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Let A is set of all real values of a for which equation x^(2)-ax+1=0 has no real roots and B is set of al real values of b for which f(x)=bx^(2)+bx+0.5gt0AAxepsilonR then AcapB=

Answer»

`{X:0ltxlt2}`
`{x:0lexlt2}`
`{x:0lexle2}`
`{x:-2ltxlt2}`

SOLUTION :`x^(2)-ax+1=0` has REAL root `
`becauseDlt0impliesa^(2)-4lt0`
`impliesaepsilon(-2,2)`
`because A={x:-2ltxlt2}`
`f(x)=bx^(2)+bx+0.5gt0AAxepsilonR`, if `bgt0`
`because Dlt0impliesb^(2)-4b.(1)/(2)lt0`
`impliesb^(2)-2blt0impliesbepsilon(0,2)`
ALSO `b=0` then `f(x)gt0AAxepsilonRbecausebepsilon[0,2]`
`becauseB={x:0lexlt2}`
`becauseAcapB={x:0lexlt2}`


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