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Qualification => GCE AS & A2 Level => Queries => Topic started by: Stay10 on May 09, 2010, 08:41:52 am

Title: CIE AS LEVEL Math..
Post by: Stay10 on May 09, 2010, 08:41:52 am
hey every1..needed sum help in math..is there any notes or reference materials tht could help me understand the range and domain of a function (basic overview)..gone through lotz of past papers nd found those question question tricky..so if any1 could explain me or any reference would be really appreciated...need it as soon as possible..nt much time left..Thanks  :)
Title: Re: CIE AS LEVEL Math..
Post by: Meticulous on May 09, 2010, 08:45:20 am
Post your questions. I will be happy to help.

Try www.astarmathsandphysics.com
Title: Re: CIE AS LEVEL Math..
Post by: Stay10 on May 09, 2010, 08:51:36 am
..Thanks 4 the site..ll go through it..but i mean like i dont know wht it means..both range nd domain..there is a question in paper 12 of oct/nov 2009..no. 4..would really appreciate the explanation..
Title: Re: CIE AS LEVEL Math..
Post by: Stay10 on May 09, 2010, 06:33:54 pm
came across another question from the paper oct/nov 2008, question 7 (i) & (ii)..w8in 4 the reply..
Title: Re: CIE AS LEVEL Math..
Post by: Freaked12 on May 10, 2010, 01:29:49 am
Quote from: Stay10 on May 09, 2010, 06:33:54 pm
came across another question from the paper oct/nov 2008, question 7 (i) & (ii)..w8in 4 the reply..

Show that A =
(? + 4)x2 ? 160x + 1600
divided by pie

So the perimeter according to the question is 80 of both the shapes so as a result
x + x + x + x + 2(pie) r
4x+ 2 (pie)r=80
2(2x+ (pie)r)=80
2x + (pie) r= 40
We make r the subject
r= 40/(pie) - (2/(pie))x

we now substitute the r value into the equation (the area of both sides)
A= x square plus (pie) r square
x square + (pie)(40/(pie) - (2/(pie))x)square
x square + pie(1600/pie^2 - 16.21x + 4/pie^2 x^2)
which if you calculate
(1600- 16.21x pie^2 + 4 x^2 + pie x^2)/pie
((4+ pie)x^2 +1600 - 160x)/ pie

2) it depends on whether u have taken r or x the subject
in this case r
so we differentiate
(2((pie) +4) x -160)/pie=0<br />pie * 0=0 so 2x(pie +4)-160=0<br />2x(pie +4)=160<br />x(pie +4)=80<br />x=80 divided by pie plus 4<br />x=11.2
Title: Re: CIE AS LEVEL Math..
Post by: Stay10 on May 10, 2010, 04:50:41 am
Thanks.. :)
Title: Re: CIE AS LEVEL Math..
Post by: Stay10 on May 10, 2010, 04:52:57 am
came across another question from the same paper..question 9..dont get the markin scheme..ll be w8in..
Title: Re: CIE AS LEVEL Math..
Post by: immortal on May 10, 2010, 08:07:17 am
9>Area of region
    Area of square-area under graph
     (2*1)-(Integration of (3x+1)1/2 by 0,1)
     2-[2/9*(3X+1)3/2],0,1
     =4/9
ii> Volume of region
     volume of cylinder-volume under graph
     Pi*2*2-Pi integration of [(3x+1)1/2]2 to 0,1
     4Pi-Pi[1/6*(3x+1)2] to 0,1
     4Pi-Pi[8/3-1/6]
     4Pi-2.5Pi   
     1.5Pi

iii> Differentiate da eqn & substitute point of x to find da gradient
     for x=1  m=3/4                                        for x=0   m=3/2
tanQ=3/4                                              tanQ=3/2
Q=36.87                                                           56.31
Thare4 Acute angel =56.31-36.87
                           =19.4
[tanQ because gradient is also =y/x]
                         
Title: Re: CIE AS LEVEL Math..
Post by: Stay10 on May 10, 2010, 09:14:48 am
Thank You ! :)
Title: Re: CIE AS LEVEL Math..
Post by: Meticulous on May 10, 2010, 09:30:04 am
That's considered as SPAM I guess. Posting with smileys only. Check with a moderator/ an administrator.
Title: Re: CIE AS LEVEL Math..
Post by: Stay10 on May 10, 2010, 04:29:38 pm
is tht so!!..well srry than..never thought this waz the case..