If a differentiable function f(x) satisfies the equation $f (x - 2) + f (x) + f (x + 2) = 0, \forall x \in R$ and $f^{'} (0) = 4$ , then select incorrect option

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$e^{f (x)} = e^{f (x + 12)}, \forall x \in R$

$\lim_{x \to 0} \frac{f^{2} (x + 12) - f (x) . f (0) - f (x + 6) . f (18) + f^{2} (18)}{x (\frac{π}{4} - \tan^{- 1} (1 - x))} = 32$

$\lim_{x \to 0} \frac{f (x) - f (0)}{\sin^{- 1} x} = 4$

$e^{f (x - 6)} = e^{f (6 - x)}, \forall x \in R$

answer is D.

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Detailed Solution

$∴ f (x - 2) + f (x) + f (x + 2) = 0 ....... (i)$
Replace x by $x + 2$ , we get
$f (x) + f (x + 2) + f (x + 4) = 0 ........ (i i)$
From (i) and (ii)
$f (x - 2) = f (x + 4)$
Replace x by $x + 2$ : $f (x) = f (x + 6)$
f(x) is periodic with period 6
and $f^{'} (x)$ is also periodic with period 6.