Basics & LogarithmsmediumFree

Basics & Logarithms: Equation Can Real Solutions Belongs

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Basics & Logarithms · easy
If x=∏n=12026nx = \prod_{n=1}^{2026} n, then the value of the expression 11log⁡2x+1log⁡3x+⋯+1log⁡2026x\frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x} + \cdots + \dfrac{1}{\log_{2026} x}} is
Basics & Logarithms · easy
Number of integral values of xx satisfying the inequality (34)6x+10−x2<2764\left(\dfrac{3}{4}\right)^{6x+10-x^2} < \dfrac{27}{64} is:
Basics & Logarithms · easy
Number of natural numbers for which the number log⁡4−x(x2−14x+45)\log_{4-x}(x^2-14x+45) is defined is:
Basics & Logarithms · easy
The product of all solutions of the equation e5(log⁡ex)2+3=x8e^{5(\log_{e}x)^{2}+3}=x^{8}, x>0x>0, is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath