Mathematics: Post your doubts here!

[Rutzaba]

syed 1996 panda
http://i1275.photobucket.com/albums/y444/Rutzaba/rutaba_zps3cbb8530.png

 

[A star]

awes

syed 1996 panda
http://i1275.photobucket.com/albums/y444/Rutzaba/rutaba_zps3cbb8530.png

ome job

 

[ZainH]

Shit happens dude
Just make sure it doesnt tomorrow.

Exactly lol, emphasis on TOMORROW.

 

[asd]

Exactly lol, emphasis on TOMORROW.

TOMORROW*
there you go.

 

[Just visiting]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf

There's a problem in 6(i) right?

can you tell me how to make Q7 iii pls

 

[ZainH]

Q. A geometric progression has the first term 4, and common ration (2x-3) , find the set of values of x for which the geometric progression has a sum to infinity.

How ?

 

[A star]

Q. A geometric progression has the first term 4, and common ration (2x-3) , find the set of values of x for which the geometric progression has a sum to infinity.

How ?

easy 2x-3<1
2x<4
x<2

 

[daredevil]

Q. A geometric progression has the first term 4, and common ration (2x-3) , find the set of values of x for which the geometric progression has a sum to infinity.

How ?

r<1 if sum to infinity exists
solve this inequality using the vlaue of r u have : - (2x-3) and u'll get ur range... i think

 

[A star]

guys i need some dif questions please dont send easy ones ( NO OFFENCE MEANT)

 

[ZainH]

easy 2x-3<1
2x<4
x<2

r<1 if sum to infinity exists
solve this inequality using the vlaue of r u have : - (2x-3) and u'll get ur range... i think

You guys make me feel dumb e_e ..

Ill bring hard questions next time \:<

 

[A star]

You guys make me feel dumb e_e ..

Ill bring hard questions next time \:<

sorry
but yeah pls i dotnhave energy left to search each past paper xD

 

[ZainH]

sorry
but yeah pls i dotnhave energy left to search each past paper xD

Haha , relax. I'll go solve some papers and be back in ~20 minutes with questions.

 

[daredevil]

You guys make me feel dumb e_e ..

Ill bring hard questions next time \:<

haha ooooh thats okay..... i get stuck on soe pretty simple questions and then i think damn wat am i doing here!! we hav our moments

 

[Manobilly]

Hey can any1 give me right answers for oct 2012 paper 13. The ms is confusing ,not all answers in there

 

[A star]

Hey can any1 give me right answers for oct 2012 paper 13. The ms is confusing ,not all answers in there

all questions :O
try www.freeexampapers.com

 

[Manobilly]

all questions :O
try www.freeexampapers.com

The marking scheme is confusing doesn't have answers for some.

 

[Maz]

The marking scheme is confusing doesn't have answers for some.

sometimes the "er" is helpful

 

[A star]

yeah most of the time you make the same mistake the students who took the exam are making so er can help

 

[Manobilly]

yeah most of the time you make the same mistake the students who took the exam are making so er can help

Okay will try that Jazak Allah

 

[beeloooo]

thanks al

i) y=0
0=x(x-2)
x=0 and x=2
a=2 (since its visible from the graph that a cant be zero)

ii) b is a max pt.
dy/dx = 3x^2 -8x +4 =0
x= 2 and x= 2/3

b=2/3 (since both turning pts are visible from the graph and b is the first turning pt so i ll take the smaller value of x) (beside a was 2)
s
iii) integrate the function with limits 0 and 2

iv) It is asking for the minimum value of dy/dx... dy/dx= 3x^2 - 8x +4
thus yu just have to find minum value of that quadratic function
Two ways to do it .. Do completing squares or just derivate it again and equate it to zero .. u shall get x substitute that x to the dy/dx equation and that will be yur anser

Thanks alot many thanks but can u just tell me when is it necessary for us to double derivate it and put it equal to zero ?