^ n - 1 C_r = ( k^2 - 3) ,. ,^n C_ r + 1 if k - Vidyadip

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 $\frac{{(n - 1)\,!}}{{(n - r - 1)\,!\,r\,!}} = \frac{{({k^2} - 3)\,n\,!}}{{(n - r - 1)\,!\,(r + 1)\,!}}$, $0 \le r \le n - 1$

$ \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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