Mathematics: Post your doubts here!

[saadgujjar]

expand (1-x+x^2)^1/2 up to terms in x^3

 

[snowbrood]

Verify that the plane with equation x-2y+2z=6 is parallel to the plane with equation r. (1,-2,2)=4 Find the perpendicular distances from the origin to each plane, and hence find the perpendicular distance between the planes.

 

[sonu1996]

Can anyone solve Question 9 of 9709_s13_qp_31
Thanks a lot for help!

 

[Haris Bin Zahid]

Please help me solve the 3rd part of the following question. I want an elaborate reply.

 

[Haris Bin Zahid]

Can anyone solve Question 9 of 9709_s13_qp_31
Thanks a lot for help!

(9-i) Let theta be denoted by "x".
4 cosx + 3 sinx = R cos (x-a)
R = (4^2 + 3^2)^(1/2) = 5
a = arctan (3/4) = 0.6435 rad. (4 d.p.)
Hence,
4 cosx + 3 sinx = 5 cos (x-0.6435)

(ii) 5 cos (x-0.6435) = 2
0<x<2pi
0-0.6435<x-0.6435<2pi -0.6435
-0.6435<x-0.6435<5.6397
cos b = 2/5 (b is the basic angle)
b=1.15928
x-0.6435 = 1.15928, 2pi - 1.15928
x = 1.80278, 5.76741
x= 1.80, 5.77 (3 s.f.)

(iii) I'll denote the integral sign by "int".
int (50/(5cos(x-0.6435))^2) = int (2/(cos(x-0.6435)^2)
= int (2 (sec(x-0.6435))^2)
2 tan(x-0.6435)

 

[suryashekhar]

The curve y =(ln x)/(x + 1)
has one stationary point.
(i) Show that the x-coordinate of this point satisfies the equation
x =(x + 1)/ln x
and that this x-coordinate lies between 3 and 4.

PLease post answer ASAP

 

[ahmed abdulla]

i)

AC has weight 20 N. The weigh is at the middle.
BC has weight 25 N. The weight is at the middle.
Moment = force * perpendicular distance.

Taking moments at B

Clockwise moment = anti clock wise moment

5 * T = 2*20 + 2*25

Make T the subject of formula..

what about the rest ?

 

[mania _ manal]

can someone please help me with may june 2010 pp31 q7 part iii) ?

 

[biscuitbiscuit]

Someone solve this please " The sixth term of an arithmetic progression is twice the third term, and the first term is 3. The sequence has ten terms. Find the common difference."

 

[gilliandunn8]

Anybody having some notes for binomial distribution. I am finding this subject really horrifying. Even my Home tutor is not able to make me understand this topic.!

 

[josephsai]

i need A level maths P3 complete note on vectors pls..

 

[yousef]

i need A level maths P3 complete note on vectors pls..

hope this helps ..

btw do u do M2 ?

 

[TheZodiac]

Someone solve this please " The sixth term of an arithmetic progression is twice the third term, and the first term is 3. The sequence has ten terms. Find the common difference."

a = 3
T6 = 2T3
3 + 5d = 2(3 + 2d)
3 + 5d = 6 + 4d
d = 3

 

[TheZodiac]

We know that the integral of tanx = -ln(cosx) + c
and this can be done by substituion method. Can anyone do this with integration by parts?
Why can't I get the same answer by using f(x) = Integral [ (1)tanx ]; I was considering tanx = u and 1 = dv.

 

[zoza7elw]

From where I can get 9709 past paper BEFORE 2000
Can anyone help me please????

 

[biscuitbiscuit]

Solve this please " The sum of the first n terms of a geometric progression is 2^(2n+1) -2. Find the first term and the common ratio."

 

[biscuitbiscuit]

Solve this please " The sum of the first n terms of a geometric progression is 2^(2n+1) -2. Find the first term and the common ratio."

 

[salvatore]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf
Could anyone please help me with question no. 5(ii)? I don't understand what to do..
Thanks!

 

[Oksey6]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
Can anyone pleaseee help me with q7. Integration.
Pleasee help needed!

 

[TheZodiac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf
Can anyone pleaseee help me with q7. Integration.
Pleasee help needed!

I pressumed you had problem with the (a) part so here's the solution.
First of all, convert dx in the du form;
u = sin2x
du/dx = 2cos2x (Differential of u)
therefore dx = du/2cos2x

y = sin^3 2x . cos^3 2x . du/2cos2x
Therefore after cancelling the cox2x, the remainder is

sin^3 2x . cos^2 x . du/2

Therefore this can also be written as:

sin^3x . (1-sin^2 2x) . du/2 (using the identity sin^2 2x + cos^2 2x = 1)

Now replace sin 2x with 'u'.

u^3 . (1-u^2) . du/2
u^3 -u^5 . du/2
Now integrate this and put the limits, you'll the the answer. But make sure to convert the limits into 'u' term.