If n Cr-1=(k2-8)( n+1 Cr), then k belongs to

Question

If n Cr-1=(k2-8)( n+1 Cr), then k belongs to (A) [-3,-2 √2) (B) (-3,3) (C) [-3,-2 √2) ∪(2 √2, 3] (D) [-2 √2, 2 √2]

Tardigrade Admin · Accepted Answer

We have, r-1 ≥ 0, r ≤ n+1 ⇒ 1 ≤ r ≤ n +1 ⇒ (1/n+1) ≤ (r/n+1) ≤ 1 text Also, k2-8 =(n !/(r-1) !(n-r+1) !) ⋅ (r !(n+1-r) !/(n+1) !) =(r/n+1) Thus, (1/n+1) ≤ k2-8 ≤ 1 ⇒ 8<(1/n+1)+8 ≤ k2 ≤ 9 ⇒ 8<k2 ≤ 9 ⇒-3 ≤ k<-2 √2 text or 2 √2<k ≤ 3 Hence, k ∈[-3,-2 √2)

Q. If $^{n} C_{r - 1} = (k^{2} - 8) (^{n + 1} C_{r})$ , then $k$ belongs to

$[- 3, - 22)$

$(- 3, 3)$

$[- 3, - 22) \cup (22, 3]$

$[- 22, 22]$

Q. If nCr−1​=(k2−8)(n+1Cr​), then k belongs to

[−3,−22​)

(−3,3)

[−3,−22​)∪(22​,3]

[−22​,22​]

We have, r−1≥0,r≤n+1 ⇒1≤r≤n+1⇒n+11​≤n+1r​≤1 Also, k2−8=(r−1)!(n−r+1)!n!​⋅(n+1)!r!(n+1−r)!​ =n+1r​ Thus, n+11​≤k2−8≤1 ⇒8<n+11​+8≤k2≤9 ⇒8<k2≤9⇒−3≤k<−22​ or 22​<k≤3 Hence, k∈[−3,−22​)

Q. If $^{n} C_{r - 1} = (k^{2} - 8) (^{n + 1} C_{r})$ , then $k$ belongs to

$[- 3, - 22)$

$(- 3, 3)$

$[- 3, - 22) \cup (22, 3]$

$[- 22, 22]$