• 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 2015 MX2 Marathon (archive) (2 Viewers)

Status
Not open for further replies.

porcupinetree

not actually a porcupine
Joined
Dec 12, 2014
Messages
664
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon

http://imgur.com/w1DCp1x

I can't use the program (forget what it's called) that everyone uses for writing/answering questions, so:

If P(x) = x^4 - 8x^3 + 30x^2 - 56x + 49 has a non-real double zero, solve the equation P(x) = 0 over C and factorise P(x) fully over R
 
Last edited:

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: HSC 2015 4U Marathon

Let z = x + iy, where x, y ∈ ℝ.

Then




Equating real and imaginary parts, x - y = 0 and y - x = 0, so x = y, i.e. the locus is the straight line y = x, which is a straight line passing through the origin with slope 1.
 

porcupinetree

not actually a porcupine
Joined
Dec 12, 2014
Messages
664
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon

Here's my working, I feel as though there's a much easier solution to part i that's right under my nose, however this is the way that I did it.

https://imgur.com/UoF6Kk9

Axio, may I ask a question? In your answer to my previous question, one of your lines of working was:
Can I ask how you concluded that P(x) was that after just looking at it? I can understand that the constant of P(x) must be the square of the constant inside the brackets, but did you use any particular method for concluding that the coefficient of x (in the brackets) was -4?

Next question:

If P(x) = x^6 + x^4 + x^2 + 1, show that the solutions of the equation P(x) = 0 are among the solutions of the equation x^8 - 1 = 0. Hence factorise P(x) fully over R.
 

Axio

=o
Joined
Mar 20, 2014
Messages
484
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon

Can I ask how you concluded that P(x) was that after just looking at it? I can understand that the constant of P(x) must be the square of the constant inside the brackets, but did you use any particular method for concluding that the coefficient of x (in the brackets) was -4?
I did it by equating the coefficients of the x terms. So if you just let -2Re(z)=n, then by looking at the brackets: (x^2 +nx+7)(x^2+nx+7) you know that the sum of the x terms will be 7nx+7nx which =-56x. Then 14n=-56, n=-4.

Here's my working, I feel as though there's a much easier solution to part i that's right under my nose, however this is the way that I did it.
Your method is good :). An alternative method would be:





 
Last edited:

Axio

=o
Joined
Mar 20, 2014
Messages
484
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon

Next question:

If P(x) = x^6 + x^4 + x^2 + 1, show that the solutions of the equation P(x) = 0 are among the solutions of the equation x^8 - 1 = 0. Hence factorise P(x) fully over R.








---

 

porcupinetree

not actually a porcupine
Joined
Dec 12, 2014
Messages
664
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon

https://imgur.com/AbcGhk4

---

z satisfies the equation |z - 2 - 2i| = √2.
a) Sketch the locus of the point P representing z on an Argand diagram.
b) Find the maximum and minimum values of |z| and the values of z for which these extremes are attained.
c) Find the range of possible values of arg z.
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: HSC 2015 4U Marathon

https://imgur.com/AbcGhk4

---

z satisfies the equation |z - 2 - 2i| = √2.
a) Sketch the locus of the point P representing z on an Argand diagram.
b) Find the maximum and minimum values of |z| and the values of z for which these extremes are attained.
c) Find the range of possible values of arg z.
a) The locus is |z - (2+2i)| = √2. This is a circle with centre (2,2), radius √2, sketched here:

http://graphsketch.com/parametric?m..._lines=1&line_width=4&image_w=850&image_h=850

b) Closest and furthest points on the locus to and from the origin lie on the line through the origin and the circle's centre, which is the line y = x. The circle's equation is (x - 2)2 + (y - 2)2 = 2. Sub. y = x and solve for x:

(x - 2)2 + (x - 2)2 = 2
⇒ (x - 2)2 = 1
⇒ (x - 2) = -1, +1
x = 1, 3.

So as y = x, the closest point on the locus to the origin is (1,1), and the furthest is (3,3).
Hence the minimum value of |z| is , occurring when z = 1+i, and the maximum value of |z| is , occurring when z = 3+3i.

c) Clearly, arg z is acute. Its extreme values occur when OP is tangent to the circle. Let the equations of the tangents be y = mx (m, the slope, will have two values, as there are two tangents).

When the tangent touches the circle, substituting y = mx into (x - 2)2 + (y - 2)2 = 2, we have

(x - 2)2 + (mx - 2)2 = 2
x2 - 4x + 4 + m2x2 - 4mx + 4 = 2
⇔ (m2+1)x2 - 4(m+1)x + 6 = 0.

For tangents, set discriminant to 0:

16(m+1)2 - 4(m2+1)(6) = 0

Solving this quadratic equation for m gives

Since m = tan(arg z), arg z = tan-1 m

⇒ the range of possible values for arg z is that

One can further show using compound angle formulae that , and by symmetry of the situation, , so
 
Last edited:

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: HSC 2015 4U Marathon

Alternatively, denoting the centre of the circle be C, point of contact be P, then

 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: HSC 2015 4U Marathon

NEXT QUESTION

Prove that all the roots of the equation



Denote the equation by . Let be a root of , where and (so ). We show that r must be greater than ½.

As is a root, we have .

Taking the modulus of both sides,





(by the extended triangle inequality)



, equality iff (since )



So we have , equality iff .

Using the geometric series sum formula on the LHS, and noting that the LHS is strictly less than the infinite sum of powers of r (since r > 0), this inequality yields

. (The left-most expression here is the value of the aforementioned infinite sum)

i.e. .

Now, we can see that if , the above inequality does not hold, as when , the LHS is equal to 1 (not greater than it), and the LHS is monotonic increasing for r > 0. Thus we cannot have , i.e. r = |ω| must be greater than ½, as required. ■
 
Last edited:

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: HSC 2015 4U Marathon



If the expression is real, find the set of all possible values for .
 

porcupinetree

not actually a porcupine
Joined
Dec 12, 2014
Messages
664
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon



If the expression is real, find the set of all possible values for .
I have got as far as saying that tanx - sinx + cosx = 1, but don't really know where to go from there
 

Ton5698

New Member
Joined
Apr 9, 2013
Messages
24
Location
Eastern Suburbs
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon

just wondering but what method did you use to input mathematics online like that Sy123?
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2015 4U Marathon

just wondering but what method did you use to input mathematics online like that Sy123?
This forum has built-in LaTeX generator

LaTeX is a code for doing math writing, on this forum if you put your LaTeX code in between [.tex] and [./tex] tags (NOT including the dots)

You can look at this to get familiar with/copy paste the LaTeX code: http://www.codecogs.com/latex/eqneditor.php
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top