#### Question

If f (x) = e^{x} sin x in [0, π], then c in Rolle's theorem is

(a) π/6

(b) π/4

(c) π/2

(d) 3π/4

#### Solution

(d) 3π/4

The given function is

*x,*we get

\[f'\left( x \right) = e^x \cos n x + \sin x e^x \]

\[ \Rightarrow f'\left( c \right) = e^c \cos c + \sin c e^c \]

\[\text{Now }, e^x cos x \text { is continuous and derivable in R } . \]

\[\text { Therefore, it is continuous on } \left[ 0, \pi \right] \text { and derivable on} \left( 0, \pi \right) . \]

\[ \therefore f'\left( c \right) = 0 \]

\[ \Rightarrow e^c \left( \cos c + \sin c \right) = 0\]

\[ \Rightarrow \left( \cos c + \sin c \right) = 0 \left( \because e^c \neq 0 \right)\]

\[ \Rightarrow \tan c = - 1\]

\[ \Rightarrow c = \frac{3\pi}{4}, \frac{7\pi}{4}, . . . \]

\[ \therefore c = \frac{3\pi}{4} \in \left( 0, \pi \right)\]

Thus,

\[c = \frac{3\pi}{4} \in \left( 0, \pi \right)\]

for which Rolle's theorem holds.

Hence, the required value of *c* is 3π/4.