Mathematics: Post your doubts here!

[Most_UniQue]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_1.pdf
8(ii) domain?

Domain of f^-1 = Range of f(x)

and you can find range by plugging -2 & 4 into the equation

 

[sea_princess]

can someone explain how to solve question 10 : iii
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf

 

[Esme]

The (ii) part of this question, wen we integrate it, i get the ans as 3square root(4x-3). The mark scheme gives a diff ans. plz explain.

The mark scheme is :

 

[Most_UniQue]

The (ii) part of this question, wen we integrate it, i get the ans as 3square root(4x-3). The mark scheme gives a diff ans. plz explain.
View attachment 8769
The mark scheme is : View attachment 8770

They have forgot to put the square root cuz C will nt be -6 if sqrt is nt used... Anser is 3 sqrt (4x-3) -6

 

[hm12]

The (ii) part of this question, wen we integrate it, i get the ans as 3square root(4x-3). The mark scheme gives a diff ans. plz explain.
View attachment 8769
The mark scheme is : View attachment 8770

its a flaw in the mark scheme in the first step its written correct but in the second last step they have missed (to the power 1/2)

 

[Thampi4]

PLs help how do i do question 10 ii) for paper 11 mj 2010

 

[Esme]

its a flaw in the mark scheme in the first step its written correct but in the second last step they have missed (to the power 1/2)

yeah even ii thought that would be the case but was just confirming.. Thanks anyways.

 

[Esme]

They have forgot to put the square root cuz C will nt be -6 if sqrt is nt used... Anser is 3 sqrt (4x-3) -6

yea i figured.. thank you.

 

[kewlryan58]

please help me with
Q:7 (ii) 9709_w11_qp_32
and if possible, provide me with summarise notes of vectors in 3-D... i will be thankul to you...!

 

[MEGUSTA_xD]

NOW ITS ALL ENDS AND TOMORROW IS THE DAY (

More 3 are left

 

[aaakhtar19]

okay correct me if I am wrong but AB=OB-OA right?

yah its perfect !!

 

[MEGUSTA_xD]

Pls post p1. Questions today

 

[rash1233]

for this you need to put 1/3 instead of dy/dx so this equation will be

1/3 = 3/(1+2x)^2

so by doing cross multiply we will get

(1+2x)^2 = 9
then open the bracket using a^2 + 2ab +b^2 formula and then simplify it! , you will get 2 values of X

Thanks!

 

[Dug]

Someone please help me with q ii . I get the integration wrong. Help ASAP Thanks


View attachment 8779

For y=5,
V = 25x
Limits are 1 and 4 if I remember correctly.
V = 25(4) - 25(1)
V = 75pi

For y=x+4/x
y^2 = (x^2 + 16/x^2 + 8) dx
V = x^3/3 - 16/x +8x
V = [4^3/3 - 16/4 +8(4)] - [1/3 - 16 + 8]
V = 64/3 - 4 +32 -1/3 +8
V = 57pi

Volume of the shaded region = 75pi - 57pi = 18pi

 

[robotic94]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_12.pdf

question 6 i

 

[hamzaarshad]

Find the range of values of k for which 8-3x-x2 <k for all values of x.
please show the working and reply at the earliest.
Thanks is advance.

 

[usmiunique]

can anyone plz help?
O/N/2011 P33 q 10(i)
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
how do we integrate the tan equation?
i cant go any farther than ((u^n+2+u^n)/(u^2+1))du. the limits are 1 and 0

 

[robotic94]

Find the range of values of k for which 8-3x-x2 <k for all values of x.
please show the working and reply at the earliest.
Thanks is advance.

are you sure the question is complete?

 

[maryamshake94]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w10_qp_12.pdf
someone please help me with question 6 part (ii), quickly!

 

[leadingguy]

can anyone plz help?
O/N/2011 P33 q 10(i)
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
how do we integrate the tan equation?
i cant go any farther than ((u^n+2+u^n)/(u^2+1))du. the limits are 1 and 0

(tanx^n+2 tanx^n)dx substitute tanx = u

(u^n+2 + u^n)dx now du = sec^2x dx

so when substituting du will give


(u^n+2 + u^n)/sec^2x du
now use identity sec^2x = 1+ tan^2x .. means (1 + U^2)

substitute this value of sec^2x in the eq. above



(u^n+2 + u^n)/ (1 + U^2)du now take U^n in numenator

U^n (U^2 + 1 )/ (1 + U^2)du now cxancel (U^2 +1)
u wil get a simple eq. (U^n)du
inegrate it
u^n+1 / n+1 now convert back into original by substiuting tanx

tanx^n+1/ n+1 put the limits
tan(45)^n+1/n+1 tan45 = 1 so... 1^n+1 = 1


it wil be as follows 1/ n+1