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If ,, then k belongs to
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a
(−∞,−2]
b
[2,∞)
c
[−3,3]
d
(3,2]
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Correct option is D
We have, n−1Cr=k2−3nCr+1⇒(n−1)!(n−1−r)!r!=k2−3n!(n−r−1)!(r+1)!⇒1=k2−3n(r+1)⇒r+1n=k2−3 [∵0≤r≤n−1⇒r+1≤n]⇒k2−3≤1⇒k2−4≤0⇒−2≤k≤2 ... (i)Again n−1Cr=k2−3nCr+1⇒ k2−3= n−1Cr nCr+1>0⇒ k2−3>0⇒k<−3or k>3 ... (ii)From (i) and (ii), we have k∈(3,2].Similar Questions
is equal to
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