If x€R then x^2 +2x +a/x^2 +4x+ 3a can take all real values if ? Answer is a€[0,1] Exerts I've already asked this and exert gave me link to.similar questions. But pls solve it. Pls solve it in the way that we take denominator to other side and multiply by y then write the new formed quadratic equation and then take D >or=0 and again get a quadratic equation and again D>or =0 Pls solve using this method Thanks

Dear Student,
Please find below the solution to the asked query:

We havey=x2+2x+ax2+4x+3ayx2+4xy+3ay=x2+2x+ayx2+4xy+3ay-x2-2x-a=0x2y-1+2x2y-1+3ay-a=0For x to be real,Discriminant022y-12-43ay-ay-1042y-12-43ay-ay-10Dividing both sides by 4, we get,2y-12-3ay-ay-104y2+1-4y-3ay2-3ay-ay+a04y2+1-4y-3ay2-4ay+a04y2+1-4y-3ay2+4ay-a0y24-3a+4ya-1+1-a0For above equation to be always true,4-3a>0 and Discriminant0Now 4-3a>0a<43Discriminant04a-12-44-3a1-a016a2+1-2a+44-3aa+10Dividing both sides by 4 we get,4a2+1-2a+4-3aa+104a2+4-8a+4a-3a2-4+3a0a2-a0aa-10a0,1Taking intersection of a<43 and a0,1, we get,a0,1 Answer

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