Mathematics Extension 1 Exam Predictions/Thoughts (3 Viewers)

[gunnerfan99]

anyone else accidentally get an inverse tan integral for the u=(cosx)^2 by subbing u^2 lol

I know 3 that did that

 

[luckystrike826]

The change of sign alone doesn’t tell you which way around the inequality is. You need to know the behaviour of the function in that domain to be able make that conclusion. You are slightly saved by the fact that x2 can be explictly shown to be greater than x1 in the particular example.

ahh i think I know what you mean. But it was given that x1 = k, and x2 = k+1/k^2, so x2> x1

 

[Skuxxgolfer]

Which past paper would you say is similar to the difficulty of this exam?

 

[akkjen]

@Trebla what would you say e4 cutoff is looking at the exam?

 

[TeheeCat]

anyone else accidentally get an inverse tan integral for the u=(cosx)^2 by subbing u^2 lol

BAHAHAHHA YES, that was one of the silly mistakes I made in this exam :') kms

 

[botselenium]

For that f(x1) < alpha < f(x2) question, I did the argument Trebla said about f(x) being increasing in the domain and how there are no stat pts after the line x=k. But I forgot to mention that the thing about the sign changing. However I said: by Newton's Method since f(x) is increasing (I showed this), each approximation can only take us closer to the root (as there are no more stat pts); hence f(x1) < alpha (the actual root) < f(x2). Do you guys think this would be enough to garner 2 marks or have I lost one for not doing the sign change thingo? Thanks in advance! =)

 

[deceased]

anyone else accidentally get an inverse tan integral for the u=(cosx)^2 by subbing u^2 lol

yes ... im deceased

 

[aduncan]

anyone else accidentally get an inverse tan integral for the u=(cosx)^2 by subbing u^2 lol

you reckon they will only take a mark off??

 

[botselenium]

Probably 2, because idk if they do carry on error within the same question

 

[fet]

my life depends on carry over error

 

[creativemaster]

You don't need to say that the function is continuous and increasing in the given domain. The question stated that K was a positive integer with x1. Since x2 is K+(1/K^2), the x2 must be greater than x1 since (1/K^2)>0. From this, it can be deduced that the root alpha must lie between x2 and x1 (with given proof of course).

 

[creativemaster]

I.e x1>alpha>x2 can not be true since x2>x1

 

[luckystrike826]

You don't need to say that the function is continuous and increasing in the given domain. The question stated that K was a positive integer with x1. Since x2 is K+(1/K^2), the x2 must be greater than x1 since (1/K^2)>0. From this, it can be deduced that the root alpha must lie between x2 and x1 (with given proof of course).

i agree but i still think f(x) is continuous is important because if it isn't continuous, there doesn't have to be a root between x2 and x1

 

[Idkwhattoput]

hey lads what do we reckon the raw mark needs to be for a state rank?

No way it'll be below 70/70. Even then, I don't think full marks will be enough. You'd need a really good 2u/4u mark aswell.

 

[ShoeMat]

No way it'll be below 70/70. Even then, I don't think full marks will be enough. You'd need a really good 2u/4u mark aswell.

They take your other subject marks into account??!!

 

[luckystrike826]

They take your other subject marks into account??!!

well 4u doubles ur 3u mark but i've not clue how everything works

 

[Drdusk]

They take your other subject marks into account??!!

Yeah to even out ties iirc. Like if 3 people tie on 100 RAW but one of them got 99 in 4u and the others got like 95, then the person who got 99 in 4u will get the higher rank.

 

[Idkwhattoput]

They take your other subject marks into account??!!

Only if there are ties. But in 3u, I'm fairly sure it is usually a tie between like 20/30 people for 100. So generally, they do need to look at 2u/4u marks.

 

[greetings]

you reckon they will only take a mark off??

Nah we’ll prob get 1/3 for it ://

 

[Idkwhattoput]

My solution to 14biii) (x1<a<x2 question) is as follows:
Consider x1:
The parabola intersects with the positive branch of the hyperbola, and the hyperbola is asymptotic (from the positive side) towards x=k. Therefore, the intersection of the parabola and the hyperbola (x=a) must lie to the right of x=k. So, x1<a.
Consider x2:
To the right of x=a (intersection), the y value of the parabola is always greater than the y value of the hyperbola. If we substitute x=x2, we find that the y value for the parabola is indeed greater than the y value of the hyperbola at x=x2 (I proved this). Therefore, x2 must lie to the right of a. So x2>a

Overall, we obtain x1<a<x2

Unfortunately it wasn't worded as good in the exam, but I think it was understandable with a bit of effort. Will this get me full marks for that part?