functions

f:R→R suchthat
f(1)=2 f(2)=8
f(x+y) -kxy=f(x)+2y2

f(x+y)*f(1/(x+y))=
1.2k
2.k
3.k2
4.k+1

7 Answers

1
Aditya ·

Is the answer k?

1
Samarth Kashyap ·

yes.........could u please expln

1
Aditya ·

Taking x=1, y=1,

f(2) - k = f(1) + 2

So, we get k=4

In f(x+y)*f(1/x+y),

Putting x=1, y=0
We get it as f(1)*f(1)=4
So the answer k.

I dont know this method is acceptable or not!

1
Samarth Kashyap ·

anyways thanks a lot

1
Philip Calvert ·

common aditya
it could even have been k^2/4 then ....
u cant guess like that ...
there has to be a method

1
Aditya ·

Yeah i know Philip, but first only i told dat dis is just an objective method. Someone pls provide some proper method.

62
Lokesh Verma ·

f:R→R suchthat
f(1)=2 f(2)=8
f(x+y) -kxy=f(x)+2y2

f(x+y) - f(x) = 2y2 + kxy
divide by y on both sides.. take limit y going to zero

you get df/dx=kx

f=kx2+c

f(1)=2=k+c
f(2)=8=4k+c

k=2, c=0

function is 2x2

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