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Author Topic: C3 and C4 DOUBTS HERE!!!  (Read 35692 times)

Offline sweetest angel

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Re: C3 and C4 Doubts Here!!!
« Reply #15 on: June 11, 2010, 10:30:03 am »
Quote from: astarmathsandphysics on June 11, 2010, 10:11:52 am
x=4 sin(2y+6)
differentiate wrt y
\frac {dx}{dy} =8 cos(2y+6)
\frac {dy}{dx}=\frac {1}{8 cos(2y+6)} =\frac {1}{8 sqrt(1-sin^2 (2y+6))} =\frac {1}{8(1-(x/4)^2))

from where did u get the square root at the end?
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Offline cooldude

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Re: C3 and C4 Doubts Here!!!
« Reply #16 on: June 11, 2010, 10:38:44 am »
Quote from: sweetest angel on June 11, 2010, 10:30:03 am
from where did u get the square root at the end?

(sinA)^2+(cosA)^2=1
therefore cos A=(1-(sinA)^2)^0.5
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Offline sweetest angel

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Re: C3 and C4 Doubts Here!!!
« Reply #17 on: June 11, 2010, 10:41:09 am »
oooh Thanks alot i got it :D
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Offline sweetest angel

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Re: C3 and C4 Doubts Here!!!
« Reply #18 on: June 12, 2010, 06:43:39 am »
hey can anyone xplain to me the domain and range of chapter 2??
i don't know how to get them in questions of state the domain and range of the function or inverse of function.
thanks :)
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"Dont fear the creation, but fear the creator who has created the creation you fear"
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 I DO WHAT I WANT WHERE I WANT WHEN I WANT if mom says ok.

Offline Saladin

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Re: C3 and C4 Doubts Here!!!
« Reply #19 on: June 12, 2010, 08:06:53 am »
Quote from: sweetest angel on June 12, 2010, 06:43:39 am
hey can anyone xplain to me the domain and range of chapter 2??
i don't know how to get them in questions of state the domain and range of the function or inverse of function.
thanks :)

Please give a question.
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Offline bilal920

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Re: C3 and C4 Doubts Here!!!
« Reply #20 on: June 13, 2010, 06:21:17 am »
i have a Question in the mok paper

How the hell did they do Q7b
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Offline Hwa1

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Re: C3 and C4 Doubts Here!!!
« Reply #21 on: June 14, 2010, 01:38:20 pm »
Does anybody have a link to the Jan 09 mark scheme (edexcel)? The file on freeexampapers is damaged and the edexcel section on xtremepapers is under maintenance :(
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Offline 7ooD

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Re: C3 and C4 Doubts Here!!!
« Reply #22 on: June 14, 2010, 09:16:02 pm »
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Offline T.Q

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Re: C3 and C4 Doubts Here!!!
« Reply #23 on: June 15, 2010, 02:14:19 pm »
i need help in this

Use the substitution x=sec\theta and integration to find the value of


\int_{2}^{3}\frac{1}{(x^2-1)^ (\frac{3}{2} } dx



giving your answer to 3 decimal places


the answer in MS is 0.094
« Last Edit: June 15, 2010, 02:19:50 pm by T.Q »
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Offline Meticulous

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Re: C3 and C4 Doubts Here!!!
« Reply #24 on: June 15, 2010, 03:42:00 pm »
Is it (x2 -1 )3/2 ..?
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Offline T.Q

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Re: C3 and C4 Doubts Here!!!
« Reply #25 on: June 15, 2010, 05:29:13 pm »
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Offline S.M.A.T

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Re: C3 and C4 Doubts Here!!!
« Reply #26 on: June 15, 2010, 07:39:46 pm »
Quote from: T.Q on June 15, 2010, 02:14:19 pm
i need help in this

Use the substitution x=sec\theta and integration to find the value of


\int_{2}^{3}\frac{1}{(x^2-1)^ (\frac{3}{2} } dx



giving your answer to 3 decimal places


the answer in MS is 0.094


THE SOLUTION IS IN THE LINK BELOW
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Offline T.Q

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Re: C3 and C4 Doubts Here!!!
« Reply #27 on: June 15, 2010, 11:11:13 pm »
thanx asiftasfiq93

+rep
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Offline M-H

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Re: C3 and C4 Doubts Here!!!
« Reply #28 on: September 10, 2010, 04:00:52 pm »
i want to know how to solve these kind of Qs:
Eg.
Q: The functions l(x), m(x), n(x) and p(x) are defined by l(x)=2x+1, m(x)=x^2 -1, n(x)=1/(x+5) and p)x)=x^3.
Find in terms of l,m,n and p the functions:
a) 4x(^2) +4x.

I need a logical way, or atleast a way which can be used for solving all Qs of this type!!! Please help!

Thanks a million in advance!!
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Offline S.M.A.T

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Re: C3 and C4 Doubts Here!!!
« Reply #29 on: September 10, 2010, 07:20:36 pm »
 4x(^2) +4x
=4((x^2)+x)
=4(((x+0.5)^2)-(0.5)^2)
=4((x+0.5)^2)-1
=4(((2x+1)^2)/4)-1
=((2x+1)^2)-1
=l^2(x)-1
=ml(x)

When the equation is quadratic, always convert it to complete square form ,then simply the equation and write in terms of l(x),m(x),....etc.

If the quadratic equation is already written in complete square form,then all you have to do is write in terms of l(x),m(x),....etc.

« Last Edit: September 10, 2010, 07:31:07 pm by asiftasfiq93 »
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