If n -1 C r - k 2-3 n C r -1- then k -A- -2-B- -2- - -3- -3-D- -3- 2-

The correct option is D (√(3),2]^n 1C_r=^nC_r+1(k^2 3)⇒ k^2 3 = ^n 1C_r/^nC_r+1= r+1/n Since 0≤ r≤ n 1 ⇒ 1≤ r+1≤ n ⇒1/n≤r+1/n≤1 ⇒1/n≤ k^2 3≤1 ⇒ 3+1/n≤ k^2≤4 ...

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Byju's Answer

Standard XII

Mathematics

Binomial Coefficients

If n -1 C r =...

Question

If $^{n - 1} C_{r} = (k^{2} - 3)^{n} C_{r + 1}$ , then $k ϵ$

(- \infty, - 2]

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[2, \infty)

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[- \sqrt{3}, \sqrt{3}]

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(\sqrt{3}, 2]

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Solution

The correct option is D $(\sqrt{3}, 2]$
$^{n - 1} C_{r} =^{n} C_{r + 1} (k^{2} - 3) \Rightarrow k^{2} - 3 = \frac{^{n - 1} C_{r}}{^{n} C_{r + 1}} = \frac{r + 1}{n} Since 0 \leq r \leq n - 1 \Rightarrow 1 \leq r + 1 \leq n \Rightarrow \frac{1}{n} \leq \frac{r + 1}{n} \leq 1 \Rightarrow \frac{1}{n} \leq k^{2} - 3 \leq 1 \Rightarrow 3 + \frac{1}{n} \leq k^{2} \leq 4 \Rightarrow \sqrt{3 + \frac{1}{n}} \leq k \leq 2 as n \to \infty \Rightarrow \sqrt{3} < k \leq 2 \Rightarrow k ϵ (\sqrt{3}, 2]$

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If n−1Cr=(k2−3)nCr+1, then k ϵ

The correct option is D (√3,2]n−1Cr=nCr+1(k2−3)⇒k2−3=n−1CrnCr+1=r+1nSince 0≤r≤ n−1⇒1≤r+1≤n⇒1n≤r+1n≤1⇒1n≤k2−3≤1⇒3+1n≤ k2≤4⇒√3+1n≤ k≤2as n→∞⇒√3<k≤ 2⇒kϵ(√3,2]

If $^{n - 1} C_{r} = (k^{2} - 3)^{n} C_{r + 1}$ , then $k ϵ$