If `f(x) = (x^2 - 1)^(n + 1) (x^2 + x+ 1)n in N` and f(x) has a local extremum at x = 1, then n =

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to find the value of \( n \) such that the function

\[ f(x) = (x^2 - 1)^{n + 1} (x^2 + x + 1)^n \]

has a local extremum at \( x = 1 \).

### Step 1: Find the derivative of \( f(x) \)

To find the local extremum, we first need to differentiate \( f(x) \). We will use the product rule and chain rule for differentiation.

Let \( u = (x^2 - 1)^{n + 1} \) and \( v = (x^2 + x + 1)^n \).

Using the product rule, we have:

\[ f'(x) = u'v + uv' \]

Now, we need to compute \( u' \) and \( v' \).

### Step 2: Differentiate \( u \)

Using the chain rule:

\[ u' = (n + 1)(x^2 - 1)^n \cdot (2x) \]

### Step 3: Differentiate \( v \)

Using the chain rule:

\[ v' = n(x^2 + x + 1)^{n - 1} \cdot (2x + 1) \]

### Step 4: Substitute \( u' \) and \( v' \) back into \( f'(x) \)

Now substituting back into the product rule:

\[ f'(x) = (n + 1)(x^2 - 1)^n \cdot (2x)(x^2 + x + 1)^n + (x^2 - 1)^{n + 1} \cdot n(x^2 + x + 1)^{n - 1} \cdot (2x + 1) \]

### Step 5: Set \( f'(1) = 0 \)

To find the local extremum at \( x = 1 \), we set \( f'(1) = 0 \).

Calculating \( f'(1) \):

1. Calculate \( (1^2 - 1) = 0 \) and \( (1^2 + 1 + 1) = 3 \).
2. Substitute into \( f'(x) \):

\[ f'(1) = (n + 1)(0)^n \cdot (2 \cdot 1)(3)^n + (0)^{n + 1} \cdot n(3)^{n - 1} \cdot (2 \cdot 1 + 1) \]

Both terms contain \( (1^2 - 1)^n \) or \( (1^2 - 1)^{n + 1} \), which is zero. Thus, we need to analyze the conditions under which this derivative can equal zero.

### Step 6: Analyze the conditions for \( n \)

The derivative \( f'(1) = 0 \) implies that we need to consider the other factors in the derivative. The factor \( (x^2 - 1)^n \) means that for \( n \) to be a natural number, \( n \) must be greater than or equal to 1, since \( n \) cannot be negative or zero (as \( f(x) \) would not be defined).

### Conclusion

Thus, for \( f(x) \) to have a local extremum at \( x = 1 \), the value of \( n \) must be:

\[ n = 1 \]