Key Concepts and Formulas

• Condition for a Point to Lie Inside an Angle: If lines $y=m_1x$ and $y=m_2x$ with $m_1 < m_2$ form an angle for $x>0$, a point $(x_0, y_0)$ lies inside this angle if $m_1x_0 < y_0 < m_2x_0$ and $x_0>0$.
• Solving Quadratic Inequalities: To solve $ax^2 + bx + c > 0$ or $ax^2 + bx + c < 0$, find the roots of $ax^2 + bx + c = 0$. If $a>0$, then $ax^2 + bx + c > 0$ when $x$ is outside the roots and $ax^2 + bx + c < 0$ when $x$ is between the roots.
• Intersection of Intervals: To find the values that satisfy multiple conditions, find the intersection of the intervals representing each condition.

Step-by-Step Solution

1. Identify the Given Information

We are given the lines $y = \frac{x}{2}$ and $y = 3x$, and the point $(a, a^2)$. We are also given that $x > 0$. The slopes of the lines are $m_1 = \frac{1}{2}$ and $m_2 = 3$.

2. Verify the Order of Slopes

We need to ensure that $m_1 < m_2$ to correctly apply the condition for a point to lie inside the angle. Since $\frac{1}{2} < 3$, the condition is satisfied.

3. Apply the Condition for the Point to Lie Inside the Angle

For the point $(a, a^2)$ to lie inside the angle formed by the lines, we must have: $\frac{1}{2}a < a^2 < 3a$ Also, we need $a > 0$ since the angle is formed for $x > 0$.

4. Solve the Inequality $a^2 > \frac{1}{2}a$

We need to find the values of $a$ for which $a^2 > \frac{1}{2}a$.

• Step 4a: Rearrange the inequality. Subtract $\frac{1}{2}a$ from both sides: $a^2 - \frac{1}{2}a > 0$
• Step 4b: Factor the expression. Factor out $a$: $a\left(a - \frac{1}{2}\right) > 0$
• Step 4c: Find critical points. The critical points are $a = 0$ and $a = \frac{1}{2}$.
• Step 4d: Determine the solution interval. Since the parabola opens upwards, the inequality is satisfied when $a < 0$ or $a > \frac{1}{2}$. Thus, $a \in (-\infty, 0) \cup \left(\frac{1}{2}, \infty\right)$.

5. Solve the Inequality $a^2 < 3a$

We need to find the values of $a$ for which $a^2 < 3a$.

• Step 5a: Rearrange the inequality. Subtract $3a$ from both sides: $a^2 - 3a < 0$
• Step 5b: Factor the expression. Factor out $a$: $a(a - 3) < 0$
• Step 5c: Find critical points. The critical points are $a = 0$ and $a = 3$.
• Step 5d: Determine the solution interval. Since the parabola opens upwards, the inequality is satisfied when $0 < a < 3$. Thus, $a \in (0, 3)$.

6. Combine the Conditions

We have three conditions:

1. $a \in (-\infty, 0) \cup \left(\frac{1}{2}, \infty\right)$
2. $a \in (0, 3)$
3. $a > 0$

We need to find the intersection of these intervals. Since $a>0$, we can ignore the $(-\infty, 0)$ part of the first interval. Intersecting $\left(\frac{1}{2}, \infty\right)$ with $(0, 3)$ gives $\left(\frac{1}{2}, 3\right)$. Intersecting this with $a>0$ gives $\left(\frac{1}{2}, 3\right)$. Therefore, $a \in \left(\frac{1}{2}, 3\right)$.

Common Mistakes & Tips

• Dividing by a Variable: Avoid dividing inequalities by variables without considering their sign, as it can change the inequality sign.
• Quadratic Inequality Sign: Remember that $ax^2+bx+c$ has same sign as $a$ outside the roots and opposite sign between the roots (when the roots are real).
• Combining Intervals: Use a number line to visualize the intersection of intervals to avoid errors.

Summary

We found the range of $a$ by applying the condition for a point to lie inside the angle formed by two lines. We solved the resulting quadratic inequalities and found the intersection of the solution intervals, also considering the condition $a>0$. The final interval for $a$ is $\left(\frac{1}{2}, 3\right)$.

Final Answer The final answer is $\boxed{\left( {{1 \over 2},3} \right)}$, which corresponds to option (C).