JEE 2024 — Mathematics PYQ
JEE | Mathematics | 2024n−1Cr=(k2−8)−nRr+1 if and only if:
Choose the correct answer:
- A.
2\sqrt{3}<k\leq 2\sqrt{2}
- B.
2\sqrt{2}<k\leq 3
(Correct Answer) - C.
2\sqrt{3}<k<3\sqrt{3}
- D.
2\sqrt{2}<k<2\sqrt{3}
2\sqrt{2}<k\leq 3
Explanation
\begin{aligned}
& \mathrm{Given,}^{n-1}\mathrm{C}_{r}=(k^{2}-8)^{n}\mathrm{C}_{r+1} \\
& \text{一} & & >{\frac{(n-1)!}{r!(n-r-1)!}}=(k^{2}-8){\frac{n(n-1)!}{(n-r-1)!(r+1)r!}} \\
& \Rightarrow & & \frac{r+1}{n}=k^{2}-8 \\
& \Rightarrow & & 0<k^{2}-8\leq1\Rightarrow8\leq k^{2}\leq9 \\
& \text{一} & & \cdot k\in(2\sqrt{2},3]
\end{aligned}
Explanation
\begin{aligned}
& \mathrm{Given,}^{n-1}\mathrm{C}_{r}=(k^{2}-8)^{n}\mathrm{C}_{r+1} \\
& \text{一} & & >{\frac{(n-1)!}{r!(n-r-1)!}}=(k^{2}-8){\frac{n(n-1)!}{(n-r-1)!(r+1)r!}} \\
& \Rightarrow & & \frac{r+1}{n}=k^{2}-8 \\
& \Rightarrow & & 0<k^{2}-8\leq1\Rightarrow8\leq k^{2}\leq9 \\
& \text{一} & & \cdot k\in(2\sqrt{2},3]
\end{aligned}