Given A={x:x is a root of `x^(2)-1=0`},B={x:x is a root of `x^(2)-2x+1=0`} then (a)`AnnB=B`(b)`AnnB=A`(c) `AuuB=A`(d)`AnnB=phi`

Options

A. `AnnB=B`
B. `AnnB=A`
C. `AuuB=A`
D. `AnnB=phi`

Answer

To solve the given problem, we need to evaluate the sets A and B based on the roots of the equations provided, and then determine the relationships between these sets.

### Step 1: Determine the set A
Set A is defined as the set of roots of the equation \(x^2 - 1 = 0\).

1. Solve the equation:
\[ x^2 - 1 = 0 \implies x^2 = 1 \implies x = \pm 1 \]
2. Therefore, the set A is:
\[ A = \{1, -1\} \]

### Step 2: Determine the set B
Set B is defined as the set of roots of the equation \(x^2 - 2x + 1 = 0\).

1. Solve the equation:
\[ x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x - 1 = 0 \implies x = 1 \]
2. Therefore, the set B is:
\[ B = \{1\} \]

### Step 3: Evaluate the intersection of A and B
Now we need to find the intersection of sets A and B, denoted as \(A \cap B\).

1. The intersection consists of elements that are common to both sets:
\[ A \cap B = \{1, -1\} \cap \{1\} = \{1\} \]

### Step 4: Evaluate the union of A and B
Next, we evaluate the union of sets A and B, denoted as \(A \cup B\).

1. The union consists of all elements that are in either set:
\[ A \cup B = \{1, -1\} \cup \{1\} = \{1, -1\} \]

### Step 5: Analyze the options
Now we can analyze the options provided in the question:

(a) \(A \cap B = B\)
- We found \(A \cap B = \{1\}\) and \(B = \{1\}\), so this is **true**.

(b) \(A \cap B = A\)
- We found \(A \cap B = \{1\}\) and \(A = \{1, -1\}\), so this is **false**.

(c) \(A \cup B = A\)
- We found \(A \cup B = \{1, -1\}\) and \(A = \{1, -1\}\), so this is **true**.

(d) \(A \cap B = \phi\)
- We found \(A \cap B = \{1\}\), so this is **false**.

### Conclusion
The correct options are:
- (a) \(A \cap B = B\) is true.
- (c) \(A \cup B = A\) is true.