Mathematics: Post your doubts here!

[leadingguy]

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w03_qp_3.pdf

no. 4(ii) & 8.

qstn 4 is implicit function so differentiate x and y both.


x^1/2 + y^1/2 = a^1/2 a is a constant means a figure that can be 2 ,3 , or else



differentiating
1/2x^(1/2-1)dx/dx + 1/2y^(1/2-1)dy/dx = O as "a" is a constant

1/2x^(-1/2) + 1/2y^(-1/2)dy/dx = O

make dy/dx the subject

1/2y^(-1/2)dy/dx = - 1/2x^(-1/2) cancel 1/2 on both sides.


y^(-1/2)dy/dx = - x^(-1/2) send y^(-1/2) on R.H.S

dy/dx = - x^(-1/2) /y^(-1/2) take inverse to make the powers positive

dy/dx = - (y/x)^1/2 Ans
have to go.. will do qstn 8 a little later

 

[bamteck]

qstn 4 is implicit function so differentiate x and y both.


x^1/2 + y^1/2 = a^1/2 a is a constant means a figure that can be 2 ,3 , or else



differentiating
1/2x^(1/2-1)dx/dx + 1/2y^(1/2-1)dy/dx = O as "a" is a constant

1/2x^(-1/2) + 1/2y^(-1/2)dy/dx = O

make dy/dx the subject

1/2y^(-1/2)dy/dx = - 1/2x^(-1/2) cancel 1/2 on both sides.


y^(-1/2)dy/dx = - x^(-1/2) send y^(-1/2) on R.H.S

dy/dx = - x^(-1/2) /y^(-1/2) take inverse to make the powers positive

dy/dx = - (y/x)^1/2 Ans
have to go.. will do qstn 8 a little later

Actually, I needed help for the second part !!

 

[leadingguy]

Actually, I needed help for the second part !!

the curve cuts the line y = x.

so solve simultaneously both eq.'s

√x + √y = √a

y = x

now
√x + √x = √a

2√x = √a

√x = √a/2 square both sides

x = a/4

the co-ordinates of p are x,y

(a/4 , a/4 ) as shown above.

now in first part we have found the dy/dx put value of x and y in that to find the gradient at the point p.

dy/dx = -√(y/x)



dy/dx = -√(a/4/a/4)

dy/dx = -1

now form eq.

y = mx + c

a/4 = -1(a/4) + c

solve to get c = 1/2a



now eq. will be

y= -x + 1/2a Ans

 

[leadingguy]

Actually, I needed help for the second part !!

x3 − x − 2
(x − 1)(x2+ 1)
.
(i) Express f(x) in the form

remember both powers of numenator and denominator are equal in this part, so this is an improper fraction.



denominator when expanded in this fraction becomes

x^3 + x - x2 - 1

now divide the numenator for denominator.



as we always do in polinomial fractions.

it is shown in the attachment below.


now 1 + ( x^2 - 2x - 1 )/ (x-1)(x^2 + 1)

1 is the constant known as A here.

now rest is a proper fration.

B/(x-1) + Cx+D(X^2 + 1) = x^3 - X -2

taking lcm

B(X^2+1) + Cx+D(X-1) = X^3 - X -2

take x= 1


2B + 0 = - 2

B = -1



x= 0

B(X^2+1) + Cx+D(X-1) = X^3 - X -2

-1 -D = -1 -1 - 2

D = 3

x= -1


B(X^2+1) + Cx+D(X-1) = X^3 - X -2

-2 + (-C +3)(-2) = -1 +1 -2

-2 + 2C -6 = -2
2c -8 = -2

2c = -2+ 8
c = 3


hope these values are correct have not checked the m.s yet.

 

[bamteck]

x3 − x − 2
(x − 1)(x2+ 1)
.
(i) Express f(x) in the form

remember both powers of numenator and denominator are equal in this part, so this is an improper fraction.



denominator when expanded in this fraction becomes

x^3 + x - x2 - 1

now divide the numenator for denominator.



as we always do in polinomial fractions.

it is shown in the attachment below.


now 1 + ( x^2 - 2x - 1 )/ (x-1)(x^2 + 1)

1 is the constant known as A here.

now rest is a proper fration.

B/(x-1) + Cx+D(X^2 + 1) = x^3 - X -2

taking lcm

B(X^2+1) + Cx+D(X-1) = X^3 - X -2

take x= 1


2B + 0 = - 2

B = -1



x= 0

B(X^2+1) + Cx+D(X-1) = X^3 - X -2

-1 -D = -1 -1 - 2

D = 3

x= -1


B(X^2+1) + Cx+D(X-1) = X^3 - X -2

-2 + (-C +3)(-2) = -1 +1 -2

-2 + 2C -6 = -2
2c -8 = -2

2c = -2+ 8
c = 3


hope these values are correct have not checked the m.s yet.













View attachment 14858

Thank you very much. It was a great help from your side
ALLAH bless you my dear

 

[parthrocks]

Thank you very much. It was a great help from your side
ALLAH bless you my dear

Sorry for my late reply but I was not present here!!ANd my friend leadingguy has done with all the steps!!

 

[bamteck]

Sorry for my late reply but I was not present here!!ANd my friend leadingguy has done with all the steps!!

hey there parthrocks help me with Nov 2004 p1 no. 9 & p3 no. 5, 6
Thanks

 

[shanky631]

when x=4sinu

plz can someone help me. thanks in advance.

 

[leadingguy]

View attachment 14874

when x=4sinu

plz can someone help me. thanks in advance.

I may solve this if u post the ans as well. why do not u post it with the question????

it will help u in geting ur queries solved soon!

 

[Snowberry]

Stuck with those 3. If anyone could help, it'll be greatly appreciated. Thanks!

http://i.imgur.com/xx9SH.jpg
http://i.imgur.com/Z3VP3.jpg
http://i.imgur.com/Gd0qk.jpg

 

[shanky631]

I may solve this if u post the ans as well. why do not u post it with the question????

it will help u in geting ur queries solved soon!

ans= 8pie

 

[leadingguy]

Stuck with those 3. If anyone could help, it'll be greatly appreciated. Thanks!

http://i.imgur.com/xx9SH.jpg
http://i.imgur.com/Z3VP3.jpg
http://i.imgur.com/Gd0qk.jpg

will be great if u post the ans as well. for these. so then I may reply easily.

 

[shanky631]

ans= 8pie

and sorry there are limits 4 to -4.

 

[Snowberry]

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w05_ms_3.pdf

This is the link to the paper from which the questions have been cut out. Thanks a lot.

 

[Albert Einstein]

Kuld u help me to solve the last question.. plzz

 

[Snowberry]

will be great if u post the ans as well. for these. so then I may reply easily.

Hope you could solve them!

 

[leadingguy]

View attachment 14874

when x=4sinu

plz can someone help me. thanks in advance.

x = 4sinu so dx= 4cosu.du

replace the x in eq. with 4sinu


√(16 - (4sinu)^2) .dx

√(16 - (4sinu)^2).4cosu.du as dx = 4cosu.du


√(16 - 16sin^2u).4cosu.du

√(16 (1 - sin^2u).4cosu.du

simplify

4√(1 - sin^2u). 4cosu .du remember that √(1 - sin^2u) = cosu

4.cosu.4cosu.du

16cos^2u.du
now integrate this term.

it cannot be integrated directedly as it is cos^2u so use identity cos2x

cos^2u - sin^2u = cos2u

this will give u cos^2u = (1+cos2x)/2



substitute this in place of cos^2u , it will be then

16cos^2u.du

16(1+cos2x)/2.du

8(1+cos2x).du

now integrate this,

8(U + sin2x/2)

the limits are 4 qnd -4 we cannot use them directedly as the term x is also in the form of "u" so convert the limits using this formula.

x = 4sinu

first limit is 4 , means x = 4

4 = 4sinu

u will be π/2
second limit -4 will become 3/2π

now equate using limits.

8(U + sin2x/2)

8[(U + sin2x/2) - (U + sin2x/2) ]

8[( π/2 + sinπ) - ( 3/2π + sin3π)]

8 [π/2 + 0 -3/2π + 0 ]

8π ans

 

[leadingguy]

Stuck with those 3. If anyone could help, it'll be greatly appreciated. Thanks!

http://i.imgur.com/xx9SH.jpg
http://i.imgur.com/Z3VP3.jpg
http://i.imgur.com/Gd0qk.jpg

first link first part.


sin^2u = x

dx = 2sinu.cosu.du

coming back to question. substitute the value sin^2x in place of x

√(sin^2u/(1-sin^2u))dx

remove square root to simplify

sinu/(√(1-sin^2u)) dx .................. √(1- sin^2u) = cosu

so

sinu / cosu.dx ............... dx = 2sinu.cosu.du (proven above )

sinu/cosu * 2sinu.cosu.du

simplify

/2sin^2u.du this is the fraction proven ANS



2sin^2u.du

sin^2u can not be integratd directly
use identity cos2x = cos^2x - sin^2x

sin^2u = (1-cos2u)/2


now substitute

2sin^2u.du


2(1-cos2u)/2.du

simplify
(1-cos2u).du

now we can integrate these terms directly.

u -sin2u/2

put limits

the first one is 1/4
conert this in the form of u

the formula given is x= sin^2u

so 1/4 = sin^2u

u will be 30 degrees or pi/6

second limit is O convert it using same formula
we will get

O

now using these two limits solve ur obtained expression as follows

u -sin2u/2

pi/6 -sin2.(pi/6)/2 - 0 +sin2(O)/2

pi/6 -√(3/4) Ans

 

[nightrider1993]

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w03_qp_3.pdf

no. 4(ii) & 8.

 

[nightrider1993]

Stuck with those 3. If anyone could help, it'll be greatly appreciated. Thanks!

http://i.imgur.com/xx9SH.jpg
http://i.imgur.com/Z3VP3.jpg
http://i.imgur.com/Gd0qk.jpg

Question 5

Question 6