Mathematics: Post your doubts here!

[David Hussey]

can someone help me in http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
q3i
q6ii
7i
this is one crazy paper :/

 

[SitiPutri]

can someone help me in http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
q3i
q6ii
7i
this is one crazy paper :/

Sorry not an A2 student so cant really help much. But this is for Q3i

 

[sumeru]

can someone help me in http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
q3i
q6ii
7i
this is one crazy paper :/

For 7(ii)...i have attatched a solved one yesterday..so look above you'll find it.....and for 7(i) use quadratic equation to find the roots...

 

[daredevil]

can someone help me in http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
q3i
q6ii
7i
this is one crazy paper :/

 

[daredevil]

can someone help me in http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
q3i
q6ii
7i
this is one crazy paper :/

for 7i
either use the quadratic formula such that
a=1 , b=2(sqrt)3 i c= -4

or put (x+iy) in place of z and solve it u wil get the answers.

 

[David Hussey]

Jazakallahu khair guys! Thanks

 

[Fahm Deen]

October/November 2010/63 6(ii and iii)
Please give the solution with full explanation ASAP.
Very urgent!!

 

[istudent]

View attachment 39727

thanks man

 

[istudent]

t

View attachment 39729 this is a better diagram:

thanks man

 

[SitiPutri]

October/November 2010/63 6(ii and iii)
Please give the solution with full explanation ASAP.
Very urgent!!

Not really neat, though

 

[Snowysangel]

If you have cos^2, sin^2...you can't integrate them directly..so at first you have to conver them to integrate

Can such a question come in As?

 

[sumeru]

Can such a question come in As?

No, don't worry.....you'll have to study them in A2 ...

 

[asd]

Oct Nov 2007, Q7, part (ii) Why are we supposed to use the radian mode instead of degree mode?
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

 

[daredevil]

thankkss alot! and can u also do part ii ??

 

[daredevil]

Oct Nov 2007, Q7, part (ii) Why are we supposed to use the radian mode instead of degree mode?
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

yaar i asked a friend and she said that she doesnt know the logic of this but apparently all differential questions are solved in radians.

did u know that?? A star syed1995 ??

 

[Thought blocker]

Where is genius ?

 

[asd]

thankkss alot! and can u also do part ii ??

http://www.netcomuk.co.uk/~jenolive/vect18e.html
Here's a detailed method and I'm getting sqrt[5] using this. I did method 2.

 

[asd]

But for the same question daredevil asked, why doesn't my method work?
Here's the method sumeru:

Since they ask for the shortest distance from the point p to the line that lies ON the plane, so we can use the normal vector of the plane as the directional vector for the line equation PX, where X is a point on the plane where a line from P and the line (given in the question) on the plane intersect.
So my line equation for PX becomes: <0,2,4> +b<2,-2,3>
Now, if I put into the plane, 2x-2y+3z=1 this point, <2t, 2-2t, 4+3t> to find the value of b, and then by substituting this value into the line equation of PX, am I finding the the position vector for X where the line from P intersect the line in the equation on the plane?
Why is thisi method wrong?

 

[daredevil]

 

[sumeru]

Oct Nov 2007, Q7, part (ii) Why are we supposed to use the radian mode instead of degree mode?
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

Well, in our P3, we mostly use the angles in radians, unless specified in the questi

But for the same question daredevil asked, why doesn't my method work?
Here's the method sumeru:

Since they ask for the shortest distance from the point p to the line that lies ON the plane, so we can use the normal vector of the plane as the directional vector for the line equation PX, where X is a point on the plane where a line from P and the line (given in the question) on the plane intersect.
So my line equation for PX becomes: <0,2,4> +b<2,-2,3>
Now, if I put into the plane, 2x-2y+3z=1 this point, <2t, 2-2t, 4+3t> to find the value of b, and then by substituting this value into the line equation of PX, am I finding the the position vector for X where the line from P intersect the line in the equation on the plane?
Why is thisi method wrong?

Are you giving me the method or asking whether it is correct or not?