MCQ
$^{n - 1}{C_r} = ({k^2} - 3)\,.{\,^n}{C_{r + 1}}$ if $k \in $
- A$[ - \sqrt 3 ,\,\sqrt 3 ]$
- B$( - \infty ,\, - 2)$
- C$(2,\,\infty )$
- ✓$(\sqrt 3 ,\,2)$
✓
Answer
Correct option: D.
$(\sqrt 3 ,\,2)$
d
(d) We have $\frac{{(n - 1)\,!}}{{(n - r - 1)\,!\,r\,!}} = \frac{{({k^2} - 3)\,n\,!}}{{(n - r - 1)\,!\,(r + 1)\,!}}$, $0 \le r \le n - 1$
(d) We have $\frac{{(n - 1)\,!}}{{(n - r - 1)\,!\,r\,!}} = \frac{{({k^2} - 3)\,n\,!}}{{(n - r - 1)\,!\,(r + 1)\,!}}$, $0 \le r \le n - 1$
$ \Rightarrow $${k^2} = \frac{{r + 1}}{n} + 3,\,\frac{1}{n} \le \frac{{r + 1}}{n} \le 1$
==>${k^2} \in \left[ {\frac{1}{n} + 3,\,4} \right]\,,\,n \ge 2$
$k \in \left[ { - 2,\, - \sqrt {\frac{1}{n} + 3} } \right] \cup \left[ {\sqrt {\frac{1}{n} + 3} ,\,2} \right];\,n \ge 2$.
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