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Hardest ever question 8 (1 Viewer)

FinalFantasy

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1989 Q8
a) Find all values of θ with 0≤θ≤2 pi such that sin θ-√(3) cosθ=1.
b)The difference between a real number r and the greatest integer less than or equal to r is called the fractional part of r, F(r). Thus F(3.45)=0.45.
Note that for all real numbers n, 0≤F(r)<1,
i) Let @=2136 log _(base 10)_ 2
Given that F(@)=7.0738... x 10^-5
observe that F(2@)=14.1476... x 10^-5
F(3@)=21.2214... x 10^-5

(a) Use your calculator to show that
log_(base 10)_ 1.989<F(4223@)<log_(base 10)_1.990
(b) Hence calculate an integer M such that the ordinary decimal representation of 2^M begins with 1989. Thus 2^M=1989... .

ii) Let r be a real number and let m and n be non-zero integers with m*n.
(a) Show that if F(mr)=0, then r is rational.
(b) Show that if F(mr)=F(nr), then r is rational.
iii) Suppose that b is an irrational number. Let N be a positive integer and consider the fractional parts F(b), F(2b), ..., F((N+1)b).
(a) Show that these N+1 numbers F(b), ..., F((N+1)b) are all distinct.
(b) Divide the interval 0≤x<1 into N subintervals each of length 1/N and show that there must be integers m and n with m*n and 1≤m, n≤N+1 such that
F((m-n)b)<1/N.
iv) Given that log_(base 10)_2 is irrational, choose any integer N such that 1/N<log_(base 10)_(1990\1989);
note that in (i), F(@)<log_(base 10)_(1990\1989).
Use (iii) to decide whether there exists another integer M such that 2^M=1989... .

lol, hope no typing mistakes:p
 

A l

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Question 8 b), was the one that no one got out. By the way, the a)s and b)s that Final Fantasy typed out under each roman numeral are meant to be noted alpha and beta. They are like sub-questions of part b) of question 8. In other words, here's how question 8 looked like, just to minimise any confusion (I think Final Fantasy missed a few bits).

Note: means not equal to. The actual sign somehow did not come out in the post.

Question 8

a) Find all values of θ with 0≤θ≤2 pi such that sin θ-√(3) cosθ=1.

b)The difference between a real number r and the greatest integer less than or equal to r is called the fractional part of r, F(r). Thus F(3.45)=0.45.
Note that for all real numbers r, 0≤F(r)<1,

i) Let a =2136 log _(base 10)_ 2

Given that F(a)=7.0738... x 10^-5

observe that
F(2a)=14.1476... x 10^-5
F(3a)=21.2214... x 10^-5

(α) Use your calculator to show that
log_(base 10)_ 1.989 < F(4223a) < log_(base 10)_ 1.990.
(β) Hence calculate an integer M such that the ordinary decimal representation of 2^M begins with 1989. Thus 2^M=1989... .

ii) Let r be a real number and let m and n be non-zero integers with m ≪ n.
(α) Show that if F(mr)=0, then r is rational.
(β) Show that if F(mr)=F(nr), then r is rational.

iii) Suppose that b is an irrational number. Let N be a positive integer and consider the fractional parts F(b), F(2b), ..., F((N+1)b).

(α) Show that these N +1 numbers F(b), ..., F((N+1)b) are all distinct.
(β) Divide the interval 0≤x<1 into N subintervals each of length 1/N and show that there must be integers m and n with mn and 1 ≤ m, nN+1 such that F((m-n)b)<1/N.

iv) Given that log_(base 10)_2 is irrational, choose any integer N such that 1/N < log_(base 10)_ (1990/1989);
note that in (i), F(a) < log_(base 10)_ (1990/1989)
Use (iii) to decide whether there exists another integer M such that 2^M=1989....

I've got the solution to question 8 b), do you want it to be posted? Or do you want to work it out yourselves? It's fairly long.
 
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justchillin

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it is indeed a cruel question, I rekon the examiner was trying to be arrogant by giving his exam an "unanswerable" question. Could you post up the solns please...
 
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Affinity

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justchillin said:
What year do u guyz rekon had the toughest question...
That's a dumb question... everyone doing the HSC knows that the hardest one is this years
 

Antwan23q

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A l said:
b)The difference between a real number r and the greatest integer less than or equal to r is called the fractional part of r, F(r). Thus F(3.45)=0.45.
Note that for all real numbers r, 0≤F(r)<1,
Wtf is a fractional part?
 

thunderdax

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As soon as I saw the title of this thread i thought 1989 too. Dumbest question ever.
 

Stefano

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"e is trancedental, therefore e is irrational" :D

I wonder if we would get marks for that? Haha!

PS. Well written buchanan
 

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