If `(a,a^(2))` falls inside the angle made by the lines `y=(x)/(2), x gt 0 and y=3x, x gt 0`, then a belongs to the interval

Options

A. `(-3,-1/2)`
B. `(0,1/2)`
C. `(3,oo)`
D. `(1/2,3)`

Answer

To solve the problem, we need to determine the interval of values for \( a \) such that the point \( (a, a^2) \) lies between the lines \( y = \frac{x}{2} \) and \( y = 3x \) in the first quadrant.

### Step 1: Analyze the first line \( y = \frac{x}{2} \)

We start by substituting \( (a, a^2) \) into the equation of the first line:

\[ a^2 > \frac{a}{2} \]

### Step 2: Rearrange the inequality

To solve this inequality, rearrange it:

\[ a^2 - \frac{a}{2} > 0 \]

Multiply through by 2 to eliminate the fraction:

\[ 2a^2 - a > 0 \]

### Step 3: Factor the inequality

Factoring out \( a \):

\[ a(2a - 1) > 0 \]

### Step 4: Determine the critical points

The critical points are \( a = 0 \) and \( a = \frac{1}{2} \). We need to test the intervals determined by these points: \( (-\infty, 0) \), \( (0, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \).

- For \( a < 0 \): \( a(2a - 1) < 0 \)
- For \( 0 < a < \frac{1}{2} \): \( a(2a - 1) < 0 \)
- For \( a > \frac{1}{2} \): \( a(2a - 1) > 0 \)

Thus, the solution for the first inequality is:

\[ a \in \left( \frac{1}{2}, \infty \right) \]

### Step 5: Analyze the second line \( y = 3x \)

Now, we substitute \( (a, a^2) \) into the equation of the second line:

\[ a^2 < 3a \]

### Step 6: Rearrange the inequality

Rearranging gives:

\[ a^2 - 3a < 0 \]

### Step 7: Factor the inequality

Factoring out \( a \):

\[ a(a - 3) < 0 \]

### Step 8: Determine the critical points

The critical points are \( a = 0 \) and \( a = 3 \). We need to test the intervals determined by these points: \( (-\infty, 0) \), \( (0, 3) \), and \( (3, \infty) \).

- For \( a < 0 \): \( a(a - 3) > 0 \)
- For \( 0 < a < 3 \): \( a(a - 3) < 0 \)
- For \( a > 3 \): \( a(a - 3) > 0 \)

Thus, the solution for the second inequality is:

\[ a \in (0, 3) \]

### Step 9: Combine the intervals

Now, we combine the results from both inequalities:

1. From the first line: \( a \in \left( \frac{1}{2}, \infty \right) \)
2. From the second line: \( a \in (0, 3) \)

The intersection of these two intervals is:

\[ a \in \left( \frac{1}{2}, 3 \right) \]

### Final Answer

Thus, the value of \( a \) belongs to the interval:

\[ \boxed{\left( \frac{1}{2}, 3 \right)} \]