Published by:
CGP EDU Academic Team
Published on: August 14, 2026
If x, |x + 1|, |x – 1| are three terms of an A.P., then find the number of possible values of x
Text Solution
Verified by ExpertsThe correct answer is:
2
(2)
Sol. since x , |x + 1|, |x – 1| are in A.P.
so 2 |x + 1| = x + |x – 1| .... (i)
Case-I If x < – 1, then (i) becomes
– 2 (x + 1) = x – (x – 1) ⇒ x = 
Case-II If –1 ≤ x ≤ 1, then (i) becomes
2 (x + 1) = x – (x – 1) ⇒ x = – 1/2 then series 
, 
, 
Case-III If x ≥ 1, then (i) becomes
2 (x + 1) = x + x – 1
2 = – 1 impossible.
──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
Match the set P in column one with its super set Q in column II
Column – Ι (set P) Column – ΙΙ (set…
If ≤ 4, then the least and the highest values of 4x 2 are:
If ≥ 1 and is an odd integer then number of possible values of α is
If log a b = 2; log b c = 2 and log 3 c = 3 + log 3 a then (a + b + c) equals
The sum of the solutions of the equation 9 x – 6 · 3 x + 8 = 0 is
The expression: reduces to