Let f(x) = (x^2 - 1)^n (x^2 + x + 1). Then f( | vedclass

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(D) Given $f(x) = (x^2 - 1)^n (x^2 + x + 1)$.For $f(x)$ to have a local extremum at $x = 1$,$f'(1)$ must be $0$ and the sign of $f'(x)$ must change ac

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Both $(A)$ and $(C)$

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(D) Given $f(x) = (x^2 - 1)^n (x^2 + x + 1)$.For $f(x)$ to have a local extremum at $x = 1$,$f'(1)$ must be $0$ and the sign of $f'(x)$ must change across $x = 1$.First,calculate the derivative $f'(x)$ using the product rule:$f'(x) = n(x^2 - 1)^{n-1}(2x)(x^2 + x + 1) + (x^2 - 1)^n(2x + 1)$.Factor out $(x^2 - 1)^{n-1}$:$f'(x) = (x^2 - 1)^{n-1} [2nx(x^2 + x + 1) + (x^2 - 1)(2x + 1)]$.For $x$ near $1$,let $x = 1 + h$ where $h$ is small.Then $(x^2 - 1) = ((1+h)^2 - 1) = (1 + 2h + h^2 - 1) = 2h + h^2 \approx 2h$.So,$f'(1+h) \approx (2h)^{n-1} [2n(1)(3) + (0)(3)] = (2h)^{n-1} (6n)$.For a local extremum,the sign of $f'(x)$ must change at $x = 1$. This happens if the power of $(x-1)$ in $f'(x)$ is odd.Since $(x^2 - 1)^{n-1} = (x-1)^{n-1}(x+1)^{n-1}$,the power of $(x-1)$ is $n-1$.For the sign to change,$n-1$ must be an odd integer.Thus,$n-1 = 1, 3, 5, \dots$,which implies $n = 2, 4, 6, \dots$.Therefore,$n$ must be an even integer.Both $n=2$ and $n=4$ are even integers.Thus,the correct option is $(D)$.

Let $f(x) = (x^2 - 1)^n (x^2 + x + 1)$. Then $f(x)$ has a local extremum at $x = 1$ when:

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