Let gx=fx-1. If fx+f1-x=2∀x∈R, then gx is symmetrical about

fx-1+f1-x-1=0So, gx+g1-x=0.Replacing x by x+12, We get g12+x+g12-x=0.So, it is symetrical about 12,0.

Let gx=fx-1. If fx+f1-x=2∀x∈R, then gx is symmetrical about

fx-1+f1-x-1=0So, gx+g1-x=0.Replacing x by x+12, We get g12+x+g12-x=0.So, it is symetrical about 12,0.

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$Let g (x) = f (x) - 1 . If f (x) + f (1 - x) = 2 \forall x \in R, then g (x)$ is symmetrical about

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the origin

$the line x = \frac{1}{2}$

$the point (1, 0)$

$the point (\frac{1}{2}, 0)$

answer is D.

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$\begin{array}{l} f (x) - 1 + f (1 - x) - 1 = 0 \\ So, g (x) + g (1 - x) = 0 . \\ Replacing x by x + \frac{1}{2}, We get g (\frac{1}{2} + x) + g (\frac{1}{2} - x) = 0 . \\ So, it is symetrical about (\frac{1}{2}, 0) . \end{array}$

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Let gx=fx-1. If fx+f1-x=2∀x∈R, then gx is symmetrical about

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$Let g (x) = f (x) - 1 . If f (x) + f (1 - x) = 2 \forall x \in R, then g (x)$ is symmetrical about