2011 hsc mx1 marathon (1 Viewer)

[blackops23]

heres a question:

View attachment 23277

well AC/AB = AB/BC

therefore 1+x/x = x/1

therefore x+1 = x^2

Part (ii) looks scary..

 

[apollo1]

thats ext 2 induction

aah sorry bout that ill delete it.

 

[apollo1]

well AC/AB = AB/BC

therefore 1+x/x = x/1

therefore x+1 = x^2

Part (ii) looks scary..

its not scary.

 

[blackops23]

its not scary.

post a solution?

 

[apollo1]

given that a < 0 find the solutions to the inequality:



leaving answers in terms of a.

this is definitely 3U btw.

 

[blackops23]

given that a < 0 find the solutions to the inequality:



leaving answers in terms of a.

this is definitely 3U btw.

-3/a < x < -2/a and 0 < x < 1/a --> I'll suicide if I'm wrong.

 

[apollo1]

-3/a < x < -2/a and 0 < x < 1/a --> I'll suicide if I'm wrong.

lol thats actually wrong.

 

[blackops23]

lol thats actually wrong.

given that a<0 -- woops didnt read

EDIT: 1/a <= x < 0 or -2/a < x <= -3/a

apollo brah if im still wrong please show me how to do it correctly

btw i graphed it and equated the limits of the inequality

 

[barbernator]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2002exams/pdf_doc/mathemat_ext1_02.pdf HSC 2002 paper, Q7 b) ii) I integrated twice for the final part and subbed in x=-1, which gave me the correct RHS, yet my LHS came to 0 which wasn't the correct solution. ideas?

 

[hup]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2002exams/pdf_doc/mathemat_ext1_02.pdf HSC 2002 paper, Q7 b) ii) I integrated twice for the final part and subbed in x=-1, which gave me the correct RHS, yet my LHS came to 0 which wasn't the correct solution. ideas?

haha i just did that paper
you might be forgetting constants

 

[apollo1]

given that a<0 -- woops didnt read

EDIT: 1/a <= x < 0 or -2/a < x <= -3/a

apollo brah if im still wrong please show me how to do it correctly

btw i graphed it and equated the limits of the inequality

yeh thats right.

 

[barbernator]

AHH yes of course my bad!

 

[barbernator]

thanks hup! i reckon the rest of that paper was easy

 

[barbernator]

this one caught me out in the trials, integral of pi(1+tany)^2 dy

 

[nightweaver066]

this one caught me out in the trials, integral of pi(1+tany)^2 dy

New question:

 

[blackops23]

New to binomials here, for times when you have to integrate binomial expansions, what do you do about constants? do you evaluate them? if so how?

+ how do you do binomial induction? If there is such a thing....?

 

[nightweaver066]

New to binomials here, for times when you have to integrate binomial expansions, what do you do about constants? do you evaluate them? if so how?

Integrate with the limits 0 to what ever you want to sub in for x.

 

[blackops23]

Integrate with the limits 0 to what ever you want to sub in for x.

so I'm guessing in binomials with integration, you have to evaluate a DEFINITE integral?

 

[nightweaver066]

so I'm guessing in binomials with integration, you have to evaluate a DEFINITE integral?

I guess.. i'm not too sure though. If you don't do definite integrals, you'll always end up with a constant that i wouldn't know how to get rid of.

I've always used limits whenever solving binomial problems involving integration.

 

[hup]

I guess.. i'm not too sure though. If you don't do definite integrals, you'll always end up with a constant that i wouldn't know how to get rid of.

I've always used limits whenever solving binomial problems involving integration.

there is only one variable x and the constant so just let x = 0

definite integrals wont work if you hav to integrate twice