If `f^(prime)(x)=|x|-{x},` where {x} denotes the fractional part of `x ,` then `f(x)` is decreasing in `(-1/2,0)` (b) `(-1/2,2)` `(-1/2,2)` (d) `(1/2,oo)`

Options

A. `((-1)/(2),0)`
B. `((-1)/(2),2)`
C. `((-1)/(2),2)`
D. `((1)/(2),oo)`

Answer

1

Explanation

`f(X) =|x|-{x}-|x|-(x-[x])=|x|-x+[x]`
For x `in (-1//2,0)`
`f(X) =-x-x-1=-2x-1`
Also for `-1/2ltxlt0 or 0lt-2xlt1 or -1lt-2xlt0`
or `f(x) lt0,f(X)` decreases in `(-1//2,0)`
Similary we can chech for other given option say for `x in (-1//2,2)`
`f(x)={{:((ix-x-x),(-1)/(2)ltxlt0),(x-x+0,0lex1),(x-x+1,1lexlt2):}`
Here f(X) decreases only in `(-1//2,0)` otherwise f(X) in other intervals is constant