Question

In [0, 1] Lagranges Mean Value theorem is NOT applicable to

• A. $f(x) = \begin{cases} \frac{1}{2} - x & x < \frac{1}{2} \\ \left( \frac{1}{2} - x \right)^2 & x \ge \frac{1}{2} \end{cases}$
• B. $f(x) = \begin{cases} \frac{\sin x}{x} , & x \neq 0 \\ 1, & x = 0 \end{cases}$
• C. $f (x) = x | x |$
• D. $f (x) = | x |$

Answer 1

The Correct Option is A

There is only one function in option (a) whose critical point $\frac{1}{2} \in (0, 1).$ It can be easily seen that functions in options (b), (c) and (d) are continuous on [0, 1] and differentiable in (0, 1).
Now for $f(x) = \begin{cases}
\left( \frac{1}{2} - x \right) & x < \frac{1}{2} \\
\left( \frac{1}{2} - x \right)^2 & x \ge \frac{1}{2}
\end{cases}$
Here $f' \left( \frac{1^-}{2} \right) = - 1$ and
$f'(1/2^+) = - 2 \left( \frac{1}{2} - \frac{1}{2} \right) = 0$
$\therefore f'\left(\frac{1^{-}}{2}\right) \ne f'\left(1/2^{+}\right) $
$\therefore$ f is not differentiable at $1/2 \in (0,1)$
$\therefore$ LMV is not applicable for this function in [0, 1]