Let `p(x) =x^6-x^5-x^3-x^2-x` and `alpha, beta, gamma, delta` are the roots of the equation `x^4-x^3-x^2-1=0` then `P(alpha)+P(beta)+P(gamma)+P(delta)=`
A. `4`
B. `6`
C. `8`
D. `12`
Correct Answer - A
We have,
`A(alpha, beta)^(-1)=1/e^(beta) [(e^(beta) cos alpha,-e^(beta) sin alpha,0),(e^(beta) sin alpha,e^(beta) cos alpha,0),(0,0,1)]`
`=A(-alpha, -beta)`
Correct Answer - B::C
`(b,c)` `alpha,beta,gamma,delta` are in `H.P.`
`implies(1)/(alpha)`, `(1)/(beta)`, `(1)/(gamma)`, `(1)/(delta)` are in `A.P.`
Let `d` be the common difference of the `A.P.`
Since `alpha`, `gamma` are roots of...
Correct Answer - D
`(d)` We have `betaalpha+gammaalpha+alphabeta=0`
`Delta=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma,gammaalpha,alphabeta),(gammaalpha,alphabeta,betagamma),(alphabeta,betagamma,gammaalpha):}|`
`=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma+gammaalpha+alphabeta,gammaalpha,alphabeta),(gammaalpha+alphabeta+betagamma,alphabeta,betagamma),(alphabeta+betagamma+gammaalpha,betagamma,gammaalpha):}|` [using `C_(1)toC_(1)+C_(2)+C_(3)`]
`=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(0,gammaalpha,alphabeta),(0,alphabeta,betagamma),(0,betagamma,gammaalpha):}|=0` [all zero property].
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