In [0, 1] Lagrange's mean value theorem is NOT applicable to

Hindi

5. Continuity and Differentiation

medium

In $[0, 1]$ Lagrange's mean value theorem is $ NOT$ applicable to

$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{2} - x,\,\,\,\,\,\,\,\,\,x < \frac{1}{2}} \\
{{{\left( {\frac{1}{2} - x} \right)}^2},\,x \geqslant \frac{1}{2}}
\end{array}} \right.$

$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\sin x}}{x}\,\,x \ne 0} \\
{1,\,\,\,\,\,\,\,\,x = \frac{1}{2}}
\end{array}} \right.$

$f(x) = x|x|$

$f(x) = |x|$

(IIT-2003)

Solution

(a) The function defined in option $ (a) $ is not differentiable at $x = \frac{1}{2}$.

Standard 12

Mathematics

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