Question

In \[\left( {0,1} \right)\] mean value theorem is not applicable to
A. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\left( {\dfrac{1}{2} - x} \right);x < \dfrac{1}{2}} \\
  {{{\left( {\dfrac{1}{2} - x} \right)}^2};x \geqslant \dfrac{1}{2}}
\end{array}} \right.\]
B. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\left( {\dfrac{{\sin x}}{x}} \right);x \ne 0} \\
  {1;x = 0}
\end{array}} \right.\]
C. \[f\left( x \right) = x\left| x \right|\]
D. \[f\left( x \right) = \left| x \right|\]

Answer 1

Hint: In this question, we need to find the which option that has a function for which mean value theorem is not applicable for the interval \[\left( {0,1} \right)\]. For this, we will use the concept such as for LaGrange’s mean value theorem, the function should be continuous and differentiable in the given interval.

Complete step-by-step solution:
Consider the function \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\left( {\dfrac{1}{2} - x} \right);x < \dfrac{1}{2}} \\
  {{{\left( {\dfrac{1}{2} - x} \right)}^2};x \geqslant \dfrac{1}{2}}
\end{array}} \right.\]
Here, we will find the left-hand and right-hand derivatives.
Let us find the Left-hand derivative.
 \[f'{\left( {\dfrac{1}{2}} \right)^ - } = \dfrac{d}{{dx}}\left( {\dfrac{1}{2} - x} \right)\]
So, we get
\[f'{\left( {\dfrac{1}{2}} \right)^ - } = - 1\]
Now, Right hand derivative is \[f'{\left( {\dfrac{1}{2}} \right)^ + } = \dfrac{d}{{dx}}{\left( {\dfrac{1}{2} - x} \right)^2}\]
Thus, we get
\[f'{\left( {\dfrac{1}{2}} \right)^ + } = -2\left( {\dfrac{1}{2} - \dfrac{1}{2}} \right)\]
By simplifying, we get
\[f'{\left( {\dfrac{1}{2}} \right)^ + } = 0\]
That means the left-hand side derivative is not equal to the right-hand side derivative as \[ - 1 \ne 0\].
So, it means that \[f\left( x \right)\] is not differentiable at \[x = \dfrac{1}{2}\].
Thus, it has been proved that LaGrange’s mean value theorem is not applicable here.
Therefore, we can say that in \[\left( {0,1} \right)\] mean value theorem is not applicable to \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\left( {\dfrac{1}{2} - x} \right);x < \dfrac{1}{2}} \\
  {{{\left( {\dfrac{1}{2} - x} \right)}^2};x \geqslant \dfrac{1}{2}}
\end{array}} \right.\].

Therefore, the correct option is (A).

Additional information: LaGrange’s mean value theorem (also referred to as the First Mean Value Theorem) enables the increment of a function on an interval to be expressed by the value of the derivative at an intermediate point of the section. The mean value theorem of Lagrange has a wide range of applications in mathematics, computational mathematics, and some other fields.

Note: Many students make mistakes in finding the derivative. Also, from the given function they may be confused with the left-hand derivative and right-hand derivative. They may forget to mark signs while calculating left-hand and right-hand derivatives.