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
