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Basics & Logarithms: Find Number Positive Integers Satisfying

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

Question
Find the number of positive integers satisfying ∣x2−1∣+∣x2−4∣≤3|x^2-1| + |x^2-4| \leq 3 is
Solution
Answer: 2
Step 1: Let u=x2≥0u = x^2 \geq 0. Case 1: u≥4u \geq 4: (u−1)+(u−4)≤3  ⟹  2u≤8  ⟹  u≤4(u-1)+(u-4) \leq 3 \implies 2u \leq 8 \implies u \leq 4. So u=4u = 4, i.e., x=±2x = \pm 2. Case 2: 1≤u<41 \leq u < 4: (u−1)+(4−u)=3≤3(u-1)+(4-u) = 3 \leq 3. True for all u∈[1,4)u \in [1,4). Case 3: 0≤u<10 \leq u < 1: (1−u)+(4−u)≤3  ⟹  5−2u≤3  ⟹  u≥1(1-u)+(4-u) \leq 3 \implies 5-2u \leq 3 \implies u \geq 1. Contradicts u<1u < 1. Step 2: Solution is x2∈[1,4]x^2 \in [1,4], i.e., x∈[−2,−1]∪[1,2]x \in [-2,-1] \cup [1,2]. Positive integers in [1,2][1, 2]: {1,2}\{1, 2\}. Answer: 2
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