## Question

f(x+y) = f(x) + f(y) + 2xy - 1 " x,y. f is differentiable and f'(0) = cos a. Prove that f(x) > 0 " x Î R.

Joshi sir comment

f(x+y) = f(x) + f(y) + 2xy - 1

so f(0) = 1      obtain this by putting x = 0 and y = 0

now f (x+y) = f ' (x)  + 2y         on differentiating with respect to x

take x = 0, we get f ' (y) = f '(0) + 2y

or f ' (x) = cosα + 2x

now integrate this equation within the limits 0 to x

we get f(x) = x2 + cosα x + 1

its descreminant is negative and coefficient of x2 is positive so f(x) > 0