Mathematics: Post your doubts here!

[Anum96]

Oh yes sorry

(3/(3-2x)) + ((-x-2)/(x^2+4))
i.e.
View attachment 57730

Okay so here it is:

Bring the denominators up.

3(3 + 2x)^(-1) + (-x – 2)[(x^2 + 4)^(-1)]

Expand.

3(3 + 2x)^(-1) + [-x (x^2 + 4)^(-1)] – [2((x^2 + 4)^(-1)]

You need to separately expand :

(3 + 2x)^(-1) & (x^2 + 4)^(-1)

So,

(3 + 2x)^(-1) = 3^(-1)[1 +2x/3]^(-1)

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

1/3 (1 + (-1)(2x/3) + [(-1)(-1-1)]/2! (2x/3)^2

1/3 ( 1-2x/3+(-1)(-2)/2!(2x/3)^2

1/3 ( 1 – 2x/3 + 2/2! (4x^2/9)

1/3 ( 1 – 2x/3 + 4x/9)

1/3 – 2x/9 + 4x^2/27

Similarly,

(x^2 + 4)^(-1)

Is same as

(4 + x^2)^(-1)

4^(-1)[(1 + x^2/4)^(-1)]

¼ [1 + (-1)(x^2/4)

¼ [1 – x^2/4]

¼ - x^2/16

Plug back in.

3(1/3 – 2x/9 + 4x^2/27) – x(¼ - x^2/16) -2(¼ - x^2/16)

Simplify;

1 – 2x/3 + 4x^2/9 – x/4 + x^3/16 – ½ + x^2/8

Combine the like terms and solve.

1 – ½ - 2x/3 –x/4 + 4x^2/9 + x^2/8 [ term with power 3 is not needed]

½ - 11x/12 + 41x^2/72 Ans.

Okay, Im guessing I messed up somewhere Whats the answer?

 

[nehaoscar]

Okay so here it is:

Bring the denominators up.

3(3 + 2x)^(-1) + (-x – 2)[(x^2 + 4)^(-1)]

Expand.

3(3 + 2x)^(-1) + [-x (x^2 + 4)^(-1)] – [2((x^2 + 4)^(-1)]

You need to separately expand :

(3 + 2x)^(-1) & (x^2 + 4)^(-1)

So,

(3 + 2x)^(-1) = 3^(-1)[1 +2x/3]^(-1)

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

1/3 (1 + (-1)(2x/3) + [(-1)(-1-1)]/2! (2x/3)^2

1/3 ( 1-2x/3+(-1)(-2)/2!(2x/3)^2

1/3 ( 1 – 2x/3 + 2/2! (4x^2/9)

1/3 ( 1 – 2x/3 + 4x/9)

1/3 – 2x/9 + 4x^2/27

Similarly,

(x^2 + 4)^(-1)

Is same as

(4 + x^2)^(-1)

4^(-1)[(1 + x^2/4)^(-1)]

¼ [1 + (-1)(x^2/4)

¼ [1 – x^2/4]

¼ - x^2/16

Plug back in.

3(1/3 – 2x/9 + 4x^2/27) – x(¼ - x^2/16) -2(¼ - x^2/16)

Simplify;

1 – 2x/3 + 4x^2/9 – x/4 + x^3/16 – ½ + x^2/8

Combine the like terms and solve.

1 – ½ - 2x/3 –x/4 + 4x^2/9 + x^2/8 [ term with power 3 is not needed]

½ - 11x/12 + 41x^2/72 Ans.

Okay, Im guessing I messed up somewhere Whats the answer?

Thanks a lot! I'm sorry actually i seem to have typed out the question wrong in the image!
It's like this (3/(3-2x)) + ((-x-2)/(x^2+4)) and i accidentally typed the first denominator as 3+2x
I'm really sorry but i got the procedure
the answer is
1/2 + 5x/12 +41x^2/72

 

[Anum96]

Thanks a lot! I'm sorry actually i seem to have typed out the question wrong in the image!
It's like this (3/(3-2x)) + ((-x-2)/(x^2+4)) and i accidentally typed the first denominator as 3+2x
I'm really sorry but i got the procedure
the answer is
1/2 + 5x/12 +41x^2/72

Hahaha. Oops. Its okay. Youll get 5x/12 as the middle term with that.
You're welcomee!

 

[nehaoscar]

9093 English AS
Any tips on how to write a commentary?
Like the layout and perhaps a list of features and effects to look for?
Can anyone provide me with sample commentaries if you have done in school? (preferably with marks)

Also tips on paper 1 and paper 2 as well to get an A
Thanks in advance

 

[musiclover gurl]

Hello? When y=0 , (x-2)^2 = -3, and that means no real roots for that eq. 2 values for x must be in terms of i

Hi ***amd***
Thanks for replying and yeah its good !

 

[musiclover gurl]

Show that x^2 - x + 1 and x^2 + x + 1 are positive for all real x.
Help please?

 

[Anum96]

Show that x^2 - x + 1 and x^2 + x + 1 are positive for all real x.
Help please?

Put any value of x you'll get a positive result. Why? Because the squared term of both the expressions is positive. You can even use your sketching skills to show.

 

[***amd***]

Show that x^2 - x + 1 and x^2 + x + 1 are positive for all real x.
Help please?

find the min value, and show that it is a cup up graph.
proved

 

[holoholo]

View attachment 57739
How do you find the midpoint of each ?

 

[Anum96]

View attachment 57739
How do you find the midpoint of each ?

1 plus 100 /2
101 plus 150 /2
151 plus 200 /2
and so on....

 

[holoholo]

first midpoint is 50.5 not 50

 

[Anum96]

first midpoint is 50.5 not 50

Oops I corrected. Sorreh!

 

[holoholo]

where did 1 come from ? isn't it 0<=x<=100 so shouldn't it be 0+100/2 ?

 

[***amd***]

where did 1 come from ? isn't it 0<=x<=100 so shouldn't it be 0+100/2 ?

that means there are 101 numbers, i.e. 100 numbers, from 1 to 100 (that excludes Zero), and 101 numbers including 'Zero'.

 

[Anum96]

where did 1 come from ? isn't it 0<=x<=100 so shouldn't it be 0+100/2 ?

Yup. include 0. You'll get 101.

 

[holoholo]

How about this one ? Will the first midpoint be 3 or 2.5 ?

 

[Anum96]

View attachment 57741
How about this one ? Will the first midpoint be 3 or 2.5 ?

3 ^_^

 

[holoholo]

Can you show me how to do it for this one ?

 

[Anum96]

View attachment 57742
Can you show me how to do it for this one ?

0.1 plus 0.5 / 2
0.6 plus 1 / 2
1.1 plus 2 /2
2.1 plus 3 / 2
and 3.1 plus 4.5 / 2

 

[holoholo]

if it was 2.8<=x<3, is the midpoint 2.8+2.9/2 ?