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Basics & Logarithms: Find Number Single Digit Positive Integers Satisfying

JEE Maths question with a full step-by-step solution.

Question
Find the number of single digit positive integers satisfying ∣3xx2−4∣≥1\left|\dfrac{3x}{x^2-4}\right| \geq 1 is
Solution
Answer: 3
Step 1: The condition requires x≠±2x \neq \pm 2. Case A: x2>4x^2 > 4, i.e., ∣x∣>2|x| > 2:
3∣x∣≥x2−4  ⟹  x2−3∣x∣−4≤0  ⟹  (∣x∣−4)(∣x∣+1)≤0  ⟹  ∣x∣≤43|x| \geq x^2-4 \implies x^2-3|x|-4 \leq 0 \implies (|x|-4)(|x|+1) \leq 0 \implies |x| \leq 4
 Combined with ∣x∣>2|x| > 2: x∈[−4,−2)∪(2,4]x \in [-4,-2) \cup (2,4]. Case B: x2<4x^2 < 4, i.e., ∣x∣<2|x| < 2:
3∣x∣≥4−x2  ⟹  x2+3∣x∣−4≥0  ⟹  (∣x∣+4)(∣x∣−1)≥0  ⟹  ∣x∣≥13|x| \geq 4-x^2 \implies x^2+3|x|-4 \geq 0 \implies (|x|+4)(|x|-1) \geq 0 \implies |x| \geq 1
 Combined with ∣x∣<2|x| < 2: x∈[−2,−1]∪[1,2)x \in [-2,-1] \cup [1,2). Step 2: Total solution set: [−4,−1]∪[1,4][-4,-1] \cup [1,4] (excluding x=±2x = \pm 2). Single digit positive integers in [1,4][1,4] excluding x=2x=2: {1,3,4}\{1, 3, 4\}. Answer: 3
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