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$Let f (x + y) = f (x) f (y) and f (x) = 1 + (\sin 2 x) g (x) where g (x) is$ $continuous. Then f^{1} (x) equals$

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a

$f (x) g (0)$

b

$2 f (x) g (0)$

c

$2 g (0)$

d

$2 f (0)$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{f (x) f (h) - f (x)}{h} \\ = f (x) \lim_{h \to 0} \frac{f (h) - 1}{h} = f (x) \lim_{h \to 0} \frac{1 + (\sin 2 h) g (h) - 1}{h} \\ = f (x) \lim_{h \to 0} \frac{\sin 2 h}{h} \underset{h \to 0}{\lim g} (h) = 2 f (x) g (0) \end{array}$

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