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Let g(x)=f(x)-1. If f(x)+f(1-x)=2AAx in ...

Let `g(x)=f(x)-1.` If `f(x)+f(1-x)=2AAx in R ,` then `g(x)` is symmetrical about. the origin (b) `t h el in ex=1/2` the point (1,0) (d) the point `(1/2,0)`

A

the origin

B

the line `x = (1)/(2)`

C

the point (1, 0)

D

the point `((1)/(2), 0)`

Text Solution

Verified by Experts

The correct Answer is:
D
`f(x)-1+f(1-x)-1=0`
So, `g(x)+g(1-x)=0`.
Replacing x by `x + (1)/(2)`,
we get `g((1)/(2)+ x)+g((1)/(2)-x)=0`
So, it is symmetrical about `((1)/(2), 0)`
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