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A new trick in mathematics

Forum Expert
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21 Oct 2010 19:54:18 IST
Posts: 1958
21 Oct 2010 19:54:18 IST
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A new trick in mathematics
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

Hiii ! After getting respond in my article.......I am submitting this article....to let you all know the general method of solving question of this given format...

1) f(x+y) = f(x).f(y)   where f'(0) = 1.....find f(3)

2) f(2x+y) = 2f(x) + f(y) ...where f(0)=3 f'(0) = 2 ...find f(4)...

3) f(x-y) = f(x)/f(y) ...given f'(0) = 1....find f(3) ...

Since last two-or three years...these types of questions were invisible in JEE..but may be in future there is a chance of such question....

Method :

Step 1 :

It is clear from all the equations that...x and y are independent variable ...means...if you are substituting x=1 ..then there is no boundation on y .....y is entirely independent ..you can substitute any thing....

hence the expression lead us to a conclusion that...dy/dx = 0
Because no change in y if x is going to change

Step 2:

Differentiate the given equation both side...

like  f(x+y) = f(x).f(y) ......given f'(0) = 1

f'(x+y)(1 + dy/dx) = f'(x).f(y) + f(x).f'(y).dy/dx

put dy/dx = 0

then f'(x+y) = f'(x).f(y)

Put x= 0 then

f'(y) = f'(0) .f(y)    given f'(0) = 1
so f'(y) = f(y)

This is analogous to ...dy/dx = y

so  ln(y) = x + C ...

(If any condition is given then find C ...otherwise consider it as 0)

y = e^x

f(x) = e^x

f(3) = e^3 .....f(e) = e^e ...and f([:D]) = e^([:D]  ...

So thats all ....Now solve the other two examples that i have given....
Or may be you can refer to any book where such problems or given...try with this method...you will get your answer ....

And ...Comment on this...did you get the method...or not ?

Regards
Yagya

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Scorching goIITian

Joined: 27 May 2010 11:07:15 IST
Posts: 223
21 Oct 2010 21:42:12 IST
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thanx sir! i`ve properly understood the method...

Forum Expert
Joined: 19 Feb 2009 13:24:30 IST
Posts: 1958
21 Oct 2010 21:48:36 IST
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good

New kid on the Block

Joined: 21 Oct 2010 22:15:19 IST
Posts: 4
21 Oct 2010 22:19:53 IST
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vey very helpful sir, u have done a very nice job, but can you also post some good content on how to solve some tough type of functional equations????

Forum Expert
Joined: 19 Feb 2009 13:24:30 IST
Posts: 1958
21 Oct 2010 23:39:49 IST
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Veenu....Can u please submit questions....some tough type....i will shuffle the question wid this method ..and will give u the solution...So ..please submit a question...

Hot goIITian

Joined: 6 Oct 2010 15:37:54 IST
Posts: 148
22 Oct 2010 00:03:03 IST
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thanx sir but i have a query , here we got dy/dx=y then how can we take it to be zero at the start of the question?

Forum Expert
Joined: 19 Feb 2009 13:24:30 IST
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22 Oct 2010 03:10:34 IST
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Hii abhishek.....in last we get f'(y) = f(y) ....I have converted it into new variable....dy/dx = y ...It has nothing to do with the original consideration...look...it can be done this way toof'(y) = f(y) ...f'(y)/f(y) = K ...Integrate ...ln[f(y)] = Ky ....f(y) = e^(Ky) ....or f(x) = e^(kx) ..........( like in integration...we take z^2=t ..and so on...similarly in last i have taken f'(y) = dy/dx and f(y) = y) ...

Hot goIITian

Joined: 7 Jul 2009 08:39:47 IST
Posts: 164
22 Oct 2010 07:03:50 IST
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Thank you very much sir. Its very easy..

Blazing goIITian

Joined: 6 Mar 2010 10:45:20 IST
Posts: 380
22 Oct 2010 08:19:04 IST
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thnx sir, dy/dx = y then how ln(y) = x+c ??

Hot goIITian

Joined: 7 Jul 2009 08:39:47 IST
Posts: 164
22 Oct 2010 08:23:22 IST
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sir, what are the answers of the other two questions???

Blazing goIITian

Joined: 20 Aug 2010 11:20:32 IST
Posts: 342
22 Oct 2010 09:36:12 IST
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i always luvd calculus...thnks!

Cool goIITian

Joined: 19 Oct 2009 23:09:31 IST
Posts: 92
22 Oct 2010 10:19:30 IST
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it's very useful sir:-)thnks a ton...@ThIrStY of IIT separate the variables nd then integrate.....

Forum Expert
Joined: 19 Feb 2009 13:24:30 IST
Posts: 1958
22 Oct 2010 12:51:13 IST
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Blazing goIITian

Joined: 6 Mar 2010 10:45:20 IST
Posts: 380
22 Oct 2010 13:15:31 IST
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thnkyou sir, i got the trick..

Hot goIITian

Joined: 23 Sep 2010 20:19:09 IST
Posts: 114
22 Oct 2010 21:10:35 IST
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really nice

Cool goIITian

Joined: 22 Mar 2010 17:43:22 IST
Posts: 63
23 Oct 2010 09:09:41 IST
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thank you sir.really this is nice

Cool goIITian

Joined: 7 Oct 2010 17:41:16 IST
Posts: 89
23 Oct 2010 11:05:15 IST
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very nice sir,i got the trick in da first read......dese type of Q's really boggled me out before.....bt da way u hv explained...just fantastic....thanx a lot sir.

New kid on the Block

Joined: 29 Jun 2009 13:21:07 IST
Posts: 12
23 Oct 2010 11:41:20 IST
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thnx sir...these questions...i kind of never knew how to proceed...but nw it seems so easy....eager to know sm more tricks in othr subs as well...

Blazing goIITian

Joined: 10 Jul 2010 10:42:22 IST
Posts: 487
27 Oct 2010 19:11:41 IST
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as always...u r gr8 sir.......thnxxx

Blazing goIITian

Joined: 10 Feb 2010 19:56:25 IST
Posts: 830
28 Oct 2010 16:52:40 IST
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good

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