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#1 Posted 9:32pm 07-02-10 aits question[b]1[/b]..THE POSITIVE REAL FUNCTION SATISFIES [frac]F(X)[/]F(Y)[/frac]≤7[p](X-Y)[p]2[/p][/p] FOR X , Y BELONG TO ITS DOMAIN ,F(3/2)=[frac]Π[/]4[/frac] THEN 1.F(X)=[sqrt]X[/sqrt]+X[p]3[/p] 2.F'(7)=0 3.F(X)=∫sin[p]-1[/p][sqrt]t[/sqrt] dt +∫ cois[p]-1[/p][sqrt]t[/sqrt] dt (in 3--limit for sin[p]-1[/p][sqrt]t[/sqrt] is from 0 to sin[p]2[/p]x while for second one 0 to cos[p]2[/p]x) |
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#2 Posted 2:40pm 08-02-10 Re: aits questionLemma: If a differentiable function satisfies for all x,y in its domain, the relation [im]http://codecogs.izyba.com/gif.latex?%7Cf%28x%29%20-%20f%28y%29%20%7C%20%5Cle%20C%20%7Cx-y%7C%5E2[/im] for some C>0, then f(x) is a constant. Proof: [im]http://codecogs.izyba.com/gif.latex?0%20%5Cle%20%5Cleft%7C%20%5Cfrac%7Bf%28x+h%29%20-%20f%28x%29%7D%7Bh%7D%20%5Cright%7C%20%5Cle%20Ch[/im] By Sandwich theorem f'(x) = 0. Here taking logarithms on both sides we get (this is permitted since range of f is the +ve real numbers) [im]http://codecogs.izyba.com/gif.latex?%7C%5Cln%20f%28x%29%20-%20%5Cln%20f%28y%29%7C%20%5Cle%202%20%28x-y%29%5E2[/im] So if g(x) = ln f(x) then g(x) and hence f is a constant. So Option (2) is correct Dunno abt Option 3 |
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