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Question

Mathematics Question on Binomial theorem

If n−1Cr=(k2−3) nCr+1^{n-1}C_r =(k^2 -3) \, ^nC_{r+1}then k belongs to

A

(−∞,−2](-\infty,-2]

B

[2,−∞,)[2,-\infty,)

C

[−3,3][-\sqrt 3,\sqrt 3]

D

(\sqrt{3, 2}]

Answer

(\sqrt{3, 2}]

Explanation

Solution

Given, n−1Cr=(k2−3) nCr+1^{n-1}C_r=(k^2-3) \, ^nC{r+1}
=            n−1Cr=(k2−3)nr+1 n−1Cr \, \, \, \, \, \, \, \, \, \, \, ^{n-1}C_r=(k^2-3)\frac{n}{r+1} \, ^{n-1}C_r
=            k2−3=r+1n \, \, \, \, \, \, \, \, \, \, \, k^2-3=\frac{r+1}{n}
\hspace20mm [since, n \ge r \Rightarrow \, \, \frac{r+1}{n} \le 1 \, and \, n,r >0]
⇒          0<k2−3≤1\Rightarrow \, \, \, \, \, \, \, \, \, \, 0< k^2-3\le 1
⇒           3<k2≤4  ⇒   k∈[−2,−3)∪(3,2)]\Rightarrow \, \, \, \, \, \, \, \, \, \, \, 3 < k^2 \le 4 \, \, \Rightarrow \, \, \, k \in [-2,-\sqrt 3) \cup (\sqrt 3,2)]