More questions from the ends of papers (1 Viewer)

[CM_Tutor]

Grimreaper, 3(a) should definitely be approached from root theory, rather than by trying to solve f'(x) = 0. Whilst you know two of the roots of f'(x) = 0, you don't know the third, and you don't want to have to try to solve the general cubic.

Shkspeare, you wanted a clue for 3(b)(i) - Sorry I didn't notice your request earlier.

You have alpha + beta = -1, and (alpha * beta)² = 1, correct? This second equation leads to two possible values of alpha * beta - take each in turn. One must be excluded, the other leads to the conclusion that alpha³ = 1, where alpha is not real.

 

[AGB]

Originally posted by CM_Tutor
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... ", 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?

much better thank you

i can do the 'feelgood questions', but i think i would struggle with a couple of others in an exam...

 

[DcM]

hey CM

its cool to have all ur questions to think about..

just wondering...cos i have assessments coming up on:
locus problems of complex no.
conics
polynomials..

wondering if it wud be too much hassle if u can start a new thread and give questions on those topics =)

if not then dw..thanks anywayz

 

[DcM]

(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)

RHS = 2sin(a/2)cos(a/2)/2sin(a/2)
= cos(a/2)
= LHS ............( u can start from RHS yea?)

Similarly, for Prove: cos(a / 2) * cos(a / 4) = sin a / 4sin(a / 4)

RHS = 2 sin(a/2)cos(a/2) / 4sin(a/4)
= 4sin(a/4)cos(a/4)cos(a/2) / 4sin(a/4)
= cos(a/2) cos(a/4)
= LHS

 

[DcM]

for 1 (b)
i got up to the part where u used the assumption to make
LHS = sin a/ 2^k sin (a/2^k) * cos (a/2^k+1)

gimme a hint of wat to do next...like wat does cos (a/2^k+1) equal?

 

[maniacguy]

Originally posted by CM_Tutor
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 n³ + 2n is a multiple of 3 for all positive integers n?

Hmm.... Yes, there is a more elegant method, but I would say that induction is still 'easier' in the sense that it requires less thought. In an exam scenario where time to think is at a premium that's valuable. Depends on how fast you urn through an inductive algorithm, I suppose.

(Quick check you mean the same method I'm thinking of - write
(1-i*x_1) in mod-arg form and then do the multiplication?)


As for the other:

n^3 + 2n
= n^3 - n + 3n
= n(n^2-1) + 3n
= n(n-1)(n+1) + 3n

Clearly one of n, (n-1), (n+1) must be divisible by 3 (3 consecutive integers) and 3n is divisible by 3, hence their sum is divisible by 3.

 

[maniacguy]

Originally posted by CM_Tutor
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... ", 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.

Well, my comment on 1(a) being a 'feelgood mark' was meant for now - by this time the double angle formulae for sin and cos should be engraved on the inside of your eyelids. Seeing an alpha and an alpha/2 in the expression therefore...

I agree that some of my other comments might be a little overstated, although I think that 1(a),1(b),2(a),2(b) are all approachable for the better 3u students, and in the case of 1(a) and 2(a) for the better 2u students...

(this means that the other questions are at 4u level, so not leading in much... in the case of 3(a) this is more because the quartic sum of roots etc. isn't taught to the 3u students than for real difficulty standard.)

Also bear in mind all of my comments (ever) are relative to the HSC. It's easier than working out which schools are doing which topics. Anyone who hasn't done Polynomials, yet, for instance, might feel I'm underplaying q3 a bit...

 

[CM_Tutor]

Originally posted by DcM
hey CM

its cool to have all ur questions to think about..

just wondering...cos i have assessments coming up on:
locus problems of complex no.
conics
polynomials..

wondering if it wud be too much hassle if u can start a new thread and give questions on those topics =)

if not then dw..thanks anywayz

Will see what I can do...

I don't know whether you tried them, but there were a few questions involving compex numbers and locus in the thread "Complex Numbers and Geometry" that I started - one still unanswered, and all worth working through (there are only very brief answers at the moment). Grey Council also posted some good complex number locus questions, to which I added another.

I'll definitely post some conics.

 

[CM_Tutor]

DcM, your answer to 1(a) is fine - yes, you can work on RHS first (In fact, you can work on either side, or even both, just do it one side at a time).

As for 1(b), I have the induction hypothesis as:
cos(a / 2) * cos(a / 4) * ... * cos(a / 2^k) = sin a / 2^ksin(a / 2^k) ______(**)
And the statement to prove for n = k + 1 as:
cos(a / 2) * cos(a / 4) * ... * cos(a / 2^k+1) = sin a / 2^k+1sin(a / 2^k+1)

Then , LHS = cos(a / 2) * cos(a / 4) * ... * cos(a / 2^k+1)
= [cos(a / 2) * cos(a / 4) * ... * cos(a / 2^k)] * cos(a / 2^k+1)
= [sin a / 2^ksin(a / 2^k)] * cos(a / 2^k+1), using the induction hypothesis (**)
= sin a * cos(a / 2^k+1) / 2^ksin(a / 2^k)

I would then note from 1(a) that cos(@ / 2) = sin @ / 2sin(@ / 2), and I'd let @ = a / 2^k - As I said above, 1(a) is not only a feelgood mark, it's also the realisation necessary for the induction, and so it is a logical lead in to it.

 

[CM_Tutor]

Originally posted by maniacguy
Hmm.... Yes, there is a more elegant method, but I would say that induction is still 'easier' in the sense that it requires less thought. In an exam scenario where time to think is at a premium that's valuable. Depends on how fast you urn through an inductive algorithm, I suppose.

I have the worked solutions - as the examiner wrote it, the induction proof is over double the length, and I think that thinking about shorter methods is a desirable goal of Extn 2 Maths - I always favour thinking your way out of a problem is better than the 'bash through' approach, where one can be reasonably found. I guess this turns on how much insight a student has developed, as to whether the otherwise approach is easier.

(Quick check you mean the same method I'm thinking of - write
(1-i*x_1) in mod-arg form and then do the multiplication?)

I'll PM you - don't want to explain the otherwise approach while they are working.

As for the other:

n^3 + 2n
= n^3 - n + 3n
= n(n^2-1) + 3n
= n(n-1)(n+1) + 3n

Clearly one of n, (n-1), (n+1) must be divisible by 3 (3 consecutive integers) and 3n is divisible by 3, hence their sum is divisible by 3.

Absolutely - although the question was more directed at current Extn 2 students, but anyway...

 

[CM_Tutor]

Originally posted by maniacguy
Well, my comment on 1(a) being a 'feelgood mark' was meant for now - by this time the double angle formulae for sin and cos should be engraved on the inside of your eyelids. Seeing an alpha and an alpha/2 in the expression therefore...

I agree that some of my other comments might be a little overstated, although I think that 1(a),1(b),2(a),2(b) are all approachable for the better 3u students, and in the case of 1(a) and 2(a) for the better 2u students...

(this means that the other questions are at 4u level, so not leading in much... in the case of 3(a) this is more because the quartic sum of roots etc. isn't taught to the 3u students than for real difficulty standard.)

Also bear in mind all of my comments (ever) are relative to the HSC. It's easier than working out which schools are doing which topics. Anyone who hasn't done Polynomials, yet, for instance, might feel I'm underplaying q3 a bit...

I agree with you that 1(a) really is a feelgood mark - all I was saying was that that is not its only purpose, as it also illustrates the method needed for the induction. Have a look at the question DcM asked above about 1(b), and my answer, and I'm sure there are plenty of others who would need the hint. I agree that seeing alpha and alpha / 2 should trigger the double angle formulae, but it is not that straight forward for alpha / 2^k and alpha / 2^k+1, unless you have much experience. Remember that we bring more insight and experience ot a question than will a current student.

Agreed that 1(a), 1(b), 2(a) and 2(b) are approachable for better 3u students, but they are also necessary for where the problems are going. I disagree about 1(a) and 2(a) for 2u students - neither the expansion of sin(alpha + beta), nor inverse trig is in the 2u syllabus.

As for the overall difficulty of the problems, they are from the ends of half-yearlies. I feel that they are challenging enough for the moment - I agree that HSC questions 7 and esp. 8 will be harder, but they may also be more cross-topic, and we're about 200 days from the exam. There's plenty of time to reach that standard yet. (I realise that you did make your comments in the context of 'not always talking right now' - I'm just not sure that current students (who have not done an HSC) will necessarily appreciate how much they'll develop over the course of the year. )

 

[maniacguy]

Originally posted by CM_Tutor
Agreed that 1(a), 1(b), 2(a) and 2(b) are approachable for better 3u students, but they are also necessary for where the problems are going. I disagree about 1(a) and 2(a) for 2u students - neither the expansion of sin(alpha + beta), nor inverse trig is in the 2u syllabus.

You're kidding me, aren't you?
Grrr.... what have those bastards done to the syllabus?

I remember doing those in Year 10, and I'm sure we didn't do 3u stuff in Year 10. 2u maybe, but not 3u...

Besides, it's in the 2u Fitzpatrick, unless my memory fails me...

 

[CM_Tutor]

Originally posted by maniacguy
You're kidding me, aren't you?
Grrr.... what have those bastards done to the syllabus?

I remember doing those in Year 10, and I'm sure we didn't do 3u stuff in Year 10. 2u maybe, but not 3u...

Besides, it's in the 2u Fitzpatrick, unless my memory fails me...

No, I'm not kidding you. They aren't in the syllabus, and haven't been as far back as the early 80's - I can't speak for the syllabus before the one that started in about '83.

They also aren't in the copy of 2u Fitzpatrick that I have, unless you count the formulae like sin(90 - x), and similar with 180 +/- x, and 360 +/- x, and the radian equivalents.

Sorry.

 

[CM_Tutor]

We still haven't got all the answers to these ...

Do people want some hints?

 

[CM_Tutor]

Do people want me to post the answers to those questions that have not yet been answered? Alternately, are people still working on them?

 

[Grey Council]

I will give ALL of these questions a shot (whether they've been done or not) as soon as the holidays begin. I haven't had my first 4u assessment, thats why i haven't already tried doing them. ^_^

I'd say that wait for at least the holidays to end, then post up the solutions. Who knows, some people might actually do some work (ever though you've had your assessments now).

 

[Affinity]

maybe post the answers as attachments?

 

[nike33]

3)

alpha = w beta = z

z+w = -b/2
z²+4zw+w² = c
2zw(z+w) = -d
(zw)² = e

ai)

2zw(z+w)= -d
+-2sqr(e)(-b/2) = -d
+-sqr(e)b = -d
eb² = d²

ii)

(z+w)²+2zw= c
(-b/2)²-d(-b/2) = c
b²/4 + 2d/b = c
b³ + 8d = 4cb

b) f(x) = x⁴+2x³+3x²+2x+1

x² [(x+1/x)²+2(x+1/x)+1] = 0

solve quadratic let u = x+1/x or watever

and [x²+x+1]² = 0

[x²+x+1] = 0

hence z is a cube root of 1 as it satisfies [x²+x+1] now, as conj(z) = w as real co-effs then w is also a cube root of 1 (as they occur in conjugates) (given roots are complex)

 

[nike33]

ahhh i just typed answers for questiosn1/2 and it said ur not logged in blah blah ><

 

[CM_Tutor]

Nike33, your answer to 3(a) is fine, but with 3(b), how do you know that f(x) = x⁴ + 2x³ + 3x² + 2x + 1?

I know it is true, but in an exam, you would need to provide evidence to support such an assertion...