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Q> Find the value of 'a' for which ax2 + (a - 3)x + 1 < 0 for at aleast one positive real x.

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21
Shubhodip ·

if a<0, then you of course have one such positive x, at least x→ +∞

If a>0, you must have D>0, and also atleast one of the roots must be positive..atleast one of the roots will be positive if a-3<0 i.e a<3(note that here both the roots will be positive with the conditions)

and D>0 gives (a-1)(a-9)>0, a>9 or a<1, we had a<3 so that 0<a<1

so a \in (-∞, 1) - {0}

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