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Mathematics: Post your doubts here!

ahmed abdulla

ahmed abdulla

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TheZodiac said:
Separate the variables and this would lead to an equation of
1/y dy = 6x/x^2 +4 dx
Integrate it and then put the values x and y which would result in your getting 'c'. Then express the equation in terms of x for y
Click to expand...
I did the same way bro , but the answer in marking scheme is different !
 
daredevil

daredevil

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simohayak said:
I have a doubt regarding a differentiation sum of the following form:
The equation : (X^3 + XY^2 + aY^2 - 3aX^2=0) is given. Where a is a positive constant. The maximum point on the curve is M. How do i find the x coordinate of M? in terms of "a"
Click to expand...

rearrange the equation to make y the subject
find dy/dx [differentiate the equation]

find the double derivative [differentiate the answer of the precious differentiation]

put the double derivative equal to 0 and find x in terms of a [which means that make x the subject of the equation]

hope u get it (Y)
 
salvatore

salvatore

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daredevil

daredevil

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salvatore said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_3.pdf
Please help me with qn no. 3

I used the squaring method to solve this and got the critical values as x=1 and x=5/3. I don't understand why only x=1 is accepted and not 5//3. Could anyone please explain that? In addition, I'll really appreciate if you could show how to solve the question using graphical method.

Thanks a lot
Click to expand...
here's a tip
when you have a modulus |...| on one side do not use the squaring method. try this instead: take the term in the modulus as a positive and a negative and solve them separately and then see what answers you get. also..... take care that your answer (value of x) is an INEQUALITY and not an EQUALITY as you previously stated. then plot them on a no. line and your answer will be the overlapping region.
Practice this method because our sir told us that after a lot of experience he has come to realize that the questions with a modulus on one side alone sometimes give the wrong answers or rather confusing answers (as in your case) when solved with the 'squaring on both sides' method.
Please NOTE that it MAY work on some cases and is not ALWAYS confusing or wrong but its better to be safe and use the positive/negative method instead for modulus at one side of the inequality.
Use the squaring on both sides method when there are modulus on both sides of the inequality. That will ALWAYS work.

If you don't get it (the method that I described) then please get back to me and i'll see if I can further explain.

:) (Y)
 
salvatore

salvatore

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daredevil said:
here's a tip
when you have a modulus |...| on one side do not use the squaring method. try this instead: take the term in the modulus as a positive and a negative and solve them separately and then see what answers you get. also..... take care that your answer (value of x) is an INEQUALITY and not an EQUALITY as you previously stated. then plot them on a no. line and your answer will be the overlapping region.
Practice this method because our sir told us that after a lot of experience he has come to realize that the questions with a modulus on one side alone sometimes give the wrong answers or rather confusing answers (as in your case) when solved with the 'squaring on both sides' method.
Please NOTE that it MAY work on some cases and is not ALWAYS confusing or wrong but its better to be safe and use the positive/negative method instead for modulus at one side of the inequality.
Use the squaring on both sides method when there are modulus on both sides of the inequality. That will ALWAYS work.

If you don't get it (the method that I described) then please get back to me and i'll see if I can further explain.

:) (Y)
Click to expand...
Thank you for your help mate..

I kinda get what you said, here's what I did:
|x-2| < 3-2x
-(x-2) < 3-2x or (x-2) < 3-2x
-x+2 < 3-2x or x-2 < 3-2x
x < 1 or x < 5/3

The marking scheme gave the answer as x<1 only. I don't know what to do next.. please show me how to continue from here.

Thanks a lot :)
 
daredevil

daredevil

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salvatore said:
Thank you for your help mate..

I kinda get what you said, here's what I did:
|x-2| < 3-2x
-(x-2) < 3-2x or (x-2) < 3-2x
-x+2 < 3-2x or x-2 < 3-2x
x < 1 or x < 5/3

The marking scheme gave the answer as x<1 only. I don't know what to do next.. please show me how to continue from here.

Thanks a lot :)
Click to expand...
upload_2013-12-23_12-37-56.png

If you don't get it them tell me :)
 
daredevil

daredevil

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salvatore said:
Thanks a lot..

Just one more doubt: you ALWAYS have to use the overlapping region? Even if it is the > sign?
Click to expand...
yeah you do. the < or > just tell you what region you mark on your number line. its not your answer.your answer is whatever region is overlapping. sometimes there may be an ambiguity there too because it's A-friggin-level ryt?! so the best way is to check all three regions using a value for x from each region and see which regions satisfy your ORIGINAL inequality.
 
salvatore

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daredevil said:
yeah you do. the < or > just tell you what region you mark on your number line. its not your answer.your answer is whatever region is overlapping. sometimes there may be an ambiguity there too because it's A-friggin-level ryt?! so the best way is to check all three regions using a value for x from each region and see which regions satisfy your ORIGINAL inequality.
Click to expand...
Got it!
Thanks man.. jazakallah
 
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panoramafolks

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simohayak said:
I have a doubt regarding a differentiation sum of the following form:
The equation : (X^3 + XY^2 + aY^2 - 3aX^2=0) is given. Where a is a positive constant. The maximum point on the curve is M. How do i find the x coordinate of M? in terms of "a"
Click to expand...

Do tell me if my answer is correct.

Capture.PNG
 
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studyresourcecentre

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Mathematics solved papers:)
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