Question

The set of real values of x for which \[{\log _{0.2}}\dfrac{{x + 2}}{x} \leqslant 1\] is
(A) \[\left( { - \infty , - \dfrac{5}{2}} \right] \cup \left( {0, + \infty } \right)\]
(B) \[\left[ {\dfrac{5}{2}, + \infty } \right)\]
(C) \[\left( { - \infty ,0} \right) \cup \left( {\dfrac{8}{5}, + \infty } \right)\]
(D) None of these

Answer 1

Hint: Here it is given that inequality relation of logarithmic function. In this question, we have to find the real value of specificity on the basis of given inequality.
We first need to consider the condition for which the inequality is valid then solving the inequality using the logarithm formula we can find the required specificity.

Formula used: In mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$, must be raised, to produce that number $x$.
By definition, if \[{\log _a}x = b\], then \[x = {a^b}\].

Complete step-by-step solution:
It is given that, \[{\log _{0.2}}\dfrac{{x + 2}}{x} \leqslant 1\].
We need to find out the set of real values of x for which \[{\log _{0.2}}\dfrac{{x + 2}}{x} \leqslant 1\] is true.
Now we know that, \[\log x\] is only valid for \[x > 0\], then \[\dfrac{{x + 2}}{x} > 0\] must be true.
This implies \[x > 0\]or, \[x < - 2 - - - - (i)\]
Now, \[{\log _{0.2}}\dfrac{{x + 2}}{x} \leqslant 1\]
That is, \[\dfrac{{x + 2}}{x} \leqslant {\left( {0.2} \right)^1}\]
Using the definition of logarithm
\[ \Rightarrow \dfrac{{\left( {x + 2} \right)}}{x} \leqslant 0.2\]
Simplifying we get,
\[ \Rightarrow x + 2 \leqslant 0.2x\]
Rearranging the terms we get,
\[ \Rightarrow 0.8x \leqslant - 2\]
Solving for x we get,
\[ \Rightarrow x \leqslant - \dfrac{2}{{0.8}}\]
Then \[x \leqslant - \dfrac{{2 \times 10}}{8}\]
\[x \leqslant - \dfrac{5}{2} - - - - - (ii)\]
From i) and ii) we get,
\[x \in \left( { - \infty , - \dfrac{5}{2}} \right] \cup \left( {0, + \infty } \right)\]
Hence, the set of real values of x is \[\left( { - \infty , - \dfrac{5}{2}} \right] \cup \left( {0, + \infty } \right)\].

$\therefore $ Option (A) is the correct answer.

Note: Interval:
In mathematics, an interval is a set of real numbers that contains all real numbers lying between any two numbers of the set. For example, the set of numbers x satisfying \[0 \leqslant x \leqslant 1\] is an interval which contains \[0,1\] and all numbers in between.
In some contexts, an interval may be defined as a subset of the extended real numbers, the set of all real numbers augmented with \[ - \infty \] and \[ + \infty \].
In this interpretation, the notations \[\left[ { - \infty ,b} \right],\left( { - \infty ,b} \right]\left[ {a, + \infty } \right],\left[ {a, + \infty } \right)\] are all meaningful and distinct. In particular, \[\left( { - \infty , + \infty } \right)\] denotes the set of all ordinary real numbers, while \[\left[ { - \infty , + \infty } \right]\] denotes the extended reals.