Let R be the interior region between the lines 3x - y + 1 = 0 and x...

Let, $P ≡ \left({a}^{2}, a + 1\right)$, ${L}_{1}: 3x - y + 1 = 0$ and ${L}_{2}: x + 2y - 5 = 0$.

Now, origin and $P$ lies on the same side with respect to line ${L}_{1}$.

$⇒ {L}_{1} ( 0 ) \cdot {L}_{1} ( P ) > 0$

$⇒ 3\left({a}^{2}\right) - ( a + 1 ) + 1 > 0$

$⇒ 3{a}^{2} - a > 0$

$⇒ a\left(3a - 1\right) > 0$

$⇒ a \in ( - \infty , 0 ) \cup \left(\dfrac{1}{3}, \infty\right) . . . \left(i\right)$

Now, ${L}_{2}: x + 2y - 5 = 0$

Origin and $P$ lies on same side with respect to ${L}_{2}$

$⇒ {L}_{2} ( 0 ) \cdot {L}_{2} ( P ) > 0$

$⇒ {a}^{2} + 2 ( a + 1 ) - 5 < 0$

$⇒ {a}^{2} + 2a - 3 < 0$

$⇒ ( a + 3 ) ( a - 1 ) < 0$

$∴ a \in ( - 3, 1 ) . . . \left(ii\right)$

Using intersection of \left(i\right)\text{and}\left(ii\right)$,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/20af40c9-608a-4ef2-a447-b8e2a6e5b72d.png" style="width: 470px; height: 174px;"/></p><p>$⇒ a \in ( - 3, 0 ) \cup \left(\dfrac{1}{3}, 1\right)$