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

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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