find no of fns...

harsh 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....

a^b = b^a

Now raise both sides to the power \frac{1}{ab}

We get a^{\frac{1}{a}} = b^{\frac{1}{b}}

f(x)^{1/f(x)}

ye kaise aaya??

the interchage that u did b4 this, I got it.......

tapan, 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

@priyam: usually in functional equations, we do not assume differentiability unless given. Of course, since JEE is objective ....

PROPHET SIR's solution : 3rd line, interchangig part samjh aa gaya, but how di u get deduction Sir?

maine to differentiation wala kiya tha [3] usse bhi ban gaya tha.. [1]

two functions..... :) me got the same.. :)but from diff method :)

lekin yaha 3rd line nahi samjhme aaya.. :(

First 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

no of intersections of

y=x^x
y=2x

:)

hmm [12] either answer is 1,2 or 3 plz say whether ans is within my options

Hint: Can you prove that we must have f(x) = constant.

[We will worry about finding that constant later]

f(y)logf(x)=logf(x)f(y)
f(x) exponential type ka aur y linear type ka