Question

In $[0,2]$, Lagrange's mean value theorem is not applicable to

• A. $f\left(x\right)=\{\begin{array}{cc}x,& x<\frac{1}{2}\\ \frac{1}{2}{(\frac{1}{2}+x)}^2,& x\ge \frac{1}{2}\end{array}$
• B. $f\left(x\right)=\{\begin{array}{cc}\frac{tanx}{x},& x\ne 0\\ 1,& x=0\end{array}$
• C. $f\left(x\right)=(x^2-4x+3)|x-1|$
• D. $f\left(x\right)=|3x-1|$

Answer 1

Answer: (B), (D)

$f\left(x\right)=\{\begin{array}{cc}\frac{tanx}{x},& x\ne 0\\ 1,& x=0\end{array}$
$f\left(x\right)=|3x-1|$

For Lagrange's mean value theorem to be applicable the function must be continuous and differentiable for all values of $x$ for which the function is defined.
For option A :
$f\left(x\right)=\{\begin{array}{cc}x,& x<\frac{1}{2}\\ \frac{1}{2}{(\frac{1}{2}+x)}^2,& x\ge \frac{1}{2}\end{array}$
We need to check continuity at $x=\frac{1}{2}$
$LHL={lim}_{x\to {\frac{1}{2}}^{-}}f\left(x\right)=\frac{1}{2}$
$RHL={lim}_{x\to {\frac{1}{2}}^{+}}f\left(x\right)=\frac{1}{2}$
$f\left(\frac{1}{2}\right)=\frac{1}{2}$
$\Rightarrow$ f(x) is continuous at $x=\frac{1}{2}$
We need to check differentiability at $x=\frac{1}{2}$
$LHD=1$ and $RHD=1$
Hence, $f\left(x\right)$ is differentiable at $x=\frac{1}{2}$
For option A, conditions of Lagrange's mean value theorem are valid.
For option B,
$f\left(x\right)$ is discontinuous at $x=\frac{\pi }{2}$
For option C,
$f\left(x\right)$ is continuous at $x=1$
$LHL=RHL=f\left(1\right)=0$
For option C, conditions of Lagrange's theorem are valid.
For option D,
$f\left(x\right)=3x-1;x\ge \frac{1}{3}$
$=1-3x;x<\frac{1}{3}$
$\therefore LHD=3$ and $RHD=-3$ at $x=\frac{1}{3}$
Hence Lagrange 's theorem is not valid .