Let `f(x) = (x^2 - 1)^(n+1) * (x^2 + x + 1)`. Then f(x) has local extremum at `x = 1`, when n is (A) `n = 2` (B) `n = 4` (C) `n = 3 ` (D)` n = 5 `

Options

A. a maxima at x=1 if n is odd
B. a maxima at x =1 if n is even
C. a minima x=1 if n is even
D. a minima at x =2 if n is even

Explanation

To find the value of \( n \) for which the function \( f(x) = (x^2 - 1)^{n+1} (x^2 + x + 1) \) has a local extremum at \( x = 1 \), we will follow these steps:

### Step 1: Differentiate the function
We need to find the first derivative \( f'(x) \) using the product rule. Let:
- \( u = (x^2 - 1)^{n+1} \)
- \( v = (x^2 + x + 1) \)

Using the product rule, we have:
\[ f'(x) = u'v + uv' \]

### Step 2: Calculate \( u' \) and \( v' \)
1. **Differentiate \( u \)**:
\[ u' = (n+1)(x^2 - 1)^n \cdot (2x) = 2x(n+1)(x^2 - 1)^n \]

2. **Differentiate \( v \)**:
\[ v' = 2x + 1 \]

### Step 3: Substitute \( u' \) and \( v' \) into \( f'(x) \)
Now substituting back into the product rule:
\[ f'(x) = 2x(n+1)(x^2 - 1)^n (x^2 + x + 1) + (x^2 - 1)^{n+1}(2x + 1) \]

### Step 4: Factor out common terms
Notice that \( (x^2 - 1)^n \) is a common factor:
\[ f'(x) = (x^2 - 1)^n \left[ 2x(n+1)(x^2 + x + 1) + (x^2 - 1)(2x + 1) \right] \]

### Step 5: Evaluate \( f'(1) \)
To find the extremum at \( x = 1 \), we need to evaluate \( f'(1) \):
1. Calculate \( (1^2 - 1)^n = 0^n = 0 \) (for \( n > 0 \)).
2. The first term becomes \( 0 \), thus we need to check the second part:
\[ f'(1) = (1^2 - 1)^n \left[ 2(1)(n+1)(1^2 + 1 + 1) + (1^2 - 1)(2(1) + 1) \right] \]
This simplifies to:
\[ f'(1) = 0 \cdot \text{(anything)} = 0 \]

### Step 6: Check the condition for local extremum
For \( f'(1) = 0 \) to indicate a local extremum, we need to ensure that \( (x^2 - 1)^n \) does not change sign around \( x = 1 \). This is true if \( n \) is a positive integer.

### Step 7: Conclusion
Since \( n \) must be greater than 0, we check the options:
- (A) \( n = 2 \) ✓
- (B) \( n = 4 \) ✓
- (C) \( n = 3 \) ✓
- (D) \( n = 5 \) ✓

All options satisfy the condition for local extremum at \( x = 1 \).

### Final Answer
All options (A), (B), (C), and (D) are correct.

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