A) \[[-\sqrt{3},\,\sqrt{3}]\]
B) \[(-\infty ,\,-2)\]
C) \[(2,\,\infty )\]
D) \[(\sqrt{3},\,2)\]
Correct Answer: D
Solution :
We have\[\frac{(n-1)\,!}{(n-r-1)\,!\,r\,!}=\frac{({{k}^{2}}-3)\,n\,!}{(n-r-1)\,!\,(r+1)\,!}\], \[0\le r\le n-1\] \[\Rightarrow \]\[{{k}^{2}}=\frac{r+1}{n}+3,\,\frac{1}{n}\le \frac{r+1}{n}\le 1\]Þ \[{{k}^{2}}\in \left[ \frac{1}{n}+3,\,4 \right]\,,\,n\ge 2\] \[k\in \left[ -2,\,-\sqrt{\frac{1}{n}+3} \right]\cup \left[ \sqrt{\frac{1}{n}+3},\,2 \right];\,n\ge 2\].
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