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Check out this website: http://patrickjmt.com/. It has various video tutorials.Alhamdolillah these videos helped alot. If u can give us the link for videos for summation of series it will be greatly admired.
w/s. thanks. But i did not understand the last two steps. Please explain.
AoA! If you've studied curve sketching in F.Maths, this is quite a typical question.Hey Guys can you help me out with Oct/Nov 2004 Paper 1, Question 10. Please and thank you
Parts (ii) and (iv):HELP GUYS!!!
I was wondering if you guys could help me with http://www.xtremepapers.com/CIE/International A And AS Level/9231 - Further Mathematics/9231_s05_qp_1.pdf . Question number 10. Particuarly part 2 and part 4. Thanks so much in advance
AoA!
Re: The Further Maths Thread
Thank you, Nibz!
Here's the first problem:
If the roots of the equation x^4 - px^3 + qx^2 - pq x + 1 = 0 are α, ß, Γ and delta(D) show that:
( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D) = 1.[/quoteerh... not to interrupt but abcd= -1 not +1Re: The Further Maths Thread
Oops...;p I guess I had allocated a misleading notation for the sum "α + ß + Γ + D." I should've used the symbol "S(1)" instead of just "S" to represent the sum, to avoid any type of confusion.
So, wherever I wrote "S," I indeed meant S(1). Accordingly, the product "( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D)" in the question lead to the expression "[S(1)]^4 - (∑ α)*[S(1)]^3 + (∑ αß)*[S(1)]^2 - (∑ αßΓ)*S(1) + αßΓD" upon simplification.
Then, from the equation "x^4 - px^3 + qx^2 - pq x + 1 = 0," ∑ α = p = S(1); ∑ αß = q; ∑ αßΓ = pq; αßΓD = 1
Leading to, ( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D) = [p]^4 - (p)*[p]^3 + (q)*[p]^2 - (pq)*p + 1
= p^4 - p^4 + q*p^2 - q*p^2 + 1
= 1
Note that none of S(2), S(3) or S(4) is involved in any of the expressions I wrote.
Hopin' I removed your confusion now....
P.S.
To be honest, I don't know of any further maths book covering this "elegant & exceptionally-simple + shortcut" method. However, the past-papers are full of questions on it. Just have a look at some of the questions about polynomials there (along with the corresponding er's/ms's), and you'll master this method very soon....!
therefore the whole thing supposed to be -1
since abcd= -e/a
)
They were not published by the CIE. So just use examiner reports to solve the papers before 2008.
Let x = cos θ
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