First slide
Differentiability
Question

 If f:RR such that f(x+y)f(xy)2y36x2yy4x,yR then 

Difficult
Solution

 Put x+y=u;xy=v

f(u)f(v)(x+y)3(xy)3∣≤uv24f(u)f(v)u3v3uv24|g(u)g(v)|uv24,  g(x)=f(x)x3

limuv|g(u)g(v)u-v|limuvuv23=0     g'x0g'x=0gx is a constant function. let it be k.    therefore fx=x3+k   1f'x=3x2 =0 x=0 , which is point of inflection      

 

 

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