Basics & LogarithmsmediumFree

Basics & Logarithms — JEE Maths practice question

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

Question
The equation ∣x+1∣∣x−1∣=a2−2a−3|x+1||x-1| = a^2 - 2a - 3 can have real solutions for xx, if aa belongs to:
A(−∞,−1]∪[3,∞)(-\infty, -1] \cup [3, \infty)correct
B[1−5,1+5][1-\sqrt{5}, 1+\sqrt{5}]
C[1−5,−1]∪[3,1+5][1-\sqrt{5}, -1] \cup [3, 1+\sqrt{5}]
D[−1,3][-1, 3]
Solution
Step 1: Rewrite the equation
∣x+1∣∣x−1∣=∣x2−1∣=a2−2a−3|x+1||x-1| = |x^2-1| = a^2 - 2a - 3
 Step 2: Apply the non-negativity condition Since ∣x2−1∣≥0|x^2-1| \geq 0 and can take any non-negative value, a solution exists iff:
a2−2a−3≥0  ⟹  (a+1)(a−3)≥0a^2 - 2a - 3 \geq 0 \implies (a+1)(a-3) \geq 0
a∈(−∞,−1]∪[3,∞)a \in (-\infty, -1] \cup [3, \infty)
 Answer: (1)
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