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Mathematics
In [0,1] Lagranges Mean Value theorem is NOT applicable to
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Q. In $[0,1]$ Lagranges Mean Value theorem is NOT applicable to
Continuity and Differentiability
A
$f(x)=\begin{cases}\frac{1}{2}-x & x<\frac{1}{2} \\ \left(\frac{1}{2}-x\right)^{2} & x \geq \frac{1}{2}\end{cases}$
B
$f(x)=\begin{cases}\frac{\sin x}{x}, & x \neq 0 \\ 1, & x=0\end{cases}$
C
$f(x)=x|x|$
D
$f ( x )=| x |$
Solution:
The function defined in $(a)$ is not differentiable at $x = 1/2$.