f(x).........................!!!

i know...all r easy once you get f(x)=x³

I'm really sorry guys [2][2][2]....I made a big mistake...in post#5 & #7
I have made the changes!!

i 2 feel que needs mor clarity

I dont get the definiton of g only

how can u solve for g if you dont know g!

bhai log : wat abt post #7? uska ans kisiko aaya kya?

can u explain the third step a bit more?

what happened to r

(x-y).f(x+y) - (x+y).f(x-y) = 4xy(x2 - y2) for all x,y R , and f(1)=1.
x-y=t
x+y=r

tf(r)-rf(t)=(r²-t²)tr

f(r)/r-f(t)/t=(r²-t²)

f(t)/t-t²=f(r)/r-r²

hence it is equal for all values of r or t

that is LHS is independent of r and
RHS is independent of T

thus, it is constant.

thus

f(t)/t-t² = k

f(t)=tk+t³

f(1)=1 .. then k=0

neone got it?

@subhash...

proceeding after post #12 ??? ......i mean proving that x³ is the only soln.

wat abot the longer method.....its no tooo long...needs some integration!!

then what?

yaar krish...

u got the fn y=x³ others are easy... ;)

so 1)c
2)a

Q5.

The area bounded by the curves g(x) and \phi(x) , is
(a) (2+√3) sq.units
(b) (2\pi+√3) sq.units
(c) (1/(2+√3)) sq.units
(d) (1/(2\pi+√3)) sq.units

is one of the solutions but a longer method would be there to prove it is the only one

yeah !!!...actually i did not work out the function, but i worked out that it is odd...
dats enuf to answer both i think...

how did u work out the exact function priyam ?? plzz xpln yaar..

Q3.

Define the function g(x) ≡ (f(x))^1/3 + (f(y))^2/3 = 0
(a) x³ + y⁶ + 1 = 0
(b) x = y²
(c) x + y² = 0
(d) none of these

Q1.

The function at x=0,attains
(a) local maximum
(b) local minimum
(c) point of inflexion
(d) none of these