The roots of the equation xCr n−1Cr n−1Cr−1 x+1Cr nCr nCr−1 x+2Cr n+1Cr n+1Cr−1=0 are
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
x = n
b
x = n + 1
c
x = n - 1
d
x = n - 2
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
xCr n−1Cr nCr x+1Cr nCr n+1Crx+2Cr n+1Cr n+2Cr=0 (1)⇒x!r!(x−r)!(n−1)!r!(n−r−1)!n!r!(n−r)!(x+1)!r!(x+1−r)!n!r!(n−r)!(n+1)!r!(n−r+1)!(x+2)!r!(x+2−r)!(n+1)!r!(n+1−r)!(n+2)!r!(n−r+2)!=0Taking x!r!(x−r)! common from C1, we have quadratic equation in x.Now in (1), if we put x = n - 1, C1 and C2 are the same; hence, x = n - 1 is one root of the equation.If we put x = n, then C1 and C3 are same. Hence, x = n is the other root.