Published by:
CGP EDU Academic Team
Published on: August 14, 2026
If 
exists and finite (n, k ∈ N), then the least value of 4k + n 
2 is :
Text Solution
Verified by ExpertsThe correct answer is:
21
(21)
Sol. 
= 
= 
limit exists, if coff. of x 2 is zero.
⇒ n 2 + 
– k = 0 ⇒ 4k = 5n 2
so the possible value match that is n = 2, k = 5
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