The solution set of the inequality \(\displaystyle\frac{x - 3}{x + 2} \le 0\), where \(x\) is a real number, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \((-2, 3]\)
A quotient is negative when the numerator and the denominator have opposite signs, and it is zero when the numerator is zero.
The numerator is zero at \(x = 3\) and the denominator is zero at \(x = -2\). Checking the sign on each part of the number line:
\(x \lt -2\): both negative, quotient positive
\(-2 \lt x \lt 3\): numerator negative, denominator positive, quotient negative
\(x \gt 3\): both positive, quotient positive
At \(x = 3\) the quotient is 0, so 3 is included. At \(x = -2\) the quotient is not defined, so \(-2\) is excluded.
So the solution set is \((-2, 3]\).
Why the other options are incorrect
Option (2) \([-2, 3]\) — Incorrect. At \(x = -2\) the denominator is zero, so \(-2\) cannot be in the solution set.
Option (3) \((-2, 3)\) — Incorrect. At \(x = 3\) the quotient is 0, which satisfies \(\le 0\), so 3 must be included.
Option (4) \((-\infty, -2) \cup [3, \infty)\) — Incorrect. On these parts the quotient is positive or zero; this is the solution of \(\ge 0\).