If f (x) = ex sin x in [0, $\pi$], then c in Rolle's theorem is

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3$\pi$/4 The given function is f\left( x \right) = e^x \sin x. Differentiating the given function with respect to x, we get f'\left( x \right) = e^x \cos n x + \sin x e^x \Rightarrow f'\left( c \right) = e^c \cos c + \sin c e^c \text{Now }, e^x cos x \text { is continuous and derivable in R } . \text { Therefore, it is continuous on } \left[ 0, $\pi$ \right] \text { and derivable on} \left( 0, $\pi$ \right) . \therefore f'\left( c \right) = 0 \Rightarrow e^c \left( \cos c + \sin c \right) = 0 \Rightarrow \left( \cos c + \sin c \right) = 0 \left( \because e^c \neq 0 \right) \Rightarrow \tan c = - 1 \Rightarrow c = 3$\pi$/4, 7$\pi$/4, . . . \therefore c = 3$\pi$/4 \in \left( 0, $\pi$ \right) Thus, c = 3$\pi$/4 \in \left( 0, $\pi$ \right)for which Rolle's theorem holds. Hence, the required value of c is 3$\pi$/4.

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