1.

Let p,q,r in R satisfies [p q r][(1,8,7),(9,2,3),(7,7,7)] = [0 0 0] .....(i) {:(,"List-I",,"List-I"),((P),"If the point M(p.q.r) with reference to (i) lies on the curve" 2x+y+z=1 then (7p+q+r) "is equalto",(1),-2),((Q),"Let" omega(ne1) "cube root of unity with"lm(omega) gt0.If p=2 "with q and r satisfying" (i) then(3/(omega^(p))+1/(omega^(q))+3/(omega^(r))) "is equal to ",(2),7),((R),"Let q=6 with p and r satisfying"(i). ifalpha and beta"are roots of quadratic equation " px^(2)+qx+r=0 " then" Sigma_(n=0)^(oo) (1/(alpha)+1/(beta))^(n) " is equal to ",(3),6):}

Answer»


Solution :(P) `p=1,Q=6,r=-7`
`therefore (7p+q+r)=6`
(Q) `OMEGA =(-1)/(2)+isqrt(3)/(2)`
`p=2,q=12,r=-14`
`implies(3)/(omega^(p))+(1)/(omega^(q))+(3)/(omega^(r))=(3)/(omega^(2))+(1)/(omega^(12))+(3)/(omega^(-14))`
`= 3omega+1+3omega^(2)=-3+1=-2`
(R ) `p=1,q=6,r=-7`
`impliesx^(2)+6x-7=0`
`OVERSET(^^^)(alpha beta)`
`therefore (1)/(alpha)+(1)/(beta)=(alpha+beta)/(alphabeta)=(-6)/(-7)=(6)/(7)`
`impliessum_(n-0)^(oo)`
`((1)/(alpha)+(1)/(beta))^(n)=1+(6)/(7)+((6)/(7))^(2)+.....x`
`=(1)/(1-(6)/(7))=7`


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