Match the following Column I to Column II

Answer

`(A)to(q,r,s),(B)to(p),(C)to(q)`

Explanation

`Ato(q,r,s),Bto(p),Cto(q)`
(A)`:'(d+a-b)^(2)+(d+b-c)^(2)=0`
which is possible only when
`d+a-b=0,d+b-c=0`
`impliesb-a=c-b`
`implies2b=a+c`
`:.a,b` and `c` are in AP ……….i
`:'a(b-c)+b(c-a)+c(a-b)=0`
`:.x=1` is a root of
`a(b-c)x^(2)+b(c-a)x+c(a-b)=0`........ii
Given roots[Eq (ii) ] are equal.
`:.1xx1=(c(a-b))/(a(b-c))`
`impliesa(b-c)=c(a-b)`
or `b=(2qc)/(a+c)`
`:.a,b` and c are in HP ............iii
From Eq. i and ii we get
`a=b=c`
`:.a, b` and c are in AP, GP and HP
(B)`:'x^(3)-3x^(2)+3x-1=0`
`implies(x-1)^(3)=0`
`:.x=1,1,1`
`implies` Common root `x=1`
`:.a(1)^(2)+b(1)+c=0`
`impliesa+b+c=0`
(C)Given `bx^(2)+(sqrt((a+c)^(2)+4b^(2)))x+(a+c)ge0`
`=:.Dle0`
`implies(a+c)^(2)+4b^(2)-4b(a+c)le0`
`implies(a+c-2b)^(2)le0`
or `(a+c-2b)^(2)=0`
`:.a+c=2b`
Hence a, b and c are in AP.