Mathematics: Post your doubts here!

[Alice123]

Alice123
Thanks 2325590381028302802 times.
Jazaki Allahu Khairan

Just one more request, I realize I've been a burden, sorry about that
When do we add +c when integrating, and when not?

NO PROBLEM, jus remember me in your prayers
We do not add c when limits are given, like when u r finding area or volume. if limits r not given, we add a c

 

[Maz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_12.pdf
Please help me sketch the graph for question no. 4(ii), I'm confused!
Thanks

forgive me the uneven graph but actually the curve downwards is equal to the curve upwards

 

[GorgeousEyes]

GorgeousEyes
8iii) Simple integration:
View attachment 24746

After that put in the value of x = -1 and find "c" i guess you could do that.
And sorry about the stupid drawing

Thanks for ur help

 

[sma786]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 1, if they squared the equation, why did they only square the '6'? why not the entire bracket? (2x-3) and then why they divided by 2 ?!

 

[GorgeousEyes]

please jun010 variant 12 :no.10 ii and no.11 iv , v

 

[daredevil]

Q 6i)
and
Q9ii)

yelps for help

 

[GorgeousEyes]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 1, if they squared the equation, why did they only square the '6'? why not the entire bracket? (2x-3) and then why they divided by 2 ?!

(6)^2 /(2x-3)^2
so make the integ . --> 36 (2x-3)^-2 -- > 36(2x-3)^-1 / -2
and by using the limits 2 and 3 --> -6-(-18)= 12 pi

 

[Maz]

@GorgeousEyes
9i) if OAB is a straight line then OA is a part of OB,
Therefore, by comparing the two vectors we can that OB is twice of OA
p = 2

unit vector = each vector/ magnitude of vector
= (2i+ j + k)/(2^2 +1 +1)[square root]
= (2i+ j + k)/[square root]6

ii) Remember that the dot product of perpendicular vectors is '0'
therefore, OA.AB = 0
(pi + j + k) . ( [4-p]i, j, [p-1]k) = 0
p*(4-p) + (1*1) + (1*(p-1)) = 0
5p - p^2 = 0
solve quadricatically p= 0 or 5

iii) if you can roughly draw the diagram you will realise that AB = OC
therefore c = b-a
I hope you can work that out.

[i may have made some calculation errors in ii but the methods generally correct]

And next time could you please give the link of qp please

 

[A star]

Alice123
Thanks 2325590381028302802 times.
Jazaki Allahu Khairan

Just one more request, I realize I've been a burden, sorry about that
When do we add +c when integrating, and when not?

when integrating with limits we donot add c.
when integrating without limits we add c

 

[A star]

Q 6i)
and
Q9ii)

yelps for help

when scaler product less then 1 we sayy acute if greater then 1 then obtuse
other is simple integration

 

[PANDA-]

when integrating with limits we donot add c.
when integrating without limits we add c

I don't really understand the meaning of limits to be honest.

 

[Sarah22]

r u sure about c3?cn u tell me it is in which year

Thank you soo much.. The paper is Oct/Nov 2005, question 8 & 9

 

[iKhaled]

I don't really understand the meaning of limits to be honest.

in integration limits r the values u put on the top of that sign ∫ and on the bottom of it

 

[GorgeousEyes]

@GorgeousEyes
9i) if OAB is a straight line then OA is a part of OB,
Therefore, by comparing the two vectors we can that OB is twice of OA
p = 2

unit vector = each vector/ magnitude of vector
= (2i+ j + k)/(2^2 +1 +1)[square root]
= (2i+ j + k)/[square root]6

ii) Remember that the dot product of perpendicular vectors is '0'
therefore, OA.AB = 0
(pi + j + k) . ( [4-p]i, j, [p-1]k) = 0
p*(4-p) + (1*1) + (1*(p-1)) = 0
5p - p^2 = 0
solve quadricatically p= 0 or 5

iii) if you can roughly draw the diagram you will realise that AB = OC
therefore c = b-a
I hope you can work that out.

[i may have made some calculation errors in ii but the methods generally correct]

And next time could you please give the link of qp please

Thanks

 

[Sarah22]

1. May/June 2006, question 7
2. Oct/Nov 2006, question 10-Part (v)
It would be great help.. Thank you..

 

[A star]

1. May/June 2006, question 7
2. Oct/Nov 2006, question 10-Part (v)
It would be great help.. Thank you..

link would be nice

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_11.pdf 9iii) how to find the value of D the ms just states the answer not the method pls reply asap those who can as ii want to finish maths and do phy for rest of the nite

 

[falcon678]

link would be nice

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf 9iii) how to find the value of D the ms just states the answer not the method pls reply asap those who can as ii want to finish maths and do phy for rest of the nite

on it!

plus coordinates of B?

 

[Maz]

I don't really understand the meaning of limits to be honest.

Limits mean the the values of x between which the graph stretches.
For instance if the dy/dx is given as = 3x^2, the integration of this will give us y=x^3. it is correct however dy/dx= 3x^2 can also be gotten from y=x^3+2 or =x^3+4, there are multiple answers therefore we add "+c".
I hope that makes sense.

 

[A star]

on it!

plus coordinates of B?

just explain the last part of this question thankyou

 

[sma786]

(6)^2 /(2x-3)^2
so make the integ . --> 36 (2x-3)^-2 -- > 36(2x-3)^-1 / -2
and by using the limits 2 and 3 --> -6-(-18)= 12 pi

so why do we divide by -2?