Mathematics: Post your doubts here!

[iKhaled]

can someone pls explain to me question 11(ii) may june 2004 paper 3 mathematics
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s04_qp_3.pdf

i would really appreciate it if someone could draw for me the diagram of that question

 

[PANDA-]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_42.pdf

guys question 7 part (i) anyone?
thanks

Got it

Now at maximum speed, acceleration is 0, obviously because it can no longer go faster.
So you must differentiate to get the acceleration function, and equate it to 0, to get t.
First expand the velocity formula... from k(60t^2 - kt^3) to 60kt^2 - kt^3..
v = 60kt^2 - kt^3 ... Differentiate this.
a = 120kt - 3kt^2
120kt-3kt^2 = 0
120kt = 3kt^2

Divide both sides by kt

120 = 3t
t = 40

Now go back to the velocity function, and replace v with 6.4, as given in the question, and t with 40.

k(60(40)^2 - 40^3) = 6.4
k = 6.4/32000
k= 0.0002

 

[Alice123]

can someone pls explain to me question 11(ii) may june 2004 paper 3 mathematics
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s04_qp_3.pdf

i would really appreciate it if someone could draw for me the diagram of that question

i can do that without diag.. wait im doing it

 

[iKhaled]

i can do that without diag.. wait im doing it

thanks..i am still trying to do it ow and failing to answer it arghh :/

 

[Alice123]

d1=PQ=OQ-OP=(-2 4 1)-(1 0 -1)=(-3 4 1)
d2=PR=OR-OP=(4 2 1)-(1 0 -1)=(3 2 2)
Find the cross product of d1 and d2
ie ( 6 9 -18) it can also be written as (2 3 -6)
Now
rn=an
(x y z)(2 3 -6)=(3 5 -6)(2 3 -6) where a is OS( position vector of S)
the equatn is 2x +3y-6z=8

this is a very useful formula, try to remember it
modulus of (ax1+ by1+cz1-d)/root of (a^2+b^2+c^2) see winter 2012...its given there
mod of{(2(3)+3(5)-(-6)-8)}/root of(2^2+3^2+6^2)
and the ans comes to 7 units
if u dnt understand feel free to ask iKhaled

 

[rosogolla993]

wats up !! evry1

 

[rosogolla993]

Mashallah jajak allah khair

 

[Alice123]

iKhaled

 

[rosogolla993]

fi Amanillah

 

[GorgeousEyes]

Please Nov2012 v.13 no.6 i . no.7 ii , 8 iii ) , 9 all , 10 iii
11 ) ii and iv urgentlyyy .

 

[rosogolla993]

i need help in functions in p1 as

 

[rosogolla993]

is it really necessary to draw quadrants beside the trigometry sums ???

 

[Alice123]

@rosogolla

 

[rosogolla993]

how to draw function graphs like for example june 2001 variant paper1 no.3 ii and iii

 

[rosogolla993]

still dint get it y is it soo hard

 

[PANDA-]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_13.pdf

Q4 ... i)
Q7 ... i)
Q8 ... i)

If anyone could please explain these 3 questions.
PhyZac syed1995

 

[iKhaled]

d1=PQ=OQ-OP=(-2 4 1)-(1 0 -1)=(-3 4 1)
d2=PR=OR-OP=(4 2 1)-(1 0 -1)=(3 2 2)
Find the cross product of d1 and d2
ie ( 6 9 -18) it can also be written as (2 3 -6)
Now
rn=an
(x y z)(2 3 -6)=(3 5 -6)(2 3 -6) where a is OS( position vector of S)
the equatn is 2x +3y-6z=8

this is a very useful formula, try to remember it
modulus of (ax1+ by1+cz1-d)/root of (a^2+b^2+c^2) see winter 2012...its given there
mod of{(2(3)+3(5)-(-6)-8)}/root of(2^2+3^2+6^2)
and the ans comes to 7 units
if u dnt understand feel free to ask iKhaled

i dont understand y did u find the cross product of d1 and d2..can u pls explain this part?

 

[Maz]

Please Nov2012 v.13 no.6 i . no.7 ii , 8 iii ) , 9 all , 10 iii
11 ) ii and iv urgentlyyy .

I'll answer your questions one by one
6i) gf(x) = 1
f(x) = 1/2 x +1/6 n [take that n as pi please ]
g(1/2 x + 1/6 n) = 1
cos ( 1/2 x + 1/6 n ) = 1
1/2 x + 1/6 n = cos-1( 1)
1/2 x + 1/6 n = 0
1/2 x = -1/6 n
x = 2(- 1/6 n )
x = -n/3

 

[rosogolla993]

i dont knw how to upload a data

 

[PANDA-]

For 4i
sin2x=2sinxcosx and cos2x=2cos^2x-1 these r given in the formula booklet... hope these help

Doesn't really make a difference.
Thank you for the help nonetheless.