If `x^n-1` is divisible by `x-k` then the least positive integral value of k is(a) 1 (b) 2 (c) 3 (d) 4

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to determine the least positive integral value of \( k \) such that \( x^n - 1 \) is divisible by \( x - k \).

### Step-by-Step Solution:

1. **Understanding the Condition of Divisibility**:
- The expression \( x^n - 1 \) is divisible by \( x - k \) means that when we substitute \( x = k \) into \( x^n - 1 \), the result should be zero. This follows from the Factor Theorem.

2. **Setting Up the Equation**:
- According to the Factor Theorem, if \( x - k \) is a factor of \( x^n - 1 \), then:
\[ k^n - 1 = 0 \]
- This simplifies to:
\[ k^n = 1 \]

3. **Finding Values of \( k \)**:
- The equation \( k^n = 1 \) implies that \( k \) must be a root of unity. The only positive integral value of \( k \) that satisfies this equation for any positive integer \( n \) is \( k = 1 \).
- This is because \( 1^n = 1 \) for any integer \( n \).

4. **Conclusion**:
- Therefore, the least positive integral value of \( k \) such that \( x^n - 1 \) is divisible by \( x - k \) is:
\[ \boxed{1} \]