 0.01440…\ \ \ \ \ text{(from (ii))}`
 

`:.\ P\ text{(earns at most 16 points)}`

`= 1 – (0.0688… + 0.0144…)`

`= 0.916…`

`= 0.92\ \ \ text{(to 2 d.p.)}`

Mechanics, EXT2* M1 2005 HSC 5c

A particle moves in a straight line and its position at time  `t`  is given by

`x = 5 + sqrt 3 sin3t - cos 3t.`

  1. Express  `sqrt 3 sin3t − cos 3t`  in the form  `R sin(3t - alpha)`  where  `alpha`  is in radians.  (2 marks)

  2. The particle is undergoing simple harmonic motion. Find the amplitude and the centre of the motion.  (2 marks)

  3. When does the particle first reach its maximum speed after time  `t = 0`?  (1 mark)

Show Answers Only
  1. `2 sin ( 3t – pi/6)`
  2. `text(Amplitude) = 2\ text(units),\ \ \ text(Centre of the motion at)\ \ x = 5`
  3. `pi/18`
Show Worked Solution

i.    `x = 5 + sqrt 3 sin 3t – cos 3t`

`sqrt 3 sin 3t – cos 3t` `= R sin (3t – alpha)`
  `= R sin 3t cos alpha – R cos 3t sin alpha`

 
`=> R cos alpha = sqrt 3\ \ \ \ \ R sin alpha = 1`

`R^2 = sqrt 3^2 + 1^2 = 4`

`R = 2`

`=> 2 cos alpha` `= sqrt3`
`cos alpha` `= (sqrt3)/2`
`alpha` `= pi/6`

 
`:.\ 2 sin ( 3t – pi/6) = sqrt 3 sin 3t – cos 3t`

 

ii.  `x` `= 5 + sqrt 3 sin 3t – cos 3t`
  `= 5 + 2 sin (3t – pi/6)`

 
`text(Amplitude) = 2\ text(units)`

`text(Centre of motion at)\ \ x = 5`

 

iii.  `text(Solution 1)`

`x = 5 + 2 sin (3t – pi/6)`

`dot x = 6 cos (3t – pi/6)`

 
`text(Maximum speed occurs when)`

`cos (3t – pi/6)` `= 1`
`3t – pi/6` `= 0`
`3t` `= pi/6`
`t` `= pi/18`

 
`text(Solution 2)`

`text(Maximum speed occurs at the)`

`text(centre of motion,)\ \ x = 5`

`5 + 2 sin (3t – pi/6)` `= 5`
`2 sin (3t – pi/6)` `= 0`
`sin (3t – pi/6)` `= 0`
`3t – pi/6` `= 0`
`t` `= pi/18`

Plane Geometry, EXT1 2005 HSC 5b

Two chords of a circle, `AB`  and  `CD`, intersect at  `E`. The perpendiculars to  `AB`  at  `A` and  `CD`  at  `D`  intersect at  `P`. The line  `PE`  meets  `BC`  at  `Q`, as shown in the diagram.

  1. Explain why  `DPAE`  is a cyclic quadrilateral.  (1 mark)
  2. Prove that  `/_ APE = /_ ABC.`  (2 marks)
  3. Deduce that  `PQ`  is perpendicular to  `BC.`  (1 mark)
Show Answers Only

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

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

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

Show Worked Solution

(i)

 

  

`/_ PAE = /_ PDE = 90°\ \ text{(given)}`

`:. DPAE\ \ text(is a cyclic quadrilateral)`

`text{(opposite angles are supplementary)}`

 

(ii)  `text(Prove)\ \ /_ APE = /_ ABC`

`/_ ABC = /_ ADE`

`text{(angles in the same segment on arc}\ AC text{)}`

`text(S) text(ince)\ \ DPAE\ \ text(is a cyclic quad)\ \ text{(from (i))}`

`/_ APE = /_ ADE`

`text{(angles in the same segment on arc}\ AE text{)}`

`:.\ /_ APE = /_ ABC\ \ text(…  as required.)`

 

(iii)   `/_ APE = /_ ABC\ \ \ text{(from part (ii))}`

`/_ AEP = /_ BEQ\ \ \ text{(vertically opposite angles)}`

`:.\ Delta APE\ \ text(|||)\ \ Delta QBE\ \ text{(equiangular)}`

`:.\ /_ EQB = /_ EAP = 90°`

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

`:.\ PQ _|_ BC\ \ text(…  as required.)`

Calculus, EXT1 C3 2005 HSC 5a

Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve  `y = sin 2x`, the `x`-axis and the line  `x = pi/8`  is rotated about the `x`-axis.  (3 marks)

Show Answers Only

`pi/16 (pi – 2)\ \ \ text(u³)`

Show Worked Solution
`y` `= sin 2x`
`y^2` `= sin^2 2x` 

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

COMMENT: Michael Wells (1st in state Ext2) would derive this formula in his working from the `sin^2x+cos^2=1` identity in ~5 seconds every time he used it in an exam.
  

`:. V` `=pi int_0^(pi/8) y^2 \ dx`
  `= pi int_0^(pi/8) sin^2 2x \ dx`
  `= pi/2 int_0^(pi/8) 1 – cos\ 4x\ dx`
  `= pi/2 [x – 1/4 sin\ 4x]_0^(pi/8)`
  `= pi/2 [(pi/8 – 1/4 sin\ pi/2) – 0]`
  `= pi/2 (pi/8 – 1/4)`
  `= pi/2 ((pi – 2)/8)`
  `= pi/16 (pi – 2)\ \ \ text(u³)`

Quadratic, EXT1 2005 HSC 4c

The points  `P (2ap, ap^2)`  and  `Q (2aq, aq^2)`  lie on the parabola  `x^2 = 4ay`.

The equation of the normal to the parabola at  `P`  is  `x + py = 2ap + ap^3`  and the equation of the normal at  `Q`  is similarly given by  `x + qy = 2aq + aq^3.`

  1. Show that the normals at  `P`  and  `Q`  intersect at the point  `R`  whose coordinates are
    1. `(–apq[p + q], a[p^2 + pq + q^2 + 2]).`  (2 marks)

  2. The equation of the chord  `PQ`  is
    1. `y = 1/2 (p + q) x - apq.` (Do NOT show this.)
  4. If the chord  `PQ`  passes through  `(0, a)`, show that  `pq = –1.`  (1 mark)

  6. Find the equation of the locus of  `R`  if the chord  `PQ`  passes through  `(0, a).`  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `y = x^2/a + 3a`
Show Worked Solution

(i)   `text(Show)\ \ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`

 

ext1 2005 4c

`text(Equations of normals through)\ \  P and Q`

`x + py` `= 2ap + ap^3` `\ \ text{… (1)}`
`x + qy` `= 2aq + aq^3` `\ \ text{… (2)}`

 

`text{Multiply (1)} xx q\ ,\ \ text{(2)} xx p`

`qx + pqy` `= 2apq + ap^3q` `\ \ text{… (3)}`
`px + pqy` `= 2apq + apq^3` `\ \ text{… (4)}`

 

`text{Subtract  (4) – (3)}`

`x (p – q)` `= apq^3 – ap^3q`
  `= apq (q^2 – p^2)`
  `= apq (q – p) (q + p)`
  `= -apq (p -q) (p + q)`
`x` `= -apq [p + q]`

 

`text(Substitute)\ \ x = -apq [p + q]\ \ text{into  (1)}`

`py – apq[p + q]` `= 2ap + ap^3`
`py` `= 2ap + ap^3 + apq (p + q)`
`y` `= 2a + ap^2 + aq (p + q)`
  `= 2a + ap^2 + apq + aq^2`
  `= a[p^2 + pq + q^2 + 2]`

 

`:.\ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`

`text(…  as required.)`

 

(ii)  `PQ\ \ text(is)\ \ y = 1/2 (p + q)x – apq`

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

`a` `= 1/2 (p + q)0 – apq`
`a` `= -apq`
`:. pq` `= -1\ \ text(…  as required)`

 

(iii)   `R\ \ text(has coordinates)`

`x` `= -apq [p + q]`
`x` `= a(p + q)\ \ \ text{(using part (ii))}`
 `x/a` `=(p + q)`

 

`y` `= a [p^2 + pq + q^2 + 2]`
  `= a (p^2 – 1 + q^2 + 2)`
  `= a (p^2 + q^2 +1)`
  `= a [(p + q)^2 – 2pq + 1]`
  `= a [(p + q)^2 + 3]`
  `= a [(x/a)^2 + 3]`
  `= x^2/a + 3a`

 

`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = x^2/a + 3a`

Plane Geometry, EXT1 2005 HSC 3d

In the circle centred at  `O`  the chord  `AB`  has length  `7`. The point  `E`  lies on  `AB`  and  `AE`  has length  `4`. The chord  `CD`  passes through  `E`.

Let the length of  `CD`  be  `l`  and the length of  `DE`  be  `x`.

  1. Show that  `x^2 - lx + 12 = 0.`  (2 marks)
  2. Find the length of the shortest chord that passes through  `E.`  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `4 sqrt 3\ text(units.)`
Show Worked Solution

(i)   `text(Show)\ \ x^2 – lx + 12 = 0`

`AB` `= 7`
`:.\ EB` `= 7 – 4 = 3`

 

`AE xx EB = ED xx CE`

`text{(intercepts of intersecting chords)}`

`4 xx 3` `= x (l – x)`
`12` `= xl – x^2`

 

`:.\ x^2 – lx + 12 = 0\ \ text(…  as required.)`

 

(ii)   `x^2 – lx + 12 = 0\ \ text(has a solution)`

MARKER’S COMMENT: A majority of students tried to differentiate this equation, not realising that `l` is a variable, NOT a constant.

`text(When)\ \ Delta >= 0`

`b^2 – 4ac` `>= 0`
`(-l)^2 – 4 * 1 * 12` `>= 0`
`l^2 – 48` `>= 0`
`l^2` `>= 48`
`l` `>= sqrt 48`
  `>= 4 sqrt 3`

 

`:.\ text(The shortest chord that could pass through)`

`E\ \ text(is)\ \ 4 sqrt 3\ text(units.)`

Calculus, EXT1 C2 2005 HSC 3b

  1. By expanding the left-hand side, show that
  2. `qquad sin(5x + 4x) + sin(5x-4x) = 2 sin (5x) cos(4x)`  (1 mark)

  3. Hence find  `int sin(5x) cos (4x)\ dx.`  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `-1/18 cos(9x)-1/2 cos(x) + c`
Show Worked Solution

i.   `sin (5x + 4x) + sin (5x-4x) = 2 sin(5x) cos(4x)`

`text(LHS)` `= sin (5x) cos (4x)-sin(4x) cos (5x) + sin (5x) cos (4x)+ sin (4x) cos (5x)`
  `= 2 sin (5x) cos (4x)\ \ text(…  as required)`

 

ii.  `int sin (5x) cos (4x)\ dx`

`= 1/2 int 2 sin (5x) cos (4x)\ dx`

`= 1/2 int sin (5x + 4x) + sin (5x-4x)\ dx`

`= 1/2 int sin (9x) + sin (x)\ dx`

`= 1/2 [-1/9 cos(9x)-cos(x)] + c`

`= -1/18 cos(9x)-1/2 cos(x) + c`

Polynomials, EXT1 2005 HSC 3a

  1. Show that the function  `g(x) = x^2 - log_e (x + 1)`  has a zero between  `0.7`  and  `0.9.`  (1 mark)
  2. Use the method of halving the interval to find an approximation to this zero of  `g(x)`, correct to one decimal place.  (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `0.7\ \ text{(to 1 d.p.)}`
Show Worked Solution
(i)    `g(x)` `= x^2 – log_e (x + 1)`
`g(0.7)` `= 0.7^2 – log_e (0.7 + 1)`
  `= 0.49 – log_e 1.7`
  `= -0.04…`
`g(0.9)` `= 0.9^2 – log_e (0.9 + 1)`
  `= 0.81 – log_e 1.9`
  `= 0.168…`

 

`:.\ text(S)text(ince the sign changes, a zero exists)`

`text(between 0.7 and 0.9.)`

 

(ii)  `text(Halving the interval)`

`g(0.8)` `= 0.8^2 – log_e (0.8 + 1)`
  `= 0.64 – log_e 1.8`
  `= 0.0522…`

 

`:.\ text(S)text(ince)\ \ g(0.8) > 0,\ \ text(the zero exists)`

`text(between 0.7 and 0.8.)`

♦♦♦ Part (ii) was “very poorly done”.

 

`g(0.75)` `= 0.75^2 – log_e 1.75`
  `= 0.00288…`

 

`=>text(S)text(ince)\ \ g(0.75) > 0,\ \ text(the zero exists)`

`text(between 0.7 and 0.75.)`

`:.\ text{The zero will be  0.7  (to 1 d.p.)}`

Linear Functions, EXT1 2005 HSC 1f

The acute angle between the lines  `y = 3x + 5`  and  `y = mx + 4`  is  `45°`. Find the two possible values of  `m`.  (2 marks)

Show Answers Only

`1/2\ or -2`

Show Worked Solution
`y_1` `= 3x + 5\ \ ,` `\ \ m_1` `= 3`
`y_2` `= mx + 4\ \ ,` `\ \ m_2` `= m`
♦ This question surprised markers by being poorly done.
`tan 45°` `= |(m_1 – m_2)/(1 + m_1 m_2)|`
`1` `= |(3 – m)/(1 + 3m)|`

 

`:. (3 – m)/(1 + 3m)` `= 1` `\ \ \ text(or)\ \ \ \ \ ` `-((3 – m)/(1 + 3m))` `= 1`
`3 – m` `= 1 + 3m`   `m – 3` `= 1 + 3m`
`4m` `= 2`   `2m` `= -4`
`m` `= 1/2`   `m` `= -2`

 

`:.\ m = 1/2 or -2` 

Calculus, EXT1 C2 2005 HSC 1d

Using the substitution  `u = 2x^2 + 1`, or otherwise, find  `int x (2x^2 + 1)^(5/4)\ dx.`  (3 marks)

Show Answers Only

`1/9 (2x^2 + 1)^(9/4) + c`

Show Worked Solution

 `u = 2x^2 + 1`

`(du)/(dx)` `= 4x`
`du` `= 4x\ dx`
`dx` `=(du)/(4x)` 

 

`:.\ int x (2x^2 + 1)^(5/4)\ dx`

`= int x * u^(5/4) * (du)/(4x)`

`= 1/4 int u^(5/4)\ du`

`= 1/4 * 4/9\ u^(9/4) + c`

`= 1/9\ u^(9/4) + c`

`= 1/9 (2x^2 + 1)^(9/4) + c`

Linear Functions, EXT1 2005 HSC 1b

Sketch the region in the plane defined by  `y <= |\ 2x + 3\ |.`   (2 marks)

Show Answers Only

Show Worked Solution

EXT1 2005 1b

`y <= |2x + 3|`

♦ This question surprised the markers by being poorly done.

`text(Consider)\ \ (0, 0)`

`0 <= |\ 0 + 3\ |\ \ =>\ \ text(correct.)`

 

`:.\ text(Shaded area defines)\ \ y <= |2x + 3|`

Calculus, 2ADV C3 2007 HSC 10b

The noise level, `N`, at a distance `d` metres from a single sound source of loudness `L` is given by the formula

`N = L/d^2.`

Two sound sources, of loudness `L_1` and `L_2` are placed `m` metres apart.
 

The point `P` lies on the line between the sound sources and is `x` metres from the sound source with loudness `L_1.`

  1. Write down a formula for the sum of the noise levels at `P` in terms of `x`.  (1 mark)

  2. There is a point on the line between the sound sources where the sum of the noise levels is a minimum.

     

    Find an expression for `x` in terms of `m`, `L_1` and `L_2` if `P` is chosen to be this point.  (4 marks)

Show Answers Only
  1. `N = L_1/x^2 + L_2/(m-x)^2`
  2. `x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`
Show Worked Solution
i.  

`N = L/d^2`

`text(Noise from)\ L_1` `= L_1/x^2`
`text(Noise from)\ L_2` `= L_2/(m-x)^2`
`:. N` `= L_1/x^2 + L_2/(m-x)^2`

 

ii.  `N = L_1\ x^-2 + L_2 (m – x)^-2`

`(dN)/(dx)` `= -2 L_1 x^-3 + -2 L_2 (m – x)^-3 xx d/(dx) (m – x)`
  `= (-2 L_1)/x^3 + (2 L_2)/(m – x)^3`

 

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

`(2 L_1)/x^3` `= (2 L_2)/(m – x)^3`
`2 L_1 (m – x)^3` `= 2 L_2\ x^3`
`L_1 (m – x)^3` `= L_2\ x^3`
`root 3 L_1 (m – x)` `= root 3 L_2\ x`
`root 3 L_1\ m – root 3 L_1\ x` `= root 3 L_2\ x`
`root 3 L_2\ x + root 3 L_1\ x` `= root 3 L_1\ m`
`x (root 3 L_2 + root 3 L_1)` `= root 3 L_1\ m`
`x` `= (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`

 

`(dN)/(dx)` `= -2 L_1\ x^-3 + 2 L_2 (m – x)^-3`
`(d^2N)/(dx^2)` `= 6 L_1\ x^-4 – 6 L_2 (m – x)^-4 xx -1`
  `= (6 L_1)/x^4 + (6 L_2)/(m – x)^4 > 0`

 

`:.\ text(A minimum occurs when)`

`x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`

Calculus in the Physical World, 2UA 2007 HSC 10a

An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are  `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for  `t >= 6`.
 


 

  1. Using Simpson’s rule, estimate the distance travelled between  `t = 0`  and  `t = 4`.  (2 marks)
  2. The object is initially at the origin. During which time(s) is the displacement of the object decreasing?  (1 mark)
  3. Estimate the time at which the object returns to the origin. Justify your answer.  (2 marks)
  4. Sketch the displacement, `x`, as a function of time.  (2 marks)
Show Answers Only
  1. `~~ 6\ \ text(units)`
  2. `t > 5\ \ text(seconds)`
  3. `7.2\ \ text(seconds)`

 

 

 

 

 

 

 

 

Show Worked Solution

(i)

`text(Distance travelled)`

`= int_0^4 (dx)/(dt)\ dt`

`~~  h/3 [y_0 + 4y_1 + y_2]`

`~~ 2/3 [0 + 4 (1) + 5]`

`~~ 2/3 [9]`

`~~ 6\ \ text(units)`

 

(ii)  `text(Displacement is reducing when the velocity is negative.)`

`:. t > 5\ \ text(seconds)`

 

(iii)  `text(At)\ B,\ text(the displacement) = 6\ text(units)`

`text(Considering displacement from)\ B\ text(to)\ D.`

`text(S)text(ince the area below the graph from)`

`B\ text(to)\ C\ text(equals the area above the)`

`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`

`text(in displacement from)\ B\ text(to)\ D.`

 

`text(Considering)\ t >= 6`

`text(Time required to return to origin)`

`t` `= d/v`
  `= 6/5`
  `= 1.2\ \ text(seconds)`

 

`:.\ text(The particle returns to the origin after 7.2 seconds.)`

(iv)

 2UA HSC 2007 10ai

Probability, 2ADV S1 2007 HSC 9b

A pack of 52 cards consists of four suits with 13 cards in each suit.

  1. One card is drawn from the pack and kept on the table. A second card is drawn and placed beside it on the table. What is the probability that the second card is from a different suit to the first?  (1 mark)

  2. The two cards are replaced and the pack shuffled. Four cards are chosen from the pack and placed side by side on the table. What is the probability that these four cards are all from different suits?  (2 marks)

Show Answers Only
  1. `13/17`
  2. `0.105\ \ \ text{(to 3 d.p.)}`
Show Worked Solution

i.  `text(After 1st card is drawn)`

`text(# Cards left from another suit) = 39`

`text(# Cards left in pack) = 51`

`:. P\ text{(2nd card from a different suit)}`

`= 39/51`

`= 13/17`

 

ii.  `P\ text{(all 4 cards from different suits)}`

`= 52/52 xx 39/51 xx 26/50 xx 13/49`

`= 2197/(20\ 825)`

`= 0.1054…`

`= 0.105\ \ \ text{(to 3 d.p.)}`

Calculus, EXT1* C3 2007 HSC 9a

2007 9a
  

The shaded region in the diagram is bounded by the curve  `y = x^2 + 1`, the `x`-axis, and the lines  `x = 0`  and  `x = 1.`

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

Show Answers Only

`(28 pi)/15\ \ text(u³)`

Show Worked Solution
`V` `= pi int_0^1 y^2\ dx`
  `= pi int_0^1 (x^2 + 1)^2\ dx`
  `= pi int_0^1 x^4 + 2x^2 + 1\ dx`
  `= pi [1/5 x^5 + 2/3 x^3 + x]_0^1`
  `= pi[(1/5 + 2/3 + 1) – 0]`
  `= pi [3/15 + 10/15 + 1]`
  `= (28 pi)/15\ \ text(u³)`

Plane Geometry, 2UA 2007 HSC 8b

In the diagram, `AE` is parallel to `BD`, `AE = 27`, `CD = 8`, `BD = p`, `BE = q` and `/_ABE`, `/_BCD` and `/_BDE` are equal.

Copy or trace this diagram into your writing booklet.

  1. Prove  that `Delta ABE\ text(|||)\ Delta BCD`.  (2 marks)
  2. Prove  that `Delta EDB\ text(|||)\ Delta BCD`.  (2 marks)
  3. Show that `8`, `p`, `q`, `27` are the first four terms of a geometric series.  (1 mark)
  4. Hence find the values of `p` and `q`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solution)}`
  2. `text(Proof)\ \ text{(See Worked Solution)}`
  3. `text(Proof)\ \ text{(See Worked Solution)}`
  4. `p = 12, q = 18`
Show Worked Solution

(i)

`text(Prove)\ Delta ABE\ text(|||)\  Delta BCD`

`/_ ABE = /_ BCD\ \ \ text{(given)}`

`/_ EAB = /_ DBC\ \ \ text{(corresponding angles,}\ AE\ text(||)\ BD text{)}`

`:. Delta ABE\ text(|||)\ Delta BCD\ \ \ text{(equiangular)}`

 

(ii)  `text(Prove)\ Delta EDB\ text(|||)\ Delta BCD`

`/_ EDB = /_ BCD\ \ \ text{(given)}`

`/_ CDB = 180° – (/_ BCD + /_ DBC)\ \ \ text{(Angle sum of}\ Delta BCD text{)}`

`/_ EBD` `= 180° – (/_ ABE + /_ DBC)\ \ \ text{(}/_ ABC\ text{is a straight angle)}`
  `= 180° – (/_ BCD + /_ DBC)\ \ \ text{(}/_ ABE = /_ BCD,\ text{given)}`
  `= /_ CDB`

`:. Delta EDB\ text(|||)\ Delta BCD\ \ \ text{(equiangular)}`

 

(iii)  `text(In a GP,)\ \ r = T_n/T_(n-1)`

`text(If)\ \ \ 8, p, q, 27\ \ \ text(are 1st 4 terms of a GP)`

`=> p/8 = q/p = 27/q .`

`text(S)text(ince corresponding sides of similar)`

`text(triangles are in the same ratio)`

`(BD)/(DC) = (EB)/(BD) = (AE)/(EB)`

`p/8 = q/p = 27/q .`

 

`:. 8, p, q, 27\ \ text(are 1st 4 terms of a GP.)`

 

(iv)  `8, p, q, 27`

`a = 8`

`text(Using)\ \ T_n = ar^(n-1)`

`T_4=ar^3`

`27` `= 8 xx r^3`
`r^3` `= 27/8`
`r` `= 3/2`

 

`T_2` `=ar`
`:.p` `= 8 * 3/2`
  `= 12`

 

`T_3` `=ar^2`
`:.q` `= 8 * (3/2)^2`
  `= 18`

Calculus, EXT1* C1 2007 HSC 8a

One model for the number of mobile phones in use worldwide is the exponential growth model,

`N = Ae^(kt)`,

where `N` is the estimate for the number of mobile phones in use (in millions), and `t` is the time in years after 1 January 2008.

  1. It is estimated that at the start of 2009, when  `t = 1`, there will be 1600 million mobile phones in use, while at the start of 2010, when  `t = 2`, there will be 2600 million. Find `A` and `k`.  (3 marks)

  2. According to the model, during which month and year will the number of mobile phones in use first exceed 4000 million?  (2 marks)

Show Answers Only
  1. `A = (12\ 800)/13, k = log_e\ 13/8`
  2. `text(November 2010)`
Show Worked Solution

i.  `N = Ae^(kt)`

`text(When)\ t = 1,\ N = 1600`

`:. 1600` `= Ae^k`
`A` `= 1600/e^k`

 
`text(When)\ t = 2, N = 2600`

`:. 2600` `= A e^(2k)`
  `= 1600/e^k xx e^(2k)`
  `= 1600 e^k`
`e^k` `= 2600/1600 = 13/8`
`log_e e^k` `= log_e\ 13/8`
`k` `= log_e\ 13/8`
  `=0.4855…`
  `=0.49\ \ \ text{(to 2 d.p.)}`

 

`:. A` `= 1600/(e^(log_e\ 13/8)`
  `= 1600/(13/8)`
  `= (12\ 800)/13`

 

ii.  `text(Find)\ \ t\ \ text(such that)\ N > 4000`

`(12\ 800)/13 xx e^(kt)` `> 4000`
`e^(kt)` `> (13 xx 4000)/(12\800)`
`log_e e^(kt)` `> log_e\ 65/16`
`kt` `> log_e\ 65/16\ \ \ \ (k = log_e\ 13/8)`
`t` `> (log_e\ 65/16)/(log_e\ 13/8`
`t` `> 2.887…\ \ text(years)`
  `>\ text{2 years and 10.6 months (approx)}`

 

`:.\ text(The number of mobile phones in use will exceed)`

`text(4000 million in November 2010.)`

Calculus, 2ADV C4 2007 HSC 7b

2ua 2007 7b
 

The diagram shows the graphs of  `y = sqrt 3 cos x`  and  `y = sin x`. The first two points of intersection to the right of the `y`-axis are labelled  `A`  and  `B`.

  1. Solve the equation  `sqrt 3 cos x = sin x`  to find the `x`-coordinates of  `A`  and  `B`.  (2 marks)

  2. Find the area of the shaded region in the diagram.  (3 marks)

Show Answers Only
  1. `pi/3, (4 pi)/3`
  2. `4\ text(u²)`
Show Worked Solution
i.  `sqrt 3 cos x` `= sin x`
`tan x` `= sqrt 3`

 

`text(S)text(ince)\ tan\ pi/3 = sqrt 3 and tan\ text(is)`

`text(positive in 1st/3rd quadrants:)`

`x` `= pi/3 , pi + pi/3`
  `= pi/3, (4 pi)/3`

 

`:. text(The)\ x text(-coordinates of)\ A and B`

`text(are)\ \ x = pi/3, (4 pi)/3.`

 

ii.  `text(Shaded Area)`

`= int_(pi/3)^((4 pi)/3) sin\ x\ dx – int_(pi/3)^((4 pi)/3) sqrt 3\ cos\ x\ dx`

`= int_(pi/3)^((4 pi)/3) sin\ x – sqrt 3\ cos\ x\ dx`

`= [-cos\ x – sqrt 3\ sin\ x]_(pi/3)^((4 pi)/3)`

`= [(-cos\ (4 pi)/3 – sqrt 3\ sin\ (4 pi)/3) – (-cos\ pi/3 – sqrt 3\ sin\ pi/3)]`

`= [(1/2 + sqrt 3 xx (sqrt 3)/2) – (- 1/2 – sqrt 3 xx (sqrt 3)/2)]`

`= [2 – (-2)]`

`= 4\ text(u²)`

Quadratic, 2UA 2007 HSC 7a

  1. Find the coordinates of the focus, `S`, of the parabola  `y = x^2 + 4`.  (2 marks)
  2. The graphs of  `y = x^2 + 4`  and the line  `y = x + k`  have only one point of intersection, `P`. Show that the `x`-coordinate of `P` satisfies.
    1. `x^2 - x + 4 - k = 0`.  (1 mark)
  3. Using the discriminant, or otherwise, find the value of `k`.  (1 mark)
  4. Find the coordinates of `P`.  (2 marks)
  5. Show that `SP` is parallel to the directrix of the parabola.  (1 mark)
Show Answers Only
  1. `(0, 4 1/4)`

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

  3. `15/4`

  4. `(1/2, 4 1/4)`

  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   `y` `= x^2 + 4`
`x^2` `= y – 4`

`text(Using)\ (x – x_0) = 4a (y – y_0)`

`x_0` `= 0`
`a` `= 1/4`
`y_0` `= 4`

`(x – 0) = 4 xx 1/4 xx (y – 4)`

`text(Vertex)` `= (0, 4)`
`:.\ text(Focus)` `= (0, 4 1/4)`

 

(ii)  `y` `= x^2 + 4` `\ \ …\ \ (1)`
`y` `= x + k` `\ \ …\ \ (2)`

 

`text(Solving simultaneously)`

`x^2 + 4` `= x + k`
`x^2 – x + 4 – k` `= 0`

 

(iii)  `text(If only 1 point of intersection,)`

`b^2 – 4ac` `= 0`
`(–1)^2 – 4 xx 1 xx (4 – k)` `= 0`
`1 – 16 + 4k` `= 0`
`4k` `= 15`
`k` `= 15/4`

 

(iv)  `text(Finding)\ \ P`

`x^2 – x + 4 – 15/4 = 0`

`x^2 – x + 1/4 = 0`

`(x – 1/2)^2 = 0`

`x = 1/2`

`text(When)\ x = 1/2`

`y` `= (1/2)^2 + 4`
  `= 4 1/4`

`:. P\ text(has coordinates)\ \  (1/2, 4 1/4)`

 

(v)  `text(Focus)\ (S)` `= (0, 4 1/4)`
`P` `= (1/2, 4 1/4)`

 

`:.\ text(S)text(ince)\ y text(-values are the same,)\ SP\ text(is parallel)`

`text(with the)\ x text(-axis.)`

`=>text(Directrix has the equation)\ \ y = 3 3/4`

`:. SP\ text(is parallel with the directrix.)`

Calculus, 2ADV C3 2007 HSC 6b

Let  `f (x) =x^4 - 4x^3`.

  1. Find the coordinates of the points where the curve crosses the axes.  (2 marks)

  2. Find the coordinates of the stationary points and determine their nature.  (4 marks)

  3. Find the coordinates of the points of inflection.  (1 mark)

  4. Sketch the graph of  `y = f (x)`, indicating clearly the intercepts, stationary points and points of inflection.  (3 marks)

Show Answers Only
  1. `(0, 0)\ , \ (4, 0)`
  2. `text(Minimum S.P. at)\ (3,\ text(-27))`
  3. `(0, 0) and (2, 16)`
  4.  
Show Worked Solution

i.  `f (x) = x^4 – 4x^3`

`text(Cuts)\ x text(-axis when)\ f(x) = 0`

`x^4 – 4x` `= 0`
`x^3 (x – 4)` `= 0`

`x = 0 or 4`

`:. text(Cuts the)\ x text(-axis at)\ (0, 0)\ ,\ (4, 0)`

 

`text(Cuts the)\ y text(-axis when)\ x = 0`

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

 

ii.   `f(x) = x^4 – 4x^3`

`f prime (x) = 4x^3 – 12x^2`

`f″ (x) = 12x^2 – 24x`

 

`text(S.P.’s when)\ f prime (x) = 0`

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

`x = 0 or 3`

`text(When)\ x = 0`

`f(0) = 0`

`f″(0) = 0`

`text(S)text(ince concavity changes, a P.I.)`

`text(occurs at)\ (0, 0)`

 

`text(When)\ x = 3`

`f (3)` `= 3^4 – 4 xx 3^3`
  `= -27`
`f″ (3)` `= 12 xx 3^2 – 24 xx 3`
  `= 36 > 0`

 

`:. text(Minimum S.P. at)\ (3,\ text(–27))`

 

iii.  `text(P.I. when)\ f″(x) = 0`

`12x^2 – 24x` `= 0`
`12x(x – 2)` `= 0`

`x = 0 or 2`

`text(P.I. at)\ (0, 0)\ \ \ text{(from(ii))}`

 

`text(When)\ x = 2`

`text(S)text(ince concavity changes, a P.I.)`

`text(occurs when)\ x = 2`

`f (2)` `= 2^4 – 4 xx 2^3`
  `= 16`

 

`:. text(P.I.’s at)\ (0, 0) and (2, 16)`

 

iv.

 2UA HSC 2007 6b

L&E, 2ADV E1 2007 HSC 6a

Solve the following equation for `x`:

`2e^(2x) - e^x = 0`.  (2 marks)

Show Answers Only

`x = ln\ 1/2`

Show Worked Solution

`text(Solution 1)`

`2e^(2x) – e^x = 0`

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

`2X^2 – X` `= 0`
`X (2X – 1)` `= 0`

 
`X = 0 or 1/2`

 
`text(When)\ e^x = 0\  =>\ text(no solution)`

`text(When)\ e^x = 1/2`

`ln e^x` `= ln\ 1/2`
`:. x` `= ln\ 1/2`

 

`text(Solution 2)`

`2e^(2x)-e^x` `=0`
`2e^(2x)` `=e^x`
`ln 2e^(2x)` `=ln e^x`
`ln 2 +ln e^(2x)` `=x`
`ln 2 + 2x` `=x`
`x` `=-ln2`
  `=ln\ 1/2`

Calculus, EXT1* C1 2007 HSC 5b

A particle is moving on the `x`-axis and is initially at the origin. Its velocity, `v` metres per second, at time `t` seconds is given by

`v = (2t)/(16 + t^2).`

  1. What is the initial velocity of the particle?  (1 mark)

  2. Find an expression for the acceleration of the particle.  (2 marks)

  3. Find the time when the acceleration of the particle is zero.  (1 mark)

  4. Find the position of the particle when `t = 4`.  (3 marks)

Show Answers Only
  1. `0`
  2. `{2(16 – t^2)}/(16 + t^2)^2`
  3. `4\ text(seconds)`
  4. `log_e 2\ \ text(metres)`
Show Worked Solution

i.  `v = (2t)/(16 + t^2)`

`text(When)\ t` `= 0`
`v` `= 0`

`:.\ text(Initial velocity is 0.)`

 

ii.  `a = d/(dt) ((2t)/(16 + t^2))`
 

`text(Using quotient rule)`

`u` `= 2t` `v` `= 16 + t^2`
`u prime` `= 2` `v prime` `= 2t`
`(dv)/(dt)` `= (u prime v – uv prime)/v^2`
  `= {2(16 + t^2) – 2t * 2t}/(16 + t^2)^2`
  `= (32 + 2t^2 – 4t^2)/(16 + t^2)^2`
  `= {2(16 – t^2)}/(16 + t^2)^2`

 

iii.  `text(Find)\ t\ text(when)\ (dv)/(dt) = 0`

`{2 (16 – t^2)}/(16 + t^2)^2` `= 0`
`2 (16 – t^2)` `= 0`
`t^2` `= 16`
`t` `= 4\ ,\ t >= 0`

 

`:.\ text(The acceleration is zero when)`

`t = 4\ text(seconds.)`

 

iv.  `v = (2t)/(16 + t^2)`

`x` `= int v\ dt`
  `= int (2t)/(16 + t^2)`
  `= log_e (16 + t^2) + c`

 

`text(When)\ \ t = 0\ ,\ x = 0`

`0 = log_e (16 + 0) + c`

`c = -log_e 16`

`:. x = log_e(16 + t^2) – log_e 16`
 

`text(When)\ t = 4,`

`x` `= log_e (16 + 4^2) – log_e 16`
  `= log_e 32 – log_e 16`
  `= log_e (32/16)`
  `= log_e 2\ \ text(metres)`

 

`:.\ text(When)\ t = 4\ , \ text(the position of the)`

`text(particle is)\ log_e 2 \ text(metres.)`

Plane Geometry, 2UA 2007 HSC 5a

In the diagram, `ABCDE` is a regular pentagon. The diagonals `AC` and `BD` intersect at `F`.

Copy or trace this diagram into your writing booklet.

  1. Show that the size of `/_ABC` is `108°`.  (1 mark)
  2. Find the size of `/_BAC`. Give reasons for your answer.  (2 marks)
  3. By considering the sizes of angles, show that `Delta ABF` is isosceles.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `36°`
  3. `text(See Worked Solutions)`
Show Worked Solution
(i)

`text(Sum of all internal angles`

`= (n – 2) xx 180°`

`= (5 – 2) xx 180°`

`= 540°`

`:. /_ABC` `= 540/5= 108°`

 

(ii)  `BA = BC`

`text{(equal sides of a regular pentagon)}`

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

`/_BAC` `= 1/2 (180 – 108)\ \ \ text{(base angle of}\ Delta BAC text{)}`
  `= 36°`

 

(iii)  `text(Consider)\ Delta BCD and Delta ABC`

`BC = CD = BA`

`text{(equal sides of a regular pentagon)}`

`/_BCD = /_ABC = 108°`

`text{(internal angles of a regular pentagon)}`

`:. Delta BCD -= Delta ABC\ \ \ text{(SAS)}`

`:. CBF` `= 36°` `text{(corresponding angles in}`
     `text{congruent triangles)}`
`/_FBA` `= 108 – 36`
  `= 72°`
`/_BFA` `= 180 – (72 + 36)\ \ \ \ text{(angle sum of}\ Delta ABF text{)}`
  `= 72°`

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

Trigonometry, 2ADV T1 2007 HSC 4c

 
An advertising logo is formed from two circles, which intersect as shown in the diagram.

The circles intersect at `A` and `B` and have centres at `O` and `C`.

The radius of the circle centred at `O` is 1 metre and the radius of the circle centred at `C` is `sqrt 3` metres. The length of `OC` is 2 metres.

  1. Use Pythagoras’ theorem to show that  `/_OAC = pi/2`.  (1 mark)

  2. Find  `/_ ACO`  and  `/_ AOC`.  (2 marks)

  3. Find the area of the quadrilateral  `AOBC`.  (1 mark)

  4. Find the area of the major sector  `ACB`.  (1 mark)

  5. Find the total area of the logo (the sum of all the shaded areas).  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `/_ ACO = pi/6\ ,\ /_ AOC = pi/3`
  3. `sqrt 3\ \ text(m²)`
  4. `(5 pi)/2\ text(m²)`
  5. `((19 pi + 6 sqrt 3)/6)\ text(m²)`
Show Worked Solution

i.

`text(In)\ Delta AOC`

`AO^2 + AC^2` `= 1^2 + sqrt 3^2`
  `=1 + 3`
  `= 4`
  `= OC^2`

 
`:. Delta AOC\ \ text(is right-angled and)\ \ /_OAC = pi/2`

 

ii.  `sin\  /_ACO` `= 1/2`
`:. /_ACO` `= pi/6`
`sin\  /_AOC` `= sqrt 3/2`
`:. /_AOC` `= pi/3`

 

iii.  `text(Area)\ AOBC`

`= 2 xx text(Area)\ Delta AOC`

`= 2 xx 1/2 xx b xx h`

`= 2 xx 1/2 xx 1 xx sqrt 3`

`= sqrt 3\ \ text(m²)`

 

iv.  `/_ACB = pi/6 + pi/6 = pi/3`

`:. /_ACB\ text{(reflex)}` `= 2 pi – pi/3`
  `= (5 pi)/3`

 
`text(Area of major sector)\ ACB`

`= theta/(2 pi) xx pi r^2`

`= {(5 pi)/3}/(2 pi) xx pi(sqrt 3)^2`

`= (5 pi)/6 xx 3`

`= (5 pi)/2\ text(m²)`

 

v.  `/_AOB = pi/3 + pi/3 = (2 pi)/3`

`:. /_AOB\ text{(reflex)}` `= 2 pi – (2 pi)/3`
  `= (4 pi)/3`

`text(Area of major sector)\ AOB`

`= {(4 pi)/3}/(2 pi) xx pi xx 1^2`

`= (2 pi)/3\ text(m²)`

 

`:.\ text(Total area of the logo)`

`= (5 pi)/2 + (2 pi)/3 + text(Area)\ AOBC`

`= (15 pi + 4 pi)/6 + sqrt 3`

`= ((19 pi + 6 sqrt 3)/6)\ text(m²)`

Probability, 2ADV S1 2007 HSC 4b

Two ordinary dice are rolled. The score is the sum of the numbers on the top faces.

  1. What is the probability that the score is 10?  (2 marks)

  2. What is the probability that the score is not 10?  (1 mark)

Show Answers Only
  1. `1/12`
  2. `11/12`
Show Worked Solution
i. 2UA HSC 2007 4b

`text{P (score = 10)}`

`= 3/36`

`= 1/12`

 

ii.  `text{P (score is not ten)}`

`= 1 – text{P (score is ten)}`

`= 1 – 1/12`

`= 11/12`

Geometry and Calculus, EXT1 2005 HSC 7b

Let  `f(x) = Ax^3 - Ax + 1`,  where  `A > 0`.

  1. Show that  `f(x)`  has stationary points at
    1. `x = +- (sqrt 3)/3.`  (1 mark)

  2. Show that  `f(x)`  has exactly one zero when
    1. `A < (3 sqrt 3)/2.`  (2 marks)

  3. By observing that  `f(-1) = 1`, deduce that  `f(x)`  does not have a zero in the interval  `-1 <= x <= 1`  when
    1. `0 < A < (3 sqrt 3)/2.`  (1 mark)

  4. Let  `g(theta) = 2 cos theta + tan theta`,  where  `-pi/2 < theta < pi/2.`
  6. By calculating  `g prime (theta)`  and applying the result in part (iii), or otherwise, show that  `g(theta)`  does not have any stationary points.  (3 marks)
  7. Hence, or otherwise, deduce that  `g(theta)`  has an inverse function.  (1 mark)
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)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)   `f(x) = Ax^3 – Ax + 1\ \ ,\ \ \ A > 0`

`f prime (x) = 3Ax^2 – A`

`text(S.P.’s when)\ \ f prime (x) = 0`

`3Ax^2 – A` `= 0`
`A (3x^2 – 1)` `= 0`
`3x^2` `= 1`
`x^2` `= 1/3`
`x` `= +- 1/sqrt3 xx (sqrt 3)/(sqrt 3)`
  `= +- (sqrt 3)/3\ \ text(…  as required)`

 

(ii)   `text(Consider)\ \ f(x)\ \ text(with only 1 zero.)`

`f(0) = 1`

`:.\ f(x)\ \ text(passes through)\ \ (0, 1)\ \ text(and has)`

`text(turning points either side at)\ \ x = +- (sqrt 3)/3.`

 

`text(Given 1 zero)`

`=> f((sqrt 3)/3) > 0`

`A ((sqrt 3)/3)^3 – A ((sqrt 3)/3) + 1` `> 0`
`A((3 sqrt 3)/27) – A ((9 sqrt 3)/27)` `> -1`
`A ((-6 sqrt 3)/27)` `> -1`
`A` `< 27/(6 sqrt 3)`
`A` `< 9/(2 sqrt 3) xx (sqrt 3)/(sqrt 3)`
`A` `< (9 sqrt 3)/6`
 `A` `< (3 sqrt 3)/2\ \ text(…  as required.)`

 

(iii)  `text(S)text(ince 1 zero occurs when)\ \ A < (3 sqrt 3)/2\ \ text{(part (ii))}`

`=> text(Minimum TP when)\ \ x = (sqrt3)/3`

`=> text(Maximum TP when)\ \ x = -(sqrt 3)/3\ \ text{(see graph)}`

 

`text(S)text(ince)\ \ f(-1) = 1`

`:. f(x)\ \ text(cannot have a zero for)\ \ \ -1 <= x <= 1`

 

(iv)   `g(theta) = 2 cos theta + tan theta,\ \ \ \ \ \ -pi/2 < theta < pi/2`

`g prime (theta) = -2 sin theta + sec^2 theta`

 

`text(S.P.’s when)\ \ g prime (theta) = 0`

`-2 sin theta + sec^2 theta` `= 0`
`-2 sin theta + 1/(cos^2 theta)` `= 0`
`-2 sin theta + 1/((1 – sin^2 theta))` `= 0`
`(-2 sin theta (1 – sin^2 theta) + 1)/((1 – sin^2 theta))` `= 0`
`-2 sin theta (1 – sin^2 theta) + 1` `= 0`
 `-2 sin theta + 2 sin^3 theta + 1` `= 0`
 `2 sin^3 theta – 2 sin theta + 1` `= 0\ \ text(…  (1))`

 

`text(Let)\ \ x = sin theta and A = 2`

`text{Equation (1) becomes}`

`Ax^3 – 2x + 1 = 0`

`text(S)text(ince)\ \ A = 2`

`0 < A < (3 sqrt 3)/2`

`text(S)text(ince)\ -pi/2` `< \ \ \ theta` `< pi/2`
`-1` `< sin theta` `< 1`
`-1` `< \ \ \ x` `< 1`

 

`:.\ text{Using Part (iii)} => g prime (theta)\ \ text(has no zeros and)`

`text(therefore)\ \ g (theta)\ \ text(has no S.P.’s for)\ \ -pi/2 < theta < pi/2.`

 

(v)   `text(Given)\ \ g(theta)\ \ text(has no stationary points, it will be)`

`text(either an increasing or decreasing function in the)`

`text(range)\ \ (-pi)/2 < theta < pi/2\ \ text(and therefore it has an inverse)`

`text(function.)`

Quadratic, 2UA 2005 HSC 10a

2005 10a

The parabola  `y = x^2`  and the line  `y = mx + b`  intersect at the points  `A(α,α^2)`  and  `B(β, β^2)`  as shown in the diagram.

  1. Explain why  `α + β = m`  and  `αβ = –b`.  (1 mark)
  2. Given that
  3. `(α − β)^2 + (α^2 − β^2)^2 = (α − β)^2[1 + (α + β)^2]`, show that the distance  `AB = sqrt((m^2 + 4b)(1 + m^2)).`  (2 marks)
  4. The point  `P(x, x^2)`  lies on the parabola between  `A`  and  `B`. Show that the area of the triangle  `ABP`  is given by  `1/2(mx − x^2 + b)sqrt(m^2 + 4b).`  (2 marks)
  5.  
  6. The point  `P`  in part (iii) is chosen so that the area of the triangle  `ABP`  is a maximum.
  7. Find the coordinates of  `P`  in terms of  `m`.  (2 marks)
  8.  
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `P(m/2, m^2/4)`
Show Worked Solution

(i)   `text(Instersection)`

`y = x^2` `\ \ …\ (1)`
`y = mx + b` `\ \ …\ (2)`

`text(Substitute)\ y = x^2\ text(into)\ \ (2)`

`x^2 = mx + b`

`x^2 − mx − b = 0`

`text(We know the intersection occurs when)`

`x = α\ \ text(and)\ \ x = β`

`:.α\ \ text(and)\ \ β\ \ text(are roots of)\ \ x^2 − mx − b = 0`

`:.α + β` `= (−b)/a` `= m`  
`αβ` `= c/a` `= −b` `\ \ …text(as required)`

 

 

(ii)   `A(α, α^2),\ \ B(β, β^2)`

`AB` `= sqrt((α − β)^2 + (α^2 − β^2)^2)`
  `= sqrt((α − β)^2[1 + (α + β)^2])`
  `= sqrt([(α + β)^2 − 4αβ][1 + (α + β)^2])`
  `= sqrt([m^2 − 4(−b)][1 + m^2])`
  `= sqrt((m^2 + 4b)(1 + m^2))\ \ …text(as required)`

 

(iii) `⊥\ text(distance of)\ \ (x, x^2)\ \ text(from)\ \ \ y=mx+b`

`text(i.e.)\ \ mx – y+b=0`

`= |(ax_1 + by_1 + c)/(sqrt(a^2 + b^2))|`

`= |(mx − 1(x^2) + b)/(sqrt(m^2 + (−1)^2))|`

`= |(mx − x^2 + b)/(sqrt(m^2 + 1))|`

 

`text(Area of)\ ΔABP`

`= 1/2 xx AB xx h`

`= 1/2 xx sqrt((m^2 + 4b)(1 + m^2)) xx (mx − x^2 + b)/(sqrt(m^2 + 1))`

`= 1/2 (mx − x^2 + b)sqrt(m^2 + 4b)`

 

(iv) `A` `= 1/2 sqrt(m^2 + 4b)(mx − x^2 + b)`
  `(dA)/(dx)` `= 1/2 sqrt(m^2 + 4b)(m − 2x)` 
  `(d^2A)/(dx^2)` `= 1/2 sqrt(m^2 + 4b)(−2)`
    `= −sqrt(m^2 + 4b)` 

 

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

`1/2sqrt(m^2 + 4b)(m − 2x)` `= 0`
`m-2x` `=0`
`2x` `= m`
`x` ` = m/2`

`text(When)\ x = 2`

`(d^2A)/(dx^2) = −sqrt(m^2 + 4b) < 0\ \ \ text{(constant)}`

`:.\ text(Maximum when)\ x = m/2`

`:.P\ text(is)\ (m/2, m^2/4)` 

Mechanics, EXT2* M1 2015 HSC 14b

A particle is moving horizontally. Initially the particle is at the origin `O` moving with velocity `1 text(ms)^(−1)`.

The acceleration of the particle is given by  `ddot x = x − 1`, where `x` is its displacement at time  `t`.

  1. Show that the velocity of the particle is given by  `dot x = 1 − x`.  (3 marks)

  2. Find an expression for `x` as a function of `t`.  (2 marks)

  3. Find the limiting position of the particle.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `x = 1 − e^(-t)`
  3. `x = 1`
Show Worked Solution

i.   `ddot x = d/(dx)(1/2 v^2) = x − 1`

`1/2 v^2` `= int ddot x\ dx`
  `= int x − 1\ dx`
  `= 1/2x^2 − x + c`

 
`text(When)\ \ x = 0, \ v = 1:`

`1/2·1^2` `= 0 − 0 + c`
`c` `= 1/2`

 

`:.1/2  v^2` `= 1/2x^2 − x + 1/2`
`v^2` `= x^2 − 2x + 1`
  `= (x − 1)^2`
`:.dot x` `= ±(x − 1)`

 
`text(S)text(ince)\ \ dotx = 1\ \ text(when)\ \ x = 0,`

`dot x = 1 − x\ \ …\ text(as required)`

 

ii.   `(dx)/(dt)` `= 1 − x\ \ \ text{(from (i))}`
  `(dt)/(dx)` `= 1/(1 − x)`
  `t` `= int 1/(1 − x)\ dx`
    `= -ln(1 − x) + c`

 

`text(When)\ \ t = 0, x = 0:`

♦ Mean mark 49%.
`0` `= -ln1 + c`
`:.c` `= 0`
`t` `= -ln(1 − x)`
`-t` `= ln(1 − x)`
`1 − x` `= e^(-t)`
`:.x` `= 1 − e^(-t)`

 

♦ Mean mark 42%.
iii.   `text(As)\ t` `→ ∞`
  `e^(-t)` `→ 0`
  `x` `→ 1`

 

`:.\ text(Limiting position is)\ \ x = 1`

Calculus, EXT1 C2 2015 HSC 13d

Let  `f(x) = cos^(-1)\ (x) + cos^(-1)\ (-x)`, where  `-1 ≤ x ≤ 1`.

  1. By considering the derivative of  `f(x)`, prove that  `f(x)`  is constant.  (2 marks)

  2. Hence deduce that  `cos^(-1)\ (-x) = pi - cos^(-1)\ (x)`.  (1 mark)

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

i.  `text(Prove)\ f(x)\ text(is a constant)`

`f(x)` `= cos^(-1)(x) + cos^(-1)(-x), \ \ -1 ≤ x ≤ 1`
`f′(x)` `= (-1)/sqrt(1 – x^2) + (-1)/sqrt(1 -(-x)^2) xx d/(dx) (-x)`
  `= (-1)/sqrt(1 – x^2) + 1/sqrt(1 – x^2)`
  `= 0`

 
`:.\ text(S)text(ince)\ \ f′(x) = 0,  f(x)\ text(must be a constant.)`
 

♦ Mean mark 35%.
ii.   `f(0)` `= cos^(−1)(0) + cos^(−1)(0)`
    `= pi/2 + pi/2`
    `= pi`

 
`:.f(x) = pi`

`pi = cos^(−1)(x) + cos^(−1)(−x)`

`:.cos^(−1)(−x) = pi – cos^(−1)(x)\ \ …text(as required)`

Financial Maths, 2ADV M1 2007 HSC 3b

Heather decides to swim every day to improve her fitness level.

On the first day she swims 750 metres, and on each day after that she swims `100` metres more than the previous day. That is, she swims 850 metres on the second day, 950 metres on the third day and so on.

  1. Write down a formula for the distance she swims on the `n`th day.  (1 mark)

  2. How far does she swim on the 10th day?  (1 mark)

  3. What is the total distance she swims in the first 10 days?  (1 mark)

  4. After how many days does the total distance she has swum equal the width of the English Channel, a distance of 34 kilometres?  (2 marks)

Show Answers Only
  1. `T_n = 650 + 100n`
  2. `1650\ text(metres)`
  3. `12\ text(km)`
  4. `34\ text(km)`
Show Worked Solution
i.  `T_1` `= 750`
`T_2` `= 850`

`=>  text(AP)\ text(where)\ a = 750\ ,\ d = 100`

`:. T_n` `= a + (n-1) d`
  `= 750 + (n – 1) 100`
  `= 750 + 100n – 100`
  `= 650 + 100n`

 

ii.  `T_10` `= 650 + 100 xx 10`
  `= 1650\ text(metres)`

 

`:.\ text(She swims 1650 metres on the 10th day.)`

 

iii.  `S_n = n/2 [2a + (n – 1) d]`

`:. S_10` `= 10/2 [2 xx 750 + (10 -1) 100]`
  `= 5 [1500 + 900]`
  `= 12\ 000`

 

`:.\ text(She swims 12 km in the first 10 days.)`

 

iv.  `text(Find)\ n\ text(such that)\ S_n = 34\ text(km)`

`n/2 [2 xx 750 + (n – 1) 100]` `= 34\ 000`
`n/2 [1500 + 100n – 100]` `= 34\ 000`
`n/2 [1400 + 100n]` `= 34\ 000`
`700n + 50n^2` `= 34\ 000`
`50 n^2 + 700n – 34\ 000` `= 0`
`50 (n^2 + 14 n – 680)` `= 0`

 

`text(Using the quadratic formula)`

`n` `= {-b +- sqrt(b^2 – 4ac)}/(2a)`
  `= {-14 +- sqrt(14^2 – 4 xx 1 xx (-680))}/(2 xx 1)`
  `= (-14 +- sqrt 2916)/2`
  `= (-14 +- 54)/2`
  `= 20 or -34`
  `= 20\ ,\ n > 0`

 

`:.\ text(Her total distance equals 34 km after 20 days.)`

Calculus, 2ADV C3 2007 HSC 2c

The point  `P (pi, 0)`  lies on the curve  `y = x sinx`. Find the equation of the tangent to the curve at  `P`.  (3 marks)

Show Answers Only

`y = – pi x + pi^2`

Show Worked Solution

`y = x sin x`

`(dy)/(dx)` `= x xx d/(dx) (sin x) + d/(dx) x xx sin x`
  `= x  cos x + sin x`

 
`text(When)\ \ x = pi`

`(dy)/(dx)` `= pi xx cos pi + sin pi`
  `= pi (-1) + 0`
  `= – pi`

 
`text(Equation of line,)\ \ m = – pi,\ text(through)\ P(pi, 0):`

`y – y_1` `= m(x – x_1)`
`y – 0` `= – pi(x – pi)`
`:. y` `= – pi x + pi^2`

Combinatorics, EXT1 A1 2015 HSC 13b

Consider the binomial expansion
 

`(2x + 1/(3x))^18 = a_0x^(18) + a_1x^(16) + a_2x^(14) + …`
 

where `a_0, a_1, a_2`, . . . are constants.

  1. Find an expression for `a_2`.  (2 marks)

  2. Find an expression for the term independent of `x`.  (2 marks)

Show Answers Only
  1. `(\ ^18C_2 · 2^(16))/(3^2)`
  2. `(\ ^(18)C_9 · 2^9)/(3^9)`
Show Worked Solution

i.   `text(Need co-efficient of)\ x^(14)`

`text(General term of)\ (2x + 1/(3x))^(18)`

`T_k` `= \ ^(18)C_k(2x)^(18 − k) · (1/(3x))^k`
  `= \ ^(18)C_k · 2^(18 − k) · x^(18 − k) · 3^(−k) · x^(−k)`
  `= \ ^(18)C_k · 2^(18 − k) · 3^(−k) · x^(18 − 2k)`

 

`a_2\ text(occurs when:)`

`18 − 2k` `= 14`
`2k` `= 4`
`k` `= 2`

 

`:.a_2` `= \ ^(18)C_2 · 2^(18 − 2) · 3^(−2)`
  `= (\ ^(18)C_2 · 2^(16))/(3^2)`

 

ii.  `text(Independent term occurs when:)`

`18 − 2k` `= 0`
`2k` `= 18`
`k` `= 9`

 
`:.\ text(Independent term)`

`= \ ^(18)C_9 · 2^(18− 9) · 3^(−9)`

`= (\ ^(18)C_9 · 2^9)/(3^9)`

Mechanics, EXT2* M1 2015 HSC 13a

A particle is moving along the `x`-axis in simple harmonic motion. The displacement of the particle is `x` metres and its velocity is `v` ms`\ ^(–1)`. The parabola below shows `v^2` as a function of `x`.

2015 13a

  1. For what value(s) of `x` is the particle at rest?  (1 mark)

  2. What is the maximum speed of the particle?  (1 mark)

  3. The velocity `v` of the particle is given by the equation

         `v^2 = n^2(a^2 − (x −c)^2)`  where `a`, `c` and `n` are positive constants.

     

    What are the values of `a`, `c` and `n`?  (3 marks)

Show Answers Only
  1. `3\ text(or)\ 7`
  2. `V = sqrt11\ text(m/s)`
  3. `a = 2, c = 5,`
  4. `n = (sqrt11)/2`
Show Worked Solution

i.   `text(Particle is at rest when)\ v^2 = 0`

`:. x = 3\ \ text(or)\ \ 7`

 

ii.   `text(Maximum speed occurs when)`

`v^2` `= 11`
`v` `= sqrt11\ text(m/s)`

 

iii.  `v^2 = n^2(a^2 − (x − c)^2)`

♦ Mean mark 41%.

`text(Amplitude) = 2`

`:.a = 2`

`text(Centre of motion when)\ x = 5`

`:.c = 5`

`text(S)text(ince)\ \ v^2 = 11\ \ text(when)\ \ x = 5`

`11` `= n^2(2^2 − (5 − 5)^2)`
  `= 4n^2`
`n^2` `= 11/4`
`:.n`  `= sqrt11/2`

Trig Ratios, EXT1 2015 HSC 12d

A kitchen bench is in the shape of a segment of a circle. The segment is bounded by an arc of length 200 cm and a chord of length 160 cm. The radius of the circle is `r` cm and the chord subtends an angle `theta` at the centre `O` of the circle.
 

2015 12d
 

  1. Show that  `160^2 = 2r^2 (1 - cos\ theta)`.  (1 mark)
  2. Hence, or otherwise, show that  `8 theta^2 + 25 cos\ theta - 25 = 0`.  (2 marks)
  3. Taking  `theta_1 = pi`  as a first approximation to the value of  `theta`, use one application of Newton’s method to find a second approximation to the value of  `theta`. Give your answer correct to two decimal places.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `2.57\ \ text{(to 2 d.p.)}`
Show Worked Solution

(i)   `text(Using the cosine rule)`

`a^2` `= b^2 + c^2 − 2bc\ cos\ A`
`160^2` `= r^2 + r^2 −2 xx r xx r xx cos\ theta`
  `= 2r^2 − 2r^2\ cos\ theta`
  `= 2r^2(1 − cos\ theta)\ \ …\ text(as required)`

 

(ii)   `text(Show)\ \ 8theta^2 + 25\ cos\ theta − 25 = 0`

`text(Arc length)` `= 200`
`theta/(2pi) xx 2pir` `= 200`
`rtheta` `= 200`
`r` `= 200/theta`

 

`text(Substitute)\ \ r = 200/theta\ text{into part (i) equation}`

`2 xx (200/theta)^2(1 – cos\ theta)` `= 160^2`
`80\ 000(1 – cos\ theta)` `= 25\ 600\ theta^2`
`25(1 – cos\ theta)` `= 8theta^2`
`8theta^2 + 25\ cos\ theta – 25` `= 0\ \ …\ text(as required.)`

 

(iii)   `f(theta)` `= 8theta^2 + 25\ cos\ theta – 25`
  `f′(theta)` `= 16theta – 25\ sin\ theta`
  `f(pi)` `= 8pi^2 + 25\ cos\ pi – 25`
    `= 28.9568…`
  `f′(pi)` `= 16pi – 25\ sin\ pi`
    `= 50.2654…`
`theta_2` `= pi – (f(pi))/(f′(pi))`
  `= pi – (28.9568…)/(50.2654…)`
  `= 2.5655…`
  `= 2.57\ \ text{(to 2 d.p.)}`

Trig Ratios, EXT1 2015 HSC 12c

A person walks 2000 metres due north along a road from point `A` to point `B`. The point `A` is due east of a mountain `OM`, where `M` is the top of the mountain. The point `O` is directly below point `M` and is on the same horizontal plane as the road. The height of the mountain above point `O` is `h` metres.

From point `A`, the angle of elevation to the top of the mountain is 15°.

From point `B`, the angle of elevation to the top of the mountain is 13°.
 

Trig Ratios, EXT1 2015 HSC 12c
 

  1. Show that `OA = h\ cot\ 15°`.  (1 mark)
  2. Hence, find the value of  `h`.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `910\ text{m  (nearest metre)}`
Show Worked Solution

(i)   `text(Show)\ \ OA = h\ cot\ 15^@` 

Trig Ratios, EXT1 2015 HSC 12c Answer1

`text(In)\ \ Delta MOA,`

`tan\ 15^@` `= h/(OA)`
`OA` `= h/(tan\ 15^@)`
  `= h\ cot\ 15^@\ \ …text(as required)`

 

(ii)   `text(In)\ \ ΔMOB`

`tan\ 13^@` `= h/(OB)`
`OB` `= h/(tan\ 13^@)`
  `= h\ cot\ 13^@`

 

Trig Ratios, EXT1 2015 HSC 12c Answer2 
 

`text(In)\ \ ΔAOB`

`OA^2 + AB^2` `= OB^2`
`OB^2 − OA^2` `= AB^2`
`(h\ cot\ 13^@)^2 − (h\ cot\ 15^@)^2` `= 2000^2`
`h^2[(cot^2\ 13^@ − cot^2\ 15^@)]` `= 2000^2`
`h^2` `= (2000^2)/(cot^2\ 13^@ − cot^2\ 15^@)`
`:. h` `= sqrt((2000^2)/(cot^2\ 13^@ − cot^2\ 15^@))`
  `= 909.704…`
  `= 910\ text{m  (nearest metre)}`

Quadratic, EXT1 2015 HSC 12b

The points  `P(2ap, ap^2)`  and  `Q(2aq, aq^2)`  lie on the parabola  `x^2 = 4ay`.

The equation of the chord  `PQ`  is given by  `(p + q)x- 2y- 2apq = 0`.  (Do NOT prove this.)

  1. Show that if  `PQ`  is a focal chord then  `pq = –1`.  (1 mark)
  2. If  `PQ`  is a focal chord and  `P`  has coordinates  `(8a, 16a)`, what are the coordinates of  `Q`  in terms of  `a`?  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(−a/2, a/16)`
Show Worked Solution

(i)   `text(Show if)\ PQ\ text(is a focal chord)\ pq = −1`
 

Quadtratic, EXT1 2015 HSC 12b Answer1

 
`(p + q)x – 2y – 2apq = 0`

`text(If)\ PQ\ text(is a focal chord, it passes)\ (0, a)`

`(p + q)0 − 2a − 2apq` `= 0`
`2apq` `= -2a`
`pq` `= (−2a)/(2a)`
  `= −1\ \ …\ text(as required)`

 

(ii)  `text(If)\ P(2ap, ap^2) = (8a, 16a)`

`⇒ 2ap` `= 8a`
`p` `= 4`
`text(Using)\ \ q` `= −1/p\ \ \ text{(from part (i))}`
`q` `= −1/4`
`:.Q(2aq, aq^2)` `= (2a(−1/4), a(−1/4)^2)`
  `= (−a/2, a/16)`

Trig Calculus, EXT1 2015 HSC 10 MC

The graph of the function  `y = cos\ (2t − pi/3)`  is shown

2015 10 mc

What are the coordinates of the point `P`?

  1. `((5pi)/(12), 0)`
  2. `((2pi)/3, 0)`
  3. `((11pi)/(12), 0)`
  4. `((7pi)/6, 0)`
Show Answers Only

`C`

Show Worked Solution

`y = cos\ (2t − pi/3)`

`cos\ x = 0\ text(when)\ x = pi/2\ text(or)\ (3pi)/2`

`:.2t − pi/3` `= pi/2`    `text(or)` `2t − pi/3` `= (3pi)/2`
`2t` `= (5pi)/6`   `2t` `= (11pi)/6`
`t` `= (5pi)/12`   `t` `= (11pi)/12`

`text(From the graph,)\ P\ text(is where)\ y = cos\ (2t − pi/3)`

`text(hits the)\ xtext(-axis the second time.)`

`:.P\ text(has the coordinates)\ ((11pi)/12, 0)`

`⇒ C`

Mechanics, EXT2* M1 2015 HSC 9 MC

Two particles oscillate horizontally. The displacement of the first is given by  `x = 3\ sin\ 4t`  and the displacement of the second is given by  `x = a\ sin\ nt`. In one oscillation, the second particle covers twice the distance of the first particle, but in half the time.

What are the values of `a` and `n`?

  1. `a = 1.5,\ \ n = 2`
  2. `a = 1.5, \ \ n = 8`
  3. `a = 6,\ \ n = 2`
  4. `a = 6, \ \ n = 8`
Show Answers Only

`D`

Show Worked Solution
`x_1` `= 3\ sin\ 4t`
`x_2` `= a\ sin\ nt`

 

`x_2\ text(has twice the amplitude of)\ x_1`

`:. a = 2 xx 3 = 6`

`x_2\ text(has a period)\ (T)\ text(that is half of)\ x_1`

`T(x_1)` `= (2 pi)/n = (2pi)/4`
`:.T(x_2)` `= 1/2 xx (2 pi)/4 = (2pi)/8`
`:.n` `= 8`

 
`=> D`

Geometry and Calculus, EXT1 2015 HSC 5 MC

What are the asymptotes of `y = (3x)/((x + 1)(x + 2))`

A.    `y = 0,` `x = −1,` `x = −2`
B.    `y = 0,` `x = 1,` `x = 2`
C.    `y = 3,` `x = −1,` `x = −2`
D.    `y = 3,` `x = 1,` `x = 2`
Show Answers Only

`A`

Show Worked Solution

`y = (3x)/((x + 1)(x + 2))`

`text(Asymptotes at)\ x = −1\ text(and)\ x = −2`

`text(As)\ x → ∞, y → 0^+`

`text(As)\ x → −∞, y → 0^−`

`:.\ text(Horizontal asymptote at)\ \ y = 0`

`⇒ A`

Statistics, STD2 S4 2015 HSC 28e

The shoe size and height of ten students were recorded.

\begin{array} {|l|c|c|}
\hline \rule{0pt}{2.5ex} \text{Shoe size} \rule[-1ex]{0pt}{0pt} & \text{6} & \text{7} & \text{7} & \text{8} & \text{8.5} & \text{9.5} & \text{10} & \text{11} & \text{12} & \text{12} \\
\hline \rule{0pt}{2.5ex} \text{Height} \rule[-1ex]{0pt}{0pt} & \text{155} & \text{150} & \text{165} & \text{175} & \text{170} & \text{170} & \text{190} & \text{185} & \text{200} & \text{195} \\
\hline
\end{array}

  1. Complete the scatter plot AND draw a line of fit by eye.  (2 marks)
     
     
  2. Use the line of fit to estimate the height difference between a student who wears a size 7.5 shoe and one who wears a size 9 shoe.  (1 mark)

  3. A student calculated the correlation coefficient to be 1 for this set of data. Explain why this cannot be correct.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `13\ text{cm  (or close given LOBF drawn)}
  3. `text(A correlation co-efficient of 1 would)`
    `text(mean that all data points occur on the)`
    `text(line of best fit which clearly isn’t the case.)`
Show Worked Solution

i.    
      2UG 2015 28e Answer

ii.   `text{Shoe size 7½ gives a height estimate of 162 cm (see graph)}`

`text{Shoe size 9 gives a height estimate of 175 cm (see graph)}`

`:.\ text(Height difference)` `= 175-162`
  `= 13\ text{cm  (or close given LOBF drawn)}`

 

iii.   `text(A correlation co-efficient of 1 would mean)`

♦ Mean mark (c) 39%.

`text(that all data points occur on the line of best)`

`text(fit which clearly isn’t the case.)`

Algebra, STD2 A1 2015 HSC 28d

The formula  `C = 5/9 (F-32)`  is used to convert temperatures between degrees Fahrenheit `(F)` and degrees Celsius `(C)`.

Convert 3°C to the equivalent temperature in Fahrenheit.  (2 marks)

Show Answers Only

`37.4\ text(degrees)\ F`

Show Worked Solution
`C` `= 5/9(F-32)`
`F-32` `= 9/5C`
`F`  `= 9/5C + 32`

 
`text(When)\ \ C = 3,`

`F` `= (9/5 xx 3) + 32`
  `= 37.4\ text(degrees)\ F`

Measurement, 2UG 2015 HSC 28c

Three equally spaced cross-sectional areas of a vase are shown.

2UG 2015 29c

Use Simpson’s rule to find the approximate capacity of the vase in litres.  (3 marks)

Show Answers Only

`4\ text(litres)`

Show Worked Solution

2UG 2015 29c Answer 

`V` `≈ h/3[y_0 + 4y_1 + y_2]`
  `≈ 15/3[45 + 4(180) + 35]`
  `≈ 5[800]`
  `≈ 4000\ text(cm³)`
  `~~4\ text(litres)\ \ \ text{(1 cm³ = 1 mL)}`