English: {}^{n-1}C_{r}=(k^{2}-8)^{n}C_{r+1} if and only if: (A) 2sqrt…

Question ID: #1562
JEE Main27 January Shift 1, 2024Algebra
English: $~^{n-1}C_{r} = (k^{2}-8) ~^{n}C_{r+1}$ if and only if:

(A) $2\sqrt{2}\lt k\le3$    (B) $2\sqrt{3}\lt k\le3\sqrt{2}$    (C) $2\sqrt{3}\lt k<3\lt \sqrt{3}$    (D) $2\sqrt{2}\lt k\lt 2\sqrt{3}$

Hindi: $~^{n-1}C_{r} = (k^{2}-8) ~^{n}C_{r+1}$ यदि और केवल यदि:

(A) $2\sqrt{2}\lt k\le3$    (B) $2\sqrt{3}\lt k\le3\sqrt{2}$    (C) $2\sqrt{3}\lt k\lt 3\sqrt{3}$    (D) $2\sqrt{2}\lt k\lt 2\sqrt{3}$

Gujarati: $~^{n-1}C_{r} = (k^{2}-8) ~^{n}C_{r+1}$ જો અને માત્ર જો:

(A) $2\sqrt{2}\lt k\le3$    (B) $2\sqrt{3}\lt k\le3\sqrt{2}$    (C) $2\sqrt{3}\lt k\lt 3\sqrt{3}$    (D) $2\sqrt{2}\lt k\lt 2\sqrt{3}$


Hint:

  • English: Use the property $\frac{^{n-1}C_r}{^nC_{r+1}} = \frac{r+1}{n}$ and the boundary condition $0 < \frac{r+1}{n} \le 1$ to establish inequalities for $k$.
  • Hindi: $k$ का परिसर ज्ञात करने के लिए गुणधर्म $\frac{^{n-1}C_r}{^nC_{r+1}} = \frac{r+1}{n}$ और शर्त $0 < \frac{r+1}{n} \le 1$ का उपयोग करें।
  • Gujarati: $k$ નો વિસ્તાર શોધવા માટે ગુણધર્મ $\frac{^{n-1}C_r}{^nC_{r+1}} = \frac{r+1}{n}$ અને શરત $0 < \frac{r+1}{n} \le 1$ નો ઉપયોગ કરો.

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