yes!
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Mathematics: Post your doubts here!
- Thread starter XPFMember
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yes!
wait doing
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A star i have to go right now but i'll do it and post later today 
i have done a bit of it actually
just the equations are complicated otherwise its just like any other such question
just do it with a clear head and fresh start and u'll be able to do it
i have done a bit of it actually
just the equations are complicated otherwise its just like any other such question
just do it with a clear head and fresh start and u'll be able to do it
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Nov2012 v.12 no.11
no.8 (b) please .
no.8 (b) please .
f
or the first part put y =0 and take out value of a it wud be 2
than fr 2nd part use 2 as the value of y and than put inthe quation aftr simplifying diffrentiate it and get the value of b
or the first part put y =0 and take out value of a it wud be 2
than fr 2nd part use 2 as the value of y and than put inthe quation aftr simplifying diffrentiate it and get the value of b
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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
- Messages
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please gve the link.
- Messages
- 398
- Reaction score
- 685
- Points
- 103
- Messages
- 1,321
- Reaction score
- 490
- Points
- 93
- Messages
- 398
- Reaction score
- 685
- Points
- 103
thanks
no noproblem go i did it anywayA star i have to go right now but i'll do it and post later today
i have done a bit of it actually
just the equations are complicated otherwise its just like any other such question
just do it with a clear head and fresh start and u'll be able to do it
- Messages
- 398
- Reaction score
- 685
- Points
- 103
I want nov.012 no.7 ii please .
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link please + varient
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http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf
Question 10, part (ii). Do give an explanation as well. Thanks.
Question 10, part (ii). Do give an explanation as well. Thanks.
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Does anyone have may/june 2013 paper 1 variant 12 with marking scheme ? Please i need it urgently
Please i need it urgently 
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AB = b -a = (k-1 , -k, 2k - 2)
Unit vector = vecter/ magnitude
Therefore, (k-1, -k, 2k-2)/[sqrt] (k-1)^2 + (-k)^2 + ((2k -2)^2
Since they have said that the vector AB is the unit vector, the magnitude would equal 1.
Equate the magnitude you should get the answer.
you will have it tmrw ,dont be impatientDoes anyone have may/june 2013 paper 1 variant 12 with marking scheme ?
- Messages
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it will cost u nearly half a million dollars/euros/pounds in bribeDoes anyone have may/june 2013 paper 1 variant 12 with marking scheme ?