If Rolle's theorem holds for the function f(x | vedclass

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(B) For Rolle's theorem to hold on the interval $[-1, 1]$,we must have $f(-1) = f(1)$.Given $f(x) = 2x^3 + bx^2 + cx$,we calculate:$f(1) = 2(1)^3 + b(

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(B) For Rolle's theorem to hold on the interval $[-1, 1]$,we must have $f(-1) = f(1)$.Given $f(x) = 2x^3 + bx^2 + cx$,we calculate:$f(1) = 2(1)^3 + b(1)^2 + c(1) = 2 + b + c$$f(-1) = 2(-1)^3 + b(-1)^2 + c(-1) = -2 + b - c$Equating $f(1) = f(-1)$:$2 + b + c = -2 + b - c$$2c = -4 \implies c = -2$Also,Rolle's theorem states there exists $c' \in (-1, 1)$ such that $f'(c') = 0$. Here,$c' = \frac{1}{2}$.$f'(x) = 6x^2 + 2bx + c$$f'\left(\frac{1}{2}\right) = 6\left(\frac{1}{4}\right) + 2b\left(\frac{1}{2}\right) + c = 0$$\frac{3}{2} + b + c = 0$Substituting $c = -2$:$\frac{3}{2} + b - 2 = 0 \implies b - \frac{1}{2} = 0 \implies b = \frac{1}{2}$Finally,we calculate $2b + c$:$2\left(\frac{1}{2}\right) + (-2) = 1 - 2 = -1$.

If Rolle's theorem holds for the function $f(x) = 2x^3 + bx^2 + cx$ on the interval $x \in [-1, 1]$ at the point $x = \frac{1}{2}$,then the value of $2b + c$ is:

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