Mathematics: Post your doubts here!

[parthrocks]

Yes, pls..?



The book states the answer is 2^5/4 x^-1 y^5/4 .. but I dunno how to reach at this answer..

I am sorry in a bit hurry if I get time will for sure solve it...bye...

 

[soumayya]

Haseefa

 

[shanky631]

can someone plz show me how to integrate (1/(x-2(x)^0.5) using substitution method where x=u^2??? plz

 

[Haseefa]

Haseefa

I understood it now Jazakillah Khair

 

[Silent Hunter]

Aslamoalikum
need some help in exponential and logs thingy
what does basically 'e' stand for in exponential functions? i mean what it is and its significance?

JazakAllah and Thank you

 

[bamteck]

Aslamoalikum
need some help in exponential and logs thingy
what does basically 'e' stand for in exponential functions? i mean what it is and its significance?

JazakAllah and Thank you

base for natural logarithms and instead of 10, is 2.71828 function. ln on calculator gives log to this base.

 

[bamteck]

Please someone help me for no. 4(ii), 5 & 7
smzimran if possible !

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

 

[shanky631]

doubt in the following questions. plz someone solve it.

smzimran,
nightrider1993

 

[parthrocks]

doubt in the following questions. plz someone solve it.
View attachment 14832
smzimran,
nightrider1993

i HOPE U UNDERSTOOD!!

 

[parthrocks]

doubt in the following questions. plz someone solve it.
View attachment 14832
smzimran,
nightrider1993

THIS IS ANOTHER FOR YOU!!GT IT?
we did one silly mistake when doing it in school i guess??

 

[shanky631]

i HOPE U UNDERSTOOD!!

hey thanks dude got it.

 

[nightrider1993]

doubt in the following questions. plz someone solve it.
View attachment 14832
smzimran,
nightrider1993

Hey sorry i read your question just now. but i think parthrock's already answered it pretty well

 

[shanky631]

Hey sorry i read your question just now. but i think parthrock's already answered it pretty well

hey no problem.

 

[bamteck]

Hey sorry i read your question just now. but i think parthrock's already answered it pretty well

nightrider1993 or parthrocks Please help me for the above question that I posted

 

[bamteck]

Which question?

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

no. 4(ii) & 8.

 

[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