Question

If $ ^{n-1}C_r = (k^2- 3)(^nC_{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, $\,{}^{n-1} C_r = \,{}^nC_{r+1}(k^2 - 3)$
$\Rightarrow k^2 - 3 = \frac{^{n-1}C_r}{^n C_{r+1}} = \frac{r+1}{n}$
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$
$\sqrt{3} < k \le 2$
$\therefore k \in (\sqrt{3}, 2]$