11I'm really sorry guys [2][2][2]....I made a big mistake...in post#5 & #7
I have made the changes!!
Let f(x) be a differentiable funtion satisfying
(x-y).f(x+y) - (x+y).f(x-y) = 4xy(x2 - y2) for all x,y \in R , and f(1)=1.
Then , ...........................
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11
1I dont get the definiton of g only
how can u solve for g if you dont know g!
1(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
11@subhash...
proceeding after post #12 ??? ......i mean proving that x3 is the only soln.
33wahi.. hua...
what is not done in a line or in mind is longer method :p
11Q4Define the function (x) ≡ (f(x))2/3 + (g(x))2/3 = 12
are these 2/3s multiplied or is it powers
11wat abot the longer method.....its no tooo long...needs some integration!!
33waise great one shubhas ...
i think this should be the only one... but let me try... :)
11Q5.
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
11is one of the solutions but a longer method would be there to prove it is the only one
33;) guess...
wild one... just saw the question and replied.. that
have a method (lengthy one) will try to post that...
1yeah !!!...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..
11
Define the function \phi(x) ≡ (f(x))2/3 + (f(y))2/3 = 12
(a) x6 + y6 + 1 = 12
(b) x3 = y2
(c) x2 + y2 = 0
(d) none of these
11Q3.
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
11Q1.
The function at x=0,attains
(a) local maximum
(b) local minimum
(c) point of inflexion
(d) none of these