Let f(x) = x^3 + ax^2 + bx + c where a, b, c | vedclass

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(B) Given $f(x) = x^3 + ax^2 + bx + c$. Since $f(2) = 0$,we have $8 + 4a + 2b + c = 0$ ... $(i)$. The derivative is $f'(x) = 3x^2 + 2ax + b$. Since $f

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$\frac{-14}{3}$

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(B) Given $f(x) = x^3 + ax^2 + bx + c$. Since $f(2) = 0$,we have $8 + 4a + 2b + c = 0$ ... $(i)$. The derivative is $f'(x) = 3x^2 + 2ax + b$. Since $f(x)$ has critical points at $x = 1$ and $x = -\frac{1}{3}$,$f'(1) = 0$ and $f'(-\frac{1}{3}) = 0$. $f'(1) = 3 + 2a + b = 0$ ... (ii). $f'(-\frac{1}{3}) = 3(-\frac{1}{3})^2 + 2a(-\frac{1}{3}) + b = \frac{1}{3} - \frac{2a}{3} + b = 0$,which implies $1 - 2a + 3b = 0$ ... (iii). Solving (ii) and (iii): $b = -2a - 3$. Substituting into (iii): $1 - 2a + 3(-2a - 3) = 0 \Rightarrow 1 - 2a - 6a - 9 = 0 \Rightarrow -8a = 8 \Rightarrow a = -1$. Then $b = -2(-1) - 3 = -1$. From $(i)$,$8 + 4(-1) + 2(-1) + c = 0 \Rightarrow 8 - 4 - 2 + c = 0 \Rightarrow c = -2$. Thus,$f(x) = x^3 - x^2 - x - 2$. We need to evaluate $\int_{-1}^1 (x^3 - x^2 - x - 2) dx$. Since $x^3$ and $-x$ are odd functions,their integral over $[-1, 1]$ is $0$. So,$\int_{-1}^1 f(x) dx = \int_{-1}^1 (-x^2 - 2) dx = -2 \int_0^1 (x^2 + 2) dx$. $= -2 [\frac{x^3}{3} + 2x]_0^1 = -2 (\frac{1}{3} + 2) = -2 (\frac{7}{3}) = -\frac{14}{3}$.

Let $f(x) = x^3 + ax^2 + bx + c$ where $a, b, c$ are real numbers. If $f(x)$ has a local minimum at $x = 1$ and a local maximum at $x = -\frac{1}{3}$ and $f(2) = 0$,then $\int_{-1}^1 f(x) dx$ equals

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