Key Concepts and Formulas

• Position of a Point Relative to a Line: For a line $Ax + By + C = 0$, a point $(x_1, y_1)$ lies on the same side of the line as the origin $(0,0)$ if $Ax_1 + By_1 + C$ has the same sign as $A(0) + B(0) + C = C$.

• Quadratic Inequalities: Solving inequalities of the form $ax^2 + bx + c > 0$ or $ax^2 + bx + c < 0$ involves finding the roots of the quadratic equation $ax^2 + bx + c = 0$ and then analyzing the sign of the quadratic expression in the intervals determined by these roots.

Step-by-Step Solution

Step 1: Analyze the first line and apply the condition.

The first line is $3x - y + 1 = 0$. We want the point $(a^2, a+1)$ to lie on the same side of this line as the origin $(0,0)$. The constant term of the line is 1, which is positive. Therefore, we need $3(a^2) - (a+1) + 1 > 0$.

$3a^2 - a - 1 + 1 > 0$ $3a^2 - a > 0$ $a(3a - 1) > 0$ This inequality holds when $a < 0$ or $a > \frac{1}{3}$.

Step 2: Analyze the second line and apply the condition.

The second line is $x + 2y - 5 = 0$. We want the point $(a^2, a+1)$ to lie on the same side of this line as the origin $(0,0)$. The constant term of the line is -5, which is negative. Therefore, we need $a^2 + 2(a+1) - 5 < 0$.

$a^2 + 2a + 2 - 5 < 0$ $a^2 + 2a - 3 < 0$ $(a+3)(a-1) < 0$ This inequality holds when $-3 < a < 1$.

Step 3: Find the intersection of the two solution sets.

We need to find the values of $a$ that satisfy both inequalities:

1. $a < 0$ or $a > \frac{1}{3}$
2. $-3 < a < 1$

Combining these inequalities, we get:

• $-3 < a < 0$
• $\frac{1}{3} < a < 1$

Therefore, the solution set is $a \in (-3, 0) \cup \left(\frac{1}{3}, 1\right)$.

Common Mistakes & Tips

• Be careful with the sign of the constant term when determining which side of the line the origin lies on.
• Remember to consider both cases when solving quadratic inequalities.
• Always check your solution by plugging in values from the intervals you found into the original inequalities.

Summary

The problem involves finding the values of $a$ for which the point $(a^2, a+1)$ lies in the region between two lines, on the same side as the origin. This requires using the concept of the position of a point relative to a line. We set up two inequalities based on the given lines and the origin, solved each inequality, and then found the intersection of the solution sets. The final solution set is $(-3, 0) \cup \left(\frac{1}{3}, 1\right)$.

Final Answer

The final answer is $(-3,0) \cup\left(\frac{1}{3}, 1\right)$, which corresponds to option (B).