Hii friends

Here is the solution of the two question :-

1) f(2x+y) = 2f(x) + f(y) ...given f'(0) = 2  f(0) = 3

2) f(x-y) = f(x)/f(y)     f'(0) = 1

f(2x+y) = 2.f(x) + f(y) ...........;f'(0) = 2 & f(1) = 3    find  f(3)

Again solve by differentiating both side ..

f'(2x+y) [ 2 + dy/dx ] = 2.f'(x) + f'(y) .dy/dx 

Now put dy/dx = 0   because y and x are independent ...

f'(2x+y) (2) = 2.f'(x)

f'(2x+y) = f'(x)

Put   x=0

f'(y) = f'(0)     =>     f'(y) = 2 

So  f(y) = 2y + C   ( simply integratre f'(y) = 2 )

f(1) = 2 + C   => C = 1    ( given f(1) =3)

So   f(y) = 2y + 1

So f(3) = 7    ans 

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

You have to use different way here....differentiate with respect tp y...i.e dx/dy

so  f'(x-y) [ dx/dy - 1 ]  =  [f(y)*f'(x).dx/dy  -  f'(y).f(x) ] / [f(y)]^2

Put dx/dy = 0

-f'(x-y) = -f(x).f'(y) / f(y)^2

put  y=0

-f'(x) = - f(x) * 1    [ f'(0) = 1 f(0) = 1 ]

f'(x) = f(x)

Integrate    ln(f(x)) =   x

so  f(x) = e^x .....................ans   f(2) = e^2    ans 

You can check it like this ...

f(x-y) = e^(x-y) = e^x/e^y   = f(x)/f(y)

regards

Yagya

Dear Pranav

That was one hell of the mistake sorry !

Actually I have picked up the question randomly....take it as f(1)=3 ......not f(0) = 3 

Sorry !

IF ur 1st answer is right ...then 0=2 .. Ur solution is wrong!!!!! stop giving fake knwledge to people ...it hurts...

U just showed in solution of the 1st problem that f(y) = 2y + 1 ... right?

that means f ( x) = 2x+1 .... that means LHS  = f( 2x + y ) = 2 ( 2x +y ) +1 = 4x + 2y +1.... equation 1

 RHS = 2f( x) + f(y) = 2 ( 2x +1) + 2y+1 = 4x + y+3...... equation 2

since equation 1 is not same as equation 2 ... ur procedure opposes the question itself!!! so please be careful

I have choosen random question.....just to show the steps of my method...

The constant that i have taken  f'(0) = 1  f'(1) = 3 or whatever  ...is random ...and ..i have only choosen it to show that how will u calculate the constant

If my students are able to understand the steps ..how to solve it ...i will be happy........

And this trick is no more incorrect....I have checked it solving all the numericals.....

And if any one is trying to explain his method by taking an arbitary example   like f(x) = 1/x .....then questioning like f(x) = 1/x is not possible is stupidious in my sense...!