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

Answer 1

(d) 3π/4

The given function is $f\left(x\right)=e^x\sin x$.

Differentiating the given function with respect to x, we get

$$\begin{gathered} f\text{'}\left(x\right)=e^x\cos x+\sin x{e}^x \\ \Rightarrow f\text{'}\left(c\right)=e^c\cos c+\sin c{e}^c \\ \mathrm{Now},{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}\mathrm{is}\mathrm{continuous}\mathrm{and}\mathrm{derivable}\mathrm{in}\mathrm{R}. \\ \mathrm{Therefore},\mathrm{it}\mathrm{is}\mathrm{continuous}\mathrm{on}\left[0,\mathrm{\pi }\right]\mathrm{and}\mathrm{derivable}\mathrm{on}\left(0,\mathrm{\pi }\right). \\ \therefore f\text{'}\left(c\right)=0 \\ \Rightarrow e^c\left(\cos c+\sin c\right)=0 \\ \Rightarrow \left(\cos c+\sin c\right)=0\left(\because e^c\ne 0\right) \\ \Rightarrow \tan c=-1 \\ \Rightarrow c=\frac{3\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4},... \\ \therefore c=\frac{3\mathrm{\pi }}{4}\in \left(0,\mathrm{\pi }\right) \end{gathered}$$

Thus, $c=\frac{3\mathrm{\pi }}{4}\in \left(0,\mathrm{\pi }\right)$ for which Rolle's theorem holds.

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