A2 Maths p3 Doubt. Plz solve (Urgent!)

[Khan Boi]

Anybody just do Q10) bi and bii. It would be a great help! Thank you

 

[ParicJ]

Hope this help.

 

[Khan Boi]

Thanks alot! It did!

Hope this help.

 

[GCE As and a level]

Hope this help.

HEY GUYS
how are u ??
PLZ i need a help in 9709 Mathematics Paper 3

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
Q7
help is highly appreciated
THANK YOU

 

[Khan Boi]

HEY GUYS
how are u ??
PLZ i need a help in 9709 Mathematics Paper 3

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
Q7
help is highly appreciated
THANK YOU

Bro, sorry couldn't upload my working but i can give you some tips.
Take cos^3(2x) as (cos2x)(cos^2(2x)
Convert the cos^2(2x) into (1-sin^2(2x)) using the identity
Differentiate u (sin2x) to get (cos2x) and make dx the subject
And then replace all your sin2x and dx with u and du.
And finally proceed with normal integration.

ii) Whatever value you get for A. Multiply it by 40 and equate it with k(pi) to get the value for k.
I am guess pi will cancel out.

 

[$$AK$$]

Bro, sorry couldn't upload my working but i can give you some tips.
Take cos^3(2x) as (cos2x)(cos^2(2x)
Convert the cos^2(2x) into (1-sin^2(2x)) using the identity
Differentiate u (sin2x) to get (cos2x) and make dx the subject
And then replace all your sin2x and dx with u and du.
And finally proceed with normal integration.

ii) Whatever value you get for A. Multiply it by 40 and equate it with k(pi) to get the value for k.
I am guess pi will cancel out.

Hey man
could u plz help me with this question in 9709 mathematics p3 ??

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
Q 9

Your help is highly appreciated
WISH U BEST OF LUCK

 

[GCE As and a level]

Bro, sorry couldn't upload my working but i can give you some tips.
Take cos^3(2x) as (cos2x)(cos^2(2x)
Convert the cos^2(2x) into (1-sin^2(2x)) using the identity
Differentiate u (sin2x) to get (cos2x) and make dx the subject
And then replace all your sin2x and dx with u and du.
And finally proceed with normal integration.

ii) Whatever value you get for A. Multiply it by 40 and equate it with k(pi) to get the value for k.
I am guess pi will cancel out.

Thnx man

 

[Khan Boi]

Hey man
could u plz help me with this question in 9709 mathematics p3 ??

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
Q 9

Your help is highly appreciated
WISH U BEST OF LUCK

Differentiate the power and then differentiate what's inside the bracket
You can do it by using Quotient Rule to differentiate (1-x)/(1+x)
Multiply both the differentials to obtain an equation in terms of dy/dx
The thing here is to split (1+x) or (1-x) into (1+x)^0.5 * (1+x)^0.5 or (1-x)^0.5 * (1-x)^0.5 respectively, to cancel out the denominator
Hopefully, now you'll get the answer

Hope this helps. Sorry can't upload the working of it..

 

[Harsh Poddar]

I've made this site that can be used to quickly access past papers to all the subjects of IGCSE, and AS and A level students. Try it out and please give feedback. Already being used in 52 countries by 900 users.

IGCSE: gopapers.net/igcse.html
A & AS Level: gopapers.net/alevel.html

 

[GCE As and a level]

Bro, sorry couldn't upload my working but i can give you some tips.
Take cos^3(2x) as (cos2x)(cos^2(2x)
Convert the cos^2(2x) into (1-sin^2(2x)) using the identity
Differentiate u (sin2x) to get (cos2x) and make dx the subject
And then replace all your sin2x and dx with u and du.
And finally proceed with normal integration.

ii) Whatever value you get for A. Multiply it by 40 and equate it with k(pi) to get the value for k.
I am guess pi will cancel out.

Thnx man
It was helpful

could u plz help me also with this
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_33.pdf
Q7

 

[$$AK$$]

Differentiate the power and then differentiate what's inside the bracket
You can do it by using Quotient Rule to differentiate (1-x)/(1+x)
Multiply both the differentials to obtain an equation in terms of dy/dx
The thing here is to split (1+x) or (1-x) into (1+x)^0.5 * (1+x)^0.5 or (1-x)^0.5 * (1-x)^0.5 respectively, to cancel out the denominator
Hopefully, now you'll get the answer

Hope this helps. Sorry can't upload the working of it..

THNX MAN

 

[$$AK$$]

Differentiate the power and then differentiate what's inside the bracket
You can do it by using Quotient Rule to differentiate (1-x)/(1+x)
Multiply both the differentials to obtain an equation in terms of dy/dx
The thing here is to split (1+x) or (1-x) into (1+x)^0.5 * (1+x)^0.5 or (1-x)^0.5 * (1-x)^0.5 respectively, to cancel out the denominator
Hopefully, now you'll get the answer

Hope this helps. Sorry can't upload the working of it..

Hey man
How are u ?

Could u plz help me with this:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf
Q4(ii)( how did you find the initial value ??) and Q6(ii) ( the curve)

Thnx in advance
Help is highly appreciated

 

[Khan Boi]

Thnx man
It was helpful

could u plz help me also with this
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_33.pdf
Q7

Bro, really don't know how to solve that question. I am just hoping that it will never come again

 

[Harsh Poddar]

Thnx man
It was helpful

could u plz help me also with this
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_33.pdf
Q7

Use (a+bi) in place of z and (a-bi) in place of z* and you shall be able to get the answer.

Also check this out:

I've made this site that can be used to quickly access past papers to all the subjects of IGCSE, and AS and A level students. Try it out and please give feedback. Already being used in 52 countries by 900 users.

IGCSE: gopapers.net/igcse.html
A & AS Level: gopapers.net/alevel.html

 

[Khan Boi]

Hey man
How are u ?

Could u plz help me with this:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf
Q4(ii)( how did you find the initial value ??) and Q6(ii) ( the curve)

Thnx in advance
Help is highly appreciated

Q4)ii Take any suitable value for initial and you will end up with the answer. Since x is less than 90 degrees so take a value below 90 degrees. Btw, it's radians so do the conversions

Q6)ii The best thing i do with random curves is take random values of x and put it in the equation to get corresponding value of y. And just connect the dots.

 

[GCE As and a level]

Use (a+bi) in place of z and (a-bi) in place of z* and you shall be able to get the answer.

Also check this out:

man could u plz solve it and send me the solution plz

 

[GCE As and a level]

Bro, really don't know how to solve that question. I am just hoping that it will never come again

Thnx man
Dont worry
If u got the answer share it with me plz and even me if i got the answer i will share it with u

 

[GCE As and a level]

Q4)ii Take any suitable value for initial and you will end up with the answer. Since x is less than 90 degrees so take a value below 90 degrees. Btw, it's radians so do the conversions

Q6)ii The best thing i do with random curves is take random values of x and put it in the equation to get corresponding value of y. And just connect the dots.

man can u plz explain them more in details plz
and if u can solve them and send me the answer
BEST OF LUCK
THNX

 

[$$AK$$]

Q4)ii Take any suitable value for initial and you will end up with the answer. Since x is less than 90 degrees so take a value below 90 degrees. Btw, it's radians so do the conversions

Q6)ii The best thing i do with random curves is take random values of x and put it in the equation to get corresponding value of y. And just connect the dots.

Thnx man
They are helpful
Inshallah u will get A* in mathematics

 

[Mohamed ibrahim]

The complex question is so so hard but you should think about i, and use it, if you're really smart you will get it but everyone in my center wasnt able to solve it