Question

$\left|\left(\frac{x}{\left[x+1\right]}\right)\right|<{10}^{-5}$ hold if

• A. $-{10}^5<x+1<{10}^{-4}$
• B. $-{\left(100001\right)}^{-1}<x<{\left(99999\right)}^{-1}$
• C. $\left(\frac{1}{10000}\right)<x<1$
• D. ${\left(99999\right)}^{-1}<x<{\left(100001\right)}^{-1}$

Answer 1

The correct option is B

$-{\left(100001\right)}^{-1}<x<{\left(99999\right)}^{-1}$

Explanation for the correct option

Solve the inequality:

Formula:

Absolute value formula: if $\left|a\right|\le b$ then, $-b\le a\le b$

The given inequality is,

$\left|\left(\frac{x}{\left[x+1\right]}\right)\right|<{10}^{-5}$

$\Rightarrow$ $\left|\left(\frac{x+1-1}{\left[x+1\right]}\right)\right|<{10}^{-5}$

$\Rightarrow$ $\left|\left(1-\frac{1}{\left[x+1\right]}\right)\right|<{10}^{-5}$

$\Rightarrow$ $-{10}^{-5}<1-\frac{1}{x+1}<{10}^{-5}$ $\left[Byabsolutevalueformula\right]$

$\Rightarrow$ $-1-{10}^{-5}<-\frac{1}{x+1}<-1+{10}^{-5}$ $\left[Subtracted1fromeachterm\right]$

$\Rightarrow$ $\left(1-{10}^{-5}\right)<\frac{1}{\left(x+1\right)}<\left(1+{10}^{-5}\right)$ $\left[Multipliedeachtermby-1\right]$

$\Rightarrow$ $\left(1-\frac{1}{100000}\right)<\frac{1}{\left(x+1\right)}<\left(1+\frac{1}{100000}\right)$

$\Rightarrow$ $\left(\frac{99999}{100000}\right)<\frac{1}{\left(x+1\right)}<\left(\frac{100001}{100000}\right)$

$\Rightarrow$ $\left(\frac{100000}{100001}\right)<\left(x+1\right)<\left(\frac{100000}{99999}\right)$ $\left[Takeresiprocal\right]$

$\Rightarrow$ $\left(\frac{100000}{100001}-1\right)<\left(x+1-1\right)<\left(\frac{100000}{99999}-1\right)$ $\left[Subtracted1fromeachterm\right]$

$\Rightarrow$ $\left(\frac{-1}{100001}\right)<x<\left(\frac{1}{99999}\right)$

$\Rightarrow$ $-{\left(100001\right)}^{-1}<x<{\left(99999\right)}^{-1}$

Hence, the correct option is $B)-{\left(100001\right)}^{-1}<x<{\left(99999\right)}^{-1}$.