Mathematics: Post your doubts here!

[smzimran]

is there any way at all that you could help me?

Maybe i can take a pic but can't do it right now, tomorow morning ?

 

[miss irfan]

Maybe i can take a pic but can't do it right now, tomorow morning ?

all right one confusion..if i get -5x+3y+4z=2 as the equation of plane and the marking scheme says 5x-3y-4z=-2. will my answer be considered wrong?

 

[hm12]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf
please explain question 10 part (iii) please

 

[takeurtime]

http://www.xtremepapers.com/CIE/ind...l/9709 - Mathematics/&file=9709_s10_qp_31.pdf

June 2011 v1 .. no. 9 ALL pleasee!! :'( ... also no.4 i ...and 7 ii which part should i shade ?? ...

 

[smzimran]

all right one confusion..if i get -5x+3y+4z=2 as the equation of plane and the marking scheme says 5x-3y-4z=-2. will my answer be considered wrong?

No, its absolutely correct!

 

[angelicsuccubus]

Got it for the integral of 1/(a square+x square) the integral Wud b 1/a tan inverse of (x/a) +c

what?! is that a rule?

 

[angelicsuccubus]

How Wud u integrate 1/(1+X^2) ?

[ln (1+x^2)]/2x

Its a rule .. integration of 1/x is lnx and in this case you just go back and divide by the differenciation of the inner term hence the division by 2x

 

[hm12]

please someone reply to my post above

 

[ALM&B]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_32.pdf
Q. 8 (ii) from the paper. For some reason I'm getting ln25 + ln8. Please help!

Sorry for replying late :$
no u made something wrong!
yes it is (-3ln (1) + 2 ln 5 ) - (-3ln (2) + 2 ln 4)
which gives you -->
2 ln 5 + 3 ln 2 - 2 ln4
then -->
ln 25 + 3 ln2 - 4ln2
-->
ln 25 - ln 2 so , ln (25/2)

 

[angelicsuccubus]

I hate vectors

 

[ALM&B]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf
please explain question 10 part (iii) please

first, when they told you the tangent at point P passes through the origin and if you draw a straight line ..to find the gradient you will do
(y-0)/(x-0) and so you will the gradient for the tangent at P = y/x

then, from the dy/dx equation you got above, make it equal to y/x
so ..
e^-x (-x^2+2x) = y/x
and multiply the x.. you will end up with y= e^-x (2x^2-x^3) ..this is your y!
substitute this y in the equation y=x^2e^-x to get
x^2e^-x = e^-x(2x^2 - x^3)
cancel the e^-x

you will get
x^2 = 2x^2 - x^3
-1x^2 = -1x
-1x=-1
so x=1

 

[Mobeen]

can anybody post the link or something for any notes of vectors? please that would be great help cause i dont know anything about them !

 

[sea_princess]

I didn't make these, all thanks goes to the members here

 

[Mobeen]

I didn't make these, all thanks goes to the members here

thanks !

 

[GoodRobot]

wth? don't you just divide it by 2 after getting 380.8?

nope,380 is greater than 360 so is incorrect.

 

[kinal]

hi...could any1 help me?O/n 2011 Number 10 part ii b
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_33.pdf​

 

[Abdulrab]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf Q 10 part iii how to do it !! i know its easy but i dont get it

 

[redapple20]

Indeed

 

[Aarjit]

[ln (1+x^2)]/2x

Its a rule .. integration of 1/x is lnx and in this case you just go back and divide by the differenciation of the inner term hence the division by 2x

This is how you do it:

∫1/(1 + x²) dx

suppose x = tan Ө so Ө = tan^(-1) x

so 1 + x² = 1 + tan² Ө = sec² Ө

and dx/dӨ = sec² Ө

dx = sec² Ө . dӨ

So ∫1/(1 + x²) dx = ∫(1/sec² Ө) . sec² Ө dӨ

= ∫ 1 dӨ

= Ө + k

= tan^(-1) x + k

credit: http://malaysia.answers.yahoo.com/question/index?qid=20090903194558AAkgNsU

 

[redapple20]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf Q 10 part iii how to do it !! i know its easy but i dont get it

The tangent of line of a point on the curve is has dy/dx. Same as the differential of that curve on that point