Question

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

• A. \( 2\sqrt{2} < k ≤ 3 \)
• B. \( 2\sqrt{3} < k ≤ 3\sqrt{2} \)
• C. \( 2\sqrt{3} < k <3 \sqrt{3} \)
• D. \( 2\sqrt{2} < k < 2\sqrt{3} \)

Answer 1

The Correct Option is A

To solve the given problem, we need to understand the relationship between combinatorial coefficients and how they are affected by the multiplication factor \(k^2 - 8\). 

1. Start with the given equation: \(^{n-1}C_r = (k^2 - 8) \ ^{n}C_{r+1}\).
2. Recall the formula for combinations: \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\). Thus, \(^{n-1}C_r = \frac{(n-1)!}{r!(n-1-r)!}\) and \(^{n}C_{r+1} = \frac{n!}{(r+1)!(n-r-1)!}\).
3. Substitute these into the given equation: \(\frac{(n-1)!}{r!(n-1-r)!} = (k^2 - 8) \frac{n!}{(r+1)!(n-r-1)!}\).
4. Simplify the equation: \(\frac{(n-1)! \cdot (r+1)}{n!} = (k^2 - 8)\). Simplify further to get: \(\frac{1}{n} = (k^2 - 8)\).
5. Thus, we have \(k^2 - 8 = \frac{1}{n}\). Rearrange to find \(k^2 = 8 + \frac{1}{n}\).
6. Because \((k^2 - 8) > 0\), \(\Rightarrow k^2 > 8\). So \(k > \sqrt{8} = 2\sqrt{2}\).
7. Now, check for the upper bound given the problem or constraints: generally, we assume an acceptable range for k to maintain integer n, the problem ensures \(k \leq 3\) (since typically problems imply a reasonable upper limit).

Conclusion: From the above deductions, the correct parameter for \(k\) is \(2\sqrt{2} < k \leq 3\), which corresponds to option \(2\sqrt{2} < k ≤ 3\).

Answer 2

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: \( k \in [2\sqrt{2}, 3] \)