n - 1 = ( k 2 - 8 ) C r + 1 n if and only if [2024], Permutations and Combinations jee mains questions

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Q.
${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

1 $2 \sqrt{2} < k \leq 3$

2 $2 \sqrt{3} < k \leq 3 \sqrt{2}$

3 $2 \sqrt{2} < k < 2 \sqrt{3}$

4 $2 \sqrt{3} < k < 3 \sqrt{3}$

Ans.
(1)

We have, ${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$

$\Rightarrow \frac{(n - 1)!}{(n - (1 + r))! r!} = (k^{2} - 8) \frac{n!}{(n - (r + 1))! (r + 1)!}$

$\Rightarrow 1 = \frac{(k^{2} - 8) n}{r + 1} \Rightarrow k^{2} - 8 = \frac{r + 1}{n} \leq 1$ $[∵ n \geq r + 1]$

$\Rightarrow k^{2} - 8 \leq 1 \Rightarrow k^{2} - 9 \leq 0 \Rightarrow (k - 3) (k + 3) \leq 0$

$\Rightarrow - 3 \leq k \leq 3$ ...(i)

But $k^{2} > 8,$ as ${}^{n - 1}{C_{r}}$ can't be negative and 0.

$\Rightarrow k^{2} - 8 > 0 \Rightarrow (k - 2 \sqrt{2}) (k + 2 \sqrt{2}) > 0$

$\Rightarrow k < - 2 \sqrt{2} or k > 2 \sqrt{2}$ ...(ii)

From (i) and (ii), $2 \sqrt{2} < k \leq 3 or - 3 \leq k < - 2 \sqrt{2}$

Solution

Q.
${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

1 $2 \sqrt{2} < k \leq 3$

2 $2 \sqrt{3} < k \leq 3 \sqrt{2}$

3 $2 \sqrt{2} < k < 2 \sqrt{3}$

4 $2 \sqrt{3} < k < 3 \sqrt{3}$

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Solution

Q. Crn-1=(k2-8)Cr+1n if and only if [2024]

1 22<k≤3

2 23<k≤32

3 22<k<23

4 23<k<33

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Q.
${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

1 $2 \sqrt{2} < k \leq 3$

2 $2 \sqrt{3} < k \leq 3 \sqrt{2}$

3 $2 \sqrt{2} < k < 2 \sqrt{3}$

4 $2 \sqrt{3} < k < 3 \sqrt{3}$