The lines `x + y=|a| and ax-y = 1` intersect each other in the first quadrant. Then the set of all possible values of a is the interval:

Options

A. `[1,oo)`
B. `(-1,oo)`
C. `(-1,1)`
D. `(0,oo)`

Explanation

To solve the problem of finding the set of all possible values of \( a \) for which the lines \( x + y = |a| \) and \( ax - y = 1 \) intersect in the first quadrant, we can follow these steps:

### Step 1: Find the intersection point of the lines
We have the equations:
1. \( x + y = |a| \) (Equation 1)
2. \( ax - y = 1 \) (Equation 2)

From Equation 2, we can express \( y \) in terms of \( x \):
\[ y = ax - 1 \]

Now, substitute this expression for \( y \) into Equation 1:
\[ x + (ax - 1) = |a| \]
This simplifies to:
\[ x + ax - 1 = |a| \]
\[ (1 + a)x = |a| + 1 \]
Thus, we can solve for \( x \):
\[ x = \frac{|a| + 1}{1 + a} \quad \text{(provided \( 1 + a \neq 0 \))} \]

### Step 2: Calculate \( y \)
Now, substitute the value of \( x \) back into the equation for \( y \):
\[ y = a \left( \frac{|a| + 1}{1 + a} \right) - 1 \]
\[ y = \frac{a(|a| + 1)}{1 + a} - 1 \]
To combine the terms, we find a common denominator:
\[ y = \frac{a(|a| + 1) - (1 + a)}{1 + a} \]
\[ y = \frac{a|a| + a - 1 - a}{1 + a} \]
\[ y = \frac{a|a| - 1}{1 + a} \]

### Step 3: Conditions for intersection in the first quadrant
For the intersection point \( (x, y) \) to be in the first quadrant, both \( x \) and \( y \) must be greater than or equal to 0.

1. **Condition for \( x \)**:
\[ \frac{|a| + 1}{1 + a} > 0 \]
The denominator \( 1 + a \) must be positive:
\[ 1 + a > 0 \implies a > -1 \]

2. **Condition for \( y \)**:
\[ \frac{a|a| - 1}{1 + a} \geq 0 \]
Again, \( 1 + a > 0 \) must hold. Now we analyze the numerator:
- If \( a \geq 0 \): \( a|a| = a^2 \)
\[ a^2 - 1 \geq 0 \implies a^2 \geq 1 \implies a \geq 1 \text{ or } a \leq -1 \]
- If \( a < 0 \): \( a|a| = -a^2 \)
\[ -a^2 - 1 \geq 0 \implies -a^2 \geq 1 \implies a^2 \leq -1 \text{ (not possible)} \]

### Step 4: Combine conditions
From the conditions derived:
- \( a > -1 \)
- \( a \geq 1 \)

The valid range for \( a \) is:
\[ a \in [1, \infty) \]

### Conclusion
The set of all possible values of \( a \) for which the lines intersect in the first quadrant is:
\[ \boxed{[1, \infty)} \]