HSC 2015 MX2 Marathon (archive) (2 Viewers)

[FrankXie]

Re: HSC 2015 4U Marathon

NEXT QUESTION

 

[porcupinetree]

Re: HSC 2015 4U Marathon

NEXT QUESTION

Here's my go:

√(x^2 + y^2) = √[(x-4)^2 + (y+2)^2]
Simplification gives:
y = 2x - 5

|z| + |z - 2i| => |2z - 2i| = 2|z - i| (triangle inequality)
Clearly, equality is reached when x = 0
Sub x=0 into y = 2x - 5 = 2*0 - 5 = -5.
|-5| + |-5i+2i| = |-5| + |-3i| = 5 + 3 = 8.

I feel like my third last line ('clearly, equality...') is not very (for want of a better word) good. I tried solving it algebraically but I got nowhere, I just ended up with three square root terms, which, if x = 0 is subbed in, produces a correct result, however it just looked like a big mess. Is there a better way?

 

[dan964]

Re: HSC 2015 4U Marathon

hmm I am not sure whether that's right consider whether |z| lies on the
x-axis.
You get |z| = 2.5??
|z-2i| approx. 3 point something
which is less than 8 when added together.
|z-2i| is the distance from (0,2) to line not (0,-2)

so you actually consider the distance between
the line Y=2X+5 and (0,0)
and (0,2) added together

 

[braintic]

Re: HSC 2015 4U Marathon

Here's my go:

√(x^2 + y^2) = √[(x-4)^2 + (y+2)^2]
Simplification gives:
y = 2x - 5

|z| + |z - 2i| => |2z - 2i| = 2|z - i| (triangle inequality)
Clearly, equality is reached when x = 0
Sub x=0 into y = 2x - 5 = 2*0 - 5 = -5.
|-5| + |-5i+2i| = |-5| + |-3i| = 5 + 3 = 8.

I feel like my third last line ('clearly, equality...') is not very (for want of a better word) good. I tried solving it algebraically but I got nowhere, I just ended up with three square root terms, which, if x = 0 is subbed in, produces a correct result, however it just looked like a big mess. Is there a better way?

Geometry beats algebra

 

[porcupinetree]

Re: HSC 2015 4U Marathon

hmm I am not sure whether that's right consider whether |z| lies on the
x-axis.
You get |z| = 2.5??
|z-2i| approx. 3 point something
which is less than 8 when added together.
|z-2i| is the distance from (0,2) to line not (0,-2)

so you actually consider the distance between
the line Y=2X+5 and (0,0)
and (0,2) added together

Oh my goodness, I'm stupid. Yes, I made the mistake that it was from (0,-2) not (0,2)

 

[Kaido]

Re: HSC 2015 4U Marathon

Find the values that the real numbers a and b can take for z=1 to be a root of the complex equation: iz^2+(ia-1)z+(i-b)=0
(starting easy as i'm not sure if everyone has begun the polynomials topic)

Note: Doesn't require '4U polynomial' knowledge

 

[PhysicsMaths]

Re: HSC 2015 4U Marathon

Find the values that the real numbers a and b can take for z=1 to be a root of the complex equation: iz^2+(ia-1)z+(i-b)=0
(starting easy as i'm not sure if everyone has begun the polynomials topic)

Note: Doesn't require '4U polynomial' knowledge

Let P(z) = iz^2 + (ia-1)z + (i-b)
For z = 1 to be a root, P(1) = 0
∴ P(1) = i + ia-1 + i-b = 0
i(2+a) - (1+b) = 0
∴ a = -2, b = -1

 

[PhysicsMaths]

Re: HSC 2015 4U Marathon

Write in simplest form:
(sinθ - icosθ)^-8

 

[Kaido]

Re: HSC 2015 4U Marathon

(sinθ - icosθ)^-8 = 1/(i(cosθ+isinθ))^8
(then by DeMoivre's theorem) = 1/(cos8θ+isin8θ)

 

[enigma_1]

Re: HSC 2015 4U Marathon

(sinθ - icosθ)^-8 = 1/(i(cosθ+isinθ))^8
(then by DeMoivre's theorem) = 1/(cos8θ+isin8θ)

OMG THETA caps

 

[Kaido]

Re: HSC 2015 4U Marathon

Let P(x) be a polynomial with degree of 1996. If P(n) = 1/n for n = 1,2,3...1997, Find the value of P(1998)

Edit: Not sure if it is technically under the 4U syllabus, so just ignore it (By all means, try it if you wish)

 

[InteGrand]

Re: HSC 2015 4U Marathon

Let P(x) be a polynomial with degree of 1996. If P(n) = 1/n for n = 1,2,3...1997, Find the value of P(1998)

Edit: Not sure if it is technically under the 4U syllabus, so just ignore it (By all means, try it if you wish)

Let .

We are given that for .

is a 1997th degree polynomial, and has its 1997 roots given to us, so can be factorised as using the given roots, where .

Now, , so the leading coefficient is , constant term is –1.

Therefore, using product of roots of (which is ), we have

, so

.

Substituting x = 1998,







.

Solving for P(1998) yields .

 

[porcupinetree]

Re: HSC 2015 4U Marathon

NEXT QUESTION

The equation x^3 + 3x^2 - 4x + 5 = 0 has roots A, B, Y.

Evaluate A^3 + B^3 + Y^3

 

[Kaido]

Re: HSC 2015 4U Marathon

so since A.B.Y are the roots of x^3 + 3x^2 - 4x + 5 = 0, then: A,B,Y are values of x
-> Rearranging: x^3=-3x^2 +4x -5
so, A^3 + B^3 + Y^3 = -3A^2 +4A -5 - 3B^2 +4B -5 - 3C^2 +4C -5 = -3(A^2+B^2+C^2) +4(A+B+C) -15

->A^2+B^2+C^2 = (A+B+Y)^2 - 2(AB +BC +AC) = 17

therefore A^3 + B^3 + Y^3 = -3(17) +4(-3) -15
=-78 (not good at subtraction)

(Prob mistake somewhere, hella hard to type in this shifty font)

Edit: corrected the mistake, thanks braintic.

 

[braintic]

Re: HSC 2015 4U Marathon

so since A.B.Y are the roots of x^3 + 3x^2 - 4x + 5 = 0, then: A,B,Y are values of x
-> Rearranging: x^3=-3x^2 +4x -5
so, A^3 + B^3 + Y^3 = -3A^2 +4A -5 - 3B^2 +4B -5 - 3C^2 +4C -5 = -3(A^2+B^2+C^2) +4(A+B+C) -15

->A^2+B^2+C^2 = (A+B+Y)^2 - 2(AB +BC +AC) = 1
-
therefore A^3 + B^3 + Y^3 = -3(1) +4(-3) -15
=-30

(Prob mistake somewhere, hella hard to type in this shifty font)

Yep, slight error.

AB+AC+BC = -4
So A^2+B^2+C^2 = 3^2 -2(-4) = 17

 

[Kaido]

Re: HSC 2015 4U Marathon

Yep, slight error.

AB+AC+BC = -4
So A^2+B^2+C^2 = 3^2 -2(-4) = 17

wallah im stoned on chocolates

 

[dan964]

Re: HSC 2015 4U Marathon

Next question
(I) Using Demoivre's find expressions. for sin 5x and cos 5x

 

[Kaido]

Re: HSC 2015 4U Marathon

-> (cosθ+isinθ)^5: (c=cosθ, s=sinθ. too lazy bro)
=(by Binomial expansion/Pascals) c^5+5c^4(is)+10c^3(is)^2+10c^2(is)^3+5c(is)^4+(is)^5
(grouping real and im.) = c^5 -10c^3s^2 +5cs^4 + yeah
therefore equating: cos5x=c^5-10c^3(1-c^2)+5c(1-c^2)^2
=c^5-10c^3+10c^5+5c-10c^3+5c^5
=16c^5-20c^3+5c
Same thing for sin5x but group the sin and yeah,

(as beforementioned, am stoned, home alone, ceebs)

 

[dan964]

Re: HSC 2015 4U Marathon

hence show that cos 9 and sin 9 satisfy the equation
16x^5-20x^3+5x-1/sqrt(2) = 0
angles are in degrees please convert to radians

 

[FrankXie]

Re: HSC 2015 4U Marathon

Here's my go:

√(x^2 + y^2) = √[(x-4)^2 + (y+2)^2]
Simplification gives:
y = 2x - 5

|z| + |z - 2i| => |2z - 2i| = 2|z - i| (triangle inequality)
Clearly, equality is reached when x = 0
Sub x=0 into y = 2x - 5 = 2*0 - 5 = -5.
|-5| + |-5i+2i| = |-5| + |-3i| = 5 + 3 = 8.

I feel like my third last line ('clearly, equality...') is not very (for want of a better word) good. I tried solving it algebraically but I got nowhere, I just ended up with three square root terms, which, if x = 0 is subbed in, produces a correct result, however it just looked like a big mess. Is there a better way?

good try but i'm afraid it is invalid.