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mesa, next time put a term test question; let the pain begin :P
let a=arctan(1/(2n-1)) => tana= 1/(2n-1)) a belongs to [0, pie/2]
b=arctan(1/(2n+1)) => tanb= 1/(2n+1) b belongs to [0, pie/2]
a>b a-b belongs to [0, ]
c=arctan(1/2n^2) => tanc=1/2n^2 c belongs to [0, pie/2 ]

Tan(a-b)=(Tana-Tanb)/(1+TanaTanb)
replace tana and tanb and put it in the equailty = c
and we get the solution


b) arctan(1/2) = arctan(1) - arctan(1/3)
arctan(1/8) = arctan(1/3) - arctan (1/5)
ect ect
arctan(1/2n^2) = arctan(1/(2n-1)) - arctan(1/(2n+1))
simplify you get: Sn=pie /4 - arctan(1/(2n+1))

c) Sn= Pie/4 - lim arctan= pie/4 - 0


NEXT
physics if you may
@jadf24: nice job, you got most of it right. review your solution again, because you've just done a grave mistake in part a): you found the range but you were asked to find the domain of definition.
better here than anywhere else. parts b) and c) are correct, I'm thinking you're ready for the heavy set. I'll start a new thread labeled "Kasti Quest - Math 4 LS&GS" for the math questions and "Kanj Unlimited - Physics for GS" for the physics questions.
mesa177 wrote@jadf24: nice job, you got most of it right. review your solution again, because you've just done a grave mistake in part a): you found the range but you were asked to find the domain of definition.
better here than anywhere else. parts b) and c) are correct, I'm thinking you're ready for the heavy set. I'll start a new thread labeled "Kasti Quest - Math 4 LS&GS" for the math questions and "Kanj Unlimited - Physics for GS" for the physics questions.
Let the Fun Begin ! Haha :)
Competition is a must or else it would feel like homework. Plus a time limit :)
Let's first see the answer: what's the domain of definition?
I'll have to go with Df of c: IR
Df of a : IR - {1/2)
Df of b: IR - {-1/2)
Though I’m not so sure


P.S.: The new curriculum doesn't emphasize on inverse of trigonometric functions. Hence, they rarely appear on exams anymore. However, our school curriculum does include this lesson yet does not concentrate and emphasize on them as much as they do on other lessons such as conics, irrational functions ect.. So I'm no expert in inverse of trigonometric functions.
jadf24 wroteI'll have to go with Df of c: IR
Df of a : IR - {1/2)
Df of b: IR - {-1/2)
Though I’m not so sure


P.S.: The new curriculum doesn't emphasize on inverse of trigonometric functions. Hence, they rarely appear on exams anymore. However, our school curriculum does include this lesson yet does not concentrate and emphasize on them as much as they do on other lessons such as conics, irrational functions ect.. So I'm no expert in inverse of trigonometric functions.
That's the problem with the new curriculum, it doesn't emphasize on the subjects that really matter (i.e. the ones that you're actually gonna need when you go to a university. Conics, though very interesting, are rarely used by most majors. Trigonometry and inverse trigonometric functions, however, are a core subject especially in the domain of engineering. Of course sequences, sets, probability, derivatives, integrals, and ordinary diffrential equations (ODE) and partial diffrential equations (PDE) are just as important.

As for the domain of definition, you could have easily deduced it from your solution in part b): you didn't resort to 0 because in arctan(1/2x^2) it cannot be defined, nor did you resort to irrational and rational fractions or negative integers and for a good reason: the equation is not defined on such sets. The equation given holds true to only one set: natural numbers with the exclusion of 0 => domain of definition is IN*.

If you have covered the subject of conics, then I'll try to find you a question in that field (check out question IV in the 2005 offical exams round 1, it was a good question about conics). Also, you need to focus on sequences since there'spractically always a question about them. I'll also search for a juicy one.

PS: What about physics? what did you cover up so far? Mechancis? Electricity?