Question

In $\left[0,1\right]$ Lagrange’s mean value theorem is NOT applicable to

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

Answer 1

The correct option is A

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

Explanation for the correct option.

Find the required function.

A function is applicable for Lagrange’s mean value theorem in interval $\left[a,b\right]$ when it is continuous and differentiable at every point in that interval.

For the function $f\left(x\right)=\left\{\begin{array}{l}\left(\frac{1}{2}-x\right)x<\frac{1}{2}\\ {\left(\frac{1}{2}-x\right)}^{2}x\ge \frac{1}{2}\end{array}\right.$ differentiation is given as:

$f\text{'}\left(x\right)=\left\{\begin{array}{l}-1x<\frac{1}{2}\\ -2\left(\frac{1}{2}-x\right)x\ge \frac{1}{2}\end{array}\right.$

So it can be clearly seen that the left hand derivative and right hand derivative at $x=\frac{1}{2}$is not same. So the function is not differentiable at $x=\frac{1}{2}$ which is in the interval $\left[0,1\right]$.

So Lagrange’s mean value theorem is not applicable to the function $f\left(x\right)=\left\{\begin{array}{l}\left(\frac{1}{2}-x\right)x<\frac{1}{2}\\ {\left(\frac{1}{2}-x\right)}^{2}x\ge \frac{1}{2}\end{array}\right.$in the interval $\left[0,1\right]$.

Hence, the correct option is A.

Explanation for the incorrect options.

Check whether the function is continuous and differentiable in the interval.

All functions in option B, C, and D are defined as:

$f\left(x\right)=\left\{\begin{array}{l}\frac{\sin x}{x}x\ne 0\\ 1x=0\end{array}\right.$, $f\left(x\right)=x\left|x\right|$, and $f\left(x\right)=\left|x\right|$ respectively.

All the functions are continuous and differentiable in the interval $\left[0,1\right]$.

So Lagrange’s mean value theorem is applicable to all the three functions.

Hence, options B,C,D are incorrect.