MCQ
In  $[0, 1]$ Lagrange's mean value theorem is $ NOT$  applicable to
  • $f(x) = \left\{ {\begin{array}{*{20}{c}}
      {\frac{1}{2} - x,\,\,\,\,\,\,\,\,\,x < \frac{1}{2}} \\ 
      {{{\left( {\frac{1}{2} - x} \right)}^2},\,x \geqslant \frac{1}{2}} 
    \end{array}} \right.$
  • B
    $f(x) = \left\{ {\begin{array}{*{20}{c}}
      {\frac{{\sin x}}{x}\,\,x \ne 0} \\ 
      {1,\,\,\,\,\,\,\,\,x = \frac{1}{2}} 
    \end{array}} \right.$
  • C
    $f(x) = x|x|$
  • D
    $f(x) = |x|$

Answer

Correct option: A.
$f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\frac{1}{2} - x,\,\,\,\,\,\,\,\,\,x < \frac{1}{2}} \\ 
  {{{\left( {\frac{1}{2} - x} \right)}^2},\,x \geqslant \frac{1}{2}} 
\end{array}} \right.$
a
(a) The function defined in option $ (a) $ is not differentiable at $x = \frac{1}{2}$.

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