Q.

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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a

the origin

b

the line x=12

c

the point (1,0)

d

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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