For an integer K,if the point P(K^2, K+1) and | vedclass

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(B) Let $L_1(x, y) = x+2y-5$ and $L_2(x, y) = 3x-y+1$. The origin $O(0,0)$ gives $L_1(0,0) = -5  0$. For the point $P(K^2, K+1

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$2$

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(B) Let $L_1(x, y) = x+2y-5$ and $L_2(x, y) = 3x-y+1$. The origin $O(0,0)$ gives $L_1(0,0) = -5  0$. For the point $P(K^2, K+1)$ to lie in the same region as the origin,it must satisfy $L_1(K^2, K+1)  0$. $1$) $L_1(K^2, K+1) = K^2 + 2(K+1) - 5 = K^2 + 2K - 3 $(K+3)(K-1) $2$) $L_2(K^2, K+1) = 3K^2 - (K+1) + 1 = 3K^2 - K > 0$. $K(3K-1) > 0 \implies K \in (-\infty, 0) \cup (1/3, \infty)$. Taking the intersection of both conditions: $K \in (-3, 0) \cup (1/3, 1)$. Since $K$ is an integer,the possible values for $K$ are $\{-2, -1\}$. Thus,there are $2$ such points $P$.

For an integer $K$,if the point $P(K^2, K+1)$ and the origin $O(0,0)$ lie in the same region between the lines $x+2y-5=0$ and $3x-y+1=0$,then the possible number of such points $P$ is

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