Question
In [0,1] Lagranges Mean Value theorem is NOT applicable to
A.
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{2} - x}&{x < \frac{1}{2}} \\
{{{\left( {\frac{1}{2} - x} \right)}^2}}&{x \geqslant \frac{1}{2}}
\end{array}} \right.\]
B.
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\sin x}}{x},} \\
{1,}
\end{array}} \right.\begin{array}{*{20}{c}}
{x \ne 0} \\
{x = 0}
\end{array}\]
C.
$$f\left( x \right) = x\left| x \right|$$
D.
$$f\left( x \right) = \left| x \right|$$
Answer :
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{2} - x}&{x < \frac{1}{2}} \\
{{{\left( {\frac{1}{2} - x} \right)}^2}}&{x \geqslant \frac{1}{2}}
\end{array}} \right.\]
Solution :
There is only one function in option (A) whose critical point $$\frac{1}{2} \in \left( {0,1} \right)$$ for the rest of the parts critical point $$0 \notin \left( {0,1} \right).$$ It can be easily seen that functions in options (B), (C) and (D) are continuous on [0, 1] and differentiable in (0, 1).
\[{\text{Now}}\,{\text{for}}\,f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\left( {\frac{1}{2} - x} \right),}&{x < \frac{1}{2}} \\
{{{\left( {\frac{1}{2} - x} \right)}^2},}&{x \geqslant \frac{1}{2}}
\end{array}} \right.\]
$$\eqalign{
& {\text{Here}}\,f'\left( {\frac{{{1^ - }}}{2}} \right) = - 1\,{\text{and}}\,f'\left( {\frac{{{1^ + }}}{2}} \right) = - 2\left( {\frac{1}{2} - \frac{1}{2}} \right) = 0 \cr
& \therefore f'\left( {\frac{{{1^ - }}}{2}} \right) \ne f'\left( {\frac{{{1^ + }}}{2}} \right) \cr
& \therefore f\,{\text{is not differentiable at }}\frac{1}{2} \in \left( {0,1} \right) \cr
& \therefore LMV{\text{ is not applicable for this function in}}\,\left[ {0,1} \right] \cr} $$