2011 hsc mx1 marathon (3 Viewers)

apollo1

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



leaving answers in terms of a.

this is definitely 3U btw.
 
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blackops23

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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
 
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apollo1

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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

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

blackops23

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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

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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.
 
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nightweaver066

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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

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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
 

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