Let R be the interior region between the lines `3x-y-1=0 and x+2y-5=0` containing the origin. The set of all values of a for which the points `(a^2,a+1)` lies in R is

Options

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

Answer

To solve the problem, we need to determine the region \( R \) defined by the lines \( 3x - y - 1 = 0 \) and \( x + 2y - 5 = 0 \) that contains the origin, and then find the set of values of \( a \) for which the point \( (a^2, a + 1) \) lies within this region.

### Step 1: Find the equations of the lines

The first line is given by:
\[ 3x - y - 1 = 0 \implies y = 3x - 1 \]

The second line is given by:
\[ x + 2y - 5 = 0 \implies 2y = 5 - x \implies y = \frac{5 - x}{2} \]

### Step 2: Determine the intersection of the lines

To find the intersection point of the two lines, we set \( y \) from both equations equal to each other:
\[ 3x - 1 = \frac{5 - x}{2} \]

Multiplying through by 2 to eliminate the fraction:
\[ 6x - 2 = 5 - x \]
\[ 6x + x = 5 + 2 \]
\[ 7x = 7 \implies x = 1 \]

Now substituting \( x = 1 \) back into either equation to find \( y \):
\[ y = 3(1) - 1 = 2 \]
Thus, the intersection point is \( (1, 2) \).

### Step 3: Determine the region \( R \)

To find the region \( R \) that contains the origin, we need to check the positions of the lines relative to the origin \( (0, 0) \).

For the first line:
\[ 3(0) - 0 - 1 < 0 \quad \text{(True, since } -1 < 0\text{)} \]
For the second line:
\[ 0 + 2(0) - 5 < 0 \quad \text{(True, since } -5 < 0\text{)} \]

Since both inequalities are satisfied, the region \( R \) is below both lines.

### Step 4: Determine conditions for the point \( (a^2, a + 1) \)

We need to find the conditions under which the point \( (a^2, a + 1) \) lies in the region \( R \).

1. For the first line \( 3x - y - 1 \leq 0 \):
\[ 3(a^2) - (a + 1) - 1 \leq 0 \]
Simplifying:
\[ 3a^2 - a - 2 \leq 0 \]

2. For the second line \( x + 2y - 5 \leq 0 \):
\[ a^2 + 2(a + 1) - 5 \leq 0 \]
Simplifying:
\[ a^2 + 2a + 2 - 5 \leq 0 \implies a^2 + 2a - 3 \leq 0 \]

### Step 5: Solve the inequalities

**First inequality:**
\[ 3a^2 - a - 2 \leq 0 \]
Factoring:
\[ (3a + 4)(a - 1) \leq 0 \]
The critical points are \( a = -\frac{4}{3} \) and \( a = 1 \). Testing intervals, we find:
\[ -\frac{4}{3} \leq a \leq 1 \]

**Second inequality:**
\[ a^2 + 2a - 3 \leq 0 \]
Factoring:
\[ (a + 3)(a - 1) \leq 0 \]
The critical points are \( a = -3 \) and \( a = 1 \). Testing intervals, we find:
\[ -3 \leq a \leq 1 \]

### Step 6: Find the intersection of the intervals

The intervals from both inequalities are:
1. From \( 3a^2 - a - 2 \leq 0 \): \( -\frac{4}{3} \leq a \leq 1 \)
2. From \( a^2 + 2a - 3 \leq 0 \): \( -3 \leq a \leq 1 \)

The intersection of these two intervals is:
\[ -\frac{4}{3} \leq a \leq 1 \]

### Final Answer

The set of all values of \( a \) for which the point \( (a^2, a + 1) \) lies in the region \( R \) is:
\[ \boxed{[-\frac{4}{3}, 1]} \]