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Quadratic, 2UA 2012 HSC 16c

The circle  `x^2+(y-c)^2=r^2`,  where  `c>0`  and  `r>0`,  lies inside the parabola  `y=x^2`.  The circle touches the parabola at exactly two points located symmetrically on opposite sides of the  `y`-axis, as shown in the diagram.

2012 16c

  1. Show that  `4c=1+4r^2`.   (2 marks)
  2. Deduce that  `c>1/2`.    (1 mark)
Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `text{Proof  (See Worked Solutions)}`
Show Worked Solution

(i)    `text(We need to find the intersection of)`

`x^2+(y-c)^2` `=r^2\ \ \ text{… (1)}`
`y` `=x^2\ \ \ text{… (2)}`

`text(Subst.)\ y=x^2\ text{into (1)}`

`y+(y-c)^2` `=r^2`
`y+y^2-2cy+c^2` `=r^2`
`y^2+y(1-2c)+(c^2-r^2)` `=0`

 

`text(S)text(ince circle touches symmetrically)`

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: Few students were successful in this part, with those that solved the simultaneous equations correctly often not realising that the  `Delta=0`  was a critical element in solving the problem.

`=>\ \ 1\ ytext(-value only)`

`=>Delta=b^2-4ac=0`

`(1-2c)^2-4xx1xx(c^2-r^2)` `=0`
`1-4c+4c^2-4c^2+4r^2` `=0`
`:.\ 4c` `=1+4r^2\ \ \ text(… as required)`

 

(ii)   `text(Deduce)\ c>1/2`

`text(Given)\ y^2+y(1-2c)+(c^2-r^2)`

`y=(-(1-2c)+-sqrt((1-2c)^2-4xx1xx(c^2-r^2)))/2`

`text(S)text(ince)\ y>0\ \ text(and)\ \ b^2-4ac=0`

♦♦♦ If you found this part difficult, don’t despair, it is a beast. With a mean mark of 0%, it ranks as one of the hardest questions in the history of the course.
`=>\ (-(1-2c)+-0)/2` `>0`
`(-1+2c)/2` `>0`
`2c` `>1`
`c` `>1/2\ \ text(… as required)`

 

Calculus, 2ADV C3 2008 HSC 10b

 

The diagram shows two parallel brick walls  `KJ`  and  `MN`  joined by a fence from  `J`  to  `M`.  The wall  `KJ`  is  `s`  metres long and  `/_KJM=alpha`.  The fence  `JM`  is  `l`  metres long.

A new fence is to be built from  `K`  to a point  `P`  somewhere on  `MN`.  The new fence  `KP`  will cross the original fence  `JM`  at  `O`.

Let  `OJ=x`  metres, where  `0<x<l`.

  1. Show that the total area,  `A`  square metres, enclosed by  `DeltaOKJ`  and  `DeltaOMP`  is given by
     
         `A=s(x-l+l^2/(2x))sin alpha`.   (3 marks)

  2. Find the value of  `x`  that makes  `A`  as small as possible. Justify the fact that this value of  `x`  gives the minimum value for  `A`.  (3 marks)

  3. Hence, find the length of  `MP`  when  `A`  is as small as possible.    (1 mark)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `l/sqrt2`
  3. `(sqrt2-1)s\ \ text(metres)`
Show Worked Solution
i.

`A=text(Area)\  Delta OJK+text(Area)\ Delta OMP`

♦♦♦ Low mean marks highlighted (although exact data not available before 2009).

`text(Using sine rule)`

`text(Area)\ Delta OJK=1/2\ x s sin alpha` 

`text(Area)\ DeltaOMP =>text(Need to find)\ \ MP`

`/_OKJ` `=/_MPO\ \ text{(alternate angles,}\ MP\ text(||)\ KJtext{)}`
`/_PMO` `=/_OJK=alpha\ \ text{(alternate angles,}\ MP\ text(||)\ KJtext{)}`

 
`:.\ DeltaOJK\ text(|||)\ Delta OMP\ \ text{(equiangular)}`

`=>x/s` `=(l-x)/(MP)\ ` ` text{(corresponding sides of similar triangles)}`
`MP` `=(l-x)/x *s`  
`text(Area)\ Delta\ OMP` `=1/2 (l-x)* MP * sin alpha`
  `=1/2 (l-x)*((l-x))/x* s  sin alpha`
`:. A`  `=1/2 x*s sin alpha+1/2 (l-x)*((l-x))/x* s sin alpha`
  `=s sin alpha(1/2 x+1/2 (l-x)*((l-x))/x)`
  `=s sin alpha(1/2 x+(l-x)^2/(2x))`
  `=s sin alpha(1/2 x+l^2/(2x)-l+1/2 x)`
  `=s(x-l+l^2/(2x))sin alpha\ \ \ \ text(… as required)`

 

ii.   `text(Find)\ x\ text(such that)\ A\ text(is a minimum)`

MARKER’S COMMENT: Students who could not complete part (i) are reminded that they can still proceed to part (ii) and attempt to differentiate the result given.
Note that `l` and `alpha` are constants when differentiating. 
`A` `=s(x-l+l^2/(2x))sin alpha`
`(dA)/(dx)` `=s(1-l^2/(2x^2))sin alpha`

`text(MAX/MIN when)\ (dA)/(dx)=0`

`s(1-l^2/(2x^2))sin alpha` `=0`
`l^2/(2x^2)` `=1`
`2x^2` `=l^2`
`x^2` `=l^2/2`
`x` `=l/sqrt2,\ \ \ x>0`

 

`(d^2A)/(dx^2)=s((l^2)/(2x^3))sin alpha`

`text(S)text(ince)\ \ 0<alpha<90°\ \ =>\ sin alpha>0,\ \ l>0\ \ text(and)\ \  x>0`

`(d^2A)/(dx^2)>0\ \ \ =>text(MIN at)\ \ x=l/sqrt2`

 

iii.   `text(S)text(ince)\ \ MP=((l-x))/x s\ \ text(and MIN when)\ \ x=l/sqrt2`

`MP` `=((l-l/sqrt2)/(l/sqrt2))s xx sqrt2/sqrt2`
  `=((sqrt2 l-l))/l s`
  `=(sqrt2-1)s\ \ text(metres)`

 

`:.\ MP=(sqrt2-1)s\ \ text(metres when)\ A\ text(is a MIN.)`

Calculus, 2ADV C3 2009 HSC 9b

An oil rig,  `S`,  is 3 km offshore. A power station,  `P`,  is on the shore. A cable is to be laid from `P`  to  `S`.  It costs $1000 per kilometre to lay the cable along the shore and $2600 per kilometre to lay the cable underwater from the shore to  `S`.

The point  `R`  is the point on the shore closest to  `S`,  and the distance  `PR`  is 5 km.

The point  `Q`  is on the shore, at a distance of  `x`  km from  `R`,  as shown in the diagram.


  

  1. Find the total cost of laying the cable in a straight line from  `P`  to  `R`  and then in a straight line from  `R`  to  `S`.   (1 mark)

  2. Find the cost of laying the cable in a straight line from  `P`  to  `S`.  (1 mark)

  3. Let  `$C`  be the total cost of laying the cable in a straight line from  `P`  to  `Q`,  and then in a straight line from `Q`  to  `S`.
     
    Show that  `C=1000(5-x+2.6sqrt(x^2+9))`.    (2 marks)

  4. Find the minimum cost of laying the cable.    (4 marks)

  5. New technology means that the cost of laying the cable underwater can be reduced to $1100 per kilometre.

     

    Determine the path for laying the cable in order to minimise the cost in this case.    (2 marks)

Show Answers Only
  1. `$12\ 800`
  2. `$15\ 160`
  3. `text{Proof (See Worked Solutions)}`
  4. `$12\ 200`
  5. `P\ text(to)\  S\ text(in a straight line.)`
Show Worked Solution
♦♦ Although specific data is unavailable for question parts, mean marks were 35% for Q9 in total.
i.   `text(C)text(ost)` `=(PRxx1000)+(SRxx2600)`
  `=(5xx1000)+(3xx2600)`
  `=12\ 800`

 
`:. text(C)text(ost is)\   $12\ 800`

 

ii.   `text(C)text(ost)=PSxx2600`

`text(Using Pythagoras:)`

`PS^2` `=PR^2+SR^2`
  `=5^2+3^2`
  `=34`
`PS` `=sqrt34`

 

`:.\ text(C)text(ost)` `=sqrt34xx2600`
  `=15\ 160.474…`
  `=$15\ 160\ \ text{(nearest dollar)}`

 

iii.  `text(Show)\ \ C=1000(5-x+2.6sqrt(x^2+9))`

`text(C)text(ost)=(PQxx1000)+(QSxx2600)`

`PQ` `=5-x`
`QS^2` `=QR^2+SR^2`
  `=x^2+3^2`
`QS` `=sqrt(x^2+9)`
`:.C` `=(5-x)1000+sqrt(x^2+9)\ (2600)`
  `=1000(5-x+2.6sqrt(x^2+9))\ \ text(…  as required)`

 

iv.   `text(Find the MIN cost of laying the cable)`

`C` `=1000(5-x+2.6sqrt(x^2+9))`
`(dC)/(dx)` `=1000(–1+2.6xx1/2xx2x(x^2+9)^(–1/2))`
  `=1000(–1+(2.6x)/(sqrt(x^2+9)))`

`text(MAX/MIN when)\ (dC)/(dx)=0`

IMPORTANT: Tougher derivative questions often require students to deal with multiple algebraic constants. See Worked Solutions in part (iv).

`1000(–1+(2.6x)/(sqrt(x^2+9)))=0`

`(2.6x)/sqrt(x^2+9)` `=1`
`2.6x` `=sqrt(x^2+9)`
`(2.6)^2x^2` `=x^2+9`
`x^2(2.6^2-1)` `=9`
`x^2` `=9/5.76`
  `=1.5625`
`x` `=1.25\ \ \ \ (x>0)`
MARKER’S COMMENT: Check the nature of the critical points in these type of questions. If using the first derivative test, make sure some actual values are substituted in.

`text(If)\ \ x=1,\ \ (dC)/(dx)<0`

`text(If)\ \ x=2,\ \ (dC)/(dx)>0`

`:.\ text(MIN when)\ \ x=1.25`

`C` `=1000(5-1.25+2.6sqrt(1.25^2+9))`
  `=1000(122)`
  `=12\ 200` 

 

`:.\ text(MIN cost is)\  $12\ 200\ text(when)\   x=1.25`

 

v.   `text(Underwater cable now costs $1100 per km)`

`=>\ C` `=1000(5-x)+1100sqrt(x^2+9)`
  `=1000(5-x+1.1sqrt(x^2+9))`
`(dC)/(dx)` `=1000(–1+1.1xx1/2xx2x(x^2+9)^(-1/2))`
  `=1000(–1+(1.1x)/sqrt(x^2+9))`

 
`text(MAX/MIN when)\ (dC)/(dx)=0`

`1000(–1+(1.1x)/sqrt(x^2+9))` `=0`
`(1.1x)/sqrt(x^2+9)` `=1`
`1.1x` `=sqrt(x^2+9)`
`1.1^2x^2` `=x^2+9`
`x^2(1.1^2-1)` `=9`
`x^2` `=9/0.21`
`x` `~~6.5\ text{km (to 1 d.p.)}`
  `=>\ text(no solution since)\ x<=5`

 
`text(If we lay cable)\ PR\ text(then)\ RS`

MARKER’S COMMENT: Many students failed to interpret a correct calculation of  `x>5`  as providing no solution.

`=>\ text(C)text(ost)=5xx1100+3xx1000=8500`

`text(If we lay cable directly underwater via)\ PS`

`=>\ text(C)text(ost)=sqrt34xx1100=6414.047…`

`:.\ text{MIN cost is  $6414 by cabling directly from}\ P\ text(to)\ S`.

Calculus, 2ADV C3 2011 HSC 10b

A farmer is fencing a paddock using  `P`  metres of fencing. The paddock is to be in the shape of a sector of a circle with radius  `r`  and sector angle `theta`  in radians, as shown in the diagram.

2011 10b

  1. Show that the length of fencing required to fence the perimeter of the paddock is
      
       `P=r(theta+2)`.    (1 mark)

  2. Show that the area of the sector is  `A=1/2 Pr-r^2`.    (1 mark) 

  3. Find the radius of the sector, in terms of  `P`, that will maximise the area of the paddock.    (2 marks)

  4. Find the angle  `theta`  that gives the maximum area of the paddock.    (1 mark)

  5. Explain why it is only possible to construct a paddock in the shape of a sector if
     
         `P/(2(pi+1)) <\ r\ <P/2`   (2 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `text{Proof  (See Worked Solutions)}`
  3. `r=P/4`
  4. `2\ text(radians)`
  5. `text{Proof  (See Worked Solutions)}`
Show Worked Solution

i.   `text(Need to show)\ \ P=r(theta+2)`

`P` `=(2xxr)+theta/(2pi)xx2pir`
  `=2r+ r theta`
  `=r(theta+2)\ \ text(…  as required)`

 

ii.   `text(Need to show)\ \ A=1/2 Pr-r^2`.

`text(S)text(ince)\ \ P=r(theta+2)\ \ \ =>\ theta=(P-2r)/r`

♦ Mean mark 41%.
TIP: Area of a sector `=pi r^2 xx` the percentage of the circle that the sector angle accounts for, i.e. `theta/(2 pi)` radians. `:. A=pi r^2“ xx theta/(2 pi)=1/2 r^2 theta`.
`:. A` `=1/2 r^2 theta`
  `=1/2 r^2*(P-2r)/r`
  `=1/2(Pr-2r^2)`
  `=1/2 Pr-r^2\ \ text(…  as required)`

 

iii.  `A=1/2 Pr-r^2`

♦♦ Mean mark 29%
MARKER’S COMMENT: The second derivative test proved much more successful and easily proven in this part. Make sure you are comfortable choosing between it and the 1st derivative test depending on the required calcs.

`(dA)/(dr)=1/2 P-2r`

`text(MAX or MIN when)\ \ (dA)/(dr)=0`

`1/2 P-2r` `=0`
`2r` `=1/2 P`
`r` `=P/4`

 
`(d^2A)/(dr^2)=-2\ \ \ \ =>text(MAX)`

`:.\ text(Area is a MAX when)\ r=P/4\ text(units)`

 

iv.   `text(Need to find)\ theta\ text(when area is a MAX)\ =>\ r=P/4`

♦♦♦ Mean mark 20%
`P` `=r(theta+2)`
  `=P/4(theta+2)`
`4P` `=P(theta+2)`
`theta+2` `=4`
`theta` `=2\ text(radians)`

 

v.  `text(For a sector to exist)\ \ 0<\ theta\ <2pi, \ \ text(and)\ \ theta=(P-2r)/r`

♦♦♦ Mean mark 4%. A BEAST!
MARKER’S COMMENT: When asked to ‘explain’, students should support their answer with a mathematical argument.
`=>(P-2r)/r` `>0`
`P-2r` `>0`
`r` `<P/2`
`=>(P-2r)/r` `<2pi`
`P-2r` `<2r pi`
`2r pi+2r` `>P`
`2r(pi+1)` `>P`
`r` `>P/(2(pi+1))`

 

`:.P/(2(pi+1)) <\ r\ <P/2\ \ text(…  as required)`

Mechanics, EXT2* M1 2009 HSC 6a

Two points, `A` and `B`,  are on cliff tops on either side of a deep valley. Let `h` and `R` be the vertical and horizontal distances between `A` and `B` as shown in the diagram. The angle of elevation of `B` from `A` is  `theta`,  so that  `theta=tan^-1(h/R)`.
 

2009 6a
 

At time `t=0`,  projectiles are fired simultaneously from `A` and `B`.  The projectile from `A` is aimed at `B`, and has initial speed `U` at an angle of  `theta`  above the horizontal. The projectile from `B` is aimed at `A` and has initial speed `V` at an angle  `theta`  below the horizontal.

The equations of motion for the projectile from `A` are

`x_1=Utcos theta`   and   `y_1=Utsin theta-1/2 g t^2`,

and the equations for the motion of the projectile from `B` are

`x_2=R-Vtcos theta`   and   `y_2=h-Vtsin theta-1/2 g t^2`,     (DO NOT prove these equations.)

  1. Let `T` be the time at which  `x_1=x_2`.
     
    Show that  `T=R/((U+V)\ cos theta)`.   (1 mark)

  2. Show that the projectiles collide.    (2 marks)

  3. If the projectiles collide on the line  `x=lambdaR`,  where  `0<lambda<1`,  show that
     
         `V=(1/lambda-1)U`.    (1 mark)

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

i.   `text(Let)\ \ x_1=x_2\ \ text(at)\ \ t=T`

`UTcos theta` `=R-VTcos theta`
`UTcos theta+VTcos theta` `=R`
`Tcos theta\ (U+V)` `=R`

 
 `:. T=R/((U+V)\ cos theta)\ \ text(… as required)`

 

ii.   `text(If particles collide,)\ \  y_1=y_2\ \ text(at)\ \ t=T`

♦♦♦ Mean mark data not available although “few” students were able to complete this part.
MARKER’S COMMENT: Better responses showed  `y_1=y_2`,  or  `y_1-y_2=0`  which eliminated  `–½ g t^2`  from the algebra. Substituting `tan theta=h//R` was a key element in completing this proof.
`y_1` `=UTsin theta-1/2 g T^2`
  `=(UR sin theta)/((U+V)\ cos theta)-1/2 g T^2`
`y_2` `=h-VTsin theta-1/2 g T^2`
  `=Rtan theta-(VR sin theta)/((U+V)\ cos theta)-1/2 g T^2`
  `=R(((sin theta/cos theta)(U+V) cos theta-V sin theta)/((U+V)\ cos theta))-1/2 g T^2`
  `=R/((U+V)\ cos theta)(Usin theta+Vsin theta-Vsin theta)-1/2 g T^2`
  `=(UR sin theta)/((U+V)\ cos theta)-1/2 g T^2`
  `=y_1`

 

`:.\ text(Particles collide since)\  y_1=y_2\ text(at)\ t=T`.

 

iii.  `text(Particles collide on line)\ \ x=lambdaR,\ text(i.e.  where)\ \ t=T`

MARKER’S COMMENT: Students must show sufficient working in proofs to convince markers they have derived the answer themselves.
`lambdaR` `=UTcos theta`
  `=U R/((U+V)\ cos theta)cos theta`
  `=(UR)/((U+V))`
`(U+V)(lambdaR)` `=UR`
`U+V` `=(UR)/(lambdaR)`
`V` `=U/lambda-U`
  `=(1/lambda-1)U\ \ text(… as required)`

Mechanics, EXT2* M1 2010 HSC 6b

A basketball player throws a ball with an initial velocity  `v`  m/s at an angle of  `theta`  to the horizontal. At the time the ball is released its centre is at  `(0,0)`, and the player is aiming for the point `(d,h)`  as shown on the diagram. The line joining  `(0,0) `  and  `(d,h)`  makes an angle  `alpha`  with the horizontal, where  `0<alpha<pi/2`.
 

6b
 

Assume that at time  `t`  seconds after the ball is thrown its centre is at the point  `(x,y)`,  where

`x=vtcos theta`

`y=vt sin theta-5 t^2`.      (DO NOT prove this.)

  1. If the centre of the ball passes through  `(d,h)`  show that
     
         `v^2=(5d)/(cos theta sin theta-cos^2 theta tan alpha)`    (3 marks)

    (2) What happens to  `v`  as  `theta\ ->pi/2` ?    (1 mark)

  2. For a fixed value of  `alpha`,  let  `F(theta)=cos theta sin theta-cos^2 theta tan alpha`.
     
    Show that  `F prime(theta)=0`   when   `tan2theta tan alpha=-1`  (2 marks)

  3. Using part (a)(ii)* or otherwise show that  `F prime(theta)=0`,   when  `theta=alpha/2+pi/4`.    (1 mark)

     

    *Please note for the purposes of this question, part (a)(ii) showed that  when  `tanA tanB=-1`,  then  `A-B=pi/2` 

  4. Explain why  `v^2`  is a minimum when  `theta=alpha/2+pi/4`     (2 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}` 
  2. (1)  `v->oo`

     

    (2)  `v->oo`

  3. `text{Proof  (See Worked Solutions)}`
  4. `text{Proof  (See Worked Solutions)}` 
  5. `text{Explanation  (See Worked Solutions)}`
Show Worked Solution

i.   `text(Ball passes through)\ \ (d,h).`

`text(Find)\ t\ text(when)\ x=d :`

♦ Mean mark 43%
IMPORTANT: It is critical to use the relationship `tan alpha=h/d` to achieve the required proof.
`d` `=vtcos theta`
`:. t` `=d/(vcos theta)`

 

`text(S)text(ince)\ \ y=h\ \ text(when)\ \ t=d/(vcos theta)`

`h` `=vtsin theta-5t^2`
  `=v(d/(vcos theta))sin theta-5(d^2)/(v^2cos^2 theta)`
`(5d^2)/(v^2cos^2 theta)` `=(dsin theta)/cos theta-h`
`(5d^2)/v^2` `=dsin theta cos theta-cos^2 thetaxxh`
`v^2/(5d^2)` `=1/(dsin theta cos theta-cos^2 thetaxxh)`
`v^2` `=(5d^2)/(dsin theta cos theta-cos^2 thetaxxh)`
  `=(5d)/(cos theta sin theta-cos^2 thetaxxh/d)`
  `=(5d)/(cos theta sin theta-cos^2 theta tan alpha)\ \ text(… as required)`

 

 

ii. (1)  `text(As)\ \  theta->alpha`

♦ Mean marks of 37% and 32% for parts (ii)(1) and (ii)(2) respectively.

`costheta sin theta-cos^2 theta tan alpha->0`

`v->oo`

ii. (2)  `text(As)\ \  theta->pi/2`

`cos theta sin theta-cos^2 theta tan theta->0`

`v->oo`

 

iii.  `text(Show)\ \ tan 2thetatan alpha=–1\ \ text(when)\ \ F prime(theta)=0`.

`F(theta)` `=costheta sin theta-cos^2theta tan alpha`
`F prime(theta)` `=cos theta costheta+sin theta (–sin theta)-2cos theta (–sin theta) tan alpha`
  `=cos^2 theta-sin^2 theta+sin 2theta tan alpha`
  `=cos 2theta+sin 2theta tan alpha`

 

`text(When)\ F prime(theta)=0`

♦♦ Mean mark part (iii) 21%
MARKER’S COMMENT: Many students failed to treat  `tan alpha` as a constant when differentiating.
`cos 2theta+sin2thetatan alpha` `=0`
`sin2thetatan alpha` `=-cos2theta`
`tan 2theta tan alpha` `=-1\ \ text(… as required)`

 

iv.  `text(Show that)\ \ F prime(theta)=0,\ \ text(when)\ \ theta=alpha/2+pi/4`

`text(If)\ \ tanA tanB=–1\ => A-B=pi/2`

`text{(Given in part (a)(ii) – see note in question)}`

 

`text(S)text(ince)\ \ tan2theta tan alpha=–1\ \ text{(see part (iii))}`

♦♦♦ Mean mark 14%
MARKER’S COMMENT: Linking part (a)(ii) to this solution was a feature of the more efficient and successful approaches.
`=>\ \ 2theta-alpha` `=pi/2`
`2theta` `=alpha+pi/2`
`theta` `=alpha/2+pi/4`

 

 `text(Given that)\ \ Fprime(theta)=0\ \ text(when)\ \ tan 2thetatan alpha=–1`

`:. Fprime(theta)=0\ \ text(when)\ \ theta=alpha/2+pi/4\ \ text(… as required)`

 

v.   `text(S)text(ince)\ \ v^2=(5d)/(F(theta)),`

`=>\ v^2\ \ text(is a MIN when)\ \  F(theta)\ \ text(is a MAX)`

`text(We know)\ \ F prime(theta)=0\ \ text(when)\ \ theta=alpha/2+pi/4`

♦♦♦ Mean mark 9%
MARKER’S COMMENT: Only a small minority of students  explained correctly that  `v^2`  is a MIN when  `F(theta)`  is a MAX.

`:.\ text(MAX or MIN when)\ \ theta=alpha/2+pi/4`

 

`F prime(theta)=cos 2theta+sin 2theta tan alpha`

`F ″(theta)=-2sin2theta+2cos2theta tan alpha`

`text(When)\ \ theta=alpha/2+pi/4`

`F″(alpha/2+pi/4)=-2sin(alpha+pi/2)+2cos(alpha+pi/2)tan alpha`

`text(S)text(ince)\ \ 0<alpha<pi/2`

`=> tan alpha>0`

`=>sin(alpha+pi/2)>0`

`=>cos(alpha+pi/2)<0`

`:. F″(alpha/2+pi/4)<0\ \ text(i.e.  MAX)`

`:.\ v^2\ \ text(is a minimum.)`  

Mechanics, EXT2* M1 2013 HSC 13c

Points `A` and `B` are located `d` metres apart on a horizontal plane. A projectile is fired from `A` towards `B` with initial velocity `u` m/s at angle `alpha` to the horizontal.

At the same time, another projectile is fired from `B` towards `A` with initial velocity `w` m/s at angle `beta` to the horizontal, as shown on the diagram.

The projectiles collide when they both reach their maximum height.
 

2013 13c
 

The equations of motion of a projectile fired from the origin with initial velocity  `V` m/s at angle  `theta`  to the horizontal are

`x=Vtcostheta`   and   `y=Vtsintheta-g/2 t^2`.        (DO NOT prove this.)

  1. How long does the projectile fired from  `A`  take to reach its maximum height?  (2 marks)

  2. Show that  `usinalpha=w sin beta`.    (1 mark)

  3. Show that  `d=(uw)/(g)sin(alpha+beta)`.    (2 marks)

Show Answers Only
  1. `(u)/(g) sin alpha\ \ text(seconds)`
  2. `text{Proof  (See Worked Solutions)}`
  3. `text{Proof  (See Worked Solutions)}`
Show Worked Solution

i.   `y=Vtsintheta-g/2 t^2`

`text(Projectile)\ A\   => V=u,\ \ theta=alpha`

`y` `=ut sinalpha- (g)/2 t^2`
`doty` `=u sinalpha- g t`

 

`text(Max height when)\  doty=0`

`0` `=usinalpha-g t`
`g t` `=usinalpha`
`t` `=(u)/(g)sinalpha`

 
`:.\ text(Projectile from)\ A\ text(reaches max height at)`

 `t=(u)/(g)sin alpha\ \ text(seconds)`

 

ii.  `text(Show that)\ \ usin alpha=wsin beta`

IMPORTANT: Part (i) in this example leads students to a very quick solution. Always look to previous parts for clues to direct your strategy.

`text(Projectile)\ B\ =>V=w,\ \ theta=beta`

`y` `=wt sin beta- (g)/2 t^2`
`doty` `=wsin beta-g t`

 

`text(Max height when)\ \ doty=0`

`t=(w)/(g) sin beta`

`text{Projectiles collide at max heights}`

`text(S)text(ince they were fired at the same time)`

`(u)/(g) sin alpha` `=(w)/(g) sin beta`
`:.\ usin alpha` `=wsin beta\ \ text(… as required)`

 

iii  `text(Show)\ \ d=(uw)/(g) sin (alpha+beta) :`

`text(Find)\ x text(-values for each projectile at max height)`

`text(Projectile)\ A`

`x_1` `=utcos alpha`
  `=u((u)/(g) sin alpha)cos alpha`
  `=(u^2)/(g) sin alpha cos alpha`

 

`text(Projectile)\ B`

♦ Mean mark 39%
IMPORTANT: Students should direct their calculations by reverse engineering the required result. `sin(alpha+beta)` in the proof means that  `sin alpha cos beta+“ cos alpha sin beta`  will appear in the working calculations.
`x_2` `=wt cos beta`
  `=w((w)/(g) sin beta)cos beta`
  `=(w^2)/(g) sin beta cos beta`
`d` `=x_1+x_2`
  `=(u^2)/(g) sin alpha cos alpha+(w^2)/(g) sin beta cos beta`
  `=(u)/(g)(u sin alpha)cos alpha+(w)/(g) (w sin beta) cos beta`
  `=(u)/(g)(w sin beta) cos alpha+(w)/(g) (usin alpha)cos beta\ \ text{(part (ii))}`
  `=(uw)/(g) (sin alpha cos beta+cos alpha sin beta)`
  `=(uw)/(g) sin (alpha+beta)\ \ text(… as required)`

 

Calculus, 2ADV C3 2012 HSC 16b

The diagram shows a point  `T`  on the unit circle  `x^2+y^2=1`  at an angle  `theta`  from the positive  `x`-axis, where  `0<theta<pi/2`.

The tangent to the circle at  `T`  is perpendicular to  `OT`, and intersects the  `x`-axis at  `P`,  and the line  `y=1`  intersects the  `y`-axis at  `B`.
 

 
 

  1. Show that the equation of the line  `PT`  is  `xcostheta+ysin theta=1`.    (2 marks)

  2. Find the length of  `BQ`  in terms of  `theta`.    (1 mark)

  3. Show that the area,  `A`,  of the trapezium  `OPQB`  is given by 
     
         `A=(2-sintheta)/(2costheta)`    (2 marks)

  4. Find the angle  `theta`  that gives the minimum area of the trapezium.    (3 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `(1-sin theta)/cos theta`
  3. `text{Proof  (See Worked Solutions)}`
  4. `theta=pi/6\ \ text(radians)`
Show Worked Solution
i.    

`text(Find)\  T:`

♦♦ Mean mark 20%

`text(S)text(ince)\ \ cos theta=x/1\ \ \ text(and)\ \ \  sin theta=y/1`

`:. T\ (cos theta, sin theta)`

`text(Gradient of)\ OT=sin theta/cos theta`

`:.\ text(Gradient)\ PT=-cos theta/sin theta\ \ text{(} _|_ text{lines)}`

`text(Equation of)\ PT\ text(where)`

`m=-cos theta/sin theta,\ \ text(and through)\ \ (cos theta, sin theta)`

`text(Using)\ \ y-y_1` `=m(x-x_1)`
`y-sin theta` `=-cos theta/sin theta(x-cos theta)`
`y sin theta-sin^2 theta` `=-x cos theta+cos^2 theta`
`x cos theta+y sin theta` `=sin^2 theta+cos^2 theta`
`x cos theta+y sin theta` `=1\ \ \ \ \ text(… as required)`

 

ii.   `text(Find)\ Q:`

 `Q\ => text(intersection of)\ xcos theta+y sin theta=1\ \ text(and)\ \ y=1`

`x cos theta+sin theta` `=1`
`x cos theta` `=1-sin theta`
`x` `=(1-sin theta)/cos theta`

 
`:.\ text(Length of)\ BQ\ text(is)\ \ (1-sin theta)/cos theta\ text(units)`

 

iii.  `text(Show Area)\ OPQB=(2-sin theta)/(2cos theta)`

`A=1/2h(a+b)\ \ text(where)\ \ h=OB=1\ \   a=OP\ \  text(and)`  

                `b=BQ=(1-sin theta)/cos theta`

`text(Find  length)\ OP:`

`P => xcos theta+ysin theta=1 \ text(cuts)\ \ x text(-axis)`

♦♦ Mean mark 24%
`xcos theta` `=1`
`x` `=1/cos theta`
`=>text(Length)\ OP` `=1/cos theta`

 

`text(Area)\ OPQB` `=1/2xx1(1/cos theta+(1-sin theta)/cos theta)`
  `=1/2((2-sin theta)/cos theta)`
  `=(2-sin theta)/(2cos theta)\ \ text(u²)\ \ \ text(… as required)`

 

iv.  `text(Find)\ theta\ text(such that Area)\ OPQB\ text(is a MIN)`

`A` `=(2-sin theta)/(2cos theta)`
`(dA)/(d theta)` `=(2cos theta(-cos theta)-(2- sin theta)(-2 sin theta))/(4cos^2 theta)`
  `=(4 sin theta-2sin^2 theta-2 cos^2 theta)/(4 cos^2 theta)`
  `=(4sin theta-2(sin^2 theta+cos^2 theta))/(4 cos^2 theta)`
  `=(4sin theta-2)/(4cos^2 theta)`
  `=(2sin theta-1)/(2 cos^2 theta)`
Mean mark 19%
IMPORTANT: Look for any opportunity to use the identity `sin^2 theta“+cos^2 theta=1` → often a key to simplifying difficult trig equations.
 

`text(MAX or MIN when)\ (dA)/(d theta)=0`

`=>2sin theta-1` `=0`
`sin theta` `=1/2`
`theta` `=pi/6\ \ \ \ \0<theta<pi/2` 

 
`text(Test for MAX/MIN)`

IMPORTANT: Is the 1st or 2nd derivative test easier here? Examiners have often made one significantly easier than the other.

`text(If)\ theta=pi/12\ \ (dA)/(d theta)<0`

`text(If)\ theta=pi/3\ \ (dA)/(d theta)>0\ \ =>text(MIN)`

`:.\text(Area)\ OPQB\ text(is a MIN when)\ theta=pi/6`.

Calculus, 2ADV C3 2013 HSC 14b

Two straight roads meet at  `R`  at an angle of 60°.  At time  `t=0`  car  `A`  leaves  `R`  on one road, and car  `B`  is 100km from  `R`  on the other road.  Car  `A`  travels away from  `R`  at a speed of 80 km/h, and car  `B`  travels towards  `R`  at a speed of 50 km/h.
 

2013 14b

The distance between the cars at time  `t`  hours is  `r`  km.

  1. Show that  `r^2=12\ 900t^2-18\ 000t+10\ 000`.   (2 marks)

  2. Find the minimum distance between the cars.    (3 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `61\ text(km)`
Show Worked Solution

i.   `text(Need to show)\  r^2=12\ 900t^2-18\ 000t+10\ 000`

♦♦ Mean mark 26%

`RB=100-50t`

`RA=80t`

`text(Using the cosine rule)`

`r^2` `=(RB)^2+(RA)^2-2(RB)(RA)cos/_R`
  `=(100-50t)^2+(80t)^2-2(100-50t)(80t)cos60`
  `=10\ 000-10\ 000t+2500t^2+6400t^2-8000t+4000t^2`
  `=12\ 900t^2-18\ 000t+10\ 000\ \ \ \ text(… as required)` 

 

ii.   `text(Max/min when)\ (dr^2)/(dt)=0`

♦♦ Mean mark 27%
ALGEBRA TIP: Finding the derivative of `r^2` (rather than making `r` the subject), makes calculations much easier. ENSURE you apply the test to confirm a minimum.

`(dr^2)/(dt)=25\ 800t-18\ 000=0`

`t=(18\ 000)/(25\ 800)=30/43`

`text(When)\  t=30/43,\   (d^2(r^2))/(dt^2)=25\ 800>0\ \ =>text(MIN)`

`:.\ text(Minimum distance when)\  t=30/43\ text(hr)`

`text(Find)\  r\ text(when)\ t=30/43`

`r^2` `=12\ 900(30/43)^2-18\ 000(30/43)+10\ 000`
  `=3720.9302…`
`:.\ r` `=sqrt3702.9302…`
  `~~60.99942…`
  `~~61\ text{km  (nearest km)}`

 

`:.\ text(MIN distance between the cars is 61 km.)`

Calculus, EXT1* C1 2013 HSC 16b

Trout and carp are types of fish. A lake contains a number of trout. At a certain time, 10 carp are introduced into the lake and start eating the trout. As a consequence, the number of trout,  `N`,  decreases according to

`N=375-e^(0.04t)`,

where `t` is the time in months after the carp are introduced.

The population of carp,  `P`,  increases according to  `(dP)/(dt)=0.02P`.

  1. How many trout were in the lake when the carp were introduced?    (1 mark)

  2. When will the population of trout be zero?    (1 mark)

  3. Sketch the number of trout as a function of time.     (1 marks)

  4. When is the rate of increase of carp equal to the rate of decrease of trout?    (3 marks)

  5. When is the number of carp equal to the number of trout?    (2 marks)

Show Answers Only
  1. `text(374 trout)`
  2. `text{148 months (nearest month)}`
  3.  
     
    2UA 2013 HSC 16b Answer
  4. `text{After 80 months (nearest month)}`
  5. `text{After 135 months (nearest month)}`
Show Worked Solution

i.    `text(Carp introduced at)\ \ t=0`

MARKER’S COMMENT: A number of students did not equate `e^0=1` in this part.

`N=375-e^0=374`

`:.\ text(There was 374 trout when carp were introduced.)`

 

ii.    `text(Trout population will be zero when)`

NOTE: The last line of the solution isn’t necessary but is included as good practice as a check that the answer matches the exact question asked.
`N` `=375-e^(0.04t)=0`
`e^(0.04t)` `=375`
`0.04t` `=ln375`
`t` `=ln375/0.04`
  `=148.173 …`
  `=148\ text{months (nearest month)}`

 
`:.\ text(After 148 months, the trout population will be zero.)`
 

iii.   2UA 2013 HSC 16b Answer

♦♦ Mean mark 33% for part (iii)

 

iv.   `text(We need) \ |(dN)/(dt)|=(dP)/(dt)`

`text(Given)\ N=375-e^(0.04t)`

`(dN)/(dt)=-0.04e^(0.04t)`
  

`text(Find)\ P\ text(in terms of)\  t`

`text(Given)\ (dP)/(dt)=0.02P`

`=> P=Ae^(0.02t)`

♦♦♦ Mean mark 20%
COMMENT: Students who progressed to `0.2e^(0.02t)“=0.04e^(0.04t)` received 2 full marks in this part. Show your working!

 `text(Find)\ A\ \ =>text(when)\  t=0,\ P=10`

`10` `=Ae^0`
`:.A` `=10`
`=>P` `=10xx0.02e^(0.02t)`
  `=0.2e^(0.02t)`

 
`text(Given that)\ \ (dP)/(dt)=|(dN)/(dt)|`

`0.2e^(0.02t)` `=0.04e^(0.04t)`
`5e^(0.02t)` `=e^(0.04t)`
`e^(0.04t)/e^(0.02t)` `=5`
`e^(0.04t-0.02t)` `=5`
`lne^(0.02t)` `=ln5`
`0.02t` `=ln5`
`t` `=ln5/0.02`
  `=80.4719…`
  `=80\ text{months (nearest month)}`

 

v.   `text(Find)\ t\ text(when)\ N=P`

`text(i.e.)\ \ 375-e^(0.04t)` `=10e^(0.02t)`
`e^(0.04t)+10e^(0.02t)-375` `=0`

`text(Let)\ X=e^(0.02t),\ text(noting)\ \ X^2=(e^(0.02t))^2=e^(0.04t)`

♦♦♦ Mean mark 4%!
MARKER’S COMMENT: Correctly applying substitution to exponentials to form a solvable quadratic proved very difficult for almost all students.
`:.\ X^2+10X-375` `=0`
`(X-15)(X+25)` `=0`

`X=15\ \ text(or)\ \ –25`

 
`text(S)text(ince)\  X=e^(0.02t)`

`e^(0.02t)` `=15\ \ \ \ (e^(0.02t)>0)`
`lne^(0.02t)` `=ln15`
`0.02t` `=ln15`
`t` `=ln15/0.02`
  `=135.4025…`
  `=135\ text(months)`

Calculus, 2ADV C3 2011 HSC 7b

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

`v=8-8e^(-2t)`,

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

  1. Show that the particle is initially at rest.     (1 mark)

  2. Show that the acceleration of the particle is always positive.     (1 mark)

  3. Explain why the particle is moving in the positive direction for all  `t>0`.     (2 marks)

  4. As  `t->oo`, the velocity of the particle approaches a constant.

     

    Find the value of this constant.     (1 mark) 

  5. Sketch the graph of the particle's velocity as a function of time.     (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text(See Worked Solutions.)`
  4. `8\ text(m/s)`
  5. `text(See sketch in Worked Solutions)`
Show Worked Solution

i.   `text(Initial velocity when)\ \ t=0`

`v` `=8-8e^0`
  `=0\ text(m/s)`
 
`:.\ text(Particle is initially at rest.)`

 

 

MARKER’S COMMENT: Students whose working showed `e^(-2t)` as `1/e^(2t)`, tended to score highly in this question.

ii.   `a=d/(dt) (v)=-2xx-8e^(-2t)=16e^(-2t)`

`text(S)text(ince)\  e^(-2t)=1/e^(2t)>0\ text(for all)\  t`.

`=>\ a=16e^(-2t)=16/e^(2t)>0\ text(for all)\  t`.
 

`:.\ text(Acceleration is positive for all)\ \ t>0`.
 

iii.  `text{S}text{ince the particle is initially at rest, and ALWAYS}`

♦♦♦ Mean mark 22%
COMMENT: Students found part (iii) the most challenging part of this question by far.

`text{has a positive acceleration.`
 

`:.\ text(It moves in a positive direction for all)\ t`.
 

iv.   `text(As)\ t->oo`,  `e^(-2t)=1/e^(2t)->0`

`=>8/e^(2t)->0\  text(and)`

`=>v=8-8/e^(2t)->8\ text(m/s)`
 

`:.\ text(As)\ \ t->oo,\ text(velocity approaches 8 m/s.)`

 

IMPORTANT: Use previous parts to inform this diagram. i.e. clearly show velocity was zero at  `t=0`  and the asymptote at  `v=8`. 
v.   

Calculus in the Physical World, 2UA 2011 HSC 7b Answer

Financial Maths, 2ADV M1 2008 HSC 9b

Peter retires with a lump sum of $100 000. The money is invested in a fund which pays interest each month at a rate of 6% per annum, and Peter receives a fixed monthly payment `$M` from the fund. Thus the amount left in the fund after the first monthly payment is   `$(100\ 500-M)`.

  1. Find a formula for the amount, `$A_n`, left in the fund after `n\ ` monthly payments.   (2 marks)

  2. Peter chooses the value of `M` so that there will be nothing left in the fund at the end of the 12th year (after 144 payments). Find the value of `M`.   (3 marks)

Show Answers Only
  1.  `100\ 000(1.005^n)-M((1.005^n-1)/0.005)`
  2. `$975.85`
Show Worked Solutions

i.    `r=(1+0.06/12)=1.005`

MARKER’S COMMENT: Students who developed this answer from writing the calculations of `A_1`, `A_2`, `A_3` to then generalising for `A_n` were the most successful.
`A_1` `=(100\ 500-M)`
  `=100\ 000(1.005)^1-M`
`A_2` `=A_1 (1.005)-M`
  `=[100\ 000(1.005^1)-M](1.005)-M`
  `=100\ 000(1.005^2)-M(1.005)-M`
  `=100\ 000(1.005^2)-M(1+1.005)`
`A_3` `=100\ 000(1.005^3)-M(1+1.005^1+1.005^2)`

`\ \ \ \ \ vdots`

`A_n` `=100\ 000(1.005^n)-M(1+1.005+\ …\ +1.005^(n-1))`
  `=100\ 000(1.005^n)-M((a(r^n-1))/(r-1))`
  `=100\ 000(1.005^n)-M((1(1.005^n-1))/(1.005-1))`
  `=100\ 000(1.005^n)-M((1.005^n-1)/0.005)`

 

ii.  `text(Find)\ M\  text(such that)\ \ $A_n=0\ \ text(when)\ \ n=144`

`A_144=100\ 000(1.005)^144-M((1.005^144-1)/0.005)=0`

`M((1.005^144-1)/0.005)` `=100\ 000(1.005^144)`
`M(1.005^144-1)` `=500(1.005^144)`
`M` `=(500(1.005^144))/(1.005^144-1)`
  `=975.85`

 

`:. M=$975.85\ \ text{(nearest cent).}`

Proof, EXT2* P2 2013 HSC 14a

  1. Show that for  `k>0,\ \ 1/(k+1)^2-1/k+1/(k+1)<0`.    (1 mark)

  2. Use mathematical induction to prove that for all integers  `n>=2`,
     
        `1/1^2+1/2^2+1/3^2+\ …\ +1/n^2<2-1/n`.   (3 marks)

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

i.  `text(Prove that for)\ k>0,\ \ 1/(k+1)^2-1/k+1/(k+1)<0`

♦♦ Mean mark 25%.
MARKER’S COMMENT: Student algebra was poor in creating a common denominator and justifying why the final expression is always negative proved challenging.
`text(LHS)` `=(k-(k+1)^2+k(k+1))/(k(k+1)^2)`
  `=(k-(k^2+2k+1)+k^2+k)/(k(k+1)^2)`
  `=(-1)/(k(k+1)^2)`

 
`text(S)text(ince)\ \ k>0,\ text(we know)\ \ k(k+1)^2>0`

`:.\ (-1)/(k(k+1)^2)<0\ \ text(… as required)`

 

ii.  `text(Prove)\ \ 1/1^2+1/2^2+1/3^2+\ …\ +1/n^2<2-1/n\ \ text(for integer)\ \ n>=2`

`text(If)\ \ n=2`

`text(LHS)=1/1^2+1/2^2=5/4`

`text(RHS)=2-1/2=3/2 > text(LHS)`

`:.\ text(True for)\ \ n=2`

 
`text(Assume true for)\ \ n=k`

`text(i.e.)\ \ 1/1^2+1/2^2+\ …\ +1/k^2<2-1/k`

 
`text(Prove true for)\ \ n=k+1`

`text(i.e.)\ 1/1^2+1/2^2+\ …\ +1/k^2+1/(k+1)^2<2-1/(k+1)`

♦♦ Mean mark of 26%.
IMPORTANT: Students have to carefully examine part (i) of this question to provide the information required to prove true for `n=k+1`.
`text(LHS)` `=1/1^2+1/2^2+\ …\ +1/k^2+1/(k+1)^2`
  `<2-1/k+1/(k+1)^2`
  `< underbrace{(1/(k+1)^2-1/k+1/(k+1))}_text(<0  from part i)+2-1/(k+1)`
  `<2-1/(k+1)`

  
`=> text(True for)\ n=k+1`

`:.\ text(S)text(ince it is true for)\ n=2, text(by PMI, true for integral)\  n>=2`.

Calculus, 2ADV C4 2010 HSC 3b

  1. Sketch the curve  `y=lnx`.   (1 mark)

  2. Use the trapezoidal rule with 3 function values to find an approximation to `int_1^3 lnx\ dx`   (2 marks) 

  3. State whether the approximation found in part (ii) is greater than or less than the exact value of  `int_1^3 lnx\ dx`. Justify your answer.   (1 mark)

Show Answer Only
  1. `text(See Worked Solutions for sketch.)`
  2. `1.24\ \ text(u²)`
  3. `text(See Worked Solutions)`
Show Worked Solutions
i. 2010 3b image - Simpsons
MARKER’S COMMENT: Many students failed to illustrate important features in their graph such as the concavity, `x`-axis intercept and `y`-axis asymptote (this can be explicitly stated or made graphically clear).

 

ii.    `text(Area)` `~~h/2[f(1)+2xxf(2)+f(3)]`
  `~~1/2[0+2ln2+ln3]`
  `~~1/2[ln(2^2 xx3)]`
  `~~1/2ln12`
  `~~1.24\ \ text(u²)`    `text{(to 2 d.p.)}`

 

iii. 2010 13b image 2 - Simpsons

 

♦♦♦ Mean mark 12%.
MARKER’S COMMENT: Best responses commented on concavity and that the trapezia lay beneath the curve. Diagrams featured in the best responses.

`text(The approximation is less because the sides)`

`text{of the trapezia lie below the concave down}`

`text{curve (see diagram).}`

 

Probability, 2ADV S1 2013 HSC 15d

Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.

  1. Find the probability that Pat wins the game on the first throw.     (1 mark)

  2. What is the probability that Pat wins the game on the first or on the second throw?     (2 marks)

  3. Find the probability that Pat eventually wins the game.     (2 marks)

Show Answers Only
  1. `1/36`
  2. `(2521)/(46\ 656)\ \ text(or)\ \ 0.054`
  3. `36/71`
Show Worked Solutions

i.   `P\ text{(Pat wins on 1st throw)}=P(W)`

`P(W)` `=P\ text{(Pat throws 2 sixes)}`
  `=1/6 xx 1/6`
  `=1/36`

 

ii.  `text(Let)\ P(L)=P text{(loss for either player on a throw)}=35/36`

`P text{(Pat wins on 1st or 2nd throw)}` 

♦♦ Mean mark 33%
MARKER’S COMMENT: Many students did not account for Chandra having to lose when Pat wins on the 2nd attempt.

`=P(W) + P(LL W)`

`=1/36\ + \ (35/36)xx(35/36)xx(1/36)`

`=(2521)/(46\ 656)`

`=0.054\ \ \ text{(to 3 d.p.)}`

 

iii.  `P\ text{(Pat wins eventually)}`

`=P(W) + P(LL\ W)+P(LL\ LL\ W)+ … `

`=1/36\ +\ (35/36)^2 (1/36)\ +\ (35/36)^2 (35/36)^2 (1/36)\ +…`

 
`=>\ text(GP where)\ \ a=1/36,\ \ r=(35/36)^2=(1225)/(1296)`

♦♦♦ Mean mark 8%!
 COMMENT: Be aware that diminishing probabilities and `S_oo` within the Series and Applications are a natural cross-topic combination.

 
`text(S)text(ince)\ |\ r\ |<\ 1:`

`S_oo` `=a/(1-r)`
  `=(1/36)/(1-(1225/1296))`
  `=1/36 xx 1296/71`
  `=36/71`

 

`:.\ text(Pat’s chances to win eventually are)\  36/71`.