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

Correct option is 4the point 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=12

the point (1,0)

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 . If fa+x+fa-x=0 then fx is symmetric about a,0

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

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