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****Leaving Certificate: Higher Level Maths Discussion****
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chatterboxxx95 wrote: »Oh holy sh!te is amortisation proof 100% definitely on the course! ! Our teaher never even mentioned it to us
It's on the syllabus, says so on the bottom right of page 25:
http: //www .ncca.ie/en/Curriculum_and_Assessment/Post-Primary_Education/Project_Maths/Syllabuses_and_Assessment/LC_Maths_for_examination_in_2014 .pdf0 -
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(This is me testing myself to see if I can remember :pac:)
I believe surjective means that a horizontal line can intersect it more than once, all y values have to have at least one corresponding x value (so, it can't have asymptotes or whatnot), think something along the lines of a cubic equation
Then injective means that it has to have only one, or no, corresponding y values, so a horizontal line will either intersect it once or not at all, not all y values need to have an x value
Then bijective is essentially a combination of the two, it means every y value has one, and only one, x value, and every y value is corresponding to an x value, something along the lines of say, a straight line graph, a horizontal line will cut it only once :P
Can someone confirm if all this is correct...?
Yes this is right but don't forget this ; in order for it to be surjective the range must equal the codomain I.e. every y value has one x value. But in injective the range does not have to equal the codomain. I.e some y values may not have an x value. Then in order for it to be bijective it must be both in ejective and surjective! Therefore the range must equal the codomain (as in surjective) And remember. A function can only have an inverse if and only if it is bijective
(makesure you know how to get the inverse) 0 -
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2014lchelp wrote: »Yes this is right but don't forget this ; in order for it to be surjective the range must equal the codomain I.e. every y value has one x value. But in injective the range does not have to equal the codomain. I.e some y values may not have an x value. Then in order for it to be bijective it must be both in ejective and surjective! Therefore the range must equal the codomain (as in surjective) And remember. A function can only have an inverse if and only if it is bijective (makesure you know how to get the inverse)
thanks man this makes sense now0 -
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2014lchelp wrote: »Yes this is right but don't forget this ; in order for it to be surjective the range must equal the codomain I.e. every y value has one x value. But in injective the range does not have to equal the codomain. I.e some y values may not have an x value. Then in order for it to be bijective it must be both in ejective and surjective! Therefore the range must equal the codomain (as in surjective) And remember. A function can only have an inverse if and only if it is bijective (makesure you know how to get the inverse)
To get the inverse you let the function equal y and rearrange it to have x in terms of y and them sub in x for y, correct?
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Deriving the Amortisation formula is 100% on the course. I've a feeling there will be a lot of financial maths tomorrow too.
What's amortisation I feckin hate financial mathsTo get the inverse you let the function equal y and rearrange it to have x in terms of y and them sub in x for y, correct?
You got it
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chatterboxxx95 wrote: »Holy mother of god...thank you people who confirmed that proof is on the course.
Having heart attacks here because realising there's a growing list of stuff my teacher never taught us. It's times like these I utterly fúcking dispair.
A lot of that list are important because they need to be understood beforehand as opposed to done from scratch on the day. But in the bigger picture of the whole exam they aren't worth too much marks compared to the more intuitive questions, so don't fret the 'small' stuff! 0 -
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If someone has these handy could they post it up please2014lchelp wrote: »Our teachers gave us a list of learning stuff and I'd recommend to learn before tomorrow;
How to construct root 2 and 3
Proof of de moivres theorem by induction
Proof by contradiction (there's about 8 examples) incl.
Prove the sum of 2 odd numbers is always even
Prove of geometric series by induction
Differentiation by first principles formula
I think that's it if you have these down at least if one of these comes up you'll feel more at ease with it . Hope this helps!
Im hoping de moivres proof and proof by contradiction come up.0 -
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c0unterpart wrote: »I've never been so lost in my life
That is the easiest way to think about it!
Do you know what the horizontal line test is?0 -
c0unterpart wrote: »I've never been so lost in my life
I find the easiest way to remember is just that a SURjective function can be cut at OVER one point by a horizontal line... a horizontal line can hit it more than once.
so any horizontal line has to hit it AT LEAST once.
As opposed to injective which is where a horizontal line will hit it AT MOST once.
this isnt much help if any like, theoretical questions come up, but its how I remember.
I admit its a lot less helpful to anyone who doesnt do french0 -
If someone has these handy could they post it up please
Im hoping de moivres proof and proof by contradiction come up.
Just a guess at sum of 2 odd numbers...
2m + 1 and 2n +1 with m and n integers, are both odd as they aren't divisible by 2
their sum is
2m + 1 + 2n + 1 = 2m + 2n + 2 = 2(m+n+2) which is even since it is divisible by 2
Hence the sum of two odd numbers is even0 -
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Yeah, you have to derive the Amortization formula.
Its pretty much just 'simplifying' a geometric series.
Here's a nice explanation if you forgot how to do it: en.wikipedia.org/wiki/Amortization_calculator#Derivation_of_the_formula
Paper 1 tomorrow I am so excited!
Good Luck everyone!0
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