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How was Math P1??????

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First part of the question was pretty easy because first term = sum of the first term so you just had to plug 1 into the equation they provided. Then you could have found out d by figuring out Sn1, Sn2, Sn3 and figuring out the common difference between their sums. Didn't need an equation or anything.
No way! I got thiisss close to doing that.. but it didnt hit me properly.. damnit!
 
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really...??? i thought second part was more tricky than the first one...
I must be crazy then .. cause I spent so long tryna figure out the first part and couldn't do it.. but the second part I think I formed a quadratic equation pretty quickly that gave a=27.. which worked when I tested it.
 
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Let me see how far I can recall.. ;)

tanx + 1/tanx = (sinx/cosx)+(cosx/sinx) = (sin^2 x + cos^2 x)/sinx.cosx = 1/sinx.cosx Q.E.D

now,
solve for 2/sinx.cosx = (an expression with tanx); 0 < x < 180
>> 2*(tanx + 1/tanx) = (RHS) [from the first part]
factorize to get (tanx - 1)(tanx + 2) = 0
>> tanx = 1 ; tanx = -2
>> x =45 ; x = 180 + (-63.43...)
= 116.56 ~ 116.6
 
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the second term if a GP is 9 less than the first term. The sum of the second and third terms is 30. Given that all terms are positive*, find the value of the first term.

a.r = a - 9 ; a.r + a.r^2 = 30
>> r = (a - 9)/a

>> 30 = (a - 9) + a.[(a-9)/a]^2
>> 30 = a - 9 + (a-9)^2/a
>> 39.a = (a^2) + a^2 - 18.a + 81
>> 2.a^2 - 57.a + 81 = 0
>> factorize to get (a - 27)(2.a - 3) = 0

a = 27 or a = 3/2
since r= -5 when a= 3/2, the terms become negative*(see question);

Ans: a = 27
 
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the second term if a GP is 9 less than the first term. The sum of the second and third terms is 30. Given that all terms are positive*, find the value of the first term.

a.r = a - 9 ; a.r + a.r^2 = 30
>> r = (a - 9)/a

>> 30 = (a - 9) + a.[(a-9)/a]^2
>> 30 = a - 9 + (a-9)^2/a
>> 39.a = (a^2) + a^2 - 18.a + 81
>> 2.a^2 - 57.a + 81 = 0
>> factorize to get (a - 27)(2.a - 3) = 0

a = 27 or a = 3/2
since r= -5 when a= 3/2, the terms become negative*(see question);

Ans: a = 27
Yes man..got it all right! ;) Only hope they dont eat my method marks!! :'(
 
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I KNOW! It's like I get home and realize the answer and I couldn't stop thinking about it and feeling totally stupid >.<
i know the feeling... :p do you do p3 as well....??? if so how did it go....??? :) i found it REALLY HARD!!!!!!!!!! :(
 
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No, I do P6 :)
thats statistics 1... same here... plus A2 as well... p7 included... thats the messed up part... :S
i hope p6 will help to compensate on the marks i've lost already... cos so far i havnt got less than 38 outta 50 for that... :) some kind of hope there for me i gz... :)
GOOD LUCK for thursday then... :) oh btw... do you do accounts or BS by any chance...? :)
 
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thats statistics 1... same here... plus A2 as well... p7 included... thats the messed up part... :S
i hope p6 will help to compensate on the marks i've lost already... cos so far i havnt got less than 38 outta 50 for that... :) some kind of hope there for me i gz... :)
GOOD LUCK for thursday then... :) oh btw... do you do accounts or BS by any chance...? :)
Yeah I'm hoping to cover up in S1 too. No, I do chem, physics, biology.. Good luck to you too!
 
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The paper quite extended, but I managed it in 1 1/4 hours, leaving 30 minutes for revision. Here are some of the answers:

(volume of rotation about x-axis): 12 pi
binomial theorem: a= 5
GP>> first term (a)= 27
AP>> a = 9, d= 2
unit vector scalar multiple: 1/sqr(61)
(coordinates): A(4,6), B(2,2)
functions: g-1(x) = (8/x)+3; undefined when x=0

0<k<64/9

Yes! Think I got 100% :p And it was pretty easy! Didn't even take an hour! No offense to anyone who couldn't finish! :)
 
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so nobody did 13 here.. I did it today.. somewhat harder than the previous one.. but I managed to answered it all.. may I get A ! Amin
i did 13. i dk y i just got really confused with the ques about refelection in y-axis. y did thy giv that B coordinates :S
 
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