The range of f(x)=x−[x]1−[x]+x(where [.] is G.I.F ) is

Let y=f(x)={x}1+{x}0≤x<12x<1+xx1+x<12⇒y∈0,12or y+yx=x⇒{x}=y1−y⇒0≤y1−y<1 ⇒y∈0,12

The range of f(x)=x−[x]1−[x]+x(where [.] is G.I.F ) is

Let y=f(x)={x}1+{x}0≤x<12x<1+xx1+x<12⇒y∈0,12or y+yx=x⇒{x}=y1−y⇒0≤y1−y<1 ⇒y∈0,12

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The range of $f (x) = \frac{x - [x]}{1 - [x] + x}$ (where [.] is G.I.F ) is

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$[0, \frac{1}{2}]$

$[0, 1]$

$(0, \frac{1}{2}]$

$[0, \frac{1}{2})$

answer is D.

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Detailed Solution

Let $y = f (x) = \frac{{x}}{1 + {x}}$

$0 \leq \{x\} < 1$

$2 \{x\} < 1 + \{x\}$

$\frac{\{x\}}{1 + \{x\}} < \frac{1}{2}$

$\Rightarrow y \in [0, \frac{1}{2})$

$y + y \{x\} = \{x\}$

$\Rightarrow {x} = \frac{y}{1 - y}$

$\Rightarrow 0 \leq \frac{y}{1 - y} < 1$

$\Rightarrow y \in [0, \frac{1}{2})$

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The range of f(x)=x−[x]1−[x]+x(where [.] is G.I.F ) is

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The range of $f (x) = \frac{x - [x]}{1 - [x] + x}$ (where [.] is G.I.F ) is