Answer :
To determine which of the given inequalities has no solution, we'll analyze each conjunction step-by-step.
### Option 1: [tex]\((x + 1 \leq 1) \cap (x + 1 \geq 1)\)[/tex]
This represents the intersection of the inequalities [tex]\(x + 1 \leq 1\)[/tex] and [tex]\(x + 1 \geq 1\)[/tex].
1. Solve [tex]\(x + 1 \leq 1\)[/tex]:
[tex]\[ x + 1 \leq 1 \implies x \leq 0 \][/tex]
2. Solve [tex]\(x + 1 \geq 1\)[/tex]:
[tex]\[ x + 1 \geq 1 \implies x \geq 0 \][/tex]
Combining these, we have:
[tex]\[ x \leq 0 \quad \text{and} \quad x \geq 0 \implies x = 0 \][/tex]
So, this inequality has a solution: [tex]\(x = 0\)[/tex].
### Option 2: [tex]\((x + 1 < 1) \cap (x + 1 > 1)\)[/tex]
This represents the intersection of the inequalities [tex]\(x + 1 < 1\)[/tex] and [tex]\(x + 1 > 1\)[/tex].
1. Solve [tex]\(x + 1 < 1\)[/tex]:
[tex]\[ x + 1 < 1 \implies x < 0 \][/tex]
2. Solve [tex]\(x + 1 > 1\)[/tex]:
[tex]\[ x + 1 > 1 \implies x > 0 \][/tex]
Combining these, we have:
[tex]\[ x < 0 \quad \text{and} \quad x > 0 \][/tex]
These inequalities cannot both be satisfied at the same time because there is no number [tex]\(x\)[/tex] that is both less than [tex]\(0\)[/tex] and greater than [tex]\(0\)[/tex] simultaneously.
So, this inequality has no solution.
### Option 3: [tex]\((x + 1 < -1) \cap (x + 1 < 1)\)[/tex]
This represents the intersection of the inequalities [tex]\(x + 1 < -1\)[/tex] and [tex]\(x + 1 < 1\)[/tex].
1. Solve [tex]\(x + 1 < -1\)[/tex]:
[tex]\[ x + 1 < -1 \implies x < -2 \][/tex]
2. Solve [tex]\(x + 1 < 1\)[/tex]:
[tex]\[ x + 1 < 1 \implies x < 0 \][/tex]
Combining these, we have:
[tex]\[ x < -2 \quad \text{and} \quad x < 0 \][/tex]
Since [tex]\(x < -2\)[/tex] is more restrictive than [tex]\(x < 0\)[/tex], the combined condition is effectively [tex]\(x < -2\)[/tex].
So, this inequality has solutions for all [tex]\(x < -2\)[/tex].
### Conclusion:
The inequality that has no solution is:
[tex]\[ (x + 1 < 1) \cap (x + 1 > 1) \][/tex]
Therefore, the second option has no solution.
### Option 1: [tex]\((x + 1 \leq 1) \cap (x + 1 \geq 1)\)[/tex]
This represents the intersection of the inequalities [tex]\(x + 1 \leq 1\)[/tex] and [tex]\(x + 1 \geq 1\)[/tex].
1. Solve [tex]\(x + 1 \leq 1\)[/tex]:
[tex]\[ x + 1 \leq 1 \implies x \leq 0 \][/tex]
2. Solve [tex]\(x + 1 \geq 1\)[/tex]:
[tex]\[ x + 1 \geq 1 \implies x \geq 0 \][/tex]
Combining these, we have:
[tex]\[ x \leq 0 \quad \text{and} \quad x \geq 0 \implies x = 0 \][/tex]
So, this inequality has a solution: [tex]\(x = 0\)[/tex].
### Option 2: [tex]\((x + 1 < 1) \cap (x + 1 > 1)\)[/tex]
This represents the intersection of the inequalities [tex]\(x + 1 < 1\)[/tex] and [tex]\(x + 1 > 1\)[/tex].
1. Solve [tex]\(x + 1 < 1\)[/tex]:
[tex]\[ x + 1 < 1 \implies x < 0 \][/tex]
2. Solve [tex]\(x + 1 > 1\)[/tex]:
[tex]\[ x + 1 > 1 \implies x > 0 \][/tex]
Combining these, we have:
[tex]\[ x < 0 \quad \text{and} \quad x > 0 \][/tex]
These inequalities cannot both be satisfied at the same time because there is no number [tex]\(x\)[/tex] that is both less than [tex]\(0\)[/tex] and greater than [tex]\(0\)[/tex] simultaneously.
So, this inequality has no solution.
### Option 3: [tex]\((x + 1 < -1) \cap (x + 1 < 1)\)[/tex]
This represents the intersection of the inequalities [tex]\(x + 1 < -1\)[/tex] and [tex]\(x + 1 < 1\)[/tex].
1. Solve [tex]\(x + 1 < -1\)[/tex]:
[tex]\[ x + 1 < -1 \implies x < -2 \][/tex]
2. Solve [tex]\(x + 1 < 1\)[/tex]:
[tex]\[ x + 1 < 1 \implies x < 0 \][/tex]
Combining these, we have:
[tex]\[ x < -2 \quad \text{and} \quad x < 0 \][/tex]
Since [tex]\(x < -2\)[/tex] is more restrictive than [tex]\(x < 0\)[/tex], the combined condition is effectively [tex]\(x < -2\)[/tex].
So, this inequality has solutions for all [tex]\(x < -2\)[/tex].
### Conclusion:
The inequality that has no solution is:
[tex]\[ (x + 1 < 1) \cap (x + 1 > 1) \][/tex]
Therefore, the second option has no solution.
