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Mathematics: Post your doubts here!

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

JazakAllah and Thank you :)
 
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Aslamoalikum
need some help in exponential and logs thingy :p
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. :p
 
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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 :)
 
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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 !! :p
 
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Actually, I needed help for the second part !! :p


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
 
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Actually, I needed help for the second part !! :p
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.













untitled.JPG
 
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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 :)
 
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