The point `(a^(2), a)` lies between the straight lines `x+y=6` and `x+y=2` for

Options

A. all values of a
B. no value of a
C. `|2a-3|lt 1`
D. `|2a+5|gt 1`

Answer

To determine the values of \( a \) for which the point \( (a^2, a) \) lies between the lines \( x + y = 6 \) and \( x + y = 2 \), we will follow these steps:

### Step 1: Identify the inequalities
The point \( (a^2, a) \) lies between the lines \( x + y = 6 \) and \( x + y = 2 \). This means:
1. The point must satisfy the inequality for the line \( x + y = 6 \):
\[ a^2 + a < 6 \]
2. The point must also satisfy the inequality for the line \( x + y = 2 \):
\[ a^2 + a > 2 \]

### Step 2: Solve the first inequality
We start with the first inequality:
\[ a^2 + a < 6 \]
Rearranging gives:
\[ a^2 + a - 6 < 0 \]
Next, we factor the quadratic:
\[ (a - 2)(a + 3) < 0 \]
Now, we find the critical points, which are \( a = 2 \) and \( a = -3 \).

### Step 3: Determine the intervals for the first inequality
We analyze the sign of \( (a - 2)(a + 3) \):
- For \( a < -3 \), both factors are negative, so the product is positive.
- For \( -3 < a < 2 \), one factor is negative and the other is positive, so the product is negative.
- For \( a > 2 \), both factors are positive, so the product is positive.

Thus, the solution for the first inequality is:
\[ -3 < a < 2 \]

### Step 4: Solve the second inequality
Now we solve the second inequality:
\[ a^2 + a > 2 \]
Rearranging gives:
\[ a^2 + a - 2 > 0 \]
Factoring this quadratic gives:
\[ (a - 1)(a + 2) > 0 \]
The critical points are \( a = 1 \) and \( a = -2 \).

### Step 5: Determine the intervals for the second inequality
We analyze the sign of \( (a - 1)(a + 2) \):
- For \( a < -2 \), both factors are negative, so the product is positive.
- For \( -2 < a < 1 \), one factor is negative and the other is positive, so the product is negative.
- For \( a > 1 \), both factors are positive, so the product is positive.

Thus, the solution for the second inequality is:
\[ a < -2 \quad \text{or} \quad a > 1 \]

### Step 6: Find the intersection of the two solutions
Now we combine the solutions from both inequalities:
1. From the first inequality: \( -3 < a < 2 \)
2. From the second inequality: \( a < -2 \) or \( a > 1 \)

The intersection of these intervals is:
- For \( a < -2 \): The valid range is \( -3 < a < -2 \).
- For \( a > 1 \): The valid range is \( 1 < a < 2 \).

### Final Result
Thus, the values of \( a \) for which the point \( (a^2, a) \) lies between the lines \( x + y = 6 \) and \( x + y = 2 \) are:
\[ a \in (-3, -2) \cup (1, 2) \]