Question

If ${}^{n-1}{C}_r=(k^2-3{)}^n{C}_{r+1}$, then $kϵ$

• A. $(-\infty ,-2]$
• B. $[2,\infty )$
• C. $[-\sqrt{3},\sqrt{3}]$
• D. $(\sqrt{3},2]$

Answer 1

The correct option is D $(\sqrt{3},2]$

${}^{n-1}{C}_r{=}^n{C}_{r+1}(k^2-3)\Rightarrow k^2-3=\frac{{}^{n-1}{C}_r}{{}^n{C}_{r+1}}=\frac{r+1}{n}\text{Since}0\le r\le n-1\Rightarrow 1\le r+1\le n\Rightarrow \frac{1}{n}\le \frac{r+1}{n}\le 1\Rightarrow \frac{1}{n}\le k^2-3\le 1\Rightarrow 3+\frac{1}{n}\le k^2\le 4\Rightarrow \sqrt{3+\frac{1}{n}}\le k\le 2\text{as}n\to \infty \Rightarrow \sqrt{3}<k\le 2\Rightarrow kϵ(\sqrt{3},2]$