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Q.

If $^{n - 1} C_{r} = {(k^{2} - 3)}^{n} C_{r + 1}$ then $k \in$

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a

$(- 2, - 2]$

b

$(\sqrt{3}, 2]$

c

$[2, \infty)$

d

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

answer is D.

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Detailed Solution

$\begin{array}{l} ^{n - 1} C_{2} = (k^{2} - 3) {\frac{n}{r + 1}}^{n - 1} C_{r} \\ \Rightarrow k^{2} - 3 = \frac{r + 1}{n} [∵ n - 1 \geq r \Rightarrow \frac{r + 1}{n} \leq 1 and r \geq 0] \\ \Rightarrow 0 < k^{2} - 3 \leq 1 \Rightarrow 3 < k^{2} \leq 4 \\ k \in (\sqrt{3}, 2] \end{array}$

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If n−1Cr=(k2−3)nCr+1then k∈