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ANAM V. RANGA REDDY Grade: Upto college level
        

Suppose f is a function that satisfies the equation f (x+y) = f(x)+f(y)+x2y+xy2 for all real real numbers x and y.if limit {f(x)/x} as x →0 is equal to 1, then


a) f(x)>0 for x>0 and f(x)<0 for x<0              2)  f1(0)=1    


3)f11(0)=1  4)f111(x) is a constant function


 

6 years ago

Answers : (1)

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
										

Dear Aman V Ranga Reddy,


Ans:-f(x+y)=f(x)+f(y)+x²y+y²x


Differntiating With Respect to X keeping y constant we get, f′(x+y)=f′(x)+2xy+y²...............(1)


Now limit{f(x)/x) as x tends to 0 is=1 Hence f′(0)=1.............(2)


Putting x=0 in eq (1) we get


f′(y)=f′(0)+y²=1+y²


f′(y)=1+y² (which is true as we put y=0 it satifies eq 1)


f′′(y)=2y


f′′′(y)=2


Hence options 2 and 4 are correct


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All the best Aman V Ranga Reddy !!!


 



Regards,


Askiitians Experts
Soumyajit Das IIT Kharagpur

6 years ago
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