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(B) Given $f(x) = \ln(g(x))$,we have $g(x) = e^{f(x)}$.Taking the derivative,$g^{\prime}(x) = e^{f(x)} \cdot f^{\prime}(x)$.Since $e^{f(x)} > 0$ for a

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$1$

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(B) Given $f(x) = \ln(g(x))$,we have $g(x) = e^{f(x)}$.Taking the derivative,$g^{\prime}(x) = e^{f(x)} \cdot f^{\prime}(x)$.Since $e^{f(x)} > 0$ for all $x \in R$,the sign of $g^{\prime}(x)$ is the same as the sign of $f^{\prime}(x)$.$f^{\prime}(x) = 2010(x-2009)(x-2010)^2(x-2011)^3(x-2012)^4$.The critical points are $x = 2009, 2010, 2011, 2012$.We analyze the sign change of $f^{\prime}(x)$ across these points:- At $x = 2009$: $(x-2009)$ changes from negative to positive. $f^{\prime}(x)$ changes from negative to positive. This is a local minimum.- At $x = 2010$: $(x-2010)^2$ is always non-negative. $f^{\prime}(x)$ does not change sign. This is a point of inflection.- At $x = 2011$: $(x-2011)^3$ changes from negative to positive. $f^{\prime}(x)$ changes from positive to negative (since the overall sign changes). This is a local maximum.- At $x = 2012$: $(x-2012)^4$ is always non-negative. $f^{\prime}(x)$ does not change sign. This is a point of inflection.Thus,$g(x)$ has a local maximum only at $x = 2011$. The number of such points is $1$.

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