Question

If for \[f\left( x \right) = 2x - {x^2}\], Legranges’s theorem satisfies in \[\left[ {0,1} \right]\] , then find the value of \[c\] in \[\left[ {0,1} \right]\].
A. \[c = 0\]
B. \[c = \dfrac{1}{2}\]
C. \[c = \dfrac{1}{4}\]
D. \[c = 1\]

Answer 1

Hint: In this question, we need to find the value of \[c\] in the interval \[\left[ {0,1} \right]\]. We have been given a function \[f\left( x \right) = 2x - {x^2}\]. For this, we need to use Legranges’s theorem.

Formula used: We will use the Legranges’s theorem for solving this example.
If a function \[f\left( x \right)\] is a continuous function on the interval \[\left[ {a,b} \right]\] and also differentiable on \[\left( {a,b} \right)\] then there exists a point \[c\] in this interval \[\left( {a,b} \right)\]such that the derivative of the function \[f\left( x \right)\] at the point \[c\] is the difference of the function values at these particular points, divided by the difference of the point values. Mathematically, it is expressed as
\[f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{\left( {b - a} \right)}}\]

Complete step-by-step answer:
We know that \[f\left( x \right) = 2x - {x^2}\]
Also, the interval is \[\left[ {a,b} \right] \equiv \left[ {0,1} \right]\]
That means, \[a = 0;b = 1\]
First, we will find the values of \[f\left( a \right)\] and \[f\left( b \right)\]
That means, we need to find the values of \[f\left( 0 \right)\] and \[f\left( 1 \right)\].
Now, consider \[f\left( x \right) = 2x - {x^2}\]
Put \[x = 0\] in the above equation.
Thus, we get
\[
  f\left( 0 \right) = 2\left( 0 \right) - {\left( 0 \right)^2} \\
   \Rightarrow f\left( 0 \right) = 0 \\
 \]
Put \[x = 1\] in \[f\left( x \right) = 2x - {x^2}\].
Thus, we get
\[
  f\left( 1 \right) = 2\left( 1 \right) - {\left( 1 \right)^2} \\
   \Rightarrow f\left( 1 \right) = 2 - 1 \\
   \Rightarrow f\left( 1 \right) = 1 \\
 \]
Let us determine the value of \[f'\left( c \right)\].
So, first, we will find the derivative of the given function.
Thus, we get
\[
  f'\left( x \right) = \dfrac{d}{{dx}}\left( {2x - {x^2}} \right) \\
   \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {2x} \right) - \dfrac{d}{{dx}}\left( {{x^2}} \right) \\
   \Rightarrow f'\left( x \right) = 2 - 2x \\
 \]
Let us find the value of \[f'\left( c \right)\]
Put \[x = c\]in \[f'\left( x \right) = 2 - 2x\]
Hence, we get
\[f'\left( c \right) = 2 - 2c\]
Now, we will use Legranges’s theorem.
\[f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{\left( {b - a} \right)}}\]
Here, \[\left[ {a,b} \right] \equiv \left[ {0,1} \right]\]
So, \[f'\left( c \right) = \dfrac{{f\left( 1 \right) - f\left( 0 \right)}}{{\left( {1 - 0} \right)}}\]
Thus, we get
\[
  2 - 2c = \dfrac{{f\left( 1 \right) - f\left( 0 \right)}}{{\left( {1 - 0} \right)}}_{}^{} \\
   \Rightarrow 2 - 2c = \dfrac{{1 - 0}}{{\left( 1 \right)}} \\
   \Rightarrow 2 - 2c = 1 \\
   \Rightarrow 2 - 1 = 2c \\
 \]
By simplifying, we get
\[
   \Rightarrow 1 = 2c \\
   \Rightarrow \dfrac{1}{2} = c \\
   \Rightarrow c = \dfrac{1}{2} \\
 \]
Hence, the value of \[c\]is \[\dfrac{1}{2}\] if for \[f\left( x \right) = 2x - {x^2}\] , Legranges’s theorem satisfies in \[\left[ {0,1} \right]\].
Therefore, the correct option is (B).

Note: . Many students make mistakes in writing the formula of Legranges’s theorem. Specifically, they may write \[{f}'\left( c \right)=\dfrac{f\left( a \right)-f\left( b \right)}{\left( b-a \right)}\] instead of \[{f}'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{\left( b-a \right)}\]. This gives the wrong final result. Also, the derivative of the function plays a significant role in finding the value of c.