Mathematics: Post your doubts here!

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Hello,
could you help me please? I'm having a hard time with number 5 (iii). I have got 2sec^2X - 1 - 2 secXtanX. There is a sign problem. :( :$
Please help me.
 

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iKhaled

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Mustafa Malik said:
Hello everyone
I am new to this part of Xtremepapers cuz i hv just finished my O levels n in these vacations i wanna study for A levels .Although i created a thread but there was no reply. Actually i need the name of the best books for Math A levels that most renown skools follow.
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hey man, use this site will prepare u very well for the maths papers

http://www.examsolutions.net/
 
waleedsmz

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Mess said:
Hello,
Click to expand...
Mess said:
could you help me please? I'm having a hard time with number 5 (iii). I have got 2sec^2X - 1 - 2 secXtanX. There is a sign problem. :( :$
Please help me.
Click to expand...


Similar to the question above that one, you'd simply have to expand ( sec + tan ) ^2
sec^2 + 2sectan + tan^2
sec^2 + 2sectan + sec^2 - 1
What's the problem?
 
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waleedsmz said:
Similar to the question above that one, you'd simply have to expand ( sec + tan ) ^2
sec^2 + 2sectan + tan^2
sec^2 + 2sectan + sec^2 - 1
What's the problem?
Click to expand...

Thank you so much.........I had not link that part with the question above. Thank you. <3 ^^
 
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Fatima18 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
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Fatima18 said:
Guys, number 7 ii and 8ii pls. Thanx!..
Click to expand...
As for 7ii)
You have the derivative of the equation, also known as "y dash". To get the original curve you simply need to integrate. So...
5 becomes 5x and -8/x^2 becomes +8/x. However, there is the constant of integration, so the final thing is. y = 5x + 8/x + c.
Point P is on the curve thus it satisfies the curve's equation. Substitute with its coordinates to find c.

As for 8ii)
First you find the intersections with the x-axis ( By putting y = 0 ). Those will be the limits of integration. The volume needed has a specific formula which is PI *int(y^2 dx)
y^2 = 8x - x^2
int ( 8x - x^2 dx ) = 4x^2 - x^3 / 3
Substitute with the limits and don't forget to multiply by PI.
 
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waleedsmz said:
As for 7ii)
You have the derivative of the equation, also known as "y dash". To get the original curve you simply need to integrate. So...
5 becomes 5x and -8/x^2 becomes +8/x. However, there is the constant of integration, so the final thing is. y = 5x + 8/x + c.
Point P is on the curve thus it satisfies the curve's equation. Substitute with its coordinates to find c.

As for 8ii)
First you find the intersections with the x-axis ( By putting y = 0 ). Those will be the limits of integration. The volume needed has a specific formula which is PI *int(y^2 dx)
y^2 = 8x - x^2
int ( 8x - x^2 dx ) = 4x^2 - x^3 / 3
Substitute with the limits and don't forget to multiply by PI.
Click to expand...

Thanx a lot! :D...However , I didn't understand something.The required volume is pi*y^2*dx....so before we integrate, dont we square the equation 8x -x^2????
 
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Fatima18 said:
Thanx a lot! :D...However , I didn't understand something.The required volume is pi*y^2*dx....so before we integrate, dont we square the equation 8x -x^2????
Click to expand...
The equation itself is root(8x-x^2) so when it is squared, the root is just omitted and that's it. :)
 
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