• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

HSC 2013-14 MX1 Marathon (archive) (4 Viewers)

Status
Not open for further replies.

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Apply this to the second thing of each term




Consider:



Expand LHS and RHS and equate the coefficient of x^(n-2)

Yep well done, alternatively you can consider the co-efficient of 1/x^2 in the expansion:

 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

lol hsc 2000 we did this one in class today.
The symmetry argument is really elegant, but the more obvious way which is kinda what I was leaning to, involved calculus and tangents (i.e. not very elegant)
 
Joined
Apr 1, 2011
Messages
1,012
Location
District 12
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

The symmetry argument is really elegant, but the more obvious way which is kinda what I was leaning to, involved calculus and tangents (i.e. not very elegant)
is the symmetry argument basically saying that since pi/4 is the projection angle that leads to the longest range (once you prove that of course), it will also hit the ground at pi/4 in the opposite direction, hence they are perpendicular?
 
Last edited:

Makematics

Well-Known Member
Joined
Mar 26, 2013
Messages
1,829
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

is the symmetry argument basically saying that since pi/4 is the projection angle that leads to the longest range (once you prove that of course), it will also hit the ground at pi/4 in the opposite direction, hence they are perpendicular?
i believe theta=pi/4 does not give max range. iirc it was alpha/2 + pi/4. you can prove this by differentiation or by inspection. the hsc 2000 sorta guides you through it in 4 parts for a total of 8 marks.
 
Joined
Apr 1, 2011
Messages
1,012
Location
District 12
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

i believe theta=pi/4 does not give max range. iirc it was alpha/2 + pi/4. you can prove this by differentiation or by inspection. the hsc 2000 sorta guides you through it in 4 parts for a total of 8 marks.
Oh oops, my bad then
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread













 
Last edited:

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Consider the individual graphs of and

Now there will always be one solution as they cross somewhere in the 3rd quadrant no matter what, however the values of 'a' and 'b' determine how many times they cross in the first quadrant.

As for has , then once this curve reaches a gradient equal to the gradient of that is a gradient of 'b', then after that there is no possible way for the two curves to intersect. Thus we deal with this condition, and so we find when has gradient 'b':





as we are only considering the first quadrant and so

Now if there is only 1 solution, then at we must see that is:





Similarly, if there are 2 real solutions, then at we must see that is:





Lastly, for 3 real solutions, then at we must see that is:





Now the equation can be re-arranged to and thus and . It follows then that for this equation and there are 2 real solutions.

Now note that this doesn't mean there are 2 real solutions and 1 imaginary solution, it means that there are 2 DISTINCT real solutions, with one of these solutions being a double root.

Hence if we let the roots be and then we can say:



Expanding and equating co-efficients gives:

by equating

by equating constants.

Thus solving these simultaneously we get:







Now we know that the double root must be in the first quadrant as that is where the two curves and will be tangent to one another. Thus and so we get and

Hence

So the roots are -6 and a double root at 3.
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Consider the individual graphs of and

Now there will always be one solution as they cross somewhere in the 3rd quadrant no matter what, however the values of 'a' and 'b' determine how many times they cross in the first quadrant.

As for has , then once this curve reaches a gradient equal to the gradient of that is a gradient of 'b', then after that there is no possible way for the two curves to intersect. Thus we deal with this condition, and so we find when has gradient 'b':





as we are only considering the first quadrant and so

Now if there is only 1 solution, then at we must see that is:





Similarly, if there are 2 real solutions, then at we must see that is:





Lastly, for 3 real solutions, then at we must see that is:





Now the equation can be re-arranged to and thus and . It follows then that for this equation and there are 2 real solutions.

Now note that this doesn't mean there are 2 real solutions and 1 imaginary solution, it means that there are 2 DISTINCT real solutions, with one of these solutions being a double root.

Hence if we let the roots be and then we can say:



Expanding and equating co-efficients gives:

by equating

by equating constants.

Thus solving these simultaneously we get:







Now we know that the double root must be in the first quadrant as that is where the two curves and will be tangent to one another. Thus and so we get and

Hence

So the roots are -6 and a double root at 3.
Yep nice work.
 

AnimeX

Member
Joined
Aug 11, 2012
Messages
588
Gender
Male
HSC
N/A
Re: HSC 2013 3U Marathon Thread

Can I assume max range is when theta = 45 degrees? (have no idea how to prove..)

Additionally can I assume the parabolic symmetry? v up = v down?

=S, not sure how to do without assumptions.
 
Last edited:

Makematics

Well-Known Member
Joined
Mar 26, 2013
Messages
1,829
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Can I assume max range is when theta = 45 degrees? (have no idea how to prove..)

Additionally can I assume the parabolic symmetry? v up = v down?

=S, not sure how to do without assumptions.
i believe theta=pi/4 does not give max range. iirc it was alpha/2 + pi/4. you can prove this by differentiation or by inspection. the hsc 2000 sorta guides you through it in 4 parts for a total of 8 marks.
.
trust me if you need help do the 2000 hsc last q, it guides u through it
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Can I assume max range is when theta = 45 degrees? (have no idea how to prove..)

Additionally can I assume the parabolic symmetry? v up = v down?

=S, not sure how to do without assumptions.
You can't make those assumption as there is a plane at an angle to the x-axis.
 

AnimeX

Member
Joined
Aug 11, 2012
Messages
588
Gender
Male
HSC
N/A
Re: HSC 2013 3U Marathon Thread

You can't make those assumption as there is a plane at an angle to the x-axis.
Haha, can you dumb this down for me?

Why is the plane at an angle to the x-axis? how does the question infer this and what does this mean?
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Given prove that:

Considering the identity:



Then expanding both sides, equate co-efficient of on both sides knowing that , the result comes immediately.

=======


 

Makematics

Well-Known Member
Joined
Mar 26, 2013
Messages
1,829
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Haha, can you dumb this down for me?

Why is the plane at an angle to the x-axis? how does the question infer this and what does this mean?
look at the original question where sy drew a diagram. the plane is at angle of alpha to the x -axis, the question tells you that.
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 4)

Top