#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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, thena) 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 Grade:Upto college level

## 1 Answers Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
11 years ago

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,

Askiitians Experts
Soumyajit Das IIT Kharagpur

## ASK QUESTION

Get your questions answered by the expert for free