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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
```
7 years ago

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
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Aman V Ranga Reddy !!!

Regards,
```
7 years ago
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