Help In Understanding. (1 Viewer)

[Skeptyks]

ya

 

[GoldyOrNugget]

In the markers comments (for extension 2 at least), they've complained about long conclusions before. All you need to do is 'and thus, the statement is true for all integers n >= 1 by induction." or something equivalent.

 

[RishBonjour]

I just say "hence proved by induction for n> whatever" if the marker is having a bad day - won't take marks off.

 

[Shadowless]

@Shadow - you do NOT have to write that long conclusion for induction. In the words of my maths teacher, you do "not get marks for a memorised sentence".

It's a misconception that you NEED to write the statement. While there's nothing wrong with the statement, it's unnecessary.

Do you mean the 'long conclusion' by Skeptyks? And if not, the one I issued myself? Or is my one just fine the way it is? Concise and straight to the point? And if not either his or mine, what do you propose I should write instead?

 

[Skeptyks]

i think they mean your one

 

[shravan_872]

can someone please tell me all the circle theorems we will need to know for tomorrow?

 

[Shadowless]

Question:

Wow. I don't know about you guys, but I reckon this question is pretty difficult, and with all the pressure of being in the hall and stuff, I probably wouldn't be able to complete this in the exam. Do you guys reckon a similar sort of question will appear in this year's paper?

 

[Skeptyks]

There's always polynomials but I don't think you can tell whether it'll be an easy or hard one

also how do you do that question haha?

 

[shongaponga]

There's always polynomials but I don't think you can tell whether it'll be an easy or hard one

also how do you do that question haha?

i) Find first derivative and stationary points - it'll fall out to the result needed.

ii) Since stationary points are at x = +- √3/3, and A > 0 .'. f(√3/3) > 0. Sub in x = √3/3 into the equation for f(x), manipulate --> state that it must be greater than 0 since f(√3/3) is greater than 0. It falls out from there in a few lines.

iii) If f(-1) = 1, and the only zero occurs when A < (3√3)/2 [from(ii)], then the zero must occur when x < -1. Hence there is no zero in the given interval. I'd draw the graph here of f(x) to support argument.

iv) First derivative --> stationary points occur g'(@) = 0. After some manipulation the equation falls into the form Ax^3 - Ax + 1 = 0. So .'. 0 < A < (3√3)/2. Now since -pi/2 < @ < pi/2, then -1 < sin@ < 1. But from (iii), Ax^3 - Ax +1 has no zeros in the interval -1 < x < 1 when 0 < A < (3√3)/2. Hence g'(@) =/= 0 and g(@) has no stationary points for the domain.

v) Since g(@) has no stationary points, and g'(@) > 0, then g(@) is monotonic increasing. Hence it has an inverse for the domain.

Sorry I didn't tex this, don't have the time atm

 

[Shadowless]

There's always polynomials but I don't think you can tell whether it'll be an easy or hard one

also how do you do that question haha?

Answers: (Going to assume you know how to do the first part.)

 

[GoldyOrNugget]

I think questions of that difficulty, and harder, will definitely come up. Look at, say, q5a from 2011, or any q7s from past years. If you can't get them, you just have to steal as many marks as possible with partial solutions and working-out.

 

[shongaponga]

Personally I think the recent papers haven't been nearly as hard as the mid-late 90's. Apart from 2008.

 

[Skeptyks]

Thanks but I still don't understand how you would get to the idea of substituing x = root 3 / 3 and letting it be >0

 

[Shadowless]

I think questions of that difficulty, and harder, will definitely come up. Look at, say, q5a from 2011, or any q7s from past years. If you can't get them, you just have to steal as many marks as possible with partial solutions and working-out.

Hmm... I didn't really find 2011 that difficult... as the questions that really stump me are the ones where you're meant to intuitively know when to draw graphs and how to interpret it and stuff. =( Other than that I'd say my other weakness in 3U would be the infamous "perms & combs", speaking of which I will post up a few more tonight.

 

[GoldyOrNugget]

The fuck? I found mid-to-late 90's trivial. Across 1995-1999 I was finishing half an hour early and getting consistent 98%+s. Then 2000 kicked my ass, and everything recent has been a struggle to even finish all the questions, not to mention avoiding silly mistakes.

 

[Shadowless]

The fuck? I found mid-to-late 90's trivial. Across 1995-1999 I was finishing half an hour early and getting consistent 98%+s. Then 2000 kicked my ass, and everything recent has been a struggle to even finish all the questions, not to mention avoiding silly mistakes.

LMAO you must wish you did the HSC in the pre 21st century era don't you?

 

[Shadowless]

Here's an easy question to kick off the night...

Question: How many integers, greater than 999 but not greater than 5000, can be formed with the digits 0, 1, 2, 3, 4, 5 if repetition of digits is allowed?

 

[v1]

OP, for first problem,
solving for l in x^2 + lx + 12 = 0 and finding minimum value for l using dl/dx is pretty easy to understand, also works

 

[Shadowless]

OP, for first problem,
solving for l in x^2 + lx + 12 = 0 and finding minimum value for l using dl/dx is pretty easy to understand, also works

I know. That's what I did on my first attempt, but it was way more work relative to what the answers had. Hence, I wanted to understand the answers, as it will make me more efficient in tomorrow's exams. However, if something of a similar disposition arises in tomorrow's exams and I somehow forget what I learnt today, then I would probably take the long way and solve for minimum through differentiation, as it makes more sense to me that way.