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More questions from the ends of papers (1 Viewer)

CM_Tutor

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There has been a discussion of some of the harder stuff that can turn up at the ends of papers, so I thought I'd post some stuff from later questions of some 4u half-yearlies I have. I've skipped the motion, as people probably haven't done it yet, but please advise if I'm wrong...

1. (For this question, I'm going to use a in place of alpha, for the sake of clarity.)

(a) Provided sin(a / 2) <> 0, show that cos(a / 2) = sin a / 2sin(a / 2)
Similarly, show that if sin(a / 2) and sin(a / 4) are both <> 0, cos(a / 2) * cos(a / 4) = sin a / 4sin(a / 4)

(b) Prove by mathematical induction that if n is a positive integer, and sin(a / 2<sup>n</sup>) <> 0, then
cos(a / 2) * cos(a / 4) * ... * cos(a / 2<sup>n</sup>) = sin a / 2<sup>n</sup>sin(a / 2<sup>n</sup>)

(c) Hence, deduce that a = sin a / {[cos(a / 2)] * [cos(a / 4)] * [cos(a / 8)] * ... } and that
pi = 1 / {(1/2) * sqrt(1/2) * sqrt[(1/2) + (1/2)sqrt(1/2)] * sqrt{(1/2) + (1/2)sqrt[(1/2) + (1/2)sqrt(1/2)]} * ...}

2. (a) Let @ = tan<sup>-1</sup>x + tan<sup>-1</sup>y. Show that tan@ = (x + y) / (1 - xy)

(b) If tan<sup>-1</sup>x + tan<sup>-1</sup>y + tan<sup>-1</sup>z = pi / 2, show that xy + yz + zx = 1

(c) Let W<sub>n</sub> = tan<sup>-1</sup>x<sub>1</sub> + tan<sup>-1</sup>x<sub>2</sub> + ... + tan<sup>-1</sup>x<sub>n</sub>, where n is a positive integer.
Show, by mathematical induction or otherwise, that tan W<sub>n</sub> = - Im(w<sub>n</sub>) / Re(w<sub>n</sub>)
where w<sub>n</sub> = (1 - ix<sub>1</sub>)(1 - ix<sub>2</sub>)...(1 - ix<sub>n</sub>)

3. The quartic polynomial f(x) = x<sup>4</sup> + bx<sup>3</sup> + cx<sup>2</sup> + dx + e has two double zeroes at alpha and beta, where the coefficients b, c, d, and e are real, but the zeroes alpha and beta may be complex.

(a) Express b, c, d and e in terms of the double zeroes alpha and beta, and hence show that:
(i) d<sup>2</sup> = b<sup>2</sup>e
(ii) b<sup>3</sup> + 8d = 4bc

(b) Now, suppose that b = 2 and e = 1, and that the double zeroes alpha and beta are both non-real complex numbers.
(i) Show that alpha and beta are cube roots of 1.
(ii) Write f(x) as the product of polynomials irreducible over the real numbers.
 
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Calculon

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(a) Provided sin(a / 2) <> 0, show that cos(a / 2) = sin a / 2sin(a / 2)
Similarly, show that if sin(a / 2) and sin(a / 4) are both <> 0, cos(a / 2) * cos(a / 4) = sin a / 4sin(a / 4)

cos(a/2)=sina/2sin(a/2)

cos(a/2) = 2sin(a/2)cos(a/2)/2sin(a/2)
cos(a/2)= cos(a/2)
LHS = RHS
 
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CM_Tutor

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When you are asked to prove something, you should never rewrite it until you get a true statement, and then assert that it must be true. This is a technique of 'proof' that is really frowned upon in Extn 2, as it isn't really a proof. Otherwise, the following would be valid:

1 = -1
1<sup>2</sup> = (-1)<sup>2</sup>
1 = 1
So, LHS = RHS

A proof should work on one side at a time, ie
LHS =
=
=
...
= RHS
or start from a provably true statement, like 2 > 1, or sin 2x = 2sin x * cos x
or be based on the structure of contradiction.

Calculon, for 1(a) you have the right idea, so let's look at a more valid way of expressing it. You have worked to a provably true statement, so work backwards from there, ie

cos(a / 2) = 2sin(a / 2)cos(a / 2) / 2sin(a / 2), as sin(a / 2) <> 0, given
= sin(2 * a / 2) / 2sin(a / 2)
= sin a / 2sin(a / 2), as required

or alternately,

sin a = sin[2 * (a / 2)] = 2sin(a / 2)cos(a / 2)
So, sin a / 2sin(a / 2) = cos(a / 2), on division by 2sin(a / 2), as sin(a / 2) <> 0 (given).
 

Estel

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CM_Tutor are you allowed to manipulate both sides independent of each other and establish LHS=RHS?
 

Xayma

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Of course you could always work backwards on another bit of paper and then copy it down the write way. Thats how all those hard to factorise polynomials get done in one line in my tests ;)
 

Affinity

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It's usually more convenient to do:

a = b (true statement)
ma = mb
...
LHS = RHS

than
LHS=
=
...
= RHS
 

Affinity

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and regarding estel's question..

yes, you could do:

RHS=
=
...
= K

then

LHS =
=
...
= K

then conclude LHS=RHS
 

Xayma

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As long as you dont start raising things to even powers ;) Edit: or 0 or timising by 0
 

maniacguy

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Sprinkling some comments (note: when I say 'this should be easy', I'm not always talking about now. It should definitely be trivial by the time of the exam in seven months, but if you take a bit of time right now, don't worry about it... much... :D):

(WARNING: There will also be hints. If you don't want those, don't read this yet!)

1.
(a) This should be easy. It's the feelgood mark to start off.
(b) This is an induction that wouldn't be out of place in a 3u exam.
(c) The part that's meant to be tricky.
Hint: recall that limit(sin(x)/x,x->0) = 1, and so we can take x = a/2<sup>n</sup>
Hint: notice in the second part that you can take the 1/2 in the denominator of the RHS over to the LHS.

2.
(a) Straightforward again.
(b) Not hard, but not trivial (i.e. I wouldn't give it to a 2u student, but probably would to a 3u student - I may be selling 2u students short here, but on balance probably not. I may also be overestimating 3u students, but doubt it.)
(c) Induction is the easiest way to go. Anyone who's memorised the formula for tan(a<sub>1</sub>+a<sub>2</sub>+...+a<sub>n</sub>), you're not allowed to quote it. Also, take this opportunity to examine why you are memorizing that thing. Deriving is acceptable, but memorizing is most likely a waste of time. PM me if you come up with any amazing uses for it.

(Yes, I did derive it back then, and no, I didn't bother memorizing it because I couldn't work out a use for the thing that was worth memorizing it for! One way is to work out tan(n*x) and then generalize it, before using induction to prove yourself right. This is actually a much nicer way to derive the formula though!)

3.
(a) - will be fairly standard as a polynomials HSC question. Maybe slightly on the harder side, but then again, maybe not (depends on which question it gets chucked in, I guess. As a q 7/8 I would call it a bit easy). Make sure you're comfortable with it.
(b) Somewhat longer, if not harder (although it is a bit harder than (a)). Hint: alpha and beta must be complex conjugates.
 

Affinity

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xyama: actually, you can raise expressions to even powers when working forward (ie, not the 'backward' method), the operations you need to becareful of are those such as squareroots

by the way,
q1 is from 2003 q7, where did the other 2 questions come from?
 
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CM_Tutor

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Originally posted by Estel
CM_Tutor are you allowed to manipulate both sides independent of each other and establish LHS=RHS?
Yes, in just the way Affinity suggested,
LHS =
=
= ... = K
RHS =
=
= ... = K = LHS
Originally posted by Xayma
Of course you could always work backwards on another bit of paper and then copy it down the write way. Thats how all those hard to factorise polynomials get done in one line in my tests ;)
Yes, you can, and that is a common way around this problem - esp useful with harder 3u inequalities that seem to pop up at the ends of trials.
Originally posted by Affinity
It's usually more convenient to do:

a = b (true statement)
ma = mb
...
LHS = RHS

than
LHS=
=
...
= RHS
I disagree. Strongly. This method of 'proof' is flawed, unless you know that the 'true' statement a = b is actually true. In the case of 1(a), above, you do not know it is true, because you are trying to prove it!. Such a method starts with the assumption that the result you're after is true, and demonstrate that this do not produce a contradiction. This does not constitute a proof, despite the fact that such 'proofs' are commonly offered. This method can produce all sorts of problems - most won't come up in the HSC, but that does not make it any more valid.

I strongly recommend that everyone banishes this method of 'proof' from their repetoire immediately, or sooner if possible. You are, of course, free to take this advice, or ignore it as you please. Just be warned - there are examiners out there that won't accept it. :)
Originally posted by Xayma
As long as you dont start raising things to even powers ;) Edit: or 0 or timising by 0
These aren't the only times when this runs into problems.
 

CM_Tutor

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Originally posted by maniacguy
Sprinkling some comments (note: when I say 'this should be easy', I'm not always talking about now. It should definitely be trivial by the time of the exam in seven months, but if you take a bit of time right now, don't worry about it... much... :D):
I agree with the implied comment from maniacguy - these questions would not be 7 or 8 in a trial or HSC, but they would be the ends of half yearlies. I know this because they are from the ends of half yearlies! I probably should have made this clearer in my introductory comment at the start of this thread - I figured that half yearlies were more useful to you at the moment. :)
2.
(a) Straightforward again.
(b) Not hard, but not trivial (i.e. I wouldn't give it to a 2u student, but probably would to a 3u student - I may be selling 2u students short here, but on balance probably not. I may also be overestimating 3u students, but doubt it.)
(c) Induction is the easiest way to go. Anyone who's memorised the formula for tan(a<sub>1</sub>+a<sub>2</sub>+...+a<sub>n</sub>), you're not allowed to quote it. Also, take this opportunity to examine why you are memorizing that thing. Deriving is acceptable, but memorizing is most likely a waste of time. PM me if you come up with any amazing uses for it.
2(b) is not suitable for 2u, as inverse trig is not in their syllabus. I agree it could be a 3u question, but since it is used for (c) - a 4u quesion involving complex numbers - it is not out of place here.

2(c) - I wouldn't agree that induction is the easiest way to go - it certainly is not the quickest, as there is an 'otherwise' method that's much quicker (and more elegant) - but induction is certainly the obvious way to go. :)

It's always worth looking for otherwise methods as alternatives to induction - there is often a much quicker method. For example, can anyone see a 2 line way to prove that n<sup>3</sup> + 2n is a multiple of 3 for all positive integers n?
Originally posted by Affinity
by the way,
q1 is from 2003 q7, where did the other 2 questions come from?
That may be true, but it wasn't my source.
Question 1 is Q6(b) - the last question - on the SGS 1999 4u half yearly paper
Question 2 is Q6(b) - the last question - on the SGS 2002 Extn 2 half yearly paper
Question 3 is Q5(b) - the second last question - on the SGS 1998 4u half yearly
 

Affinity

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hey CM
what did you think I mean by

"a=b (true statement)"?

it means you start from something that is true.. and not the meaning you suggested

And it's completely valid..
 

grimreaper

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on question 3(a) should we just do it by considering the sum of the roots, etc where the roots are a,a,b,b?? Or should we use the derivative at that point already?
 

AGB

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Re: Re: More questions from the ends of papers

Originally posted by maniacguy
Sprinkling some comments (note: when I say 'this should be easy', I'm not always talking about now. It should definitely be trivial by the time of the exam in seven months, but if you take a bit of time right now, don't worry about it... much... :D):

(WARNING: There will also be hints. If you don't want those, don't read this yet!)

1.
(a) This should be easy. It's the feelgood mark to start off.
(b) This is an induction that wouldn't be out of place in a 3u exam.
(c) The part that's meant to be tricky.
Hint: recall that limit(sin(x)/x,x->0) = 1, and so we can take x = a/2<sup>n</sup>
Hint: notice in the second part that you can take the 1/2 in the denominator of the RHS over to the LHS.

2.
(a) Straightforward again.
(b) Not hard, but not trivial (i.e. I wouldn't give it to a 2u student, but probably would to a 3u student - I may be selling 2u students short here, but on balance probably not. I may also be overestimating 3u students, but doubt it.)
(c) Induction is the easiest way to go. Anyone who's memorised the formula for tan(a<sub>1</sub>+a<sub>2</sub>+...+a<sub>n</sub>), you're not allowed to quote it. Also, take this opportunity to examine why you are memorizing that thing. Deriving is acceptable, but memorizing is most likely a waste of time. PM me if you come up with any amazing uses for it.

(Yes, I did derive it back then, and no, I didn't bother memorizing it because I couldn't work out a use for the thing that was worth memorizing it for! One way is to work out tan(n*x) and then generalize it, before using induction to prove yourself right. This is actually a much nicer way to derive the formula though!)

3.
(a) - will be fairly standard as a polynomials HSC question. Maybe slightly on the harder side, but then again, maybe not (depends on which question it gets chucked in, I guess. As a q 7/8 I would call it a bit easy). Make sure you're comfortable with it.
(b) Somewhat longer, if not harder (although it is a bit harder than (a)). Hint: alpha and beta must be complex conjugates.
hey is this true?? (could someone like CM or affinity answer please?? "feel good question", "straightforward again", "fairly standard" etc etc.... is this actually true?? like are these questions considered to be easy??
 

CM_Tutor

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Originally posted by Affinity
hey CM
what did you think I mean by

"a=b (true statement)"?

it means you start from something that is true.. and not the meaning you suggested

And it's completely valid..
Sorry, Affinity, I have misinterpreted what you were saying. I thought you were referring to the type of approach Calculon had taken above, starting from the statement we are trying to prove, ie
cos(a/2)=sina/2sin(a/2)

cos(a/2) = 2sin(a/2)cos(a/2)/2sin(a/2)
cos(a/2)= cos(a/2)
LHS = RHS
rather than the approach that I had taken, starting from a provably true statement, ie
sin a = sin[2 * (a / 2)] = 2sin(a / 2)cos(a / 2)
So, sin a / 2sin(a / 2) = cos(a / 2), on division by 2sin(a / 2), as sin(a / 2) <> 0 (given).
You are quite right, the method you are discussing is valid, and I should have read what you wrote more carefully. Sorry. I hope that you would agree that the method I am trying to discourage is not valid.
 
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CM_Tutor

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AGB, as it says above, questions 1 and 3 are from the last part of the last question of recent SGS 4u half yearlies. Question 2 is from the last part of the second to last question. These are not easy questions, as indicated by their placement in the exams.

Having said that, difficult questions will include relatively straight forward parts. I don't think 1(a) or 2(a) are difficult, and by the time you get to the trials, a strong student won't have a problem with them.

I think maniacguy's comments are a little overstated, and I've already made some comment on the comments made about Q2. However, when read in the context of the introductory comment "note: when I say 'this should be easy', I'm not always talking about now. It should definitely be trivial by the time of the exam in seven months, but if you take a bit of time right now, don't worry about it... much... :D", they aren't unreasonable (IMO).

Take the comment on 1(a) being a 'feelgood mark' as an example. As you can see from my post above, this can be done in 2 - 3 lines, so its hardly difficult. In that sense, it is an easy mark. However, it's also a hint for the induction to follow, as the same basic technique is needed, and so it's a logical lead in to 1(b). The induction in 1(b) would not be out of place in a 3u paper, but it wouldn't be anywhere near the start. It's an example of 'Harder 3u', which makes up around half of the 4u paper, and it's a necessary lead in to the totally 4u part (c) - I know it's Harder 3u again, but it would not be reasonable in a 3u paper.

Does that make you feel any better?
 

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