Question

Value of $c$ using Rolle's theorem for $f\left(x\right)=\{\begin{array}{cc}{x}^2+1& ,\text{when}0\le x\le 1\\ 3-x& ,\text{when}1<x\le 2\end{array}$ is:

• A. $1$
• B. $0$
• C. $-1$
• D. does not exist

Answer 1

The correct option is D does not exist

$f\left(x\right)=\{\begin{array}{cc}{x}^2+1& 0\le x\le 1\\ 3-x& 1<x\le 2\end{array}$

$f^1\left(x\right)=\{\begin{array}{cc}2x& 0\le x\le 1\\ -1& 1<x\le 2\end{array}$

$f^1\left(1^{-}\right)\ne f^1\left(1^{+}\right)$

$L\cdot H\cdot S\ne R\cdot H\cdot S$

So $f^{\prime }\left(x\right)$ is not continuous.

$\therefore$ Rolle's theorem can't be applied.