Question

If ${}^{n-1}{C}_r=(K^2-3){}^n{C}_{r+1}$ , then $K$ lies in the interval

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

Answer 1

Answer: (D)

By the given conditions.
$\frac{\left(n-1\right)!}{r!\left(n-r-1\right)!}$
$=\left(k^2-3\right)\frac{n!}{\left(r+1\right)!\left(n-r-r\right)!}$
where $0<r\le n-1$
$\Rightarrow 1=\left(k^2-3\right)\cdot \frac{n}{r+1}$
$\Rightarrow k^2-3=\frac{r+1}{n}$
Since $n>0$ and $r+1\le n$
$\therefore \frac{n}{r+1}>0$
and $1\le \frac{n}{r+1}$
$\therefore 0<\frac{n}{r+1}\le 1$
$\Rightarrow 0<k^2-3\le 1$
$\Rightarrow 3<k^2\le 4$
$\Rightarrow \sqrt{3}<\left|k\right|\le 2$
$\Rightarrow \sqrt{3}<k\le 2$
or $\sqrt{3}<k\le 2$ i.e. $-2\le k<\sqrt{3}$
$\Rightarrow k\in (\sqrt{3,}2]$ or $k\in \left[-2,\sqrt{3}\right]$