Question

For
$\left| {\dfrac{x}{{x + 1}}} \right| < {\left( {10} \right)^{ - 5}}$ hold if
$\left( 1 \right)$ $- {(10)^{ - 5}} < x + 1 < {(10)^{ - 4}}$
$\left( 2 \right)$ $- {\left( {100001} \right)^{ - 1}} < x < {\left( {99999} \right)^{ - 1}}$
$\left( 3 \right)$ $\dfrac{1}{{10000}} < x < 1$
$\left( 4 \right)$ ${(99999)^{ - 1}} < x < {(100001)^{ - 1}}$

Answer 1

We have to find the value of $x$ form the given expression $\left| {\dfrac{x}{{x + 1}}} \right| < {\left( {10} \right)^{ - 5}}$ . We solve this question using the concept of solving the linear equations , the concept of linear inequality and the concept of splitting of the modulus function . First we will simplify the left hand side of the given expression such that we will make the expression in terms of $x$ only by simplifying the terms . For that we will add $\pm 1$ in the numerator of the given expression and then we will solve the expression by taking the denominator separately under the terms of the numerator . Then splitting the modulus and changing the sign of the inequality , and hence we would compute the range of $x$ for the given expression .

Complete answer: Given :
$\left| {\dfrac{x}{{x + 1}}} \right| < {\left( {10} \right)^{ - 5}}$

Now , on adding $\pm 1$ in the numerator , we get the expression as :
$\left| {\dfrac{{x + 1 - 1}}{{x + 1}}} \right| < {\left( {10} \right)^{ - 5}}$

Now , we know that on splitting the modulus function , we add $\pm$ to the given expression . Also , we know that as the given expression is of linear inequality , we would also obtain a linear inequality after splitting the modulus function .

So , on splitting the modulus function , we get the expression as :
$- {\left( {10} \right)^{ - 5}} < \dfrac{{x + 1 - 1}}{{x + 1}} < {\left( {10} \right)^{ - 5}}$

Now , taking the denominator separately under the numerator , we get the expression as :
$- {\left( {10} \right)^{ - 5}} < \dfrac{{x + 1}}{{x + 1}} - \dfrac{1}{{x + 1}} < {\left( {10} \right)^{ - 5}}$

On further simplifying , we get the above expression as :
$- {\left( {10} \right)^{ - 5}} < 1 - \dfrac{1}{{x + 1}} < {\left( {10} \right)^{ - 5}}$

Subtracting $1$ from the inequality , we get the expression as :
$- 1 - {\left( {10} \right)^{ - 5}} < \dfrac{{ - 1}}{{x + 1}} < {\left( {10} \right)^{ - 5}} - 1$

As , we know that when we change the sign of the initial condition of the inequality the signs of the inequality also changes .

Using this property of inequality .
Now multiplying the inequality by $\left( { - 1} \right)$ , we get the expression as :
$1 - {\left( {10} \right)^{ - 5}} < \dfrac{1}{{x + 1}} < {\left( {10} \right)^{ - 5}} + 1$
$1 - \dfrac{1}{{{{\left( {10} \right)}^5}}} < \dfrac{1}{{x + 1}} < \dfrac{1}{{{{\left( {10} \right)}^5}}} + 1$

On further simplifying the expression , we get
$\dfrac{{99999}}{{{{\left( {10} \right)}^5}}} < \dfrac{1}{{x + 1}} < \dfrac{{100001}}{{{{\left( {10} \right)}^5}}}$

Again using the property of the inequality of change in sign .

Taking the reciprocal of the inequality , we get the expression as :
$\dfrac{{{{\left( {10} \right)}^5}}}{{100001}} < x + 1 < \dfrac{{{{\left( {10} \right)}^5}}}{{99999}}$

Subtracting $1$ from the inequality , we get the expression as :
$\dfrac{{ - 1}}{{100001}} < x < \dfrac{1}{{99999}}$

We also , know that the inverse of a function can be represented as :
$\dfrac{1}{a} = {a^{ - 1}}$

Using this formula , we get the value of x as :
$- {\left( {100001} \right)^{ - 1}} < x < {\left( {99999} \right)^{ - 1}}$

Hence , the value of x for the given expression $\left| {\dfrac{x}{{x + 1}}} \right| < {\left( {10} \right)^{ - 5}}$ is $- {\left( {100001} \right)^{ - 1}} < x < {\left( {99999} \right)^{ - 1}}$ .
Thus , the correct option is $\left( 2 \right)$ .

Note:
Modulus function : It is a function which always gives a positive value when applied to a function irrespective of the values of the function . The graph of a modulus function is a V shaped graph where the tip is the point of contact on the graph . We add $\pm$ for removing the modulus function as we don’t know the value was taken as negative or positive , so to remove errors while solving we add $\pm$ sign and solve it for two cases separately .
Example : The value of a mod function is as given below
$\left| { - 1} \right| = 1$
$\left| 1 \right| = 1$
We get the value as $1$ for both $+ 1$ or $- 1$ .