Question

Value of ${}^{\prime }{c}^{\prime }$ of Rolle's Theorem for

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

• A. $1$
• B. $0$
• C. $1.5$
• D. Does not exist

Answer 1

The correct option is D Does not exist

$f\left(x\right)=\{\begin{array}{cc}{x}^2+1,& 0\le x\le 1\\ 3-x,& 1<x\le 2\end{array}$
Checking continuity at $x=1$
$L.H.L.=$
$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\left(\right(1-h{)}^2+1)=2$
$R.H.L.=$
$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}(3-(1+h\left)\right)=2$
and at $x=1,f\left(x\right)=2$
$\Rightarrow L.H.L.=R.H.L.=f\left(1\right)$
Thus $f\left(x\right)$ is continuous in $[0,2]$
Checking differentiability,
$f^{\prime }\left(x\right)=\{\begin{array}{cc}2x,& 0\le x\le 1\\ -1,& 1<x\le 2\end{array}$
Clearly $f\left(x\right)$ is not differentiable at $x=1$ hence Rolle's theorem cannot be applied .