• 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 2016 MX2 Marathon (archive) (1 Viewer)

Status
Not open for further replies.

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2016 4U Marathon

Note is the distance between a point on the circle centered on the origin with radius 12 and the circle with center and radius 5. Looking at this diagram:


We can see that is the shortest distance whilst is the longest.

Don't you have to justify something before using the diameter/radius despite it's obviousness?
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: HSC 2016 4U Marathon

Don't you have to justify something before using the diameter/radius despite it's obviousness?
It seems to take too long to prove to make it part of that question. So we can make it the next question.



 
Last edited:

Chensanity

New Member
Joined
Apr 1, 2015
Messages
1
Gender
Male
HSC
N/A
Re: HSC 2016 4U Marathon

Help :(

Find the roots of the following equation: (hint divide by x^2 and reduce to quadratic)
3x^4-x^3+4x^2-x+3=0
 

internalsarekey

New Member
Joined
Aug 19, 2015
Messages
16
Gender
Male
HSC
2016
Re: HSC 2016 4U Marathon

b) The polynomial equationx4+ 4x3−2x2−12x−3 = 0 has rootsα,β,γandδ.Find the polynomial equation with roots (α+ 1), (β+ 1), (γ+ 1) and (δ+ 1).(c) Hence, or otherwise, solve the equationx4+ 4x3−2x2−12x−3 = 0.
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2016 4U Marathon

b) The polynomial equationx4+ 4x3−2x2−12x−3 = 0 has rootsα,β,γandδ.Find the polynomial equation with roots (α+ 1), (β+ 1), (γ+ 1) and (δ+ 1).(c) Hence, or otherwise, solve the equationx4+ 4x3−2x2−12x−3 = 0.
Use _^_ notation to denote a power.

Replace x with x-1 for the equation with roots (α+ 1) etc.

(x-1)^4 + 4(x-1)^3 - 2(x-1)^2 - 12(x-1) - 3 = 0
With the help of the binomial theorem this is simplified to
x^4-8x^2+4=0
Use completing the square to factorise said expression:
x^4-8x^2+16=12
(x^2-4)^2=12
x^2-4=+-2sqrt(3)
x^2=4+-2sqrt(3)
Hence:
x=sqrt(4+2sqrt3)
x=sqrt(4-2sqrt3)
x=-sqrt(4+2sqrt3)
x=-sqrt(4-2sqrt3)

If one can see ahead and identify that:
(1+sqrt3)^2 = 1 + 2sqrt3 + 3 = 4 + 2sqrt(3)
and similarly (1-sqrt3)^2 = 4 - 2sqrt(3)

The roots can be rewritten as:
x=1+sqrt3
x=1-sqrt3
x=-1+sqrt3
x=-1-sqrt3

Now, note that these are the roots (α+ 1) etc. because we already established these are the roots to the polynomial equation which we just solved.

Hence:
α=sqrt3
β=-sqrt3
γ=-2+sqrt3
δ=-2-sqrt3
 

hedgehog_7

Member
Joined
Dec 13, 2015
Messages
50
Gender
Male
HSC
2016
Re: HSC 2016 4U Marathon

The equation x^3 + 2x + 1 = 0 has the roots a,b,c

Find the monic cubic equation with roots (b+c)/a^2 , (a+c)/b^2 and (a+b)/c^2
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2016 4U Marathon

Not sure if this is a next question or SOS so, one way of approaching this:

a+b+c=0. So b+c=-a. Hence a root (b+c)/a^2 just turns into -a/a^2 --> -1/a. And similarly other roots -1/b and -1/c. So we're finding the polynomial with THOSE roots.
 

glittergal96

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
Re: HSC 2016 4U Marathon

Here is a good exercise to test understanding of the connection between integration and differentiation:

 

Ambility

Active Member
Joined
Dec 22, 2014
Messages
336
Gender
Male
HSC
2016
Re: HSC 2016 4U Marathon

Here is a good exercise to test understanding of the connection between integration and differentiation:

The 2016 cohort hasn't supposed to have done integration at this level, has it?
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2016 4U Marathon

Lol multidimensional calculus was never a part of something like 4u. That's why glittergal gave reference to the partial derivative
 

glittergal96

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
Re: HSC 2016 4U Marathon

The 2016 cohort hasn't supposed to have done integration at this level, has it?
Having thought about this a little more, indeed the question requires a little more than can be expected of HS students.

If any want to still give it a crack, the following tools from outside curriculum are the basic things you might need:

-The epsilon-delta definition of limits.
-The mean value theorem.
-The extreme value theorem.
(there are several alternate routes though, some of which use different things)

It is first year calculus basically.

The problem in higher dimensions is a bit harder.

P.s. There really should be a university marathon, it is a shame the actual tertiary maths subforums are so dead.
 
Last edited:

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: HSC 2016 4U Marathon

Having thought about this a little more, indeed the question requires a little more than can be expected of HS students.

If any want to still give it a crack, the following tools from outside curriculum are the basic things you will need:

-The epsilon-delta definition of limits.
-The mean value theorem.
-The extreme value theorem.
(there are several alternate routes though, some of which use different things)

It is first year calculus basically.

The problem in several variables is a bit harder.
Can't it also be done using the Leibniz rule for differentiating under the integral sign and also multivariable chain rule? Or is this assuming too many rules for what you had in mind?
 

glittergal96

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
Re: HSC 2016 4U Marathon

Can't it also be done using the Leibniz rule for differentiating under the integral sign and also multivariable chain rule? Or is this assuming too many rules for what you had in mind?
Finding an expression for this quantity is precisely a derivation of one form of the Leibniz rule. (And so assuming a weaker version of the Leibniz rule as a starting point is cheating a little.)

What I was looking for was such a derivation based on first principles. (And the fundamental theorem of calculus of course).
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top