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Q.

If  n−1Cr=k2−3nCr+1, then k is belongs to

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a

(−∞,−2]

b

[2,∞)

c

[−3,3]

d

(3,2]

answer is D.

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Detailed Solution

∵k2−3= n−1Cr nCr+1= n−1Crnr+1n−1Cr=r+1n -----(i)∵ 0≤r≤n−1⇒ 1≤r+1≤n⇒1n≤r+1n≤1⇒ 1n≤k2−3≤1⇒ 3+1n≤k2≤4When n→∞,3
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