Mathematics: Post your doubts here!

[Binyamine]

hey i meant to integrate not differentiate.

Aie...You wanted to integrate and i explained how to differentiate.lol.

So we have to integrate 1/(2x-1)ᶺ2. we will re write it as (2x-1)ᶺ (-2)

When we integrate, we add one to the power, divide by the new power and divide by the derivative of the function.

= [( 2x - 1 ) ^ (-1 )] / [(-1) (2)]
= -1 / 2(2x -1 )
= -1/ (4x -2 )
or
= 1/ (2-4x)

 

[yousef]

Assalamoalaikum Wr Wb!

Post your doubts here. Make sure you give the link to the question paper when posting your doubts.


May Allah give us all success in this world as well as the HereAfter...Aameen!!

And oh yeah, let me put up some links for A level Maths Notes here.

Check this out! - Nice website, with video tutorials for everything! MUST CHECK

My P1 Notes! - Only few chapters available at the moment!

Maths Notes by destined007

Some Notes for P1 and P3 - shared by hamidali391

A LEVEL MATHS TOPIC WISE NOTES

MATHS A LEVEL LECTURES

compiled pastpapers p3 and p4 - by haseebriaz

Permutations and Combinations (my explanation)- P6

Permutations and Combinations - P6

Vectors - P3

Complex No. max/min IzI and arg(z) - P3

Sketcing Argand Diagrams - P3 (click to download..shared by ffaadyy)

Range of a function. - P1

can u please answer this question>>> give that expansion of (1+ax)^n begins 1+36x +576x^2... find "a" and "n">>>

 

[chris123]

Hi,

Do u have alll the past papers for 2012? I can't find some 2012 on this site.
Also I need further maths (all modules including mechanics) and chemosytry and physics.

Thanks

 

[yousef]

Hi,

Do u have alll the past papers for 2012? I can't find some 2012 on this site.
Also I need further maths (all modules including mechanics) and chemosytry and physics.

Thanks

http://olevel.sourceforge.net/papers/ >>>> hope that helps bro ... dont forget me in ur prayers

 

[magnesium]

ANY EASY WAY TO ATTEMPT THIS QUESTION!
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_33.pdf
Q4
PLZ HELP!!!!!!!!!!!!!!

 

[PhyZac]

Assalamu Alikum


Solve the equation

cos(q+60) = 2 sin q

giving all solutions in the interval 0 =< q >= 360 [5 marks]

 

[iKhaled]

ANY EASY WAY TO ATTEMPT THIS QUESTION!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_33.pdf
Q4
PLZ HELP!!!!!!!!!!!!!!

dx/dt = 1/x - x/4

make this into one fraction first:

dx/dt = (4-x^2)/4x then dt/dx = 4x/(4-x^2) and dt = 4x/(4-x^2) dx so t = ∫4x/(4-x^2) dx integrate this and u will get an answer of t = -2ln(4-x^2) + C ( if u dont know how to integrate it then tell me) anyway after u have integrated this u will sub the values of x = 1 when t = 0 and u will get C = 2ln 3 so..

t = -2ln(4-x^2) + 2ln3
t - 2ln3 = -2ln(4-x^2)
1/2t + ln3 = ln(4-x^2)
e^1/2t . e^ln3 = 4-x^2
3e^1/2t - 4 = x^2
x^2 = 4-3e^(1/2t)

i hope u got it..any questions about it pls ask me!

 

[iKhaled]

Assalamu Alikum


Solve the equation

cos(q+60) = 2 sin q

giving all solutions in the interval 0 =< q >= 360 [5 marks]

cos ( q + 60) = 2sinq
cosqcos60 - sinqsin60 = 2sinq
1/2cosq - √3/2 sinq = 2sinq
1/2cosq = 2sinq + √3/2 sinq
tan q = 0.1745
q = 9.9 and 189.9

check if its correct in the mark scheme or something and tell me !

 

[PhyZac]

cos ( q + 60) = 2sinq
cosqcos60 - sinqsin60 = 2sinq
1/2cosq - √3/2 sinq = 2sinq
1/2cosq = 2sinq + √3/2 sinq
tan q = 0.1745
q = 9.9 and 189.9

check if its correct in the mark scheme or something and tell me !

Oh thank you so much...!!!! Jazak Allah khairan!!!

Yes correct and i understood the method...May Allah reward you for helping me..

 

[Silver Wing]

Thank you for your help
We did it this way, and we THINK we got it correct (since we came second)
Let the 3 doors be A ( leading to safety in 2hrs)
B ( bringing man back in 3hrs) n
C ( " " " 5hrs)

possible pathway - probability - time taken
1. A - (1/3) - 2hrs
2. B,A - (1/3).(1/3)= 1/9 - 5hrs
3. C,A - 1/9 - 7hrs
4. B,C,A or C,B,A - 1/27 - 10hrs


so expected time= one with max probability= 2hrs

What do you think?

P(T=A) = P(T=B) = P(T=C) = 1/3
Options avalible are:
A, p = 1/3
BA, p = 1/6
CA, p= 1/6
BCA, p = 1/6
CBA, p = 1/6

Construct a prabability density function table:
x * 2hrs * 5hrs * 7hrs * 10hrs *
P(T=x)* 1/3 * 1/6 * 1/6 * 1/3 *

E(X) = sigma(xp)
= (2* 1/3) + (5* 1/6) + (7* 1/6) + (10* 1/3)
= 6 hrs

Hope this has helped

 

[iKhaled]

Oh thank you so much...!!!! Jazak Allah khairan!!!

Yes correct and i understood the method...May Allah reward you for helping me..

Ameen ya rab..

 

[Syed Mohammad Ali]

please can you solve me da followin quest ?
I will be grateful
Q:find the values of k for which equation (k2 X2) +2kX +1 = 0 have no roots.

 

[Syed Mohammad Ali]

can uh make meh understand da followin questions , i will be grateful to ya buddy...
Q : Find the values of k for which the equation k2X2 + 2kX +1 =0 have no roots.
Q 2 : Solve the following inequality :
X(X-2)<5

 

[Martee100]

Maths As Co Ordinate Geometry Questions

http://www.scribd.com/doc/121689460/AS-Mathematics-Co-ordinate-Geometry-Questions

Add this to the list

 

[Raiyan3]

can uh make meh understand da followin questions , i will be grateful to ya buddy...
Q : Find the values of k for which the equation k2X2 + 2kX +1 =0 have no roots.
Q 2 : Solve the following inequality :
X(X-2)<5

For question no. 2 you multiply inside the brackets first and then take the 5 from the other side, it becomes
X^2-2X-5<O
Now it becomes quadratic equation. Now you can solve that, using a calculator aswell.

 

[PhyZac]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.. iKhaled

5 The parametric equations of a curve are
x = ln(tan t), y = sin2t, (it is sin square t )
where 0 < t <1/2π.

(i) Express
dy/dx
in terms of t. [4]

(ii) Find the equation of the tangent to the curve at the point where x = 0.

 

[Syed Mohammad Ali]

For question no. 2 you multiply inside the brackets first and then take the 5 from the other side, it becomes
X^2-2X-5<O
Now it becomes quadratic equation. Now you can solve that, using a calculator aswell.

it give complex roots...

 

[Kandinsky]

For the question x(x-2) < 5:
x^2 - 2x - 5 < 0

x^2 - 2x + 1 - 6 < 0
(x-1)^2 - 6 < 0
(x-1)^2 < 6
|x-1| < sqrt(6)
hence,
1-sqrt(6) < x < sqrt(6)+1

 

[iKhaled]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.. iKhaled

5 The parametric equations of a curve are
x = ln(tan t), y = sin2t, (it is sin square t )
where 0 < t <1/2π.

(i) Express
dy/dx
in terms of t. [4]

(ii) Find the equation of the tangent to the curve at the point where x = 0.

w alikom al salam..

here is the solution to the question

(i) you have to know that whenever u have a parametric equation, dy/dx = dy/dt X dt/dx

dy/dt = 2sin t. cos t
dy/dt = 2sintcost

dx/dt = (1/tan t) x sec^2 t
dx/dt = (cos t / sin t ) x 1/cos^2 t
dx/dt = 1/sin t cos t

dy/dx = 2sintcost X sintcost
dy/dx = 2sin^2 t cos^2 t

(ii) here we need to find the equation of the tangent at the point where x is 0 so first lets find y so we have a coordinate and we know that dy/dx is out tangent

x = ln(tan t )
0 = ln (tan t)
e^0 = tan t
t = tan^-1(1)
t= 1/4π

dy/dx = 2sin^2(1/4π)cos^2(1/4π)
m(gradient) = 1/2

y = sin^2(1/4π)
y= 1

y-1 = 1/2(x-0)
2y-x = 2

is this how it is in the mark scheme? if its correct pls tell me and i hope u get it! any questions feel free to ask me

 

[PhyZac]

w alikom al salam..

here is the solution to the question

(i) you have to know that whenever u have a parametric equation, dy/dx = dy/dt X dt/dx

dy/dt = 2sin t. cos t
dy/dt = 2sintcost

dx/dt = (1/tan t) x sec^2 t
dx/dt = (cos t / sin t ) x 1/cos^2 t
dx/dt = 1/sin t cos t

dy/dx = 2sintcost X sintcost
dy/dx = 2sin^2 t cos^2 t

(ii) here we need to find the equation of the tangent at the point where x is 0 so first lets find y so we have a coordinate and we know that dy/dx is out tangent

x = ln(tan t )
0 = ln (tan t)
e^0 = tan t
t = tan^-1(1)
t= 1/4π

dy/dx = 2sin^2(1/4π)cos^2(1/4π)
m(gradient) = 1/2

y = sin^2(1/4π)
y= 1

y-1 = 1/2(x-0)
2y-x = 2

is this how it is in the mark scheme? if its correct pls tell me and i hope u get it! any questions feel free to ask me

Jazak Allah Khairan..thank you so much, May Allah reward you for the help ur providing other students with.

I fully understood the first part. Alhamdulilah. And your answer is exactly similar to markscheme

coming to the second part of the question..

t=1/4pi ..................i got it
m=1/2 ....................i got it

but in markscheme the equation seem different..... the "c"

here is a copy: