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# Year 13 Maths Help Thread

1. (Original post by RDKGames)
It is indeed 256. Also it's the area under the curve from -5 to 2, and above it from 2 to 3 (reference to the sketch).
But you can just work it out as -5 to 3 like ive done?
2. (Original post by kiiten)
But you can just work it out as -5 to 3 like ive done?
Yes that would give you the whole area enclosed by the curve and the x-axis, but your reference to the sketch is technically incorrect.
3. Ughh guys ive gone wrong again :/ ques 4.b)

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4. (Original post by kiiten)
Ughh guys ive gone wrong again :/ ques 4.b)

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If the area of the 2 sections are equal, then the integral between 0 and k will be 0, as the net area above the x axis will be 0, as it will cancel with the part of the curve that is below the x axis.
5. (Original post by B_9710)
If the area of the 2 sections are equal, then the integral between 0 and k will be 0, as the net area above the x axis will be 0, as it will cancel with the part of the curve that is below the x axis.
thanks
6. (Original post by kiiten)
thanks
Ah crap, just realised what I've said about the previous question is wrong, sorry. Anyway, k=root(8) in case you want it validated.
7. (Original post by RDKGames)
Ah crap, just realised what I've said about the previous question is wrong, sorry. Anyway, k=root(8) in case you want it validated.
Ahh - i got the notif but never saw your post so its fine. Thanks
8. (Original post by kiiten)
Ahh - i got the notif but never saw your post so its fine. Thanks
Also your method would still work if you equate the area from 2 to k to -4 rather that positive 4, because the area is negative.
9. (Original post by RDKGames)
Also your method would still work if you equate the area from 2 to k to -4 rather that positive 4, because the area is negative.
Eh? where did the 4 come from??
10. (Original post by kiiten)
Eh? where did the 4 come from??
That's the area from 0 to 2.
11. can 4/x^-1/2 be written in a different form?
12. (Original post by kiiten)
can 4/x^-1/2 be written in a different form?
Of course.
13. (Original post by kiiten)
can 4/x^-1/2 be written in a different form?
14. (Original post by RDKGames)
Of course.
(Original post by Zacken)

Thanks (sorry for asking easy ques - ive forgotten a lot of maths since the exams :3 )
15. Ughhh where am i going wrong. You have to find the coordinates of the minimum. Ive completely forgotten what you have to do (x is only 1/2)

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16. (Original post by kiiten)
Ughhh where am i going wrong. You have to find the coordinates of the minimum. Ive completely forgotten what you have to do (x is only -1/2 not 2 but why?)

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Complete the square. Or you could differentiate and find the value of x that gives f'(x)=0, then plug in to find f at that point.
17. (Original post by B_9710)
Complete the square. Or you could differentiate and find the value of x that gives f'(x)=0, then plug in to find f at that point.
i already found 2 values of x but they're wrong. why?
18. (Original post by kiiten)
i already found 2 values of x but they're wrong. why?
If f(x) = 6x^2 - 6x + 3 then f ' (x) = 12x - 6 so f ' (x) = 0 only when 12x - 6 = 0, not sure how you got two values of x from that.
19. (Original post by Zacken)
If f(x) = 6x^2 - 6x + 3 then f ' (x) = 12x - 6 so f ' (x) = 0 only when 12x - 6 = 0, not sure how you got two values of x from that.
yeah i forgot to differentiate. Thanks
20. (Original post by kiiten)
yeah i forgot to differentiate. Thanks
You even wrote "differentiation" on top!

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