Mathematics: Post your doubts here!

[Komail Sabba']

Can anyone explain to me how to solve these questions? Thank You.

Intergrate the following functions with respect to x.
1. x^2 cos x
2. In (x-1)
3. (In x)^2

Use the product rule:
1- u= x^2 and dv/dx= cosx
2- u= ln(x-1) and dv/dx= 1
3- u= (ln x)^2 and dv/dx= 1

 

[TheJDOG]

Variant 42 - O/N - 2014
Q4 - ii) Need explanation!
View attachment 52529

Here you go, any more doubts? Just tell me =)

 

[PlutoHuman]

Can someone do this question? Thank you!

 

[Jennifer4678]

Hey!! Can someone please help me with M/J 2012 P33 Q10. Been stuck here for so long thanks!

 

[TheJDOG]

View attachment 52635
Can someone do this question? Thank you!

Done if I have any mistake please tell me

 

[TheJDOG]

Hey!! Can someone please help me with M/J 2012 P33 Q10. Been stuck here for so long thanks!

Hey, I have just solved this for you it's attached here hopefully I'm correct

 

[Jennifer4678]

Hey, I have just solved this for you it's attached here hopefully I'm correct

Thanks so much!!

 

[Jennifer4678]

Can someone help me with this question! After getting the equation, I don't know how to rearrange it to get the final answer

 

[TheJDOG]

Can someone help me with this question! After getting the equation, I don't know how to rearrange it to get the final answer

Ok ok I'll solve it for you gimme 5 minutes

 

[TheJDOG]

Can someone help me with this question! After getting the equation, I don't know how to rearrange it to get the final answer

Here you go

 

[Jewel504]

Length of chord + Arc length

Thanks a lot

 

[DeadlYxDemon]

Here you go, any more doubts? Just tell me =)

Thank you so much!

 

[nehaoscar]

How do you do part (iii) ?
I always get confused on how to do it when it's a bracket within a bracket... please help!

 

[Komail Sabba']

View attachment 52686

How do you do part (iii) ?
I always get confused on how to do it when it's a bracket within a bracket... please help!

Have you tried using

(1+x)^n = 1 + nx + (n(n-1)/2!) x^2 +........... ?

 

[nehaoscar]

Have you tried using

(1+x)^n = 1 + nx + (n(n-1)/2!) x^2 +........... ?

It says hence find the co-eff
By doing the expansion
5C4 x 1 x (x + 3x^2)^4
I get the co-eff to be 405

Then in part ii, the co-eff is 270
And in part i, the co-eff is 81

What they have done is added 405 and 270 to give co-eff = 675
But then why not add 81 to it ... and why add 270 to it ?

 

[hamzashariq]

http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_31/

Q5(ii). Really need help with this.

 

[Debonny1]

It says hence find the co-eff
By doing the expansion
5C4 x 1 x (x + 3x^2)^4
I get the co-eff to be 405

Then in part ii, the co-eff is 270
And in part i, the co-eff is 81

What they have done is added 405 and 270 to give co-eff = 675
But then why not add 81 to it ... and why add 270 to it ?

How did you expand (x+3x^2)^4?

 

[Ram97]

View attachment 52686

How do you do part (iii) ?
I always get confused on how to do it when it's a bracket within a bracket... please help!

From your above results you can see that x^8 is when (x+3x^2)^4 and (x+3x^2)^5
Similarly in (1+(x+3x^2))^5 x^8 would be at when the expression has power of 4 and 5
So 5C4 (1) x (x+3x)^4 + 5C5 (x+3x^2)^5
Therefore 450x^8 + 270x^8 = 675x^8

Hope you understand

 

[nehaoscar]

How did you expand (x+3x^2)^4?

Since you only need the coefficient of x^8
You can skip the full expansion and only do (3x^2)^4 Since no other terms in the expansion will give x^8
So it gives 81x^8

 

[nehaoscar]

From your above results you can see that x^8 is when (x+3x^2)^4 and (x+3x^2)^5
Similarly in (1+(x+3x^2))^5 x^8 would be at when the expression has power of 4 and 5
So 5C4 (1) x (x+3x)^4 + 5C5 (x+3x^2)^5
Therefore 450x^8 + 270x^8 = 675x^8

Hope you understand

Ohhh! this makes so much sense now! the mark scheme just confused me as hell as to why they just added part i and ii ...
Thanks a lot!