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Question

If f'(x)=|x|{x} where x denotes the fractional part of x, then f(x) is decreasing in

A
(12,0)
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B
(12,2)
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C
(1,2)
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D
(12,)
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Solution

The correct option is A (12,0)
f(x)=|x|{x}=|x|(x[x])=|x|x+[x]
For x(12,0),
f(x)=xx1=2x1
As,
12<x<0
or 0<2x<1
or 1<2x1<0
or f(x)<0,

f(x) decreases in (12,0)
Similarly we can check for other given options say for x(12,2),
f(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪(x)x1,12<x<0xx+0,0x<1xx+1,1x<2...

Here, f(x) decreases only in (12,0), otherwise f(x) in other intervals is constant.

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