Timing in the MX2 exam (2 Viewers)

[Trev]

I remember because of this slut that used to be in my maths class:
T(rest of name) =2t/(1-t²)
Sucks = 2t/(1+t²)
Cock = (1-t²)/(1+t²)
First two both have 2t on top; third one has the numerator same as first denominator and it's denominator same as second denominator... lol.

 

[Stefano]

I remember because of this slut that used to be in my maths class:
T(rest of name) =2t/(1-t²)
Sucks = 2t/(1+t²)
Cock = (1-t²)/(1+t²)
First two both have 2t on top; third one has the numerator same as first denominator and it's denominator same as second denominator... lol.

Brilliant!

 

[ishq]

Nice.
But there is a problem with me and acronyms and shortenings - i feel all proud when i've brought something down to a few letters, all memorised, and then in the exam i completely screw up because i cant remember what is what.
Due to the fact that I have only one brain cell.

Anyhow, did the SBHS 2005 Trial - and GUESS WHAT KFUNK - it had the same question that I couldnt do in the Grammar one. So I gave this arrogant laugh and said "you cant fool me, you silly poo" and did it.

However, the paper got back at me - with 20 marks which I couldnt solve. So, does ANYBODY have the solutions? Or will I have to trudge to school tomorrow? OR shall I post up the 5 questions?

 

[KFunk]

Hey, Nice work . I think I've got the solutions to that paper somewhere under piles and piles of other papers but post up the questions anyway and if people can't be bothered answering them I'll see if I can be bothered scanning the relevant parts of the solutions... that is, if I can even work my scanner or find it under the many mountains of paper.

 

[ishq]

people! lollies if you solve these!

1. given that the integral of f(x) from 0 to a = the integral of f(a-x) from o to a, evaluate the integral of ln(tanx) from o to pi/2.

2. If for a complex number z it is given that conjugate of z = z, where z is not equal to 0, determine the locus of z.

3. A complex number z is such that arg(z+2) = pi/6 and arg(z-2) = 2pi/3. Find z, expressing your answer in the form a+ib where a and b are real.

4. Consider the polynomial P(x) = x^5 -ax +1. By considering turning points on the curve P(x), prove that P(x) = 0 has three distinct roots if a > 5[(1/2)^8/5]

5. HOw many sets of 5 quartets can be formed from 5 violinists, 5 viola players, 5 cellists and 5 pianists if each quartet is to consist of one player of each instrument.


And that is all which had me confunded (trademark - harry potter)
Please help!
Merci

 

[ishq]

OH!!!
and please integrate: 2-|x| from -3 to 3.

How the hell do you integrate absolute values?

 

[KFunk]

2. If for a complex number z it is given that conjugate of z = z, where z is not equal to 0, determine the locus of z.

If Z = x + iy then Z(bar) = x - iy

If Z = Z(bar) then im(Z) = y = 0, ∴ the locus of Z is y = 0

 

[Stefano]

people! lollies if you solve these!

1. given that the integral of f(x) from 0 to a = the integral of f(a-x) from o to a, evaluate the integral of ln(tanx) from o to pi/2.

2. If for a complex number z it is given that conjugate of z = z, where z is not equal to 0, determine the locus of z.

3. A complex number z is such that arg(z+2) = pi/6 and arg(z-2) = 2pi/3. Find z, expressing your answer in the form a+ib where a and b are real.

4. Consider the polynomial P(x) = x^5 -ax +1. By considering turning points on the curve P(x), prove that P(x) = 0 has three distinct roots if a > 5[(1/2)^8/5]

5. HOw many sets of 5 quartets can be formed from 5 violinists, 5 viola players, 5 cellists and 5 pianists if each quartet is to consist of one player of each instrument.

And that is all which had me confunded (trademark - harry potter)
Please help!
Merci

No lollies for me, I can't do any of those.

 

[Antwan23q]

3. A complex number z is such that arg(z+2) = pi/6 and arg(z-2) = 2pi/3. Find z, expressing your answer in the form a+ib where a and b are real.

for number 3, that is solved in http://community.boredofstudies.org/showthread.php?t=86802

because from the u minus the first from the second.

so you get
arg (z+2) - arg(z-2) = pi/2

 

[ishq]

I got that far too. But how do you find the complex number z from there?

 

[KFunk]

people! lollies if you solve these!

1. given that the integral of f(x) from 0 to a = the integral of f(a-x) from o to a, evaluate the integral of ln(tanx) from o to pi/2.

I might be wrong in this but using what they give you:

∫ ln(tanx) dx = ∫ ln(tan{π/2 -x}) dx (both integrals between 0 and π/2)

∫ ln(tanx) dx = ∫ ln(1/tanx) dx ... (since tan(π/2 - x) = 1/tanx)

∫ ln(tanx) dx = - ∫ ln(tanx) dx ... ( 'cause ln(1/tanx) = -ln(tanx)

so 2 ∫ ln(tanx) dx = 0

∴ ∫ ln(tanx) dx = 0

 

[KFunk]

5. HOw many sets of 5 quartets can be formed from 5 violinists, 5 viola players, 5 cellists and 5 pianists if each quartet is to consist of one player of each instrument.

My reasoning may be shaky but in the first ensemble you have 5 cellist options, 5 pianist options, 5 violist options and 5 violinist options. For the next you have 4x4x4x4 then 3x3x3x3 and so on giving:

5⁴ + 4⁴ + 3⁴ + 2⁴ + 1 = 979

(i'm not sure if this is right so correct me if you think I'm wrong)

 

[ishq]

that makes sense...
and its intelligent...

 

[KFunk]

4. Consider the polynomial P(x) = x^5 -ax +1. By considering turning points on the curve P(x), prove that P(x) = 0 has three distinct roots if a > 5[(1/2)^8/5]

P(x) = x⁵ - ax + 1 (I'm going to assume that a is positive)

P'(x) = 5x⁴ - a, so x = ±(a/5)^1/4 when P'(x) = 0

Here I used a bit of BS intuition... it only has 2 real inflection points indicating that it has a cubic shape where P(x) --> ∞ as x--> ∞ and vice versa for -∞. P(x) passes through the point (0,1) no matter what so you know that P(x) must cut the x-axis before the negative inflection point (there will always be a root there) so in order for there to be three roots the y-coordinate of the positive inflection point must be less than zero... Sorry about the shonky reasoning .

Anyhow, if the positive inflection point must have a y value less than zero then:

P[(a/5)^1/4] < 0

(a/5)^5/4 - a(a/5)^1/4 + 1 < 0

a^5/4(1/5)^5/4 - a^5/4(1/5)^1/4 < -1

a^5/4[(1/5)^1/4 - (1/5)^5/4] > 1

a^5/4[1 - 1/5] > 5^1/4

a^5/4 > 5^5/4/4

&there4; a > 5(1/2)^8/5

 

[belmomoo]

those questions were from sbhs 05 trial paper

 

[noah]

OH!!!
and please integrate: 2-|x| from -3 to 3.

How the hell do you integrate absolute values?

because the integral of x is x^2/2 you can ignore the absolute value signs as the out come will be positive anyway

therefore

int ( 2 - abs(x) ) from -3 to 3

2 int ( 2 - x ) from 0 to 3 (even function)

= 2[2x - x^2/2 ] from 0 to 3

= 2[(6 - 9/2) - (0)]

= 2[3/2]

= 3



this can be confirmed by considering the area.

 

[blackfriday]

geez thats pretty sweet.

i got 80-odd/120 in that paper, and im just wondering if that would have been enough for a band e4.

 

[justchillin]

Do you mind me asking what mark you got in the 2000 paper? I perceive you as a good benchmark coz I know that you'd be searching for ~97+/100 aligned.


...or in the case of the 1993 paper, notably the most difficult, 120/120...

2000: 105/120 (2hrs 56mins) if u want exact figures. But u can never be sure because the book I use doesnt give marking scale for every part...
Edit: sorry for this being completely off the topic but I dont come here much anymore, on account of latin eating hours away (scaling better be worth it!)

 

[Antwan23q]

we do not miss you, care to read the rest of the posts?

 

[danza108]

geez thats pretty sweet.

i got 80-odd/120 in that paper, and im just wondering if that would have been enough for a band e4.

Your talkin about the Sydney High 4u trial right? I personally thought Kourty went really soft, all of Q7 andQ 8, cept for hte projectile was free marks. Normally the High trials are impossible in q7 and 8, look at 2002.

low 80s came a rank of 60s in our trial. and last year they had 44 band E4s, I think in the reports, mid 80s for the trial was projected to band E4. if thats anything to go by. Top got 118.5