341jee, aieee vagera ke liye theek hey. par bina assumptions solve karna, uska to alag hi mazaa hey
23 Answers
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.......
341
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