Q. In [0, 1] Lagranges Mean value theorem is NOT applicable to
A) $f(x) = \left\{ \begin{array}{r}
\frac{1}{2} - x,\ \ \ \ \ x < \frac{1}{2} \\
\left( \frac{1}{2} - x \right)^{2},\ \ \ x \geq \frac{1}{2}
\end{array} \right.\$
B) $f(x) = \left\{ \begin{array}{r}
\frac{\sin x}{x},\ \ \ \ \ x \neq 0 \\
1,\ \ \ \ \ \ x = 0
\end{array} \right.\$
C) $f(x) = x|x|$
D) $f(x) = |x|$
Explanation
(a)
For Lagrange’s Mean value theorem we know, $f(x)$ should be continuous in [a, b] and differentiable in] a, b [.
$(a)\ \ \ Given,f(x) = \left\{ \begin{matrix}
\frac{1}{2} - x,\ \ \ x < \frac{1}{2} \\
\left( \frac{1}{2} - x \right)^{2},\ x \geq \frac{1}{2}
\end{matrix} \right.\$
Which is clearly not differentiable at $x = \frac{1}{2};as\ RHD$ at $(x = 1/2) = - 1\ and\$LHD at $(x = 1/2) = 0 \Rightarrow$Lagrange’s Mean Value is not applicable.
Where option, (b), (c), (d) are continuous and differentiable.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.