The minimum value of the function f(x)=12{−x}−{x} occurs at x equals to (where {.}represents fractional part function)

f(x)=12{−x}−(1−{−x})={−x}+12{−x}−1⇒f(x)≥2⋅12−1.(AM≥GM) ⇒fmin=2−1 which occurs when {−x}=12⇒−x=n+12⇒x=n−12,n∈I

The minimum value of the function f(x)=12{−x}−{x} occurs at x equals to (where {.}represents fractional part function)

f(x)=12{−x}−(1−{−x})={−x}+12{−x}−1⇒f(x)≥2⋅12−1.(AM≥GM) ⇒fmin=2−1 which occurs when {−x}=12⇒−x=n+12⇒x=n−12,n∈I

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The minimum value of the function $f (x) = \frac{1}{2 {- x}} - {x}$ occurs at $x$ equals to (where ${.}$ represents fractional part function)

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$n + \frac{1}{\sqrt{2}}, n \in I$

$n - \frac{1}{\sqrt{2}}, n \in I$

$n + \frac{1}{2 \sqrt{2}}, n \in I$

$n + \frac{1}{2}, n \in I$

answer is B.

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Detailed Solution

$f (x) = \frac{1}{2 {- x}} - (1 - {- x}) = {- x} + \frac{1}{2 {- x}} - 1 \Rightarrow f (x) \geq 2 \cdot \sqrt{\frac{1}{2}} - 1. (A M \geq G M)$ $\Rightarrow \underset{\min}{f} = \sqrt{2} - 1 which occurs when {- x} = \frac{1}{\sqrt{2}} \Rightarrow - x = n + \frac{1}{\sqrt{2}} \Rightarrow x = n - \frac{1}{\sqrt{2}}, n \in I$