Question

A function f is defined by $f\left(x\right)=e^x\sin x$ in $[0,\pi ]$.

Which of the following is not correct?

• A. $f$is continuous in $[0,\pi ]$
• B. $f$is differentiable in $[0,\pi ]$
• C. $f\left(0\right)=f\left(\pi \right)$
• D. Roll's theorem is not true in$[0,\pi ]$

Answer 1

The correct option is D

Roll's theorem is not true in$[0,\pi ]$

The explanation for the correct answer:

Step 1: Finding continuity of the function:

Given, $f\left(x\right)=e^x\sin x$

Since $\sin x$is continuous all $x\in \left[0,\mathrm{\pi }\right]$ and $e^x$is continuous all $x\in R$

Thus, $f\left(x\right)$ is continuous in $\left[0,\mathrm{\pi }\right]$

Step 2: Finding differentiability of the function:

We have $f\text{'}\left(x\right)=e^x\left(\cos x\right)+\sin x\left(e^x\right)$

Since $\sin x$ and $\cos x$ is differentiable all $x\in \left[0,\mathrm{\pi }\right]$ and $e^x$is differentiable all $x\in R$

Thus $f\left(x\right)$ is differentiable in $\left[0,\mathrm{\pi }\right]$

Step 3: Finding $f\left(0\right)$and $f\left(\mathrm{\pi }\right)$:

$$\begin{gathered} f\left(0\right)=e^0\sin \left(0\right) \\ =\left(1\right)\left(0\right) \\ =0 \\ f\left(\mathrm{\pi }\right)=e^{\mathrm{\pi }}\sin \left(\mathrm{\pi }\right) \\ =e^{\mathrm{\pi }}\left(0\right) \\ =0 \end{gathered}$$

Thus $f\left(0\right)=f\left(\mathrm{\pi }\right)$.

Therefore, all three conditions are true.

So the function is satisfied all the conditions of rolle's theorem in $[0,\pi ]$.

Hence, option (D) is the correct option.