Mathematics: Post your doubts here!

[Silver Wing]

thanx but how do u kno dat domain is all real numbers and what are real numbers

The real numbers are all the numbers including rational and irrational ones, however they do not include imaginary numbers. These are square root of negative numbers, they include the symbol "i" equivalent to square root of negative one, they do not exist as cartseian coordinates and a graph cannot contain them, they are only plotted as points on an argand diagram, between if you are not an A2 student, then you do not have to know about imaginary numbers. In addition, the domain is all real numbers (negative infinity < x < positive infinity) because if you substitute any value of x, then the graph would give a valid value for y.

Hope I have helped you

 

[Dug]

thanx but how do u kno dat domain is all real numbers and what are real numbers

Because there is no real value of 'x' for which the function cannot produce a corresponding real value of 'y'.
The set of real numbers is a union of all rational and irrational numbers.

If the function was something like f(x) = x^2/(x+1), then the domain would've been as follows:

x + 1 = 0
x = -1

Domain: All real numbers except -1. f(-1) is undefined since division by zero is attempted. You can observe this phenomenon in tan curves too.

 

[Iffat]

The real numbers are all the numbers including rational and irrational ones, however they do not include imaginary numbers. These are square root of negative numbers, they include the symbol "i" equivalent to square root of negative one, they do not exist as cartseian coordinates and a graph cannot contain them, they are only plotted as points on an argand diagram, between if you are not an A2 student, then you do not have to know about imaginary numbers. In addition, the domain is all real numbers (negative infinity < x < positive infinity) because if you substitute any value of x, then the graph would give a valid value for y.

Hope I have helped you

thanx

 

[Iffat]

Because there is no real value of 'x' for which the function cannot produce a corresponding real value of 'y'.
The set of real numbers is a union of all rational and irrational numbers.

If the function was something like f(x) = x^2/(x+1), then the domain would've been as follows:

x + 1 = 0
x = -1

Domain: All real numbers except -1. f(-1) is undefined since division by zero is attempted. You can observe this phenomenon in tan curves too.

thanx

 

[7devilsallaround]

taking sub u=2x+3:
du/dx = 2
dx=du/2

x=(u-3)/2

insert values in original eq:
=> ⌡{u-3)/2}/u du/2
take 1/4 outside: 1/4 ⌡(u-3)/u du
simplify: 1/4 ⌡1 - (3/u) du
integrate: 1/4 [u - 3lnu] + k
replace u by x: 1/4[(2x+3) - 3ln|2x+3|] +k
simplify: x/2 + 3/4 -3/4 ln|2x+3| + k

mine includes one more term too..are u sure that answer is correct??

That's exactly how I did it!
Guess its a misprint or something. The books can be wrong sometimes. //shrugs
Thanks!

 

[salvatore]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdfNo. 5 (ii)

Please help me solve the question above.. thanks!

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdfNo. 5 (ii)

Please help me solve the question above.. thanks!

6x+k = 7√x
Let y = √x
6x^2 - 7y + k = 0

b^2 - 4ac = 0 (Since its a tangent)
(-7)^2 - 4(6)(k) = 0
k = 49/24

 

[salvatore]

6x+k = 7√x
Let y = √x
6x^2 - 7y + k = 0

b^2 - 4ac = 0 (Since its a tangent)
(-7)^2 - 4(6)(k) = 0
k = 49/24

Thank you for your reply.
I don't understand how you got 6x^2.. isn't it 6x in the equation?

 

[Dug]

Thank you for your reply.
I don't understand how you got 6x^2.. isn't it 6x in the equation?

Sorry, that's a typo. I meant 6y^2.

 

[Silent Hunter]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf

question 7 ? help needed in last part of this question thank you

 

[salvatore]

Sorry, that's a typo. I meant 6y^2.

I'm really sorry for bothering, but i still don't understand how you solved it.. especially the part where you got 6y^2.
Please help me understand it.
Thanks

 

[Jspake]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
I completely don't understand how to solve question no. 7.
Please explain the solution.. thanks a lot

 

[Dug]

I'm really sorry for bothering, but i still don't understand how you solved it.. especially the part where you got 6y^2.
Please help me understand it.
Thanks

Don't be. I only used the substitution to make u realize that the equation was actually quadratic. Do you still not get it?

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
I completely don't understand how to solve question no. 7.
Please explain the solution.. thanks a lot

a)
Sn = n^2 +8n
S1 = a1, since the sum of the first term will be only the first term.
So proceeding to find a1:
S1 = 1 + 8 = 9

Now we know that,
a1 = a
a2 = a + d

S2 = 4 + 16 = 20

Also, S2 = a1 + a2
S2 = a + (a + d)
20 = 9 + (9 + d)
d = 2

b)
According to the statement:
a2 = a1 - 9
ar - a = - 9
a(r-1) = -9
a = -9/(r-1) ------ i

And,
a2 + a3 = 30
ar + ar^2 = 30 -------- ii

We have a system of equations.

Put i in ii, you will get 'r' and consequently 'a'.

 

[PhyZac]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.................


Can anyone solve this, please?

Integrate 4/(x+1)^2

 

[Dug]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.................


Can anyone solve this, please?

Integrate 4/(x+1)^2

Walaikum AsSalam Warahmatullahi Wabarakatohu

⌡4/(x+1)^2 dx
= 4⌡(x+1)^-2 dx
= 4 [(x+1)^-2+1 /-1] + C
= -4/(x+1) + C

 

[PhyZac]

Walaikum AsSalam Warahmatullahi Wabarakatohu

⌡4/(x+1)^2 dx
= 4⌡(x+1)^-2 dx
= 4 [(x+1)^-2+1 /-1] + C
= -4/(x+1) + C

Jazaka Allah Khairan....... Thank you so much !! May Allah reward you for your help Ameen, And In Shaa Allah you get the highest results in this world and hereafter Ameen!

 

[Dug]

Jazaka Allah Khairan....... Thank you so much !! May Allah reward you for your help Ameen, And In Shaa Allah you get the highest results in this world and hereafter Ameen!

Wa iyyakum. Thanks for this awesome Dua, for the second time!! May He grant you success in this world and the Hereafter and have mercy on you and your family. May He make these exams easy for you.

 

[RadzMau]

Hi. Can someone pls help me for this question?
-> When the polynomial P(x) is divided by (x-1), the remainder is 5 and when P(x) is divided by (x-2), the remainder is 7. Find the remainder when P(x) is divided by (x-1)(x-2).

 

[Dug]

Delete yours now ...cuz it have my qoute...And SO SORRy!

Haha. It's ok, not your fault.