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