Let `f(x)=log(2x-x^2)+sin(pix)/2dot` Then which of the following is/are true? Graph of `f` is symmetrical about the line `x=1` Maximum value of `fi s1.` Absolute minimum value of `f` does not exist. none of these

Options

A. Graph of f is symmeterical about the line x=1
B. maximum value of f is 1
C. absolute mimumum value of f does not exist
D. none of these

Answer

1,2,3

Explanation

`f(X)=log2x-x^(2)+sin(pix)/(2)`
`=log1-(x-1)^(2)+sin(pix)/(2)`
`f(1-x)=log1-(1-x-1)^(2)+sin pi(1-x)/(2)`
`=log(1-x^(2))+cos(pix)/(2)`
`f(1-x)=log(1-x-1)^(2)+sin(pi(1-x))/(2)`
`log(1-x^(2))+cos(pix)/(2)`
Also `f(1+x)=log1-(x+1-1)^(2)+sin(pi(1+x))/(2)`
`=log(1-x^(2))+cos(pix)/(2)`
Hence function is symmetrical about line x =1
Also f(x) =1 Alsof(x)=1
Aslo for domain fo hte fucntion `2x-x^(2)gt0 or x in (0,2)`
For `xgt1` f(x) decrease hecne x =1 is point of maxima
Also maxima value of the funnction is 1
Further f(X)`rarr oo, when x rarr` 2 hence absoulte minimum value of does not exit