Question

The set of all real $x$ satisfying the inequality $\dfrac{{3 - |x|}}{{4 - |x|}} \geqslant 0$ is
A.$[ - 3,3] \cup ( - \infty , - 4) \cup (4,\infty )$
B.$( - \infty , - 4) \cup (4,\infty )$
C.$( - \infty , - 3) \cup (4,\infty )$
D.$( - \infty , - 3) \cup (3,\infty )$
E.$( - 3,3) \cup (4,\infty )$

Answer 1

Since we are given the ratio to be greater than or equal to zero, we will make two cases where either both the numerator and denominator are greater than zero or both the numerator and denominator are less than zero. Also, we are given equal to sign, then we know that since the denominator can never be zero, we will write numerators as greater than or equal to zero or less than or equal to zero. After that, we know
$|x| \leqslant a \Rightarrow - a \leqslant x \leqslant a \Rightarrow x \in [ - a,a]$
$|x| \geqslant a \Rightarrow ( - \infty < x \leqslant - a) \cup (a \leqslant x < \infty ) \Rightarrow x \in ( - \infty , - a] \cup [a,\infty )$
Also, we know that if we have “AND” condition for two inequalities, we use “INTERSECTION” i.e. $\cap$
If we have “OR” condition for two inequalities, we use “UNION” i.e. $\cup$
After solving the two conditions and finding the real $x$ which satisfy the conditions individually, we will then combine the conditions using $\cup or \cap$ as the condition will be.

Complete answer:
We are given,
$\dfrac{{3 - |x|}}{{4 - |x|}} \geqslant 0$
Which means ,either
CASE 1 - $3 - |x| \geqslant 0$ and $4 - |x| > 0$ OR
CASE 2 - $3 - |x| \leqslant 0$ and $4 - |x| < 0$
Now, solving for real $x$ in CASE 1, we get
$3 - |x| \geqslant 0$ and $4 - |x| > 0$
$\Rightarrow |x| \leqslant 3$ and $|x| < 4$
$\Rightarrow x \in [ - 3,3]$ and $x \in [ - 4,4]$
Now, Since there is “AND” in between two sets, we will use “INTERSECTION”
$\therefore x \in [ - 3,3] \cap [ - 4,4]$
$\Rightarrow x \in [ - 3,3]$ (Using intersection properties)
So, for CASE 1, $x \in [ - 3,3]$ -----(1)
Now, solving for real $x$ in CASE 2, we get
$3 - |x| \leqslant 0$ and $4 - |x| < 0$
$\Rightarrow |x| \geqslant 3$ and $|x| > 4$
$\Rightarrow x \in ( - \infty , - 3] \cup [3,\infty )$ and $x \in ( - \infty , - 4) \cup (4,\infty )$
Now, since there is “AND” in between two sets, we will use “INTERSECTION”
$\therefore x \in ( - \infty , - 3] \cup [3,\infty ) \cap \\{ ( - \infty , - 4) \cup (4,\infty )\\}$
$\Rightarrow x \in ( - \infty , - 4) \cup (4,\infty )$ (Using Intersection Properties)
So, for CASE 2, $x \in ( - \infty , - 4) \cup (4,\infty )$ ----(2)
Now, we have either CASE 1 or CASE 2 i.e. we have “OR” in between the two conditions, and hence we will combine the two conditions using “UNION” i.e. $\cup$
From (1) and (2), we get
$x \in [ - 3,3]$ or $x \in ( - \infty , - 4) \cup (4,\infty )$
$\Rightarrow x \in [ - 3,3] \cup ( - \infty , - 4) \cup (4,\infty )$
$\therefore x \in ( - \infty , - 4) \cup [ - 3,3] \cup (4,\infty )$
Hence, we get, if $\dfrac{{3 - |x|}}{{4 - |x|}} \geqslant 0$ , then $x \in ( - \infty , - 4) \cup [ - 3,3] \cup (4,\infty )$
Therefore, option (A) is the correct answer.

Note:
We need to be very thorough with the “UNION” and “INTERSECTION” properties. Also, we need to keep in mind that the denominator can never be equal to zero and so not putting the equality sign with the denominator. While solving for the modulus inequalities, we need to keep in mind that Infinity is never included and only open brackets will come with infinity. And, when we have two cases where either of the cases is possible then we use “UNION” and when we have two cases where both are compulsory conditions, then we use “INTERSECTION”.