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Mathematics Question on Application of derivatives

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

A

f(x)={12−xx<12 (12−x)2x≥12f(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)={sin⁡xx,x≠0 1,x=0f(x) = \begin{cases} \frac{\sin x}{x} , & x \neq 0 \\\ 1, & x = 0 \end{cases}

C

f(x)=x∣x∣f (x) = x | x |

D

f(x)=∣x∣f (x) = | x |

Answer

f(x)={12−xx<12 (12−x)2x≥12f(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}

Explanation

Solution

There is only one function in option (a) whose critical point 12∈(0,1).\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)={(12−x)x<12 (12−x)2x≥12f(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′(1−2)=−1f' \left( \frac{1^-}{2} \right) = - 1 and
f′(1/2+)=−2(12−12)=0f'(1/2^+) = - 2 \left( \frac{1}{2} - \frac{1}{2} \right) = 0
∴f′(1−2)≠f′(1/2+)\therefore f'\left(\frac{1^{-}}{2}\right) \ne f'\left(1/2^{+}\right)
∴\therefore f is not differentiable at 1/2∈(0,1)1/2 \in (0,1)
∴\therefore LMV is not applicable for this function in [0, 1]