If the points `(a,a^(2)) and (1,2)` lie in the same angular region between the lines `3x+4y-1=0` and `2x+y-3=0` , then

Options

A. `a lt -3` or , `a gt 1`
B. `a in [-3,1]`
C. `a lt (1)/(4)` or , ` a gt -1`
D. none of these

Explanation

To solve the problem, we need to determine the range of \( a \) such that the points \( (a, a^2) \) and \( (1, 2) \) lie in the same angular region between the lines \( 3x + 4y - 1 = 0 \) and \( 2x + y - 3 = 0 \).

### Step 1: Determine the position of the point \( (1, 2) \) with respect to the lines.

1. **Substituting \( (1, 2) \) into the first line \( 3x + 4y - 1 = 0 \)**:
\[ 3(1) + 4(2) - 1 = 3 + 8 - 1 = 10 > 0 \]
This means the point \( (1, 2) \) is on the positive side of the line.

2. **Substituting \( (1, 2) \) into the second line \( 2x + y - 3 = 0 \)**:
\[ 2(1) + 2 - 3 = 2 + 2 - 3 = 1 > 0 \]
This means the point \( (1, 2) \) is also on the positive side of the second line.

### Step 2: Determine the conditions for the point \( (a, a^2) \).

1. **Substituting \( (a, a^2) \) into the first line \( 3x + 4y - 1 = 0 \)**:
\[ 3a + 4a^2 - 1 > 0 \implies 4a^2 + 3a - 1 > 0 \]

2. **Substituting \( (a, a^2) \) into the second line \( 2x + y - 3 = 0 \)**:
\[ 2a + a^2 - 3 > 0 \implies a^2 + 2a - 3 > 0 \]

### Step 3: Solve the inequalities.

1. **For the inequality \( 4a^2 + 3a - 1 > 0 \)**:
- Find the roots using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} = \frac{-3 \pm \sqrt{9 + 16}}{8} = \frac{-3 \pm 5}{8} \]
- Roots are:
\[ a_1 = \frac{2}{8} = \frac{1}{4}, \quad a_2 = \frac{-8}{8} = -1 \]
- The quadratic opens upwards, so the solution is:
\[ a < -1 \quad \text{or} \quad a > \frac{1}{4} \]

2. **For the inequality \( a^2 + 2a - 3 > 0 \)**:
- Find the roots:
\[ a = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm 4}{2} \]
- Roots are:
\[ a_1 = 1, \quad a_2 = -3 \]
- The quadratic opens upwards, so the solution is:
\[ a < -3 \quad \text{or} \quad a > 1 \]

### Step 4: Combine the conditions.

- From the first inequality, we have \( a < -1 \) or \( a > \frac{1}{4} \).
- From the second inequality, we have \( a < -3 \) or \( a > 1 \).

Combining these results:
- For \( a < -1 \): The stricter condition is \( a < -3 \).
- For \( a > \frac{1}{4} \): The stricter condition is \( a > 1 \).

### Final Answer:
Thus, the ranges of \( a \) such that both points lie in the same angular region are:
\[ a < -3 \quad \text{or} \quad a > 1 \]