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Mathematics: Post your doubts here!

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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
do u know about oct/nov 2012 mark schemes and thresholds?
 
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Hello,
Can Anyone pls help me with this question, i got one point as pi over 6 (30 degrees) but i cannot get the other point.
 

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can sum1 plz help me asap
-Express 4x2-12x+3 in the form (ax+b)2+c where a, b and c are constants and a is greater than 0
-The line y=5x-3 is a tangent to the curve y=kx2-3x+5 at the point A. Find the cordinates of A
 
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the deravitive(dx/dϴ) of sin2ϴ is 2cos2ϴ. therefore 2cos2ϴ=(x+1)cos2ϴ
this means (x+1)=2 and thus x=1
hope that helped
I think the question is a differential equation.

sin2thetadx/dtheta=(x+1)cos2theta
can ne1 help?
so if sin2theta in your notation is sin(2*theta), not sin^2(theta), then

dx/ (x+1) = cot(2*theta) d(theta)
integrating both parts

ln|x+1| = 1/2* ln|sin(2*theta)| + C
x + 1 = sqrt(C* sin(2*theta))
and hence x = sqrt(C*sin(2theta)) - 1
Let us check by substitution:
suppose theta = a
dx/da = 2*C*cos(2a)/ (2* sqrt(C*sin(2a))) = sqrt(C)*cos(2a)/ sqrt(sin2a);

then, sin(2a)* dx/da = sqrt(C)* cos(2a) * sqrt(sin 2a) = cos(2a) * (x+1). It is true.
 
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can sum1 plz help me asap
-Express 4x2-12x+3 in the form (ax+b)2+c where a, b and c are constants and a is greater than 0
-The line y=5x-3 is a tangent to the curve y=kx2-3x+5 at the point A. Find the cordinates of A

- 4[ x^2 - 3x + 3/4]
4 [ (x)^2 - 2(x)(3/2) + (3/2)^2 - (3/2)^ + 3/4]
4 [ (x-3/2)^2 - 9/4 + 3/4]
4(x-3/2)^2 -9 + 3
4(x-3/2)^2 -6
 
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can anyone help me out for these questions?
nov2011 paper 11 question 10 part ii) how do we get the gradient..
nov 2011 paper 12 question 8 part i)
 
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