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If (alpha,alpha^2) lies inside the trian...

If `(alpha,alpha^2)` lies inside the triangle formed by the lines `2x+3y-1=0,x+2y-3=0,5x-6y-1=0` , then `2alpha+3alpha^2-1>0` `alpha+2alpha^2-3<0` `alpha+2alpha^2-3<0` (d) `6alpha^2-5a+1>0`

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The correct Answer is:
`-(3)/(2)ltalpha lt-1uu(1)/(2)ltalphalt1`
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