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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.   \(f(\abs{x})\).`   (2 marks)

  3.    \(f(x)-x\).   (2 marks)

Show Answers Only

i.   


ii.       
       

iii.
           

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))`
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`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.)`

Functions, EXT1′ F1 2011 HSC 3a

  1.  Draw a sketch of the graph
     
    `quad y = sin\ pi/2 x`   for  `0 < x < 4.`    (1 mark)

  2.  Find  `lim_(x -> 0) x/(sin\ pi/2 x).`    (1 mark)

  3.  Draw a sketch of the graph
     
         `quad y = x/(sin\ pi/2 x)`  for  `0 < x < 4.`    (2 marks)

  4.  

    (Do NOT calculate the coordinates of any turning points.)  

Show Answers Only

i. 

ii.  `2/pi`

iii. 

Show Worked Solution
i.  

 

ii.   `lim_(x->0) x/(sin\ pi/2 x)` `= 2/pi lim_(x->0) (pi/2 x)/(sin\ pi/2 x)`
  `= 2/pi xx 1`
  `= 2/pi`

 

iii.   

Complex Numbers, EXT2 N2 2011 HSC 2d

  1. Use the binomial theorem to expand  `(cos theta + i sin theta)^3.`  (1 mark)

  2. Use de Moivre’s theorem and your result from part (i) to prove that
     
        `cos^3 theta = 1/4 cos 3 theta + 3/4 cos theta.`  (3 marks)

  3. Hence, or otherwise, find the smallest positive solution of
     
        `4 cos^3 theta - 3 cos theta = 1.`  (2 marks)

Show Answers Only
  1. `cos^3 theta + 3 i cos^2 theta sin theta – 3 cos theta sin^2 theta – i sin^3 theta`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `theta = (2 pi)/3`
Show Worked Solution

i.   `(cos theta + i sin theta)^3`

`=sum_(k=0)^3 \ ^3C_k (cos theta)^(3-k) (i sin theta)^k`

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

`= cos^3 theta + 3 i cos^2 theta sin theta- 3 cos theta sin^2 theta – i sin^3 theta`

 

ii.  `text(Using De Moivre’s Theorem)`

`(cos theta + i sin theta)^3 = cos 3 theta + i sin 3 theta`
 

`text(Equate real parts)`

`cos 3 theta` `= cos^3 theta – 3 cos theta sin^2 theta`
`cos 3 theta` `= cos^3 theta – 3 cos theta (1 – cos^2 theta)`
`cos 3 theta` `= 4 cos^3 theta – 3 cos theta`
`4 cos^3 theta`  `=cos 3 theta+3cos theta`
`:.cos^3 theta` `= 1/4 cos 3 theta + 3/4 cos theta\ \ \ text(… as required)`

 

iii.   `text(If)\ \ \ 4 cos^3 theta – 3 cos theta = 1`

`=>cos 3 theta = 1\ \ \ \ text{(from part (ii))}`

`3 theta` `= 2 k pi`
`:. theta` `= (2 k pi)/3`

 

`:.\ text(Smallest positive solution occurs when)`

`theta = (2 pi)/3\ \ \ \ text{(i.e. when}\ k = 1 text{)}`

Complex Numbers, EXT2 N2 2011 HSC 2c

Find, in modulus-argument form, all solutions of  `z^3 = 8.`  (2 marks)

Show Answers Only

`2text(cis)\ 0,\ 2text(cis)\ (2pi)/3,\ 2text(cis)((-2pi)/3)`

Show Worked Solution

`z^3 = 8`

`z^3 = 8 [cos (2 k pi) + i sin (2 k pi)],\ \ k=0, +-1, +-2`

`z = 2[cos ((2 k pi)/3) + i sin ((2 k pi)/3)]\ \ \ text{(De Moivre)}`
 

`text(When)\ \ k = 0,`

`z` `= 2 (cos 0 + i sin 0)`
  `= 2text(cis)\ 0`

 

`text(When)\ \ k = 1,`

`z` `= 2 [cos ((2 pi)/3) + i sin((2 pi)/3)]`
  `=2text(cis)\ (2pi)/3`

 
`text(When)\ \ k = -1,`

`z` `= 2 [cos ((-2 pi)/3) + i sin ((-2 pi)/3)]`
  `=2 text(cis)((-2 pi)/3)`

Complex Numbers, EXT2 N2 2011 HSC 2b

On the Argand diagram, the complex numbers  `0, 1 + i sqrt 3 , sqrt 3 + i`  and  `z`  form a rhombus.
 


 

  1. Find  `z`  in the form  `a + ib`, where  `a`  and  `b`  are real numbers.  (1 mark)

  2. An interior angle, `theta`, of the rhombus is marked on the diagram.

     

    Find the value of `theta.`  (2 marks)

Show Answers Only
  1. `(1 + sqrt 3) + i(1 + sqrt 3)`
  2. `(5 pi)/6`
Show Worked Solution
i.   `z` `= 1 + i sqrt 3 + sqrt 3 + i`
  `= (1 + sqrt 3) + i (1 + sqrt 3)`

 

ii.   `text(arg)\ z = tan^-1 ((1 + sqrt 3)/(1 + sqrt 3)) = pi/4`

`text(arg)\ (sqrt 3 + i) = tan^-1 (1/sqrt 3) = pi/6`

`text(Difference) = pi/4 – pi/6 = pi/12`
 

`=>\ text(Opposite angles of a rhombus are equal)`

`=>\ text(The diagonals of a rhombus bisect the angles)`

`:.theta` `= pi – 2 xx pi/12`
  `= (5 pi)/6\ \ text{(angle sum of triangle)`

Calculus, EXT2 C1 2011 HSC 1d

Find  `int cos^3 theta\ d theta`  (3 marks)

Show Answers Only

`sin theta-(sin^3 theta)/3 + c`

Show Worked Solution
`int cos^3 theta\ d theta` `= int cos^2 theta cos theta\ d theta`
  `= int (1-sin^2 theta) cos theta\ d theta`
  `= int (cos theta-sin^2 theta cos theta) d theta`
  `= sin theta-(sin^3 theta)/3 + c`

Calculus, EXT2 C1 2011 HSC 1c

  1. Find real numbers `a, b` and `c` such that 
      
       `1/(x^2 (x - 1)) = a/x + b/x^2 + c/(x - 1).`  (2 marks)

  2. Hence, find  `int 1/(x^2 (x - 1))\ dx`  (2 marks)

Show Answers Only
  1. `a = −1,\ \ \ b = −1,\ \ \ c = 1`
  2. `log_e\ ((x – 1)/x) + 1/x + c`
Show Worked Solution

i.   `1/(x^2 (x – 1)) = a/x + b/x^2 + c/(x – 1)`

`1 = ax (x – 1) + b (x – 1) + cx^2`

`1 = ax^2 + cx^2 – ax + bx – b`

`1=(a+c)x^2+(b-a)x-b`
 

`text(Equating coefficients:)`

`=>0 = a + c,\ \ \ b – a = 0,\ \ \ -b = 1`

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

 

ii.   `int 1/(x^2 (x – 1)) \ dx` `= int(-1/x – 1/x^2 + 1/(x – 1))\ dx`
  `= -log_e x + 1/x + log_e (x – 1) + c`
  `= log_e\ ((x – 1)/x) + 1/x + c`

Polynomials, EXT2 2012 HSC 15b

Let `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`, where `k` is real.

Let  `α = x + iy`, where  `x`  and  `y`  are real.

Suppose that  `α`  and  `iα`  are zeros of  `P(z)`, where  `bar α ≠ iα`.

  1. Explain why  `bar α`  and  `-i bar α`  are zeros of  `P(z)`.   (1 mark)

  2. Show that  `P(z) = z^2(z − k)^2 + (kz − 1)^2`.   (1 mark)

  3. Hence show that if  `P(z)`  has a real zero then
    1. `P(z) = (z^2 + 1)(z+ 1)^2` or  `P(z) = (z^2 + 1)(z − 1)^2.`   (2 marks)

  4. Show that all zeros of  `P(z)`  have modulus `1`.   (2 marks)
  5. Show that  `k = x − y`.   (1 mark)
  6. Hence show that  `-sqrt2 ≤ k ≤ sqrt2`.   (2 marks) 
Show Answers Only
  1. `text{Proof (See Worked Solutions.)}`
  2. `text{Proof (See Worked Solutions.)}`
  3. `text{Proof (See Worked Solutions.)}`
  4. `text{Proof (See Worked Solutions.)}`
  5. `text{Proof (See Worked Solutions.)}`
  6. `text{Proof (See Worked Solutions.)}`
Show Worked Solution

(i)   `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz +1`

`P(α) = P(iα) = 0, \ \ bar α ≠ iα.`

`text(S)text(ince)\ \ P(z)\ text(has real coefficients,)`

`=>\ text(Its zeros occur in conjugate pairs.)`

`text(i.e.)\ \ P(α) = 0\ text(and)\ P(bar α) = 0`

`bar α` `= x − iy`
`bar(i α)` `=bar (i(x+iy))`
  `=bar (ix-y)`
  `=-y-ix`
  `=-i(x-iy)`
  `=-i barα`

 

`:. -i bar α\ text(is a zero of)\ P(z)\ text(as it is the conjugate of)\ iα.`

 

(ii)   `H(z)` `= z^2(z − k)^2 + (kz − 1)^2`
    `= z^2 (z^2 − 2kz + k^2) + (k^2z^2 − 2kz + 1)`
    `= z^4 − 2kz^3 + k^2z^2 + k^2z^2 − 2kz +1`
    `= z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`
    `= P(z)` 

 

(iii)  `text(If)\ z\ text(is real then)\ z^2(z − k)^2\ text(and)\ (kz −1)^2\ text(are real and)`

`text(either positive or zero.)`

♦♦♦ Mean mark part (iii) 11%.

`:.text(If)\ \ P(z) = z^2(z − k)^2 + (kz − 1)^2 = 0`

`=>(z − k)^2 = 0\ text(and)\ (kz − 1)^2 = 0`

`text(When)\ \ z = k,\ \ \ k^2 − 1 = 0\ \ text(or)\ k = ±1.`

`text(If)\ k = 1`

`P(z)` `= z^2(z − 1)^2 + (z − 1)^2`
  `= (z^2 + 1)(z − 1)^2.`

`text(If)\ k = -1`

`P(z)` `= z^2(z + 1)^2 + (-z − 1)^2`
  `= (z^2 + 1)(z + 1)^2`

  

(iv)  `text(Product of the roots) = e/a=1`

♦♦♦ Mean mark part (iv) 20%.
`:. α *bar α *iα *(-i barα)`  `=1`
`(α bar α)^2` `=1`
`(|α|)^4` `=1`
`|α|` `=1`

`:.\ text(All zeros have modulus 1.)`

 

(v)  `text(Sum of zeroes)\ = -b/a=-(-2k)/1 = 2k`

♦♦♦ Mean mark part (v) 20%.
`:. 2k` `=α + bar α + iα + (-i bar α )`
  `=x + iy + x − iy + (-y + ix) − i(x − iy)`
  `=2x − y + ix − ix − y`
  `=2x − 2y`
`:. k` `=x-y`

 

 

(vi)  `text(S)text(ince)\ \ |α| = 1\ \ \ text{(part (iv))}`

♦♦♦ Mean mark part (vi) 2%.
`=>x^2 + y^2` `=1`
`text(Substitute)\ \ y=x-k\ \ \ text{(part (v))}`
`x^2 + (x − k)^2` `=1`
`2x^2 − 2kx + k^2 − 1` `=0`

 

`text(For a real solution to exist), Δ ≥ 0`

`4k^2 − 8(k^2 − 1) ` `≥ 0`
`-4k^2 + 8` `≥ 0`
`k^2` `≥ 2`

 

`:. -sqrt2 ≤ k ≤ sqrt2`

Integration, EXT2 2012 HSC 14a

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

Show Answers Only

` 2log_e x + 1/2log_e(x^2 + 4) + c` 

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

 `a + b = 3, \ c = 0, \ 4a = 8`

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

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

Harder Ext1 Topics, EXT2 2012 HSC 13b

The diagram shows  `ΔS′SP`. The point  `Q`  is on  `S′S` so that  `PQ`  bisects  `∠S′PS`. The point  `R`  is on  `S′P`  produced so that  `PQ\ text(||)\ RS`.

Harder Ext1 Topics, EXT2 2012 HSC 13b 

  1. Show that  `PS = PR`.   (1 mark)
  2. Show that  `(PS)/(QS) = (PS′)/(QS′)`.   (2 marks)

 

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution
(i)   `∠S′PQ` `= ∠PRS = α` `text{(corresponding angles,}\ QP\ text(||)\ SR)`
  `∠QPS` `= ∠PSR = α` `text{(alternate angles,}\ QP\ text(||)\ SR)`
  `:. ∠PSR` `= ∠PRS = α`  

 

`:. ΔPSR\ text{is isosceles (two angle equal)}`

`:. PS = PR\ \ \ text{(opposite equal angles in}\ Delta PSR text{)}`

 

(ii)  `text(Draw a line through)\ S′\ text(parallel to)\ QP`

Harder Ext1 Topics, EXT2 2012 HSC 13b Answer

`:.(PS′)/(PR)` `= (QS′)/(QS)\ \ \ text{(parallel lines preserve ratios)}`
`(PS′)/(QS′)` `= (PR)/(QS)`
`text(S)text(ince)\ \ PS = PR\ \ \ text{(from part (i))}`
`(PS′)/(QS′)` `= (PS)/(QS)\ \ \ text(… as required)`

Mechanics, EXT2 M1 2012 HSC 13a

An object on the surface of a liquid is released at time  `t = 0`  and immediately sinks. Let  `x`  be its displacement in metres in a downward direction from the surface at time  `t`  seconds.

The equation of motion is given by

`(dv)/(dt) = 10 − (v^2)/40`,

where  `v`  is the velocity of the object.  

  1. Show that  `v = (20(e^t − 1))/(e^t + 1)`.   (4 marks)

  2. Use  `(dv)/(dt) = v (dv)/(dx)`  to show that  
     
         `x = 20\ log_e(400/(400 − v^2))`   (2 marks)

     

  3. How far does the object sink in the first 4 seconds?   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `text(See Worked Solutions.)`
  3. `40log_e((e^4 + 1)/(2e^2))\ text(m)`
Show Worked Solution
i.   `(dv)/(dt)` `= 10 − (v^2)/40`
  `(dv)/(dt)` `= (400 − v^2)/40`
  `dt` `= 40/(400 −v^2)\ dv`
  `int dt` `=int 40/(400 −v^2)\ dv`
  `t` `= int (1/(20 + v) + 1/(20 − v))\ dv`
    `= log_e(20 + v) − log_e(20 − v) + c`

 

`text(When)\ \ t = 0, v = 0\ \ => \ c = 0`

`t=` ` log_e((20 + v)/(20 − v))`
`e^t=` ` (20 + v)/(20 − v)`
`20e^t-ve^t=` ` 20 + v`
`v+ ve^t=` ` 20e^t − 20`
`v(1+e^t)=` `20(e^t − 1)`
`v=` ` (20(e^t − 1))/(e^t + 1)\ \ \ \ text(… as required)`

 

ii.   `v (dv)/(dx)` `= 10 − (v^2)/40`
  `(40v\ dv)/(400 − v^2)` `= dx`
`int dx` `= int (40v)/(400 − v^2)\ dv`
 `x` `= -20log_e(400 − v^2) + c`

 

`text(When)\ \ x = 0, v = 0\ \ \ => c = 20log_e 400`

`:.x` `= 20log_e400 − 20log_e(400 − v^2)`
  `= 20log_e((400)/(400 − v^2))\ \ \ \ text(… as required)`

 

iii.  `text(When)\ \ t = 4,\  v = (20(e^4 − 1))/(e^4 + 1)`

`x` `= 20log_e[400/(400 −((20(e^4 − 1))/(e^4 + 1))^2)]`
  `= 20log_e[((e^4 + 1)^2)/((e^4 + 1)^2 − (e^4 − 1)^2)]`
  `= 20log_e(((e^4 + 1)^2)/((e^4+1 + e^4 − 1)(e^4 + 1 − e^4 + 1)))`
  `= 20log_e(((e^4 + 1)^2)/(4e^4))`
  `= 40log_e\ (e^4 + 1)/(2e^2)\ \ text(metres)`

Calculus, EXT2 C1 2012 HSC 12c

For every integer  `n ≥ 0`  let  `I_n = int_1^(e^2)(log_e x)^n\ dx`.

Show that for  `n ≥ 1,`
 
     `I_n = e^2 2^n − nI_(n − 1)`.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`I_n = int_1^(e^2)(log_e x)^n\ dx`

`u` `= (log_e x)^n,` `v′` `= 1`
`u′` `= n (log_e x)^(n − 1) xx 1/x` `v` `= x`
`I_n` `=int_1^(e^2) 1*(ln x)^n\ dx`
  `= [x (log_e x)^n]_1^(e^2) − int_1^(e^2)n/x(log_e x)^(n − 1) xx x\ dx`
  `= [e^2(log_e e^2)^n − 1*ln\ 1] − n int_1^(e^2)(log_e x)^(n − 1)\ dx`
  `= [e^2(2 log_ee)^n-0 ]− nI_(n − 1)`
 
`= e^2 2^n − nI_(n − 1)`

Conics, EXT2 2012 HSC 12b

The diagram shows the ellipse  `(x^2)/(a^2) + (y^2)/(b^2) = 1`  with  `a > b`. The ellipse has focus  `S`  and eccentricity  `e`. The tangent to the ellipse at  `P(x_0, y_0)`  meets the `x`-axis at  `T`. The normal at  `P`  meets the `x`-axis at  `N`.  

Conics, EXT2 2012 HSC 12b

  1. Show that the tangent to the ellipse at  `P`  is given by the equation
    1. `y − y_0 = -(b^2x_0)/(a^2y_0)(x − x_0)`.   (2 marks) 
  2. Show that the `x`-coordinate of  `N`  is  `x_0e^2`.   (2 marks)
  3. Show that  `ON xx OT = OS^2`  (2 marks) 
Show Answers Only

(i)   `text(See Worked Solutions.)`

(ii)  `text(See Worked Solutions.)`

(iii)  `text(See Worked Solutions.)`

Show Worked Solution

(i)   `(x^2)/(a^2) + (y^2)/(b^2) = 1`

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

 

`text(At)\ P(x_0, y_0),\ \ m_(tan)= -(b^2x_0)/(a^2y_0)`

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

`y − y_0 = -(b^2x_0)/(a^2y_0)(x − x_0)`

 

(ii)  `text(At)\ P(x_0, y_0),\ \ m_(norm) = (a^2y_0)/(b^2x_0)`

`:.text(Equation of normal at)\ P\ text(is)`

`y − y_0 = (a^2y_0)/(b^2x_0)(x − x_0)`

`N\ \ text(occurs when)\ \ y=0`

`0-y_0=` ` (a^2y_0)/(b^2x_0)(x − x_0)`
`-b^2x_0y_0=` ` a^2y_0x − a^2x_0y_0`
`a^2y_0x=` ` a^2x_0y_0 − b^2x_0y_0`
`a^2x=` ` x_0(a^2 − b^2)`
`x=` ` (x_0(a^2 − b^2))/(a^2)\ \ \ \ \ text{(using}\ \ a^2e^2=a^2-b^2 text{)}`
`=`  `x_0e^2`

 

(iii) `ON = x_0e^2`

`OS=ae\ \ \ \ text{(given}\ \ S(ae,0) text{)}`

`T\ text(is the)\ x text(-axis intercept of the tangent)\ PT`

`0-y_0` `= -(b^2x_0)/(a^2y_0)(x − x_0)`
`a^2y_0^2` `=b^2x_0x-b^2x_0^2`
`b^2x_0x` `=b^2x_0^2+a^2y_0^2`
`x` `=(b^2x_0^2+a^2y_0^2)/(b^2x_0)`
   
`text(S)text(ince)\ P(x_0,y_0)\ text(lies on the ellipse,)`
`=> (x_0^2)/(a^2) + (y_0^2)/(b^2)` ` = 1`
`b^2 x_0^2+a^2y_0^2` `=a^2b^2`

 

`:.x=OT` `=(a^2b^2)/(b^2x_0)=a^2/x_0`

 

`:.ON xx OT` `= x_0e^2 xx a^2/x_0`
  `=a^2e^2`
  `=OS^2`