Question:

\(^{n-1}C_r = (k^2 - 8)  ^{n}C_{r+1}\) if and only if:

Updated On: Dec 31, 2024
  • \( 2\sqrt{2} < k ≤ 3 \)

  • \( 2\sqrt{3} < k ≤ 3\sqrt{2} \)

  • \( 2\sqrt{3} < k <3 \sqrt{3} \)

  • \( 2\sqrt{2} < k < 2\sqrt{3} \)
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The Correct Option is A

Solution and Explanation

Given: \(n−1C_r = (k^2 − 8) ^nC_{r+1}\)
We know: \(n−1C_r = (k^2 − 8) ^nC_{r+1}\)
For this expression to hold, \( k^2 − 8 \) must be positive:
\( k^2 − 8 > 0 \Rightarrow k > 2\sqrt{2} \text{ or } k < -2\sqrt{2} \)
Thus, \( k \in (-\infty, -2\sqrt{2}) \cup (2\sqrt{2}, \infty) \)
Next, we check the range \( -3 \le k \le 3 \) to satisfy the constraint. Combining both conditions: \( k \in [2\sqrt{2}, 3] \)

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