The set of all real values of lambda for whic | vedclass

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(A) Let the lines be $L_1: x - y = 0$, $L_2: x + y - 2 = 0$, and $L_3: x + 3 = 0$. The vertices of the triangle are the intersection points of these l

Vedclass · Accepted Answer

$(-\infty, -2] \cup [0, \infty)$

Vedclass · Answer

(A) Let the lines be $L_1: x - y = 0$, $L_2: x + y - 2 = 0$, and $L_3: x + 3 = 0$. The vertices of the triangle are the intersection points of these lines: $L_1 \cap L_2: x = 1, y = 1 \implies (1, 1)$. $L_1 \cap L_3: x = -3, y = -3 \implies (-3, -3)$. $L_2 \cap L_3: x = -3, y = 5 \implies (-3, 5)$. The point $P(\lambda, \lambda^2)$ lies on the parabola $y = x^2$. For $P$ to be inside the triangle, it must satisfy the inequalities defined by the lines: $1) \lambda - \lambda^2 \le 0 \implies \lambda(\lambda - 1) \ge 0 \implies \lambda \in (-\infty, 0] \cup [1, \infty)$. $2) \lambda + \lambda^2 - 2 \le 0 \implies (\lambda + 2)(\lambda - 1) \le 0 \implies \lambda \in [-2, 1]$. $3) \lambda + 3 \ge 0 \implies \lambda \ge -3$. Combining these, the point $P$ is inside the triangle for $\lambda \in [-2, 0] \cup \{1\}$. The point $P$ does not lie inside the triangle if $\lambda \in (-\infty, -2) \cup (0, 1) \cup (1, \infty)$. Given the options, the set of values is $(-\infty, -2] \cup [0, \infty)$.

The set of all real values of $\lambda$ for which the point $P$ with coordinates $(\lambda, \lambda^2)$ does not lie inside the triangle formed by the lines $x - y = 0$, $x + y - 2 = 0$, and $x + 3 = 0$ is:

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