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(B) Let $P = (a^2, a+1)$.For line $L_1: 3x-y+1=0$,the origin $(0,0)$ gives $L_1(0,0) = 3(0)-0+1 = 1 > 0$.Since $P$ lies on the same side as the origin

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

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(B) Let $P = (a^2, a+1)$.For line $L_1: 3x-y+1=0$,the origin $(0,0)$ gives $L_1(0,0) = 3(0)-0+1 = 1 > 0$.Since $P$ lies on the same side as the origin,$L_1(a^2, a+1) > 0$.$3(a^2) - (a+1) + 1 > 0$ $\Rightarrow 3a^2 - a > 0$ $\Rightarrow a(3a-1) > 0$.Thus,$a \in (-\infty, 0) \cup (\frac{1}{3}, \infty) \dots (i)$.For line $L_2: x+2y-5=0$,the origin $(0,0)$ gives $L_2(0,0) = 0+2(0)-5 = -5 Since $P$ lies on the same side as the origin,$L_2(a^2, a+1) $a^2 + 2(a+1) - 5 Thus,$a \in (-3, 1) \dots (ii)$.Taking the intersection of $(i)$ and $(ii)$:$a \in (-3, 0) \cup (\frac{1}{3}, 1)$.

Let $R$ be the interior region between the lines $3x-y+1=0$ and $x+2y-5=0$ containing the origin. The set of all values of $a$,for which the points $(a^2, a+1)$ lie in $R$,is :

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