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Basics & Logarithms: Find Number Negative Integers Satisfying

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

Question
Find the number of negative integers satisfying −1≤x2−5x+4x2−4≤1-1 \leq \dfrac{x^2-5x+4}{x^2-4} \leq 1 is
Solution
Answer: 0
Domain: x≠±2x \neq \pm 2. Left inequality: x2−5x+4x2−4+1≥0  ⟹  2x2−5xx2−4≥0  ⟹  x(2x−5)(x−2)(x+2)≥0\dfrac{x^2-5x+4}{x^2-4} + 1 \geq 0 \implies \dfrac{2x^2-5x}{x^2-4} \geq 0 \implies \dfrac{x(2x-5)}{(x-2)(x+2)} \geq 0 Sign analysis with critical points −2,0,2,52-2, 0, 2, \frac{5}{2}: Solution: x∈(−∞,−2)∪[0,2)∪[52,+∞)x \in (-\infty,-2) \cup [0,2) \cup [\frac{5}{2},+\infty). Right inequality: x2−5x+4x2−4−1≤0  ⟹  8−5xx2−4≤0  ⟹  5x−8(x−2)(x+2)≥0\dfrac{x^2-5x+4}{x^2-4} - 1 \leq 0 \implies \dfrac{8-5x}{x^2-4} \leq 0 \implies \dfrac{5x-8}{(x-2)(x+2)} \geq 0 Sign analysis with critical points −2,85,2-2, \frac{8}{5}, 2: Solution: x∈(−2,85]∪(2,+∞)x \in (-2, \frac{8}{5}] \cup (2,+\infty). Intersection: [0,85]∪[52,+∞)[0, \frac{8}{5}] \cup [\frac{5}{2}, +\infty). Negative integers in [0,85]∪[52,+∞)[0, \frac{8}{5}] \cup [\frac{5}{2},+\infty): none. Answer: 0
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