Let $R$ be the interior region between the lines $3 x-y+1=0$ and $x+2 y-5=0$ containing the origin. The set of all values of $a$, for which the points $\left(a^2, a+1\right)$ lie in $R$, is : [JEE Main 2024, 27 Jan (Shift 2)]

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Correct answer: c) $(-3,0) \cup\left(\frac{1}{3}, 1\right)$. $3a−{a}^{2}>0$ ${a}^{2}+2a−3 \(\Rightarrow \text{ }a\in \left(−3,0\right)\cup \left(\frac{1}{3},1\right)$

Let \(R\) be the interior region between the lines \(3 x-y+1=0\) and \(x+2 y-5=0\) containing the origin. The set of all…

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