Published by:
CGP EDU Academic Team
Published on: August 13, 2026
The least integer a, for which
1 + log 5 (x 2 + 1) ≤ log 5 (ax 2 + 4x + a) is true for all x ∈ R is -
Text Solution
Verified by ExpertsThe correct answer is:
B
Inequality ax 2 + 4x + a > 0 is possible if a > 0 and 16 – 4a 2 < 0 ⇒ a > 2
log 5 5 + log 5 (x
2 + 1) ≤ log 5 (ax 2 + 4x + a)
5(x 2 + 1) ≤ ax 2 + 4x + a
(a – 5) x 2 + 4x + (a – 5) > 0 it hold if a – 5 > 0 and 16 – 4
(a – 5) 5 ≥ 0
a > 5 and a ≤ 3 or, a ≥ 7 ⇒ a ≥ 7 ⇒ a ∈ [7, ∞ )
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