First slide

Combinations

Question

If $^{n - 1} C_{r} = {(k^{2} - 3)}^{n} C_{r + 1}$ , then $k$ belongs to

Moderate

A

$(- \infty, - 2]$
B

$[2, \infty)$
C

$[- \sqrt{3}, \sqrt{3}]$
D

$(\sqrt{3}, 2]$

Solution

We have,
$\begin{array}{l} ^{n - 1} C_{r} = {(k^{2} - 3)}^{n} C_{r + 1} \\ \Rightarrow \frac{(n - 1)!}{(n - 1 - r)! r!} = (k^{2} - 3) \frac{n!}{(n - r - 1)! (r + 1)!} \\ \Rightarrow 1 = \frac{(k^{2} - 3) n}{(r + 1)} \\ \Rightarrow \frac{r + 1}{n} = k^{2} - 3 [∵ 0 \leq r \leq n - 1 \to r + 1 \leq n] \end{array}$
$\Rightarrow k^{2} - 3 \leq 1 \Rightarrow k^{2} - 4 \leq 0 \Rightarrow - 2 \leq k \leq 2$ (i)
Again, $^{n - 1} C_{r} = {(k^{2} - 3)}^{n} C_{r + 1}$
$\Rightarrow k^{2} - 3 = \frac{^{n - 1} C_{r}}{^{n} C_{r} + 1} > 0$
$\Rightarrow k^{2} - 3 > 0 \Rightarrow k < - \sqrt{3} or k > \sqrt{3}$ (ii)
From (i) and (ii), we have $k \in (\sqrt{3}, 2]$

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