Question

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

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

Answer 1

Answer: (A)

Given, ${}^{n-1}{C}_r={}^n{C}_{r+1}(k^2-3)$
$\Rightarrow k^2-3=\frac{{}^{n-1}{C}_r}{{}^n{C}_{r+1}}=\frac{r+1}{n}$
As $0\le r\le n-1$
$\Rightarrow 1\le r+1\le n$
$\Rightarrow \frac{1}{n}\le \frac{r+1}{n}\le 1$
$\Rightarrow \frac{1}{n}\le k^2-3\le 1$
$\Rightarrow 3+\frac{1}{n}\le k^2\le 1$
$\Rightarrow \sqrt{3+\frac{1}{n}}\le 2$
As $n\to \infty$
$\sqrt{3}<k\le 2$
$\therefore k\in (\sqrt{3},2]$