the roots of the equations `|{:(.^(x)C_(r),,.^(n-1)C_(r),,.^(n-1)C_(r-1)),(.^(x+1)C_(r),,.^(n)C_(r),,.^(n)C_(r-1)),(.^(x+2)C_(r),,.^(n+1)C_(r),,.^(n+1)C_(r-1)):}|=0`

Options

A. `x=n`
B. `x=n+1`
C. `x=n-1`
D. `x=n-2`

Answer

A, C

Explanation

`|{:(.^(x)C_(r),,.^(n-1)C_(r),,.^(n)C_(r)),(.^(x+1)C_(r),,.^(n)C_(r),,.^(n+1)C_(r)),(.^(x+2)C_(r),,.^(n+1)C_(r),,.^(n+2)C_(r)):}|=0`
`|{:((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)!)):}|=0`
Taking `(x!)/(r!(x-r)!)` common from `C_(1)` we have quadratic equations in x.
Now in (1) if we put `x=n-1,C_(1)` and `C_(2)` are the same hence x=n-1 is one root of the equation.
If we put x=n then `C_(1) `and `C_(3)` are same .hence x=n is the other root.