Question

In [0, 1] Lagrange's Mean Value theorem is NOT applicable to

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

Answer 1

The correct option is A $f\left(x\right)=\{\begin{array}{cc}\frac{1}{2}-x,& x<\frac{1}{2}\\ (\frac{1}{2}-x{)}^2& x\ge \frac{1}{2}\end{array}$

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

Left hand derivative$\Rightarrow {lim}_{h\to 0}f\frac{(a-h)-f\left(a\right)}{h}$

$=\Rightarrow {lim}_{h\to 0}\frac{(\frac{1}{2}-\frac{1}{2}+h)-(\frac{1}{2}-\frac{1}{2})}{h}\Rightarrow 1$

Right hand derivative$\Rightarrow {lim}_{h\to 0}f\frac{(a+h)-f\left(a\right)}{h}$

$=\Rightarrow {lim}_{h\to 0}\frac{{(\frac{1}{2}-\frac{1}{2}-h)}^2-{(\frac{1}{2}-\frac{1}{2})}^2}{h}\Rightarrow h\Rightarrow 0$

L.H.D$\ne$R.H.D

Hence, $f\left(x\right)$ is not differentiable at $x=\frac{1}{2}$

Hence, $f\left(x\right)$ will not follow Lagrange's theorem.