If ^{n-1}C_r = (k^2 - 3) cdot ^nC_{r+1},then | vedclass

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(C) We are given the equation: $^{n-1}C_r = (k^2 - 3) \cdot ^nC_{r+1}$.Using the identity $^nC_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$,we can write $^nC_

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$[\sqrt{3}, 2]$

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(C) We are given the equation: $^{n-1}C_r = (k^2 - 3) \cdot ^nC_{r+1}$.Using the identity $^nC_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$,we can write $^nC_{r+1} = \frac{n}{r+1} \cdot ^{n-1}C_r$.Substituting this into the given equation:$^{n-1}C_r = (k^2 - 3) \cdot \frac{n}{r+1} \cdot ^{n-1}C_r$.Assuming $^{n-1}C_r 
eq 0$,we divide both sides by $^{n-1}C_r$:$1 = (k^2 - 3) \cdot \frac{n}{r+1}$.Rearranging for $k^2 - 3$:$k^2 - 3 = \frac{r+1}{n}$.Since $0 \le r \le n-1$,the ratio $\frac{r+1}{n}$ lies in the interval $(0, 1]$.Thus,$0 Adding $3$ to all parts:$3 Taking the square root,we get $k \in [-2, -\sqrt{3}) \cup (\sqrt{3}, 2]$.Comparing this with the given options,the interval $[\sqrt{3}, 2]$ is a subset of the possible values for $k$.

If $^{n-1}C_r = (k^2 - 3) \cdot ^nC_{r+1}$,then an interval containing the values of $k$ is

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