Test: JEE Previous Year Questions- Binomial Numbers - JEE MCQ

(1+x)^m (1−x)ⁿ = (1 + mx + m(m−1)x²/2!)(1 − nx + n(n−1)x²/2! ) 
= 1 + (m−n)x + [(n² − n)/2 − mn + (m2 − m)/2] x² 
Given m − n = 3 or n = m − 3
Hence (n² − n)/2 − mn + (m² − m)/2 = 6
⇒ [(m−3)(m−4)]/2 − m(m−3) + (m² − m)/2 = −6
⇒ m2 − 7m + 12 − 2m2 + 6m + m² − m + 12 = 0
⇒ − 2m + 24 = 0
⇒ m = 12.

We know that, the coefficients in a binomial expansion is obtained by replacing each variable by unit in the given expression.
Therefore, sum of the coefficients in (a+b)^n
= 4096=(1+1)ⁿ
⇒ 4096=(2)ⁿ
⇒ (2)¹²=(2)ⁿ
⇒ n=12
Here n is even, so the greatest coefficient is nCn/2  i.e., ¹²C₆ = 924

It is given that T6​+T5​=0.
Hence nC4aⁿ−4b⁴−nC5aⁿ−5b⁵=0
nC4​aⁿ−4b⁴= nC5aⁿ−5b⁵
nC4​a = nC5b
n!b/(n−4)!4! = n!b/(n−5)!5!
= a/(n−4) = b/5​
Therefore a/b=(n−4)/5

(1+x)¹² (1+x¹²) (1+x²⁴)
= [C₀ + C₁x² + C₂x⁴ + C₃x⁶ + C₄x⁸ +.......C₁₂x²⁴) (1 + x¹² + x²⁴ + x³⁶)
= x²⁴(¹²C₀ + ¹²C₆ + ¹²C₁₂)
= 1 + ¹²C₆ + 1
= ¹²C₆  + 2

Formula, 

(n - 1)!/{r! × (n - 1 - r)!} = (k² - 3) × n!/(r + 1)!(n - r - 1)!
or, (n - 1)!/r! = (k² - 3) × n!/(r + 1)!
or, (n - 1)!/r! = (k² - 3) × n(n - 1)!/(r + 1)r!
or, 1/1 = (k² - 3) × n/(r + 1)
or, (r + 1)/n = (k² - 3)
we know, r and n are integers so, (r + 1)/n  (0, 1]
So, 0 < (r + 1)/n ≤ 1
or, 0 < k² - 3 ≤ 1
or, 3 < k² ≤ 4
or, √3 < k ≤ 2 , -2 ≤ k < -√3
Hence, k  (√3, 2]

(n - 1)!/{r! × (n - 1 - r)!} = (k² - 3) × n!/(r + 1)!(n - r - 1)!
or, (n - 1)!/r! = (k² - 3) × n!/(r + 1)!
or, (n - 1)!/r! = (k² - 3) × n(n - 1)!/(r + 1)r!
or, 1/1 = (k² - 3) × n/(r + 1)
or, (r + 1)/n = (k² - 3)
we know, r and n are integers so, (r + 1)/n  (0, 1]
so, 0 < (r + 1)/n ≤ 1
or, 0 < k² - 3 ≤ 1
or, 3 < k² ≤ 4
or, √3 < k ≤ 2 , -2 ≤ k < -√3
hence, k  (√3, 2]