# If f(x + y) = f(x) + f(y) – 2xy – 1 for all x and y. If f'(0) exists and f'(0) = -sin$$\alpha$$, then find f{f'(0)}.

## Solution :

f'(x) = $$\displaystyle{\lim_{h \to 0}}$$ $$f(x+h) – f(x)\over h$$

= $$\displaystyle{\lim_{h \to 0}}$$ $${f(x)+f(h)-2xh-1} – f(x)\over h$$

= $$\displaystyle{\lim_{h \to 0}}$$ -2x + $$\displaystyle{\lim_{h \to 0}}$$ $$f(h)-1\over h$$ = -2x + $$\displaystyle{\lim_{h \to 0}}$$ $$f(h) – f(0)\over h$$

[Putting x = 0 = y in the given relation we find f(0) = f(0) + f(0) + 0 – 1 $$\implies$$ f(0) = 1]

$$\therefore$$ f'(x) = -2x + f'(0) = -2x – sin$$\alpha$$

$$\implies$$  f(x) = -$$x^2$$ – (sin$$\alpha$$).x + c

f(0) = – 0 – 0 + c $$\implies$$ c = 1

$$\therefore$$  f(x) = -$$x^2$$ – (sin$$\alpha$$).x + 1

so, f{f'(0)} = f(-sin$$\alpha$$) = -$$sin^2\alpha$$ + $$sin^2\alpha$$ + 1 = 1