If $(α, α^{2})$ lies inside the triangle formed by the lines $2 x + 3 y - 1 = 0, x + 2 y - 3 = 0, 5 x - 6 y - 1 = 0$ , then

2 α + 3 α^{2} - 1 > 0

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α + 2 α^{2} - 3 > 0

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α + 2 α^{2} - 3 < 0

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6 α^{2} - 5 α + 1 > 0

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Solution

The correct options are
A $2 α + 3 α^{2} - 1 > 0$
B $α + 2 α^{2} - 3 < 0$
D $6 α^{2} - 5 α + 1 > 0$
we want $(α, α^{2})$ to be above $2 x + 3 y - 1 = 0$
$∴ 2 α + 3 α^{2} - 1 > 0$
$(α, α^{2})$ to be above $5 x - 6 y - 1 = 0$
$∴ 5 α - 6 α^{2} - 1 < 0$ $⟹ 6 α^{2} - 5 α + 1 > 0$
$(α, α^{2})$ to be below $x + 2 y - 3 = 0$
$∴ α + 2 α^{2} - 3 < 0$

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