If the point (a, a^2) lies inside the angle f | vedclass

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(B) The lines are $L_1: x - 2y = 0$ and $L_2: 3x - y = 0$. For the point $(a, a^2)$ to lie inside the angle,it must satisfy the inequalities formed by

Vedclass · Accepted Answer

$(\frac{1}{2}, 3)$

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(B) The lines are $L_1: x - 2y = 0$ and $L_2: 3x - y = 0$. For the point $(a, a^2)$ to lie inside the angle,it must satisfy the inequalities formed by the lines such that it lies on the same side of $L_1$ as the origin (if applicable) or specifically between the rays. Given $x > 0$,we have $a > 0$. For the point to be below $y = 3x$,we need $a^2 For the point to be above $y = \frac{x}{2}$,we need $a^2 > \frac{a}{2}$,which implies $a(a - \frac{1}{2}) > 0$. Since $a > 0$,we have $a > \frac{1}{2}$. Combining these,we get $\frac{1}{2} < a < 3$.

If the point $(a, a^2)$ lies inside the angle formed by the lines $y = \frac{x}{2}$ $(x > 0)$ and $y = 3x$ $(x > 0)$,then $a$ belongs to:

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