InterviewSolution
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InterviewSolution
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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)`A. `((-1)/(2),0)`B. `((-1)/(2),2)`C. `((-1)/(2),2)`D. `((1)/(2),oo)` |
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Answer» Correct Answer - 1 `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 |
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