If (lambda^2, lambda+1), lambda in mathbb{Z} | vedclass

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(C) Let the lines be $L_1: x+2y-5=0$ and $L_2: 3x-y+1=0$. The origin $(0,0)$ satisfies $L_1(0,0) = -5  0$. For a point $P(\lam

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$2$

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(C) Let the lines be $L_1: x+2y-5=0$ and $L_2: 3x-y+1=0$. The origin $(0,0)$ satisfies $L_1(0,0) = -5  0$. For a point $P(\lambda^2, \lambda+1)$ to be in the region containing the origin,it must satisfy $L_1(P)  0$. For $L_1$: $\lambda^2 + 2(\lambda+1) - 5 For $L_2$: $3(\lambda^2) - (\lambda+1) + 1 > 0$ $\Rightarrow 3\lambda^2 - \lambda > 0$ $\Rightarrow \lambda(3\lambda-1) > 0$ $\Rightarrow \lambda \in (-\infty, 0) \cup (1/3, \infty)$. Taking the intersection of $\lambda \in (-3, 1)$ and $\lambda \in (-\infty, 0) \cup (1/3, \infty)$,we get $\lambda \in (-3, 0) \cup (1/3, 1)$. Since $\lambda \in \mathbb{Z}$,the possible integer values for $\lambda$ are $\lambda = -2, -1$. Thus,there are $2$ such points.

If $(\lambda^2, \lambda+1), \lambda \in \mathbb{Z}$ belongs to the region between the lines $x+2y-5=0$ and $3x-y+1=0$ which includes the origin,then the possible number of such points is

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