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