Let f be a differential function such that `f(x)=f(2-x)` and `g(x)=f(1 +x)` then (1) g(x) is an odd function (2) g(x) is an even function (3) graph of f(x) is symmetrical about the line x= 1 (4) `f'(1)=0`

Options

A. g(x) is an odd function
B. g(x) is an even function
C. graph of `f(x)` is symmetrical about the line x = 1
D. `f'(1)=0`

Answer

B, C, D

Explanation

We have `f(x)=f(2-x)`
Replacing x by `f(1+x)`, we get
`f(1+x)=f(1-x)`
Hence graph of f(x) is symmetrical about the line x = 1.
Also `g(x)=f(1+x)=f(1-x)=g(-x)`
`therefore` g(x) is an even function.
Further `f'(1+x)=-f'(1-x)`
`therefore" "f'(1)=-f'(1)`
`therefore" "f'(1)=0`