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

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