Mathematics: Post your doubts here!

[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 ?

 

[sma786]

no problemo butt ummm *ahem ahem not a boy ahe ahem*

Thnku*girl and a boy sorry for that

 

[sma786]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
Ok so in question 1, the markscheme says they take the 4th term, why? (1/x) comes when 5C1 ryt, thats where there comes (1/x) so why the 4th term?!

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
Ok so in question 1, the markscheme says they take the 4th term, why? (1/x) comes when 5C1 ryt, thats where there comes (1/x) so why the 4th term?!

no u got that wrong..... see that x is in both the values 2x and 1/x also ..... now do it

 

[Maz]

can you tell me how to make Q7 iii pls

since g(x) = x-2
and hg = f
replace the "g" part in "f" by x.
You get h(x) = x^2 +3

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
Ok so in question 1, the markscheme says they take the 4th term, why? (1/x) comes when 5C1 ryt, thats where there comes (1/x) so why the 4th term?!

for these type of equations use this formulae
NCr (a)n-r br equate their powers like -1=5-r-r this means r=3 solve it with this value

 

[PANDA-]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_11.pdf
Question 1, all of it.

 

[shafayat]

agh.. I am tensed :/ I DONT FUKIN WANNA BE ....UGHHH

 

[shafayat]

yeah but in (ii) It's asking about a different equation , not the one i already sketched. and it's for one mark , do i have to sketch that as well?

NO... so u didn't get what I said
after the first question .. u found there are 2 solutions for one revolution, (I.e. 0- 2pie) den u find how many of these 0-2pie revolutions are there in 0-20pie ! .. there are 10 of these revolution right ? ... so if one 0-2pie revolution has 2 sol how many does 10 revolutions have ? 20 !!!!

 

[MustafaMotani]

thanks al


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 ?

double derivate was not required in this question .... actually double derivate is just used to tell the nature the of the function ... however i wanted minimum vale dy/dx ... take it this way
you have to find minimum value of the function f(x) = 3x^2 -8x +4

You can find it in two ways either express it in (x-h)^2 +k form where k will be the minimum value
or
just derivate the function and equate it to zero to get the x-cordinate of the minimum point .. and then insert that value of x in the minimum value of the function..

it just happend to be that f(x) that we are talking about was dy/dx ... so derivative of that function was double derivative ..

Hope i m clear..