1.

Consider three distinct real numbers a,b,c in a G.P with a^2+b^2+c^2=t^2 and a+b+c =alpha t .The sum of the common ratio and its reciprocal is denoted by S. Complete set of alpha^2 is

Answer»

`(1/3,3)`
`[1/3,3]`
`(1/3,3)-{1}`
`(-oo,1/3)cap (3,oo)`

Solution :a,b,c are in G.P. HENCE,a,ar,`ar^(2)` are in G.P. So,
`(a^(2)+a^(2)R^(2)+a^(2)r^(4))/((a+ar+ar^(2))^(2))=t^(2)/(alpha^(2)t^(2))=1/alpha^(2)`
`alpha^(2)=(r^(2)+r+1)/(r^(2)-r+1)`
LET `alpha^(2)=y`,
`thereforey=(r^(2)+r+1)/(r^(2)-r+1)`
`(y-1)r^(2)-r(y+1)+(y-1)=0`
For REAL r,
`(y+1)^(2)-4(y-1)^(2)ge0`
`rArr1/3leyle3`
But `ne1//3,1,3``(becauserne1,-1,0)`
`therefore1/3ltylt3andyne1`
`alpha^(2)in(1/3,3)-{1}`


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