The origin belongs to a region between the li | vedclass

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(B) Let $L_1(x, y) = x+2y-5$ and $L_2(x, y) = 3x-4y+5$. The origin $(0,0)$ gives $L_1(0,0) = -5$ and $L_2(0,0) = 5$. Since they have opposite signs,th

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$3$

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(B) Let $L_1(x, y) = x+2y-5$ and $L_2(x, y) = 3x-4y+5$. The origin $(0,0)$ gives $L_1(0,0) = -5$ and $L_2(0,0) = 5$. Since they have opposite signs,the origin lies between the lines.For a point $P = ((\alpha-1)^2, \alpha)$ to lie in the same region as the origin,$L_1(P)$ must have the same sign as $L_1(0,0)$,i.e.,$L_1(P)  0$.Condition $1$: $(\alpha-1)^2 + 2\alpha - 5 Condition $2$: $3(\alpha-1)^2 - 4\alpha + 5 > 0$ $\Rightarrow 3(\alpha^2 - 2\alpha + 1) - 4\alpha + 5 > 0$ $\Rightarrow 3\alpha^2 - 10\alpha + 8 > 0$ $\Rightarrow (3\alpha-4)(\alpha-2) > 0$ $\Rightarrow \alpha  2$.Combining the conditions: $\alpha \in (-2, 2) \cap (\alpha  2) \Rightarrow \alpha \in (-2, \frac{4}{3})$.Since $\alpha \in \mathbb{Z}$,the possible values are $\alpha \in \{-1, 0, 1\}$.Thus,there are $3$ such points.

The origin belongs to a region between the lines $x+2y-5=0$ and $3x-4y+5=0$. The number of points in that region of the form $((\alpha-1)^2, \alpha)$,where $\alpha \in \mathbb{Z}$,is

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