1harsh bhaiya!! again a great one...
btw, jus on first look , it can be easily said f(x)=2 satisfies...
but dat doesnt make any sense unless derived or proved....
21hmmmm....... yeah!!!
sir jus a min. I'll post a fundamental dbt dat i hav in hyperbola
341a^b = b^a
Now raise both sides to the power \frac{1}{ab}
We get a^{\frac{1}{a}} = b^{\frac{1}{b}}
21f(x)^{1/f(x)}
ye kaise aaya??
the interchage that u did b4 this, I got it.......
341jee, aieee vagera ke liye theek hey. par bina assumptions solve karna, uska to alag hi mazaa hey
33wasie mere us method ke piche.. Nishant bhaiya bhi piche pade the ek baar... :D
341tapan, can you explain what u need to understand.which deduction?
In case you mean, how do we conclude that f(x) is a constant. just take logs of both the equations [allowed as f(x)>0). You will see that f(x)/(1-f(x)) is a constant
oh, i had written putting x=y before that. I have now edited it to swapping x and y
33pata hai thats not 100% correct... but 99.99% sahi hi hota hai...[3]
par jaldi banta hai usse... :P
341@priyam: usually in functional equations, we do not assume differentiability unless given. Of course, since JEE is objective ....
21PROPHET SIR's solution : 3rd line, interchangig part samjh aa gaya, but how di u get deduction Sir?
33maine to differentiation wala kiya tha [3] usse bhi ban gaya tha.. [1]
11i coulnt get the ques
there r many functions for which this would be satisfied for some some value
PLEASE EXPLAIN CLEARLY WHAT DO U NEED [7]
33two functions..... :) me got the same.. :)but from diff method :)
lekin yaha 3rd line nahi samjhme aaya.. :(
341First note that f(x): \methabb{R^+} \rightarrow \mathbb{R^+}
Now, putting x = y, we get f(x)^{f(x)} = 2f(x) \Rightarrow f(x)^{f(x)-1} = 2
Swapping x and y, we get f(x)^{f(y)} = f(y)^{f(x)}) \Rightarrow f(x)^{\frac{1}{f(x)}} = \text{ a constant}
From these two we get \frac{f(x)}{1-f(x)} = \text{a constant} so that f(x) = c where c>0.
Now, from the 1st eqn, we see that c is a solution for c^c = 2c
c=2 is one solution. there is one more lying between (0,1). No other solutions exist.
So, if you have multiple choice, f(x) = 2 is the one to tick
33:P
mere dimaag ka upaj tha.. khali samay me kuch kuch kar diye....
acha const nahi aega.. :P
1hmm [12] either answer is 1,2 or 3 plz say whether ans is within my options
341Hint: Can you prove that we must have f(x) = constant.
[We will worry about finding that constant later]
11f(y)logf(x)=logf(x)f(y)
f(x) exponential type ka aur y linear type ka
