Mathematics: Post your doubts here!

[parthrocks]

Can anyone help me in this question?

Thanks in advance.

View attachment 14038

same here how to solve for this!?????????????????????????????????anyone

 

[celoth]

Can anyone help me in this question?

Thanks in advance.

View attachment 14038

Let f(x) = (x^2 - 1)Q(x) + ax + b
f(x) = (x-1)(x+1)Q(x) + ax + b

we know that (x+1) is a factor,
f(-1) = 0,
b-a = 0, this is equation 1,

(x-1) gives remainder of 4,
f(1) = 4,
a+b = 4 , this is equation 2,

solve simultaneous equation 1 and 2,
then you can find a and b.

 

[parthrocks]

hye b

Let f(x) = (x^2 - 1)Q(x) + ax + b
f(x) = (x-1)(x+1)Q(x) + ax + b

we know that (x+1) is a factor,
f(-1) = 0,
b-a = 0, this is equation 1,

(x-1) gives remainder of 4,
f(1) = 4,
a+b = 4 , this is equation 2,

solve simultaneous equation 1 and 2,
then you can find a and b.

hey bro but why did u substitute f(-1) in the remainder part it is supposed to be in the original equation thats f(x) and hey also thats what we call remainder theorem right?

 

[celoth]

hye b

hey bro but why did u substitute f(-1) in the remainder part it is supposed to be in the original equation thats f(x) and hey also thats what we call remainder theorem right?

f(-1) means we substitute the x with -1 in here:
f(x) = (x-1)(x+1)Q(x) + ax + b
every x above will turn into -1

 

[JulyMei]

Let f(x) = (x^2 - 1)Q(x) + ax + b
f(x) = (x-1)(x+1)Q(x) + ax + b

we know that (x+1) is a factor,
f(-1) = 0,
b-a = 0, this is equation 1,

(x-1) gives remainder of 4,
f(1) = 4,
a+b = 4 , this is equation 2,

solve simultaneous equation 1 and 2,
then you can find a and b.

Thank you!

 

[VelaneDeBeaute]

AoA,
This is best done using substitution, and not by parts:
View attachment 14035

cos^3 x
= (cos^2 x)(cos x)
= (1-sin^2 x)(cos x)
= cos x - (sin^2 x)(cos x)

after integration,
it become:
= -sin x - (1/3)(sin^3 x) + c

I started the same way - (I didn't know it was called substitution, i had been calling it by-parts)

The correct answer, or the one at the back of the book is,
1/3sinx (3cos^2 x + 2sin^2 x)

 

[celoth]

I started the same way - (I didn't know it was called substitution, i had been calling it by-parts)
Anyway, i'm afraid, both of you have got the wrong answers!
The correct answer, or the one at the back of the book is,
1/3sinx (3cos^2 x + 2sin^2 x)

it is the same.....

sin x - (1/3)(sin^3 x)
=(1/3)(sin x)(3-sin^2 x)
=(1/3)(sin x)(3cos^2 x + 3sin^2 x - sin^2 x)
= (1/3)(sin x)(3cos^2 x + 2sin^2 x)

 

[smzimran]

I started the same way - (I didn't know it was called substitution, i had been calling it by-parts)
Anyway, i'm afraid, both of you have got the wrong answers!
The correct answer, or the one at the back of the book is,
1/3sinx (3cos^2 x + 2sin^2 x)

It is not the wrong answer , just that you have got an answer in a different form, both forms are correct!

 

[VelaneDeBeaute]

celoth and smzimran, Oh! Sorry! I didn't quite work it out!
Thankyou very much then!

 

[leadingguy]

on smzimran's behalf... no problem

 

[VelaneDeBeaute]

I sense something fishy!
Anyway,

 

[smzimran]

I sense something fishy!
Anyway,

Nothing fishy, just my personal secretary!

 

[InnocentAngel]

Hello frens could someone please help me with these
Quest. 1 Complex numbers
Given the complex number, Z= -√(2i) ,find the argument of Z.
ANS : arg(Z)= -π/2

Quest.2 Complex N0.
Given Z= cos π/6 +cos π/3 .Find argument of Z and modulus of Z.
ANS : arg(Z)= π/6 and mdulus of Z = 1

Thanks in advance.

 

[Amy Bloom]

Hello frens could someone please help me with these
Quest. 1 Complex numbers
Given the complex number, Z= -√(2i) ,find the argument of Z.
ANS : arg(Z)= -π/2

Quest.2 Complex N0.
Given Z= cos π/6 +cos π/3 .Find argument of Z and modulus of Z.
ANS : arg(Z)= π/6 and mdulus of Z = 1

Thanks in advance.

Sorry i could only solve for question 1. See attached file.

 

[celoth]

Hello frens could someone please help me with these
Quest. 1 Complex numbers
Given the complex number, Z= -√(2i) ,find the argument of Z.
ANS : arg(Z)= -π/2

Quest.2 Complex N0.
Given Z= cos π/6 +cos π/3 .Find argument of Z and modulus of Z.
ANS : arg(Z)= π/6 and mdulus of Z = 1

Thanks in advance.

Question 2:
since Z = cos π/6 + i cos π/3
therefore, arg(Z) = π/6
or you can try to sketch this thing out on cartesian plane.

mod Z
= √ ( cos^2 π/6 + cos^2 π/3 )
= √ ( (√3 / 2)^2 + (1/2)^2 )
= 1

 

[parthrocks]

http://www.mediafire.com/view/?o2cp74d0881nbb9#
this is the link to differentitaing trignometric function chapter 6 in p2 and p3 i have a doubt in exercise 6a question number 6,question 7 and question 8 part a and part e.......
and also question 9 part b and f........................plzzzzzzzzzzzzzz reply asap...its just a few question.....plz attach scan!!
smzimran
minato112
celoth
plzzz help me asap

 

[shanky631]

plz someone solve this.

 

[Amy Bloom]

View attachment 14240
plz someone solve this.

(i) Differentiate the equation and equate the equation to zero, u shall have a value of x which u substitute in y to have the y-coordinate
(ii) Differentiate the equation u got in (i). this is d2y/dx2. Replace the value of x u got in (i) in d2y/dx2. If the value of the latter is negative , the point was negative, else positive.

Hey can u tell me how u upload an image?

 

[shanky631]

(i) Differentiate the equation and equate the equation to zero, u shall have a value of x which u substitute in y to have the y-coordinate
(ii) Differentiate the equation u got in (i). this is d2y/dx2. Replace the value of x u got in (i) in d2y/dx2. If the value of the latter is negative , the point was negative, else positive.

Hey can u tell me how u upload an image?

there is an option beside post reply which is upload a file. then click on it and select the image u want to upload.

 

[shanky631]

(i) Differentiate the equation and equate the equation to zero, u shall have a value of x which u substitute in y to have the y-coordinate
(ii) Differentiate the equation u got in (i). this is d2y/dx2. Replace the value of x u got in (i) in d2y/dx2. If the value of the latter is negative , the point was negative, else positive.

Hey can u tell me how u upload an image?

can u show me the working. i understand the concept but make errors while solving.