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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the origin

the line x=12

the point (1,0)

the point 12,0

answer is D.

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Detailed Solution

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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