If `(t, t^(2))` falls inside the angle made by the lines `y = x//2, x gt 0` and `y = 3x, x gt 0,` where `t in (a, b)` then find the value of `2a + b`.

Explanation

To solve the problem, we need to determine the range of values for \( t \) such that the point \( (t, t^2) \) lies within the angle formed by the lines \( y = \frac{x}{2} \) and \( y = 3x \) in the first quadrant.

### Step 1: Analyze the first line \( y = \frac{x}{2} \)
We want to find when the point \( (t, t^2) \) is above the line \( y = \frac{x}{2} \).

Substituting \( (t, t^2) \) into the line equation gives:
\[ t^2 > \frac{t}{2} \]
Rearranging this inequality:
\[ t^2 - \frac{t}{2} > 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 2t^2 - t > 0 \]
Factoring out \( t \):
\[ t(2t - 1) > 0 \]
This inequality holds when \( t < 0 \) or \( t > \frac{1}{2} \).

### Step 2: Analyze the second line \( y = 3x \)
Next, we want to find when the point \( (t, t^2) \) is below the line \( y = 3x \).

Substituting \( (t, t^2) \) into the line equation gives:
\[ t^2 < 3t \]
Rearranging this inequality:
\[ t^2 - 3t < 0 \]
Factoring out \( t \):
\[ t(t - 3) < 0 \]
This inequality holds when \( 0 < t < 3 \).

### Step 3: Combine the inequalities
Now we have two conditions:
1. \( t < 0 \) or \( t > \frac{1}{2} \) (from the first line)
2. \( 0 < t < 3 \) (from the second line)

The valid range for \( t \) that satisfies both conditions is:
\[ \frac{1}{2} < t < 3 \]

### Step 4: Identify \( a \) and \( b \)
From the range \( \frac{1}{2} < t < 3 \), we identify:
- \( a = \frac{1}{2} \)
- \( b = 3 \)

### Step 5: Calculate \( 2a + b \)
Now we calculate \( 2a + b \):
\[ 2a + b = 2 \left(\frac{1}{2}\right) + 3 = 1 + 3 = 4 \]

Thus, the final answer is:
\[ \boxed{4} \]