Question:
The set of real values of \(x\) for which \(\log_{0.2}\frac{x+2}{x} \le 1\) is:
\(x\) के वास्तविक मानों का समुच्चय जिसके लिए \(\log_{0.2}\frac{x+2}{x} \le 1\) है:
Options:
- \((-\infty, -5/2] \cup (0, \infty)\)
- \([-5/2, \infty)\)
- \((-\infty, -2) \cup (0, \infty)\)
- None of these
- \((-\infty, -5/2] \cup (0, \infty)\)
- \([-5/2, \infty)\)
- \((-\infty, -2) \cup (0, \infty)\)
- इनमें से कोई नहीं
Correct Answer: a)
Explanation:
Since the base \(0.2 = 1/5 < 1\), the inequality direction reverses:
\[\frac{x+2}{x} \ge (0.2)^1 = \frac{1}{5}\]
Also, the argument of the logarithm must be strictly positive:
\[\frac{x+2}{x} > 0 \implies x \in (-\infty, -2) \cup (0, \infty)\]
Solving \(\frac{x+2}{x} – \frac{1}{5} \ge 0\):
\[\frac{5(x+2) – x}{5x} \ge 0 \implies \frac{4x + 10}{5x} \ge 0 \implies \frac{2x + 5}{x} \ge 0\]
This gives \(x \in (-\infty, -5/2] \cup (0, \infty)\).
Taking the intersection with \((-\infty, -2) \cup (0, \infty)\):
\[x \in (-\infty, -5/2] \cup (0, \infty)\]
Hence, the correct option is (a).
लघुगणक का आधार \(0.2 < 1\) होने के कारण असमिका का चिह्न पलट जाता है:
\[\frac{x+2}{x} \ge 0.2 = \frac{1}{5}\]
साथ ही \(\frac{x+2}{x} > 0 \implies x \in (-\infty, -2) \cup (0, \infty)\)।
\[\frac{x+2}{x} – \frac{1}{5} \ge 0 \implies \frac{4x + 10}{5x} \ge 0 \implies x \in (-\infty, -5/2] \cup (0, \infty)\]
दोनों का उभयनिष्ठ: \(x \in (-\infty, -5/2] \cup (0, \infty)\)।
अतः सही विकल्प (a) है।