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The point $(a^{2}, a + 1)$ lies in the angle between the lines $3 x - y + 1 = 0$ and $x + 2 y - 5 = 0$ containing the origin if

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$a \in (- 3, 0) \cup (\frac{1}{3}, 1)$

$a \in (- 3, \frac{1}{3})$

$a \in (- \infty, 3) \cup (\frac{1}{3}, 1)$

$a \in (\frac{1}{3}, \infty)$

answer is A.

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Detailed Solution

$3 a^{2} - a - 1 + 1 > 0$ $a^{2} + 2 a + 2 - 5 < 0$

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