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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