SATHEE: Binomial Theorem 2 Question 5

Binomial Theorem 2 Question 5

6.

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

(a) $(- \infty, - 2)$

(b) $(2, \infty)$

(c) $(- \sqrt{3}, \sqrt{3})$

(d) $(- \sqrt{3}, 2)$

Show Answer

Answer:

Correct Answer: 6. (d)

Solution:

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

$\Rightarrow^{n - 1} C_{r} = (k^{2} - 3) \frac{n}{r + 1}^{n - 1} C_{r}$

$\Rightarrow k^{2} - 3 = \frac{r + 1}{n}$

[since, $n \geq r \Rightarrow \frac{r + 1}{n} \leq 1$ and $n, r > 0$ ]

$\Rightarrow 0 < k^{2} - 3 \leq 1 \Rightarrow 3 < k^{2} \leq 4$

$\Rightarrow k \in [- 2, - \sqrt{3}) \cup (\sqrt{3}, 2]$

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