$$g(x)$$ is symmetrical about Let $$g(x)=f(x)$$−$$1$$. If $$f(x)+f(1$$−$$x)=2$$ ∀x∈R, then

Correct option is 4the point 12,0

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

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the origin

the line x=12

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the point 12,0

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detailed solution

Correct option is D

f(x)−1+f(1−x)−1=0 So, g(x)+g(1−x)=0. Replacing x by x+12, we get g12+x+g12−x=0 So, it is symmetrical about 12,0.

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g(x) is symmetrical about Let g(x)=f(x)−1. If f(x)+f(1−x)=2 ∀x∈R, then

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