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Mechanics, EXT2 2006 HSC 5c

A particle,  `P`, of mass  `m`  is attached by two strings, each of length  `l`, to two fixed points,  `A`  and  `B`, which lie on a vertical line as shown in the diagram.

The system revolves with constant angular velocity  `omega`  about  `AB`. The string  `AP`  makes an angle  `alpha`  with the vertical. The tension in the string  `AP`  is  `T_1`  and the tension in the string  `BP`  is  `T_2`  where  `T_1 >= 0`  and  `T_2 >= 0`. The particle is also subject to a downward force,  `mg`, due to gravity.

  1. Resolve the forces on  `P`  in the horizontal and vertical directions.  (2 marks)

  2. If  `T_2 = 0`, find the value of  `omega`  in terms of  `l, g`  and  `alpha.`  (1 mark)
Show Answers Only
  1. `text(Vertical):\ (T_1 – T_2) cos alpha = mg\ \ \ \ \ text(Horizontal):\ T_1 + T_2 = ml omega^2`
  2. `omega = sqrt (g/(l cos alpha))`
Show Worked Solution
(i)   

`text(Resolving the forces vertically)`

`T_1 cos alpha = mg + T_2 cos alpha\ \ \ …\ (1)`

`text(Resolving the forces horizontally)`

` T_1 sin alpha+ T_2 sin alpha = mr omega^2\ \ \ …\ (2)`

`r = l sin alpha`

`:. (T_1 + T_2) sin alpha = m l sin alpha omega^2`

`T_1 + T_2 = m l omega^2`

 

(ii)   `text(Substitute)\ \ T_2 = 0\ \ text{into (1) and (2)}`

`T_1` `= (mg)/cos alpha\ \ \ …\ (1)`
`T_1 sin alpha` `= m r omega^2`
  `=ml sin alpha omega^2\ \ \ \ text{(since}\ \ r=l sin alpha text{)}`
`T_1` `=ml omega^2\ \ \ …\ (2)`

 

`:. m l omega^2` `= (mg)/(cos alpha)`
`omega^2` `= g/(l cos alpha)`
`:.omega` `= sqrt(g/(l cos alpha))`

Conics, EXT2 2006 HSC 4c

Let  `P(p, 1/p), Q(q, 1/q)`  and  `R(r, 1/r)`  be three distinct points on the hyperbola  `xy = 1.`

  1. Show that the equation of the line,  `l`, through  `R`, perpendicular to  `PQ`, is  `y = pqx - pqr + 1/r.`  (2 marks)
  2. Write down the equation of the line,  `m`, through  `P`, perpendicular to  `QR.`  (1 mark)
  3. The lines  `l`  and  `m`  intersect at  `T.`
  4. Show that  `T`  lies on the hyperbola.  (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `y = qrx – pqr + 1/p`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   `text(Gradient)\ \ PQ =` `(1/p – 1/q)/(p – q)`
`­=` `((q – p)/(pq))/(p – q)`
`­=` `-1/(pq)`

 

`=>text(Gradient of perpendicular) = pq`

`:.\ text(Equation of)\ \ l,\ \ text(through)\ \ R(r, 1/r)`

`y – 1/r` `= pq (x – r)`
`y – 1/r` `= pqx – pqr`
`:.y` `= pqx – pqr + 1/r`

  

(ii)  `text(Equation of)\ \ m,\ \ text(through)\ \ P(p, 1/p),\ \ text(is)`

`y = qrx – pqr + 1/p`

 

(iii)  `T\ \ text(occurs at the intersection of)`

MARKER’S COMMENT: Many students had difficulty simplifying  `1/p-1/r -: (p-r).` Be clear on how to do this. 

`y = pqx – pqr + 1/r\ \ \ …\ (1)`

`y = qrx – pqr + 1/p\ \ \ …\ (2)`

`text{Subtract (1) – (2)}`

`(pq – qr)x + 1/r – 1/p` `=0`
`q(p – r)x` `= 1/p – 1/r`
`q(p – r)x` `= (r – p)/(pr)`
`x` `= -1/(pqr)`

 

`text{Substitute into (1)}`

`y = (-pq)/(pqr) – pqr + 1/r = -pqr`

`:.T ((-1)/(pqr), -pqr)`

 

`text(Substituting into)\ \ xy = 1`

`text(LHS)` `=(-1)/(pqr) xx (-pqr)`
  `=1`
  `=\ text(RHS)`

`:. T\ \ text(lies on the hyperbola)`

Functions, EXT1′ F2 2006 HSC 4a

The polynomial  `p(x) = ax^3 + bx + c`  has a multiple zero at 1 and has remainder 4 when divided by  `x + 1`. Find  `a, b`  and  `c`.  (3 marks)

Show Answers Only

`a = 1,\ \ b = -3,\ \ c = 2`

Show Worked Solution
`p(x)` `= ax^3 + bx + c`
`p′(x)` `= 3ax^2 + b`

 

`text(S)text(ince 1 is a multiple root,)`

`p prime (1)` `= 3ax^2 + b=0\ \ \ …\ (1)`
`p(1)`  `= a + b + c = 0\ \ \ …\ (2)`
`p(–1)` ` = -a – b + c = 4\ \ \ …\ (3)`

 
 `text{Add (2) + (3)}`

`2c = 4,\ \ c = 2`

`text{Subtract (1) – (2)}`

`2a – c = 0,\ \ a=1`

`text{Substituting into (1)}`

`3 + b = 0,\ \ b = -3`

 

`:. a = 1,\ \ b = -3,\ \ c = 2`

Conics, EXT2 2006 HSC 3b

The diagram shows the graph of the hyperbola

`x^2/144 - y^2/25 = 1.`

  1. Find the coordinates of the points where the hyperbola intersects the `x`-axis.  (1 mark)
  2. Find the coordinates of the foci of the hyperbola.  (2 marks)
  3. Find the equations of the directrices and the asymptotes of the hyperbola.  (2 marks)
Show Answers Only
  1. `(12, 0),\ (– 12, 0)`
  2. `(13, 0),\ (– 13, 0)`
  3. `text(Directrices:)\ \ x = +- 144/13;\ \ text(Asymptotes:)\ \ y = +- 5/12 x`
Show Worked Solution

(i)  `text(Intersection when)\ \ y=0`

`x^2/144 + 0` `=1`
`x` `= +-12`

 

`:. xtext(-axis intersection at)\ \ (12, 0),\ (– 12, 0)`

 

(ii)  `a=12,\ \ b = 5`

`text(Using)\ \ \ b^2` `=a^2(e^2-1)`
`25` `= 144 (e^2 – 1)`
`25/144` `= e^2 – 1`
`e^2` `= 169/144`
`e` `= 13/12`

 

`:.S(ae,0)-=(13,0)`

`:. S′(– ae,0)-=(– 13,0)`

 

(iii)   `text(Directrices when)\ \ \ x` `=+-a/e`
    `=+-144/13`
  `text(Asymptotes when)\ \ \ y` `=+-b/a x`
    `=+- 5/12 x`

Functions, EXT1′ F1 2006 HSC 3a

The diagram shows the graph of  `y =f(x)`. The graph has a horizontal asymptote at  `y =2`.
 


 

Draw separate one-third page sketches of the graphs of the following:

  1.  `y = (f(x))^2`  (2 marks)

  2.  `y = 1/(f(x))`  (2 marks)

  3.  `y = x\ f(x)`  (2 marks)

Show Answers Only

i.

ii. 

iii. 

Show Worked Solution
i.   

 

ii.   

 

iii.   

Conics, EXT2 2007 HSC 7b

In the diagram the secant  `PQ`  of the ellipse  `x^2/a^2 + y^2/b^2 = 1`  meets the directrix at  `R`. Perpendiculars from  `P`  and  `Q`  to the directrix meet the directrix at  `U`  and  `V`  respectively. The focus of the ellipse which is nearer to  `R`  is at  `S`.

Copy or trace this diagram into your writing booklet.

  1. Prove that
    1. `(PR)/(QR) = (PU)/(QV).`  (1 mark)

  2. Prove that
    1. `(PU)/(QV) = (PS)/(QS).`  (1 mark)

  3. Let  `/_ PSQ = phi,\ \ /_ RSQ = theta and /_ PRS = alpha.`
  4. By considering the sine rule in  `Delta PRS and Delta QRS`, and applying the results of part (i) and part (ii),
  5. show that  `phi = pi - 2 theta.`  (2 marks)

  6. Let  `Q`  approach  `P`  along the circumference of the ellipse, so that  `phi -> 0.`
  7. What is the limiting value of  `theta?`  (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. `theta -> pi/2`
Show Worked Solution

(i)   `text(In)\ \ Delta PUR and Delta QVR`

`/_ PUR` `= /_ QVR=90^@`
`/_ PRU` `= /_ QRV\ \ \ \ \ text{(common angle)}`
`:.\ Delta PUR\ text(|||)\ Delta QVR\ \ \ \ \ text{(equiangular)}` 
`:.\ (PR)/(QR)` `= (PU)/(QV)\ \ \ \ ` `text{(corresponding sides in similar}`
`text{triangles are proportional)}`

 

(ii)  `(SP)/(PU) = e and (SQ)/(QV) = e`

`(SP)/(PU)` `= (SQ)/(QV)`
`:.\ (PU)/(QV)` `= (PS)/(QS)` 

 

(iii)  `text(In)\ \ Delta PRS`

`(PS)/(sin alpha)` `= (PR)/(sin(phi + theta))`
`(PS)/(PR) ` `= (sin alpha)/(sin (phi + theta))`

 

`text(In)\ \ Delta QRS`

`(QS)/(sin alpha)` `= (QR)/(sin theta)`
`(QS)/(QR)` `= (sin alpha)/(sin theta)`

 

`text(Using)\ \ (PR)/(QR) = (PU)/(QV) = (PS)/(QS)\ \ \ text{(from parts (i) and (ii))}`

`=>(PS)/(PR)` `= (QS)/(QR)`
`(sin alpha)/(sin (phi + theta))` `= (sin alpha)/(sin theta)`

 

`:.\ sin (phi + theta) = sin theta`

`:.\ phi + theta = pi – theta\ \ \ or\ \ phi + theta = theta`

`:.\ phi = pi – 2 theta,\ \ \ \ \ (phi ≠ 0)`

 

(iv)   `text(As)\ \ phi`  `-> 0`
  `pi – 2 theta` `-> 0`
  `:.theta` `-> pi/2`

Mechanics, EXT2 M1 2007 HSC 6b

A raindrop falls vertically from a high cloud. The distance it has fallen is given by

`x = 5 log_e ((e^(1.4 t) + e^(-1.4 t))/2)`

where `x` is in metres and `t` is the time elapsed in seconds.

  1. Show that the velocity of the raindrop, `v` metres per second, is given by
     
         `v = 7 ((e^(1.4 t) - e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))`  (2 marks)

  2. Hence show that  `v^2 = 49 (1 - e^(-(2x)/5)).`  (2 marks)

  3. Hence, or otherwise, show that  `ddot x = 9.8 - 0.2v^2.`  (2 marks)

  4. The physical significance of the 9.8 in part (iii) is that it represents the acceleration due to gravity.

     

    What is the physical significance of the term  `–0.2 v^2?`   (1 mark)

  5. Estimate the velocity at which the raindrop hits the ground.   (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. `-0.2v^2\ \ text(is the air resistance)`
  5. `7\ \ text(ms)^-1`
Show Worked Solution
i.    `x` `= 5 log_e ((e^(1.4t) + e^(-1.4t))/2)`
  `v=dx/dt` `= (5(1.4e^(1.4t) – 1.4e^(-1.4t)))/(e^(1.4t) + e^(-1.4t))`
    `= 7 ((e^(1.4 t) – e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))`

 

ii.    `v^2` `=49 ((e^(1.4 t) – e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))^2`
    `=49 ((e^(2.8 t) + e^(-2.8 t)-2)/(e^(2.8 t) + e^(-2.8 t)+2))`
    `=49 ((e^(2.8 t) + e^(-2.8 t)+2-4)/(e^(2.8 t) + e^(-2.8 t)+2))`
    `=49 (((e^(1.4t) + e^(-1.4t))^2-2^2)/((e^(1.4t) + e^(-1.4t))^2))`
    `=49 (1 – (2/(e^(1.4t) + e^(-1.4t)))^2)`

 

`text(S)text(ince)\ \ x` `= 5 log_e ((e^(1.4 t) + e^(-1.4 t))/2)`
`e^(x/5)` `=(e^(1.4 t) + e^(-1.4 t))/2`

 

`:. v^2` `=49 (1 – (e^(- x/5))^2)`
  `=49(1-e^(- (2x)/5))`

 

iii.   `ddotx` `=d/(dx) (1/2 v^2)`
    `=49/2 xx  2/5 xx e^(-(2x)/5)`
    `=49/5 e^(-(2x)/5)`
    `=49/5 (1- v^2/49),\ \ \ \ text{(from part (ii))}`
    `=9.8 – 0.2v^2`

 
iv.  `-0.2v^2\ \ text(is the wind resistance acting on the rain drop.)`
 

v.   `text(Terminal velocity occurs when)\ \ ddot x=0`

`9.8 – 0.2v^2` `=0`
`v^2` `=49`
`v` `=7,\ \ \ \ (v > 0)`

 
`:.\ text(The rain drop hits the ground travelling at)\ \ 7\ \ text(ms)^-1`

Functions, EXT1′ F2 2012 HSC 8 MC

The following diagram shows the graph  `y = P′(x)`, the derivative of a polynomial  `P(x)`.
 

Polynomials, EXT2 2012 HSC 8 MC
 

Which of the following expressions could be  `P(x)`?

  1. `(x − 2)(x − 1)^3`
  2. `(x + 2)(x − 1)^3`
  3. `(x − 2)(x + 1)^3`
  4. `(x + 2)(x + 1)^3` 
Show Answers Only

`B`

Show Worked Solution

`P′(x)\ text(has a double root at)\ x = 1`

`=>P(x)\ text(will have a triple root at)\ x = 1.`

`text(Consider)\ \ P′(−5/4) = 0`

`text(The gradients goes from negative to positive.)`

`=>\ text(Local minimum when)\ \ x=-5/4`

`=> x text(-intercept must be)\ <-5/4`

`:. (x + 2)\ text(could be a factor of)\ \ P(x).`

`=>B`

Mechanics, EXT2 2012 HSC 7 MC

A particle  `P`  of mass  `m`  attached to a string is rotating in a circle of radius  `r`  on a smooth horizontal surface. The particle is moving with constant angular velocity  `ω`. The string makes an angle  `α`  with the vertical. The forces acting on  `P`  are the tension  `T`  in the string, a reaction force  `N`  normal to the surface and the gravitational force  `mg`.

Mechanics, EXT2 2012 HSC 7 MC

Which of the following is the correct resolution of the forces on  `P`  in the vertical and horizontal directions?

  1. `T cosα+ N = mg\ \ \ text(and)\ \ \ T sinα = mrω^2`
  2. `T cosα− N = mg\ \ \ text(and)\ \ \ T sinα = mrω^2`
  3. `T sinα+ N = mg\ \ \ text(and)\ \ \ T cosα = mrω^2`
  4. `T sinα− N = mg\ \ \ text(and)\ \ \ T cosα = mrω^2` 
Show Answers Only

`A`

Show Worked Solution

`text(Resolving the forces vertically,)`

`T cosα + N = mg.`

`text(Resolving the forces horizontally,)`

`T sinα = mrω^2.`

`=>A`

Conics, EXT2 2012 HSC 6 MC

What is the eccentricity of the hyperbola  `(x^2)/6 − (y^2)/4 = 1`?

  1. `sqrt10/2`
  2. `sqrt15/3`
  3. `sqrt3/3`
  4. `sqrt13/3` 
Show Answers Only

`B`

Show Worked Solution
`b^2` `= a^2(e^2 − 1)\ \ \ text(where)\ \  a^2 = 6, b^2 = 4`
`a^2e^2` `=a^2+b^2`
`e^2` `= 1 + (b^2)/(a^2)= 1 + 4/6= 5/3`
`:.e` `= sqrt5/sqrt3`
  `= sqrt15/3`

`=>B`

Graphs, EXT2 2012 HSC 4 MC

The graph  `y =ƒ(x)`  is shown below.

Graphs, EXT2 2012 HSC 4 MC

Which of the following graphs best represents  `y = [f(x)]^2`?

Graphs, EXT2 2012 HSC 4 MC ab 

Graphs, EXT2 2012 HSC 4 MC cd

Show Answers Only

`A`

Show Worked Solution

`text(By elimination)`

`[f(x)]^2=0\ \ text(when)\ \ f(x)=0,\ \ text(i.e. at)\ x= 0 and 2`

`=>\ text{Not (C) or (D)`

`text(When)\ \ f(x)=0,\ \ [f(x)]^2\ \ text(has minimum turning)`

`text{points and not cusps as in (B).}` 

`=>A`

Graphs, EXT2 2012 HSC 2 MC

The equation  `x^3 – y^3 + 3xy + 1 = 0`  defines  `y`  implicitly as a function of  `x`.

What is the value of  `(dy)/(dx)`  at the point  `(1, 2)`?

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

`D`

Show Worked Solution
`3x^2 − 3y^2y′ + 3xy′ + 3y` `= 0`
`y′(3x − 3y^2)` `= −3x^2 − 3y`
`y′` `= (−3x^2 − 3y)/(3x − 3y^2)`
  `=(x^2 + y)/(y^2− x)`

`text(At)\ \ P(1,2),`

`(dy)/(dx)= (1 + 2)/(4 − 1)=1`

`=>D`

Mechanics, EXT2 2007 HSC 3d

A particle  `P`  of mass  `m`  undergoes uniform circular motion with angular velocity  `omega`  in a horizontal circle of radius  `r`  about  `O`. It is acted on by the force due to gravity,  `mg`, a force  `F`  directed at an angle  `theta`  above the horizontal and a force  `N`  which is perpendicular to  `F`, as shown in the diagram.

  1. By resolving forces horizontally and vertically, show that
    1. `N = mg cos theta - m r omega^2 sin theta.`   (3 marks)

  2. For what values of  `omega`  is  `N > 0?`   (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `g/r cot theta`
Show Worked Solution
(i)      

`text(Resolving forces vertically)`

`F sin theta + N cos theta` `= mg`
`N cos theta` `=mg-F sin theta\ \ \ …\ (1)`
`text(Resolving forces horizontally)`
`F cos theta – N sin theta` `= m r omega^2`
`N sin theta` `=F cos theta-m r omega^2\ \ \ …\ (2)`

 

`text{Multiply (1)}\ xx cos theta and text{(2)}\ xx sin theta`

`N cos^2 theta` `=mg cos theta -F sin theta cos theta\ \ \ …\ (3)`
`N sin^2 theta` `=F cos theta sin theta-m r omega^2 sin theta\ \ \ …\ (4)`

 

`text{Add (3) + (4)}`

`:.N` `=mg cos theta-F sin theta cos theta + F cos theta sin theta-m r omega^2 sin theta`
  `=mg cos theta – m r omega^2 sin theta`

 

(ii)  `text(When)\ \ N > 0,`

`mg cos theta` `>m r omega^2 sin theta`
`omega^2` `<(g cos theta)/(r sin theta)`
  `<g/r cot theta`

Functions, EXT1′ F1 2007 HSC 3a

The diagram shows the graph of  `y = f(x)`. The line  `y = x`  is an asymptote.

Draw separate one-third page sketches of the graphs of the following:

  1.   `f(-x).`   (1 mark)

  2.   `f(|\ x\ |).`   (2 marks)

  3.    `f(x) - x.`   (2 marks)

Show Answers Only
  1.  

     

  2.  
  3.  
Show Worked Solution
i.  
MARKER’S COMMENT: In part (ii), a significant number of students graphed  `y=|f(x)|`.
ii.

 

iii. 

Complex Numbers, EXT2 N2 2007 HSC 2d

The points  `P,Q`  and  `R`  on the Argand diagram represent the complex numbers  `z_1, z_2`  and  `a`  respectively.

The triangles  `OPR`  and  `OQR`  are equilateral with unit sides, so  `|\ z_1\ | = |\ z_2\ | = |\ a\ | = 1.`

Let  `omega = cos­ pi/3 + i sin­ pi/3.`

  1. Explain why  `z_2 = omega a.`   (1 mark)

  2. Show that  `z_1 z_2 = a^2.`   (1 mark)

  3. Show that  `z_1` and `z_2`  are the roots of  `z^2 - az + a^2 = 0.`   (2 marks)

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

i.  `text(S) text(ince)\ \ Delta ORQ\ \ text(is equilateral, each angle is)\ \ pi/3\ \ text(radians.)`

`text(From)\ \ R(a):`

`Q(z_2)\ \ text(is an anticlockwise rotation through)\ \ pi/3.`

`:. z_2 = a(cos\ pi/3+i sin\ pi/3)=omega a.`

 

ii.  `text(Solution 1)`

`text(Similarly,)\ \ a` `=z_1 omega`
`z_1` `=a/omega`
`:z_1z_2` `=a/omega xx omega a`
  `=a^2`

 

`text(Solution 2)`

`P(z_1)\ \ text(is a clockwise rotation of)\ \ R(a)\ \ text(through)\ \ pi/3.`

`:.z_1` `= bar omega a.`
`:. z_1 z_2` `= bar omega a xx omega a`
  `=a^2(cos­ pi/3 – i sin­ pi/3) xx (cos­ pi/3 + i sin­ pi/3)`
  `=a^2(cos^2­ pi/3 + sin^2­ pi/3)`
  `= a^2`

 

iii.  `z^2-az + a^2 = 0`

`text(Let the roots be)\ \  alpha and beta.`

`alpha + beta` `=-b/a=a`
`alpha beta` `=c/a=a^2`

 

`z_1 z_2` `= a^2\ \ \ \ \ text{(part (ii))}`
`z_1 + z_2` `=bar omega a + omega a`
  `=(cos­ pi/3 + i sin­ pi/3 + cos­ pi/3-i sin­ pi/3) a`
  `=2 cos ­ pi/3 xx a`
  `=2 xx 1/2 xx a`
  `=a`

 

`:.\ z_1 and z_2\ \ text(are the roots of)\ \  z^2-az + a^2 = 0.`

Complex Numbers, EXT2 N1 2007 HSC 2b

  1. Write  ` 1 + i`  in the form  `r (cos theta + i sin theta).`  (2 marks)

  2. Hence, or otherwise, find  `(1 + i)^17`  in the form  `a + ib`, where  `a`  and  `b`  are integers.  (3 marks)

Show Answers Only
  1. `sqrt 2 ( cos­ pi/4 + i sin­ pi/4)`
  2. `256 + 256i`
Show Worked Solution
i.  
`|\ 1+i\ |` `=sqrt(1^2+1^2)=sqrt2`
`text(arg)(1+i)` `=pi/4`
`:. 1 + i =` `sqrt 2 (cos­ pi/4 + i sin­ pi/4)`

 

ii.   `(1 + i)^17` `=(sqrt 2)^17 (cos\ pi/4 + i sin\ pi/4)^17`
  `=2^8 sqrt 2 (cos­ (17 pi)/4 + i sin­ (17 pi)/4)\ \ \ \ text{(De Moivre)}`
  `=2^8 sqrt 2 (cos­ pi/4 + i sin­ pi/4)`
  `=2^8 sqrt2(1/sqrt2 + 1/sqrt2 i)`
  `=2^8 (1 + i)`
  `=256 + 256 i`

Calculus, EXT2 C1 2007 HSC 1e

It can be shown that

`2/(x^3 + x^2 + x + 1) = 1/(x + 1) - x/(x^2 + 1) + 1/(x^2 + 1).`   (Do NOT prove this.)
 

Use this result to evaluate  `int_(1/2)^2 2/(x^3 + x^2 + x + 1)\ dx.`  (4 marks)

 

Show Answers Only

`tan^-1 2 – tan^-1­ 1/2`

Show Worked Solution

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

`=int_(1/2)^2 (1/(x + 1) – x/(x^2 + 1) + 1/(x^2 + 1)) dx`

`=[log_e(x + 1) – 1/2 log_e (x^2 + 1) + tan^-1 x]_(1/2)^2`

`=[(log_e 3 – 1/2 log_e 5 + tan^-1 2) – (log_e­ 3/2 – 1/2 log_e­ 5/4 + tan^-1­ 1/2)]`

`=log_e\ 3/sqrt5 -log_e (3/2 xx 2/sqrt5) + tan^-1 2 – tan^-1­ 1/2`

`=log_e (3/sqrt(5) xx sqrt (5)/3) + tan^-1 2 – tan^-1­ 1/2`

`=tan^-1 2 – tan^-1­ 1/2`

Functions, EXT1′ F1 2015 HSC 8 MC

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

A second graph is obtained from the function  `y = f(x).`


 

Which equation best represents the second graph?

  1. `y^2 = |\ f(x)\ |`
  2. `y^2 = f(x)`
  3. `y = sqrt (f(x))`
  4. `y = f(sqrt x)`
Show Answers Only

`B`

Show Worked Solution

`text(The second graph has a domain)\ \ x=a,\ \ x>=b`

`:.\ text{NOT (A) or (D)}`

`text(S)text(ince)\ \ y` `= +- sqrt (f(x))`
`y^2` ` = f(x)`

 `=>  B`

Complex Numbers, EXT2 N1 2015 HSC 5 MC

Given that  `z = 1 − i`, which expression is equal to  `z^3 ?`

  1. `sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))`
  2. `2 sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))`
  3. `sqrt 2 (cos((3 pi)/4) + i sin((3 pi)/4))`
  4. `2 sqrt 2 (cos((3 pi)/4) + i sin((3 pi)/4))`
Show Answers Only

`B`

Show Worked Solution

 HSC 2015 5MC

`z` `=1-i`
`|\ 1-i\ |` `=sqrt2`
`text{arg}(z)` `=-pi/4`
`z` `=sqrt 2 (cos(-pi/4) + i sin(-pi/4))`
`:.z^3` `=2 sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))\ \ \ \ text{(De Moivre)}`

 
`=>  B`

Conics, EXT2 2015 HSC 1 MC

Which conic has eccentricity `sqrt 13/3?`

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

`D`

Show Worked Solution

`text(S)text(ince)\ \ e > 1,\ \ =>text(hyperbola)`

`b^2` `=a^2(e^2-1)`
`e^2` `=b^2/a^2 +1=13/9`
`:. b^2/a^2` `=4/9=2^2/3^2`

`=>  D`

Functions, EXT1′ F1 2014 HSC 5 MC

Which graph best represents the curve  `y^2 = x^2 - 2x`?
 

Graphs, EXT2 2014 HSC 5 MC ab 

Graphs, EXT2 2014 HSC 5 MC cd

Show Answers Only

`C`

Show Worked Solution
`text(S)text(ince)\ \ y^2` `>0`
`x^2 – 2x` `>0\ \ \ text(which is undefined for)\ \ 0 < x < 2`

 
`=>C`

 
`text(Alternative Solution)`

Graphs, EXT2 2014 HSC 5 MC Answer

`text(Consider)\ \ y = x^2 − 2x\ \ text{(above)}`

`y` `= ±sqrt(x^2 − 2x)`

 
`text(This curve is undefined for)\ \ 0 < x < 2.`

`=> C`

Complex Numbers, EXT2 N1 2014 HSC 4 MC

Given  `z = 2(cos\ pi/3 + i sin\ pi/3)`, which expression is equal to  `(bar {:z:})^(−1)`?

  1. `1/2(cos\ pi/3 − i sin\ pi/3)`
  2. `2(cos\ pi/3 − i sin\ pi/3)`
  3. `1/2(cos\ pi/3 + i sin\ pi/3)`
  4. `2(cos\ pi/3 + i sin\ pi/3)` 
Show Answers Only

`C`

Show Worked Solution
`z` `= 2text(cis)(pi/3)`
`barz` `= 2text(cis)(−pi/3)`
`(barz)^(−1)` `= (2text(cis)(−pi/3))^(−1)`
  `= 2^(−1)text(cis)(pi/3)`
  `= 1/2text(cis)(pi/3)`
  `= 1/2(cos\ pi/3 + i sin\ pi/3)`

 
`=> C`

Conics, EXT2 2014 HSC 3 MC

What is the eccentricity of the ellipse  `9x^2 + 16y^2 = 25`?

  1. `7/16`
  2. `sqrt7/4`
  3. `sqrt15/4`
  4. `5/4` 
Show Answers Only

`B`

Show Worked Solution
`b^2` `= a^2(1 − e^2)`
`e^2` `= 1 − (b^2)/(a^2)`
  `= 1 − (25/16)/(25/9)`
  `= 1 − 9/16`
 `e^2` `= 7/16`
`:.e`  `= sqrt7/4.`

`=> B`

Polynomials, EXT2 2014 HSC 2 MC

The polynomial  `P(z)`  has real coefficients, and  `z = 2 − i`  is a root of  `P(z)`.

Which quadratic polynomial must be a factor of  `P(z)`?

  1. `z^2 −4z +5`
  2. `z^2 +4z +5`
  3. `z^2 −4z +3`
  4. `z^2 +4z +3` 
Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince coefficients are real,)`

`=>\ text{Roots (α, β) are}\ \ 2-i,\ \ 2+i`

`α+β=4`

`αβ=(2-i)(2+i)=5`

`=> A`

Volumes, EXT2 2013 HSC 8 MC

The base of a solid is the region bounded by the circle  `x^2 + y^2 = 16`. Vertical cross-sections are squares perpendicular to the `x`-axis as shown in the diagram.

Which integral represents the volume of the solid?

  1. `int_-4^4 4x^2\ dx`
  2. `int_-4^4 4 pi x^2\ dx`
  3. `int_-4^4 4 (16 - x^2)\ dx`
  4. `int_-4^4 4 pi (16 - x^2)\ dx`
Show Answers Only

`C`

Show Worked Solution

`text(Length of the cross-section base)`

`=2y=2sqrt(16-x^2)`

`text(Height of the cross-section)`

`=2y=2sqrt(16-x^2)`

`:.\ text(Area of cross-section)` `= 4y^2`
  `= 4 (16 – x^2)`

`:. V = 4 int_-4^4 16 – x^2\ dx`

`=>  C`

Integration, EXT2 2013 HSC 6 MC

Which expression is equal to  `int 1/sqrt (x^2 - 6x + 5)\ dx?`

  1. `sin^-1 ((x - 3)/2) + C`
  2. `cos^-1 ((x - 3)/2) + C`
  3. `ln (x - 3 + sqrt ((x - 3)^2 + 4)) + C`
  4. `ln (x - 3 + sqrt ((x - 3)^2 - 4)) + C`
Show Answers Only

`D`

Show Worked Solution

`x^2 – 6x + 5 = (x – 3)^2 – 4`

`:. int 1/sqrt (x^2 – 6x + 5)\ dx`

`=int 1/sqrt((x – 3)^2 – 2^2)\ dx`

`=ln (x – 3 + sqrt((x – 3)^2 – 2^2)) + C`

`=>  D`

Functions, EXT1′ F2 2013 HSC 4 MC

The polynomial equation  `4x^3 + x^2 − 3x + 5 = 0`  has roots  `alpha, beta and gamma.`

Which polynomial equation has roots  `alpha + 1, beta + 1 and gamma + 1?`

  1. `4x^3 - 11x^2 + 7x + 5 = 0`
  2. `4x^3 + x^2 - 3x + 6 = 0`
  3. `4x^3 + 13x^2 + 11x + 7 = 0`
  4. `4x^3 - 2x^2 - 2x + 8 = 0`
Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ x = α + 1`

`:. α = x – 1`

 

`text(Given that α is a zero of the equation)`

`4 (x – 1)^3 + (x – 1)^2 – 3 (x – 1) + 5=0`

`4 (x^3 – 3x^2 + 3x – 1) + (x^2 – 2x + 1) – 3x + 3 + 5=0`

`4x^3 – 11x^2 + 7x + 5=0`

`=>  A`

Conics, EXT2 2013 HSC 2 MC

Which pair of equations gives the directrices of  `4x^2 - 25y^2 = 100?`

  1. `x = +- 25/sqrt 29`
  2. `x = +- 1/sqrt 29`
  3. `x = +- sqrt 29`
  4. `x = +- (sqrt 29)/25`
Show Answers Only

`A`

Show Worked Solution
`4x^2 – 25y^2` `= 100`
`=>x^2/5^2 – y^2/2^2` `= 1`

`:. a = 5 and b = 2`

 

`text(S) text(ince)\ \ b^2` `= a^2 (e^2 – 1),`
`e =` `sqrt (a^2 + b^2)/a`
`­=` `sqrt (5^2 + 2^2)/5`
`­=` `sqrt 29/5`

`text(Directrices occur at)\ \ \ x=+- a/e`

`­:.x` `=+- 5 ÷ sqrt 29/5`
  `=+- 25/sqrt 29`

`=>  A`

Complex Numbers, EXT2 N1 2006 HSC 2b

  1. Express  `sqrt 3 - i`  in modulus-argument form.  (2 marks)

  2. Express  `(sqrt 3 - i)^7`  in modulus-argument form.  (2 marks)

  3. Hence express  `(sqrt 3 - i)^7`  in the form  `x + iy.`  (1 mark)

Show Answers Only
  1. `2 text(cis) (−pi/6)`
  2. `2^7 text(cis) ((5 pi)/6)`
  3. `64 (−sqrt 3 + i)`
Show Worked Solution
i.  
`|\ sqrt 3 – i\ |` `= sqrt ((sqrt 3)^2+1^2)`
  `=2`
`­theta` `=tan^-1(- 1/sqrt3)`
  `=- pi/6`

 
`:. sqrt 3 – i = 2 text(cis) (- pi/6)`

 

ii.   `(sqrt 3 – i)^7 =` `2^7 text(cis) (-(7 pi)/6)\ \ \ \ text{(De Moivre)}`
`­=` `128 text(cis) ((5 pi)/6)`

 

iii.  `(sqrt 3 – i)^7` `=128 (cos\ (5pi)/6 + i sin\ (5pi)/6)`
  `=128 (- sqrt 3/2 + i/2)`
  `=-64 sqrt 3 + 64i`

Complex Numbers, EXT2 N1 2006 HSC 2a

Let  `z = 3 + i`  and  `w = 2 - 5i`.  Find, in the form  `x + iy`,

  1.  `z^2.`  (1 mark)

  2.  `bar z w.`  (1 mark)

  3.  `w/z.`  (1 mark)

Show Answers Only
  1.  `8 + 6i`
  2.  `1 – 17i`
  3.  `1/10 – 17/10 i`
Show Worked Solution
i.   `z^2` `=(3 + i)^2`
  `= 9 + 6i – 1`
  `= 8 + 6i`

 

ii.  `bar{:z:} w` `=(3 – i) (2 – 5i)`
  `= 6 – 15i – 2i – 5`
  `= 1 – 17i`

 

iii.  `w/z` `=(2 – 5i)/(3 + i) xx (3 – i)/(3 – i)`
  `= (1 – 17i)/10`
  `= 1/10 – 17/10 i`

Calculus, EXT2 C1 2006 HSC 1e

Use the substitution  `t = tan\ theta/2`  to show that

`int_(pi/2)^((2 pi)/3) (d theta)/(sin theta) = 1/2 log 3.`  (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Let)\ \ t = tan\ \ theta/2,\ \ sin theta = (2t)/(1 + t^2),\ \ d theta = 2/(1 + t^2)\ dt`

`text(When)\ \ theta = pi/2,\ \ t = 1`

`text(When)\ \ theta = (2 pi)/3,\ \ t = sqrt 3`

`:.int_(pi/2)^((2 pi)/3) (d theta)/(sin theta)` `= int_1^sqrt 3 (1 + t^2)/(2t) xx2/(1 + t^2)\ dt`
  `= int_1^ sqrt 3 (dt)/t`
  `= [ln t]_1^sqrt 3`
  `= ln sqrt 3 – ln 1`
  `=ln3^(1/2)-0`
  `= 1/2 ln 3`

Calculus, EXT2 C1 2006 HSC 1c

  1. Given that  `(16x - 43)/((x - 3)^2 (x + 2))`  can be written as
     
    `qquad (16x - 43)/((x - 3)^2 (x + 2)) = a/(x - 3)^2 + b/(x - 3) + c/(x + 2)`,
     
    where  `a, b` and `c`  are real numbers, find  `a, b and c.`  (3 marks)

  2. Hence find  `int (16x - 43)/((x - 3)^2 (x + 2))\ dx.`  (2 marks)

Show Answers Only
  1. `a = 1, b = 3, c = -3`
  2. `-1/(x-3)+3ln((x-3)/(x+2)) +c`
Show Worked Solution

i.   `(16x – 43)/((x – 3)^2 (x + 2)) = a/(x – 3)^2 + b/(x – 3) + c/(x + 2)`

`16x – 43 = a (x + 2) + b (x – 3) (x + 2) + c (x – 3)^2`
 

`text(When)\ \ x = 3,\ \ 5a =5\ \ =>a=1`

`text(When)\ \ x=–2,\ \ 25c=–75\ \ =>c=–3`

`text(When)\ \ x=0`

`-43` `= 2(1) – 6b + (-3)(-3)^2`
`6b` `= 18`
`b` `=3`

 
`:.a=1, b=3, c=–3`
 

ii.   `int (16x – 43)/((x – 3)^2 (x + 2))\ dx`

`=int (1/(x – 3)^2 + 3/(x – 3) – 3/(x + 2))\ dx`

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

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

Harder Ext1 Topics, EXT2 2009 HSC 5a

In the diagram  `AB`  is the diameter of the circle. The chords  `AC`  and  `BD`  intersect at  `X`. The point  `Y`  lies on  `AB`  such that  `XY`  is perpendicular to  `AB`. The point  `K`  is the intersection of  `AD`  produced and  `YX`  produced.

Copy or trace the diagram into your writing booklet.

  1. Show that  `/_ AKY = /_ ABD.`  (2 marks)
  2. Show that  `CKDX`  is a cyclic quadrilateral.  (2 marks)
  3. Show that  `B, C and K`  are collinear.  (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)   HSC 2009 5bii

`/_ ADB = 90^@\ \ \ text{(angle in a semicircle on diameter}\ \ AB text{)}`

`text(In)\ \ Delta BDA and Delta KYA`

`/_ BAD` `= /_ KAY\ \ text{(common angle)}`
`/_ ADB` `= /_ KYA = 90^@`
`:./_ AKY` `= /_ ABD\ \ text{(angle sum of triangle)}`

 

(ii)  `text(Join)\ \ DC`

`/_ DCA` `= /_ DBA\ \ text{(angles in the same segment on chord}\ DA text{)}`
`/_ DCA` `= /_ XKD\ \ text{(both equal to}\ /_ DBA text{)}`

 

`=>text(S)text(ince)\ \ /_ DKX = /_ DCX\ \ text(are a pair of equal angles)`

`text{standing on arc}\ DX\ \ text{(in the same segment)}.`

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

 

(iii)   `/_ KDX` `= 90^@\ \ text{(} /_ ADK\ text{is a straight angle)}`

`/_ KCX`

 

`= 90^@\ \ text{(opposite angles of a cyclic}`

`text{quadrilateral are supplementary)}`
`/_ ACB` `= 90^@\ \ text{(angle in a semicircle)}`

`/_ KCX + /_ ACB = 180^@`

`:./_ BCK\ \ text(is a straight angle.)`

`:.B, C and K\ \ text(are collinear.)`

Functions, EXT1′ F1 2009 HSC 3a

The diagram shows the graph  `y = f(x).`
 


 

Draw separate one-third page sketches of the graphs of the following:

  1.  `y = 1/(f(x)) .`  (2 marks)

  2.  `y = f(x)\ f(x)`  (2 marks)

  3.  `y = f(x^2).`  (2 marks)

Show Answers Only

i.

HSC 2009 3aii

 

ii.  

iii.  

 

 

 

 

 

 

Show Worked Solution

i.   `text(Vertical asymptotes at)\ \ x=0 and 4`

`text(Horizontal asymptote at)\ \ y=-1/3`
 

HSC 2009 3aii

 

ii.  `y=f(x)\ f(x) = [f(x)]^2`

   HSC 2009 3aiii

 

iii.   `y=f(x^2) =>text(even function)`

`text(When)\ \ x=±2,\ \ y=f(4)=0`
 

 

 

 

 

 

 

Graphs, EXT2 2009 HSC 3b

Find the coordinates of the points where the tangent to the curve  `x^2 + 2xy + 3y^2 = 18`  is horizontal.  (3 marks)

Show Answers Only

`(3, –3) and (–3, 3)`

Show Worked Solution
`x^2 + 2xy + 3y^2` `= 18\ \ \ \ …\ (1)`
`2x + 2 (y + x (dy)/(dx)) + 6y (dy)/(dx)` `=0`
`(dy)/(dx) (x + 3y)` `= -(x + y)`
`:.(dy)/(dx)` `= (-(x + y))/(x + 3y)`

 

`text(T)text(angent is horizontal when)`

`dy/dx=0\ \ \ =>\ \ y = -x`

`text(Substitute)\ \ y=-x\ \ text{into (1)}`

`x^2 – 2x^2 + 3x^2` `= 18`
 `x^2` `= 9`
 `:.x` `=+- 3`

 

`:.\ text(T)text(angent is horizontal at)\ \ (3, –3) and (–3, 3).`

Complex Numbers, EXT2 N2 2009 HSC 2f

  1. Find the square roots of  `3 +4i.`  (3 marks)

  2. Hence, or otherwise, solve the equation

     

        `z^2 + iz - 1 - i = 0.`  (2 marks)

Show Answers Only
  1. `+-(2 + i)`
  2. `1 or -1-i`
Show Worked Solution
i.    `z` `=sqrt(3+4i)`
  `z^2` `=3+4i`
  `(x + iy)^2` `= 3 + 4i`
  `x^2  – y^2+ 2xyi` `= 3 + 4i`
`=>x^2 – y^2 = 3,\ \ \ xy = 2`

 
`text(By inspection,)`

`text(If)\ \ x=2,\ \ \ y=1`

`text(If)\ \ x=-2,\ \ \ y=-1` 

`:.z= 2 + i,\ \ text(or)\ \ -(2+i)`

 

ii.   `z^2 + iz – 1 – i = 0`

`:.z =` `(-i +- sqrt (-1 + 4 (1 + i)))/2`
`=` `\ \ (-i +- sqrt(3 + 4i))/2`
`=` `\ \ (-i +- (2+i))/2`
`=` `\ \ 1\ \ \ text(or)\ \ \ -1-i`

Complex Numbers, EXT2 N1 2009 HSC 2c

The points  `P`  and  `Q`  on the Argand diagram represent the complex numbers  `z`  and  `w` respectively.
 


 

Copy the diagram into your writing booklet, and mark on it the following points:

  1. the point  `R`  representing  `iz.`   (1 mark)
  2. the point  `S`  representing  `bar w.`   (1 mark)
  3. the point  `T`  representing  `z + w.`   (1 mark)

Show Answers Only

i, ii, and iii.

Show Worked Solution

i, ii, and iii.

Calculus, EXT2 C1 2009 HSC 1c

Find  `int x^2/(1 + 4x^2)\ dx.`  (3 marks)

Show Answers Only

`x/4 – 1/8 tan^-1 2x + c`

Show Worked Solution
`x^2/(1 + 4x^2)` `= 1/4 xx (4x^2)/(1 + 4x^2)`
  `= 1/4 xx (1 + 4x^2)/(1 + 4x^2) – 1/4 xx 1/(1 + 4x^2)`
  `=1/4-1/4 xx 1/(1 + 4x^2)`

 

`:.int x^2/(1 + 4x^2)\ dx` `= 1/4 int 1\ dx – 1/4 int 1/(1 + 4x^2)\ dx`
  `= x/4 – 1/4 xx 1/2 tan^-1 2x + c`
  `= x/4 – 1/8 tan^-1 2x + c`

Polynomials, EXT2 2010 HSC 6c

  1. Expand  `(cos theta + i sin theta)^5`  using the binomial theorem.   (1 mark)

  2. Expand  `(cos theta + i sin theta)^5`  using de Moivre’s theorem, and hence show that

    1. `sin 5theta = 16 sin^5 theta − 20sin^3 theta + 5 sin theta`.   (3 marks)

  3. Deduce that

  4. `x = sin (pi/10)`  is one of the solutions to

    1. `16x^5 − 20x^3 + 5x − 1 = 0`.   (1 mark)

  5. Find the polynomial  `p(x)`  such that  `(x − 1) p(x) = 16x^5 − 20x^3 + 5x − 1`.   (1 mark)

  6. Find the value of  `a`  such that  `p(x) = (4x^2 + ax − 1)^2`.   (1 mark)

  7. Hence find an exact value for
    1. `sin (pi/10)`.   (1 mark)

 

 

Show Answers Only
  1. `cos^5 theta + 5i cos^4 theta sin theta − 10 cos^3 theta sin^2 theta − 10i cos^2 theta sin^3 theta`
    `+ 5 cos theta sin^4 theta + i sin^5 theta`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
  4. `16x^4 + 16x^3 − 4x^2 − 4x + 1`
  5. `a = 2`
  6. `(-1 + sqrt5)/4`
Show Worked Solution

(i)   `(cos theta + i sin theta)^5`

`=cos^5 theta + 5cos^4 theta (i sin theta) + 10 cos^3 theta (i sin theta)^2 + `

`10 cos^2 theta (isin theta)^3 + 5 cos theta (i sin theta)^4 + (i sin theta)^5`

`= cos^5 theta + 5i cos^4 theta sin theta − 10 cos^3 theta sin^2 theta − 10i cos^2 theta sin^3 theta +`

`5 cos theta sin^4 theta + i sin^5 theta`

 

(ii)  `text(Using De Moivre)`

`(cos theta + i sin theta)^5 = cos 5theta + i sin 5theta`

`text(Equating imaginary parts)`

`sin 5theta` `= 5 cos^4theta sin theta − 10cos^2 theta sin^3 theta + sin^5 theta`
  `= 5 sin theta (1 − sin^2 theta)^2 − 10 sin^3 theta (1 − sin^2 theta) + sin^5 theta`
  `= 5 sin theta (1 − 2 sin^2 theta + sin^4 theta) − 10 sin^3 theta + 10 sin^5 theta + sin^5 theta`
  `= 5 sin theta − 10 sin^3 theta + 5 sin^5 theta − 10 sin^3 theta + 11 sin^5 theta`
  `= 16 sin^5 theta − 20 sin^3 theta + 5 sin theta`

 

(iii)  `text(If)\ x = sin (pi/10)`

`16 sin^5 (pi/10) − 20 sin^3 (pi/10) + 5 sin (pi/10)` `=sin (5 xx pi/10)`
`16x^5 − 20x^3 + 5x` `= sin\ pi/2`
`16x^5 − 20x^3 + 5x` `=1`

 

`:. sin\ pi/10\ \ text(is one solution to)\ \ 16x^5 − 20x^3 + 5x − 1=0`

 

(iv)  `16x^5 − 20x^3 + 5x − 1`

`=(x-1)(16x^4 + 16x^3 − 4x^2 − 4x + 1)`

`:.p(x) = 16x^4 + 16x^3 − 4x^2 − 4x + 1`

 

(v)   `(4x^2 + ax − 1)^2`

`= 16x^4 + 4ax^3 − 4x^2 + 4ax^3 + a^2x^2 − ax − 4x^2 − ax + 1`

`= 16x^4 + 8ax^3 − 8x^2 + a^2x^2 − 2ax +1`

`text(By equating coefficients of)\ \ x^3`

`:.a = 2`

 

(vi)  `4 x^2 + 2 x − 1 = 0`

♦♦ Mean mark part (vi) 27%.
`x` `=(-2 ± sqrt(4 + 16))/8`
  `=(-1 ± sqrt5)/4`

 

`:. sin\ pi/10 = (-1 + sqrt5)/4\ \ \ \ (sin\ pi/10\ > 0)`

Conics, EXT2 2010 HSC 5a

The diagram shows two circles, `C_1`  and  `C_2`, centred at the origin with radii  `a`  and  `b`, where  `a > b`.

The point  `A`  lies on  `C_1`  and has coordinates  `(a cos theta, a sin theta)`.

The point  `B`  is the intersection of  `OA`  and  `C_2`.

The point  `P`  is the intersection of the horizontal line through  `B`  and the vertical line through  `A`.

Conics, EXT2 2010 HSC 5a

  1. Write down the coordinates of  `B`.   (1 mark)
  2. Show that  `P`  lies on the ellipse
    `(x^2)/(a^2) + (y^2)/(b^2) = 1`.   (1 mark)

  3. Find the equation of the tangent to the ellipse
    `(x^2)/(a^2) + (y^2)/(b^2) = 1`  at  `P`.   (2 marks)

  4. Assume that `A` is not on the `y`-axis.
  5. Show that the tangent to the circle  `C_1`  at  `A`, and the tangent to the ellipse
    `(x^2)/(a^2) + (y^2)/(b^2) = 1`  at  `P`, intersect at a point on the `x`-axis.   (2 marks)
Show Answers Only
  1. `B(b cos theta, b sin theta)`
  2. `text{Proof (See Worked Solutions)}`
  3. `b cos theta x + a sin theta y = ab, or`
  4. `(x cos theta)/a + (y sin theta)/b = 1`
  6. `text{Proof (See Worked Solutions)}`
Show Worked Solution

(i)   `B(b cos theta, b sin theta)`

 

(ii)  `text(Substitue)\ \ P(a cos theta, b sin theta)\ \ text(into)\ \ (x^2)/(a^2) + (y^2)/(b^2) = 1`

`text(LHS)` `= (a^2 cos^2 theta)/(a^2) + (b^2 sin^2 theta)/(b^2)`
  `= cos^2 theta + sin^2 theta`
  `= 1`
  `=\ text(RHS)`

 

`:.P\ text(lies on the ellipse.)`

 

(iii)   `text(Solution 1)`

`(x^2)/(a^2) + (y^2)/(b^2)` `= 1`
`(2x)/(a^2) + (2y)/(b^2) xx (dy)/(dx)` `= 0`
`(dy)/(dx)` `= -(b^2x)/(a^2y)`

`text(At)\ \ P(a cos theta, b sin theta)`

`(dy)/(dx)` `= (b^2a cos theta)/(a^2b sin theta)`
  `= -(b cos theta)/(a sin theta)`

 

`:.\ text(Equation of tangent at)\ P`

`y − b sin theta=` ` -(b cos theta)/(a sin theta)(x − a cos theta)`
`a sin theta y − ab sin^2 theta=` ` −b cos theta x + ab cos^2 theta`
`b cos theta x + a sin theta y=` ` ab(sin^2 theta + cos^2 theta)`
`:.b cos theta x + a sin theta y=` `ab,\ \ \ text(or)`
`(x cos theta)/a + (y sin theta)/b=` `1`

 

`text(Alternative Solution)`

`dy/dx` `=(dy)/(d theta) xx (d theta)/(dx)`
  `=b cos theta xx 1/(-a sin theta)`
  `= -(bcos theta)/(a sin theta)`

 

`text{(then use the point-gradient formula as shown above)}`

 

(iv)  `text(Finding the tangent to)\ x^2 + y^2 = 1\ text(at)\ \ A(a cos theta, a sin theta)`

`2x + 2y* (dy)/(dx)` `=0`
`(dy)/(dx)` `= (-x)/y`

 

`:.\ text(Equation of tangent)`

`y – a sin theta` `=-(a cos theta)/(a sin theta)(x − a cos theta)`
`y sin theta – a sin^2 theta` `=-x cos theta + a cos^2 theta`
`x cos theta + y sin theta` `=a(sin^2 theta + cos^2 theta)`
`x cos theta + y sin theta` `=a`
`:.y sin theta` `=a- x cos theta`

 

`text(Substitute into the equation of the tangent at)\ \ P`

`b cos theta x + a sin theta y` `=ab`
`b cos theta x+a(a- x cos theta)` `=ab`
`bx cos theta + a^2 – ax cos theta`  `=ab`
`(b – a)x cos theta` `=a(b – a)`
`x cos theta` `= a`

 

`text(When)\ x cos theta = a,\ \ y sin theta = a − a = 0\ \ =>y=0,\ \ theta≠0`

`:.\ text(Intersection occurs on the)\ x text(-axis.)`

Mechanics, EXT2 2010 HSC 4b

A bend in a highway is part of a circle of radius  `r`, centre  `O`. Around the bend the highway is banked at an angle  `α`  to the horizontal.

A car is travelling around the bend at a constant speed  `v`. Assume that the car is represented by a point  `P`  of mass  `m`. The forces acting on the car are a lateral force  `F`, the gravitational force  `mg`  and a normal reaction  `N`  to the road, as shown in the diagram.

Mechanics, EXT2 2010 HSC 4b

  1. By resolving forces, show that
    `F = mg sin α − (mv^2)/r cos α`.   (3 marks)
  2. Find an expression for  `v`  such that the lateral force  `F`  is zero.   (1 mark)
Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `sqrt(rg tan α)`
Show Worked Solution
(i) 

Mechanics, EXT2 2010 HSC 4b Answer
`text(Resolving forces vertically)`
`N cos α + F sin α=` `mg\ \ \ \ …\ (1)`
`text(Resolving forces horizontally)`
`N sin α − F cos α=` `(mv^2)/r\ \ \ \ …\ (2)`
`text{Multiply (1) by  sin α  and  (2) by  cos α}`
`N sin α\ cos α + F sin^2 α=` `mg sin α\ \ \ \ …\ (3)`
`N sin α\ cos α − F cos^2 α=` `(mv^2)/r cos α\ \ \ \ …\ (4)`
`text{Subtract  (3) – (4)}`
`F sin^2 α + F cos^2 α=` `mg sin α − (mv^2)/r cos α`
`:.F=` `mg sin α − (mv^2)/r cos α`

  

(ii)  `F = mg sin α − (mv^2)/r cos α`

`mg sin α − (mv^2)/r cos α` `=0`
`(v^2)/r cos α` `=g sin α`
`v^2` `=rg tan α`
`v` `=sqrt(rg tan α)`

Graphs, EXT2 2010 HSC 4a

  1. A curve is defined implicitly by  `sqrtx + sqrty = 1`.
  2. Use implicit differentiation to find  `(dy)/(dx)`.   (2 marks)
  3. Sketch the curve  `sqrtx + sqrty = 1`.   (2 marks)

  4. Sketch the curve  `sqrt(|\ x\ |) + sqrt(|\ y\ |) = 1`   (1 mark)

 

Show Answers Only
  1. `- sqrty/sqrtx`
  2. `text(See Worked Solutions.)`
  3. `text(See Worked Solutions.)`
Show Worked Solution
(i)   `sqrtx + sqrt y` `= 1`
  `1/(2sqrtx) + 1/(2sqrty)*(dy)/(dx)` `= 0`
  `:.(dy)/(dx)` `= – sqrty/sqrtx`

 

(ii)    Graphs, EXT2 2010 HSC 4ai

 

♦♦ Mean mark part (iii) 46%.
(iii)   Graphs, EXT2 2010 HSC 4aii

Conics, EXT2 2010 HSC 3d

The diagram shows the rectangular hyperbola  `xy = c^2`, with  `c > 0`.

Conics, EXT2 2010 HSC 3d

The points  `A(c, c)`, `R(ct, c/t)`  and  `Q(-ct, -c/t)`  are points on the hyperbola,
with  `t ≠ ±1`.

  1. The line  `l_1`  is the line through  `R`  perpendicular to  `QA`.
  2. Show that the equation of  `l_1`  is
    1. `y = -tx + c(t^2 + 1/t)`.   (2 marks)

  4. The line  `l_2`  is the line through  `Q`  perpendicular to  `RA`.
  5. Write down the equation of  `l_2`.   (1 mark)
  6. Let  `P`  be the point of intersection of the lines  `l_1`  and  `l_2`.
  7. Show that  `P`  is the point  `(c/(t^2), ct^2)`.   (2 marks)
  8. Give a geometric description of the locus of  `P`.   (1 mark)
Show Answers Only
  1. `text{Proof (Show Worked Solutions)}`
  2. `y = tx + c(t^2 − 1/t)`
  3. `text{Proof (Show Worked Solutions)}`
  4. `text{Proof (Show Worked Solutions)}`
Show Worked Solution
(i) 

Conics, EXT2 2010 HSC 3d Answer1
`m_(QA)` `= (c + c/t)/(c + ct)`
  `= (1 + 1/t)/(1 + t)`
  `= (t + 1)/(t(1 + t))`
  `= 1/t`

 

 `:.\ text(Gradient of perpendicular to)\ QA\ \ (l_1) = -t`

`:.text(Equation of)\ \ l_1`

`y − c/t` `= -t(x − ct)`
`y` `= -tx + ct^2 + c/t`
  `= -tx + c(t^2 + 1/t)\ \ \ …\ (1)`

 

(ii)   `m_(RA)` `= (c − c/t)/(c − ct)`
    `= (1 − 1/t)/(1 − t)`
    `= (t − 1)/(t(1 − t))`
    `= -1/t`

`:.m\ text(of)\ l_2 = t`

 

`:.\ text(Equation of)\ l_2`

`y + c/t` `= t(x + ct)`
`y` `= tx + ct^2 – c/t`
  `= tx + c(t^2 – 1/t)\ \ \ …\ (2)`

 

 

(iii)  `text{Solving (1) and (2) simultaneously}`

`-tx + c(t^2 + 1/t)` `= tx + c(t^2 − 1/t)`
`ct^2+c/t-ct^2+c/t` `=2tx`
`(2c)/t` `= 2tx`
`x` `= c/(t^2)`

`text{Substitute into (2)}`

`y` `= (ct)/(t^2) + ct^2 − c/t`
  ` = ct^2`

 

`:.P\ text(has coordinates)\ (c/(t^2), ct^2)`

 

(iv)  `text(Using the coordinates of)\ P`

♦♦ Mean mark part (iv) 1%.
MARKER’S COMMENT: Most students found the correct locus, although very few indicated that it was only the branch in the 1st quadrant.

`=>t^2 = c/x,\ \ t^2 = y/c`

`y/c` `= c/x`
`:.xy` `=c^2`

 

`text(S)text(ince the parameter of)\ P\ text(is)\ t^2`

`=>\ text(the locus is in the first quadrant.)`

 

`:.\ text(The locus of)\ P\ text(is the branch of the rectangular hyperbola)`

`xy = c^2\ text{in the first quadrant (excluding}\ A, t≠±1 text{)}`

Complex Numbers, EXT2 N2 2010 HSC 2c

Sketch the region in the complex plane where the inequalities  `1 ≤ |\ z\ | ≤ 2`  and  `0 ≤ z + bar z ≤ 3`  hold simultaneously.  (2 marks)

Show Answers Only

`text(See Worked Solutions.)`

Show Worked Solution
`text(Consider)\ \ \ \ ` `1≤|\ z\ |≤2`
  `1≤x^2+y^2≤4`

 
`z + bar z = (x+iy)+(x-iy)=2x`

`text(Consider)\ \ \ \ ` `0≤z + bar z≤3`
  `0≤2x≤3`
  `0≤x≤3/2`

 
Complex Numbers, EXT2 2010 HSC 2c

Complex Numbers, EXT2 N1 2010 HSC 2b

  1. Express  `-sqrt3 − i`  in modulus–argument form.  (2 marks)

  2. Show that  `(-sqrt3 − i)^6`  is a real number.  (2 marks)

Show Answers Only
  1. `2text(cis)(-(5pi)/6)`
  2. `-64`
Show Worked Solution
i.    `|-sqrt3 − i\ |` `=sqrt((-sqrt3)^2+sqrt((-1)^2))`
    `=2`

Complex Numbers, EXT2 2010 HSC 2b 

`text(From the graph)`

`text{arg}(-sqrt3-i)=- (5pi)/6\ \ \ \ text{(for}\  –pi<theta<pi text{)}`

`:.-sqrt3 − i= 2text(cis)(-(5pi)/6)`

 

`text{Alternative Solution (to find the argument)}`

`-sqrt3-i` `= 2(- sqrt3/2 − 1/2 i)`
  `=2text(cis)(-(5pi)/6)`

 

ii.   `(-sqrt3 − i)^6` `= [2text(cis)(-(5pi)/6)]^6`
    `=2^6[cos((-5pi)/6 xx6) +i sin((-5 pi)/6 xx6)]\ \ \ \ text{(De Moivre)}`
    `= 2^6[cos(-5pi) + i sin(-5pi)]`
    `= 64(-1 + 0i)`
    `= -64`

Calculus, EXT2 C1 2010 HSC 1d

Using the substitution  `t = tan\ x/2`, or otherwise, evaluate  `int_0^(pi/2) (dx)/(1 + sin\ x)`.   (4 marks)

Show Answers Only

`1`

Show Worked Solution

`t = tan\ x/2, \ dx = (2\ dt)/(1 + t^2), \ sin\ x = (2t)/(1 + t^2)`

`text(When)\ \ x=pi/2,\ \ t=tan\ pi/4=1`

`text(When)\ \ x=0,\ \ t=tan0=0`

`:.int_0^(pi/2) (dx)/(1 + sin\ x)` `=int_0^1 ((2\ dt)/(1 + t^2))/(1 + (2t)/(1 + t^2))`
  `=int_0^1 2/(1 + t^2 + 2t)\ dt`
  `=int_0^1 (2)/((1 + t)^2)\ dt`
  `=[(-2)/(1 + t)]_0^1`
  `=(-2)/2 − (-2)/1`
  `=1`

Calculus, EXT2 C1 2010 HSC 1b

Evaluate  `int_0^(pi/4) tan\ x\ dx`.   (3 marks) 

Show Answers Only

`1/2 ln 2 \ \ text(or)\ \ ln\ sqrt2`

Show Worked Solution
`int_0^(pi/4) tan\ x\ dx` `=int_0^(pi/4) (sin\ x)/(cos\ x)\ dx`
  `=[-ln\ cos\ x]_0^(pi/4)`
  `=[-ln\ cos\ pi/4 – (-ln cos 0)]`
  `=-ln\ 1/sqrt2 + ln\ 1`
  `=ln sqrt2`
  `=1/2 ln 2`

Calculus, EXT2 C1 2011 HSC 7b

Let   `I = int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx.`

  1. Use the substitution  `u = 4-x`  to show that
     
          `I = int_1^3 (sin^2(pi/8 u))/(u(4-u))\ du.`  (2 marks)

     

  2. Hence, find the value of  `I`.  (3 marks)

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

i.   `text(Let)\ \ u = 4-x,\ \ du = -dx,\ \ x = 4-u`

`text(When)\ \ x = 1,\ \ u = 3`

`text(When)\ \ x = 3,\ \ u = 1`

`I` `=int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx`
  `=int_3^1 (cos^2 (pi/8(4-u)))/((4-u)u) (-du)`
  `=-int_3^1 (cos^2(pi/2-pi/8 u))/((4-u)u)\ du`
  `=int_1^3 (sin^2(pi/8 u))/(u (4-u))\ du`

 

ii.   `text{S}text{ince}\ \ int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx = int_1^3 (sin^2(pi/8 x))/(x(4-x))\ dx`

 

`text(We can add the integrals such that)`

`2I` `=int_1^3 (cos^2(pi/8 x))/(x(4-x)) dx + int_1^3 (sin^2(pi/8 x))/(x(4-x)) dx`
  `=int_1^3 1/(x(4-x)) dx`

 

`text(Using partial fractions:)`

♦ Mean mark 36%.
`1/(x(4-x))` `=A/x+B/(4-x)`
`1` `=A(4-x)+Bx`

 
`text(When)\ \ x=0,\ \ A=1/4`

`text(When)\ \ x=0,\ \ B=1/4`

`2I` `=1/4 int_1^3 (1/x + 1/(4-x))\ dx`
  `=1/4 [log_e x-log_e (4-x)]_1^3`
  `=1/4 [log_e 3-log_e 1-(log_e 1-log_e 3)]`
  `=1/2 log_e 3`
`:.I` `=1/4 log_e 3`

Mechanics, EXT2 M1 2011 HSC 6a

Jac jumps out of an aeroplane and falls vertically. His velocity at time  `t`  after his parachute is opened is given by  `v(t)`, where  `v(0) = v_0`  and  `v(t)`  is positive in the downwards direction. The magnitude of the resistive force provided by the parachute is  `kv^2`, where  `k`  is a positive constant. Let  `m`  be Jac’s mass and  `g`  the acceleration due to gravity. Jac’s terminal velocity with the parachute open is  `v_T.`

Jac’s equation of motion with the parachute open is

`m (dv)/(dt) = mg - kv^2.`   (Do NOT prove this.)

  1. Explain why Jac’s terminal velocity  `v_T`  is given by  
     
         `sqrt ((mg)/k).`  (1 mark)

  2. By integrating the equation of motion, show that  `t`  and  `v`  are related by the equation
     
         `t = (v_T)/(2g) ln[((v_T + v)(v_T - v_0))/((v_T - v)(v_T + v_0))].`  (3 marks)

  3. Jac’s friend Gil also jumps out of the aeroplane and falls vertically. Jac and Gil have the same mass and identical parachutes.

     

    Jac opens his parachute when his speed is  `1/3 v_T.` Gil opens her parachute when her speed is  `3v_T.` Jac’s speed increases and Gil’s speed decreases, both towards  `v_T.`

     

    Show that in the time taken for Jac's speed to double, Gil's speed has halved.  (3 marks)

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

i.   `m (dv)/(dt) = mg – kv^2`

`text(As)\ \ v ->text(terminal velocity,)\ \ (dv)/(dt) -> 0`

`mg – kv_T^2` `= 0`
`v_T^2` `= (mg)/k`
`v_T` `= sqrt ((mg)/k)`

  

ii.  `m (dv)/(dt) = mg – kv^2`

`int_0^t dt` `=int_(v_0)^v m/(mg – kv^2)\ dv`
`t` `= m/k int_(v_0)^v (dv)/((mg)/k – v^2)\ \ \ \ text{(using}\ \ v_T^2= (mg)/k text{)}`
  `= (v_T^2)/g int_(v_0)^v (dv)/(v_T­^2 – v^2)`

  

♦ Mean mark part (ii) 39%.

`text(If)\ \ v_0 < v_T,\ \ text(then)\ \ v < v_T\ \ text(at all times.)`

`:. t` `=(v_T^2)/g int_(v_0)^v (dv)/(v_T­^2 – v^2)`
  `=(v_T^2)/g int_(v_0)^v (dv)/((v_T – v)(v_T + v))`
  `=(v_T^2)/(2 g v_T) int_(v_0)^v (1/((v_T – v)) + 1/((v_T + v))) dv`
  `=v_T/(2g)[-ln (v_T – v) + ln (v_T + v)]_(v_0)^v`
  `=v_T/(2g)[ln(v_T + v)-ln(v_T – v)-ln(v_T + v_0)+ln(v_T – v_0)]`
  `=v_T/(2g) ln[((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

 

`text(If)\ \ v_0 > v_T,\ \ text(then)\ \ v > v_T\ \ text(at all times.)`

`text(Replace)\ \ v_T – v\ \ text(by) – (v – v_T)\ \ text(in the preceeding)`

`text(calculation, leading to the same result.)`

 

iii.  `text{From (ii) we have}:\ \ t = v_T/(2g) ln [((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

`text(For Jac,)\ \ v_0 = V_T/3\ \ text(and we have to find the time)`

`text(taken for his speed to be)\ \ v = (2v_T)/3.`

`t` `=v_T/(2g) ln[((v_T + (2v_T)/3)(v_T – v_T/3))/((v_T – (2v_T)/3)(v_T + v_T/3))]`
  `=v_T/(2g) ln [(((5v_T)/3)((2v_T)/3))/((v_T/3)((4v_T)/3))]`
  `=v_T/(2g) ln[(10/9)/(4/9)]`
  `=v_T/(2g) ln (5/2)`

 

`text(For Gil),\ \ v_0 = 3v_T\ \ text(and we have to find)` 

♦♦♦ Mean mark part (iii) 16%.

`text(the time taken for her speed to be)\ \ v = (3v_T)/2.`

`t` `=v_T/(2g) ln [((v_T + (3v_T)/2)(v_T – 3v_T))/((v_T – (3v_T)/2)(v_T + 3v_T))]`
  `=v_T/(2g) ln [(((5v_T)/2)(-2v_T))/((-v_T/2)(4v_T))]`
  `=v_T/(2g) ln [(-5)/-2]`
  `=v_T/(2g) ln (5/2)`

 

`:.\ text(The time taken for Jac’s speed to double is)`

`text(the same as it takes for Gil’s speed to halve.)`

Mechanics, EXT2 2011 HSC 5a

A small bead of mass  `m`  is attached to one end of a light string of length  `R`. The other end of the string is fixed at height  `2h`  above the centre of a sphere of radius  `R`, as shown in the diagram. The bead moves in a circle of radius  `r`  on the surface of the sphere and has constant angular velocity  `omega > 0`. The string makes an angle of  `theta`  with the vertical.

Three forces act on the bead: the tension force  `F`  of the string, the normal reaction force  `N`  to the surface of the sphere, and the gravitational force  `mg`.

  1. By resolving the forces horizontally and vertically on a diagram, show that
    1. `F sin theta - N sin theta = m omega^2 r`
  2. and
    1. `F cos theta + N cos theta = mg.`  (2 marks)
  3. Show that
    1. `N = 1/2 mg sec theta - 1/2 m omega^2 r\ text(cosec)\ theta.`  (2 marks)
  4. Show that the bead remains in contact with the sphere if  
    1. `omega <= sqrt (g/h).`  (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)  

`text(Resolving forces horizontally)`

`text(Net force)` `= m r omega^2\ \ \ text{(towards the centre of the circle)}`
`F sin theta – N sin theta` `= mr omega^2\ \ \ \ text{… (1)}`
♦ Mean mark part (i) 50%.

 

`text(Resolving forces vertically)`

`text(Net force)` `= mg\ \ \ \ text{(gravitational force)}`
`F cos theta + N cos theta` `= mg\ \ \ \ text{… (2)}`

 

(ii)      `text{Divide (1) by}\ \ sin theta`

`F – N = mr omega^2\ text(cosec)\ theta\ \ \ \ text{… (3)}`

`text{Divide (2) by}\ \ cos theta`

`F+N = mg sec theta\ \ \ \ text{… (4)}`

`text{Subtract (4) – (3)}`

`2N` `= mg sec theta – mr omega^2\ text(cosec)\ theta`
`:.N` `= 1/2 mg sec theta – 1/2 mr omega^2\ text(cosec)\ theta`

 

(iii)  `text(When in contact with the sphere,)\ \ N >= 0.`

`1/2 mg sec theta – 1/2 mr omega^2` `>= 0`
`1/2 mg sec theta` `>= 1/2 mr omega^2\ text(cosec)\ theta`
`g sec theta` `>= r omega^2\ text(cosec)\ theta`
`omega^2` `<= (g sec theta)/(r\ text(cosec)\ theta)`
  `<= g/r tan theta,\ \ \ \ (text{since}\ \ tan theta = r/h)`
  `<= g/r xx r/h`
  `<=g/h`
`:. omega` `<= sqrt (g/h)`

Harder Ext1 Topics, EXT2 2011 HSC 4b

In the diagram,  `ABCD`  is a cyclic quadrilateral. The point  `E`  lies on the circle through the points `A, B, C`  and  `D`  such that  `AE\ text(||)\ BC`. The line  `ED`  meets the line  `BA`  at the point  `F`. The point  `G`  lies on the line  `CD`  such that  `FG\ text(||)\ BC.`

Copy or trace the diagram into your writing booklet.

  1. Prove that  `FADG`  is a cyclic quadrilateral.  (2 marks)
  2. Explain why  `/_ GFD =/_ AED.`  (1 mark)
  3. Prove that  `GA`  is a tangent to the circle through the points  `A, B, C`  and  `D.`  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `/_ GFD = /_ AED\ \ text{(alternate angles,}\ \ FG\ text(||)\ AE text{)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)  
`text(Let)\ /_ BCD` `= alpha`
`/_ FAD` `= alpha\ \ text{(exterior angle of a cyclic quadrilateral}\ ABCD text{)}`
`/_ FGC` `= pi – alpha\ \ text{(cointerior angles,}\ \ FG\ text(||)\ BC text{)}`

 `:.\ \ /_ FAD + /_ FGD = pi`

`:.\ FADG\ \ text{is a cyclic quadrilateral  (opposite angles are supplementary)}`

 

(ii)  `/_ GFD = /_ AED\ \ text{(alternate angles,}\ \ FG\ text(||)\ AE text{)}`

 

(iii)   `text(Join)\ GA`

`/_ GAD = /_ GFD\ \ text{(angles in the same segment on arc}\ GDtext{)}`

`text(S)text(ince)\  /_ GFD` `= /_ AED\ \ \ text{(part (ii))}`
`/_ GAD` `= /_ AED`

 

`:.GA\ \ text(is a tangent to the circle through)\ \ A, B, C and D.`

`text{(angle in the alternate segment equals the angle}`

`text(between)\ GA\ text(and chord)\ AD text{).}`

Conics, EXT2 2011 HSC 3d

The equation  `x^2/16 - y^2/9 = 1`  represents a hyperbola.

  1. Find the eccentricity  `e.`  (1 mark)
  2. Find the coordinates of the foci.  (1 mark)
  3. State the equations of the asymptotes.  (1 mark)
  4. Sketch the hyperbola.  (1 mark)
  5. For the general hyperbola  
  6. `x^2/a^2 - y^2/b^2 = 1`,
  8. describe the effect on the hyperbola as  `e -> oo.`  (1 mark)

 

Show Answers Only
  1. `e = 5/4`
  2. `text(Foci) (+-5, 0)`
  3. `y = +- 3/4 x`
  5. `text(The hyperbola approaches the)\ \ y text(-axis from both sides.)`
Show Worked Solution

(i)   `x^2/16 – y^2/9 = 1,\ \ \ =>a^2 = 16,\ \ \ b^2 = 9`

`b^2` `= a^2 (e^2 – 1)`
`9` `= 16 (e^2 – 1)`
`e^2` `= 1 + 9/16 `
  `= 25/16`
`:. e` `= 5/4`

 

 

(ii)   `S (ae, 0),\ \ S prime (-ae, 0)`

`S(5, 0),\ \ S prime (–5, 0)`

 

(iii)  `y = +- b/a x`

`y = +- 3/4 x`

 

(iv)   HSC 2011 3di

 

♦♦♦ Mean mark part (v) 24%.

 

(v)   `text(As)\ \ e -> oo,\ \ b -> oo\ \ text(and the asymptotes approach the)`

`y text{-axis from both sides (i.e. the gradients of the asymptotes →∞)}`

` text(Thus the hyperbola approaches the)\ \ y text(-axis from both sides.)`