Mathematics: Post your doubts here!

[The Sarcastic Retard]

part iii? can anyone draw diagram ? ty
View attachment 60786

Its up ::

https://www.xtremepapers.com/community/attachments/november2013p3_008-png.60784/

This question is different

 

[Anum96]

Anyone here knows about any software I can use to solve math papers and prepare soft copies...? I have been using ms word for it lately but it takes oh so much time :/
I need something super time saving. Also, with all the required symbols and stuff.

Rizwan Javed Physicist

 

[The Sarcastic Retard]

8(ii) and (iii)
XPFMember

 

[techgeek]

in (i), I take the left-hand side and apply double-angle identity for cos, but what follows?...confused

 

[qwertypoiu]

View attachment 60864

in (i), I take the left-hand side and apply double-angle identity for cos, but what follows?...confused

LHS

= cos(4x) + 4cos(2x)

= 2cos^2(2x) - 1 + 4*(2cos^2(x) - 1)

= 2cos^2(2x) - 1 + 4*(2cos^2(x) - 1)

= 2 *(cos(2x))^2 - 1 + 8cos^2(x) - 4

= 2*(2cos^2(x) - 1)^2 - 1 + 8cos^2(x) - 4

= 2*(4cos^4(x) - 4cos^2(x) + 1) - 1 + 8cos^2(x) - 4

= 8cos^4(x) - 8cos^2(x) + 2 - 1 + 8cos^2(x) - 4

= 8cos^4(x) - 3

 

[Binyamine]

View attachment 60864

in (i), I take the left-hand side and apply double-angle identity for cos, but what follows?...confused

I think it would be better to write also the question number and from which year it is being taken. Nevertheless against a payment of 10 USD, you can have access to all detailed worked solutions from June 2005 to November 2015 for Maths Main 9709.

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[Deepsha]

Hello im having difficulties in paper june 2013 paper 32 question 9
Im unable to shade the required region and find the greatest value of |z|.
Please help....
Thxx in advance

 

[Thought blocker]

Hello im having difficulties in paper june 2013 paper 32 question 9
Im unable to shade the required region and find the greatest value of |z|.
Please help....
Thxx in advance

 

[iSean97]

CIE M/J 2014 V42 Q2

 

[Binyamine]

Hello im having difficulties in paper june 2013 paper 32 question 9
Im unable to shade the required region and find the greatest value of |z|.
Please help....
Thxx in advance

For access to detailed worked solutions for P1, P3, M1 and S1 from Year 2005 to Year 2015 at only 10 USD.

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[Binyamine]

View attachment 60868
CIE M/J 2014 V42 Q2

For access to detailed worked solutions for P1, P3, M1 and S1 from Year 2005 to Year 2015 at only 10 USD.

GOOGLE PLAY

Maths A LEVEL Topical :: https://play.google.com/store/apps/details?id=com.alevelmathstopicaldws.com

Maths A LEVEL YEARLY :: https://play.google.com/store/apps/details?id=com.alevelmathscie9709ybydws.com

 

[Rizwan Javed]

View attachment 60868
CIE M/J 2014 V42 Q2

The time at which the two particles collide is T. So the distance covered by particle Q during this time, T can be calculated using:
s = v * t (this formula is applicable since the speed of particle Q is constant)
substitute the values to find the distance covered by Q during this time:
s = 0.75 * T = 0.75T

the total distance between A & B is 10. So if the distance covered by particle Q before it collides is 0.75T, so the distance covered by particle P during this time will be :
10-0.75T

Now use the equation of motion: s = ut + 1/2at^2 to find the distance covered by P (which is having uniform acceleration) during this time T and put it equal to 10-0.75T. Finally equate and find T.

10-0.75T = (0)(T) + 0.5(0.5)(T^2)
10-0.75T = 0.25T^2
Solve and you'll get:
either
T = 5 or T = -8(ignored)

Hence T = 5s.

(b) the speed of P immediately before collision can be found using the equation of motion ( v^2 = u^2 + 2as )
The P starts from rest, so u = 0, distance covered in 5s is: 0.25(5^2) = 6.25, and acceleration is 0.5ms^-2.
Put these values in to the equation and solve to find v (the velocity just before the collision)

v ^2 = 2 ( 0.5 ) * ( 6.25)
v = 2.5 ms^-1 <--- the speed just before collision.

 

[musiclover gurl]

Hi....I need to integrate tan 2x.
I'm confused as why i need to convert it to sin 2x/cos 2x first.
Can't I just do: (sec^2 2x)/2 + c
(The answer is -0.5 ln cos 2x +c)
Help please?

 

[Thought blocker]

Hi....I need to integrate tan 2x.
I'm confused as why i need to convert it to sin 2x/cos 2x first.
Can't I just do: (sec^2 2x)/2 + c
(The answer is -0.5 ln cos 2x +c)
Help please?

you should know differentiation of tan x = sec^2x not integration. Integration of tan x = - ln |cosx|

use reverse chain rule to get the integral of tanx

good luck

 

[Binyamine]

Hi....I need to integrate tan 2x.
I'm confused as why i need to convert it to sin 2x/cos 2x first.
Can't I just do: (sec^2 2x)/2 + c
(The answer is -0.5 ln cos 2x +c)
Help please?

 

[shingiechingz@8]

guys i need help with question #9(b) on this paper...........
please help!!!!!!!!!!!!!!!!!!!!

 

[aqibmansoor97]

I am got 70% in As level (b) and now I am getting almost 60/75 in P3 and 40/50 in S2. Will I be able to get an A ?
Thanks.

Yep that's an A for sure

 

[musiclover gurl]

you should know differentiation of tan x = sec^2x not integration. Integration of tan x = - ln |cosx|

use reverse chain rule to get the integral of tanx

good luck

Thank you!

 

[musiclover gurl]

Thanks

 

[The Sarcastic Retard]

Yep that's an A for sure

If I got 35 in stats than? xD