Question

If $^{n-1}C_r=(K^2-3)\,{}^nC_{r+1}$ , 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 $1 \le \frac{n}{r+1}$
$\therefore 0 < \, \frac{n}{r+1}\le\,1$
$\Rightarrow 0 < k^{2}-3 \le1$
$\Rightarrow 3 < k^{2} \le4$
$\Rightarrow \sqrt{3} < \left|k\right|\le2$
$\Rightarrow \sqrt{3}< k \le2$
or $\sqrt{3}< k \le2$ i.e. $-2 \le k <\sqrt{3}$
$\Rightarrow k \in (\sqrt{3, } 2]$ or $k\in \left[-2, \sqrt{3}\right]$