Question

What is the solution set of the inequality \[\dfrac{{\left[ {x + 11} \right]}}{{\left[ {x - 3} \right]}} > 0\]?
A. \[\left( { - \infty , - 11} \right) \cup \left( {3,\infty } \right)\]
B. \[\left( { - \infty , - 10} \right) \cup \left( {2,\infty } \right)\]
C. \[\left( { - 100, - 11} \right) \cup \left( {1,\infty } \right)\]
D. \[\left( { - 5,0} \right) \cup \left( {3,7} \right)\]
E. \[\left( {0,5} \right) \cup \left( { - 1,0} \right)\]

Answer 1

Hint: First, check the all-possible conditions for which the given inequality is true. Then simplify the conditions and find the solution set of the inequality.

Formula Used:
An open interval does not include its endpoints, and it is enclosed in round parentheses.
If \[a < x < b\], then \[x \in \left( {a,b} \right)\]

Complete step by step solution:
The given inequality equation is \[\dfrac{{\left[ {x + 11} \right]}}{{\left[ {x - 3} \right]}} > 0\].
Let’s solve the inequality.
Since the value of the inequality is greater than zero.
So, the values of the numerator and denominator are either positive or negative.

Simplify both cases.
Case 1: The values of the numerator and denominator are positive.
 \[x + 11 > 0\] and \[x - 3 > 0\]
Solve the above inequalities.
\[x > - 11\] and \[x > 3\] \[.....\left( 1 \right)\]

Case 2: The values of the numerator and denominator are negative.
 \[x + 11 < 0\] and \[x - 3 < 0\]
Solve the above inequalities.
\[x < - 11\] and \[x < 3\] \[.....\left( 2 \right)\]

Simplify the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[x < - 11\] and \[x > 3\]
Now convert the inequalities into an interval form.
\[x \in \left( { - \infty , - 11} \right) \cup \left( {3,\infty } \right)\]

Hence the correct option is A.

Note: Students often get confused with the conversion of an inequality to an interval notation.
The possible conversions are:
\[a < x < b \Rightarrow x \in \left( {a,b} \right)\]
\[a \le x \le b \Rightarrow x \in \left[ {a,b} \right]\]
\[a < x \le b \Rightarrow x \in \left( {a,b} \right]\]
\[a \le x < b \Rightarrow x \in \left[ {a,b} \right)\]