Mathematics: Post your doubts here!

[Wolfgangs]

Can someone tell me the final answer and working for this question please?

 

[Anum96]

Can someone tell me the final answer and working for this question please?

You should post this in some physics thread. Your previous questions were related to mechanics. That's why I could help. I don't know physics

 

[Sarosh Jameel]

plz help the answer i am getting is 1 / -y ( 1+y )

 

[Anum96]

plz help the answer i am getting is 1 / -y ( 1+y )

Yes, Im getting the same. I believe we are not wrong. There must be a problem with the question. Which year is it?

 

[Sarosh Jameel]

yeah !! i am searching for the paper year !! i will tell u

 

[Anum96]

yeah !! i am searching for the paper year !! i will tell u

Okay. Waiting

 

[Sarosh Jameel]

yeah !! i am searching for the paper year !! i will tell u

couldn't find .. i hope our answers are correct !

 

[Anum96]

couldn't find .. i hope our answers are correct !

Listen. We are wrong. Why? because if you take note of your solution. You get the limit 1 when x is zero. and limit 0.5 when x is ln2. Observe how the position changes of both the values. Before transformation we have ln2 as the larger limit but when we calcute y we get 1/2 which is the smaller value in the transformation. Therefore, When this happens, The sign of the expression can change

 

[Wolfgangs]

Can someone please help me with this? This is for Mechanics 2

 

[Sarosh Jameel]

Listen. We are wrong. Why? because if you take note of your solution. You get the limit 1 when x is zero. and limit 0.5 when x is ln2. Observe how the position changes of both the values. Before transformation we have ln2 as the larger limit but when we calcute y we get 1/2 which is the smaller value in the transformation. Therefore, When this happens, The sign of the expression can change

Ohh ! i didnt know about this points . BTW thanks (Y)

 

[Anum96]

Ohh ! i didnt know about this points . BTW thanks (Y)

 

[Sarosh Jameel]

So we just have to put a minus sign in this situation ?!?

 

[Anum96]

So we just have to put a minus sign in this situation ?!?

Oh no no. You're flipping the limits, right? So the sign will flip as well. Your negative sign will be changed to positive sign and so, You will get the correct transformation.

 

[Sarosh Jameel]

ohh ok !

 

[Hassan Ali Abid]

Yes, the book is available online, check this website:
http://www.aliensservices.com/E-books.html

As for the solved papers, it's difficult, a mere chance, that they'll be available for S2, very few students choose S2, but I'll search and let you know if I find!

Im trying to open the link but its not working for me ...anyways can you please have a look in to that book and lemm know if itcontains the probability using sets and venn diagrams

yeah please do search..it will surely be a great favour.

 

[Hassan Ali Abid]

Can anyone please explain me why the integration of (1/1+y^2) is tan inverse y and how the integration of y/1+y^2 gives .5ln(1+y^2) ???

 

[Anum96]

View attachment 57680 Can anyone please explain me why the integration of (1/1+y^2) is tan inverse y and how the integration of y/1+y^2 gives .5ln(1+y^2) ???

Look. When you have a fraction. And you have to integrate it.
Consider the denominator. differentiate it. If your differentiated expression is the same as your numerator then you can simply use log to integrate. If not then see if you can make it the same by multiplying. For example. In the above case. Your denominator is
1+y^2
Now differentiate it and you will get 2y right? BUT you have only y in numerator. Therefore you will make them same by taking out 1/2 and then simply use log. I hope Im coherent enough for you to understand.
It will become 0.5ln(1+y^2)
If its still not clear then let me know I will try and explain via working step by step.

 

[Anum96]

A

View attachment 57680 Can anyone please explain me why the integration of (1/1+y^2) is tan inverse y and how the integration of y/1+y^2 gives .5ln(1+y^2) ???

Also. Is there any substitution given in the question?

 

[Hassan Ali Abid]

Look. When you have a fraction. And you have to integrate it.
Consider the denominator. differentiate it. If your differentiated expression is the same as your numerator then you can simply use log to integrate. If not then see if you can make it the same by multiplying. For example. In the above case. Your denominator is
1+y^2
Now differentiate it and you will get 2y right? BUT you have only y in numerator. Therefore you will make them same by taking out 1/2 and then simply use log. I hope Im coherent enough for you to understand.
It will become 0.5ln(1+y^2)
If its still not clear then let me know I will try and explain via working step by step.

Great ...yeah I have recalled and understood that part.Thanks for it ..but what about the first one ....are we suppose to use any substitution ???

 

[Hassan Ali Abid]

A

Also. Is there any substitution given in the question?

NO..nothing as such is given in the question.